Machine Tools Design And Numerical Control [3 ed.] 9781259004575

Table of contents :
Title
Contents
1. INTRODUCTION TO MACHINE TOOL DRIVES AND MECHANISMS—GENERAL PRINCIPLES OF MACHINE TOOL DESIGN
1.1 Working and Auxiliary Motions in Machine Tools
1.3 Machine Tool Drives
1.4 Hydraulic Transmission and Its Elements
1.5 Mechanical Transmission and Its Elements
1.6 Technico-economical Prerequisites for Undertaking the Design of a New Machine Tool
1.7 General Requirements of Machine Tool Design
1.8 Engineering Design Process Applied to Machine Tools
1.9 Layout of Machine Tool
Review Questions
References
2. REGULATION OF SPEED AND FEED RATES
2.1 Aim of Speed and Feed Rate Regulation
2.2 Stepped Regulation of Speed: Design of Speed Box
2.3 Design of Feed Box
2.4 Machine Tool Drives Using Multiple-speed Motors
2.5 Special Cases of Gear Box Design
2.6 General Recommendaitions for Developing the Gearing Diagram
2.7 Determining the Number of Teeth of Gears
2.9 Stepless Regulation of Speed and Feed Rates
2.10 Kinematics of Machine Tools
Review Questions
References
3. DESIGN OF MACHINE TOOL STRUCTURES
3.1 Functions of Machine Tool Structures and their Requirements
3.2 Design Criteria for Machine Tool Structures
3.3 Materials of Machine Tool Structures
3.4 Static and Dynamic Stiffness
3.6 Basic Design Procedure of Machine Tool Structures
3.7 Design of Beds
3.8 Design of Columns
3.9 Design of Housings
3.10 Design of Bases and Tables
3.11 Design of Cross Rails, Arms, Saddles and Carriages
3.12 Design of Rams
3.13 Model Technique in Design of Machine Tool Structures
Review Questions
References
4. DESIGN OF GUIDEWAYS AND POWER SCREWS
4.1 Functions and Types of Guideways
4.2 Design of Slideways
4.3 Design Criteria and Calculations for Slideways
4.4 Guideways Operating under Liquid Friction Conditions
4.5 Design of Aerostatic Slideways
4.6 Design of Anti-friction Guideways
4.7 Combination Guideways
4.8 Protecting Devices for Slideways
4.9 Design of Power Screws
Review Questions
References
5. DESIGN OF SPINDLES AND SPINDLE SUPPORTS
5.1 Functions of Spindle Unit and Requirements
5.2 Materials of Spindles
5.3 Effect of Machine Tool Compliance on Machining Accuracy
5.4 Design Calculations of Spindles
5.5 Anti-friction Bearings
5.6 Sliding Bearings
Review Questions
References
6. DYNAMICS OF MACHINE TOOLS
6.1 Machine Tool Elastic System-cutting Process Closed-loop System
6.2 General Procedure for Assessing Dynamic Stability of Ees—Cutting Process Closed-Loop System
6.3 Dynamic Characteristics of Elements and Systems
6.4 Dynamic Characteristic of the Equivalent Elastic System
6.5 Dynamic Characteristic of the Cutting Process
6.6 Stability Analysis
6.7 Forced Vibrations of Machine Tools
Review Questions
References
7. CONTROL SYSTEMS IN MACHINE TOOLS
7.2 Control Systems for Changing Speeds and Feeds
7.3 Control Systems for Executing Forming and Auxiliary Motions
7.4 Manual Control Systems
7.5 Automatic Control Systems
7.6 Adaptive Control Systems
References
8. NUMERICAL CONTROL OF MACHINE TOOLS
8.2 Manual Part Programming
8.3 Computer Aided Part Programming
Review Questions
References
9. EXTENSIONS OF NUMERICAL CONTROL— CNC, DNC, MACHINING CENTRES
9.1 Distributive Numerical Control (DNC-1)
9.2 Computer Numerical Control (CNC)
9.3 Machining Centres
9.4 Direct Numerical Control (DNC-2)
9.5 CNC Programming
Review Questions
Index

Citation preview

Contents

Machine Tool Design and Numerical Control Third Edition

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About the Author Dr. N. K. Mehta retired as Professor from IIT Roorkee (former University of Roorkee) in 2010 after serving in the Department of Mechanical and Industrial Engineering for almost 40 years. He also served as Counselor (Science and Technology) at the Embassy of India in Moscow from 1995 to 1998. Dr. Mehta has done extensive research in the areas of Machine Tool Design, Machining Science and Computer aided Manufacturing. He has more than 120 research papers to his credit and has supervised 12 PhD Thesis and over 50 M Tech dissertations. In addition, Dr. Mehta was Translation Editor of the Soviet Journal of Structural Mechanics and Design of Structures for four years and has translated nine text books and monographs in English. He was also the Convenor of the First International and Twenty second AIMTDR Conference in 2006. Dr. Mehta’s contribution to teaching and research has been widely acknowledged and he has been the recipient of numerous awards and honours such as G. C. Sen Memorial Prize for Best Research Paper at the Tenth AIMTDR conference in 1982, the A. N. Khosla Research Prize and silver medal in 1984, Member Program Advisory Committee of Mechanical Engineering and Robotics of the Department of Science and Technology, Govt. of India and Member Core Advisory Group of R&D in Machine Tool sector constituted by the

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Machine Tool Design and Numerical Control Third Edition

N K Mehta Formerly Professor Indian Institute of Technology, Roorkee Uttarakhand

Tata McGraw Hill Education Private Limited NEW DELHI

New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

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Tata McGraw-Hill Published by the Tata McGraw Hill Education Private Limited, 7 West Patel Nagar, New Delhi 110 008. Machine Tool Design and Numerical Control (3e) Copyright © 2012, by Tata McGraw Hill Education Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, Tata McGraw Hill Education Private Limited. ISBN (13): 978-1-25-900457-5 ISBN (10): 1-25-900457-0 Vice President and Managing Director—MHE: Ajay Shukla Head—Higher Education Publishing and Marketing: Vibha Mahajan Manager: Sponsoring—SEM & Tech Ed.: Shalini Jha Editorial Researcher: Harsha Singh Copy Editor: Preyoshi Kundu Sr Production Manager: Satinder S Baveja Production Executive: Anuj K. Shriwastava Marketing Manager—Higher Ed.: Vijay Sarathi General Manager—Production: Rajender P Ghansela Production Manager: Reji Kumar Information contained in this work has been obtained by Tata McGraw-Hill, from sources believed to be reliable. However, neither Tata McGraw-Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGraw-Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGraw-Hill and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Tej Composers, WZ-391, Madipur, New Delhi 110063, and printed at Cover Printer:

Contents

In Memory of My Parents

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Contents About the Author

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Preface

xi

1. INTRODUCTION TO MACHINE TOOL DRIVES AND MECHANISMS— GENERAL PRINCIPLES OF MACHINE TOOL DESIGN 1.1 Working and Auxiliary Motions in Machine Tools 1.3 1.4 1.5 1.6 1.7 1.8 1.9

1

1

3 Machine Tool Drives 19 Hydraulic Transmission and Its Elements 23 Mechanical Transmission and Its Elements 34 Technico-economical Prerequisites for Undertaking the Design of a New Machine Tool 52 General Requirements of Machine Tool Design 54 Engineering Design Process Applied to Machine Tools 57 Layout of Machine Tool 60 Review Questions 65 References 67

2. REGULATION OF SPEED AND FEED RATES

68

Aim of Speed and Feed Rate Regulation 68 Stepped Regulation of Speed: Design of Speed Box 69 Design of Feed Box 90 Machine Tool Drives Using Multiple-speed Motors 95 Special Cases of Gear Box Design 98 General Recommendaitions for Developing the Gearing Diagram 105 Determining the Number of Teeth of Gears 108 118 2.9 Stepless Regulation of Speed and Feed Rates 124 2.10 Kinematics of Machine Tools 140 Review Questions 163 References 166 2.1 2.2 2.3 2.4 2.5 2.6 2.7

3. DESIGN OF MACHINE TOOL STRUCTURES 3.1 Functions of Machine Tool Structures and their Requirements 167 3.2 Design Criteria for Machine Tool Structures 167

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3.3 Materials of Machine Tool Structures 170 3.4 Static and Dynamic Stiffness 175 182 3.6 Basic Design Procedure of Machine Tool Structures 188 3.7 Design of Beds 197 3.8 Design of Columns 213 3.9 Design of Housings 216 3.10 Design of Bases and Tables 219 3.11 Design of Cross Rails, Arms, Saddles and Carriages 221 3.12 Design of Rams 222 3.13 Model Technique in Design of Machine Tool Structures 224 Review Questions 228 References 231 4. DESIGN OF GUIDEWAYS AND POWER SCREWS 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

Functions and Types of Guideways 233 Design of Slideways 234 Design Criteria and Calculations for Slideways 243 Guideways Operating under Liquid Friction Conditions 251 Design of Aerostatic Slideways 265 Design of Anti-friction Guideways 267 Combination Guideways 273 Protecting Devices for Slideways 274 Design of Power Screws 276 Review Questions 285 References 287

5. DESIGN OF SPINDLES AND SPINDLE SUPPORTS 5.1 5.2 5.3 5.4 5.5 5.6

233

Functions of Spindle Unit and Requirements 288 Materials of Spindles 289 Effect of Machine Tool Compliance on Machining Accuracy 290 Design Calculations of Spindles 294 Anti-friction Bearings 303 Sliding Bearings 310 Review Questions 332 References 333

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6. DYNAMICS OF MACHINE TOOLS

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335

6.1 Machine Tool Elastic System-cutting Process Closed-loop System 335 6.2 General Procedure for Assessing Dynamic Stability of Ees—Cutting Process Closed-Loop System 336 6.3 Dynamic Characteristics of Elements and Systems 339 6.4 Dynamic Characteristic of the Equivalent Elastic System 340 6.5 Dynamic Characteristic of the Cutting Process 352 6.6 Stability Analysis 366 6.7 Forced Vibrations of Machine Tools 378 Review Questions 383 References 385 7. CONTROL SYSTEMS IN MACHINE TOOLS 7.2 7.3 7.4 7.5 7.6

386 Control Systems for Changing Speeds and Feeds 386 Control Systems for Executing Forming and Auxiliary Motions Manual Control Systems 397 Automatic Control Systems 410 Adaptive Control Systems 415 References 418

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8. NUMERICAL CONTROL OF MACHINE TOOLS

419 419

8.2 Manual Part Programming 440 8.3 Computer Aided Part Programming Review Questions 486 References 493

465

9. EXTENSIONS OF NUMERICAL CONTROL— CNC, DNC, MACHINING CENTRES 9.1 9.2 9.3 9.4 9.5

494

Distributive Numerical Control (DNC-1) 495 Computer Numerical Control (CNC) 495 Machining Centres 498 Direct Numerical Control (DNC-2) 501 CNC Programming 504 Review Questions 538

Index

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Preface About the Book The first edition of the book was published in 1984 under the title, Machine Tool Design, when the subject was gaining popularity as a specialised course in engineering institutions in the country. The motivation for writing the book was to provide a basic text for undergraduate students that would also serve as a useful reference for postgraduate students and practicing engineers. The revision of the book for the second edition published in 1996 was undertaken with the limited objective of incorporating the advances in numerical control in the intervening years. Therefore, the chapter on ‘Numerical Control of Machine Tools’ was substantially modified and a new chapter on Extensions of Numerical Control – CNC, DNC, Machining Centers’ was added. To adequately, reflect the updated content, the title of the book was changed to Machine Tool Design and Numerical Control. In today’s ‘user-friendly age’, the revision for the third edition has been undertaken primarily to make the book more reader friendly and the changes are mostly based on the feedback from the reviewers and a survey carried out by the publisher. The important new features of this edition are summarised below: • A subsection on ‛Calculation of Machining Time’ has been added in Chapter 1. The highlight of this section is the inclusion of calculation of machining time of grinding operations which is usually not covered in text books. • A major section on ‛Kinematics of Machines Tools’ has been added in Chapter 2, wherein the gearing diagrams of lathe, drilling machine and milling machine have been discussed to give the reader a better understanding of the finer practical aspects of gear box design. A new attractive feature of this section is the discussion on thread cutting operation on lathe and operations using indexing head on milling machine based on fundamental principles, as distinct from the usual thumb rule type approach in most of the existing books. It is felt that this section will serve as useful base material for formulating design projects and independent assignments for final year students of mechanical and production engineering disciplines • In the second edition the design procedure of machine tool gear boxes was terminating with the calculation of gear teeth. This has been extended in the present edition, to its logical conclusion by adding a subsection on ‛Determination of Shaft and Gear Dimensions’ in Chapter 2. • A subsection on ‛Design of Lathe Bed’ has been added in Chapter 3 giving the detailed procedure supported with a solved example to provide practical illustration of the theoretical aspects for one specific case. This material will be helpful in formulating design projects and assignments not only for beds of various machine tools but also for other structural elements of machine tools such as bases, columns, tables etc. • A large number of solved examples have been added, especially in Chapters 1–3 in support of the elaboration of the new topics added in these chapters. In addition, new review questions have also been added in almost all the chapters.

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• A major curtailment has been undertaken in Chapter 8 on ‛Numerical Control of Machine Tools’. Previous edition contained a detailed description of the hardware of NC technology spread over seven subsections. Most of this technology has now become obsolete. It has, therefore, been thoroughly condensed and retained in one subsection only to the extent necessary for understanding the functioning/operation of NC machine tools. • Enhanced pedagogy includes Solved Examples: 25 Review Questions: 130 Computer Programs for NC, CNC and DNC: 12

Structure of the Book The book is organised into 9 chapters. Chapter 1 is an introductory chapter that provides a review of the concepts of working and auxiliary motions in machine tools and calculation of machining time of various operations. It also gives an overview of the elements of hydraulic and mechanical transmissions employed in machine tools. A section on layout of machine tools is unique to this book and is extremely relevant in the context of increasing emphasis on modularity and reconfigurability in CNC machines. Chapter 2 deals with the description of the laws of stepped regulation of speed and feed in machine tools and goes on to provide in rigorous detail the procedures for the design of gear boxes for stepped control of speed and feed, covering the whole gamut of issues from selecting the optimum structural diagram and speed chart to the finalisation of gearing diagram and determination of shaft and gear dimensions. A separate section is devoted to design of gear boxes with multiple speed motors and special gear boxes with overlapping speeds, broken geometric progression, etc., which have been supported with multiple diagrams. Stepless regulation of speed and feed rates by electrical, hydraulic and mechanical methods is discussed in great detail. To strengthen the understanding of kinematics of machining operations, thread cutting on lathe and operations using indexing head on milling machine are described from first principles, as distinct from the thumb rule approach presented in most of the existing books. In Chapter 3, the functions and requirements of the machine tool structures are discussed along with the design criteria and their application to individual structural elements such as beds, bases, columns etc. Aspects of design related to selection of the shape of structural elements and their strengthening with ribs and stiffeners are discussed in detail with lot of supporting data. In view of the complexity of their configuration and force system, it is seldom possible to analytically arrive at an exact design solution for structural elements of machine tools. Model techniques are therefore an essential part of the validation of their design and the fundamentals of these techniques are discussed at the end of the chapter. The description of the functions and classification of guide ways are dealt in Chapter 4. The design criteria of slideways are discussed and the detailed procedure of slideways design for stiffness and wear resistance based on average and maximum pressure is presented. Selection of slideway profiles and techniques of clearance adjustment and protection are presented and explained with the help of simple sketches. The design of hydrodynamic guideways, hydrostatic guideways, aerostatic guideways and anti-friction guideways is described in detail, supported with analysis as well as the relevant design data and curves. Design of sliding friction and rolling friction power screws is also included in this chapter. Chapter 5 discusses the functions and requirements of machine tool spindles and analyses the effect of the compliance of spindles and their supports on machining accuracy. A major portion of this chapter is devoted to the design of sliding bearings, hydrodynamic and hydrostatic journal bearings and aerodynamic and aero-

Contents Preface

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static bearings, supported with analysis and the relevant design data and curves. Issues specific to machine tools such as functional requirements, appropriate combinations of bearings for different machine tools and pre loading of bearing are discussed in detail. The thrust of Chapter 6 is to study the dynamic behaviour of a machine system for which the latter can be looked upon as a closed loop system in which the machine tool elastic system (MTES) and cutting process (CP) are the interacting elements. The dynamic cutting force models of Tlusty, Tobias and Kudinov are discussed and compared. Stability analysis of single and multiple degree of freedom systems with and without mode coupling is described. Regenerative chatter and the response of MTES-CP system under forced vibrations are also discussed. Dynamics of machine tools is a difficult topic, but by adopting a logical approach based on fundamental principles of control theory, it has been made easy to understand. Chapter 7 discusses the functions, requirements and classification of machine tool controls and goes on to describe the speed and feed changing mechanisms with simple centralised control, preselective control and selective control. For manual control systems, anthropometric and functional anatomy data has been systematically compiled for ergonomic design of control members such as push buttons, knobs, toggles, cranks, levers, hand wheels, etc., and also for the location of displays and control members. The highlight of this chapter is the detailed compilation of data for ergonomic design of control members which is not only unique to this book but also sets it apart from any other text book on machine tool design. Chapter 8 elaborates on the fundamental concepts of numerical control and classification of numerically controlled machine tools. It provides an overview of the NC hardware technology to the extent necessary for understanding the functioning and operation of NC machine tools. A major portion of this chapter is devoted to manual part programming for point-to-point, positioning- cum-straight cut and continuous path systems. The concept of computer aided part programming has been discussed and the APT programming system has been covered in reasonable detail. Both the manual and APT part programming systems have been illustrated with sample problems with step-by-step explanation of the part programs. The concluding chapter, i.e., Chapter 9 deals with the extensions of numerical control, namely computer numerically controlled (CNC) machine tools, machining centres and direct numerical control (DNC). A major portion of this chapter is devoted to CNC part programming for machining centres as well as turning centres. The programming concepts of tool diameter compensation, tool length off set, etc., are taken one ata-time and illustrated with suitable programming examples. Advanced programming features such as mirror imaging and canned cycles are discussed and illustrated with complete programs for sample parts. In addition, the present edition also contains new and improved solved examples, computer programs and chapter-end review questions to help students understand the concepts in a better way.

Acknowledgements I would like to thank my M.Tech student Ganesh Jagdale and Ph.D. student Vikas Upadhyay for their assistance in collecting and collating the reference material and for agreeing to be the sounding boards who helped me to fine tune the new material for this edition from the viewpoint of a student. I would like to take this opportunity to express my deep gratitude to Shri Rajesh Kumar for typing the text and drawing the technical figures for the new material of this edition and for always being by my side and providing technical and secretarial assistance to me over the years much beyond the call of duty. I would also like to thank the following reviewers for reviewing this book:

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Dinesh Khanduja National Institute of Technology, Kurukshetra Haryana

L. Bhaskara Rao Vellore Institute of Technology University, Vellore Tamil Nadu

A. P. Harsha Institute of Technology, Banaras Hindu University, Varanasi Uttar Pradesh

A. Venugopal National Institute of Technology, Warangal Andhra Pradesh

Ashish Banerjee Jadavpur University, Kolkata West Bengal

T. Rangaswamy Malnad College of Engineering, Hassan Karnataka

I am especially thankful to the editorial team of Tata McGraw Hill for their sincere help and guidance during the development of this book. I would like to personally thank Ms Harsha Singh, Ms Preyoshi Kundu and Mr In the end, I must confess that the response received from the readers for the earlier editions is both humbling and gratifying. I am grateful to them for their continuous support of my modest contribution to the teaching and practise of machine tool design and numerical control. I sincerely hope that the changes incorporated in the current edition will add more value to the book and will continue to provide useful service to students, teachers and practising engineers for many more years to come. N K MEHTA

Introduction to Machine Tool Drives and Mechanisms

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INTRODUCTION TO MACHINE TOOL DRIVES AND MECHANISMS— GENERAL PRINCIPLES OF MACHINE TOOL DESIGN

The machine tool is a machine that imparts the required shape to a workpiece with the desired accuracy by removing metal from the workpiece in the form of chips. In view of the extremely vast range of shapes that are in practise imparted to various industrial components, there exists a very large nomenclature of machine 1. By the degree of automation into (i) machine tools with manual control, (ii) semi-automatic machine tools, and (iii) automatic machine tools. 2. By weight into (i) light-duty machine tools weighing up to 1 t, (ii) medium-duty machine tools weighing up to 10 t, and (iii) heavy-duty machine tools weighing greater than 10 t. 3. By the degree of specialisation into (i) general-purpose machine tools—which can perform various operations on workpieces of different shapes and sizes, (ii) single-purpose machine tools—which can perform a single operation on workpieces of a particular shape and different sizes, and (iii) special machine tools—which can perform a single operation on workpieces of a particular shape and size.

1.1

WORKING AND AUXILIARY MOTIONS IN MACHINE TOOLS

For obtaining the required shape on the workpiece, it is necessary that the cutting edge of the cutting tool should move in a particular manner with respect to the workpiece. The relative movement between the workpiece and cutting edge can be obtained either by the motion of the workpiece, the cutting tool, or by a combination of the motions of the workpiece and cutting tool. These motions which are essential to impart the required shape to the workpiece are known as working motions into two categories: 1. Drive motion or primary cutting motion 2. Feed motion

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Machine Tool Design and Numerical Control

Working motions in machine tools are generally of two types: rotary and translatory. Working motions of some important groups of machine tools are shown in Fig. 1.1.

Fig. 1.1 Working motions for some machine tools

1. For lathes and boring machines drive motion—rotary motion of workpiece feed motion—translatory motion of cutting tool in the axial or radial direction 2. For drilling machines drive motion—rotary motion of drill feed motion—translatory motion of drill 3. For milling machines drive motion—rotary motion of the cutter feed motion—translatory motion of the workpiece 4. For shaping, planing, and slotting machines drive motion—reciprocating motion of cutting tool feed motion—intermittent translatory motion of workpiece 5. For grinding machines drive motion—rotary motion of the grinding wheel feed motion—rotary as well as translatory motion of the workpiece. Besides the working motions, a machine tool also has provision for auxiliary motions. The auxiliary motions do not participate in the process of formation of the required surface but are nonetheless necessary to clamping and unclamping of the workpiece, idle travel of the cutting tool to the position from where cutting is to proceed, changing the speed of drive and feed motions, engaging and disengaging of working motions, etc.

Introduction to Machine Tool Drives and Mechanisms

3

In machine tools, the working motions are powered by an external source of energy (electrical or hydraulic motor). The auxiliary motions may be carried out manually or may also be power-operated depending upon the degree of automation of the machine tool. In general-purpose machine tools, most of the auxiliary motions are executed manually. On the other hand, in automatic machines, all auxiliary motions are automated and performed by the machine tool itself. In between these two extremes, there are machine tools in which the auxiliary motions are automated to various degrees, i.e., some auxiliary motions are automated while others are performed manually.

1.2

PARAMETERS DEFINING WORKING MOTIONS OF A MACHINE TOOL

mary cutting motion or drive motion is known as cutting speed, while the velocity of feed motion is known as feed. The cutting speed is denoted by v and measured in the units m/min. Feed is denoted by s and measured in the following units: 1. 2. 3. 4.

mm/rev in machine tools with rotary-drive motion, e.g., lathes, boring machines, etc., mm/tooth in machine tools using multiple-tooth cutters, e.g., milling machines, mm/stroke in machine tools with reciprocating-drive motion, e.g., shaping and planing machines, and mm/min in machine tools which have a separate power source for feed motion, e.g., milling machines.

In machine tools with rotary primary cutting motion, the cutting speed is determined by the relationship, p dn m/min 1000 where d = diameter of workpiece (as in lathes) or cutter (as in milling machines), mm n = revolutions per minute (rpm) of the workpiece or cutter In machine tools with reciprocating primary cutting motion, the cutting speed is determined as v=

v=

L m/min 1000Tc

(1.1)

(1.2)

L = length of stroke, mm Tc = time of cutting stroke, min If the time of the idle stroke in minutes is denoted by Ti, the number of strokes per minute can be determined as 1 n= Tc + Ti where

Generally, the time of idle stroke Ti is less than the time of cutting stroke; if the ratio Tc /Ti is denoted by K, the expression for number of strokes per minute may be rewritten as n=

1 K = Tc (1 + Ti /Tc ) Tc (1 + K )

(1.3)

Machine Tool Design and Numerical Control

4

minute may be written as follows: v=

n ◊ L ( K + 1) 1000K

(1.4)

The feed per revolution and feed per stroke are related to the feed per minute by the relationship, sm = s ◊ n sm = feed per minute s = feed per revolution or feed per stroke n = number of revolutions or strokes per minute The feed per tooth in multiple-tooth cutters is related to the feed per revolution as follows: s = sz ◊ Z where s = feed per revolution sz = feed per tooth of the cutter Z = number of teeth on the cutter The machining time of any operation can be determined from the following basic expression:

(1.5)

where

Tm = where

1.2.1

L min sm

(1.6)

(1.7)

Tm = machining time, min L = length of machined surface, mm sm = feed per minute

Calculation of Machining Time sm is

l of a workpiece, the actual tool travel is greater on account of the need to provide an approach of D1 for safe entry of tool (on commencement of machining) and over travel of D2 for safe exit of tool (on completion of the machining cut). Generally, D1 and D2 are taken equal to 2–3 mm. The difference in the formulae of machining time corresponding tool travel. Hence, the calculation of tool travel for various operations is described below. In II at the end of cut.

Operations on Lathe (a) Turning operation on workpiece held between centres (Fig. 1.2) length of tool travel L = l + D1 + D2 + D3 where l = length of workpiece D1 = approach; generally equal to 2–3 mm D2 = over travel; generally equal to 2–3 mm D3 = t cot f; where t is depth of cut and f is principal or side cutting edge angle; for straight edged tools f = 90°, hence D3 = 0

Introduction to Machine Tool Drives and Mechanisms

f t l

D3

n

f

t

II

I

s

D2

D1

D3

L

Fig. 1.2 Turning operation on workpiece supported between centres

(b) Turning operation on workpiece clamped in chuck (Fig. 1.3) length of tool travel L = l + D1 + D3 where l = length of machined surface D1 and D3 are the same as in turning of workpiece held between centres

f t

l

D3

t n f

I

s II D1

D3

L

Fig. 1.3 Turning operation on workpiece clamped in chuck

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Machine Tool Design and Numerical Control

(c) Facing operation (Fig. 1.4) length of tool travel L = D/2 + D1 + D2 + D3 where D = diameter of workpiece D1 = approach; generally equal to 2–3 mm D2 = over travel; generally equal to 1–2 mm is essential to ensure that a protruding stem is not left attached to the face of the machined workpiece D3 = t cot f; where t is depth of cut and f is principal or side cutting edge angle; for straight edged tools f = 90°, hence D3 = 0 The length of tool travel for parting and grooving operations is determined in a similar manner. t f D3 t

II D2

n

f

L D1

D l= 2

D3

I s

Fig. 1.4 Facing operation

(d) Boring operation in partial length of workpiece; hole fd to be enlarged to fD (Fig. 1.5) length of tool travel L = l + D1 + D3 where l = length of bore D1 = approach; generally equal to 2–3 mm D3 = t cot f; where t is depth of cut and f is principal or side cutting edge angle; for straight edged tools f = 90°, hence D3 = 0

Introduction to Machine Tool Drives and Mechanisms

D3 l t f II

s

fD

fd

f n I D3

D1 L

Fig. 1.5 Boring operation in partial length of workpiece

(e) Boring operation in full length of workpiece; hole fd to be enlarged to fD (Fig. 1.6) length of tool travel L = l + D1 + D2 + D3 where l = length of bore D2 = over travel; generally equal to 2–3 mm D1 and D3 are the same as in boring operation in partial length of workpiece

D3 l

f II I

fD

fd

f n

D2

D1

D3

L

Fig. 1.6 Boring operation in full length of workpiece

t

7

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Machine Tool Design and Numerical Control

Example 1.1 Determine the machining time for turning a shaft from f 70 mm to f 64 mm over a length of 200 mm at n = 600 rpm and s = 0.4 mm/rev. The turning tool has principal cutting edge angle f = 45°. 70 - 64 = 3 mm 2 Length of travel L = 200 + t cot f + D1 + D2 Assuming D1 and D2 = 2 mm each

Depth of cut =

L = 200 + 3 ¥ 1 + 2 + 2 = 207 mm Machining Time =

207 = 0.8625 mm. 600 ¥ 0.4

Example 1.2 A ring has to be cut out from a pipe of outside diameter D = 100 mm and inside diameter d = 84 mm at 250 rpm and feed 0.14 mm/rev. Calculate the machining time. Length of travel in a pipe cutting operation is L=

D-d + D1 + D2 2

Assuming D1 = D2 = 2 mm 100 - 84 + 2 + 2 = 12 2 L 12 Machining time Tm = = 0.342 mm. = n ◊ s0 250 ¥ 0.14 L=

Operations on Drilling Machine

Drilling operation (Fig. 1.7)

length of tool travel L = l + D1 + D2 + D3 where l = height of the workpiece D1 = approach; generally equal to 2–3 mm D2 = over travel; generally equal to 2–3 mm D3 = (d/2) cot f, where d is drill diameter and 2f is the lip angle of the drill The length of tool travel for counter boring and reaming operations can be determined in a similar manner.

Introduction to Machine Tool Drives and Mechanisms

9

n I

s

f D3 D1 d 2 l

L

D2 D3

II

Fig. 1.7 Drilling operation

Example 1.3 Calculate the machining time for drilling a f 30 through hole in a 30 mm thick plate at a speed of 30 m/min and feed 0.1 mm/tooth. d cot f 2 Assuming D1 = D2 = 2 mm and f = 60° (half of lip angle) 30 L = 30 + 2 + 2 + cot 60° = 42.66 mm 2 The rpm of the drill is 1000 1000v 1000 ¥ 30 = = n= p pd p ¥ 30 Length travel L = 30 + D1 + D2 +

Feed per revolution of drill = 2 ¥ feed per tooth because a drill has two cutting teeth Therefore, so = 2 ¥ 0.1 = 0.2 mm/rev 200 1000 ¥ 0.2 = , mm/min p p 42.66 L = Machining time Tm = = 0.67 min. 200/p sm

Hence, feed per minute sm =

10

Machine Tool Design and Numerical Control

Operations on Milling Machine

In all the milling operations described below.

D1 = approach; generally equal to 2–3 mm D2 = over travel; generally equal to 2–3 mm (a) Horizontal milling machine: Plain milling operation (Fig. 1.8) length of cutter travel L = l + D1 + D2 + D3 where l = length of the workpiece D3 = BC = =

R 2 - OB 2 =

OC 2 - OB 2 =

2 Rt - t 2 =

R 2 - ( R - t )2 =

R 2 - ( R 2 + t 2 - 2 Rt )

t (D - t)

II I n

O

fD R t

B

C

A sz

D2

l

D1

D3

L

Fig. 1.8 Plain milling operation

(b) Vertical milling machine: Symmetrical face milling operation (Fig. 1.9) length of cutter travel L = l + D1 + D2 + D3 where l = length of the workpiece 2 Ê Bˆ D3 = AB = OA – OB = R – OC 2 - BC 2 = R – R 2 - Á ˜ = 0.5(D – Ë 2¯

D2 - B2 )

Introduction to Machine Tool Drives and Mechanisms

fD

I II C sz

B

R O

B

n

A

D1 D2

D3

l L

Fig. 1.9 Symmetrical face milling operation

D (Fig. 1.10) (c) Vertical milling machine: Asymmetrical face milling operation B > 2 length of cutter travel L = l + D1 + D2 + D3 where l = length of the workpiece D3 = AB =

OA2 - OB 2 =

R 2 - ( B - R)2 =

R 2 - R 2 - B 2 + 2 BR =

B ( D - B)

L D3

l II

D2

D1

I

A

R B O

FD sz

n

B

Fig. 1.10 Asymmetrical face milling operation, B >

D 2

D (d) Vertical milling machine: Asymmetrical face milling operation B < (Fig. 1.11) 2 length of cutter travel L = l + D1 + D2 + D3

11

12

Machine Tool Design and Numerical Control

l = length of the workpiece

where

OA2 - OB 2 =

D3 = AB =

R 2 - ( R - B)2 =

R 2 - R 2 - B 2 + 2 BR =

B ( D - B) I

II

R O

fD A

n

sz

B

D1

l

DZ

B

D3

L

Fig. 1.11

Asymmetrical face milling operation, B
Ps2, the ball valve opens, pressure at the piston end drops, the drained back to the reservoir and the line pressure drops. Spool and piston-type pressure valves are used mostly as bypass valves. The piston-type pressure valve has the ability to absorb minor pressure variations and is, therefore, the best from the point of view of pressure pulsations and dynamic behaviour.

PS2 Pr

3

P2 F PS1

2

P1 W 5

P0

Fig. 1.32 Schematic diagram of a compound-relief valve

5. Throttles Flow-control valves or throttles which have provision for changing the area of the constricted passage are

Machine Tool Design and Numerical Control

34

schematic diagrams of a few of the simplest throttle valves are given in Fig. 1.33.

Disc

(a)

Needle

(b)

Fig. 1.33 Throttle valves: (a) Globe valve (b) Needle valve

In all these valves, the area of the constricted passage is varied by displacing a movable member; for instance, the moving member in the globe valve (Fig. 1.33a) is a disc, and in the needle valve (Fig. 1.33b) a needle. In simple valves of this type, changes in oil temperature and pressure go uncompensated. Therefore, of the constricted passage. The chief aim of compensating for variations of oil pressure and temperature is to provide uniform travel of the machine-tool operative element. This aspect has been dealt with in Sec. 2.9.1 in which stabilisation of the motion velocity with the help of reducing valves has been discussed.

1.5

MECHANICAL TRANSMISSION AND ITS ELEMENTS

Mechanical transmission is employed for transmitting rotary as well as translatory motion to the operative element. This transmission can provide both stepped and stepless regulation of speed and feed rates. Stepless regulation is achieved through special devices called variators, which will be discussed in Sec. 2.9.3. A mechanical transmission that provides for stepped regulation of speed and feed rates is made up of elementary into the following groups: 1. 2. 3. 4. 5. 6.

lementary transmissions that transfer rotation lementary transmissions that transform rotary motion into translatory motion Devices for intermittent motion Reversing and differential mechanisms Special mechanisms and devices Couplings and clutches

1.5.1

Elementary Transmissions for Transmitting Rotary Motion

The important elementary transmissions which are used for transmitting rotary motion from one shaft to another are described below.

Introduction to Machine Tool Drives and Mechanisms

Gear Transmission

In a gear transmission, the rpm of the driven shaft is determined as n2 = n1 ◊

where

n2 n1 Z1 Z2

35

Z1 Z2

= rpm of the driven shaft = rpm of the driving shaft = number of teeth of the driving gear = number of teeth of the driven gear

The ratio Z1/Z2 is known as the transmission ratio of the gear drive and is constant for a particular gear pair.

(a)

(b)

(c)

Fig. 1.34 Gears: (a) Spur (b) Helical (c) Herringbone

Rotation is transmitted between parallel shafts by means of spur, helical and herringbone gears (Fig. 1.34). Spur gears have teeth parallel to the axis of rotation, while in helical gears the teeth are inclined with respect to the axis of rotation at an angle known as the helix angle. The herringbone gear is essentially a pair of helical gears in which the helix angle is oppositely directed. Spur gears are used in sliding gear blocks, while helical gears are preferred when the gear pairs are permanently in meshing. Transmission of rotation between inclined intersecting axes is done with the help of bevel gears. A bevel gear is shown in Fig. 1.35a. The angle between the inclined axes is generally 90° and the bevel-gear transmission (Fig. 1.35b) is commonly employed for transmitting rotation between perpendicular shafts. Transfer of rotation between skewed axes, i.e., axes that are inclined to each other but do not intersect, is achieved by means of a spiral gear transmission (Fig. 1.36a) or a worm-worm gear transmission (Fig. 1.36b). The spiral gear transmission is characterised by point contact between the meshing gears, and therefore, it cannot be employed for transmitting large torques. In machine tools, the worm-worm gear transmission is commonly employed to achieve heavy speed reduction. Also, since the contact between the worm and worm gear is along a line, this pair can transmit large torques. It should be noted that the worm-worm gear

36

Machine Tool Design and Numerical Control

transmission is irreversible and rotation may be transmitted from the worm to the worm gear, but not vice versa. The worm is, in principle, a helical screw and the rpm of the worm gear can be determined by the relationship,

(b)

(a)

Fig. 1.35 (a) Bevel gear (b) Bevel gear pair

(a)

(b)

Fig. 1.36 (a) Spiral gear pair (b) Worm-worm gear pair

Introduction to Machine Tool Drives and Mechanisms

n2 = n1 ◊ where

n2 n1 Z k

= rpm of the worm gear = rpm of the worm = number of teeth of the worm gear = number of passes of the worm

37

k Z

For a single pass worm, k = 1, for a double pass worm, k = 2. If a transmission chain consists of a number of elementary gear transmissions connected in series, the overall transmission ratio of the chain is obtained as the product of transmission ratios of the elementary transmissions. In general, the transmission ratio of a gear drive may be > 1 (speed increase) or < 1 (speed reduction), except the worm-worm gear transmission which always has a transmission ratio < 1.

Belt Transmission The belt transmission is used for transmitting rotation between shafts that are located at a considerable distance from each other. It is distinguished by smooth and jerk-free rotation which enables its application in high-speed machine tools, e.g., grinding machines. Belt transmission can be employed for transmitting rotation between parallel and skewed shafts. The most commonly used arrangements are shown in Fig. 1.37.

Fig. 1.37

Belt drives: (a) Open-belt arrangement (b) Cross-belt arrangement (c) Quarter-turned arrangement

38

Machine Tool Design and Numerical Control

The open-belt arrangement (Fig. 1.37a) is employed for transmitting motion between parallel shafts rotating in the same direction. The cross-belt arrangement (Fig. 1.37b) is used when rotation is transmitted between parallel shafts rotating in opposite directions and the quarter-turned arrangement (Fig. 1.37c) is used for transmitting rotation between skewed shafts. machine tools in which torques are of small magnitude. Flat belts are the most versatile as they can be em-

a number of V-belts (generally two to four) are used for varying the load-carrying capacity in order to avoid large bending stresses in one V-belt, which would otherwise be of unduly large dimensions. V-belts are usually employed only in the open-belt arrangement. For proper functioning of the belt drive, it is essential to provide some mechanism which keeps the belt tight during operation; this increases their cost. Other major drawbacks of the belt transmission are its relatively large dimensions and inability to guarantee constant transmission ratio due to unavoidable slip between the belt and pulleys. The rpm of the driven shaft in the belt drive may be determined by the relationship, n 2 = n1 where

n2 n1 D1 D2 x

D1(1 - x ) D2

= rpm of the driven shaft = rpm of the driving shaft = diameter of the driving pulley = diameter of the driven pulley = relative slip between belt and pulley

The value of x varies between 0.01–0.02 depending upon the belt material. The belt transmission can be employed to provide transmission ratios > 1 as well as < 1.

Chain Transmission The chain transmission (Fig. 1.38) is employed for transmitting rotation only between parallel shafts that are located at a considerable distance. The chain transmission consists of a driving sprocket, driven sprocket and chain. Chain transmission is used in machine tools when it is essential to keep the dimension of the drive within reasonable limits and also ensure transmission without slip. The rpm of the driven shaft is determined as, n2 = n1 where

n1 n2 Z1 Z2

Z1 Z2

= rpm of the driving shaft = rpm of the driven shaft = number of teeth on the driving sprocket = number of teeth on the driven sprocket

The chain transmission is also capable of providing transmission ratios < 1 and > 1.

Introduction to Machine Tool Drives and Mechanisms

39

Fig. 1.38 Chain transmission

1.5.2

Elementary Transmission for Transforming Rotary Motion into Translatory

These elementary transmissions are employed in feed mechanisms of most of the machine tools and also in the drives of machine tools having a reciprocating primary cutting motion. The important elementary transmissions that are used in machine tools for transforming rotary motion into

Slider Crank Mechanism The schematic diagram of a slider crank mechanism is shown in Fig. 1.39. The mechanism consists of a crank, connecting rod and slider. The forward and reverse strokes each take place during half a revolution of the crank. Therefore, the speeds of forward and reverse speeds in the slider crank mechanism are identical. Since metal removal occurs during one stroke (generally the forward stroke), it is desirable from the point of view of productivity to have a higher speed of the other stroke (the reverse stroke). Due to this property, the slider crank mechanism is used only in machine tools with small strokes (< 300 mm), where an increase of the reverse-stroke speed does not result in an appreciable increase of productivity, e.g., in the drive of the primary cutting motion of gear shaping machines. The length of stroke may be changed by adjusting the crank radius and is equal to L = 2R, where R is the crank radius. Crank-and-Rocker Mechanism The crank-androcker mechanism (Fig. 1.40) consists of a rotating crank which makes the rocker arm oscillate by means of a block sliding along the groove in the rocker arm. The forward cutting stroke takes place during the clockwise rotation of the crank through angle a, and the reverse (idle) stroke during rotation of the crank through angle b. Since a > b and the crank rotates with uniform speed, the idle stroke

Crank

Connecting rod

Slider

Fig. 1.39 Slider crank mechanism

Fig. 1.40 Crank-and-rocker mechanism

40

Machine Tool Design and Numerical Control

is completed faster than the cutting stroke. The length of stroke can be varied by adjusting the crank radius. With a decrease in the crank radius, the ratio of angles a/b decreases and the speeds of cutting and reverse strokes tend to become equal. The crank-and-rocker mechanism is, therefore, preferred in machine tools with large strokes (up to 1000 mm) where it can be effectively employed, e.g., in the drive of the primary cutting motion of shaping and slotting machines. The length of stroke can be calculated from the expression, Ê Lˆ L = 2 Á ˜ R mm Ë e¯ where

L = length of the rocker arm, mm e = off-set distance between the centres of rotation of the rocker arm and crank, mm R = radius of the crank, mm

Cam Mechanism The cam mechanism (Fig. 1.41) consists of a cam and a follower. The cam mechanism 1. on the periphery of a disc—disc type cam mechanism (Fig. 1.41a), 2. on the face of a disc—face type cam mechanism (Fig. 1.41b), and 3. on a cylindrical surface—drum type cam mechanism (Fig. 1.41c).

e

(a)

(b)

(c)

Fig. 1.41 Cam mechanism: (a) Disc type (b) Face type (c) Drum type

The main advantage of cam mechanisms is that the velocity of the operative element is independent of the radius changes from R1 to R2 (Fig. 1.42a) along an Archimedes’ spiral while the cam rotates through angle a, the velocity of the follower can be determined from the expression: v= where

n = rpm of the cam R1, R2 = radii, mm

n R2 - R1 m/min ◊ 360 ◊ 1000 a

Introduction to Machine Tool Drives and Mechanisms

R1

41

a R2

b

a

(a)

c

(b)

Fig. 1.42 (a) Profile of a disc-type cam (b) Development of the profile of a drum-type cam

Similarly, in face- or drum-type cam mechanisms, the speed of the follower depends upon the steepness a depicts the steep rise of the follower corresponding to the rapid advance, segment b depicts the slow rise corresponding to the working stroke and segment c the steep fall corresponding to the rapid withdrawal of the cutting tool. The speed during, say, the working stroke, can be determined by the following relationship: v= where

h b D n

h pD ◊ ◊ n m/min b 1000

= rise during the working stroke, mm = length of the working stroke, mm = diameter of the drum, mm = rpm of the drum

It should be kept in mind that cam mechanisms are costly and a new set is required whenever any change in working conditions is sought to be incorporated. Cam mechanisms are, therefore, generally used in automatic machine tools for mass production of components.

Nut-and-Screw Transmission A nut-and-screw mechanism is schematically depicted in Fig. 1.43. The screw and nut have a trapezoidal thread. When the screw, axis. The direction of movement can be reversed by reversing the rotation of the screw. The nut-and-screw transmission is compact, but has a high load-carrying capacity. Its other advantages are simplicity, ease of manufacture, and possibility of achieving slow and uniform movement of the operative member. The speed of the operative member can be found from the relationship,

Fig. 1.43

sm = t ◊ K ◊ n mm/min

Schematic diagram of a nut-and-screw transmission

42

where

Machine Tool Design and Numerical Control

sm t K n

= feed per minute of the operative member = pitch of the thread, mm = number of starts of the thread = rpm of the screw

frictional losses. This restricts its application in machine tools to feed and auxiliary motion drives. In this transmission, the sliding friction between the nut and screw is replaced by rolling friction by introducing intermediate members, such as balls and rollers. An anti-friction nut-and-screw transmission with balls as rolling members is shown in Fig. 1.44. The balls run along the thread between the screw and the nut and there is provision for their continuous recirculation. For instance, in the transmissions shown on Fig. 1.44, the balls return through an axial channel drilled in the nut (Fig. 1.44b) and through an external return chute (Fig. 1.44a). The thread of the screw and nut in this case is usually half-round and the transmission has provireaches 0.9– 0.95 as compared to 0.2–0.4 of the sliding-friction transmission. The anti-friction nut-and-screw transmission is mainly used in the feed-motion drive of precision machine tools, such as grinding, jig-boring machines, etc. It is used in numerically controlled machine tools in which backlash is extremely undesirable.

(a)

(b)

Fig. 1.44 Schematic diagram of anti-friction nut-and-screw transmission

Rack-and-Pinion Transmission A rack-and-pinion transmission is shown in Fig. 1.45. When the rotating gear (pinion) meshes with a stationary rack, the centre of the gear moves in a straight line. On the other hand, if the gear axis is stationary, then the rack executes translatory motion. The direction of motion can be reversed by reversing the rotation of the pinion. The speed of the operative member in this transmission can be found from the relationship, sm = p m ◊ Z ◊ n mm/min where

sm m Z n

= feed per minute of the operative member = module of the pinion, mm = number of teeth of the pinion = rpm of the pinion

Fig. 1.45 Rack-and-pinion transmission

Introduction to Machine Tool Drives and Mechanisms

43

Rack-and-pinion transmission is the simplest and cheapest among all types of transmissions used in reuse it in the feed as well as main drive motions of machine tools. Lack of uniformity in movement due to unavoidable meshing errors between rack-and-pinion teeth preclude its application in precision machine tools. Also, due to absence of self-locking, rack-and-pinion transmission cannot be applied for vertical movement of the operative element.

1.5.3

Devices for Intermittent Motion

In some machine tools, it is required that the relative position between the cutting tool and workpiece should change periodically. This requirement is generally essential in 1. machine tools with a reciprocating primary cutting motion, e.g., shaping machines in which the workpiece must be fed intermittently upon completion of one full stroke of the cutting tool, and 2. machine tools with reciprocating feed motion, e.g., grinding machines, in which the workpiece must be infed intermittently after each half or full stroke of the reciprocating table. In machine tools, intermittent motion of the operative element is generally obtained with the help of the mechanisms discussed as follows.

Ratchet-Gear Mechanism

The ratchet-gear mechanism is schematically shown in Fig. 1.46. It consists of a pawl mounted on an oscillating pin. During each oscillation in the anti-clockwise direction, the pawl turns the ratchet wheel through a particular angle. During the clockwise oscillation in the opposite direction, the pawl simply slides over the ratchet teeth and the latter remains stationary. The ratchet wheel is linked to the machine-tool table through a nut-and-screw transmission. Therefore, the periodic rotation of the ratchet wheel is transformed into the intermittent translatory motion of the table. For a particular nut-and-screw pair of some constant transmission ratio, the feed of the table during each oscillation depends upon the swing of the oscillating pawl. Generally, the rotation of the ratchet wheel in one stroke of the pawl should not exceed 45°. The ratchet-gear mechanism is most suitable in cases when the periodic displacement must be completed in a short time, e.g., in feed mechanisms of shaping, planing and grinding machines in which the intermittent feed Fig. 1.46 Pawl-and-ratchet mechanism motion takes place during the over travel of the cutting tool or during the reverse stroke.

Geneva Mechanism The schematic diagram of the Geneva mechanism is shown in Fig. 1.47. It consists of a driving disc which rotates continuously and a wheel with four radial slots. The arcs on the driving disc and wheel provide a locking effect against rotation of the slotted wheel, e.g., in the position shown in Fig. 1.47a, the wheel cannot rotate. As the disc continues to rotate, point A of the disc comes out of contact with the arc and immediately thereafter pin P mounted at the end of the driving arm enters the radial slot. The wheel now begins to rotate (Fig. 1.47b); when it has turned through an angle 90°, the pin comes out of the

44

Machine Tool Design and Numerical Control

radial slot and immediately thereafter point B comes in contact with the next arc of the wheel preventing its further rotation. Thus the wheel makes 1/K revolutions, where K is the number of radial slots.

Fig. 1.47 Geneva mechanism

In the Geneva mechanism, the angle of rotation of the wheel cannot be varied. Therefore, this mechanism is mainly used in turrets and single-spindle automatic machines for indexing cutting tools and in multiplespindle automatic machines for indexing spindles through a constant angle.

1.5.4 Reversing and Differential Mechanisms Reversing Mechanism Reversing mechanisms are used for changing the direction of motion of the operative member. Reversing is accomplished generally through spur and helical gears or bevel gears. A few reversing arrangements using spur and helical gears are shown in Fig. 1.48. In the arrangement of Fig. 1.48a the gears on the driving shaft are mounted rigidly, while the idle gear and the gears on driven shaft III are mounted freely. The jaw clutch is mounted on a key. Rotation may be transmitted to the driven shaft either through gears (A/B) ◊ (B/C) or through D/E depending upon whether the jaw clutch is shifted to the left to mesh with gear C or to the right to mesh with gear E. In the transmission (A/B) ◊ (B/C) the direction of rotation of the driving and driven shafts will coincide, whereas in the transmission D/E the direction of rotation of the driven shaft will be opposite to that of the driving shaft. In this arrangement, use of helical gears should be preferred. In the second arrangement shown in Fig. 1.48b, the gears on the driving shaft are again rigidly mounted, and the idle gear is free. On the driven shaft, a double cluster gear is mounted on a spline. By sliding the cluster gear, transmission to the driven shaft may again be achieved either through gears (A/B), (B/C) or through gear pair D/E. Only spur gears may be used in this reversal mechanism. In the arrangement of Fig. 1.48c gear A on the driving shaft and gear D on the driven shaft are both rigidly mounted. A quadrant with constantly meshing gears B and C can be swivelled about the axis of the driven shaft. By swivelling the quadrant with the help of a lever, transmission to the driven shaft may be achieved through (A/C) ◊ (C/D) or through (A/B) ◊ (B/C) ◊ (C/D).

Introduction to Machine Tool Drives and Mechanisms

45

and driven shafts will coincide while in the second it will be opposite. In this mechanism also only spur gears can be used. D

D

I

II

A

A

A ¥

¥

I

I

B II

II

III

III

III

¥

¥

I

C

II

B

B

III

C

C Jaw clutch

D

E

E (a)

(c)

(b)

Fig. 1.48 Reversing mechanisms: (a) using spur gears (b) using helical gears

It should be noted that in the reversing mechanisms of Fig. 1.48a and b the ratio of direct and reversal speeds will depend upon the transmission ratio of gear pairs A/C and D/E. By selecting A/C = D/E we can ensure identical speeds in both directions. However, if desired, a faster reversal speed can be achieved by selecting a larger transmission ratio for the gear pair used in the reversal train (gear pair A/C, as the transmission with the idler gear is usually employed for reversal). driving shaft and shaft II the driven shaft. In the arrangement of Fig. 1.49a, the double-cluster bevel gear is mounted on a splined shaft, and by shifting it the direction of rotation of shaft II can be changed by getting either gear B or gear C to mesh with bevel gear A which is rigidly mounted on the driving shaft. In the arrangement of Fig. 1.49b, gears B and C are freely mounted on the driven shaft, while the jaw clutch is mounted on splines. By shifting the clutch to the left or right, rotation to shaft II can be transmitted either through bevel gear pair A/B or A/C and thus the direction of rotation of the driven shaft can be reversed. C

B

A

B

C

x

x I (a)

A

I (b)

Fig. 1.49 Reversing mechanisms using bevel gears

46

Machine Tool Design and Numerical Control

Differential Mechanism Differential mechanisms are used for summing up two motions in machine tools, in which the operative member gets input from two separate kinematic trains. They are generally employed in thread-and-gear cutting machines where the machined surface is obtained as a result of the summation of two or more forming motions. A simple differential mechanism using spur or helical gears is shown in Fig. 1.50. The mechanism is essentially a planetary gear mechanism consisting of sun gear A, planetary gear B and arm C. The planetary gear is mounted on the arm which can rotate about the axis of gear A. Suppose gear A makes nA and arm C, nC revolutions per minute in the clockwise direction. The relative motion between the elements of the mechanism will remain unaffected if the whole mechanism is rotated in the anti-clockwise direction with nC revolutions per minute. Then the arm becomes stationary and the mechanism is reduced to a simple gear transmission with gear A making nA – nC revolutions per minute and gear B making nB – nC revolutions per minute. The transmission ratio of the mechanism may be written as: Z n A - nC =– B ZA nB - nC

(the minus sign denotes the external gear pair) B nB

C nC

nA A

Fig. 1.50 Differential mechanism using spur or helical gears

where ZA and ZB are the number of teeth of gear A and B, respectively. The above expression may be rewritten as follows: Z ˆ Z Ê nB = nC Á1 + A ˜ – nA ◊ A ZB ¯ Ë ZB i.e., the rpm of any one element of the differential mechanism is a function of independent motions of the remaining two elements. Differential mechanisms using a double-cluster planetary gear are shown in Fig. 1.51. The mechanisms consist of gear A, cluster gear block B–B¢ mounted on arm C and gear D. If nA, nC and nD are the rpm’s of gear A, arm C and gear D, respectively, then the transmission ratio of the kinematic train between gears A and D may be expressed as nD - nC Z Z¢ = A ◊ B n A - nC ZB ZD

(for Fig. 1.51a)

Introduction to Machine Tool Drives and Mechanisms

47

Fig. 1.51 Differential mechanisms using double-cluster planetary gears

Differential mechanisms consisting of bevel gears are shown in Fig. 1.52. These mechanisms are widely used in automobiles to provide different rotational speeds to the wheels powered by a single source. This is essential for the functioning of an automobile because, while tackling a turn, the outer wheel of the automobile must rotate faster than the inner wheel. This mechanism is also widely used in machine tools on account of its compactness. The mechanism consists of bevel gears A and D and planetary bevel gears B and C. Planetary gears can be rotated about the common axes of gears A and D 1. by means of a ring gear (Fig. 1.52a)—this differential is used in automobiles, and 2. by means of a T-shaped shaft (Fig. 1.52b)—this differential is used in machine tools.

Fig. 1.52 Differential mechanisms: (a) used in automobiles (b) used in machine tools

48

Machine Tool Design and Numerical Control

If gears A, B and D make nA, nB and nD revolutions per minute, respectively, then the transmission ratio of the kinematic train between gears A and D can be written as n A - nB Z Z =– A ◊ B nD - nB ZB ZD where ZA, ZB and ZD are the number of teeth of gears A, B and D, respectively. The minus sign indicates that gears A and D rotate in opposite directions if the rotation of the arm is stopped, i.e., nB = 0. If ZA = ZD, the expression becomes n A - nB = –1 nD - nB

wherefrom

nA + nD = 2nB In the automobile differential, the constancy of the sum nA + nD indicates that when the vehicle is taking a turn a reduction in the rpm of one wheel is accompanied by an increase in the rpm of the other. If the automobile is travelling on a straight line, nA = nD = nB, but if on a bend nA = 0, wheel D begins to rotate at twice the speed of the ring gear, i.e., nD = 2nB.

1.5.5

Special Mechanisms and Devices

Special mechanisms and devices are employed in machine tool feed boxes. These mechanisms are: 1. Gear cone with sliding key 2. Norton gear mechanism 3. Meander’s mechanism They are discussed in Sec. 2.8.2.

1.5.6 Couplings and Clutches Couplings and clutches are devices used for connecting one rotating shaft to another. If two shafts are permanently connected so that they can be disengaged only by disassembling the connecting device, the latter is known as a coupling. Devices that can readily engage shafts to transmit power and disengage them when desired are known as clutches.

Couplings

Couplings are of two types:

1. Rigid 2. Flexible Rigid couplings require that axial alignment between the connected shafts be maintained strictly. In

Introduction to Machine Tool Drives and Mechanisms

A

C

B

(b)

Fig. 1.53 (a) Rigid coupling (b) Flexible coupling

A and B with diametrical slots and an intermediate plate C with A and B. Slight misalignment between the connected shafts is compenIf there is considerable misalignment between the shafts to used. In this coupling, the shafts are connected through a Cardan or Hooke’s joint, which consists of yokes that are mounted on the ends of the shafts and a cross that provides a pivot joint between the yokes.

Fig. 1.54 Elastic coupling

49

50

Machine Tool Design and Numerical Control

or feed box.

Clutches 1. Positive-action clutches 2. Friction clutches A positive-action clutch is incapable of slipping. It can be engaged only when the shafts to be connected are stationary or are rotating at identical speed. The most commonly used positive-action clutch is the jaw shafts and is stationary, while the other is mounted on the second shaft on a key or splines and is moved into engagement. The faces of both the halves have projections, or so-called jaws and recesses such that the jaws

Fig. 1.55 Jaw clutch

A friction clutch, as the name implies, transmits torque by virtue of friction between the two halves. It can engage shafts rotating with different speeds or a rotating shaft with a stationary shaft. Friction clutches are generally not capable of transmitting large torques on account of slip. The commonly used friction clutches are discussed below. A disc-type friction clutch consists of one or more discs which are pressed against each other between the

with external splines, outer discs 3 with splines on their periphery and inner discs 4 with slines on their bore hole. The housing is rigidly mounted on one of the shafts and the sleeve on the other. Now, the discs are assembled by slipping them alternately along the splines of the housing and the hub. Thus, the outer discs rotate with the housing but are free to slide axially along its internal splines. Similarly, the inner discs rotate with the hub but can slide along its external splines. If the discs are to operate in oil, they are made of hardened steel. Since oil greatly reduces friction between discs, most clutches are operated dry. In such a case, the metal discs experience extensive wear, and therefore, one group of discs (generally outer one) is made of solid asbestos or a layer of asbestos is bonded on the metal discs.

Introduction to Machine Tool Drives and Mechanisms

51

3 1 4 2 Q 5

Fig. 1.56 Multiple-disc friction clutch

When the engaging sleeve 5 is moved towards the left, it exerts an axial force which is multiplied by the lever arrangement and applied on the friction discs. The discs get pressed against each other and the clutch gets engaged to transmit rotation between the two shafts. The lever system is so designed that it holds the clutch in engagement so that it is not necessary to continuously apply a force on the operating handle. Disc-type friction clutches have large load-carrying capacity with small overall dimensions. They are distinguished by smooth engagement and their capacity can be easily varied by increasing or decreasing the number of discs according to the requirement. Generally, the number of discs does not exceed 10–12 because otherwise there is wear between rotating discs even when the clutch is disengaged. clutch is essentially a multiple-disc friction clutch in which friction discs are pressed by an electromagnet. These clutches are particularly suitable for automatic control and are, therefore, being widely used in numerically controlled machine tools. A cone-type friction clutch is shown in Fig. 1.57. It a consists of two halves: one with an internal tapered surface is mounted on one shaft, while the other with an identical external taper is mounted on the other. One half is mounted rigidly, while the other is mounted on splines to permit axial displacement. The tapered surfaces are made of materials engaged by pressing the two halves against each other. If the contacting surfaces are made of hardened steel, half-taper angle a = 8–10°, while if the surfaces have an asbestos lining a = 12–15°. Because of the taper of the friction surfaces, a relatively small axial pressing force provides a large force normal to the contacting surfaces which holds them together once the clutch is engaged. Therefore, in cone-type friction

Fig. 1.57 Cone-type friction clutch

52

Machine Tool Design and Numerical Control

clutches, an elaborate linkage system is not required. This gives the cone clutch the advantage of simplicity. The major drawbacks that restrict the application of cone clutches in machine tools are their large dimensions and strict requirement of coaxiality between the connected shafts.

1.6

UNDERTAKING THE DESIGN OF A NEW MACHINE TOOL

general index of economic effectiveness. The economic effectiveness of a machine tool, and for that matter of any equipment, can be quantitatively expressed through the total annual cost which is represented as where

Ct C CI k

Ct = C + k ◊ CI = total annual cost = annual production cost = capital investment = factor of capital recovery along with interest, generally k = 0.15–0.2

The design and manufacture of a new machine tool can be considered economically feasible if Ctn < Cte i.e.,

Cn + kn (CI)n < Ce + ke (CI)e

(1.19)

where subscript n stands for the new machine tool and e for the existing machine tool that is sought to be replaced or updated. If the period of recovery of the capital investment is assumed to be the same in both the cases, i.e., kn = ke = k, (CI) n - (CI)e 1 < (1.20) Ce - Cn k Keeping in mind the relationship T = 1/k, where T is the period of recovery of the capital investment, (CI) n - (CI)e 75 m2, g = 1.5 Ebs = Ebsm ◊ As

where

Ebs = expenditure on building the servicing premises Ebsm = mean expenditure on building 1 m2 of servicing premises As = total area of servicing premises

Thus CI = En + Ebp + Ebs

Annual Cost of Production (C)

The annual cost of production of the designed machine tool is n

C=N◊

ÂC

i

i =1

where

C N Ci n

= annual cost of production = annual output = cost of the ith part of machine tool = number of parts in the machine tool

Machine Tool Design and Numerical Control

54

The production cost of a part includes versions are being compared, 2. wages paid to labour, 3. overheads; these cover the recurring expenditure on cutting tools, expenditure on the maintenance and expenditure on operating the equipment, etc. All cost factors need not necessarily be taken into account when two versions are being compared; it genversions. The design and manufacture and subsequent industrial application of new models of machine tools is one of the major factors in increasing productivity. Periodic renovation of production capacities is essential some initial investment; this must be recovered during the pay-back period, and for the remaining period of The search for new design and production solutions must, therefore, be based upon a thorough economic analysis on the lines discussed above.

1.7 Any machine tool should satisfy the following requirements: 1. 2. 3. 4. 5. 6.

High productivity A Simplicity of design Safety and convenience of controls Good appearance Low cost of manufacturing and operation

We shall discuss how these requirements are met in the design of machine tools. 1. Productivity Productivity of a metal cutting machine tool is given by the expression, Q= where

1 ◊h tc + tno

(1.22)

tc = machining time tno = non-productive time that includes job handling time, tool handling time, time of idle travel prior to commencement of cut, time of idle travel for guiding the tool to home position after completion of cut, set up time, inspection time and time spent on unscheduled delays, h = factor that accounts for stoppages for maintenance as well as unscheduled stoppages on account of breakdowns.

Introduction to Machine Tool Drives and Mechanisms

55

(i) Cutting down machining time: This is possible if high cutting speeds and feed rates are available on the machine tool in accordance with the latest developments in cutting tool materials and design. At the design stage itself, the machine tool must be provided with a margin to accommodate future developments so that it does not become obsolete in a short period of time. The application of stepless mechanical, hydraulic and electrical drives also helps in reducing machining time as the optimum cutting speed can be accurately set without reducing its value to the nearest available rpm on the machine tool with a stepped drive. Machining time can also be reduced by making provision for simultaneous multiple cuts and use of coolants. (ii) Cutting down non-productive time: clamping and unclamping time, and mechanising and automating machine tool controls. During the last few decades, developments in machine tool design have been largely directed at reducing the nonproductive time through automation. Hard automation in the form of automatic machines, mechanised industrialised nations, the consumers became more discerning and since the 70s the demand pattern has changed from mass produced goods to batch produced and custom built goods. This triggered a change in the manufacturing philosophy from one of hard automation to soft automation that is manifested today in the increasing proliferation of numerically controlled and computer numerically (iii) Machining with more than one tool simultaneously: This principle is employed in multiple-spindle lathes, drilling machines, etc. (iv) Improving the reliability of the machine tools to avoid break downs and adopt proper maintenance policy to prevent unscheduled stoppages and delays. 2. Accuracy The accuracy of a machine tool depends upon its geometrical and kinematic accuracy and its ability to retain this accuracy during operation. Accordingly, the ability of a machine tool to consistently methods: (i) Improving the geometrical accuracy of the machine tool: This is mainly determined by the accuracy of guiding elements, such as guideways, power screws, etc. It is also essential to ensure uniform, jerkfree movement of the traversing member of the machine tool. (ii) Improving the kinematic accuracy of the machine tool: The kinematic accuracy determines the relationship between velocities of two or more forming motions and it depends upon the length of kinematic trains and the accuracy of manufacture and assembly of components. Obviously, the kinematic accuracy of a machine tool can be improved by using as short kinematic trains as possible, and manufacturing and assembling the components with a high degree of accuracy. (iii) Increasing the static and dynamic stiffness of machine-tool structures: The greater is the static stiffness of the machine-tool structure, the smaller will be its deformation due to the cutting forces and hence the higher will be the accuracy of machining. A high dynamic stiffness reduces the vibrations during (iv) Providing accurate devices for measuring distance of travel: This concerns the accuracy of manufacture of dials, scales, verniers, optical systems, etc. The accuracy of measuring instruments is of paramount

56

Machine Tool Design and Numerical Control

importance in machine tools with automatic size control during machining, e.g., automatic machines, machine tools with adaptive controls, etc. (v) Arranging the machine tool units in such a manner that the thermal deformations during the machining operation result in the least possible change in the relative position between the tool and the workpiece. machines. 3. Simplicity of Design

Simplicity of design of machine tools determines the ease of its manufacture and

possible. The complexity of design of a machine tool depends to a large extent upon the degree of its ‘universality’. Thus a general-purpose machine tool is, as a rule, more complex than a special-purpose machine tool on its range of application, e.g., on the type of different operations that may be carried out, or on the size of parts which may be machined, etc. 4. Safety and Convenience of Controls requirements of safety and convenience of operation. Safety of controls is achieved by taking, among others, the following measures: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

Shielding the rotating and moving parts of the machine tool with hoods. Protecting the worker from chips, abrasive dust and coolant by means of screens, shields, etc. Providing reliable clamping for the tool and workpiece. Precluding the possibility of accidental pressing of push buttons and handles. Providing reliable earthing of the machine. Providing devices for safe handling of heavy workpieces. P Providing travel limiting devices for traversing machine tool members and also devices for overload protection. The convenience of machine-tool controls is intimately linked with their safety. Convenient controls will protect the worker from excessive fatigue and thus contribute towards safety. The convenience of controls also determines to a large extent the quality of the workers’ performance. Machine tool controls should be 1. The control system should be rationally selected and should be automated to as large an extent as possible. 2. The control system should be designed by giving due consideration to ergonomic principles. 5. Appearance thus facilitates better operation. It is generally conceded that a machine tool that is simple in design and safe contribute to the overall aesthetic quality of the machine tool. For instance, painting of machine tools in grey-green or green-blue colours impart a bright and pleasing appearance to the shop. Nowadays, painting of machines in different colours according to the production purpose is becoming popular, e.g., transportation facilities within the shop are painted yellow with black stripes, etc.

Introduction to Machine Tool Drives and Mechanisms

57

6. Low Cost of Manufacturing and Operation The cost of manufacturing a machine tool is determined by the complexity of its design. Therefore, factors that help in simplifying the machine tool design also contribute towards lowering its manufacturing cost. The cost can also be brought down by reducing the amount of metal required in manufacturing the machine tool. This is achieved by using stronger materials and more precise design calculations pertaining to the strength and rigidity of parts to keep the safety margins as low as possible. For instance, considerable saving of metal can be achieved by using welded steel structures instead of cast iron for heavy parts, such as beds, columns, bases, etc. It should also be noted that a reduction of the weight and dimensions of the machine tool also makes transportation and installation of the machine tool easier and cheaper, thus indirectly contributing to a further reduction of the overall cost.

1.8

ENGINEERING DESIGN PROCESS APPLIED TO MACHINE TOOLS

a man with a logical decision-making ability by which he explores all possible solutions to a given problem and arrives at an optimum after carefully analysing all the alternatives. Until a decade or so ago, any design which was technically feasible, i.e., capable of being manufactured, principally new design solutions to keep ensuring higher productivity has in recent times greatly increased the expenditure on design. Design is progressively becoming a team activity as optimum solutions can be found only by considering a large number of factors of diverse nature with which the designer may not always be well conversant. The block diagram of Fig. 1.58 shows how design is related to different engineering, economic, natural and social sciences. Social economy

Economics Physics and mathematics

Engineering Science like material science, theory of machines etc.

Design

Manufacturing sciences like welding, forging machining, etc.

Engineering aesthetics

Fig. 1.58 Block diagram depicting the influence of various sciences on design

In view of the heavy responsibilities on the designer and the large expenditure involved in designing a new machine, it is necessary to streamline the design process so that a sound design solution is achieved with minimum expenditure.

58

Machine Tool Design and Numerical Control

Fig. 1.59 Block diagram of the design process in respect of machine tools

The design process for designing a new machine tool is presented in the form of a block diagram in Fig. 1.59. It is evident from Fig. 1.43 that the design process is carried out in three important stages: 1. Design proposal 2. Preliminary design 3. Detailed design At the end of each stage, the design must be subjected to a critical feasibility analysis, and a technical report prepared and submitted to the customer. The steps involved in the design process will now be elaborated one by one. 1. Requirement The customer outlines the requirement by furnishing information about the parts for machining of which he wants the machine tool to be designed. The information should include the nomenclature of parts and their annual output, the dimensions and shapes of surfaces to be machined, materials of the parts,

Introduction to Machine Tool Drives and Mechanisms

59

lecting appropriate machining methods and cutting tools.

product does not already exist. A consideration in undertaking a new design, in the presence of available solutions, may be the need to make the product economically viable by reducing its cost. The designer must, therefore, make a preliminary assessment of the requirement to see whether it is economically feasible. If necessary, he may, in consultation with the customer, modify or expand the requirement to increase the market potential of the designed machine tool. design. The information furnished by the customer about the parts forms the basis of determining important

rate regulation, degree of mechanisation and automation to be employed on the machine tool, appearance of -

Fig. 1.603 and prevents him from deviating from the basic goal. 0

1

2

3

0

4

1

2

3

4

Cost

Cost Reliability

Reliability

Weight

Weight

Appearance

Appearance

Ease of inspection

Ease of inspection (a)

(b)

Fig. 1.60 Importance of specification items for: (a) Vacuum cleaner (b) Aircraft

3. Selection of Proper Kinematic Solution and Layout down, the designer explores the combinations of relative motions that can ensure machining of surfaces of required shapes and dimensions. The different possibilities are evaluated and those found technically feasible are selected. Kinematic solutions on the basis of basic motion combinations are now developed. All these solutions are analysed for their technical feasibility and infeasible solutions are screened. A kinematic solution correlates the motions of the workpiece and cutting tool and can be realised in a number of layouts of major machine tool units. A technical feasibility analysis, keeping in mind the constraints of the requirements and in Sec. 1.9.

60

Machine Tool Design and Numerical Control

4. Design Calculations Design calculations cover the design of the major units of the machine tools, such as the speed box, feed box, bed, spindle, etc. These calculations are done in accordance with design proceversion is selected by comparing the economic feasibility of implementation of alternatives. 5. Drawings of Components and Assemblies (including the manufacturing method to be employed). Special care should be taken during the stages of design calculations and detailed drawing to make use of standard components and assemblies as far as possible. It should be appreciated that design is essentially an iterative process. The feedback that is received after prototype fabrication and testing, and particularly after marketing the product must be carefully analysed to

and even the requirements, if these are conducive to more sound and/or economic design. These aspects are indicated in the design process (Fig. 1.59) by feedback loops.

1.9

LAYOUT OF MACHINE TOOL

The layout of the machine tool must provide the required combination of forming and setting motions that are necessary for the given machining process. The required relative motions between the cutting tool and workpiece are generally realised by means of a set of translatory and rotary motions. The layout of the machine tool will typically consist of one stationary block and a number of moving blocks divided by linear or circular guideways, the number of guideways being equal to the number of elementary motions provided on the machine tool. A particular layout is obtained by placing the stationary and moving blocks in a particular order. Different layouts are obtained by changing the order of these blocks. It is the task of the designer to analyse the various layout alternatives and select the best possible version, consistent with the constraints of the particular machine tool. The selection of a suitable layout can best be carried out by structural analysis using the Boolean-algebra technique. In this method, the machine tool structure of any complexity can be represented in the form of a combination of symbols. Let us introduce a set of symbols for this purpose. Let X, Y, Z represent the basic reciprocating displacements along the corresponding co-ordinate axes and U, V, W the additional displacements in the same directions. A, B, C represent rotary motions about axes X, Y, Z while D and E represent the additional rotary motions. sponding coordinate axes, e.g., x represents auxiliary setting motion in the x-direction, while a represents an auxiliary rotary motion about the same axis. Boolean algebra permits the consecutive linking of blocks which is represented as a conjunction (AND) and parallel linking which is represented as a disjunction (OR). The consecutive linkage of blocks may not be indicated by anything or may be indicated by a full stop (◊). Parallel blocks are written in brackets and parallel linking is indicated by a plus sign (+). The layout formula begins with the block carrying the workpiece and ends with the block carrying the cutting tool.

Introduction to Machine Tool Drives and Mechanisms

61

Some examples of machine tool layouts and their layout formulae are given in Fig. 1.61.4 Figure 1.61a shows the layout of a knee-type vertical milling machine with consecutive linking of blocks. In the layout formula XYZOCv, X Y Z O C

represents table travel represents cross-slide travel represents knee travel represents the stationary block (column) represents rotation of spindle about the Z-axis

subscript v indicates that the spindle is vertical. The lathe layout shown in Fig. 1.61b also consists of blocks linked in series. In the layout formula AOXYcwd, A represents rotation of the workpiece clamped in the spindle about the X-axis O represents the stationary block (bed) X represents carriage travel along bed guideways Y represents cross-slide travel c represents rotary setting motion of the compound slide w represents setting motion of the tool post d represents rotary setting motion of the tool post The layout of the gear-shaping machine is shown in Fig. 1.61c. In the formula DuOx(CZ)v, D u O x (CZ) v

represents rotation of workpiece about the vertical axis represents setting motion of the table represents the stationary block (columns) represents tool head travel along cross-rail guideways indicates that the cutting tool experiences two simultaneous cutting motions—translatory motion along the Z-axis and rotation about the same axis indicates that the spindle is vertical

The unit built drilling machine shown in Fig. 1.61d consists of blocks linked in parallels. In the layout formula ydO(X4A + Y 4B h + Z5Cv), y d X4A Y4Bh Z5Cv

represents setting motion of the workpiece represents rotary setting motion of the workpiece indicates that there are four spindles rotating about the X-axis in the spindle head which itself travels in the direction of the X-axis indicates that there are four spindles rotating about the Y-axis in the spindle head which itself travels in the direction of the Y-axis; subscript h indicates that the spindles are horizontal Z axis; subscript v indicates that these spindles are vertical

62

Machine Tool Design and Numerical Control

Fig. 1.61 Layout and layout formulae for: (a) Knee-type vertical milling machine (b) Lathe (c) Gear-shaping machine (d) Unit-built drilling machine

The examples give a fair idea that an appropriate layout formula can be written for a machine tool of any complexity. The formal representation of the machine tool layout in the form of a layout formula enables us to obtain all possible layout versions by a mere readjustment of the symbols. The best layout is selected by a comparative analysis of all the versions. Consider a milling machine which is a triple co-ordinate machine consisting of four blocks—one stationary which is denoted by O and three moving denoted by X, Y and Z. The spindle rotation is not a forming motion and, therefore, does not affect the layout. The different layout versions are obtained by the permuta-

Introduction to Machine Tool Drives and Mechanisms

63

tion of symbols XYZO. The total number of possible versions is = 4! = 24. These versions are tabulated in the matrix shown in Fig. 1.62a. The columns of this matrix differ in terms of location of the stationary block, while the lines (in pairs) differ by the location of the vertically moving block Z. We can, for instance, isolate a subgroup of layouts with a moving column. These versions correspond to the matrix elements of Fig. 1.62a enclosed within the thick line and are shown in Fig. 1.63.

Fig. 1.62 Matrices of layout versions: (a) Complete matrix for layout formula XYZO (b) Submatrix for layouts in which vertical displacement is absent (c) Submatrix for layouts in which the workpiece receives only horizontal displacement (d) Submatrix for layouts in which the horizontal displacement occurs in the immediate vicinity of stationary block

The selection of the best layout from among the possible alternatives is done by successive elimination of those versions which do not satisfy the requirements (constraints) necessary for optimum functioning of the designed machine tool. The constraints are also formulated in the form of generalised structural formulae. For this purpose, the following new notations need to be introduced: O

represents a moving block pertaining to cutting tool displacement

Z

represents horizontal moving block

Let us write a few requirement (constraint) statements with the help of these symbols.

64

Machine Tool Design and Numerical Control

Fig. 1.63 Layouts of vertical milling machine with a moving column

Statement 1 The machine tool is meant for machining heavy parts. Therefore, it is not desirable that the machine tool table be given vertical displacement. The requirement may be stated as R1 = Z ZOO + ZOOO + OOOO In this statement the ‘+’ sign represents the OR function; the statement can be interpreted as indicating stationary block and then one (obviously vertical) motion to the cutter or imparting one horizontal motion to the work piece followed by the stationary block and then two motions (one horizontal and one vertical) to the cutting tool or rigidly attaching the workpiece to the stationary block and imparting all three motions (two horizontal and one vertical) to the cutting tool. Statement 2 The heavy parts must be machined with a high degree of accuracy. Therefore, to prevent the weight of the part from affecting the accuracy, the workpiece should be stationary or have only one horizontal displacement. The requirement may be stated as R2 = ZOOO + O OOO

Introduction to Machine Tool Drives and Mechanisms

65

Statement 3 To prevent the weight of the moving assembly with the workpiece from affecting the machining accuracy, the horizontal moving block must be adjacent to the stationary block. The requirement may be stated as R3 = ZZOZ + ZOZ Z by the triple coordinate milling machine for which the best layout has to be selected. The layout versions which satisfy requirement 1 are shown in Fig. 1.62b, while those satisfying requirements 2 and 3 are shown in Fig. 1.62c and Fig. 1.62d, respectively. To locate the versions that simultaneously satisfy all the requirements, the requirement statements are written in the matrix form so that the stationary blocks occupy identical positions. 1

2

R1

–

R2

–

R3

–

Z ZOZ

f

f

Z ZOO –

3 +

ZOOO Z OOO

+

4 + +

OOOO OOOO

Z OZ Z XOYZ

f

YOXZ The columns in which there is no formula or which contain contradictory statements are rejected (f). Thus for the given machine tool, the optimum layout lies in column 3. A comparison of this conclusion with Fig. 1.62b, c and d immediately provides the answer that the best layouts are XOYZ and YOXZ. These layouts are shown in Fig. 1.64. From among these two versions, which are equal in all respects from the point of view of the two structures, ease of control, cost of manufacture, etc.

Fig. 1.64 Layout versions which satisfy all constraints

1.1 Determine the rpm of a lathe spindle if a workpiece of diameter 100 mm is to be turned at a cutting speed of 88 m/min.

66

Machine Tool Design and Numerical Control

1.2 A 175 mm long workpiece is to be machined on a shaper which has a cutting stroke to idle stroke velocity ratio of 0.84. Calculate the strokes/min of the shaping tool if the cutting speed is 30 m min. Assume an approach and overshoot of 5 mm. 1.3 A 40 mm hole is drilled at a speed of 30 m/min and feed of 0.1 mm/tooth. Calculate the feed per minute of the operation. 1.4 L = 200 mm and width B = 100 mm at a speed of 80 m/min and feed of 0.15 mm/tooth. Calculate the machining time, assuming an approach and overshoot of 5 mm. (Hint: Travel = L + 0.5 (D – D 2 - B 2 ) + 10) 1.5 A 120 mm diameter workpiece is being turned on a lathe at 400 rpm and a feed of 0.3 mm/rev. The tangential, radial and axial components of the cutting force were measured on a dynamometer and found to be 325, 120 and 130 kgf, respectively. Calculate the power rating of the motor assuming a 1.6 A gear pump has a delivery rate of 150 1/min at 1400 rpm. The pitch circle diameter of the gears is 120 mm and they are 50 mm wide. Calculate their module. What is the maximum pressure that the leakage losses as 0.8 each. 1.7 A piston pump, powered by a 7.0 kW motor, delivers oil at a pressure of 75 kgf/cm2. The pump rotor is mounted with an eccentricity of 15 mm and rotates at 300 rpm. It has six pistons of diameter 30 mm 1.8 In a simple internal differential mechanism the sun gear rotates at 50 rpm and the planetary gear at 500 rpm. If the number of teeth of the sun and planetary gears are 120 and 20, respectively, calculate the rpm of the arm. 1.9 In a differential mechanism consisting of two external gear pairs (Fig. 1.51a), ZA = 20, ZB = 22, Z B¢ = 20, ZD = 18. Find the rpm of the arm if gear A rotates at 6 rpm. 1.10 Write the layout formulae for shaping, planing, drilling and broaching machines. 1.11 A f 40 shaft is machined at a speed of 80 m/min and feed of 0.1 mm/rev. Calculate machining time. Given shaft length = 100 mm. Assume suitable approach and over travel and sketch the operation. 1.12 A f 72 workpiece is faced at a speed of 40 m/min and feed of 0.15 mm/rev. Sketch the operation and calculate the machine time. 1.13 A f 20 blind hole is to be drilled to a depth of 30 mm at a speed of 30 m/min and feed of 0.1 mm/teeth. Calculate the machine time, assuming a suitable approach. 1.14 A f 40 hole is to be enlarged to f 50 in a boring operation. The cutting speed and feed is 40 m/min and 0.1 mm/rev, respectively and the job diameter is f 70. Calculate the machining time if the permissible depth of cut is 2 mm. Assume a suitable approach and make a neat sketch of the operation. Job length = 70 mm. 1.15 Single pass slab milling operation is carried out on a 500 mm long job by a f 70 plain milling cutter at n = 300 rpm and feed of 200 mm/min. If the cutter has 12 teeth, determine the feed per tooth and machining time. Assume suitable approach and over travel. 1.16 A f 80 face milling cutter is used for facing a 60 mm wide and 300 mm long job. The speed is 40 m/min and the feed is 40 mm/min. Calculate the machining time if the allowance is 10 mm and the permissible depth of cut is 5 mm. Assume a suitable feed rate of accelerated travel and time for manual vertical setting of the table.

Introduction to Machine Tool Drives and Mechanisms

67

1.17 A shaping operation is carried out at 20 m/min. if the length of stroke is 250 mm and the ration of 1.18 A f 60 hole is to be ground by a f 50 grinding wheel of width 50 mm. The length of the hole is 65 mm, allowance is 0.2 mm and radial feed 0.005 mm per stroke. Transverse feed (mm per revolution of work) sl = kB, where k = 0.3. If peripheral speed of the grinding wheel and workpiece is 30 m/s and 35 m/min respectively, determine the machining time. 1.19 the periphery of a grinding wheel of diameter 450 mm and width 63 mm. Six plates are ground in one setting, placed 3 along length. Assume approach of 5 mm in width and 15 mm in length. The speed of reciprocation of the table is 15 m/min. The rest of the data is the same as in Q1.18. 1.20 A slab of 50 mm width and 200 mm length is rough machined in a symmetrical face milling operation. If the cutter diameter is 75 mm, what will be the length of travel of the cutter? 1.21 In a slab milling operation the diameter of cutter is 50 mm and number of teeth of cutter is 12. If the cutter spindle speed is 300 rpm, depth of cut is 2 mm, length of job is 500 mm and table longitudinal feed is 200 m/min what will be the feed per tooth during the milling operation? If the cutter over travel is 2 mm what is the machining time for the single pass milling operation?

References Developing the Kinematic Diagrams of Machine Tool Speed Boxes, published by Peoples’ Friendship University, Moscow, 1978, p. 59. 2. Vershinin, VD, Electrical Control Circuits of Machine Tools, published by Peoples’ Friendship University, Moscow, 1969, p. 27. 3. Pitts, G, Techniques in Engineering Design, Butterworths, London, 1973, p. 9. 4. Vragov, Yu D, “Structural analysis of machine tool layouts”, Machines and Tooling, 1972, No. 8.

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Machine Tool Design and Numerical Control

2 2.1

REGULATION OF SPEED AND FEED RATES

AIM OF SPEED AND FEED RATE REGULATION

A machining operation should be conducted at such values of cutting parameters (speed, feed, depth of cut, etc.) that ensure the minimum cost price of the machined component. The machining cost can be expressed by the equation C = Cmt + Cnpt + Ctc + Ct

(2.1)

where Cmt = (W+ E)tm represents the cost of machining time; W is the wage rate, E is cost of operating the machine tool per unit time and tm the machining time. Cnpt = (W + E)tnpt represents the cost of non-productive time; tnpt is the total time of non-productive operations, such as loading and unloading, idle travel of cutting tool to provide for approach and over travel, etc. Ctc = (W + E)ttc/Q represents the tool changing cost per component; ttc is the time required for replacing (regrinding) a blunt tool and setting the new (reground) one and Q the number of components machined during the period of tool life. Ct = T/Q represents the cost of the tool per component; T is the cost of the tool for a period equal to the tool life and can be determined as the tool cost divided by the number of permissible regrindings. If the machining cost of Eq. (2.1) is optimised, it yields a particular value of tool life which corresponds to minimum machining cost. From the generalised tool life equation T = C/vasbtc, it follows that the optimum tool life can be achieved on a particular operation only by working at optimum values of cutting speed v, feed s and depth of cut t. In order to machine a part of arbitrary diameter, the spindle rpm must be set as n = 1000 , i.e., there must be a stepless regulation of v so that any desired value of the spindle rpm may be set corresponding to the optimum cutting speed. By a similar logic, the machine tool should have provision for stepless variation of the feed rate. Economically, viable systems of stepless speed and feed rate regulation have, however, not yet in recent years mainly due to the development of numerically controlled machine tools, most of the machine tools are still designed for stepped regulation of speed and feed rate. On such machine tools only certain discrete values of the spindle rpm and feed rate are available. It may be concluded from the preceding discussion that provision of regulating the spindle rpm and feed rate is an essential requirement of machine tools to ensure economic machining of workpieces of different materials and sizes by cutting tools of different shapes and compositions.

Regulation of Speed and Feed Rates

2.2

69

STEPPED REGULATION OF SPEED: DESIGN OF SPEED BOX

In this section, the principles of designing speed boxes, i.e., gear boxes that are employed for stepped regulation of the rpm of the main drive are dealt with. Most of what is discussed in connection with speed boxes is also relevant to the design of feed boxes, but the latter have some special design features that have been discussed separately in Sec. 2.3.

2.2.1

Various Laws of Stepped Regulation

It was stated above that in stepped regulation of speed only certain discrete values of the spindle rpm are available on the machine tool. A pertinent question that arises is what should be the criterion for choosing these discrete steps? Between two extreme available values n1 and nz of the spindle rpm, the same number of z intermediate steps may be placed in a number of ways. The various series of rpm values will have different operational characteristics. Let us analyse four cases and select the most suitable law of speed range distribution.

The rpm Values Constitute an Arithmetic Progression

The realisation of arithmetic progression is based upon the idea that the difference between adjacent rpm values is constant. n1 = n n2 = n 1 + a n3 = n2 + a = n1 + 2a nz = n1 + (z – 1)a

where a is the common difference of the arithmetic progression. For a particular cutting speed which is the maximum permissible under the selected cutting conditions, the diameter range of workpieces that can be machined by a particular spindle rpm value nx can be determined as follows: Upper limit of the range, 1000v dx = p nx Lower limit of the range, 1000v dx + 1 = p nx +1 Hence, the diameter range served by this particular rpm is Ddx =

1 ˆ 1000v Ê 1 p ÁË nx nx +1 ˜¯

70

Machine Tool Design and Numerical Control

Let us calculate the rpm values and diameter range served by each rpm for the following conditions: n1 = 30 rpm, nz = 375 rpm, number of speed steps z = 12, v = 20 m/min. These values are tabulated below in Table 2.1. Table 2.1 nx

fx

dx mm

Ddx mm

n1 = 30

2.04

212

n2 = 61.4

1.51

103.7

35.1

n3 = 92.8

1.33

68.6

17.3

n4 = 124.2

1.25

51.3

10.4

n5 = 155.6

1.20

40.9

6.9

n6 = 187

1.17

34.0

4.84

n7 = 218.4

1.14

29.16

3.66

n8 = 249.8

1.12

25.5

2.90

n9 = 281.2

1.11

22.6

2.20

n10 = 312.6

1.10

20.4

1.9

n11 = 344

1.09

18.5

1.6

n12 = 375

108.3

16.9

Suppose we start machining a workpiece of diameter 212 mm. For a cutting speed of v = 20 m/min, the value n1 = 30 rpm will correspond to optimum cutting. Before we can change over to the next higher rpm value of n2 = 61.4, we must remove 108.3 mm of metal from the workpiece diameter. Assuming a permissible ing uneconomically as the actual cutting speed would be less than the permissible value. On the other hand, in changing over from an rpm value of n11 = 344 to nl2 = 375, we have to reduce the workpiece diameter only by 1.6 mm. We could actually remove more than this allowance in one pass. It follows from the above analysis that in the high rpm range some values of speed steps are redundant, whereas in the low rpm range there is clearly a need to add more steps between the calculated values.

The rpm Values Constitute a Geometric Progression n1 = n n2 = n 1 f n3 = n2 f = n1f2 nz = n z–1 f = n1f z– 1

Regulation of Speed and Feed Rates

71

where f is the geometric progression ratio. For the data used in the example above, Ên ˆ f= Á z˜ Ë n1 ¯

1/( z -1)

= 1.26

Let us calculate the rpm values and diameter range served by each rpm for the same conditions as given in the arithmetic progression problem. The tabulated values are given in Table 2.2. Table 2.2 f

nx

dx mm

Ddx mm

n1 = 30

1.26

212

42

n2 = 37.5

1.26

170

36

n3 = 47.5

1.26

134

28

n4 = 60

1.26

106

21

n5 = 75

1.26

85

18

n6 = 95

1.26

67

13

n7 = 118

1.26

54

11.5

n8 = 150

1.26

42.5

9

n9 = 190

1.26

33.5

6.5

n10 = 235

1.26

27.0

5.8

n11 = 300

1.26

21.2

4.2

n12 = 375

1.26

17.0

4.2

Before we change over from a speed n1 = 30 rpm to n2 = 37.5 rpm an allowance of 42 mm must be machined from the workpiece diameter. Again, providing for a maximum depth of cut of t = 5 mm, this allown11 = 300 rpm to n12 = 375 rpm, we have to take a depth of cut of 2.1 mm which can be easily accomplished in one pass. We thus see that in order to make the machine tool performance equally feasible in the whole rpm range, the low rpm values should be brought still closer while the high rpm values can be widened a little.

The rpm Values Constitute a Harmonic Progression

The harmonic progression is developed from the basic idea that the diameter range served by each rpm of the progression is equal, i.e., Ddx = dx – d x+ 1 =

1 ˆ 1000v Ê 1 = const p ÁË nx nx +1 ˜¯

Therefore, 1 1 – = const = C nx nx +1

72

Machine Tool Design and Numerical Control

and nx+1 =

nx 1 - Cnx

The harmonic progression can be written as follows: n1 = n

n1 1 - Cn1 n1 n2 n3 = = 1 - Cn2 1 - 2Cn1 n2 =

n2 =

nz -1 1 - Cnz -1

=

n1 1 - ( z - 1)Cn1

The values of rpm and diameter range for the same data used in the previous examples are tabulated in Table 2.3. Table 2.3 nx

fx

dx mm

x

mm

n1 = 30

1.09

212

18

n2 = 32.7

1.11

194

18

n3 = 36.2

1.11

176

18

n4 = 40.0

1.13

158

18

n5 = 45.1

1.13

140

18

n6 = 51

1.18

122

17

n7 = 60.1

1.205

105

16

n8 = 72.4

1.25

89

18

n9 = 90.6

1.34

71

18

n10 = 121.2

1.51

53

18

n11 = 183

2.1

35

18

n12 = 375

—

17

—

It may be seen that in order to change from n11 = 183 rpm to n12 = 375 rpm, the workpiece diameter must be reduced from 35 mm to 17 mm. On a slender workpiece of 35 mm, a large depth of cut cannot be taken as this would lead to the deformation of the workpiece. Assuming a permissible depth of cut of t = 2 mm on

Regulation of Speed and Feed Rates

73

in harmonic progression the rpm values in the high range are too wide apart, making this range uneconomical for exploitation.

The rpm Values Constitute a Logarithmic Progression

In this progression, the diameter range

is a function of the diameter. D dx = 2Mdxp where M

p = 0.5. dx – dx+1 = 2M ◊

d xp

2M ˆ Ê d x+1 = dx 1 - 1- p ÁË d x ˜¯ 1- p ˘ 1000v 1000v È Ê p nx ˆ = Í1 - 2 M Á ˙ Ë 1000v ˜¯ p nx +1 p nx Î ˚

nx 1 Ê p nx ˆ = = 1 – 2M Á Ë 1000v ˜¯ nx +1 f x

1- p

The progression can be written as follows: n1 = n n2 = n1f1; n3 = n2 f2;

nz = nz – 1 ◊ f z–1;

1 Ê p n1 ˆ = 1 – 2M Á Ë 1000v ˜¯ f1

1- p

1 Ê p n2 ˆ = 1 – 2M Á Ë 1000v ˜¯ f2

1 f z -1

1- p

Ê p nz -1 ˆ = 1 – 2M Á Ë 1000v ˜¯

1- p

For given values of n1, nz and z be done by successive trials with different values of M. For instance d2 = d1 – 2M d1 . Assuming M = 1, d2 = d1 – 2 d1 . Knowing d2, the next diameter is found as d3 = d2 – 2

d 2 and so on. In this manner, dl2 can be found for two values of M, say M = 1.0 and 0.5. As

it is known that d12 = 17 mm, the correct value of M is found by linear interpolation. The logarithmic progression for the data of previous examples is given in Table 2.4. It has been developed for a value of M ª 0.88. A comparison of the rpm values of Tables 2.2–2.4 reveals that in the low as well as high rpm range the values of Table 2.4 lie between the values of Table 2.2 and 2.3.

74

Machine Tool Design and Numerical Control

Table 2.4 fx

dx mm

Ddx mm

n1 = 30

1.14

212

26

n2 = 34.2

1.15

186

24

n3 = 39.4

1.16

162

22

n4 = 45.8

1.18

140

21

n5 = 54.1

1.20

119

19

nx

n6 = 64

1.22

100

18

n7 = 78.1

1.25

82

17

n8 = 97.5

1.28

65

15

n9 = 125

1.34

50

12

n10 = 167.5

1.41

38

10

n11 = 236

1.55

28

9

n12 = 361

17

range, whereas that of the harmonic progression is poorer in the high rpm range. Despite the major shortcomings discussed above, geometrical progression is commonly used in machine tool drives owing to the following advantages: 1. Constant Loss of Economic Cutting Speed in the Whole rpm Range constitute the following series:

Suppose the spindle rpm values

n1, n2, n3,…, nj, n j+1 Consider that the optimum cutting speed vopt is such that it lies between the rpm values nj and nj+1, i.e., p dn j 1000

< vopt
n1

Fig. 2.41 Svetozarov variator: (a) Driving shaft (b) Driven shaft

The spheroids are made of hardened steel (HRC = 60–65) and the discs of tekstolite. This variator has R v max = 8, Nmax = 25 kW vmax = 20 m/s and h = 0.85 (ii) Cone Variator with Spheres Supported on Shafts: This variator (Fig. 2.42) is also known as the WulfelKopp tourator. The transmission ratio is varied by changing the angular positions of the shafts, which leads to a change in the effective driving radii of the spheres acting as the intermediate members. The instantaneous transmission ratio may be determined as i=

d1 2r2 ¥ 2r1 d 2

138

Machine Tool Design and Numerical Control

if d1 = d2, then i=

r2 r1

This variator has a speed variation range R v max = 9.

r2

r1

r1

r1

r2

r2

d2 d1

I

II

I

r1

II

r1

r2

I

r2

II

n2 = n1

n2 > n1

r2

r1

n2 < n1

Fig. 2.42 Wulfel-Kopp tourator: (a) Driving shaft (b) Driven shaft

4. Variators with Axially Displaceable Cones and Intermediate Members of the Belt, Ring or Chain Type The working of this type of variator may be explained through Fig. 2.43.

II

II

I

I

n2 > n1

II

II

I

I

n2 = n1

II

II

I

I

n2 < n1

Fig. 2.43 Variator with axially displaceable cones and belt-type intermediate member: (a) Driving shaft (b) Driven shaft

Regulation of Speed and Feed Rates 139

A pair of conical pulleys is mounted both on driving shaft I as well as driven shaft II. The conical pulleys can slide axially on splines or key. A belt, chain or ring transmits rotation from shaft I to shaft II. When one pair of pulleys is moved closer, the other pair automatically moves apart. This type of variator can provide transmission ratios less than zero (speed reduction) as well as greater than zero (speed increase) as shown in Fig. 2.43. The instantaneous transmission ratio is determined as i=

r1 ◊y r2

where r1 and r2 are the effective radii of the conical pulleys on shafts I and II, respectively. The characteristics of this type of variator depend to a large extent upon the intermediate member. The commonly used intermediate members for such variators are discussed below. (i) V-belt: Both narrow and wide V-belts may be used. The narrow V-belt is used in spindle drives of R v max = 9, Nmax = 2.5 kW vmax = 18 m/s and h = 0.92 The wide V-belt is used in drives of small- and medium-sized lathes and drilling machines and in the cam-shaft drive of single-spindle automats. Variator with a wide V-belt has the following R v max = 12, Nmax = 15 kW vmax = 15 m/s and h = 0.92 The cone pulleys for variators with V-belts are made of steel or cast iron. The major shortcoming of these variators is their large size. (ii) Steel Ring: A rigid steel ring made of hardened ball bearing steel can also act as an intermediate member. In this case cone pulleys are also made of hardened ball bearing steel. This type of variator R v max = 16,

Nmax = 10 kW and

h = 0.85 – 0.9

The variator with a steel ring is used in the drive of small- and medium-sized drilling machines, grinding machines, single-spindle automats and also in the feed drive of lathes, turrets, and drilling and milling machines. This variator is simple and easily adaptable for mounting in speed boxes of machine tools. However, it suffers from unfavourable velocity distribution over the area of contact. It is also expensive because the ring and conical pulleys must be hardened to improve their wear resistance and must be manufactured very accurately to ensure proper assembly. (iii) Nylon Belt: simple nylon belt, the so-called cogged belt is now widely used in the USA. The cogged belt combines the positive features of both toothed and belt transmissions. The cogs (teeth) provide positive drive between the driving and driven shafts, i.e., the belt does not slip although the assembly is simple because of the belt elasticity. The cone pulleys for these variators are made of aluminium. (iv) Chain: drive or to provide a positive drive.

140

Machine Tool Design and Numerical Control

If the chain is used as a friction-type drive between the driving and driven shafts, then the variator design is identical to that of a variator with a V-belt. The chain has rollers on the inside (contacting) surface to reduce R v max = 7,

Nmax = 15 kW

vmax = 10 m/s

and

h = 0.9

The cone pulleys of this variator are made of low carbon steel which is case-hardened. Figure 2.44 shows the cone pulleys and chain used in a variator with positive drive between driving and

the chain teeth are formed by different number of laminas. The cone pulleys are made of low carbon steel which is case hardened. The chain laminas are made of high strength alloy steel. R v max = 10,

Nmax = 14 kW and

h = 0.9

Fig. 2.44 Cone pulley and chain used in a PIV drive

Variators with a chain are used in spindle drives of lathes, turrets, drilling and milling machines and also in feed drives of lathes and drilling machines.

2.10 KINEMATICS OF MACHINE TOOLS In this section, the gearing diagrams of existing lathe, drilling machine and milling machine have been analysed to demonstrate the validity of the basic principles of gear box design in application to industrial machine

strength of the approach of seamless integration of the structural diagram, speed chart and gearing diagram as the sequential steps in design of machine tool gear boxes.

Regulation of Speed and Feed Rates 141

Thread cutting operation on lathe and operations involving the use of indexing head on milling machine have been included in this Section as illustrative cases of complex kinematics in machining operations.

2.10.1 Gearing Diagram of Speed Box of a Thread Cutting Lathe The structural diagram and speed chart of the speed box are shown in Fig. 2.45 and the gearing diagram in Fig. 2.46. The structural formula of the 23-speed gear box is Z = 23 = 2(1) 3(2) [1 + 2(6) 2(6)]. It is a combination of combined structure and overlapping structure. As the characteristic of the last transmission group is 2(6), instead of 2(12), it indicates that there are six overlapping speed steps in the low speed kinematic train. We shall later see that one of the speed steps of the high speed kinematic train coincides with that of the low speed kinematic train. Hence, though the structural formula is capable of generating 30 speed steps, it actually produces only 23 speed steps. The various kinematic trains are described below. Motor to shaft: belt drive 142–254 142 Common kinematic train 1450 ¥ ¥ 0.985 = 789.47 ª 800 rpm; here the multiplication factor 0.985 254 accounts for belt slip Shaft I–II È 51 ˘ È 56 ˘ 800 Í ˙ Í ˙ = 1046,1315 adjusted to 1000, 1250 with an error of 4.6 and 5.2%, respectively Î 39 ˚ Î 34 ˚ Thus, 2 speeds are obtained on shaft II Shaft II–III È 38 ˘ È 29 ˘ È 21 ˘ 1000 Í ˙ Í ˙ Í ˙ = 1000, 630, 400 Î 38 ˚ Î 47 ˚ Î 55 ˚

and

È 38 ˘ È 29 ˘ È 21 ˘ 1250 Í ˙ Í ˙ Í ˙ = 1250, 800, 500 Î 38 ˚ Î 47 ˚ Î 55 ˚

Total 6 speeds are obtained on shaft III Low speed kinematic train Shaft III–IV È 45 ˘ È 22 ˘ (100, 630, 400) Í ˙ Í ˙ Î 45 ˚ Î 88 ˚

È 45 ˘ È 22 ˘ (1250, 800, 500) Í ˙ Í ˙ Î 45 ˚ Î 88 ˚

1000, 630, 400 250, 160, 100

1250, 800, 500 315, 200, 125

Total 12 speeds are obtained on shaft IV Shaft IV–V È 45 ˘ È 22 ˘ (100, 125, 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250) Í ˙ Í ˙ Î 45 ˚ Î 88 ˚

142

Machine Tool Design and Numerical Control

100, 125, 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250 25, 31.5, 40, 50, 63, 80, 100, 125, 160, 200, 250, 315 The six underlined speeds are overlapping. Therefore, effectively only 18 speeds are obtained on shaft V. The overlapping speeds are highlighted by dots in shaft V in Fig. 2.45. 65–43 45–45 I

II

III 38–38

IV

V

VI

2000 1600

45–45

1250

4

56–3

1000 800 630

51–39

500 400

29–47

315 21–55

250 200 160

22.88

125 100 80 63 50 40

22.88

31.5 25 27–54

20 16

2(1)

3(2)

2(6)

2(6)

2(1)

3(2)

2(6)

2(6)

Fig. 2.45 Structural diagram and speed chart of lathe speed box

Shaft V–VI È 27 ˘ (25, 31.5, 40, 50, 63, 80, 100, 125, 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250) Í ˙ Î 54 ˚ 12.5, 16, 20, 25, 31.5, 40, 50, 63, 80, 100, 125, 160, 200, 250, 315, 400, 500, 630 Total 18 speeds given above are obtained on shaft VI (spindle) through low speed kinematic train.

12.5

Regulation of Speed and Feed Rates 143

High Speed kinematic train of gear pair

65 and transmitted directly to shaft VI (spindle). 43 65 = 1890, 1511, 1209, 952, 755, 604 (1250, 1000, 800, 630, 500, 400) 43

These speeds are adjusted to 2000, 1600, 1250, 1000, 800, 630 with an error of 4.8%. The speed 630 obtained through the high speed kinematic train coincides with the same speed obtained through the low speed kinematic train. This overlapping speed is highlighted by a dot on shaft VI. Hence, effectively only 5 speeds are obtained through the high speed kinematic train. Thus, the total speeds obtained on the spindle is 18 + 5 = 23. Reversal of spindle rotation is implemented through the gear pair 50–24 and the transmission 36-idle 50 36 gear-38. Consequently, the reversal speeds are ¥ = 1.97 times higher. Gear 60 on shaft VI is meant for 24 38 engaging the feed box. 36

24

d2 = 254 56

51

Idle gear

50

M1

88

45

21

x 34 39

22

Brake

I

x 38 x

IV

x

II

45

29 x

22

65

27

x

x

III

47

38

60

55

x

43

54

V

x 45

88

x

d1 = 142

45

VI

N = 10 kW, n = 1450 rpm

Fig. 2.46 Gearing diagram of lathe speed box

2.10.2 Gearing Diagram of Speed Box of a Lathe with Mechanical Variator using electrical motors described in Sec. 2.9.2. At the other end, small lathes are generally equipped with mechanical variators described in Sec. 2.9.3 without a stepped type gear box. Medium sized lathes are best served by a combination of stepped gear box in conjunction with a mechanical variator shown in Fig. 2.47.

144

Machine Tool Design and Numerical Control

M2 Z = 49

Z = 88 Z = 60

Z = 69

f 192

B2

Z = 88

Z = 72

Z = 19 x

Z = 49

B1

M3

x

Z = 22

Idle gear Z = 27

M1 Z = 44

Z = 38

x

x

Z = 22

Z = 60

x

x

x

x Z = 49

B3 Z = 36

Z = 54

i = 1/2 – 2

f 200

x

N = 14 kW n = 1450 rpm Triple start

B

Z = 51

f 125

Triple start f 125 x

Z = 28 Z = 19

x

x x Z = 31

N = 1 kW n = 1410 rpm

Fig. 2.47 Gearing diagram of lathe speed box with a mechanical variator

Regulation of Speed and Feed Rates 145

1. Stepped gear box 2. Mechanical variator 3. Control of the mechanical variator The stepped gear box has electromagnets clutches M1, M2 and M3 which are used for engaging, reversing and stopping the spindle, respectively. The Svetozarov mechanical variator has a speed variation range 2 1 between and (Rn = 4). Stepped speed variation in the gear box is achieved by varying the position of 1 2 double cluster gear blocks B1(88–60) and B2 (22–49) to obtain four speed steps The block B3 is shown as a reduction of the speeds in the single kinematic train of Range 1 and the gear pair (27–54) for increase of the speeds in the three kinematic trains that provide Ranges 2, 3 and 4 described below. From the motor (N = 14 kW, n = 1450 rpm), the rotation to the spindle is provided through the mechanical variator, belt drive (200–192) and stepped gear box by the following four kinematic trains, each allowing the B1 and engaging B Ê 2 1 ˆ 200 È 19 ˘ È 72 ˘ = 3000 – 750 rpm 1450 Á to ˜ Ë1 2 ¯ 192 ÍÎ 38 ˙˚ ÍÎ 36 ˙˚ B1 to the right, engaging B

B3 to the right.

Ê 2 1 ˆ 200 È 19 ˘ È 60 ˘ È 49 ˘ È 27 ˘ 1450 Á to ˜ = 750 – 190 rpm Ë1 2 ¯ 192 ÍÎ 38 ˙˚ ÍÎ 60 ˙˚ ÍÎ 49 ˙˚ ÍÎ 54 ˙˚ B1 to the right B2 to the left and B3 to the right. 1 ˆ 200 È 19 ˘ È 60 ˘ È 22 ˘ È 27 ˘ Ê2 1450 Á to ˜ = 190 – 47 rpm Ë1 2 ¯ 192 ÍÎ 38 ˙˚ ÍÎ 60 ˙˚ ÍÎ 88 ˙˚ ÍÎ 54 ˙˚ B

B2 to the left and B3 to the right.

1 ˆ 200 È 19 ˘ È 22 ˘ È 22 ˘ È 27 ˘ Ê2 = 47 – 12 rpm 1450 Á to ˜ Ë1 2 ¯ 192 ÍÎ 38 ˙˚ ÍÎ 88 ˙˚ ÍÎ 88 ˙˚ ÍÎ 54 ˙˚ Thus, this gear box provides continuous variation of spindle rotation between 12 and 3000 rev./ min. For reversal of spindle rotation, single sliding gear 49 is made to mesh with gear 19, electromagnetic cluster M1 is disengaged and M2 is engaged. The transmission now occurs through the gear pair (19–49) and the gear train (69-idle gear-44). Henceforth, the transmission occurs through the four kinematic trains described È 19 ˘ È 69 ˘ È 19 ˘ above. The additional reversal path has a transmission ratio Í ˙ Í ˙ = 0.60 as compared to Í ˙ = 0.5 of Î 49 ˚ Î 44 ˚ Î 38 ˚ the direct path. Thus, the reversal speeds of the spindle are 20% higher than the direct speeds. The control of the mechanical variator is executed through an independent kinematic train. From the 1 kw electric motor rotation is transmitted to the drum cam through belt drive 125–125 and two, triple start worm

146

Machine Tool Design and Numerical Control

gear pairs 28–3 and 51–3. The drum cam converts the rotational movement into translatory motion of the rod carrying block B, which through the system of levers implements the tilting of the Svetozarov variator Thus, overall speed variation in the range 12–3000 rpm is achieved by this very compact speed box design.

2.10.3 Kinematics of Thread Cutting on Lathe Lathes are used primarily for machining axisymmetrical parts on which a wide variety of operations can be carried out, such as simple turning, taper turning, facing, grooving, parting off, boring, drilling, forming, thread cutting and many more. In all these operations, rotation of the spindle is the primary cutting motion or working motion and translatory motion of the tool the auxiliary motion. In all the operations, except thread cutting, the working and auxiliary motions are independent of each other and their values are set on the speed box and feed box, respectively. Thread cutting operation requires a particular kinematic relation between the spindle rotation and tool move by a distance equal to the pitch of the thread. In thread cutting lathes, satisfying this kinematic relationpitch of the thread to be cut. However, in ordinary lathe machines, the required feed rate has to be calculated and set with the help of change gears. The kinematic train in this case is very simple as shown in Fig. 2.48, but the calculations demand considerable ingenuity as will become evident from the discussion below.

x

x

ic

a

c

tp

x

tx

l2

b

l1

x

x d

Fig. 2.48 Kinematics of thread cutting operation

s = tp = 1 spindle rev. ic ◊ ix ◊ tx where

s = feed per rev. of the threading tool; tp = pitch of the thread to be cut,

(2.34)

Regulation of Speed and Feed Rates 147

ic = transmission ratio of gear pairs in constant mesh; generally ic = 1 ix = transmission ratio of change gears a, b, c, d tx = pitch of the lead screw of the lathe After calculating transmission ratio ix of the change gears from formula (2.34), the next step is to determine the number of teeth of the change gears. For a single pair of change gears, a ix = b whereas for two pairs of change gear a c ix = x b d

20, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120 20, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, Besides, the change gear sets are also provided with gears having 47, 97, 127 and 157 teeth. This necessity of having gears 47, 97, 127 and 157 in the change gear set deserves special mention. While 1. The pitch of thread and lead screw are expressed in identical units. 2. The units in which tp and tx are expressed are different; i.e., cutting metric thread when the lead screw has BSW thread and vice versa. transmission in which the pitch of the worm is expressed as tp = pm, m being the module of the worm gear. In case (1), the setting calculations are done in accordance with formula (2.34) In case (2), the pitch of lead screw and thread to be cut must be expressed in identical units. For instance, if BSW thread it to be cut on a lathe having lead screw with metric thread, the pitch of the thread to be cut

tp =

25.4 127 1 = ¥ TPI 5 TPI

To meet this situation, it is essential to have a gear with 127 teeth in the change gear set. In case (3) the transmission ratio of the change gears is determined in accordance with expression (2.34) as ix =

tp ic ◊ t x

keeping in mind that ic = 1 and tp= pm , the transmission ratio may be expressed as ix =

pm , if the lead screw has metric threads tx

(2.35)

148

Machine Tool Design and Numerical Control

p mTPI 5p mTPI if the lead screw has BSW threads (2.36) = 25.4 127 5 p may be expressed through different combinations of whole numbers. Each The quantities p and 127 combination entails a certain error and depending upon the accuracy required any of the combinations may be selected. These combinations dictate the necessity of having gears with 47, 97 and 157 teeth in the change gear set. ix =

and

For formula (2.35) 22 , error = 0.04, no special gear required. 7 157 , error = 0.05, gear 157 required p= 50

p=

For formula (2.36) p=

47 127 , error = –0.005, gear 47 required 380 5

p=

12 127 , error = 0.021, gear 97 required 97 5

p=

19 ¥ 21 , error = +0.004, gear 127 required 127 a and b must be greater than the radius of gear c. c and d must be greater than the radius of gear b.

If constraints 1 is violated, gear c cannot be mounted on shaft II, whereas if constraint 2 is violated, gear b cannot be mounted on shaft II. mz The radius of a gear of module m and number of teeth z is = . If gears a, b, c and d are assumed to have 2 a, b, c, and d teeth, respectively, then co m mc (a + b) > 2 2 mb m (c + d) > 2 2 Wherefrom, we get the condition Ê a + b > cˆ ÁË c + d > b˜¯ Suppose that for a particular operation, we get 5 9 By multiplying the numerator and denominator by 5,

(a) ix =

(2.37)

Regulation of Speed and Feed Rates 149

ix =

a 5 5 25 = ¥ = b 9 5 45

gears 25 and 45 are available in series 1 120 40 ¥ 3 = (b) ix = 127 127

(i) ix =

60 a c 40 ¥ = b d 127 20

not acceptable because c + d = 60 + 20 = 80, whereas b = 127, i.e., constraint 2 is violated as c + d < b. (ii) ix =

40 60 ¥ 20 127

acceptable because a + b = 60 + 20 = 80, c = 40, hence a + b > c c + d = 40 + 127 = 167, b = 20, hence c + d > b For cutting multiple start threads, the lathe is set for the pitch value tm = tp k where

(2.38)

tm = distance between adjacent threads of a particular start tp = distance between adjacent peaks k = number of starts of the thread

After cutting one start, the kinematic linkage between the spindle and lead screw is disconnected and the spindle with work piece is rotated precisely through an angle = 360/k. The kinematic train is again connected 10

20

2

30 40 50 60 70 80 90 100 110 120 130 140 150 160 170

1

Fig. 2.49 Special chuck for cutting multiple start threads

150

Machine Tool Design and Numerical Control

ber of starts of the thread, then accurate rotation of the work piece through 360/k can be achieved with the help of gear a; for this purpose a chalk mark is made on gear a after disconnecting the spindle from the lead screw, the gear is rotated through a/k teeth and the kinematic train is again engaged. Multiple-start threads are also cut in a spindle chuck which can be indexed through any desired angle. This chuck ( Fig. 2.49) consists of two halves 1 and 2. During the threading operation, the halves are tightened by bolts. When one start has been cut, the two halves are unfastened. Half 1 remains attached to the spindle while half 2 along with the work piece is indexed through the desired angle which can be easily read on the graduated scale.

Example 2.1 It is required to cut screw thread of pitch 3 mm on a lathe with lead screw of pitch 10 mm. Determine the required change gears. t 3 3 1 30 20 = ¥ = Transmission ratio of the change gears ix = p = ¥ t x 10 5 2 50 40 a + b > c, i.e., 30 + 50 > 20 c + d > b, i.e., 20 + 40 > 50

Example 2.2 It is required to cut screw thread of pitch 1.5 mm on a lathe with lead screw having BSW thread of 27TPI. Determine the required change gears. 1 25.4 127 = mm inch = 2 2 5¥2 tp 1.5 ¥ 5 ¥ 2 15 Transmission ratio of the change gears ix = = = 127 127 tx As no gear of 15 teeth is available in the set of change gears, ix is represented as Pitch of the lead screw =

1x =

30 1 20 30 = ¥ ¥ 2 127 40 127

a + b > c, i.e., 20 + 40 > 30 c + d > b, i.e., 30 + 127 > 40

Example 2.3 It is required to cut double start thread of pitch 4 mm on a lathe with lead screw of pitch 6 mm. Determine the required change gears. Lead of the thread to be cut = pitch of the thread x number of starts = 4 ¥ 2 = 8 mm

Regulation of Speed and Feed Rates 151

Transmission ratio of the change gears = =

Lead of thread to be cut Pitch of lead screw 8 4 4 40 20 = = ¥1= ¥ 6 3 3 30 20

a + b > c; 40 + 30 > 20 c + d > b; 20 + 20 > 30

Example 2.4 It is required to cut modular thread on the worm of a worm-worm gear pair of module 4 mm on a lathe with lead screw of pitch 6 mm. Determine the required change gears. Pitch of the worm thread = pm = 4p Transmission ratio of change gears ix =

4p 6

22 157 = 7 50 157 2 157 40 1x = ¥ ¥ = 50 3 50 60

Representing p = 3.14 =

a + b > c; 157 + 50 > 40 c + d > b; 60 + 40 > 50

2.10.4 Gearing Diagram of Drilling Machine The complete gearing diagram of a single spindle drilling machine is described below separately for the speed box and feed box.

Speed Box The structural diagram and speed chart of the speed box are shown in Fig. 2.50 and the gearing diagram in Fig. 2.51. The structural formula of the 12-speed gear box is z = 4(1)2(4)2(4), which should actually yield 16 speed steps, but one of the transmissions in the last stage is not utilised as it provides only overlapping speed steps. Thus, effectively only 12 speeds are realised in the speed box. The spindle is powered by an electric motor (N = 7.0 kW, n = 1450 rpm), through belt drive 173–173 and B1 provides 4 speeds on shaft II. Between this shaft and shaft V (spindle), three more transmissions are generated thus providing a total of 12 speeds on the spindle. Motor shaft to shaft 1: Belt drive 173–173 È173 ˘ 1450 ¥ Í ¥ 0.985 = 1428, adjusted to 1410 rpm Î173 ˙˚

152

Machine Tool Design and Numerical Control

4(1)

Motor 1450

—

I

2(4)

II

2(4)

III

IV

V 61–47

43–40 50–43

2000 1410 1000 710

36–47

500

30–53

355

29–50

250

23–60

180

21–72

125 90 63 61–47

45

20–61 4(1)

—

2(4)

31.5 2(4)

Fig. 2.50 Structural diagram and speed chart of drilling machine speed box f 173 V z = 61

IV

f 173

III

z = 21 z = 20

x

II x z = 40 z = 47 x z = 53

z = 47 z = 72 z = 61

z = 60 x

z = 43

z = 50

x x x

I

z = 36 z = 43 z = 30 z = 23 x z = 29 z = 24 z = 48 x x

Fig. 2.51 Gearing diagram of drilling machine speed box

N = 7 kW n = 1450 rpm

Regulation of Speed and Feed Rates 153

Shaft I–II (four gear block 23–60, 30–53, 43–40 and 36–47) È 23 ˘ È 30 ˘ È 36 ˘ È 43 ˘ 1410 Í ˙ Í ˙ Í ˙ Í ˙ = 540, 798, 1080, 1515 Î 60 ˚ Î 53 ˚ Î 47 ˚ Î 40 ˚ Adjusted to 500, 710, 1000, 1410 with an error of 8% Shaft II–III (through gear pair 29–50) È 29 ˘ (500, 710, 1000, 1410) Í ˙ = 290, 412, 580, 818. Î 50 ˚ Adjusted to 355, 500, 710, 1000, Shaft III–IV ( double gear pair 21–72 and 50–43) È 21 ˘ È 50 ˘ 355, 500, 710, 1000 Í ˙ Í ˙ = 103, 146, 207, 291, 413, 581, 825, 1162 Î 72 ˚ Î 43 ˚ Adjusted to 90, 125, 180, 250, 355, 500, 710, 1000 Shaft IV–V (double gear pair 20–61 and 61–47) È 20 ˘ È 61 ˘ (i) 90, 125, 180, 250 Í ˙ Í ˙ = 29.5, 41, 59, 82, 117, 162, 234, 324 Î 61 ˚ Î 47 ˚ È 61 ˘ (ii) 355, 500, 710, 100, Í ˙ = 460, 649, 921, 1297 Î 47 ˚ Total speeds on shaft V (spindle) 29.5, 41, 59, 82, 117, 162, 234, 324, 460, 649, 921, 1297 (12 speeds) Adjusted to 31.5, 45, 63, 90, 125, 180, 250, 355, 500, 710, 1000, 1410 of the standard series of Table 2.9. È 20 ˘ It is noteworthy that the transmission ratio Í ˙ has not been utilised in (ii) above with the speeds 355, Î 61 ˚ 500, 710 and 1000 as this would only have provided the speeds overlapping with the existing rpm values of 125, 180, 250 and 355 obtained in (i).

Feed Box The input to the feed box comes from the last shaft (shaft V) of the speed box. The nine steps of feed rate are obtained by means of two triple gear blocks. The structural diagram and feed ‘chart’ are shown in Fig. 2.52 and the gearing diagram in Fig. 2.53. From shaft V of the feed chart, the rotation is converted to È1˘ translatory feed motion through gear pair 36–53, worm-worm gear pair of transmission ratio Í ˙ and rack Î 60 ˚ and pinion transmission spindle of module 4 and number of teeth of pinion = 12. The structural formula of the feed box is z =3(1) 2(3) 2(3), which should actually yield 12 steps but one of the transmission in the last stage is not utilised as it provides only overlapping steps. Thus, effectively only nine steps are realised in the feed box.

154

Machine Tool Design and Numerical Control

I

II

IV 35–26

III

1.0

51–18

29–47 ¥ 29–46

30–34

V

1.3090 0.8902

35–26

0.5804 0.4496 0.3052

24–40

0.1989

34–35

18–46

0.1395 0.0949

3(1)

2(3)

2(3)

18–43 2(3)

3(1)

0.0618 2(3)

Fig. 2.52 Structural diagram and feed chart of drilling machine feed box I

II z = 46

z = 29 x z = 29

z = 47 z = 18 z = 24 z = 30

III V z = 53 z = 18 z = 36 x x

IV

x z = 46 x z = 40 z = 34

x Z = 43 Z = 35

x Z= 51 x

x Z = 18 Z = 26

Rack 60

12–m–4

VI

z = 22 z = 64

x x

WormWormgear

Fig. 2.53 Gearing diagram of drilling machine feed box

The kinematic trains are described below. Spindle-shaft I-shaft II 1¥

29 29 = 0.3890 ¥ 47 46

Shaft II–III (triple gear block 18–46, 24–40 and 30–34)

Regulation of Speed and Feed Rates 155

È 18 ˘ È 24 ˘ È 30 ˘ 0.3890 Í ˙ Í ˙ Í ˙ = 0.1522, 0.2334, 0.3432 Î 46 ˚ Î 40 ˚ Î 34 ˚ Shaft III–IV (double gear pair 34–35 and 51–18) È 34 ˘ È 51 ˘ 0.1522, 0.2334, 0.3432 Í ˙ Í ˙ = 0.1478, 0.2267, 0.3334, 0.4312, 0.6613, 0.9724 Î 35 ˚ Î18 ˚ Shaft IV–V (double gear pair 18–43 and 35–26) È 18 ˘ È 35 ˘ (i) 0.1478, 0.2267, 0.3334 Í ˙ Í ˙ = 0.0618, 0.0949, 0.1395, 0.1989, 0.3052, 0.4496 Î 43 ˚ Î 26 ˚ È 35 ˘ (ii) 0.4312, 0.6613, 0.9724 Í ˙ = 0.5804, 0.8902, 1.3090 Î 26 ˚ Total speeds on shaft IV 0.0618, 0.0949, 0.1395,0.1989,0.3052,0.4496,0.5804,0.8902,1.3090 (9 speeds) È1˘ Shaft V–VI (gear pair 36–53, worm-worm gear Í ˙ and pinion rack of module 4 and 12 teeth Î 60 ˚ È 36 ˘ È 1 ˘ 0.0618, 0.0949, 0.1395, 0.1989, 0.3052, 0.4496 , 0.5804, 0.8902,1.3090 Í ˙ Í ˙ p ¥ 4 ¥ 12 Î 53 ˚ Î 60 ˚ = 0.106, 0.162, 0.24, 0.34, 0.52, 0.765, 1.0,1.52, 2.24, mm/rev È 18 ˘ It is note worthy that the transmission ration Í ˙ has not been utilised in the last stage as this would Î 43 ˚ only have provided steps overlapping with the existing values of 0.1989, 0.3052 and 0.4496 obtained in (i).

2.10.5 Gearing Diagram of the Speed Box of a Horizontal Milling Machine The structural diagram and speed chart of the speed box of a horizontal milling machine are shown in Fig. 2.54. and the gearing diagram in Fig. 2.55. The structural formula of the speed box is z = 18 = 3(1)3(3)2(9). The spindle is powered by an electric motor (N = 7 kW, n = 1450 rpm). The 18 speed steps are obtained with the help of two triple gear blocks and one double gear block. The kinematic trains are described below. Motor Shaft (I) to shaft II (gear pair 26–54) È 26 ˘ 1450 Í ˙ = 698, adjusted to 710 Î 54 ˚ Shaft II–III (triple gear block 16–39, 19–36, 22–33) È 16 ˘ È 19 ˘ È 22 ˘ 710 Í ˙ Í ˙ Í ˙ = 291, 374, 474, adjusted to 315, 400, 500 Î 39 ˚ Î 36 ˚ Î 33 ˚ Total 3 speeds are obtained on shaft III Shaft III–IV (through gear pairs 18–47, 28–37 and 39–26) È 18 ˘ È 28 ˘ È 39 ˘ 315, 400, 500 Í ˙ Í ˙ Í ˙ = 120, 153, 191, 238, 302, 378, 472, 600, 750 Î 47 ˚ Î 37 ˚ Î 26 ˚ Adjusted to 125, 160, 200, 250, 315, 400, 500, 630, 800 Total 9 speed are obtained on shaft IV

156

Machine Tool Design and Numerical Control

I

II

III

IV

1450

82–38

V 1600 1250 1000

39–26

26–54

800 630 500

22–33

400

19–36

315

16–39

250 28–37

200

18–47

160 125 100 80

19–71

63 50 40 3(1)

3(3)

2(9) 3(1)

3(3)

31.5

2(9)

Fig. 2.54 Structural diagram and speed chart of milling machine speed box

z = 38 V

z = 26 z = 37

z = 71

x

x

z = 47

IV z = 18

z = 39

x

x

x

x

x

III z = 36

z = 28 z = 33

x

II

z = 54 z = 26

z = 16 z = 22 z = 19

x N = 7 kW n = 1450 rpm

Fig. 2.55 Gearing diagram of milling machine speed box

I

z = 19 z = 82

Regulation of Speed and Feed Rates 157

Shaft IV–V (double gear pair 19–71 and 82–38) È19 ˘ È 82 ˘ 125, 160, 200, 250, 315, 400, 500, 630, 800 Í ˙ Í ˙ Î 71 ˚ Î 38 ˚ = 33, 43, 53, 67, 84, 107, 134, 169, 214, 270, 345, 432, 540, 680, 863, 1078, 1360, 1726 Adjusted to 31.5, 40, 50, 63, 80, 100, 125, 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250, 1600 of the standard series of Table 2.9 Total 18 speeds are obtained on the spindle.

2.10.6 Indexing Head and Kinematics of Indexing Operations Milling machines are used for carrying out a wide range of operations such as machining of plane surfaces, are more suitable for operations such as gear cutting and machining of multiple grooves with the help of a ported between the indexing head centre and tail stock and clamped in a chuck or with the help of dog and face plate mounted on the indexing head spindle. The spindle is rotated manually by means of a crank wheel

Worm and worm wheel

E

Spindle

C

D

F

A

B

Index plate Crank Pin

Fig. 2.56 Schematic of indexing head

158

Machine Tool Design and Numerical Control

1. Periodic rotation of the workpiece through a certain angle. 2. Continuous rotation of the workpiece which is related to the longitudinal feed of the machine table. The construction of the indexing head is shown in Fig. 2.56. It consists of two pairs of spur gear of transmission ratio 1, one pair of bevel gears of transmission ratio 1 and a worm-worm gear transmission. If the number of starts of the worm is k and the number of teeth of the worm gear is zw, then the ratio N = zw/k is called the indexing head characteristic. Generally the transmission ration of the worm-worm gear transmission and hence the index head characteristic is equal to 40. The worm gear and the spindle of the indexing head have a common axis. Indexing of the spindle is achieved by rotating the crank wheel. Rotation from the crank is transmitted to the spindle through spur gears C–D and the worm-worm gear pair. Accurate indexing through a given angle is carried out with the help of an index plate (Fig. 2.57.) The index plate has several concentric circles with different number of holes on each side. There is an adjustable sector that can be opened to cover the required number of holes of a particular circle of the index plate. Generally, two index plates are supplied with an indexing head. I 20 19 18 17 16 15

II

Fig. 2.57 Index plate

Plate No. 1 has 20, 23, 33 and 43 holes on one side and 18, 21, 31 and 41 on the other. Plate No. 2 has 16, 19, 29, 39 and 49 holes on one side and 15, 17, 27, 37 and 47 on the other. However, index plates with other hole combinations are also available. The index plate is mounted on a hollow shaft that is concentric with the crank wheel axis and can freely rotate about it.

Indexing Based on Periodic Rotation There are three distinct types of indexing operations based on periodic rotation of the workpiece, namely plain indexing, differential indexing and angular indexing. The kinematics of each of these three modes of indexing is described below. 1. Plain indexing for periodic rotation is employed when the number of crank rotations is (i) a whole number

Regulation of Speed and Feed Rates 159

(ii) a fraction in which the denominator matches the number of holes in one of the circles of the index plate (iii) a whole number-cum fraction as described in (ii) above Suppose we have to index for n divisions on an indexing head of characteristic N = 40. Then the 40 turns through which the crank must be moved, i.e., the crank rotation R = n Example: 40 (a) If n is say 8, than R = = 5, i.e., the crank will be rotated through 5 full turns. 8 40 1 10 (b) If n is say 80, than R = = = , i.e., the crank will be rotated through 10 holes of the circle 80 2 20 with 20 holes on plate No. 1. After each indexing the adjustable sector is moved forward by 10 holes. 40 4 2 22 (c) If n is say 6, then R = =6 =6 =6 , i.e., the crank will be rotated through 6 full turns 6 6 3 33 plus 22 holes of the circle with 33 holes on plate No. 1. After each indexing the adjustable sector will be moved through 22 holes and the crank will be rotated through 6 full turns and 22 holes. 2. Differential indexing for periodic rotation is resorted to when the indexing for the required number of divisions n involves a fraction whose denominator does not match with the number of holes in any of N the circles of plates 1 and 2. In this case, indexing is carried out for nx such that nx is close to n and nx involves a fraction whose denominator matches with the number of holes in one of the circle of plates 1 or 2. When indexing is carried out for nx divisions, the error in rotation of the crank per indexing N N – . Therefore, the error in one rotation of the indexing head spindle is = n nx N ( nx - n ) ÊN Nˆ x = Á - ˜n = nx Ë n nx ¯ The above error must be compensated by imparting a corresponding additional rotation to the indexing plate in the proper direction. While rotating the crank through the requisite turn for nx divisions, the crank pin is withdrawn from the index plate which can now receive additional rotation a c through the kinematic train starting from the indexing head spindle via change gears ¥ , bevel gear b d pair E–F and spur gear pair A–B ( Fig. 2.58) The change gear ratio a c N ( nx - n ) x= ¥ = nx b d If nx > n, then the index plate should rotate in the same direction as the crank. However if nx < n, the index plate should rotate in a direction opposite to that of the crank for which an idle gear is inserted between the change gears c and d. Here, it is pertinent to remember that the spur gear pairs A–B and C–D and the bevel gear pair E–F all have transmission ratio equal to unity

160

Machine Tool Design and Numerical Control

a

x c Worm and worm wheel

Spindle

b E

C

x

D

F d A

B

Index plate Crank Pin

Fig. 2.58 Schematic of differential indexing

Example 1: It is required to make a gear of 109 teeth using an indexing head with which the following change n = 109 and no circle of 109 holes or its its multiple is available on plates 1 and 2, we will have to take recourse to differential indexing for which we select nx = 110. Therefore, indexing will involve 40 4 12 (i) rotation of crank through R = = = , i.e., through 12 holes of the circle with 33 holes on 33 110 11 plate No. 1. (ii) selection of appropriate change gears of transmission ratio x=

4 a c 40(110 - 109) 4 1 40 25 ¥ = ¥ = ¥ = = 11 5.5 2 55 50 b d 110

As nx > n, the index plate and crank rotate in the same direction and there is no need of an idle gear to be inserted between change gears c and d. (iii) check

440 4 4 40 4 ¥ 109 + 4 436 + 4 = + = = = 11 ¥ 109 11 ¥ 109 109 11 11 ¥ 109 11 ¥ 109

Regulation of Speed and Feed Rates 161

3. Angular indexing is employed when the job has to be indexed through a given angle. Since 40 turns of the crank turn the job through one complete revolution (i.e., 360°), one turn of the crank evidently turns 360 = 9 = 540¢ = 32400≤. To determine the required crank rotation for angular indexing, the it through 40 given angle is divided by 9, 540 or 32400, depending upon whether the angle is expressed in degrees, minutes and seconds, respectively. The quantity thus obtained is then treated as a whole number or a fraction or a whole number-cum-fraction and tackled as a problem of plain or differential indexing, as the case may be Example 2: (a) The job has to be indexed through 50°. The required crank rotation will be 5 50 = 5 i.e., 5 full turns and 10 holes of the circle with 18 holes on plate no. 1. 9 9 The required crank rotation will be 31 ¥ 60 + 20 1860 + 20 1880 18 1 6 =3 =3 =3 , = = 540 540 540 54 3 18 i.e., 3 full turns and 6 holes of the circle of 18 holes on plate no. 1. (c) The job has to be indexed through 4°56¢1≤ The required crank rotation will be 4 ¥ 60 ¥ 60 + 56 ¥ 6 + 1 17761 = 32400 32400 The above value can be interpreted in terms of number divisions of the jobs as 40 No. of divisions = = 72.968 ª 73 17761 32400 is equivalent to indexing of a job for 73 divisions. As no circle of 73 holes or its multiple is available on any of the index plates, this becomes a problem of differential indexing for which we select nx = 75. Hence, indexing will involve 40 8 = , i.e., through eight holes of the circle of 15 holes on plate no. 2. 75 15 (ii) selection of appropriate change gears of transmission ratio (i) rotation of crank through R =

a c 40(75 - 73) 16 4 4 40 40 ¥ = = ¥ = ¥ = 75 15 3 5 30 50 b d As nx > n, the index plate and crank rotate in the same direction and there is no need of an idle gear to be inserted between change gears c and x=

(iii) check

40 600 8 16 8 ¥ 73 + 16 584 + 16 + = = = = 15 15 ¥ 73 15 ¥ 73 15 ¥ 73 15 ¥ 73 73

162

Machine Tool Design and Numerical Control

Indexing Based on Continuous Rotation

This type of indexing is used for machining of helical

head spindle completes one rotation, the milling machining table should travel a distance equal to the pitch of the helix. First, the table of the milling machine is rotated and set at an angle equal to the helix angle. Next, rotation of the index head spindle and the longitudinal feed of the table are co-ordinated by using change gears a1, b1, c1, d1 between the lead screw of the milling machine responsible for longitudinal travel of the table and the output shaft of the bevel gear pair E–F (Fig. 2.59) such that they satisfy the relation. Ê d bˆ 1 spindle rotation ¥ N Á ◊ ˜ ¥ tx = tp Ë c a¯

Worm and worm wheel

d1

E

Spindle

C

D

F

x b1

A

x

B

c1 Index plate

a1 Lead screw of milling machine

Crank Pin

Fig. 2.59 Schematic of machining of a helical flute

N is the index head characteristic tx is the pitch of the lead screw of the milling machine tp is the pitch of the helix to be machined pD tp = tan b where b is helix angle and D is diameter of the job on which the helical groove is to be machined.

where

Regulation of Speed and Feed Rates 163

Again, it must be kept in mind that the above relation implies that the spur gear pairs A–B and C–D and the level gear pair E–F all have transmission ratio equal to unity. a c If the transmission ratio of the change gears is represented by x = 1 ¥ 1 , then the above relation may b1 d1 be written as 1 N tx = tp x Nt wherefrom x = x . tp It may be interesting to note here that as the bevel gear pair E–F is engaged in machining of the helix, it is not available for differential indexing. Hence, only such helical grooves can be machined that require only plain indexing. Further, since the table of the milling machine has to be rotated through an angle equal to the helix angle, the groove milling operation can only be carried out on a universal milling machine. Example 3: It is required to make a helical gear of teeth z =35 module m = 5 and helix angle = 22.5°. The milling machine on which the gear is machined has lead screw of pitch 6 mm and is provided with a change gear set 20, 25, 30, 35, 40, 45, 50, 55, 60, 70, 80, 90, 100. Diameter of the gear D = mz = 5 ¥ 35 = 175 mm tp =

3.14 ¥ 175 pD p175 = 1302.1 ª 1300 mm = = 0.422 tan b tan 22.5∞

x =

6¥4 30 20 Nt x 40 ¥ 6 24 = = = = ¥ 1300 tp 130 10 ¥ 13 50 65

Before cutting the gear the table of the milling machine is rotated and set at an angle of 22.5°. After mill5 1 7 40 ing one tooth, the gear blank is index for 35 division for which the crank is rotated by =1 =1 =1 35 7 49 35 turns, i.e., one full turn plus rotation through 7 holes of the circle with 49 holes on plate no. 2.

Review Questions 2.1 Find the speed steps arranged in geometric, harmonic and logarithmic progressions for the following nmin = 12 rpm, nmax = 510 rpm, z = 8. 2.2 A lathe is to be designed for machining aluminium work pieces of up to 500 mm diameter and mild steel workpieces of up to 300 mm diameter. Both HSS and cemented carbide tools are used. Determine the diameter of the smallest workpiece which may be machined on this lathe if the permissible cutting speed of the HSS-mild steel pair = 50 m/min, that of the carbide-aluminium pair = 1500 m/min and Rn = 75. 2.3 Draw the structural diagrams of a machine tool speed box for nmin = 16 rpm, nmax = 770 rpm, and f = 1.26. Which layout is best and why? 2.4 Sketch the possible speed charts for the layout selected in the previous questions if the motor runs at 1440 rpm and the input shaft has rpm 630. Select the best speed chart and justify your choice. 2.5 Sketch the gearing diagram for the optimum speed chart selected in the previous question. Determine the number of teeth of the gears assuming Zmin = 17.

164

Machine Tool Design and Numerical Control

2.6 Design the feed box of a lathe machine for the feed range 0.05–4.0 mm/rev. Given f = 1.41. Assume a suitable kinematic train between the spindle and the input shaft of the feed box, and between the output shaft of the feed box and the rack, attached to the underside of the lathe bed. 2.7 A three-stage, 18-step, speed box with nmin = 16 and f = 1.26 is powered by a 7.5 kW motor running at 1440 rpm. Calculate the dimensions of the gear-box shafts. 2.8 A four-stage, 16-step, speed box with nmin = 32 and f =1.41 is powered by a two-speed (1500/750) ac motor. Draw the structural diagram and speed chart of the speed box. 2.9 A speed box having nmin = 20 rpm, nmin = 2000 rpm and f = 1.26 is to be designed. Draw the best possible structural diagram and speed chart if the speed box has (i) a structure with overlapping steps, (ii) a structure with broken geometrical progression, and (iii) a combined structure. 2.10 Calculate the piston and pull-rod diameters to provide a maximum pull of 700 kgf at a speed of 500 mm/min. Also, determine the required pump delivery and motor rating, if the oil is to be delivered at a pressure of 40 kgf/cm2. The permissible stress of the pull-rod material is 1400 kgf/cm2. Assume the 2.11 BSW thread of 12 TPI is to be cut on a lathe with lead screw of 8 TPI. Determine the required change gears. 2.12 BSW thread of 27 TPI is to be cut on a lathe with lead screw of 6 TPI. Determine the required change gears. 2.13 Screw thread of pitch 0.25 inch is to be cut on a lathe with lead screw of pitch 8 mm. Determine the required change gears. 2.14 A double start thread of pitch 0.5 inch is to be cut on a lathe with lead screw of 6 TPI. Determine the required change gears. 2.15 Modular thread of module 3 mm is to be cut on a lathe with lead screw of 2 TPI. Determine the required change gears. 2.16 Determine the required crank rotation of the indexing head to index for (i) 33 divisions (ii) 24 divisions (iii) 5 divisions 2.17 Determine the required crank rotation and change gears for differential indexing for (i) 83 divisions (ii) 73 divisions (iii) 55 divisions 2.18 Determine the required crank rotation for indexing through (i) 3° 30¢ (ii) 42° 40¢ (iii) 34° 12¢ 2.19 machine with a lead screw of pitch 6.35 mm. Determine the required crank rotation, table setting angle and change gears. 2.20 Find the rpm values of eight-speed gear box having nmin = 12 and nmax = 509 if the steps are placed in arithmetic, geometric, harmonic and logarithmic progressions. Derive the relationship between vmax and vmin as a function of diameter for the arithmetic and geometric series.

Regulation of Speed and Feed Rates 165

2.21 Given nmin = 10, nmax = 500 and Z = 18. Write the speed steps if they lie in arithmetic, geometric, harmonic and logarithmic progressions and compare them from the standpoint of suitability for machine tool gear box. 2.22 In a machine tool gear box, the smallest and largest spindle speeds are 100 rpm and 1120 rpm, respectively. If there are 8 speeds in all, what will be the fourth speed? 2.23 nmin = 31.5, nmotor = 1440 and = 1.41. Draw the structural diagram, speed chart and gearing diagram. 2.24 Design a nine-speed gear box having nmin=100 and nmax = 630. Assume motor rpm = 1400. The design should include structural diagram, speed chart, gearing diagram and number of teeth of the gears. 2.25 Design a 12-speed gear box of a lathe having nmin = 14 and = 1.26. The motor has 16 HP at 1400 rpm. The design should include structural diagram, speed chart, gearing diagram, number of gear teeth and gear dimensions. 2.26 A 2 ¥ 3 ¥ 3 gear box is to be designed for a milling machine with 1450 rpm, 5 HP motor. The speed varies from 10 rpm with F =1.41. The design should include structural diagram, speed chart, gearing diagram, number of gear teeth and gear dimensions. 2.27 The speed chart of a 12 speed gear box is given below. Write the structural formula and draw the structural diagram of the gear box. If the gear box is powered by a 16 HP motor running at 1410 rpm, sketch the gearing diagram (without link gears) and determine the gear dimensions. Neglect power loss in transmission. Gear width is 10 times the module and Zmin = 17. 1410 1000 710 500 360 250 180 125 90 63 45 31.5 22.5

Fig. 2.60

2.28 Select an appropriate structural diagram and speed chart for a machine tool gear box with overlapping speed steps. Given nmin = 20, nmax = 1200, F = 1.41. 2.29 Draw the structural diagram, speed chart and gearing diagram for a machine tool gear box with combined structure having nmin = 16, nmax = 1400 and F = 1.41.

166

Machine Tool Design and Numerical Control

2.30 In a hydraulic circuit, the supply pressure in the cylinder is 14 kgf.cm2 and the piston area is 40 cm2 on both sides. The maximum load to be moved is 200 kgf at 150 cm/min. Calculate the hydraulic 80%. Neglect friction losses.

References 1. Vragov, Yu. D, “Probability method of determining the speed characteristics of high speed milling machines”, Stanki I Instrument, 1963, No. 6. 2. Rabinovich, A, Speed Boxes of Machine Tools, Lvov University Press, Lvov, 1968, p. 14. 3. Shvartz,VV, Elyashev, AV and Gudimenko, NN ‘Developing the Kinematic Diagrams of Machine Tool Speed Boxes’ published by Peoples’ Friendship University, Moscow, 1978, p. 39.

Design of Machine Tool Structures 167

3

DESIGN OF MACHINE TOOL STRUCTURES

FUNCTIONS OF MACHINE TOOL STRUCTURES AND THEIR REQUIREMENTS

3.1

Machine tool parts such as beds, bases, columns, box-type housings, overarms, carriages, tables, etc., are known as structures. The structures, depending upon their function, may be broadly divided into the following three groups: Group 1: Beds and bases, upon which the various subassemblies are mounted. Group 2: Box-type housings in which individual units are assembled, e.g., speed box housing, spindle head, etc. Group 3: Parts that serve for supporting and moving the workpiece and cutting tool, e.g., table, carriage, knee, tail stock, etc. Machine tool structures must satisfy the following requirements: 1. All important mating surfaces of the structures should be machined with a high degree of accuracy to provide the desired geometrical accuracy. 2. The initial geometrical accuracy of the structures should be maintained during the whole service life of the machine tool. 3. The shapes and sizes of the structures should not only provide safe operation and maintenance of the should be noted that the stresses and deformations are due to mechanical as well as thermal loading. The design features that provide for ease of manufacture, maintenance, etc., are peculiar to each structure and will, therefore, be discussed separately for different structures. However, there are two common features 1. Proper selection of material. 2. High static and dynamic stiffness. These two factors will be discussed in detail in Sections 3.3 and 3.4 after the basic criteria for designing machine tool structures are laid down in Section 3.2.

3.2

DESIGN CRITERIA FOR MACHINE TOOL STRUCTURES

Consider a simple machine tool bed with two side walls, which may be represented as a simply supported beam loaded by a concentrated force P acting at its centre (Fig. 3.1).1 The maximum normal stress acting on the beam is given by the expression:

168

Machine Tool Design and Numerical Control

smax =

M max ◊ zmax Iy P

(3.1)

Z Y

b

X h

I

Fig. 3.1 Schematic diagram of simply supported beam

where

Mmax = Pl/4 = maximum bending moment zmax = h/2 Iy = bh3/12 = moment of inertia of the beam section about the neutral axis

Upon substituting these values in Eq. (3.1), we get

smax

Pl h ◊ 3 Pl = 4 32 = 2 bh 2 bh 12 s], then

s] = or

3 Pl ◊ 2 bh 2

Vs = b ◊ h ◊ l =

3 P Ê l2 ˆ ◊Á ˜ 2 [s ] Ë h ¯

(3.2)

where Vs on Strength of Materials): dmax =

Pl 3 48 EI y

(3.3)

where E = modulus of elasticity of the beam material. d], then d] =

Pl 3 Pl 3 = 48 EI y 48 E ◊ bh3 /12

Design of Machine Tool Structures 169

Vd = bhl =

or

P Ê l2 ˆ Á ˜ 4 E [d ] Ë h ¯

2

(3.4)

where V The condition of optimum design is Vs = Vd 3 P Ê l2 ˆ P Ê l2 ˆ ÁË ˜¯ = Á ˜ 2 [s ] h 4 E [d ] Ë h ¯

i.e., wherefrom,

2

l2 6E [d ] = h [s ]

(3.5)

Equation (3.5) indicates that for every structure, there exists an optimum ratio l2/h depending upon: d], and s] and E. Consider, for instance, two beams of mild steel and cast iron with mechanical properties: for mild steel, E = 2.0 ¥ 104 kgf/mm2 s] = 14 kgf/mm2 d ] = 0.002 mm; for cast iron, E = 1.2 ¥ 104 kgf/mm2 s] = 3 kgf/mm2 d ] = 0.002 mm. For the steel beam Ê l2 ˆ 6 ¥ 2 ¥ 104 ¥ 0.002 = 17.14 ÁË ˜¯ = 14 h opt For the cast iron beam Ê l2 ˆ 6 ¥ 1.2 ¥ 104 ¥ 0.002 = 48 = ÁË ˜¯ h opt 3 The volumes of the two beams with optimum l/h values will be in the ratio VCI VMS

P 3 P ¥ (48) 2 ◊ ◊ 48 4 ¥ 1.2 ¥ 104 ¥ 0.002 2 3 = = 13.07 = 3 P P 2 ◊ ◊ 17.14 ¥ ( . ) 17 14 2 14 4 ¥ 2 ¥ 104 ¥ 0.002

i.e., if the failure of the beams is determined by the normal stresses under tensile loading, the volume of the steel beam required to withstand the same load is 13.07 times less than that of the cast iron beam. The variation of Vs and Vd for mild steel and cast iron beams with change of l 2/h is shown in Fig. 3.2. For identical beam length, the height of the steel section must be 48/17.14 = 2.80 times greater. Since the volume of the steel beam is 13.07 times less and height 2.80 times greater than that of the cast iron beam, the thickness of the mild steel beam will be 36.5 times less.

170

Machine Tool Design and Numerical Control

V l2/h values less than the optimum (corresponding to the point of intersection of Vs and Vd curves), the structure should be designed 13.07 12 from consideration of strength, while for l2/h values exceeding the optimum value, the design should be Vs for CI l2/h Vd 8 ratio for a majority of machine tools lies to the right of the point of intersection, i.e., l2/h is greater than the optimum value. Consequently, the stiffness and not the load-carrying capacity of a structure is the decisive 4 Vd 1.0 factor which determines its dimensions in most of the for MS Vs machine tools. 2 17.14 That the steel structure is lighter, deeper and thinI 0 ner than a cast iron structure of equivalent strength is 0 10 20 30 40 48 h obvious. However, since structures are mostly designed 2 Fig. 3.2 Variation of Vs and Vd with l /h ratio for from stiffness considerations, the actual economy of cast iron and mild steel sections metal consumption by using steel instead of cast iron may be much less than 13.07 times, because the steel structure must be provided with stiffening ribs. This not only increases the weight of the steel structure but also adds to the labour cost. Some of the machine tool structures are stationary while there are others which travel during operation to impart the necessary forming and auxiliary motions. The moving parts generally travel along guideways. For satisfying the requirement that the structures maintain their initial accuracy, it is essential that the mating surfaces should have a high wear resistance. Thus, the criteria for the design of machine tool structures can be summed up as

1. high static and dynamic stiffness, and 2. high wear resistance of guiding and guided surfaces.

MATERIALS OF MACHINE TOOL STRUCTURES

3.3

The elongation of a bar subjected to simple tension is given by the expression: Dl = where

P l A E

P◊l E◊A

= tensile force acting on the bar = length of the bar = area of cross section of the bar = modulus of elasticity of the bar material

The quantity P/Dl represents the stiffness of the bar and may be written as P E◊A = l Dl for two bars of equal stiffness, E1 A1 E2 A2 = l1 l2

Design of Machine Tool Structures 171

if the bars are of equal length, i.e., l1 = l2 = l, E1 A1 = E2 A2 The weights of the bars will be g 1 A1 l and g 2 A2l, where g 1 and g 2 respectively. The ratio of the weights will be Eg W1 g 1 A1 E /g = 2 1= 2 2 = W2 g 2 A2 E1g 2 E1 /g 1 i.e., the weight of the bar is inversely proportional to the quantity E/g , which is known as unit stiffness of the material in tension. The larger the unit stiffness of a material, the smaller is the weight of the structure The unit stiffness values of some engineering materials are given in Table 3.1.

therefore, rarely used in industry. A comparison of the unit rigidity values of various steels shows that these alloyed steels and a structural steel which is cheaper in comparison will be just as good. Comparative evaluation of various engineering materials can also be carried out on the basis of strength. This is illustrated below for three different types of loading. 1. Bar Subjected to Tension Consider a bar of length L subjected to a tensile load P. Let s ut be the ultimate strength of the bar material and g n, the area of cross section of the bar will be A = n ◊ . V = A ◊ L and W = A ◊ L ◊ g, respectively. ut Table 3.1

Unit rigidity in tension of some engineering materials E, kgf/cm2

g, kgf/cm3

E/g

Low carbon steel

2.0 ¥ 106

7.8 ¥ 10–3

2.56 ¥ 108

Medium carbon steel

2.1 ¥ 106

7.8 ¥ 10–3

2.69 ¥ 108

Alloyed steel

2.1 ¥ 106

7.8 ¥ 10–3

2.69 ¥ 108

Grey cast iron

1.2 ¥ 106

7.2 ¥ 10–3

1.66 ¥ 108

Duraluminium

0.75 ¥ 106

2.8 ¥ 10–3

2.68 ¥ 108

Material

P, it follows that W1 (s /g ) nPL(g 1s ut1 ) = = ut2 2 W2 nPL(g 2 /s ut 2 ) (s ut1/g 1 ) Here, the greater is the quantity s ut/g of a given material, the smaller will be the required weight of the bar to withstand a given load. Quantity s ut/g, is known as unit strength under tension.

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Machine Tool Design and Numerical Control

2. Bar Subjected to Torsion The shear stress in a bar of radius r, length L and polar moment of inertia Ip subjected to torque Mt is given by the following relation: t=

M t ◊ r M t ◊ r 2M t = 4 = IP p r /2 p r 3 tu, then for a factor of safety n, the radius of the bar

subjected to torque Mt is found from the relation, r3 = wherefrom,

2M t ◊ n pt u

Ê 2M t n ˆ r= Á Ë p ˜¯

1/3

◊ (1/tu)1/3

The area of cross section of the bar, A = p r 2 = p (2M t n/p)2/3(1/t u)2/3 and its weight, where g

W = ALg = p 1/3 (2M t n) 2/3(1/t u)2/3 Lg. bar material. Mt, it follows that W1 (1/t u1 ) 2/3 g 1 = = W2 (1/t u 2 ) 2/3 g 2

t u22/3 /g 2 t u21/3 /g 1

The quantity t u2/3/g is an index of the ability of a material to resist torsion and is known as unit strength under torsion. 3. Bar Subjected to Bending The normal stress in a bar of radius r, length L and moment of inertia about M ◊ r M ◊ r 4M = 4 = 3 I p r /4 p r where M is bending moment in the given section. sb, then for a factor of safety n, the radius of the bar subjected to bending moment M is found from the relation, s=

r3 =

4M ◊ n ps b

wherefrom, 1/3

Ê 4mn ˆ (1/sb)2/3 r= Á Ë p ˜¯ The area of cross section of the bar, Ê 4 Mn ˆ A = p r2 = p Á Ë p ˜¯ and its weight, where g

2/3

(1/sb)2/3

W = ALg = p 1/3 (4Mn)2/3(1/s b)2/3 L◊ g

Design of Machine Tool Structures 173

M, it follows that 2/3 W1 (1/s b1 ) 2/3 ◊ g 1 s b 2 /g 2 = = W2 (1/s b 2 )2/3 ◊ g 2 s b21/3/g1

The quantity sb2/3/g is an index of the ability of a material to resist bending and is known as the unit strength under bending. The commonly used materials for machine tool structures are cast iron and steel. Cast iron structures were wider application due to advances in welding technology. The choice of whether the structure should be made from cast iron or steel depends upon a number of factors which are discussed below. 1. Material Properties

The material properties of relevance are:

(i) Steel has higher strength under static and dynamic loading. (ii) The unit rigidity of steel under tensile, torsional and bending loads is higher. (iii) Cast iron has higher inherent damping properties; damping in steel structures occurs mainly in welds; if welded joints are properly designed, the damping of steel structure may approach that of cast iron. (iv) Cast iron has better sliding properties. 2. Manufacturing Problems Another important factor for deciding the choice of material concerns the problems of manufacturing that are associated with the use of steel or cast iron structures: (i) Wall thickness: For a given weight of the structure, high strength and stiffness can be achieved by using large overall dimensions and small wall thickness. Thus, walls of minimum possible thickness should be employed. Generally, reduction of wall thickness in cast iron structures is restricted by process capability and depends upon the size of the casting. These values are given in Table 3.2. Table 3.2

Recommended minimum wall thickness for cast iron structures

Size factor N, m

Thickness of external walls, mm

Thickness of internal walls and ribs, mm

0.4

6

5

0.75

8

7

1.0

10

8

1.5

12

10

1.8

14

12

2.0

16

14

2.5

18

16

3.0

20

16

3.5

22

18

4.5

25

20

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Machine Tool Design and Numerical Control

Size factor N is determined from the relationship, N=

2L + B + H 4

(3.6)

where L, B and H are length, breadth and height of the structure, respectively in metres. The wall thickness may also be determined from the following relationship: d = 10

2L + B + H mm 3

(3.7)

where L, B and H have the same meaning and units as in Eq. (3.6). thickness equal to 0.8 that of external walls is permitted. Welded structures made of steel can have much thinner walls as compared to cast structures as the technological constraints are much less. Steel structures in which the wall thickness is less than that of the cast structure by up to 50% are known as thick-walled structures. They are made of 10–12 mm thick plates and are easy to manufacture, but they are not particularly effective from the point of view of economy of metal. Thin-walled steel structures have 3–6 mm thick walls, which are thinner than the walls of the cast structure by 1.5–2 times. These structures offer greater economy of metal but are (ii) Walls of different thicknesses can be welded more easily than cast one thickness to another (if t1/t2 < Ê 1 1ˆ r = Á to ˜ t1 Ë 6 3¯ where t1

t1/t2 > 1.75, then the transition is realised

(iii) Machining allowances for cast structures are generally larger than for weld steel structures; this is essential to remove the hardened skin of casting and also to account for casting defects, such as inclusions, scales, drops, etc., that result due to falling of sand into the mould cavity. (iv) A welded structure can, if required, be easily repaired and improved. Any corrections in a cast structure

t1

t1 t1 If t < 1.5 2 r = 1 – 1 t1 6 3 r + R = t1

(

(

R

R

(

t If t1 > 1.75 2 I = 4c for CI = 5c for MS t + t 2 1 1 1 r + r= 6 – 3 2 R = r + c + t2

c t2

t2 l

(a)

(

(b)

Fig. 3.3 Transitions between adjoining walls of castings

Design of Machine Tool Structures 175

3. Economy which of them provides for a lower cost of the structure. Correct selection can be made only on the basis of a comprehensive analysis of various factors, some of which are listed below. (i) Economy of metal structure may be low, the actual metal consumption may be high. This is due to the fact that whereas holes in castings are obtained with the help of cores, those in welded structures have to be machined. This results not only in scrap but also in additional labour cost. (iii) Cost of machining. On the basis of all the considerations discussed above, the application of cast iron and steel may be speci1. Steel should be preferred for simple, heavily loaded structures which are to be manufactured in small numbers; this is due to the fact that in lightly loaded structures the higher mechanical properties of steel cannot be fully exploited. 2. Cast iron should be preferred for complex structures subjected to normal loading, when these structures are to be made in large numbers. 3. Lately, combined welded and cast structures are becoming popular. They are generally used where a portions; these complex portions are separately cast and welded to the main structure. An example is that of cast-bearing housings that are welded into the feed box.

3.4

STATIC AND DYNAMIC STIFFNESS

The machine tool is an element of a closed-loop system in which it interacts with the cutting process. The machine tool and cutting process can be represented by their respective transfer functions and the closed-loop system shown as in Fig. 3.4a. The behaviour of any of the elements of the system can be studied by disconnecting the loops and considering the individual element. Such a system for studying the behaviour of the machine tool is shown in Fig. 3.4b. The element is characterised by the ratio of output and input

f(t) Machine tool transfer function

P

y

Cutting process transfer function (a)

y(t)

force), the ratio dY/dP represents the static characteristic of the machine tool, whereas if the input is a time-variable Machine P y tool quantity (dynamic force), the same ratio of dY/dP represents the dynamic characteristic of the machine tool and is known f(t) = const, y(t) = const as its transfer function. (b) For a designer, it is necessary to examine the cumulative static and dynamic characteristics of the machine tool as Fig. 3.4 Block diagram representation of well as the corresponding characteristics of individual ele(a) machine tool-cutting process closed-loop system (b) machine tool ments of which the machine tool is made.

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Machine Tool Design and Numerical Control

3.4.1 Static Stiffness The behaviour of an individual element under static loading can be examined from the force-displacement relationship. This relationship will in general be non-linear as shown in Fig. 3.5a. The quantity dP/dY = tan a = kP1 is known as the static stiffness of the element at force P1. Y is measured in the same direction in which force P is applied. P

P

P

P2 P1 a

Y

P1 Y1

Y2

Y

Y

DP = P2 –P1 DY = Y2–Y1 (a)

(b)

(c)

Fig. 3.5 Force-displacement relationships: (a) Non-linear (b) Linear in a particular range (c) Linear

P = f (Y ) is linear in a range of force between P1 and P2 (Fig. 3.5(b)), then the stiffness in this range is constant and can be determined as KP1 – P2 =

DP DY

P = f (Y ) is linear in the whole range of variation of force (Fig. 3.5(c)), as is generally the case, then the static stiffness of the element can be expressed as P (3.8) Y The quantity which is the inverse of stiffness is known as compliance (C) and is also often employed for analyzing the behaviour of structures under static loading. K=

application of the force, then the ratio P/Y is known as the cross stiffness of the element. The machine tool consists of a large number of elements and its behaviour is of interest in as much as it affects the parameters of machining and the quality of the machined surface. This can be assessed by means of the following two quantities: 1. Static Stiffness with Respect to the Workpiece Accuracy This is a cross stiffness which is represented by the ratio K ¢ = P/Y ¢ (see Fig. 3.6a), where P is the resultant cutting force acting between the cutting tool and workpiece, and Y ¢ is the relative displacement normal to the machined surface. Obviously, the higher the value of K ¢, the less will be the relative cutter-workpiece displacement perpendicular to the machind surface, and consequently the higher will be the accuracy of machining. 2. Static Stiffness with Respect to Dynamic Stability This is also a cross stiffness which is represented by the ratio K = P/Y (see Fig. 3.6b), where P is again the resultant cutting force between the cutter and the

Design of Machine Tool Structures 177

y

y

P

P

(a)

(b)

Fig.3.6 Schematic diagram depicting the cross stiffnesses (a) influencing workpiece accuracy, (b) influencing dynamic stability

workpiece, and Y is the relative displacement normal to the surface of cut. This displacement determines the chatter vibrations (for details see Chap. 6). As already stated, a machine tool consists of a number of elements. These elements experience deformation during operation of the machine tool, and therefore, can be looked upon as elastic bodies (springs) with a certain stiffness (spring constant). The machine tool is thus reduced to a system of springs which are joined in series or parallels. Some important conclusions about the cumulative stiffness of the machine tool system may be drawn from a study of simple spring systems connected in series and parallels. Consider a set of springs of stiffness K1, K2, K3, joined in series (Fig. 3.7a). The springs are acted upon by a common force and the resultant deformation is obtained as the sum of deformations of individual springs, i.e., P P P Y = Y 1 + Y2 + Y3 = + + K1 K 2 K3 K, then P P P P = + + K1 K 2 K3 K or

1 1 1 1 = + + K1 K 2 K3 K

or

C = C1 + C2 + C3

(3.9)

(elements) is obtained by adding reduced stiffnesses. Consider a set of shafts which transmit rotation from the motor to the output shaft (Fig. 3.7b). Let the torsional compliances of the shafts be C1, C2, C3, C4 and the transmission ratio i1,2, i2,3, i3,4. Here, torsional compliance represents the angle of twist between the shaft

178

Machine Tool Design and Numerical Control

P

I

K1

i12 II

K2

i23

K1

III K3

K3

i34 IV

P (a)

Fig. 3.7

K2

(b)

(c)

Connection of elastic elements: (a) Springs connected in series (b) Series connections of shafts by gear transmissions (c) Springs connected in parallel

ends if it is loaded by a unit torque and i1,2, i2,3, i3,4 represent transmission ratios between shafts 1 and 2, 2 and 3, and 3 and 4, respectively. The resultant compliance of the system is determined as 2 2 2 2 C = C1 ◊ i 1,2 i 2,3i 3,4 + C2 ◊ i 22,3i 23,4 + C3 ◊ i3,4 + C4

(3.10)

Here it is assumed that the transmission ratios are less than 1, i.e., speed is being reduced in transmitting compliances, the contribution of shaft 4 to the overall compliance is maximum, while the contributions of shaft 3, 2 and 1 go on decreasing in that order. Therefore, for reducing the overall compliance of the system, it is best to increase the dimensions of shaft 4, i.e., improve the stiffness of that element which contributes maximum to the overall compliance and is, therefore, the weakest link in the chain. ment, the springs undergo equal deformation and the force is distributed between them. Y = Y1 = Y2 = Y3 P = P1 + P2 + P3 KY = K1Y + K2Y + K3Y K = K1 + K2 + K3

(3.11)

The important corollary that ensues from Eq. (3.11) is that for a system of springs joined in parallel, the overall stiffness of the system can be improved best by strengthening the strongest element of the chain. Let us elaborate the point by an example. Suppose in a system of two sprigs one spring contributes 90% and the other 10% to the total stiffness. Now, if the stiffness of the weaker member is increased two times, it will increase the overall stiffness in the ratio 110 : 100, whereas if the stiffness of the stronger member is increased two times, the overall stiffness will increase in the ratio 190 : 100.

3.4.2 Dynamic Stiffness system is linear, a simple harmonic force will result in harmonic displacement of the element. However, for

Design of Machine Tool Structures 179

a constant value of the applied dynamic force, the displacement will vary with a change in frequency of the force. Therefore, the ratio Pdyn Kdyn = Ydyn which is known as dynamic stiffness of the element will depend upon the frequency of the applied force. Cdyn = Ydyn/Pdyn is known as dynamic compliance or simply, receptance. Again, as in the case of static loading, we have direct receptance and cross receptance. Suppose a static force Pst is applied to an element resulting in static deformation Yst, where K is the static Pdyn equivalent to the static force is applied, the deformation will increase by, say, A times, i.e., Ydyn = Yst ◊ A the quantity A = Ydyn/Yst represents the ratio of dynamic displacement to static displacement and is known as the . The dynamic and static stiffnesses of an element are related by the expression, Kdyn =

1. damping factor r, and 2. ratio h = w /w n, where

K A

(3.12)

is the frequency of excitation and wn the natural frequency of the element. -

tion factor is given by the relationship, A1 =

1 2 2

(1 - h ) + (2rh) 2

(3.13)

changes with the rotational speed, i.e., the relationship, A2 = where

Ydyn m0 ◊ e/m

=

h2 (1 - h 2 ) 2 + (2 rh) 2

(3.14)

m0 = disbalanced mass m = total mass of rotating body e = distance between centre of gravity of disbalanced mass m0 and axis of rotation

A1 and A2 are plotted as functions of h for various values of damping factor r be reduced, and hence a high dynamic stiffness can be obtained by increasing the damping and keeping the excitation frequency as far away from the natural frequency of the element as possible. Practically, for h < 0.4 and h > 2.5, the displacement under dynamic loading can be considered equal to the displacement under static loading, irrespective of the damping factor.

180

Machine Tool Design and Numerical Control

A1

A2

4

4

3

3

r=0

r=0

r = 0.125

r = 0.125 2

r = 0.25

2

r = 0.25

r = 0.5

r = 0.5 1

1

r = 1.0

r = 1.0

r = 2.0

r = 2.0 0

Fig. 3.8

0

1

2

0

h

3

0

1

2

3

h

Variation of dynamic amplification factor as a function of h for various values of r (a) when the amplitude of the exciting force is independent of the frequency of excitation, (b) when the amplitude of the exciting force depends upon the exciting frequency

From the point of view of vibration behaviour, a machine tool may be considered a multiple degree of machine tool, the masses are connected in a particular manner, each having its natural frequency and a corn degrees of freedom can be represented by n normal modes in the form of second-order differential equations of the type mi =

d2y dt 2

+ ri

dy + k i y = ri P dt

(3.15)

then the dynamic compliance of the machine tool can be determined by solving n independent equations mi, ri, ki and ri represent the mass, damping factor, stiffness and normalising force factor respectively for the ith mode of vibration. The resultant displacement is determined as y = y1 + y2 + y3 +�+ yn and the dynamic compliance is found as the ratio y/P. The dynamic compliance of the machine tool can also be determined experimentally by applying a harmonic force on the workpiece in the direction coinciding with that of the resultant force during cutting, measuring the relative cutter-workpiece displacement in a direction perpendicular to the surface of the cut and plotting the ratio y/P in the form of a polar plot.

reader from making serious omissions during the design of machine tool structures. static and dynamic stiffnesses in different ways. Therefore, the best course would be to analyse the static and dynamic stiffnesses of the machine tool separately. The study of the dynamic behaviour of a machine tool

Design of Machine Tool Structures 181

requires sophisticated equipment and expertise, which, unfortunately, every design organisation or laboratory may not have. On the other hand, the behaviour of a machine tool structure under static loading can be easily links of the system determined from static and dynamic considerations generally differ only in the higher ments in the overall corresponding stiffnesses of the machine tools are approximately the same. Thus, it may be safely presumed that in a majority of cases, improvement in the stiffness of elements leading to a higher overall static stiffness will also result in a higher overall dynamic stiffness. The reader will do well to recall an important conclusion derived earlier from the results shown in Fig. 3.8, regarding judicial selection of the natural frequency of the element vis-a-vis the frequency of excitation. The natural frequency of an undamped spring-mass system is given by the expression, f =

K /m

(3.16)

By varying K and m, the natural frequency of a particular mode of vibration can be controlled. The frequencies of the modes of vibration correspond to natural frequencies of the elements of the machine tool. Therefore, while designing the elements of machine tools, their natural frequencies should be checked at the design stage and it should be ensured as far as possible that 1. the lowest among the natural frequencies should desirably be 2.5 times greater than the highest excitation frequency or 2. the highest among the natural frequencies should desirably be 2.5 times less than the lowest excitation frequency. They are relatively low in most of the machine tools, e.g., if a lathe spindle rotates with a maximum speed of nmax = 1200 rpm, it would produce a maximum excitation frequency of fe = 1200/60 = 20 Hz. Correspondingly, the heaviest lathe unit should have a natural frequency of f the stiffness of an element which is the same as increasing its natural frequency, will improve the dynamic variations of static and dynamic stiffnesses holds. large if small-sized grinding wheels are used and relatively small if grinding wheels of large diameters are used. Keeping in mind that the average cutting speed of grinding operations is 1800 m/min, the excitation frequency will be fe =

1000 ¥ 1800 Hz p ◊ D ◊ 60

where D = diameter of the grinding wheel in mm. fe = 19.1 Hz; if the grinding wheel diameter is D = 50 mm, the excitation frequency will be fe = 191 Hz. Grinding machines of two fundamentally different designs are manufactured keeping this point in mind. These are: 1. Light-weight construction grinding machines using large-size grinding wheels in which the frequencies of modes of vibrations are kept high by using stiff but light structures; high natural frequency at the cost of stiffness is not desirable as it results in poorer accuracy.

182

Machine Tool Design and Numerical Control

2. Heavy-weight construction grinding machines using small size grinding wheels, in which the frequencies of the modes of vibration are kept low by using heavy structures. namic behaviour. Consider, for example, a vertical milling machine on which machining is done by a face milling cutter having Z = 10 teeth. Depending upon the workpiece and cutter material, machining may be done in a wide range of spindle rpm, say between nmin = 240 and nmax = 1200. The excitation frequencies corresponding to these rpm values will be f e min =

240 ¥ 10 1200 ¥ 10 = 200 = 40 and fe max = 60 60

Designing the milling machine such that the natural frequency of none of its elements lies between 40/2.5 = 16 Hz and 200 ¥ coincides with the natural frequency of the spindle or arbor to cause resonance, a heavy mass is mounted on the spindle or arbor to reduce its natural frequency and move it out of resonance.

PROFILES OF MACHINE TOOL STRUCTURES

3.5

During the operation of the machine tool, a majority of its structures are subjected to compound loading and their resultant deformation consists of torsion, bending and tension or compression. Under simple tensile or compressive loading, the strength and stiffness of an element depend only upon the area of cross section. However, the deformation and stresses in elements subjected to torsion and bending depend, additionally, upon the shape of the cross section. A certain volume of metal can be distributed in different ways to give different values of the moment of inertia and sectional modulus. The shape that provides the maximum moment of inertia and sectional modulus will be considered best as it will ensure minimum values of stresses and deformation. The stiffness of four different sections of equal cross-sectional area is compared in Table 3.3. Table 3.3

Comparison of stiffness of different sections having equal cross-sectional area Relative value of permissible

Section

Area mm2

Weight kgf/m

Bending moment kgf.cm Stress

Torque kgf.cm Stress

Angle of twist

1

1

100

29.0

22

1

1

29

Contd.

Design of Machine Tool Structures 183

Contd.

Table 3.3

10

22

1.12

1.15

43

8.8

29.5

22

14

1.6

38.5

31.4

29.5

22

1.8

1.8

4.5

1.9

0

10

28.3

10 10 100

75

10

10

100

100

assessment seems best suited both in terms of strength and stiffness. The additional advantage that goes in its favour is the ease of proper mating with other surfaces. All considerations combined point towards the

structure have an adverse effect upon its strength and stiffness. Factors that reduce the stiffness of machine tool structures and methods of improving it will now be discussed.

3.5.1

Factors Affecting Stiffness of Machine Tool Structure and Methods of Improving It

The effect of aperture on the torsional stiffness of a box-type structure is shown in Fig. 3.9.2 that a circular hole of diameter d affects a length of approximately twice the diameter, i.e., affected length

184

Machine Tool Design and Numerical Control

l1 = 2d. An elongated aperture affects the stiffness even more and the length of the disturbed range is approximately l2 = l + d, where l is the length of the aperture. The reduction in the static and dynamic stiffness of a structure can be partially compensated by using

d I

–2

plate designs is compared in Table 3.4.

f ¥ 10

Rad

the bending stiffness due to apertures can be compensated to a large extent by using suitable cover plates. However, the effect on the torsional stiffness -

10

5

torsional stiffness must be examined theoretically or experimentally in each particular case and appropriately taken into account. However, for symmetrically placed apertures, the effects can be taken into account by multiplying the torsional stiffness with a reduction k, which is determined from the curves given in Fig. 3.10, while accounting for the affect of apertures on the torsional stiffness of structures. k 1.0

0 l1 = 2d l2 = l + d

Fig. 3.9

L0 = 0 0.05 0.1 L

Variation of angle of twist as a function of aperture shape and size

L0 L = 0 0.05 0.1

L0 L = 0 0.05 0.1

0.9 0.8 0.7 0.6 0.5 0.4

L0 0.5 0.3 0.2 L h = 0.5 B h = 2.0 B

L0 =.5 0.3 0.2 L

L0 = 05 0.3 0.2 L

0.3 L0L

0.2 0.1

b0 h

b0

L =2 B

B

L0 L

L0L L =3 B

h

h

b0

L =4 B

B

0 0.2

Fig. 3.10

0.4

0.6

0.8

0.2

0.4

0.6

0.8

0.2

0.4

0.6

0.8

b0 B

Design curves to account for the effect of aperture size on the torsional stiffness through reduction coefficient k

Design of Machine Tool Structures 185

The following guidelines can be of help: 1. Bending stiffness is affected most by apertures in walls, perpendicular to the plane of bending. Table 3.4

Effect of aperture and cover plate design on static and dynamic stiffness of box section3 Relative stiffness about

Relative natural frequency of vibrations about

Relative damping of vibrations about

X-X

Y-Y

Z-Z

X-X

Y-Y

Z-Z

X-X

Y-Y

Z-Z

100

100

100

100

100

100

100

100

100

85

85

28

90

87

68

75

89

95

89

89

35

95

91

90

112

95

165

91

91

41

97

92

92

112

95

185

2. Apertures of equal dimensions have less effect on the torsional stiffness of a structure with wider walls. 3. From among the aperture dimensions, its width has maximum effect on stiffness. smaller than the dominant one can be neglected. k k ¢ = 0.7 – 0.95; the larger value of k ¢ = 0.95 being used for apertures in wide walls when b0 = 0.5B and t0 = (0.3 – 0.5)L.

186

Machine Tool Design and Numerical Control

The stiffness of structures can be improved by using ribs and stiffeners. However, it should be noted that the effect of ribs and stiffeners depends to a large extent upon how they are arranged. A relative assessment of different stiffening arrangements in box-type structures is given in Table 3.5. Table 3.5

Effect of stiffener arrangement on the bending and torsional stiffness of box-type structures

Stiffener arrangement

Relative stiffness under Bending

Torsion

1.

1.0

1.0

2.

1.10

3.

Relative weight

Relative stiffness per unit weight under Bending

Torsion

1.0

1.0

1.0

1.63

1.1

1.0

1.48

1.08

2.04

1.14

0.95

1.79

4.

1.17

2.16

1.38

0.85

1.56

5.

1.78

3.69

1.49

1.20

3.07

6.

1.55

2.94

1.26

1.23

2.39

provement in the bending and torsional stiffness of box-type structures. The stiffness of open structures, such as lathe beds, which consist of vertical shears connected by ribs also depends to a great extent upon the arrangement of stiffeners. The torsional rigidity of open structures has been compared under different stiffener arrangements and the results are shown in Table 3.6. The results of Table 3.6 indicate that only arrangements 4 and 5 are effective in terms of stiffness-toweight ratio of the structure. Arrangement 4 consisting of two parallel shears which are connected by diagonal ribs is commonly used in machine tool beds and is known as the Warren beam.

Design of Machine Tool Structures 187

The stiffness of structures can also be improved by providing a proper arrangement of fastening bolts. The effect of bolt arrangement and stiffening ribs on the bending and torsional stiffness of a vertical column is depicted in Fig. 3.115 increased by almost 50%. Rigidity of the machine tool as a whole depends upon the rigidity with which varibetween the head stock and tail stock of a lathe with the bed, the base plate of a drilling machine with its column, etc., should be made as rigid as possible. Table 3.6

Effect of stiffener arrangement on torsional stiffness of open structure4

Stiffener arrangement

Relative torsional stiffness

Relative weight

Relative torsional stiffness per unit weight

1.

1.0

1.0

1.0

2.

1.34

1.34

1.0

3.

1.43

1.34

1.07

4.

2.48

1.38

1.80

5.

3.73

1.66

2.25

188

Machine Tool Design and Numerical Control

x y

+

++ ++ ++

+ + +

+ +

+ + +

+

+ + +

+ + +

+ + + + +

+

+

+ ++ ++ ++

+ +

+

+ +

+ + +

Flange design

+

+ + +

+ + + +

160 % 140 120 100 80 60 40 20 0 Torsional stiffness

Fig. 3.11

3.6

Bending stiffness x-direction

Bending stiffness y-direction

Effect of bolt arrangement and stiffening ribs on bending and torsional stiffnesses of columns

BASIC DESIGN PROCEDURE OF MACHINE TOOL STRUCTURES -

the structure are marked. The following forces must be taken into account. 1. Cutting Force Cutting force depends upon the workpiece material, machining parameters, wear of cutting tool, etc. To a designer, a knowledge about the nature and direction of the force and the point where it acts on the structure is often more important than a very precise knowledge of its magnitude. The cutting force is

Design of Machine Tool Structures 189

for design calculation purposes with the help of empirical formulae: Pz = k(a + 0.4c)b kgf PN = where

Px2

+

Py2

= kb(0.4a + c)

(3.17) (3.18)

= force component in the direction of longitudinal feed = force component in the direction of cross feed = force component in the direction of the cutting speed vector = unit cutting force, kgf/mm2; for structural steels, k = 120–180 kgf/mm2 depending upon the hardness of steel, and for cast iron k = 90–110 kgf/mm2 b = width of undeformed chip, mm a = thickness of undeformed chip, mm c mm

Px Py Pz k

2. Friction Force the frictional force is considered proportional to the normal force acting on the contacting surfaces. The

general guidelines: Under dry friction conditions, f = 0.2 – 0.3 Under conditions of semi-liquid friction, f = 0.03 – 0.2 Under conditions of liquid friction, f = 0.002 – 0.05 Under conditions of rolling friction, f = 0.001 for steel and f = 0.0025 for cast iron. 3. Forces of Reaction is statically indeterminate, then additional deformation equations are written, as required in the general soluacting at the centre of the supporting surface.

The effect of the dynamic nature of forces during vibrations is taken into account by means of the amtion factors for single-mass systems acted upon by two types of harmonic forces were given in Eqs (3.13) P(t) from the general expression, t

1 P(t)cos wn (t0 – t)dt A =1– P(t )

Ú 0

where

t0

(3.19)

Machine Tool Design and Numerical Control

190

A = 1 – coswn ◊ t0

(3.20)

Amax = 2. The inertial forces in the machine tool drive during speeding up and braking depend upon the stiffness of the peak torque at the time of switching on the motor can be determined from the relationship, Mmax = Mst where

I1 1 ˆ Ê A 1+ I1 + I 2 ÁË cosf ˜¯

(3.21)

Mst = starting torque under static loading I1 = moment of inertia of the rotor and related masses I2 = reduced moment of inertia of the spindle A A=1–2 cos f f = 0.5 – 0.7 -

sient processes must be determined by using analog computation techniques.

ing of two walls connected by parallel and diagonal stiffeners may be analysed as statically indeterminate thin-walled bars. Group 2 : Closed box type structures like housings of speed and feed boxes are designed for forces Group 3 : Supporting structures like tables, knees, etc., which are generally loaded normal to their base plane are analysed as plates. For each particular case of loading, the design calculations are done by applying the general principles of strength of materials and theory of elasticity. Under general conditions of compound loading, most of the machine tool structures are analysed as elements subjected to bending in two perpendicular planes and torcommon design strategy for machine tool structures can, therefore, be summed up as 1. designing for bending stiffness, 2. designing for torsional stiffness, and

should be selected. The general procedure of designing thin-walled structures for stiffness and strength will now be discussed.

Design of Machine Tool Structures 191

3.6.1 Design for Strength the basis of principal stress. The principal stress may be determined if the normal stresses due to bending and shear stresses due to torsion are known. Consider a box of arbitrary section (Fig. 3.12a) in which the principal axes of inertia are known. The maximum normal stresses in the section due to bending forces acting in xz- and xy-planes may be determined by the formulae, M z max ◊ zmax s z max = (3.22) I yy s y max = where

M y max ◊ ymax

(3.23)

I zz

My max and Mz max = maximum bending moment due to forces acting in the xy- and xz-planes, respectively Iyy, Izz = moment of inertia of the section about y-y and z-z axes, respectively z-z and y-y axes, respecymax, zmax tively

The maximum normal stress occurs at that point where the algebraic sum of the two normal stresses is maximum, i.e., s max = s z max + s y max; for instance, in a rectangular section the maximum normal stress occurs at one of the corners (Fig. 3.12b). The shearing stress in a circular section may be determined from the formula t=

M t max ◊ r IP

Z

IP

zmax d

ymax

(b)

(a)

b B

a A

The shearing stress is maximum at the periphery of the section where rmax = r = the radius of the circular section. Consequently, M t max ◊ r

(3.25)

a

Y

(3.24)

where Mt max = maximum twisting moment, kgf.mm r = distance of the point where stress is being determined from the centre of the section, mm IP = polar moment of inertia of the section, mm4

t max =

b

A

B (c)

Fig. 3.12

(a) An arbitrary section (b) Rectangular box section (c) Solid rectangular section

Machine Tool Design and Numerical Control

192

Generally, machine tool structures are not round in section and Eqs (3.24) and (3.25) are not applicable to arm (point A). The shearing stress at point A can be determined from the formula, tA = tmax =

M t max

(3.26)

a ◊ ab 2

and the stress at point B can be found as tB = ht max

(3.27)

a and h depend upon the ratio of the arms of the rectangle and their values are given in Table 3.7. a and h for different value of a/b ratio

Table 3.7 a/b

1.0

1.5

2.0

2.5

3.0

4.00

6.0

8.0

10.0

a

0.208

0.231

0.246

0.258

0.267

0.282

0.299

0.307

0.313

0.333

h

1.000

0.859

0.795

0.766

0.753

0.745

0.743

0.742

0.742

0.742

For instance, for a narrow rectangular section (a/b > 10), it may be found from Table 3.7 that a = 0.333 = 1/3. Consequently, the maximum shear stress is tmax =

3M t

(3.28)

ab 2

The shearing stress due to torsion of a bar of an arbitrary section can be determined from the following general relationship: M ◊d t= t (3.29) It where

d = width of the section where stress is being determined It = torsional moment of inertia of a non-circular section.

The maximum stress occurs where the section has maximum width, i.e., where d = dmax and it is given by the relationship, tmax = where

dmax = maximum width of the section Wt = torsional section modulus

M t max d max It

=

M t max Wt

(3.30)

Design of Machine Tool Structures 193

mined from the expression, 1 It = b Sab3 3

(3.31)

where b Table 3.8. For instance, for a narrow rectangular section b = 1.0, therefore, from Eq. (3.31), It =

b are given below in

1 3 ab 3 b for various sections

Table 3.8 Shape of section

b

rectangle

1.00

angle

1.00

channel

1.12

T

1.15

I

1.20

Keeping in mind that for a rectangular section b = dmax = constant, the maximum shearing stress is found from Eq. (3.30) as M ◊ b 3M t = tmax = t 1 3 ab 2 ab 3 i.e., the same as derived earlier in Eq. (3.28). For the section consisting of a number of narrow rectangles, dmax represents the width of that rectangle which has the maximum width amongst all rectangles. lated from the relationship, t= where

Mt kgf/min2 2 A0d

A0 = area of the solid section enclosed within the centreline of the wall, mm2 d = thickness of the wall, mm

For instance, for a rectangular section (Fig. 3.12b), A0 = (a – d)(b – d) and the shearing stress is found from Eq. (3.32) as t=

Mt 2(a - d ) (b - d )d

(3.32)

194

Machine Tool Design and Numerical Control

Having determined the normal stresses due to bending forces and the shearing stresses due to the twisting moment at a point of the section, the principal stresses at the same point may be determined from the expression, sp = s/2 + (s /2) 2 + t 2 xy- and xz-planes occur in one section and the twisting moment is also maximum in the same section, then there can be no doubt that the principal stress determined in this

should be determined in a selected number of potentially critical sections in which failure can occur. The design condition is s P max £ s] (3.33) where

s Pmax = maximum principal stress s] = permissible stress for the structure material under tension; generally, s] = 150 – 200 kgf/cm2 s] = 80 – 120 kgf/cm2

3.6.2 Design for Stiffness The design criterion for bending stiffness is d z max £ d z] d y max £ d y] where

d y max, d z max = dy dz

(3.34) xy- and xz-planes, respectively xy- and xz-planes, respectively.

method, moment-area method, etc., may be applied for determining d y max and d z max of the structure in accordance with the accepted design diagram. The design criterion for torsional stiffness is fmax [f ] £ l l

(3.35)

i.e., the maximum angle of twist per unit length of the structure fmax/l should not exceed a permissible value f]/l f] for normal accuracy machine tools is 0.5 degrees per metre length of the bed. The angle of twist of a circular section is given by the expression, Mt f = G ◊ Ip l

(3.36)

Design of Machine Tool Structures 195

where

f/l = angle of twist per unit length G = shearing modulus of the structure material Ip = polar moment of inertia of the circular section

For a rectangular section, the angle of twist is determined from the relationship,

where

Mt f = l g Gab3 a and b = larger and smaller arm of the rectangle g

(3.37)

a/b

g are given in Table 3.9. g for different values of a/b ratio

Table 3.9 a/b

1.0

1.5

2.0

2.5

3.0

4.0

6.0

8.0

10.0

g

0.141

0.196

0.229

0.249

0.263

0.281

0.299

0.307

0.313

0.333

The angle of twist for a thin rectangular section or for a section made up of thin rectangles can be calculated from the expressions, f=

Mtl GI t

M f = t GI t l

or where

(3.38)

It = torsional moment of inertia of the structure section.

The value of It is determined with the help of Eq. (3.31). The maximum value of f occurs in that section in which the twisting moment is maximum, i.e., Mt = Mtmax. The angle of twist of a thin-walled closed section is determined by the following formula: f = Mt 4GA02 l where

A0 si di n

n

Âd i =1

si

(3.39)

i

= area of the solid section enclosed within the centreline of the wall = perimeter of the ith segment of the section = thickness of the ith segment of the section = number of segments of different wall thicknesses into which the section may be divided

For instance, for the rectangular box-type section of Fig. 3.12b,

S ds = 2(a - dd+ b - d ) = 2(a + db - 2d ) i

i

196

Machine Tool Design and Numerical Control

Substituting this value in Eq. (3.36) and keeping in mind that A0 = (a – d)(b – d), we can write the following expression for the rectangular box section: M t (a + b - 2d ) f = l 2G (a - d ) 2 (b - d ) 2 ◊ d

(3.40)

f/l will again occur in that section in which the twisting moment is maximum.

Example 3.1 An angle section is welded to a channel section as shown in Fig. 3.13. Calculate by how many times this will reduce the shear stress and increase the torsional stiffness of the compound structure as compared to that of channel section. 1 It = b Sab3 3 For channel section b = 1.12, therefore, 1 It = 1.12 (2bd 3 + ad 3 ) 3 1 3 It = 1.12 d (2b + a) 3 For compound section 1 It = 1.12 bd 3 + a(2d)3] 3 1 It = 1.12 d 3 b + 8a] 3 Maximum shear stress M ◊d tmax = t max It

d

d a

b

b

Fig. 3.13

For the channel section d max = d, therefore, t max =

3M t ◊ d 3

1.12d (2b + a)

=

3M t 1.12d 2 (2b + a )

For the compound section d max = 2d , therefore, tmax =

3M t ◊ 2d 3

1.12d (3b + 8a )

=

6M t 2

1.12d (3b + 8a )

Hence, reduction in shear stress is =

6M t 2

1.12d (3b + 8a )

¥

1.12d 2 (2b + a ) 2(2b + a ) = 3M t (3b + 8a )

Design of Machine Tool Structures 197

Angle of twist per unit length M f = t GI t l For the channel section 3M t f = l 1.12Gd 3 (2b + a ) For the compound section 3M t f = l 1.12Gd 3 (3b + 8a ) Hence, improvement in torsional stiffness is 3M t 1.12Gd 3 (3b + 8a )

3.7

¥

1.12Gd 3 (2b + a ) (2b + a ) = 3M t (3b + 8a )

DESIGN OF BEDS

EIn and its angle of twist depends upon the product GIt , Modulus of elasticity E and shearing modulus G signify the role of bar material In and torsional moment of inertia It EIn and GIt, EIn and GIt are analogous to the bending and torsional stiffness of the bar. These quantities will henceforth be referred to as the reduced bending rigidity and reduced torsional rigidity. The modulus of elasticity and shearing modulus are generally constant for a particular material and their values can be found from an appropriate handbook. The bending and torsional moments of inertia of sections can be determined by the general methods known to the reader from the basic course in strength of materials. Machine tool beds are generally complicated structures of partially or fully closed box sections with stiffeners, ribs, partitions, etc. Beds are usually employed in machine tools with wall arrangements shown in Table 3.10 and are analysed as bars subjected to bending and torsion. bending in the horizontal plane, the reduced bending rigidity of a bed consisting of two walls and perpendicular stiffeners is given by the expression, EIbr = s1 ◊ EI w sin

(3.41)

The reduced bending rigidity of the bed with diagonal stiffeners is given by the relationship, EIbr = s2EL2 ◊ Aw E = modulus of elasticity of the bed material, kgf/cm2 E ◊ Ibr = reduced bending rigidity of the bed, kgf.cm2 L = length of the bed that undergoes deformation, cm Aw = area of cross section of the wall, cm2

(3.42)

198

Machine Tool Design and Numerical Control

Iw min = moment of inertia of the wall cross section in the plane of minimum rigidity in bending, cm4 s1, s2 = The values of s1 and s2 h1 = 1 +

36y

v2 3 + 4x 36y h2 = + 2 3+x v

Table 3.10

9v m ˘ È Í1 + ˙ + x )2 ˚ ( 3 Î

x=

˘ 1 È I w min + 36ym ˙ Í v ÍÎ I s min ˙˚

m=

Aw As

Commonly used bed sections and wall arrangements and their applications

Wall Arrangement

Application

1. Beds on legs or shears (a) without stiffening diagonal wall; used in lathes, turrets, etc. (b) without stiffening diagonal wall; has 30–40% higher stiffness than arrangement (a); used in multiple-tool and high production lathes. (c) with stiffening wall and provision of chip disposal through opening in rear wall; used in large-sized lathes and turrets. (d) with stiffening wall; also used in large-sized lathes and turrets. (c)

(d)

milling, slotting and boring machines

bed is also required to serve as an oil reservoir; commonly employed in grinding machines

Design of Machine Tool Structures 199

Table 3.11

1

and s2 for different stiffener arrangements

Beds with vertical stiffeners Stiffener arrangements

Beds with diagonal stiffeners

n

s1

1

2

n

s2

8/h1

2

sin a ◊ cos 2 a

4(h2 + 4h1 ) h1h1

4

32 h2

6

3

v = y=

Stiffener arrangements

12[ m + sin 3 a ]

sin a ◊ cos 2 a 12[ m + 3 sin 3 a ] sin a ◊ cos 2 a 12[ m + 6.3 sin 3 a ]

L B(n + 1) I w min Aw ◊ B 2

a = half of the angle between diagonal stiffeners s2 is determined as the arithmetic mean of values of s2, corresponding to the nearest available larger and smaller values of n available in Table 3.11. The number of stiffeners is selected from the following considerations:

equal to the width of the bed, i.e., the distance between the parallel walls. An increase in the number of stiffeners over the value obtained from the above guidelines is not rational, For bending in the vertical plane, the reduced rigidity of the bed is only determined by the walls; stiffenvertical bending of short beds, the reduced rigidity is obtained by multiplying the analytical rigidity with s3, which is given by the relationship, s3 = 1+

1 30lw max L2 Aw¢

(3.43)

200

where

Machine Tool Design and Numerical Control

Iw max = moment of inertia of the cross section in the plane of maximum rigidity against bending, cm4 A w¢ = area of the vertical portions of the wall, cm2

The reduced torsional rigidity of a bed with perpendicular stiffeners is given by the expression, GIt =

where

B 2 EI w max EI w max a1kL2 + a 2 GAw¢

(3.44)

B = width of bed, cm depends upon the number of stiffeners k a1, a2

For a bed consisting of two parallel walls and perpendicular stiffeners a1 =

1 and a2 = 2 6

Consequently, the reduced torsional rigidity of such a bed can be expressed as GIt =

B 2 EI w max kL2 2 EI w max + GAw¢ 6

(3.45)

For a bed consisting of two vertical and one horizontal walls with perpendicular stiffeners, a1 = a2 = where

e +6 12(2e + 3) 3(3e 3 + 16e 2 + 42e + 36) 5(2e + 3) 2

(3.46)

e = B/hw, hw being the height of the vertical walls

For a bed consisting of two vertical walls and an inclined wall with perpendicular stiffeners, a1 = a2 = where e1 = 1 +

e1 + 2 12(2e1 + 1) 3(5e13 + 16e12 + 14e1 + 4) 5(2e1 + 1) 2

(3.47)

B2 h w2

k= more partition (perpendicular stiffener), i.e., n > 1, the value of k is determined from Fig. 3.146 as a function of e or e1, n and l/b b is determined from the relationship,

Design of Machine Tool Structures 201

L ◊ G ◊ I st 1 = b (n + 1) BEI w max where

(3.48)

Ist represents the torsional moment of inertia of the stiffener. k

e = 2 e1 = 2

e = 1 e1 = 1

0.9 0.8 n=2 1.0

e1= 2

e=2

0.9

e = 1 e1= 1

0.8 0.7 0.6

n = 23

1.0 e1= 2

e=2 e = 1 e1= 1

0.8

0.6

n=4

1.0

e1= 2 0.8

e1 = 1e = 1

0.6

n=5 0.4 0

0.02

0.04

0.06

0.08

0.1

1/b Bed consisting of two parallel walls Bed consisting of two parallel walls and an inclined wall Bed consisting of two parallel walls and a horizontal wall

Fig. 3.14

Design curves to account for the effect of the number of stiffeners on the torsional rigidity of beds through coefficient k

202

Machine Tool Design and Numerical Control

For beds having T-shaped walls (see 1(a), Table 3.10) Ist is determined from the expression, 1 Ist = d 3s (bs + hs) 3 where

(3.49)

bs = width of stiffener, cm hs = height of stiffener, cm ds = thickness of stiffener, cm p shaped section (see 1(b), Table 3.10), then Ist is calculated from the following relation-

ship: Ist =

where

es =

bs hs

0.21d s hs3 B 2 (e s + 6) 3e 3 + 16e s2 + 42e s + 36 + s 2 12bs (2e s + 3) 8e s2 (2e s + 3) 2

(3.50)

The reduced torsional rigidity of beds with diagonal stiffeners can be determined from the expression, GIt = k1EI w max

B2

(3.51)

L2

where k1 following expression: 1 1 È = Í2 + 3d1 + 6g 1 k1 12n 2 Î -

˘ - (6g 1 - 3d1 - 1) 2 [(2n - 1)(n - 1) + n(6g 1 + 6g 2 + 2d1 - 1) - (6g 1 + 3d1 - 1)] ˙ n(2n - 1)(n - 1)(6g 1 + 6g 2 + 2d1 + 2) + (6g 1 + 3d1 - 1 + 3n)[n(6g 1 + 6g 2 + 2d1 - 1) - (6g 1 + 3d1 - 1)] ˚ (3.52)

where

d1 =

n3 a 3 EI w max L3 EI s max g1 = g2 =

n 2 EI w max L2 ◊ G ◊ Aw¢

(3.53)

an3 EI w max L3 ◊ G ◊ As¢

Ismax = moment of inertia of the stiffener in the plane of maximum rigidity against bending, cm4 a = length of the diagonal stiffener, cm A s¢ = area of the vertical projection of the stiffener, cm2 The reduced torsional rigidity of a rectangular box-type section is given by the relationship, n

GIt = G ◊ 4A 02

di

Âs i =1

i

(3.54)

Design of Machine Tool Structures 203

d and the perimeter of the box section is s, GIt = G ◊

3.7.1

4 A02 ◊ d s

(3.55)

Design of Lathe Bed

The design of lathe bed is discussed for the case of a workpiece of diameter d supported between the centres. The force diagram is shown in Fig. 3.15. The forces considered in the analysis are: 1. Cutting force components Px, Py and Pz representing the axial, radial and cutting force, respectively acting on the work piece at the given location of the cutting tool. 2. Reactions in the vertical direction R¢AZ and R¢BZ at the head stock centre A and tail stock centre B, respectively. 3. Reactions in the horizontal direction R¢AY and R¢BY acting at the headstock centre A and tailstock centre B, respectively 4. Weight of the work piece G. G 5. Reactive force each acting at A and B. 2 6. Pressing force K acting on the workpiece at the tailstock B. 7. Tightening force Px + K acting on the work piece at the headstock A. L L1

RAY

L2

RBY

PZ

R¢AZ

R¢BZ PY

PX

Px +K

K G

G 2

A

fd

B G 2

Fig. 3.15 Schematic showing the forces acting on lathe bed

204

Machine Tool Design and Numerical Control

Considering the forces in the horizontal plane and taking the moment of forces about A under conditions SMA = 0;

Py L1 – RBYL – Px

wherefrom, RBY =

d =0 2

1Ê dˆ Á Py L1 - Px ˜¯ 2 LË

Similarly, taking the moment of forces about B under conditions of equilibrium SMB = 0;

Py L2 – R AY L + Px

wherefrom, RAY =

d =0 2

1Ê dˆ ÁË Py L2 + Px ˜¯ 2 L

Considering the force Pz and G acting in the vertical plane, it may be noted that reactions R¢AZ and R¢BZ vary G with the location of the cutting tool, i.e., L1 and L2, but the reaction at each of the supports is constant and 2 independent of the location of the cutting tool. Under the conditions of equilibrium, SMA = 0;

Pz L1 – R¢BZ L +

wherefrom, R¢BZ =

G L L-G =0 2 2

Pz L1 L

Hence, net reaction at the tail stock is

RBZ = R¢BZ –

G Pz L1 G = 2 L 2

Similarly, SMB = 0;

Pz L2 – R¢AZ L –

G L L+G =0 2 2

wherefrom, R¢AZ =

Pz L2 L

Hence, net reaction at the headstock is RAZ = Pz

L2 G L 2

The forces are transmitted to the bed through the 1. head stock at a distance a, where a is the distance from the head stock centre to the front bearing of the spindle,

Design of Machine Tool Structures 205

2. saddle directly underneath the middle of the saddle, and 3. tail stock at a distance b, where b is the distance from the tail stock centre to the point where the tail stock is bolted to the bed.

Design for Strength

The design for strength takes into account the shear stresses due to torsion and the bending stresses due to bending.

Shear Stresses due to Torsion The torque distribution at the point of transmission of load to the lathe bed is shown in Fig. 3.16 assuming that the saddle is positioned at the middle point between the centres. This of the bed. Therefore, at L1 = L2 =

L , 2 Pz G - ; 2 2 Py Px d = ; + 2 2L

Pz G 2 2 Py Px d = 2 2L

RAZ =

RBZ =

RAY

RBY

At the head stock the torque transmitted to the bed is MTHS = Pz

d Ê Py Px d ˆ d + RAY h = Pz + Á + ˜ h, 2 Ë 2 2L ¯ 2

d + Py h and 2 Ê Py Px d ˆ At the tail stock the torque transmitted to the bed is MTTS = RBY h = Á ˜h Ë 2 2L ¯ Under the saddle the torque transmitted to the bed is MTS = Pz

G RAZ =R ¢AZ – – 2 RAY

RBZ =R ¢AY – G – 2 PZ

RBY

PY

PX G h

fd MTHS

MTTS

MTS

Fig. 3.16 Schematic of torque distribution in lathe bed

206

Machine Tool Design and Numerical Control

MTS = MTHS + MTTS. Therefore, the design for tensional strength and stiffness is based on the maximum value of MTS which occurs while machining a job of diameter d = 2h; hence, MTSmax =

Pz 2h + P y h = (Pz + Py)h 2

Hence, maximum shear stress t max = where

MTs max ◊ max[ ymax , zmax ] It

It is the torsional moment of inertia of the bed section.

ymax and zmax represent the distance of the outermost edge of the bed section from the neutral axis along the Y- and Z-axis, respectively.

Normal Stresses due to Bending in Vertical Direction The force diagram of the system and the bending moment diagrams due to forces acting in the Z and X directions are shown in Fig. 3.17. a

G RAZ = R AZ ¢ – – 2

RBZ = R BZ ¢ –G – 2 b

PZ

PX G fd

h c hFP

MAZ

MBZ

MSZ

MBX MAX MSX

Fig. 3.17 Forces acting on lathe bed in vertical direction and the corresponding bending moment diagrams

Design of Machine Tool Structures 207

The bending moments produced by the forces are: MAZ = RAZ ¥ a, under the head stock, MBZ = RBZ ¥ b, under the tail stock and L MSZ = (Pz – G) , under the saddle 4 The feed force component Px on the tool post, the pressing force K on the tail stock and the tightening force Px + K on the head stock, all acting at the height of centre h are held in equilibrium by equal and opposite forces in the feed pinion at the pitch line of the rack at a distance hf p below the bed surface. This results in tilting moments MAX = (Px + K) (h + hfp) = (Px + K) c under the head stock, MBX = K (h + h fp) = Kc under the tail stock and MSmax = MAX = (Px + K) c under the saddle The maximum bending moment occurs at the middle of the bed. Therefore, the maximum normal stress in the vertical direction is given by the expression, s z max =

( M s z + M s max ) Z max I yy

Substituting the expressions of MSZ and MSmax L Ê ˆ ÁË ( Pz - G ) + ( Px + K )c˜¯ zmax 4 s z max = I yy where

Iyy is the moment of inertia of the bed section about the Y-axis.

Normal Stresses due to Bending in Horizontal Direction The force diagram of the system and the bending moment diagram due to forces acting in the Y-direction are shown in Fig. 3.18. The bending moments produced by the forces are: MAY = MBY =

Py 2 Py 2

MSY = Py

¥ a, ¥ b, L 4

The maximum bending moment acts at the middle of the bed. Therefore, the maximum normal stress in the horizontal direction is given by the expression, Py s y max =

L ymax 4 I zz

208

Machine Tool Design and Numerical Control a

P R¢AY = –Y 2

b

P R¢BY = –Y 2

PY

G fd

MSY

MAY

Fig. 3.18

where

MBY

Forces acting on lathe bed is horizontal direction and the corresponding bending moment diagram

Izz is the moment of inertia of the bed section about the Z-axis.

Principal Stress and Design Criterion Having found szmax, s ymax and tmax, the maximum principal stress is found from the expression, spmax =

s z max + s y max 2

+

1 2 (s z max - s y max ) 2 + 4t max 2

The design for strength is carried out with the help of Eq. (3.33).

Design for Stiffness

Z and Y directions is found by the well-known

As regards the maximum angle of twist, it is determined by using the value of maximum torque MTSnax. As already discussed in Sec. 3.6.2, the design for bending stiffness is based on Eq. (3.34) and that for torsional stiffness on Eqs (3.35) and (3.36).

Moment of Inertia of Lathe Bed Section procedure discussed above, it is necessary to know the moments of inertia of the bed section about the Y-Y and Z-Z axes for determining s z max and s y max Z and Y directions and the torsional moment of inertia for determining t max and the maximum angle of twist.

Design of Machine Tool Structures 209

A common lathe bed usually consists of two girders connected by webs at regular interval. For ease of derivation of the expressions of moment of inertia of such a composite section is described below with the assumption that the contribution of the connecting webs to the moment of inertia of the composite section is negligible. 1. Moment of inertia about the Y-Y axis

As the composite section is symmetrical about the Z-Z axis, the IYY of the

b3

Z

3¢

3

b2

Z2

a2

2¢

2 d2

Y

Y

a1

Z1

1¢

1

b1

Z

Fig. 3.19 Composite I-section of lathe bed

For segment 1

I1 =

b1 ( z1 - a1 )3 È (z - a ) ˘ + b1 ( z1 - a1 ) Í 1 1 + a1 ˙ 12 2 Î ˚

2

È ( z - a )2 ( z + a )2 ˘ = b1(z1 – a1) Í 1 1 + 1 1 ˙ 4 Î 12 ˚ =

b1 ( z1 - a1 ) 2 z 1 + a 12 – 2a1z1 + 3z 21 + 3a 21 + 6a1 z1] 12

210

Machine Tool Design and Numerical Control

b1 ( z1 - a1 ) 4z 21 + 4a 12 + 4a1z1] 12 4b = 1 (z1 – a1)(z 21 + z1a1 + a12) 12 b1 3 = (z 1 – a 13) 3 =

Hence, for both segments 1 and 1¢ I1 + I 1¢ =

2b1 3 (z 1 – a13) 3

(3.56)

For segment 2 Segment 2 consists of two subsegments a1 b2 and a2 b2 . Applying the parallel axis theorem to both these 2 2 3 È b a3 Ê a ˆ ˘ Èb a Êa ˆ ˘ I2 = Í 2 1 + b2 a1 Á 1 ˜ ˙ + Í 2 2 + b2 a2 Á 2 ˜ ˙ Ë 2 ¯ ˚ Î 12 Ë 2¯ ˚ Î 12

Ê b a 3 b a 3 ˆ Ê b a3 b a3 ˆ = Á 2 1 + 2 1˜ +Á 2 2 + 2 2˜ Ë 12 4 ¯ Ë 12 4 ¯ Ê b a 3 + 3b2 a13 ˆ Ê b2 a23 + 3b2 a23 ˆ =Á 2 1 ˜¯ + ÁË ˜¯ Ë 12 12 =

4(b2 a13 + b2 a 23 ) b2 (a13 + a 23 ) = 12 3

Hence, for both segments 2 and 2¢ I2 + I 2¢ =

2b2 3 (a 1 + a 23) 3

(3.57)

For segment 3 Segment 3 is similar to segment 1 as regards its moment of inertia about the Y-Y axis. Therefore, based on I3 + I 3¢ =

2b3 3 (z 2 – a 23) 3

(3.58)

Y-Y and (3.58) Iyy =

2b1 3 2b 2b (z 1 – a 31) + 2 (a 31 + a 23) + 3 (z 32 – a 23) 3 3 3

I yy =

2 b1(z 31 – a 13) + b2(a 13 + a 23) + b3(z 23 – a 32)] 3

2. Moment of Inertia about the Z-Z axis 1-1¢ Q¢R¢S¢T¢

(3.59)

U¢V¢W¢X¢, i.e.,

Design of Machine Tool Structures 211

1 d 1 d1 A3 – d1 B 3 = 1 (A3 – B3) 12 12 12

I 1–1¢ =

¢

IJKL minus

MNOP, i.e., I 2– 2 ¢ =

(a1 + a2 ) 3 (C – D3) and 12

¢

QRST

the rectangle UVWX, i.e., I 3– 3¢ =

d3 3 (E – F3) 12 Z-Z

Iz–z = I1–1¢ + I2–2 ¢ + I3–3¢ Izz =

1 d1(A3 – B3) + (a1 + a2) (C3 – D3) + d3(E3 – F3)] 12

(3.60)

Z E F

Q d3

3

R I

U

X

V

W

T

M

3¢ S L

P C

a2

D 2

Y 2¢

a1 Q¢ J

d1

N

U¢

X¢

K

T¢

1¢

1 V¢ R¢

O

W¢ B A Z

Fig. 3.20 Composite I-section of lathe bed

S¢

212

Machine Tool Design and Numerical Control

3. Torsional Moment of Inertia Applying the perpendicular axis theorem, the torsional moment of inertia of the lathe bed section is found as It = IY–Y + IZ–Z

(3.61)

Example: A part of diameter 200 mm and length 1500 mm is supported between centres with a tightening load of 5000 N and machined on a lathe. The cutting force components were recorded with a dynamometer as Px = 2500 N, Py = 5000 N and Pz =10000 N with cross beams. The dimensions of the bed section (refer to Figs 3.19 and 3.20) are as follows: b1 = 110 mm, b2 = 20 mm, b3 = 115 mm, d1 = 30 mm, d2= 315 mm, d3 = 25 mm, A = 365 mm, B = 145 mm, C = 275 mm, D = 235 mm, E = 370 mm, F = 140 mm. The data related to the machine tool is as follows: Height of centres h = 300 mm, distance of the feed pinion from the bed surface hfp workpiece, check the strength of the bed if the allowable stress of the bed material is 12 N/mm2.

25 ˆ Ê 315 ˆ Ê 30 x 110 x 15 + 315 x 20 x Á + 30˜ + 115 x 25 x Á 345 + ˜ Ë 2 ¯ Ë 2 ¯ = 181.05 mm z= 30 x 110 + 315 x 20 + 25 x 115 Hence,

z1 = 181.05 mm

and

z2 = d1 + d2 + d3 – z1 = 188.95 mm

a1 = z1 – d1 = 151.05 mm

and

a2 = z2 – d3 = 163.95 mm Z-axis y=0

Iyy = 466508061.6 mm4, Izz = 419005416.7 mm4 and I t = 885513478.3 mm4. We now determine the maximum normal and shear stresses in the bed section L Ê ˆ ÁË ( Pz - G ) + ( Px + K )c˜¯ zmax 4 s z max = I yy For the composite bed section, Zmax

z1, z2 s zmax = 2.58 N/mm2 Py s ymax =

For the composite bed section ymax

A/2, E

zmax = 188.95 mm.

L ymax 4 Izz ymax = 185.0 mm s ymax = 0.207 � 0.21 N/mm2

Design of Machine Tool Structures 213

Ê d ˆ ÁË Pz + Py h˜¯ ◊ max[ ymax , zmax ] MTs ◊ max[ ymax , zmax ] 2 tmax = = It It ymax, zmax

ymax, zmax] = tmax = 0.53 N/mm2

Knowing s ymax, s zmax and tmax we determine the principal stress from the expression, s pmax =

s z max + s y max 2

+

1 2 (s z max - s y max ) 2 + 4t max 2

s pmax = 2.79 N/mm2 As s pmax is less than the allowable stress it may be concluded that the bed design is safe in terms of strength.

3.8

DESIGN OF COLUMNS

only in the plane of symmetry, e.g., drilling machines or by forces acting arbitrarily in space (e.g., vertical lathe, milling and boring machine, etc.). The principal design requirements of columns are the same as those of beds, i.e., high static and dynamic stiffness. These properties are achieved, as in the case of beds, by proper selection of the column material and its cross section. Based upon considerations discussed in Sec. 3.5, the columns are made with thin-walled circular or box sections. The number and size of holes and openings is kept as small as possible and the section is strengthened by means of stiffeners. Commonly used sections of machine tool columns and their application are shown in Table 3.12.

only marginally. Generally, the area of the column section gradually increases as we move from the top towards the base where the bending moment is maximum. From the point of view of torsional stiffness, it is desirable that, as we move away from the base, the section should progressively change from a thin-walled rectangle to a thin-walled square. 1. bending in two perpendicular planes, 2. shearing in two perpendicular planes, and 3. torsion. at the base, and plotting the bending moment diagram in both the directions.

214

Machine Tool Design and Numerical Control

Table 3.12

Commonly used column sections and their applications

Section

Application

1. Circular section with vertical stiffeners; used when the load is small and it is necessary to provide for rotation of the column, e.g., in bench and radial drilling machines.

2. Rectangular box-type section; used when the forces act only in the plane of symmetry; chief application is in vertical drilling and unit-built machine tools. Optimum a/b = 2–3

3. Square box-type section with vertical ribs and spaced horizontal stiffeners; used for columns subjected to threedimensional loading; chief application is in boring and milling machines. Optimum a/b = 1

4. Rectangular box-type section with vertical ribs and spaced horizontal stiffeners; is used in gantry type machine tools. The recommended a/b ratios are: for vertical lathe a/b = 3 –4 for planing machine a/b =2 – 3 for plano-milling machine a/b = 2 – 3

D =l where

Q l A G l

Q◊l G◊A

(3.62)

= shearing force = distance from the base of the section in which shearing deformation is being determined = area of cross-section = shearing modulus of column material

Design of Machine Tool Structures 215

l may be determined from the curves (A to B and B to C) given in Fig. 3.21.7 The angle of twist f of the column is found, as in the case of beds, from Eq. (3.39). The displacement of the guideway (point A) of a rectangular box-type section (Fig. 3.22) due to torsion may be determined from the following expressions: n direction X-X

DX = y ◊ f

n direction Y-Y

DY =

l1¢

b ◊f 2

P(l1¢ )

l 2¢ A

P(l2¢ )

C

4

h

l2¢

Y

y A

b 3

b/2 X

2

1 0.25

B

0.75

X b/2

l1¢

1.25

1.75

h b

A Y

Fig. 3.21 Variation of the coefficient of distribution of shearing displacement as a function of h/b ratio

Fig. 3.22 Cross section of a milling machine column

ation. For this purpose, the base is treated as a hollow box section simply supported at the ends. The analysis

the guideway at the top of the column does not exceed 3–5 microns per metre length in the Y-Y direction and 10–25 microns per metre length in the X-X direction. The effect of holes and apertures on the torsional stiffness of the column can be taken into account with tion in the torsional stiffness due to the openings can be neglected provided each bolt is tightened by a force P≥ where

M t (b0 + l0 ) kgf Ab fn

P = tightening force of each bolt, kgf Mt = torque acting on the column, kgf.cm

(3.63)

216

Machine Tool Design and Numerical Control

b0, l0 = width and height of aperture, cm f n = number of bolts Ab = area of cross section of one bolt The effect of ribs and stiffeners on the bending stiffness of columns is small and can be neglected. The vertical ribs also do not have any appreciable effect on the torsional stiffness.

DESIGN OF HOUSINGS

3.9

Housings are structures in which the three limiting dimensions are 2a more or less equal. Examples of housings in machine tools are the speed-box housing, feed-box housing, etc. A Housings may be split or solid. Solid housings are widely used in small and medium-sized machine tools. Split housings are easier to assemble but are less stiff. Split housings that have to be opened frequently for regulating some mechanism (e.g., the speed box of engine lathes with a cone pulley drive) are provided with a hinged cover. The stiffness of such a housing is on an average 50% less than that of a solid housing. Housing-type structures (Fig. 3.23) are designed for stiffness, Fig. 3.23 Schematic diagram of a housing-type structure A due to a force acting normal to the wall at the same point. This displacement can be found from the relationship, y = k 0 k 1 k 2 k3 ◊ where

k0 k1 k2 k3 h 2a m E P

P ◊ a 2 (1 - m ) Eh3

cm

(3.64)

= = = thickness of the loaded wall without bossing, cm = the larger dimension of the rectangular loaded wall, cm = Poisson’s ratio of the housing material = Modulus of elasticity of the housing material, kgf/cm2 = force acting normal to the loaded wall, kgf

k0 depends upon the ratio 2a : 2b : 2c of the housing dimensions and also upon the point of k0 for housings are given in Table 3.13 for various ratios of a : b : c when force P is acting at the centre of the loaded wall.

Design of Machine Tool Structures 217

Table 3.13

0

for housings8

All four edges of loaded wall 2a ¥ 2b connected to adjoining walls

Three edges of loaded wall 2a ¥ 2b connected to adjoining walls, one edge free

1:1:1

0.35

0.48

1 : 1 : 0.75

0.44

—

1 : 1 : 0.5

0.5

—

—

0.45

1 : 0.75 : 0.75

0.3

0.42

1 : 0.75 : 0.5

0.33

—

1 : 0.5 : 1

—

0.28

1 : 0.5 : 0.75

—

0.27

Ratio a : b : c

1 : 0.75 : 1

Tables 3.13–3.15 have been adopted from Ref. 8.

k0 is maximum for the k0 reduces as the k0 at the two constrained corners may be up to three times less. However, if the point of application of the force is moved towards the free corners, k0 occurs at the middle of the free edge and may be about 2.5–3 times greater than the tabulated values for load acting at the centre. Bosses are made to increase the wall thickness locally where the stiffness has suffered due to a hole or aperture. They are generally located on the inside surface of the wall (Fig. 3.24). Bossing dimensions are generally limited to D = (1.4–1.6) d and H = (2.5–3)h because any further increase does not contribute apk1 in Eq. (3.63) depends upon the ratios D/d, H/h and r/a. H/h, although the ratio D/d has little effect for values of D/d > k1 are given in Table 3.14. Ha h

P

1.8

h Ha H

r/a

0.5

1.4 R

r

a¢ 2a

Fig. 3.24

=0 0.3

1.0 1.0

1.4

1.8

2.2

H h

A typical bossing in a housing type structure and design curves for determining Ha/h as a function of H/h

218

Machine Tool Design and Numerical Control

Table 3.14

1

for housings; r/a = 1.0, D/d = 1.6

D2/2a ◊ 2b

Ha/h 0

0.02

0.04

0.06

0.08

0.10

0.12

0.14

1.2

1

0.95

0.94

0.92

0.90

0.88

0.87

0.85

1.4

1

0.88

0.83

0.78

0.75

0.72

0.69

0.67

1.6

1

0.82

0.75

0.68

0.63

0.6

0.57

0.55

2.0

1

0.75

0.65

0.58

0.53

0.48

0.45

0.41

3.0

1

0.7

0.58

0.5

0.44

0.38

0.35

0.3

Ha represents the active height of the bossing. Ratio Ha/h is determined from known H/h ratios with the help of Fig. 3.24. k1 D/d = 1.2, the values of coThe effect of D/d and r/a k1 are higher than that tabulated by 5–10%. The effect of r/a is not uniform. For average values of D2/2a ◊ 2b and Ha/h (D2/2a ◊ 2b = 0.8 and Ha/h = 1.5), the reduction of ratio r/a to 0.6 has no effect on the value of k1. At high values of D2/2a ◊ 2b and Ha/h creases by 2–5% as compared to the tabulated values. k2 = 1 + SDyi /y, where Dyi i value depends upon the ratios D/d, Ha/h and R/a¢ (see Fig. 3.24). k2 are given in Table 3.15. Reduction of D/d k2 by 5–10%. The effect of R/a¢ on the value of k2 depends greatly upon D2/2a ◊ 2b and Ha /h. For average values of these ratios (D2/2a ◊ 2b = 0.5 and Ha/h = 1.5) an increase of R/a¢ to 0.7 results in an increase of k2 by about 15%. A decrease of R/a¢ to 0.3 reduces the k2 value about 15–20%. At high values of D2/2a ◊ 2b and Ha /h, the corresponding increase and decrease are approximately 25–30%. On the other hand, for combination of low values of D2/2a ◊ 2b and Ha /h, the effect of R/a¢ on k2 is virtually negligible. 2;

Table 3.15

R/a, = 0.5, D/d = 1.6 D2 2a ◊ 2b

Ha h 0

0.02

0.04

0.06

0.08

0.10

1.2

1.0

0.99

0.99

0.98

0.98

0.98

1.4

1.0

0.96

0.94

0.92

0.90

0.84

1.6

1.0

0.93

0.90

0.88

0.86

0.84

2.0

1.0

0.92

0.88

0.85

0.82

0.80

3.0

1.0

0.90

0.85

0.80

0.76

0.74

Design of Machine Tool Structures 219

k3 = 0.8–0.9 if the stiffening ribs are cast in the immediate vicinity of the loaded hole to ness, then k3 = 0.75–0.85. The smaller values are for intersecting ribs, which form a network while the larger values are for ribs that are not interconnected. total deformation y. This is the reason why bossing of the hole is the single most vital step towards providing ers in housing-type structures is of secondary importance.

3.10 DESIGN OF BASES AND TABLES Machine tools bases are analysed as plates on an elastic foundation. The dimensions are determined from the consideration that

Fig. 3.25. The total angle of slope in a section is q = q q1 + q q2 + qM where

(3.65)

q q1 = angle of slope due to distributed force q1 q q2 = angle of slope due to distributed force q2 q M = angle of slope due to bending moment M q q1 =

q1m ◊ b k q1 k

q q2 =

q 2m ◊ b k q2 k

qM =

M ◊ m3 ◊ kM k

M

a2

a2

b2

(3.66)

Fig. 3.25

b1

Loading diagram of a machine tool base

b = width of the base, cm k = k = 125b, kgf/cm2 m = (k/4EI)1/4 E being the modulus of elasticity of plate material and I the moment of inertia of plate cross section, cm–1 Kq1, Kq2, KM = mined from the curves of Fig. 3.269. The machine tool table supports the workpiece. From a purely functional point of view, the base may be considered a stationary table. The tables that not only support the workpiece but also provide motion to it are generally of two types: 1. Rectangular 2. Circular

220

Machine Tool Design and Numerical Control

b2 KM

b1 M

q2

q1

Kq1 Kq2 0.6

4 b2/2 a2

3

a1

0.4

b1/2

0.2 1

f1= a1m,B1 = b1 m

2

3

f2= a2m,B2 = b2 m 2

–0.2

f1= 2 =3 =4 =5 0

1

f or f2 = 1.15

–0.4 B1

1 2

B1, B2

4

0

≥6 =2 =3

–0.6

=4 =5

3

Fig. 3.26 Design curves for determining Kq1, Kq2 and Km

The rectangular tables of small- and medium-size (up to 1000 mm long) are made solid with T-slots on the top surface. Large-sized rectangular tables consist of two parallel walls connected by ribs. The ribs are generally perpendicular to each other and are spaced 300 – 400 mm apart. The stiffness of a machine tool table is determined primarily by its h/B ratio, where h is the height of the table and B its width. For instance, in planing and plano-milling machines the h/B ratio varies between 0.1–0.18, the larger value being applicable to tables of small width. A ratio of h/B = 0.14–0.16 can in general be considered optimum. For design purposes, rectangular tables are treated as rectangular plates of constant thickness. The reduced stiffness of a rectangular table may be determined from the expression, Sr = where

y= IX A B L E m

EI X , kgf ◊ cm L(1 + y )

(3.67)

12(1 + m ) I X

B2 ◊ A = moment of inertia of the longitudinal section of the table about the horizontal axis passing through the centre of gravity, cm4 = total area of the ribs, cm2 = width of the table, cm = length of the table, cm = modulus of elasticity of the table material, kgf/cm2, and = Poisson’s ratio of table material

Small circular tables (diameter D < 1000 mm) are made solid. Large circular tables (D > 1000 mm) consist of two parallel plates which are connected by radial ribs. These tables generally have a vertical axis of rotation and are provided with circular guideways. Exceptionally large tables having diameter D > 7–8 m are provided with two guideways.

Design of Machine Tool Structures 221

The stiffness of circular tables depends upon the ratio h/D, where h and D are the height and diameter of the table, respectively. The optimum number of radial ribs is 10–16, the larger number being applicable to larger diameters. Circular stiffeners are often provided just below the circular guideways. They reduce the maximum pressure on the guideway surface by 15–20%. For design purposes, a circular table is treated as a solid circular plate of constant thickness mounted on an elastic foundation. The reduced stiffness of a circular table may be found from the expression, Sr = S =

where

S , kgf ◊ cm 1+y

(3.68)

Eh 2

t1 ◊ t2 , kgf ◊ cm 1 - m t1 + t2 2

8p S (1.2 + m ) G ◊ A◊n◊ D = diameter of table, cm = thickness of the top and bottom plates of the table, cm = height of the circular table (distance between centre lines of the top and bottom plates), cm = area of cross section of one radial rib, cm2 = number of ribs = shearing modulus of the table material, kgf/cm2 = modulus of elasticity of the table material, kgf/cm2 = Poisson’s ratio of table material

y = D t1, t2 h A n G E m

3.11

DESIGN OF CROSS RAILS, ARMS, SADDLES AND CARRIAGES

Cross rails and arms have two sets of guideways for supporting the cutting tool and imparting to it movement in two mutually perpendicular directions. Cross rails of machine tools with a travelling tool post b b (vertical lathes, planing machines, plano-milling machines) are supported by columns over a relatively small length and can be treated as beams subjected to three-dimensional loading. They experience torsion and bending in two planes. Cross rails have a hollow rectangular section. The h recommended height-to-width ratio of cross-rail sections h (Fig 3.27) are: 1. For vertical milling machines h/b = 1.5–2.2; the higher value being valid for heavy machine tools in which the carriage weight is 3–4 times the vertical component of the cutting force. 2. For planing machines, h/b = 1–1.5 3. For plano-milling machines, h/b =1.0

Fig. 3.27 Typical cross-rail sections

222

Machine Tool Design and Numerical Control

Cross rails are, as a rule, provided with vertical ribs along the height and horizontal stiffeners spaced at appropriate distances. The deformation of cross rails is determined in the same way as that of a bar of a closed see Eq. (3.34)–(3.40)). The warping of the walls is small and can be neglected. The arms or traverses of single-column machine tools have a section that approaches that of a uniform strength beam. They can be analysed as cantilevers subjected to compound loading, experiencing torsion and bending in one (radial-drilling machine) or two planes (single-column planh h ing machine). The arms have either a hollow rectangular or hollow elliptical section (Fig. 3.28) with a ratio of h/b = 2–3. For sections of equal weight, the stiffness of the elliptical section is 15–20% higher than that of the rectangular. Arms are also provided with ribs and stiffeners for imparting sufb b be determined with the help of formulae derived earlier for bars of thin-walled closed sections (Sec. 3.6.2) by treating

Fig. 3.28 Typical traverse sections

Structures such as saddles and carriages are meant for providing motion to the workpiece or cutting tool largely upon the working conditions of the guideways. Generally, medium-size saddles are made solid, while saddles of heavy machine tools have a closed box section. The length-to-width ratio of saddles is l/b > 1.5 – 2, the smaller l/b ratio is applicable in the case of narrow guideways and the larger in the case of the wide guideways. Saddles and carriages are designed primarily for wear resistance and deformation of the guideways. This on an elastic foundation.

3.12

DESIGN OF RAMS

Rams are used for supporting the cutting tool and imparting motion to it in machine tools with a reciprocating primary cutting motion (shaping, slotting, gear-grinding machines, A etc.). The deformation of a ram depends upon its bending and Py torsional stiffness as well as the deformation of the guideways. Rams can be analysed as beams on an elastic founPx dation subjected to three-dimensional loading so that they L H Pz experience torsion and bending in two planes. The design diagram of a ram is shown in Fig. 3.29. The Fig. 3.29 Loading diagram of a ram d xy and d xz of the ram in two directions, angle of slope q in planes xy and xz, and angle of twist f at the tool tip and the normal pressure on the guideway surfaces.

Design of Machine Tool Structures 223

d xy and d xz may be determined by the following relationships: d xy = d 0 y + q 0 yL +

Py L3 3EI zz

d xz = d 0 z + q 0 z ◊ L +

+

Px ◊ a y ◊ L2 2 EI zz

+ f0 ◊ az +

Pz ◊ a y ◊ L ◊ az G ◊ It

(3.69)

Pz ◊ a 2y ◊ L Pz ◊ L3 + f0 ◊ ay + 3EI yy G ◊ It

and f from the expression f = f0 + where

d 0y, d 0z Px, Py, Pz f0 ay, az Iyy, Izz It E, G L

Pz ◊ a y ◊ L G ◊ It

(3.70)

= A) in xy- and xz-planes, respectively = components of cutting force = angle of twist of the ram at point A = distance of tool tip from the centreline of the ram in xy- and xz-planes, respectively = moment of inertia of the ram section about y-y and z-z axes, respectively = torsional moment of inertia of the ram section = modulus of elasticity and shearing modulus of the ram material, respectively = ram overhang

The values of d 0y, d 0 Z, q 0y, q 0z and f 0 can be determined from the following expressions:

d 0y = d0z = q0y = q 0z = f0 = where

2b y k B

Py ◊ u + (Py ◊ L + Px ◊ ay)b ◊ v]

2b z k pz ◊ u + Pz ◊ L ◊ b ◊ v] B 2b y2 k B

Py ◊ v + 2(Py ◊ L + Px ◊ ay)b ◊ w]

2b z2 k Pz ◊ v + 2Pz ◊ L ◊ b ◊ w] B Pz ◊ a y G ◊ I t ◊ bt ◊ tanh(bt ◊ H )

(3.71) (3.72) (3.73) (3.74) (3.75)

H = length of guideways B = Sbi cos2 a i = width of guideways (bi = width of the ith face of guideways and ai = angle between the ith face of guideways and the neutral axis) k = 1 ¥ 10–4 cm3/kgf k B ˆ Ê by = Á Ë 4 EI zz ◊ k ˜¯

1/ 4

cm –1

224

Machine Tool Design and Numerical Control

B ˆ bz = Ê ÁË 4 EI yy ◊ k ˜¯ bt =

1/ 4

cm –1

bi (ri2 + bi2 /12) , cm–1, G ◊ It ◊ k

ri being the distance between the centre of gravity of the section and the normal passing through the middle of ith guideway u, v, w = l = bH/2. For instance, for l = 1.3; for l = 1, they are = 1.1; for l > 1.2, they are = 1.0. For most of the rams, l = bH/2 > 1.6, i.e., b t H > 2.5; therefore, u = v = w = 1 and tanh (bt H) = 1 The maximum normal stresses on the guideway surfaces are: d0 cosai k f Ê bˆ Due to torsion, s ti = 0 Á ri + i ˜ Ë k 2¯ Due to bending, sbi =

(3.76) (3.77)

The pressure on guideway surfaces is quite high and may acquire values of up to 130 kgf/cm2. Therefore, this aspect must never be ignored during the design of ram-type structures.

3.13

MODEL TECHNIQUE IN DESIGN OF MACHINE TOOL STRUCTURES -

Sections 3.7 to 3.12 for the analytical design of various machine tool structures have been derived on the basis of simplifying assumptions and can easily give errors of up to 30–50%. These models are then tested and the behaviour of the actual machine is predicted from the knowledge of model behaviour. The relationships between the actual parameter and its value determined on the model are derived on the basis of the basic equations. Two derivations are discussed below. 1. Relationship for Bending Stiffness of Structure structure that may be approximated as a beam is EI yy = where

d2y dx 2

= Mx

Iyy = moment of inertia of the beam cross section about the neutral axis E = modulus of elasticity of the beam material Mx = bending moment in any arbitrary section of the beam

(3.78)

Design of Machine Tool Structures 225

For the model, all the quantities will be represented by a subscript m and for the structure by a subscript s. Accordingly, for the model d 2 ym dX m2

=

Mm Em I yym

(3.79)

=

Ms Es I yys

(3.80)

and for the structure d 2 ys dX s2 However, ys = ly m; xs = lxm Ms = lfMm; Iyys = l 4 I yym where

l = scale of linear dimensions f = scale of forces

Substituting the values in Eq. (3.80) yields, fMm 1 d 2 ym = 2 l dX m Es ◊ l 3 I yym d 2 ym

or

dX m2

=

f l

2

◊

Mm Es ◊ I yym

(3.81)

Upon equating the r.h.s. of Eqs (3.79) and (3.81), we get 1 f 1 = 2 ◊ Em l Es wherefrom, Stiffness factor =

E f =l s E l m

Stiffness of structure E =l s Stiffness of model E

(3.82)

2. Relationship for Natural Frequency of Torsional Vibrations The basic relationship of torsional vibrations of a bar is r ∂2q ∂2q (3.83) = G ∂t 2 ∂X 2 where

r = density of bar material G = shearing modulus of the bar material

Again, if subscripts m and s are employed for the model and structure respectively, we can write ∂2q m ∂ X m2

=

rm ∂2q m Gm ∂t m2

(3.84)

226

Machine Tool Design and Numerical Control

and for the structure ∂2q s ∂ X s2

=

rs ∂2q s Gs ∂t s2

(3.85)

Let Xs = lX m; ts = tt m where

l = scale of linear dimensions t = scale of time

Upon substituting the above values in Eq. (3.85), we get 1 ∂2q m l 2 ∂X m2 ∂2q m

or

∂X m2

=

=

rs 1 ∂2q m Gs t 2 ∂t m2 l2 t2

◊

rs ∂2q m Gs ∂t m2

(3.86)

Upon equating the r.h.s of Eqs (3.84) and (3.86), we get rm l 2 rs = 2 Gm t Gs wherefrom, Frequency factor =

1 1 = t l

Torsional natural frequency of structure 1 = l Torsional natural frequency of model

Gs rm Gm rs

Gs rm Gm rs

(3.87)

The expression for natural frequency of the bending vibrations of the structure can be similarly derived as follows: Bending natural frequency of structure 1 = Bending natural frequuency of model l

Es r m Em rs

(3.88)

The results of model investigations can be applied to an actual structure provided there is similarity in 1. geometrical proportions, 2. the loading pattern, and 3. the relative magnitude of material properties. The similarity of geometrical proportions can be achieved simply by using a constant scale factor for all linear dimensions. The similarity of loading can also be achieved by employing an appropriate force factor f. of mechanical similitude upon which the whole approach of the model technique is based, the various scale factors must satisfy the following relationship: l rt –2 f–1 = 1

(3.89)

Design of Machine Tool Structures 227

where

l r t f

= scale of linear dimensions = scale of masses = time scale = force scale

The relative magnitude of the material properties of the structure and model can be assessed through the Poisson’s ratio. This means that the Poisson’s ratio of the structure and model materials should be the same. Poisson’s ratio of a material is expressed as 1E –1 (3.90) m = 2G where E and G are the modulus of elasticity and shearing modulus of the material, respectively. E/G ratio is same. The material which is commonly used for making models is polymethyl methacrylate or Perspex as it is popularly known. Perspex has a Poisson’s ratio of 0.35 as compared to 0.25 for cast iron and 0.29 for steel. Consequently, the E/G ratio for Perspex is 8% greater than that of cast iron and 5% than that of steel. of a steel structure and 8% for that of a cast iron structure. Generally, a pure shear-type of loading is rarely encountered in practise. For a general compound loading to which the majority of machine tool structures are subjected, the error may be assumed to be less than 5% for cast iron structures and 3% for steel structures. One of the major limitations to the use of Perspex models is their undesirable creep strain characteristic. For load stresses exceeding 68 kgf/cm2, the value of E reduces with time. This phenomenon raises the

2. The corresponding equivalent value of E should only be used in determining the scale factors. simultaneously. 4. The structure must be allowed to relax prior to further application of loading. The problem caused by creep of the model material (Perspex) can be alleviated by adopting a special loadmaterial as the model (Fig. 3.30)10 and hence possessing similar creep characteristics. The applied force is shared by the compensating ring and the model in inverse proportion to their stiffnesses. The model being much more stiffer than the compensating ring, its deformation does not change with time. The stiffness of the structure can be determined from the following relationship: Ks = where

d mr k sr dm

Ks = stiffness of structure Ksr = stiffness of imaginary ring made of the same material as the structure dmr = dm

(3.91)

228

Machine Tool Design and Numerical Control

Imaginary steel ring Perspex compensating ring

Screw jack

Screw jack

Perspex model

Steel machine

Fig. 3.30 Schematic diagram depicting the use of a compensating ring 2 The modulus of elasticity of Perspex depends E, kgf/cm 6 upon the frequency of the dynamic force. The vari¥ 0.0703069 ¥ 10 ation of the modulus of elasticity of Perspex with 0.75 loading frequency is shown in Fig. 3.31. While calculating the natural frequency of a structure in accordance with Eq. (3.88), the value of Em should 0.65 Probable be found from Fig. 3.3111 for the particular mode curve of vibration. For conducting dynamic tests, the Perspex mod- 0.55 el should preferably be excited by a freely suspended electromagnetic exciter because clamping of the exciter to the model can result in serious deviation 0.45 0.1 0.2 0.5 1 2 5 10 20 50 100 200 500 1000 of the model behaviour. f, Hz For making a Perspex model, the sheets are cut to the required size as per the model scale and are Fig. 3.31 Variation of the modulus of elasticity of then joined either by welding or cementing. WeldPerspex with frequency ing is done by a hot-air torch using an unplasticised

screws. Cementing is done with a paste obtained by dissolving Perspex chips in chloroform. The cemented model must be cured by heating it in an oven to compensate for the reduction of the modulus of elasticity at the joints where the cementing paste attacks the Perspex. The holding clamps and screws are then removed. Perspex models should be made in a room with controlled humidity because Perspex can absorb up to 2% moisture which lowers its modulus of elasticity by about 10%.

Review Questions 3.1 Derive the expression of optimum l/h ratio for a solid rectangular structure of height h and width b which is loaded as shown in Fig. 3.32.

Design of Machine Tool Structures 229

2d h/2 2d

P

q1

q2

d h

l 4

Fig. 3.32

l 2

l 4 l

b

Schematic diagram of simply supported beam

Fig. 3.33

3.2 A vertical milling machine is to be designed for HSS as well as cemented carbide face milling cutters. Carbide cutters have diameters between 50 to 250 mm and number of teeth between 8 to 14. The corresponding ranges for HSS cutters are 80 to 250 mm and 10–26, respectively. The minimum cutting speed is 30 m/min (HSS-mild steel pair) and the maximum is 500 m/min (carbide-aluminium pair). Determine the frequency range that must be avoided while designing the elements of the machine. 3.3 A bed subjected to torsional loading is constructed as a closed box-type structure, while a bed subjected

30

0

100

100 kgf 30 100 kgf

75 75

2000

mathematical proof to support your conclusion. 3.4 A channel section is welded to a box section as shown in Fig. 3.33. Calculate by how many times the torsional stiffness of the structure will improve and the shearing

Sketch for Prob. 3.4

100

200 kgf

3.5 A 2 m long, 1m high and 0.5 m wide lathe bed consists of two vertical walls strengthened by perpendicular or 300 diagonal stiffeners. The thickness of the walls is 50 mm, while that of the stiffeners is 25 mm. Calculate Fig. 3.34 Sketch for Prob. 3.6 the reduced bending rigidity of the beds having perpendicular and diagonal stiffeners. 3.6 A rectangular, box-type cast iron column of height 2 m, section 30 cm ¥ 30 cm and wall thickness ¥

230

Machine Tool Design and Numerical Control

3.7 The box structure of mild steel shown in Fig. 3.35 is acted on by a force P = 200 kgf acting at the middle of the vertical wall of dimensions 400 mm ¥ 300 mm. The hole of the diameter 20 mm is strengthened by a rib and an internal bossing of diameter 30 mm and height three times the wall hole is 25 microns.

6

300

f 20

P

2.0 kgf mm

1.5

2 ¥10 kgf mm

kgf mm

40 0

100 250

400 500

300

3.8 A hollow rectangular steel base of height 200 mm, width 1000 mm and wall thickness 25 mm is loaded as shown in Fig. 3.36. Calculate the maximum angle of slope. 3.9 A hollow rectangular table is 2 m long, 60 cm wide and 9 cm high. The 15 mm thick horizontal walls of the table are joined by 10 mm thick vertical stiffeners spaced at 400 mm. Calculate the reduced stiffness of the table. 3.10 A steel structure is studied by making its Perspex model on 1 : 10 scale. The natural frequency of the model in bending was found to be 150 Hz. Determine the natural frequency of the steel structure, if density of Perspex = 1.2 g/m3. 3.11 Determine the limiting torque for the structure show in Fig. 3.37. Given permissible strength of the material in shear = 1000 kgf/cm2. 3.12 a twisting moment of 1000 Kgf-m, determine the maximum shear stress and the angle of twist over a length of 100 cm. Given G = 8 ¥ 105 kgf/cm2.

Fig. 3.36 Sketch for Prob. 3.8

10

100

Fig. 3.35 Sketch for Prob. 3.7

1500

60

Fig. 3.37

20

Design of Machine Tool Structures 231

20

140

10

80

Fig. 3.38

References 1. Koenigsberger, F, Design Principles of Metal Cutting Machine Tools, Pergamon Press, Oxford, 1964, p. 45. 2. Bielfeld, J, Modellversuche on Werkzeng Machine Elementen, 3. Acherkan, N, Handbook of Machine Design, 280. 4. Bhattacharya, A, et al., “Static rigidity of lathe beds by model analysis”, J. Inst. of Engrs (India), Vol. 47, No. 11, Pt ME-1, 1967. 5. Kaminskaya, VV and Frantsuzov, FA, “Effect of mounting of single column jig boring machines on their accuracy”, Stanki I Instrument, 1960, No. 5. 6. Acherkan, N, Handbook of Machine Design, 269. 8. Kaminskaya, VV, et al., Beds and Housings of Machine Tools, Mashgiz Publishers, Moscow, 1960.

232

Machine Tool Design and Numerical Control

9. Push, VE, Design of Machine Tools, Mashinostroenie Publishers, Moscow, 1977, p. 118. structures”; 6. MTDR. Conf., Birmingham, 1965. Machine Tool Research, 1967.

Design of Guideways and Power Screws 233

4 4.1

DESIGN OF GUIDEWAYS AND POWER SCREWS

FUNCTIONS AND TYPES OF GUIDEWAYS

The basic function of guideways is to ensure that the machine tool operative element carrying the workpiece or cutting tool moves along a predetermined path, which is generally a straight line, as in lathe, drilling, boring machines, etc., or circular as in vertical turret lathes and boring mills. The major requirements that the guideways must satisfy are: 1. H 2. High accuracy of travel, which is possible only when the deviation of the actual path of travel of the operative element from the predetermined path is minimum. 3. Durability, which depends upon the ability of guideways to retain the initial accuracy of manufacturing and travel. 4. Low value of frictional forces acting on the guideway surface to ensure less wear. 5. M 6. High rigidity. 7. Good damping properties. Based upon the nature of friction between the contacting surfaces of the guideways and operative element, 1. Guideways with sliding friction—known as slideways. 2. Guideways with rolling friction—known as anti-friction ways.

4.1.1

Types of Slideways

Depending upon lubrication conditions at the interface of contacting surfaces, the friction between the sliding surfaces may be described as dry, semi-liquid and liquid. Dry friction occurs when there is no lubricant between the sliding surfaces. This is rarely if ever, encountered in machine tools. Whenever a body slides with respect to another body and there Q is a lubricant between the two, the sliding body tends to rise or The hydrodynamic force Q may, in general, be represented by the expression, Q = kh ◊ v where

Kh = a constant that depends upon the geometry of the sliding surfaces, wedge angle a, parameters of the

V

a G

Fig. 4.1 Schematic diagram of a slider

234

Machine Tool Design and Numerical Control

v = sliding velocity If the weight of the sliding body is G, the resultant normal force acting on it is N =Q–G As long as Q < G, the sliding body rests on the stationary body and the friction conditions are of semi-liquid perience metal-to-metal contact. When Q > G, the resultant normal force on the sliding body begins to act their interface experiences liquid friction. The slideways in which a permanent lubricant layer separating the sliding surfaces is obtained due to the hydrodynamic action are known as hydrodynamic slideways. At low sliding speeds, a permanent lubricant layer between the sliding surfaces can be obtained by pumping the liquid into the interface under pressure, high enough to lift the sliding body. Such slideways are known as hydrostatic slideways. Henceforth, the term slideways unaccompanied by a qualifying description would imply slideways operating under conditions of semi-liquid friction.

4.1.2 Types of Anti-friction Ways 1. roller-type anti-friction ways using cylindrical rollers, and 2. ball-type anti-friction ways using spherical balls. Both these types may or may not have provision for recirculation of the rolling elements.

4.2

DESIGN OF SLIDEWAYS

Slideways are distinguished by a relatively high coresults in considerable wear, reducing life as well as er functioning of slideways it is, therefore, imperative that the friction be kept as low as possible by ensuring that a certain minimum amount of lubricant is always present between the sliding surfaces. This may be done

Cross groove Longitudinal

sliding surfaces from time to time. One of the popular groove methods of increasing the share of liquid friction in Fig. 4.2 Location of grooves in semi-liquid friction slideways and thus reducing friction at the interface is slideways to provide lubricating grooves on the guiding surface.

the general ones explained in Sec. 4.1.

Design of Guideways and Power Screws 235

2. There must be provision for compensation of possible wear. 3. The material of the slideways should have high wear resistance.

4.2.1 Shapes of Slideways

to provide good retention of the lubricant in the interface. Closed slideways are employed when the sliding velocity is not high but it is necessary to prevent chip accumulation and ensure its easy removal.

of the sliding part, less effect of wear on machining accuracy and greater accuracy of travel. It is however,

open and closed versions, is generally 90°. Sometimes, in heavily loaded slideways, the apex angle of the on the slideway surfaces. However, this should not be made a general practise as it reduces the accuracy of precision machine tools in which accuracy is of critical importance and the loads are relatively low.

Closed slideways

(b)

(e)

Fig. 4.3 Slideway profiles: (a) Flat (b) Symmetrical V (c) Asymmetrical V (d) Dovetail (e) Cylindrical

2. where there are restrictions on height due to space limitations, e.g., in carriages, saddles, knees, etc.

236

Machine Tool Design and Numerical Control

to the following drawbacks: 1. Low rigidity because they are secured to the bed only at the ends. 3. If the operative element is guided by two cylindrical slideways, its movement is constrained unless great care is taken at the time of assembly to ensure perfect parallelness between the two. Cylindrical slideways are, however, widely used in machine tools in situations requiring small and not too frequent movement, which is possible with only one cylindrical slideway, e.g., columns of radial drilling machines, overarms of universal milling machines, lathe tailstock, sleeve, etc. In machine tools a set of two slideways is generally employed. The application of various slideway pro-

4.2.2 Materials of Slideways The most common slideway materials are cast iron and low carbon or low alloyed steels. If the slideway is integral with the bed, it is almost exclusively made from grey cast iron. This material is cheap but does not have good wear resistance when subjected to heavy loading. The wear resistance of gray iron slideways can faces improves wear resistance by at least two times. If both sliding surfaces are hardened, the improvement may be up to four times. If only one sliding surface has to be hardened, it should preferably be the guiding surface of the stationary member. Table 4.1 Sketch

Application

For Beds Planing machines

Precision lathes and turret lathes

Surface-grinding machines

General-purpose lathes and heavyduty boring machines

Contd.

Design of Guideways and Power Screws 237

Table 4.1

Contd.

Heavy-duty lathes, broaching and plano-milling machines

For Vertical Columns

Most commonly used for all types of vertical columns

closed dovetail

Knee-type milling machines, small vertical drilling machines and traverses of radial drilling machines

Same as above

For Cross Slides and Compound Rests

Closed dovetail

Cross slides and compound rests

Cross slides of heavy-duty machine tools

Contd.

238

Machine Tool Design and Numerical Control

Table 4.1

Contd.

Surface-grinding machines and small hobbing machines

Precision gear-hobbing machines

The wear resistance of spheroidal cast iron slideways is much better than that of slideways made from ordinary grey iron. However, spheroidal cast iron is expensive and economically not feasible for casting the complete bed. Therefore, even in an integral construction often only the slideway is cast from spheroidal cast iron and connected to the grey iron bed through intermediate sections. The shape of the intermediate consee slideway stiffness. The types of connecting sections and their applications are given in Table 4.2. stance, chrome plating of the slideway with a 25–50 micron layer provides a hardness of RC = 70 and considerably increases the life of slideways. The deposition of a molybdenum coating can improve the wear Steel slideways are manufactured separately and then attached to the bed. If the bed is a welded steel

the slideway surface. To ensure uniform deformation of the slideway surface, the fastening screws should be so spaced that they satisfy the condition, t £ 2h where

t = distance between adjacent screws h = height of slideway

Steel slideways have greater hardness and strength, and consequently, greater wear resistance and loadcarrying capacity. They are, therefore, used heat treatment, and 2. when the welded bed is made of low carbon steel which cannot be hardened.

Design of Guideways and Power Screws 239

Table 4.2

Type of connecting sections and their applications

Type of connection

Sketch

Application

Through one wall:

Centre and turret lathes and small-size planing machines

This connection is least rigid Thick wall

Thick wall with stiffening rib

Through two walls:

This connection is more rigid than the above

Used in beds of production and turret lathes which have double wall closed disposal through opening in the rear wall Used in beds with a covered top closed slotting, and planing machines

Direct:

This connection has maximum stiffness

Rarely used in horizontal beds, except when the table moves along the bed through a rack-and-pinion mechanism located on one side

Steel slideways are generally made from the following two groups: 1. L 60 Si27 Ni25 a hardness RC = 60–65, or nitrided to a depth of 0.5 mm and preferred when the sliding velocity is high. 60 Si27 Ni25

240

Machine Tool Design and Numerical Control

(b)

Fig. 4.4 Types of slideways: (a) Welded (b) Mechanically coupled

ever, very expensive and used only in some heavy-duty machine tools under exceptional circumstances. vertical turret lathes and large planing machines due to 1. ease of manufacture, 2. less friction, and 3. good anti-scoring and anti-corrosive properties Their use is, however, restricted by the following factors: 1. P times greater than that of a cast iron-cast iron pair. 2. Poor structural stiffness due to lower modulus of elasticity. 3. Poor wear resistance when contaminated by abrasive dirt. 4. L 5. Tendency to swell by absorbing the lubricant.

4.2.3 Methods of Adjusting Clearances in Slideways surfaces. The clearance can be considered optimum if it is neither too small to give rise to excessive frictional clearance between the mating surfaces even with the best of manufacturing technology. Moreover, the initial clearance alters in the course of slideway operation due to its wear. It is, therefore, necessary to provide devices for periodic adjustment of clearances in the slideways. Devices that are commonly used for adjustment of clearances are: 1. lat strips that are not clamped after clearance adjustment. 2. lat strips that are clamped after clearance adjustment. 3. Taper gibs. devices. However, such

-

Therefore, clearances in both directions can be adjusted by dent of each other. Hence, separate devices are required for adjusting clearances in horizontal and vertical

Design of Guideways and Power Screws 241

Fig. 4.5 Clearance adjustment by taper gibs in (a) dovetail slideways (b) flat slideways

of the pressing force depends upon the degree of tightening of each screw and requires considerable skill and experience on the part of the operator. The device has low rigidity and this coupled with lack of uniformity of pressing force prevents its application in heavily loaded and high-accuracy machine tools.

Fig. 4.6 Clearance adjustment by flat strips in (a) flat slideways (b) dovetail slideways

Fig. 4.7 Devices for clamping flat strips in (a) flat slideways (b) dovetail slideways

242

Machine Tool Design and Numerical Control

The taper gib is slightly more expensive, but combines the advantages of both good stiffness and uniform pressure distribution. Taper gibs have a taper in the range of 1 : 40 to 1 : 100. The taper angle on longer gibs is small and vice versa. The tightening of taper gibs requires great caution as over tightening can lead to considerable lateral loading resulting in large pressures and fast wear of the sideways. The methods of tightening

weakening of the guided part at the thick end of the taper. In such cases, two taper gibs are provided as shown

(a)

(b)

(c)

Fig. 4.8 Methods of tightening taper gibs

Fig. 4.9 Clearance adjustment in long slideways by means of two taper gibs

During the longitudinal travel, the tightening and adjusting screws may get loose under excessive loading.

side of the slideway, the stiffness is 1.5 times less than if the gib is placed on the unloaded side.

Design of Guideways and Power Screws 243

4.3

DESIGN CRITERIA AND CALCULATIONS FOR SLIDEWAYS

Slideways are designed for the following two parameters: 1. Wear resistance 2. Stiffness The wear resistance of slideways is governed mainly by the maximum pressure acting on the mating surfaces. This condition may be written down as pmax £ [p max Here,

pmax = maximum pressure acting on the mating surfaces [pmax] = permissible value of the maximum pressure

It will be seen during the subsequent analysis that slideway design in terms of maximum pressure is quite complicated. Sometimes, this design is replaced by a simpler procedure based upon the average pressure acting on the mating surfaces. The condition is that pav £ [pav where

pav = average pressure acting on the mating surfaces [Pav] = permissible value of the average pressure -

condition may be expressed as follows: d i £ [d i where

di [d i

f th direction ith direction

4.3.1 Design of Slideways for Wear Resistance 1. pav and pmax be known, and 2. [pav] and [pmax] be known. The values of [pav] and [pmax pav and pmax surfaces.

Forces Acting on the Mating Surfaces in a Combination of V and Flat Slideways The forces are: 1. Weight of carriage G.

The

244

Machine Tool Design and Numerical Control

2. Cutting force components Pz Py 3. Unknown forces A, B and C acting on the mating surfaces.

Z

d

The unknown forces are determined from the following equilibrium conditions:

h

B b G

Py

Sum of projections of forces on the Z-axis = 0, SZ = 0; A cos a

Pz

C

Sum of projections of forces on the Y-axis = 0, SY = 0; A sin a – b sin b

B cos b

Py

A

Y

b

Fig. 4.10

C – Pz – G

Moment of all the force about the X-axis = 0, SMx = 0; A cos a ◊

a

B cos b ◊

2

2

– Pz ◊

Schematic loading diagram for a combination of V and flat slideways under orthogonal cutting conditions

d – Py ◊ h – C ◊ 2 2

B cos b – Pz

d 2h – Py b b

B cos b – Pz

d 2h – Py ◊ – Pz – G = 0 b b

C = A cos a Substituting the value of C A cos a

B cos b

A cos a

or A cos a

B cos b

dˆ 2h G Ê Pz Á1 + ˜ + Py + Ë b¯ b 2

A cos a

B cos b =

Pz (d + b) h G + Py + 2b b 2

wherefrom,

b = 90 – a, the solution of simultaneous algebraic

A=

Pz (d + b) cos a 2b

Py

h cos a – Py sin a b

G cos a 2

B=

Pz (d + b) sin a 2b

Py

h sin a b

G sin a 2

Py cos a

Substituting the values of A and B

C: C =

Pz (b - d ) h G - Py + 2b b 2

Design of Guideways and Power Screws 245

Forces Acting on the Mating Surfaces in a Combination of Two Flat Slideways

The

of oblique cutting. The forces are: 1. 2. 3. 4.

Weight of carriage G. Cutting force components—axial Px, radial Py and in the direction of velocity vector Pz. Pulling force Q Normal forces A, B and C fA, fB and fC, where f Py

C b¢

Yp

c¢

B

Px xp

Pz

h

Z

Z

A

A

Pz

a¢

xA

fA

Y G

X ZQ Q

b

Fig. 4.11

YQ

Schematic loading diagram for a combination of two flat slideways under oblique cutting conditions

The unknown forces A, B, C and Q can be determined from the following equilibrium conditions: SX = 0; Px

f A

B

C

Q

SY = 0; B – Py SZ = 0; A

C – Pz = G

SMx = 0; Pz ◊ yp

G

2

– Py ◊ h

–

B = Py

C=

Pz y p - Py h b

+

Substituting the value of C A = Pz

G Pz y p - Py h 2 b

G 2

246

Machine Tool Design and Numerical Control

Pulling force Q

A, B, and C: Q = Px

Determination of Average Pressure

f Pz

Py

G

Having determined the forces acting on the mating surfaces, the

average pressure can be determined as A a¢ ◊ L B pB = b¢ ◊ L C pC = c¢ ◊ L pA =

where

L = length of the carriage a¢, ¢, c¢ = respective widths of the slideway faces on which forces A, B and C are acting.

Determination of Maximum Pressure establish the points of action of the resultant normal forces A, B and C xA denotes the distance between the point of action of normal force A

I and the centre of the

1. xB should be understood to denote the distance between force B slideway II and the centre of the carriage, and 2. xC the distance between force C carriage. xA, xB and xc, we have only two equilibrium conditions: and

SMy = 0; Pz xp

Pxh – AxA – CxC

QzQ

SMz = 0; Px yp

Py xp – QyQ – BxB

fB

c¢

An additional equation may be written by assuming that the moment of reactive forces A and C about the Y-axis is proportional to the width of the slideway face, i.e., Ax A a¢ = CxC c¢ The values of xA, xB and xc can now be determined by

pmax

pmin

The ratio xA/L, xB L or xC/L determines the shape of the pressure-distribution diagram and the maximum pressure on a particular face of the slideway. Let us describe the

A

slideway I which is being subjected to the normal force A. yC xA L/2 The most general case of pressure distribution along the L length of contact L corresponds to a trapezoid as shown in A acts at the centre of gravity of the trap- Fig. 4.12 Trapezoidal pressure distribution ezoid. Distance yc of the centre of gravity from the larger along slideway length; XA /L < 1/6 arm of the trapezoid can be determined as

Design of Guideways and Power Screws 247

L Ê pmax - pmin ˆ L +Á ˜¯ L ◊ 2 Ë 2 3 = L pmax + 2 pmin ◊ 3 pmax + pmin Ê pmax - pmin ˆ pmin ◊ L + Á ˜¯ L Ë 2

pmin ◊ L ◊ yc =

Consequently, xA =

L L pmax - pmin – yc = 2 6 pmax + pmin

Case I xA /L < this conclusion can be illustrated by a simple example. Suppose xA /L = pmax – pmin =

6 pmax 8

pmin

wherefrom, pmin =

0.25 pmax 1.75

It may be seen that pmin is a positive, non-zero quantity. Hence, the pressure-distribution diagram must pav ◊ L = pmin ◊ L

1 pmax – pmin L 2

wherefrom, if follows that pmax

pmin = 2pav

pmax – pmin =

12x A pav L

6x ˆ Ê pmax = pav Á1 + A ˜ Ë L ¯ xA/L

A

xA/L pmax – pmin

pmax

pmin

pmax

Case II

wherefrom, pmin

P

p pmin = – max 3

x1 L/2

xA

pmin is negative. This implies that the complete length of Fig. 4.13 Pressure distribution the guided part is not in contact with the slideway. The pressure-distrialong slideway length P, the mating when xA/L > 1/6 surfaces are relieved of load.

248

Machine Tool Design and Numerical Control

In this case, pav . L = but

1 pmax ◊ x1 2

x1 ÊL ˆ = Á - x A ˜ and therefore, Ë2 ¯ 3 pav ◊ L =

1 L pmax ◊ 3 ÊÁ - x A ˆ˜ Ë2 ¯ 2

wherefrom, 4 ◊ pav 3

pmax =

1 1- 2

xA L A

Case III xA /L pmin

pmax = pmin

In this case,

L/2

pmax = pav

pmax – pmin = pmax Therefore,

Fig. 4.14

XA/L pmin

A

pmin = 0

and the pressure-distribution diagram is a triangle ( In this case,

L/2

1 pav ◊ L = pmax ◊ L 2 Therefore,

Pressure distribution along slideway length when xA /L = 0

pmin = 0

Case IV

pmax

Hence,

pmax

pmax – pmin = 0

Fig. 4.15

xA

Pressure distribution along slideway length when xA /L = 1/6

pmax = 2pav

Permissible Values of Maximum and Average Pressure The permissible values of maximum pressure for cast iron slideways under various operating conditions are as follows: 1. When the sliding movement represents feed motion, e.g., in lathes, milling machines, etc. 2

[pmax

2. When the sliding movement represents primary cutting motion, e.g., in planing, shaping, slotting machines, etc. 2

[pmax

[pmax [pmax

2

at low sliding velocities 2

at high sliding velocities

Design of Guideways and Power Screws 249

pmax] value is reduced by 25%. 2

[pmax

The permissible values of maximum pressure for a combination of steel and cast iron are the same. When both the guided and guiding surfaces are made of steel, the permissible values of maximum pressure are taken to be 20–30% greater than those given above. The permissible average pressure is taken as half the permissible maximum pressure for the appropriate machine and its operating conditions. Having determined pmax and pav, and knowing the values of [pmax] and [pav], the slideways can be designed for wear resistance

The design procedure has been discussed above with special reference to the horizontal sliding surface on which force A is acting see applied to the remaining sliding surfaces too.

n

m a

b

c¢ of one of d Fig. 4.16

d = m cos2 a

n cos2 b

Schematic diagram of an asymmetrical V profile

4.3.2 Design of Slideways for Stiffness As explained earlier in Sec. 4.3, the design of slideways for to contact deformation of the slideway should not exceed a ations of required machining accuracy. In a lathe operation, on the dimensional accuracy of the machined workpiece. d in the radial direction results in a diametral error of 2d. The lathe slideways are, therefore, designed for stiffness primarily with a Fig. 4.17 rection for stiffness design in various machine tools can be selected in each particular case from the same underlying principle.

dFF

h

dB

Z

dA y

dC b

Schematic diagram for calculating the radial displacement of the cutting edge in a combination of two flat slideways

250

Machine Tool Design and Numerical Control

1. contact deformation dB of the vertical face of the slideway, and 2. rotation of the saddle due to unequal contact deformations dA and dc of d A - dC ◊h b

dFF = dB

d A - dC ◊h b dFV

manifested in its lowering by dC. suffer deformations dA and dB which results in dC

dv = dB sin a

h

dB

Z

dA cos a

Y

by

dv

dh dA

b

dh = dB cos a – dA sin a

Fig. 4.18

dh, and 2. rotation of the saddle due to unequal vertical lowering

Schematic diagram for calculating the radial displacement of the cutting edge in a combination of flat and V slideways

dv - dC ◊h b

dFv =

dv - dC ◊h b

k is known as contact compliance for each pair of slideway materials experimentally. However, for approximate calculations, an average value of k = 1 micron ◊ cm2 d FF = kpB

k

p A - pC ◊h b

d Fv = k pB cos a – pA sin a

kh pB sin a b

pA cos a – pC

Design of Guideways and Power Screws 251

4.4

GUIDEWAYS OPERATING UNDER LIQUID FRICTION CONDITIONS

of friction m. However, often, the major obstacle to the application of slideways in precision machine tools arises not from the high value of m, but its variation. This variation occurs 1. with change of sliding speed, and 2. with passage of time. m,

The variation of m as a function of sliding speed v is depicted 1 It is evident that in the range v < v0 an increase of v is accompanied by a decrease of m, while in the range v > v0, an increase of v is accompanied by an increase of m. The value of frictional resistance is thus not constant but depends upon the sliding velocity. Of particular interest is the fact that the initiation of a sliding movement is as a rule accompanied by a jerk due to the slip-stick effect. This phenomenon can be explained in the

0.15 0.10 0.05 0 0 Vs

Fig. 4.19

20

40 v, cm/min

60

80

Variation of coefficient of

speed v0. At the start of a movement, the guided member must friction with speed be strained to overcome frictional resistance corresponding to m static. However, as soon as the guided member commences to slide, its frictional resistance drops due to a decrease in m. The excessive force applied on the member when it was stationary is released giving rise to a

It was stated in Sec. 4.2 that slideways must be oiled from time to time to ensure their proper functioning.

sliding surfaces at a particular instant of time depends upon the time period elapsed since the last oiling. The variation of m as a function

m,

2

that the frictional resistance of the guided member does not remain constant. It may be thus seen that due to variation of m in slideways work-

0.3 0.2 0.1

altogether impossible to employ them on machine tools which re0 0 20 40 60 80 100 t, s static slideways must be employed. Fig. 4.20 Variation of coefficient of friction with time

252

Machine Tool Design and Numerical Control

4.4.1 Design of Hydrodynamic Slideways In hydrodynamic slideways, liquid friction conditions between sliding surfaces are achieved due to hydrodynamic action of the

p

capable of lifting the guided member is possible only at high sliding speeds. Hydrodynamic slideways are, therefore, used mainly where the sliding motion represents the primary cutting motion, e.g., vertical boring and turning mills and planing machines. Hydrodynamic action is possible between sliding bodies only if they are inclined to each other, i.e., they form a wedge. distribution follows a parabolic law and the hydrodynamic force over a unit width is given by the expression, Q=

6 mvL2 2 hmin

hmin

xo

Fig. 4.21

◊K

Q =

5mv 2 hmin

◊

L

Distribution of hydrodynamic pressure in a wedge of infinite width

L2 b0 ◊K 1 + (L /b0 ) kgf ◊ s

m v L = length of slideway, m 0 = width of slideway, m

m2

hmin

hmin = 0.01–0.02 mm, for heavy-duty machine tools hmin = 0.06–0.1 mm,

K its length,

0.03

2 ˆ Ê m +1 K = m 2 Á ln m 2m + 1˜¯ Ë

0.02

wherein m = x0/L The variation of K as a function of m wherefrom it can be seen that the maximum value of Kmax = 0.0267 occurs at m = 0.7. If the slideway is provided with a single long wedge along its whole length, it will generally be found that 0 1, p2

Dp2

p0

Dp0

Q=

A0

p2 – p1

the bearing clearance remains constant. In this case, 3

P

p0 h p (g - 1) = 0 m ◊ CF RC

p0

Fig. 4.32

A2

pp

p2

Schematic diagram of pad bearing using constant-gap restrictor

262

Machine Tool Design and Numerical Control

wherefrom, Ê (g - 1) ◊ m ◊ CF ˆ h= Á ˜¯ Rr Ë

1/3

pressure reaches a value close to pump pressure pp. As p0 approaches pp Q falls rapidly. Similarly, in case of the constant gap restrictor, the bearing gap is maintained constant only till pocket pressure p0 reaches a value such that g p0 approaches pump pressure pp. restrictor. Q–p0 and h–p0 relationships for various restrictors are shown -

the supply pressure from the pump or the hydraulic resistance of the restrictor, both of which change the pp ¥ Q supply to the pad bearings. The optimum viscosity of oil employed in hydrostatic slideways is determined from the condition of These losses are assessed in terms of heat generation. The optimum absolute viscosity of oil can be determined from the following relationship: h

Q

1 2 3

1 2

4 3 4

p0=pp

p0=pp

p0

p0 (a)

Fig. 4.33

(b)

Effect of pocket pressure on (a) flow (b) bearing gap: 1. Capillary restrictor 2. Orifice restrictor 3. Constant-flow restrictor 4. Constant-gap restrictor

mopt = where

A¢ =

h2 v 2

.

p0 ◊ p p ◊ CL CF ◊ A¢

kgf ◊

2

Design of Guideways and Power Screws 263

The size and shape of pockets can be assigned as per the following recommendations: A–A A B

a1

A

I a

L

(a)

a2

a2 a1

a1 b

B

a2

b

B

a2 I L

I L (b)

Fig. 4.34 Pocket geometry for (a) narrow slideways (b) wide slideways

B < 50 mm: a = 0.5a1 a1 = 0.1B B > 50 mm: a2 = 2a1 = B – 2a2

Closed Hydrostatic Slideways

h1 P h2

Hydrostatic lubrication is

employed only when 1. the tilting moment is not large, 2. the weight of the table and workpiece represents the main load, 3. the ratio of maximum load to minimum does not exceed 2, and

If any of the above conditions is violated, closed-type slideways, Fig. 4.35 Schematic diagram of must be used. In these slideways, it is possible to increase the stiffness closed hydrodynamic by raising pressure without the apprehension of overturning the guided slideways member.

264

Machine Tool Design and Numerical Control

The load capacity of a closed-type slideway with a capillary restrictor is determined from the expression, P = pp ACLCp e, k and its stiffness by K =–

3 p p ACL h0

Ck e, k

Ck(e,k)

Cp(e,k)

k=0 0.2 0.4 0.6 0.8 1.0

0.9 1.35

0.8

1.20

0.7

1.05 Cp(e,k)

0.6

0.90 Ck(e,k)

0.5

–0.6

–0.4

–0.2

0.75

0.4

0.60

0.3

0.45

0.2

0.30

0.1

0.15

0

0.2

0.4

0.6

0.8

k=1.0 0.8 0.6 0.4 0.2 k=0.0 1.0

Fig. 4.36 Design curves for computing Cp(e, K) and Ck (e, K) as a function of e

Design of Guideways and Power Screws 265

Cp e, k 1. e =

Ck e, k

h0 - h1 h1

where h0 is the initial bearing clearance, and 2. k =

A2 CL 2 CF 2 Crc 2 ◊ ◊ A1CL1 CF 1 Crc1 3

Cp e, k and Ck e, k

4.5

DESIGN OF AEROSTATIC SLIDEWAYS

The operating principle of aerostatic slideways is similar to that of hydrostatic slideways, the basic difference being that the lubricating medium in this case is compressed air instead of oil. The important positive features of aerostatic slideways are: 1. It is not necessary to collect oil. 2. The moving member can be reliably clamped in the desired position after cutting off the supply of compressed air. 3. Due to air pressure the mating surfaces are effectively protected from dust, abrasive particles and chips. negligibly small friction at all sliding velocities. to hydrostatic slideways due to compressibility of air. They also have poorer damping properties because the viscosity of air is much lower than that of oil. Owing to these limitations, aerostatic slideways are not suitable for feed and priA–A

in setting motion trains of precision machine tools, such as jigboring and jig-drilling machines. The compressed air can be supplied to the aerostatic slideway either from a central supply line or from an individual compressor. In slideways of average width, air is supplied at the interface of sliding surfaces by straight triangular microgrooves

slideways of considerable width, closed rectangular grooves of

h h t ps 60°

A

A I

L

surfaces. Keeping this in view, the depth of the triangular groove is determined from the expression t£

0.7 Bh

Fig. 4.37

Schematic diagram of an aerostatic pad bearing

Machine Tool Design and Numerical Control

266

B = width of slideways, m h = gap between sliding surfaces, m

where

curs only in the transverse direction. It may be found from the following relationship: P = B ◊ l ◊ fp k where fp k k = 17.3

l 3 h3 Bt 4

4 fp k for different values of k and pocket pressure p0. The stiffness of an open-type aerostatic slideway is given by the expression,

Bl f k, ps h f k, ps depends upon k and supply pressure ps. Approximately, K=

f k, ps

ps

fr(k) =5 2 =4

=3

1

P0 = 2

4.6 4.5 4.4 4.2 3.8 3.6 3.4 3.2

h, mkm 0.1

2.8 2.6 2.4 2.2

0.1

20

10 0 0 0.1

Fig. 4.38

0.5

1.0

1.5

2 2.5

10

k

Design curves for computing fp(k) as a function of k for different values of p0

0

Fig. 4.39

t, s

Transient process during raising and lowering of aerostatic slideways

raising the supply pressure. The compressed air from a centralised supply line is supplied at a pressure of 2

switched on, the slideway clearance attains a stable uniform value only after the passage of a certain time period. Similarly, when the air supply is shut off, the moving member takes some time before it settles on into account at the time of starting a setting motion or clamping the moving member when it has reached a desired position.

Design of Guideways and Power Screws 267

4.6 thus changing the nature of friction from sliding to rolling. They have the following positive features: 1. 2. 3. 4.

Low friction as compared to slideways. Uniformity of motion even at slow speeds due to virtual absence of the stick-slip phenomenon. High stiffness if the rolling members are preloaded. Possibility of using high velocities of motion.

These properties explain the wide application of anti-friction guideways in both feed as well as primary cutting motion trains of high-accuracy machine tools. The contact of rolling elements with guideway surfaces occurs over a point or a line. It is, therefore, necessary for the guideway material to have high contact strength to be able to transmit large loads through the it on the machined surface. This necessitates that guideway surfaces be machined with a high degree of accuracy. Cast iron guideways are rarely used as they have poor contact strength; as compared to steel guideways they permit 10 times less load when rollers are used, and 30 times less load when balls are used as rolling elements. Materials that are commonly used for making anti-friction-guideways are: 1. Spindle steels hardened to RC = 60–62. 2. Low carbon structural steel 20Cr1Mn60 Anti-friction guideways employ the same shapes as slideways. These shapes can be obtained by an ap-

employed only when the dead weight of the moving member constitutes the major load that does not change appreciably during the cutting operation.

(a)

(c)

(b)

(d)

Fig. 4.40 Open-type anti-friction ways

guideways are required to have high stiffness. Higher stiffness is achieved through preloading of rolling members. As a matter of fact, horizontal rolling members automatically experience some preloading due to

268

Machine Tool Design and Numerical Control

the weight of the moving member. However, such preloading cannot be varied to yield optimum stiffness. A see the guideway which is loaded less in order to achieve higher stiffness. The recommended preloading is 1. 2–3 microns for precision machines using spherical balls, 2. 7–10 microns for general machines tools using spherical balls, and 3. 15–25 microns for general machines tools using rollers.

(a)

(b)

(c)

Fig. 4.41 Closed-type anti-friction ways

One of the major practical problems associated with the application of anti-friction guideways is that the rolling el-

v

v

linear velocity of the centre of the rolling element is only sitates that 1. either the travel of the moving member should be restricted, or 2. there should be provision for recirculating the balls

v/2

Fig. 4.42

Velocity distribution along the height of a rolling element I/2=lg–lc

members between the stationary surface and the moving member. A limited travel anti-friction guideway is shown in The rolling elements are held in a cage of length lc such that 1 lc = lg – 2 where lc = length of the cage

I 4

Ic

I 4

Ig

Fig. 4.43

Schematic diagram of a limited travel anti-friction way

Design of Guideways and Power Screws 269

lg = length of the stationary guideway l = length of travel of the moving member is at the middle of the guideway when the moving member is in the centre position. The total travel of the moving member is made up of l/2 The schematic diagram of an anti-friction guideway with recirculation of rolling elements to provide unlimited travel is in a cage. However, in some designs rolling elements alternate with idle elements of a slightly smaller diameter which thus Fig. 4.44 Schematic diagram of an serve as a separator. unlimited travel way with The important design parameters of anti-friction guideways recirculating rolling elements are now discussed below. The frictional force acting on the faces of an anti-friction guideway is given by the expression F = nFc where

fr N req

Fc = constant component of the frictional force which is approximately equal to 0.4–0.5 kgf n = number of faces of the guideway fr

f = 0.0025 cm and for general steel ways, f = 0.001 cm req = r/k = equivalent radius of the rolling element, where r is the radius of rolling element in cm and k the k = 1.4–1.5, while for closed ways, k N = Pz + Wt + Ww = normal load on the way, kgf, where Pz is the vertical component of the cutting force, Wt the weight of the table and Ww the weight of the workpiece determined the frictional force, the required pulling force can be found from the expression, Q =F where

Px

Px = axial component of the cutting force, kgf.

anti-friction roller ways it is given by the expression, P =K◊ and for anti-friction ball ways by the expression, P = Kd2

◊d

270

Machine Tool Design and Numerical Control

= length of the roller, cm d = diameter of the rolling member, cm 2

K=

K represents a conditional stress with limiting values as given in Table 4.3. Table 4.3 K, kgf/cm2 Ground steel ways

Cast iron ways

6

0.2

150–200

13–20

Ball Roller

1. calculating load P, if the dimensions of rolling members have been selected, or 2. for selecting proper dimensions of the rolling elements P is calculated and checked to ensure that it is less than the load Pmax on the maximum loaded rolling member P = where

◊ t ◊ pmax

length of the roller or diameter of the ball, cm t = pitch of rolling members, i.e. distance between the centres of two adjacent rolling members, cm Pmax = maximum pressure acting on the guideway surface determined as per the procedure explained in Sec. 4.3.1, assuming the rolling members to be replaced by a continuous elastic surface, 2

The number of rolling elements in a guideway using rollers is determined from the expression, 16 £ Z £

q 4

and in a guideway using spherical balls from 16 £ Z £

P 3 d

Design of Guideways and Power Screws 271

Z = number of rolling elements q P = load per ball, kgf d = diameter of the spherical ball, cm It has been observed from practise that if Z ing elements and transferred to the machined surface. On the other hand, if Z violates the other inequality constraint, a large number of rolling elements remains partially under loaded and some are not loaded at all. The design of anti-friction ways for stiffness involves calculation of the elastic displacement under the working loads. They can be determined for anti-friction roller ways from the equation d = Cr ◊ q and for anti-friction ball ways from the equation d =C ◊P Cr and C depend upon the load, diameter of the rolling members, material of the Cb, mm kgf

Cr, mm – cm kgf 1.6

8

1.2

6

0.8

d=5mm

d=10mm

4 2

0.4 0 0

1 2

4

6

8

10

q kgf/cm

2

2

0

1 0

1

2

3

4

5

P kgf

Fig. 4.45 Design curves for computing the compliance coefficient for (a) roller anti-friction ways (b) ball antifriction ways. 1. Ground steel ways 2. Scraped cast iron ways

assumption that guideways and rolling members are ideally manufactured and assembled. In fact, manu5 , can often be almost as large as the elastic displacement. An increase in the magnitude of the manufacturing error results in a reduction of the guideway stiffness. The effect of the manufacturing error on the guideway stiffness can be accounted through

272

Machine Tool Design and Numerical Control

d

d

Dd

d1

x

d ma

n

d mi (a)

a

d

D=aL

di

L (b) D¢2

D¢2

D≤2

D≤2

L D2 = D¢2 + D≤2

L D2 = D¢2 + D≤2

(c) dx

d d2

X L1

L (d)

Fig. 4.46 Errors in anti-friction ways due to (a) difference in ball size (b) lack of parallelness (c) and (d) lack of straightness d, mm 8

d, mm

Fig. 4.47

m

6

6

6mm

6

4

4 2

4

2

Ideal

2

0

m 20 D 2 = 10

Dd =

0

5

10

15

20

q 25 Kgf /cm

4

0

Ideal 0

5

10

15

20

25

q kgf/ cm

Effect of manufacturing errors on stiffness of roller anti-friction ways: (a) Errors due to difference in size of elements (b) Errors due to lack of straightness

Design of Guideways and Power Screws 273

4.7

COMBINATION GUIDEWAYS

mentioned that they are distinguished by a relatively m heavy-duty machine tools. The performance and life of conventional slideimproved and the scope of their application considerably expanded by combining them with other guideways which provide relief from the normal load. The two commonly used types of combination guides are discussed below.

4.7.1 Slideways with Hydraulic Relief This slideway can be considered as intermediate between conventional slideways and hydrostatic slideways. The normal load on contacting surfaces is partially relieved by supplying oil to the interface at a suitable pressure. This pressure, however, is not large enough to lift the moving member as in hydrostatic slideways. This type of combination guideway incorporates the positive features of 2. the conventional slideway characterised by high contact strength and reliable clamping of the moving member in any desired position. The oil is supplied into the clearance between the sliding surfaces through appropriately interspaced pockets. The shape and geometry of the pockets, and the arrangement utilised for pumping oil is the same as The frictional force in a slideway with hydraulic relief is given by the relationship, Hˆ Ê F = f ◊ N Á1 - ˜ Ë N¯ where

f N = normal load on the guideways H = hydraulic relief

H = 0.7N. If the load on the slideway is not uniformly distributed along its length, the oil supply is regulated by restrictors. However, the restrictors can be dispensed with when the load is uniform and more or less constant, as is the case when the moving member is heavy and the cutting force is relatively small. Slideways with hydraulic relief are particularly effective and feasible in the latter case which explains their increasing application in tables of plano-milling, copy milling and grinding machines.

4.7.2 Slideways with Rolling Members In this type of slideways, a part of the normal load is taken up by spring-loaded rollers thus reducing friction between the sliding members. This results in 1. longer life of the slideways, and phenomenon.

274

Machine Tool Design and Numerical Control

Schematic diagrams of combination guideways subjected to vertical load and compound load are shown friction way. The share of the rolling members in the total load can be regulated by the tightening of the leaf

(b)

(a)

Fig. 4.48 Schematic diagram of combination guideways using anti-friction rolling members for (a) vertical loading (b) compound loading

The friction force in this type of combination guideway is given by the expression, F = nFc

f ◊ N1

fr ◊ N2 req

Here N1 and N2 represent the share of normal load taken up by the slideway and the anti-friction way,

relief. Their most widespread application is in rotary tables of gear-hobbing machines and vertical boring and turret lathes, in which a part of the load is taken up by a central anti-friction thrust bearing. They may also be encountered in heavy-duty horizontal-boring and jig-boring machines and medium-sized lathes.

4.8

PROTECTING DEVICES FOR SLIDEWAYS

Wear of slideway surfaces is seriously affected by dirt, abrasive particles and chips penetrating into the gap between them. The life of slideways can be considerably improved by employing devices which protect the categories: 1. Seals 2. Covers 3. Intermediate steel strips. 1. Seals

stationary guideways by

Design of Guideways and Power Screws 275

Felt seal

(a)

Ruber seal

(b)

Brass chip cleaner

(c)

(d)

Fig. 4.49 Various types of protective seals

2. Covers If the stationary guideway is shorter than the moving member by an amount, equal to the maximum travel, then the slideway surfaces are always covered by the moving member. Generally, the situation is just the reverse, i.e., the moving member is shorter in length than the stationary guideway. In such Such cover plates are employed on turret lathe carriages and grinding machine tables in which the stroke is Telescopic covers are made of 1.5–3.0 mm thick sheets of steel. In heavy-duty machine tools, the telescopic covers are also quite heavy and must be supported on rollers. In grinding machines where there is no chance

(a)

(b)

(C)

Fig. 4.50 Various types of protective covers

3. Intermediate Steel Strips Accurate thin steel strips of constant thickness may be inserted between the sliding surfaces to avoid exposing the accurately ground slideway surfaces to dirt, chips, etc. The steel strip is stretched tight by a tensile preload. Three methods of holding the protective strip are schematically depicted

276

Machine Tool Design and Numerical Control

(a)

(b)

(c)

Fig. 4.51 Various methods of holding the protective strip

It should be noted that in machine tools using covers or intermediate steel strips, seals are used as additional protection.

4.9

DESIGN OF POWER SCREWS

Power screws or translation screws utilise the helical translatory motion of a screw thread for transmitting power and motion rather than for clamping parts. Power screws are widely used in machine tools for imparting motion to the operative members due to their following distinguishing features: 1. Large reduction which makes it possible to obtain slow travel of the operative element in the feed train. 2. Simple design and compactness. 3. High load-carrying capacity. 4. Uniform and accurate movement of the operative member.

1. sliding-friction power screws. 2. rolling-friction power screws.

4.9.1 Design of Sliding-friction Power Screws Sliding-friction power screws operate under conditions of semi-liquid friction, characterised by a relatively extreme care in the selection of the power-screw material: 1. The material of the power screw must have a high surface hardness and should not undergo severe deformation during machining, heat treatment and exploitation. 2. Normal accuracy power screws which are not subjected to heat treatment are made from structural 3. Precision power screws that are not subjected to heat treatment are made from high carbon steels containing 1–1.2% carbon. 4. Precision power screws that are heat treated are made from alloyed steel 45CrlMn 70 Si 27 Ni25 The nuts of power screws are made of a softer material, the most commonly used being bronze. If accuracy requirements are not high and the power screw is well protected against foreign matter, the nut may bronze lining on the inside surface of a steel bush by centrifugal casting.

Design of Guideways and Power Screws 277

be easily machined without distortion and is, therefore, widely used in normal accuracy machines. However, the acme thread cannot be employed on precision machine tools as the radial run out results in considerable pitch error. The effect of radial run out on pitch reduces with the reduction of the included angle. Therefore,

2a = 29°

(a)

2a = 10°

(b)

(c)

Fig. 4.52 Thread profiles for sliding friction power screws

The nuts of power screws may be solid or split. Solid nuts are used when the power screw is the only traction device in the kinematic train controlling the movement of the operative member. The split nut is employed when the kinematic train includes another translatory motion transmission, e.g., rack and pinion. In such cases, one half of the split nut is rigidly attached to the carriage, while the other half has provision for

1

2 (Wedge)

1

3

3 2 (Adjusting out)

(a)

1

2 (Spring)

(b)

3

1. Axially movable half 2. Element for axial adjustament 3. Fixed half clamped to the carriage

(c)

Fig. 4.53 Methods of backlash adjustment in sliding friction power screws using (a) wedge (b) nut (c) spring

278

Machine Tool Design and Numerical Control

Power screws can be supported on sliding or anti-friction bearings. Sliding bearings are preferred for jourmental effect of screw deformation due to its heating is reduced. Mounting of one thrust bearing at each end, in long, heavily loaded power screws. stiffness of the nut and screw transmission and is used generally in numerically controlled machine tools. The

(a)

(b)

(c)

(d)

Fig. 4.54 Various methods of mounting thrust bearings in sliding friction power screws

Sliding-friction power screws are designed for the following factors: 1. 2. 3. 4.

Wear resistance Strength Stiffness Buckling stability

1. Design for Wear Resistance The wear resistance of a power screw depends primarily upon the average pressure on the working surfaces of the thread. The design condition is Pav £ [pav where

pav = average pressure on the thread surface [pav] = permissible average pressure on the thread surface

Average pressure pav is determined from the relationship, pav = where

Q t d H L z

Q◊t p ◊d ◊H ◊L◊z

2

= pulling force, kgf = pitch of thread, cm = pitch diameter of thread, cm = height of thread engagement, cm = length of nut, cm, and = number of starts of the thread

p d H represents the area of one face of thread and Lz/t represents the number of thread faces participating in the transmission.

Design of Guideways and Power Screws 279

H = 0.5

t z

2Q 2 p dL The recommended values of permissible average pressure [pav] are: pav =

[pav] [pav] [pav

2

for a steel screw-cast iron nut pair of normal accuracy feed mechanism. 2 for a steel screw-bronze nut pair of normal accuracy feed mechanism used in general-purpose machine tools.

2. Design for Strength Power screws are designed for strength in accordance with the theory of maximum shearing stress. The design condition can be written as tmax £ [t where

t max = maximum shearing stress in the screw, and [t] = permissible shearing stress of the screw material

The minor diameter cross section of a power screw is subjected to biaxial stress consisting of a tensile or compressive stress due to axial pulling force and shearing stress t due to torque. The maximum shearing stress in a bi axially stressed member is given by the expression, tmax = Normal stress, s = where

1 s 2 + 4t 2 2

4Q p d m2

dm is the minor diameter of the thread in cm. Shearing stress, t =

16 M t p d m3 3

represents the torsional modulus of the screw section. where Mt is the torque and pd m Torque Mt may be found from the following formula: Mt = where

Q ◊ d p cos a tan l + m 2 cos a - m tan l

dp = pitch circle diameter of the thread a= t = helix angle of the thread, and l = tan –1 pdp m

280

Machine Tool Design and Numerical Control

Substituting the expressions for s relationship:

t

t max =

2 p d m2

+ Q2

64 M t2 d m2

The value of the permissible shearing stress is generally taken as [t] = where

sy k

sy = yield stress of the screw material k =

3. Design for Stiffness The design for stiffness is based upon the consideration that the elastic deformation of the screw thread due to axial force and torque should not exceed a permissible value, i.e., Dt £ [Dt where

Dt = change in pitch of the screw thread [Dt] = permissible value of the change in pitch of the screw thread

The change in pitch of the screw thread due to axial pulling force Q is DtQ = where

Q◊t EAm 2

E Am =

p d m2 = cross-sectional area of the screw thread at the minor diameter, cm2 4

The angle of twist of the screw over one thread pitch due to torque Mt is q=

Mt ◊ t G ◊ I pm

Therefore, the change in pitch of the screw thread due to torque Mt is Dtt =

Mt ◊ t2 Mt ◊ t t = ◊ G ◊ I pm 2p 2p G ◊ I pm 2

G Ipm =

p d m4 = polar moment of inertia of the screw at the minor diameter, cm4 32

Generally DtQ is much greater than Dtp, therefore, the total change in pitch of the screw thread can be written as Dt =

Q◊t E ◊ Am

Design of Guideways and Power Screws 281

The permissible value of change in thread pitch [Dt] is taken as [Dt] = 0.3 ¥ tolerance on difference of the pitch of adjacent threads. 4. Design for Buckling Stability If the length of the power screw is more than 7.5–10 times its minor diameter, the screw must be designed for buckling stability. The design condition is Q £ Qc where

Q = axial compressive force Qc

The pulling force Q expression, Qc =

where

p 2 EI (kl l ) 2

p d m4 = axial moment of inertia of the screw section at the minor diameter, cm4 64 l = length of screw, cm kl = I=

kl depends upon the type of constraints at the screw supports. 1. If both ends are hinged, kl = 1. kl = 0.5. kl = 0.7. kl = 2.0.

4.9.2 Design of Rolling-friction Power Screws In a rolling-friction power screw the load between the threads of the screw and nut is not transmitted by digrooves of the screw and nut in a manner akin to their function in ball bearings. An essential feature of almost

Recirculation mechanism

Fig. 4.55 Schematic diagram of a ball recirculating power-screw assembly

282

Machine Tool Design and Numerical Control

The distinguishing features of ball-recirculating power screws are: 1. L sliding-friction power screws. 2. H 3.

riction force is virtually independent of the travel velocity and the friction at rest is very small; consequently, the stick-slip phenomenon is practically absent, ensuring uniformity of motion. 4. By preloading the assembly, clearances can be eliminated and the axial stiffness can be improved to achieve or even exceed the stiffness of a sliding friction power screw. It should, however, be noted that the axial stiffness of un preloaded ball re circulating power screw is less than that of an ordinary one. The materials of the screw and nut must satisfy the following two requirements: 1. Hardness of about RC = 60 of the working surfaces. 2. Minimum residual stresses so that the initial accuracy of transmission is maintained. Both for screws and nuts, these requirements are met by 55Cr50 Mo25

25

The nuts may also be made from bearing steel, alloyed structural steel containing approximately 1% each of Si and Cr, and case hardened steel 12Ni3Cr2Mn45 Si27 Anti-friction power screws are primarily employed in the feed mechanism of precision machine tools, such as grinding, boring and jig-boring machines. During the last decade or so their application has considerably increased, particularly in machine tools with numerical control. load-carrying capacity and are rarely used in ball-recirculating power screws of machine tools. Commonly relatively higher axial stiffness and a large load capacity. a

Fig. 4.56

(a)

(b)

(c)

(d)

Various thread profiles used in ball-recirculating power screws: (a) Square (b) Trapezoidal (c) Semi-circular (d) Ogive

Design of Guideways and Power Screws 283

rb = 0.95 – 0.97 rt tact of a = 45–60°. However, a large difference between the ball and thread radii becomes counterproductive as friction losses increase due to large difference in the velocities of their contacting surfaces. The difference screws are, therefore, of the split type so that preloading can be realised by relative axial displacement of the half nuts. loading can be achieved by merely using balls of a diameter slightly exceeding the nominal. The r /rt ratio lies in the range of 0.9–0.95. The most complicated part of a ball-recirculating power screw is the device for returning the rolling bodies. Sometimes, it is possible to avoid this device in transmissions with short strokes by making the nut so large that the balls do not fall out even when the screw occupies extreme positions. However, as earlier stated, the device for recirculation of balls is almost an essential feature of all anti-friction power screws. The

Anti-friction power screws are designed for 1. strength under static 2. cyclic loading, and 3. stiffness. 1. Design for Strength under Static Loading6 This design is based upon the condition that the maximum contact stress between the thread and ball should be less than a permissible value, i.e., s c max £ [sc

Ê PE 2 (rt - rb ) 2 ˆ s c max = 1.4 Á ˜ rt2 rb2 Ë ¯ where

P = static load acting on a single ball, kgf rt = r = radius of the ball, cm E =

2Et Eb = reduced modulus of elasticity, Et and E Et + Eb 2 and ball .

1/3

2

284

Machine Tool Design and Numerical Control

If the screw and nut are both assumed to be made of steel, then E = 2.l ¥ l06 r /rt = s cmax = 4.3 ¥ 103

Ê Pˆ ÁË d 2 ¯˜

2

. Also assuming

1/3

2

b

where

d = diameter of the ball, cm

The permissible contact stress depends upon the hardness of contacting bodies and is given by the expression, 4 RC 2 [sc 60 If the average hardness of the contacting bodies is assumed to be RC = 60, then the limiting load on each ball can be found from the inequality Ê Pˆ 4.3 ¥ 103 ◊ Á 2 ˜ Ë db ¯

1/3

¥ 104

£

wherefrom, for the minimum value of [sc] ¥ 2.5 ¥ 104

2

P £ 200d

we obtain

2

The expression for the pulling force is Q = P sin a cos lZr where

a= l = helix angle of the thread Zr = reduced number of balls that are actually loaded, i.e., total number of balls Zt minus the balls in the return passage.

Assuming a = 45°, cos l Zr = 0.7Zt and substituting for P the expression for the limiting pulling force may be written as Q £ 200d 2

1 2

◊ 1 ◊ 0.7Zt kgf

or Q £ 100Z t d 2 2. Design for Strength under Cyclic Loading This design accounts for the failure of balls due to fatigue 60Tn ◊ f ˆ k = kQ ÊÁ Ë 107 ˜¯ where

1/3

kQ kQ = 0.9 T = desired service life of the balls, hours n = average rpm of the power screw, determined as the arithmetic mean of all available rpm

Design of Guideways and Power Screws 285

f = number of loading cycles during one revolution of the screw; it may be taken as half the number of balls accommodated in a single thread The limiting pulling force due to cyclic loading is determined as 2

QCL = where

Q 100 Z t db = k k

Q

3. Design for Stiffness axial force increment to the axial displacement increment, i.e., K=

dQ dd

The axial displacement of the nut with respect to the screw occurs due to bending and shear deformation of the threads and due to contact deformation. Assuming that the displacement is predominantly the outcome of contact deformation, the expression for stiffness can be written as K = 0.66 ¥ 106Zr db ◊ d 0 where

d0 is the initial axial deformation, cm.

If the initial deformation is achieved by preloading with a force PPL, the expression for stiffness is K = 104 ¥

d ◊ PPL ◊ Z 2r

However, such a straightforward conclusion is slightly misleading. An increase in preloading undoubtedly value of the preloading force can be approximately determined from the expression, PPL = 0.3 Qlim where Qlim Of course, the overall stiffness of the power screw-nut transmission can be improved by simply increasing the screw diameter. In the end, it may be pointed out that the methods of applying preloading in anti-friction power screws are essentially the same as in sliding-friction power screws.

Review Questions 4.1 ratio at which the saddle will lift from face A. Assume The resultant cutting force makes an angle of 40° with the vertical.

see = 0.5, a = 50°.

4.2 100 mm diameter workpiece, the tangential, radial and axial components of the cutting force were found to be Pz = 300 kgf, Py = 130 kgf and Px = 120 kgf, respectively. The lathe carriage weighs 150 kgf and is 30 cm long. Calculate the slideway width assuming suitable values for h, , YQ, ZQ and 2 XP see .

286

Machine Tool Design and Numerical Control

4.3 4.4 3

2

ness = 0.02 mm. If the load is supported by six composite slider bearings of same width, determine the length of each slider. 4.5 Derive the load-capacity expressions for a rectangular, hydrostatic pad bearing fed through capillary, Hint: 4.6 A rectangular, hydrostatic pad bearing of pad area 50 ¥

2

and pocket area 10 ¥ 60 mm2 supports 2

. Also, determine the length and diameter of the required capillary restrictor if the l/d ratio of the restrictor = 20. 4.7 2

. Determine the load capacity of the bearing if the minimum

Hint: 4.8 In a turning operation the cutting force components were Pz = 300 kgf, Py = 130 kgf, Px= 120 kgf. If the carriage weighs 150 kgf, determine the pulling force for conventional slideways and ball recirculating diameter of balls is 10 mm. Assume suitable data which is not given. 4.9 Calculate the average pressure, maximum shearing stress and pitch error in the lead screw of a lathe. Given: Screw material

Structural steel

Outer diameter

70 mm

Pitch

10 mm

Effective diameter

65 mm

Length of nut

100 mm

Pz

150 kgf

Py

50 kgf

Px

45 kgf 0.2

Weight of the carriage The lead screw has standard, single-start acme thread.

50 kgf

Design of Guideways and Power Screws 287

4.10 ball diameter = 10 mm and total number of balls = 120. Both the screw and nut are made of steel. Determine the load on a single ball and the maximum pulling forces under static and dynamic loading. The screw rotates at an average of 40 rpm and must have a guaranteed service life of 10000 h before the balls can be changed. Data not given may be suitably assumed.

References 1. Sparks, CA, Design and Development of Machine Tools, J Inst. of Prod. Engrs., Stanki I Instrument, 1964, No. 6. 4. Steinberg CA, et al.,

Mashinostroenie Publishers, Moscow, 1969. Stanki I Instrument,

1966, No. 10. Design of Machine Tools, Mashinostroenie Publishers, Moscow 1977, p. 221.

5 5.1

DESIGN OF SPINDLES AND SPINDLE SUPPORTS

FUNCTIONS OF SPINDLE UNIT AND REQUIREMENTS

Functions The spindle unit of a machine tool performs the following important functions: 1. Centreing the workpiece, e.g., in lathes, turrets, boring machines, etc., or the cutting tool, as in drilling and milling machines. 2. Clamping the workpiece or cutting tool, as the case may be, such that the workpiece or cutting tool is reliably held in position during the machining operation. 3. Imparting rotary motion (e.g., in lathes) or rotary cum translatory motion (e.g., in drilling machines) to the cutting tool or workpiece.

determine the important design requirements to spindle units which are listed below. Requirements 1. The spindle should rotate with a high degree of accuracy. The accuracy of rotation is determined by the radial and axial run out of the spindle nose, and these must not exceed certain permissible

the front end. 2. The spindle unit must have high static stiffness. The stiffness of the unit is made up of the stiffness of as well as torsional stiffness. 3. The spindle unit must have high dynamic stiffness and damping. Poor dynamic stability of the spindle unit adversely affects the dynamic behaviour of the machine tool as a whole, resulting in poor surface 4. The mating surfaces that are liable to wear restrict the life of the spindle unit. These surfaces, such as journals, quills (in drilling machines), etc., must be hardened to improve their wear resistance. The spindle bearings must also be selected or designed to retain the initial accuracy during the service life of the machine tool. 5. The deformation of the spindle due to heat transmitted to it by the bearings, cutting tool, work piece, etc., should not be large, as this has an adverse effect on the machining accuracy. In case of spindles running at high rotational speeds, particular care should be taken in selecting or designing the front bearing as it is the major source of heat transmitted to the spindle. cutting tool or workpiece. The centreing is achieved by means of an external or internal taper at the

Design of Spindles and Spindle Supports 289

front end of the spindle. The spindle ends, including the taper have been standardised for the common groups of machine tools and are shown in Table 5.1. Table 5.1 Spindle ends S. No.

Figure

Application 7°7¢ 30¢¢

Lathes, turrets, single spindle automatic and semi-automatic lathes, etc.

1.

2.

Milling machines

Remark External taper 7°7¢30¢¢ is used only in heavy-duty machine tools for centreing chuck or face plate. Generally, chucks or face plates are centred by a cylindrical surface.

Steep taper 7 : 24 is necessary so that the milling arbor and cutter once located and clamped should be able to hold their position under the pulsating milling force.

Taper 7:24

Morse tapers come in 8 sizes

3.

Drilling and boring machines Morse taper

4.

Grinding machines Taper 1:3

5.2

MATERIALS OF SPINDLES

The blank for a machine tool spindle may be 1. rolled stock in the case of spindles having diameter < 150 mm, and

angle varies between 1°25¢27≤ and 1°30¢26≤. Small taper is essential as rotation is transmitted from the spindle to the drill shank by friction. In grinding machines, the concentricity of location is extremely important. Therefore, the grinding wheel is mounted on an adaptor taper.

290

Machine Tool Design and Numerical Control

2. casting (preferably obtained by centrifugal casting method) in the case of spindles having diameter > 150 mm. It should be borne in mind that if the spindle blank is cut from rolled stock, the cutting must be done by a cutting tool (shears, rotary saw, parting tool, etc.) to avoid additional distortions of the material microstructure. In machine tool spindle design, the critical design parameter is not strength but stiffness. If we compare although the strength of alloyed steels can be considerably greater than that of mild steel. Since stiffness (the main design parameter) is primarily determined by the modulus of elasticity of the material, it may be conIn the light of the preceding discussion and the requirements laid down in Sec. 5.1, the following recommendations for selecting the spindle material may be formulated 1. for normal accuracy spindles, plain carbon steels C45 and C59 (AiSi C1045 and C1050) hardened and tempered to RC = 30. 2. for above normal accuracy spindles—low alloy steel 40 Cr 1 Mn 60 Si 27 Ni25 (AiSi 5140) induction hardened to RC = Cr 1 Mn 60 Si 27 Ni 25 (AiSi 5147) is used with hardening to RC = 55–60. 3. for spindles of precision machine tools, particularly those with sliding bearings—low alloyed steel 20 Cr 1 Mn 60 Si 27 Ni 25 (AiSi 5120) case hardened to RC = 56–60 or 38 Cr 1 A1 90 Mn 45 Si 27 Ni 25 Mo 20 (EN 41) nitrided to RC = 63–68, and 4. for hollow, heavy-duty spindles—grey cast iron or, spheroidal graphite iron.

5.3

EFFECT OF MACHINE TOOL COMPLIANCE ON MACHINING ACCURACY

Consider a uniform shaft being machined between centres on a lathe (Fig. 5.1). Let KA be the stiffness of centre A and KB that of centre B. Due to radial component Py of the cutting force, centre A will be displaced by a distance yA =

PA KA

(5.1)

yB =

PB KB

(5.2)

B≤ C≤ A¢ C¢ C

A

B¢ B

and centre B by Py

x l

Here PA and PB are the forces of reaction at ends A and B, Fig. 5.1 Schematic diagram of a simple respectively. They can be determined from the following turning operation equations of static equilibrium: 1. Moment of Forces about Point B = 0, i.e., PA l = Py (l – x)

Design of Spindles and Spindle Supports 291

wherefrom, xˆ Ê PA = Py Á1- ˜ Ë l¯ 2. Moment of Forces about Point A = 0, i.e.,

(5.3)

PB ◊ l = Py ◊ x wherefrom, PB = Py

x l

(5.4)

Substituting the values of PA and PB in Eqs (5.1) and (5.2), respectively, we obtain xˆ 1 Ê yA = Py Á1 - ˜ Ë l ¯ KA x 1 yB = Py ◊ l KB

(5.5) (5.6)

Owing to the compliance of centres A and B, the workpiece occupies position A¢ B¢¢ (assuming KA > KB) and its displacement at the cutting point can be found as yx = yA + C¢C¢¢ from similar triangles A¢C ¢C ¢¢ and A¢B¢B¢¢, we have C ¢C ¢¢ x = B ¢B ¢¢ l

since

B¢B¢¢ = yB – yA C ¢C¢¢ = (yB – yA) Therefore,

x l

yx = yA + (yB – yA)

x l

(5.7)

Substituting the values of yA and yB from Eqs (5.5) and (5.6), Eq. (5.7) yields ÈÊ xˆ 1 x Ïx 1 x ˆ 1 ¸˘ Ê yx = Py ÍÁ1 - ˜ ◊ + Ì ◊ - Á1 - ˜ ˝˙ l ¯ K A l Ó l KB Ë l ¯ K A ˛˙˚ ÍÎË or

yx =

2 2 È Ê xˆ Ê xˆ ˘ Í K B Á1 - ˜ + K A Á ˜ ˙ Ël¯ ˚ K A ◊ KB Î Ë l¯

Py

(5.8)

If it is assumed that KA /KB = a, yx =

2 2 Py ÈÊ xˆ Ê xˆ ˘ ÍÁ1 - ˜ + a Á ˜ ˙ Ël¯ ˚ K A ÎË l¯

(5.9)

292

Machine Tool Design and Numerical Control

wherefrom, keeping in mind that Py/KA = yA max, we get yx y A max

2

xˆ Ê Ê xˆ = Á1 - ˜ + a Á ˜ Ë Ël¯ l¯

2

(5.10)

The variation of yx/yAmax plotted as a function of x/l for different values of a is depicted in Fig. 5.2. From the curves it may be concluded that

yx /yAmax

1. when a < 1, i.e., stiffness of the headstock centre is less than the stiffness of the tailstock centre, maximum displacement of the workpiece occurs at the headstock; and 2. when a > 1, i.e., the stiffness of the tailstock centre is less than that of the headstock, maximum displacement of the workpiece occurs at the tailstock.

3 a=4 2

a=2 1 a=1 a=0

0 0

0.2

0.4

0.6

0.8

1.0

x I

Fig. 5.2 Variation of yx /yAmax as a function of x/l

and minimum displacements of the workpiece axis. As already stated, maximum displacement occurs at the headstock or tailstock depending upon the value of yx/Py is minimum. Therefore, the location of the point of d Ê yx ˆ =0 dx ÁË Py ˜¯ From Eq. (5.9), 1 Ê 2x 2 ˆ 1 2x d Ê yx ˆ = - ˜+ l ¯ KB l2 dx ÁË Py ˜¯ K A ÁË l 2

Design of Spindles and Spindle Supports 293

which, upon equating to zero yields x0 =

KB ◊ l K A + KB

(5.11)

x0 2Ê 1 1 ˆ d 2 Ê yx ˆ + = Á ˜ >0 dx 2 ÁË Py ˜¯ l 2 Ë K A K B ¯ Substituting the value of x0 in Eq. (5.10), we obtain

or

2 2 ÈÊ KB ˆ Ê KB ˆ ˘ ymin = ÍÁ1 ˙ y A max +aÁ ˜ ˜ K A + KB ¯ Ë K A + K B ¯ ˙˚ ÍÎË a ymin = y A max 1+a

(5.12)

For a < 1: The maximum displacement occurs at the headstock centre, i.e., ymax = yAmax. Therefore, a ymax – ymin = yA max – ◊ y A max 1+a wherefrom, ymax - ymin 1 (5.13) = y A max 1+a For a > 1: The maximum displacement occurs at the tailstock centre, i.e., ymax = yBmax. Therefore, ymax – ymin = yB max –

a ◊ yA max = yA max È yB max - a ˘ Íy ˙ 1+a Î A max 1 + a ˚

as yB max /yA max = KA/KB = a, we get ymax - ymin a2 = y A max 1+a

achieved when the stiffness of headstock and tailstock centre are equal.

ymax – ymin /yAmax

(5.13) for a < 1 and by Eq. (5.14) for a > 1. The relationships represented by Eqs (5.13) and (5.14) have been plotted as functions of a and shown in Fig. 5.3. The dotted portions of the curve represent the range in which the curves are not valid. It is evident from Fig. 5.3 that the difference ymax – ymin is minimum at a = 1 and is equal to half the headstock centre displacement, i.e., y (5.15) (ymax – ymin)min = A max 2

(5.14)

3

ymax = yB(a >1)

2 1 ymax = yA(a < 1) 0 0

Fig. 5.3

1.0

2.0

Variation of

3.0

4.0

a

y max - y min as a y A max

function of a

294

Machine Tool Design and Numerical Control

If the stiffness of the saddle is given by Ks, the displacement of the cutting edge due to saddle compliance can be found as Py ys = Ks The total compliance of the machine tool can be found as the sum of compliances of the workpiece and saddle, i.e., y y y Cmt = mt = s + x Py Py Py Cmt =

or

1 1 Ê + Á1 Ks K A Ë

2

xˆ 1 Ê xˆ ˜¯ + Á ˜ l KB Ë l ¯

2

(5.16)

the machine tool compliance as expressed through Eq. (5.16). The representative total compliance values for some machine tools are given in Table 5.2. Table 5.2 S. No.

5.4

Values of compliance Cmt for different machine tools

Machine tool

Description

Compliance (micron/kgf)

1.

Lathe, workpiece clamped between centres

Height of centres 200 mm Height of centres 300 mm Height of centres 400 mm

0.75 0.52 0.38

2.

Lathe, workpiece clamped in chuck which is screwed on spindle nose

Height of centres 200 mm Height of centres 300 mm

1.5 1.2

3.

Multiple-spindle vertical, semiautomatic lathes

Height of centres 400 mm

0.5–0.7

4.

Automatic bar lathes

Screw type Multiple-spindle type

0.2–0.3 0.3–0.4

5.

Vertical milling machine

Drive motor rating = 7 kW Table size 320 ¥ 1250 mm

0.4

6.

Plano-milling machine

Table size 4.25 ¥ 1.5 m

0.4

7.

Vertical turning and boring mill

Table diameter 3 m

0.36

8.

Centreless grinding machine

1.00

DESIGN CALCULATIONS OF SPINDLES

Figure 5.4 shows the schematic diagram of a spindle. As is evident from the diagram, the spindle represents a shaft with

Design of Spindles and Spindle Supports 295

1. supported length l acted upon by driving force P2, and 2. cantilever of length c acted upon by external force P1.

P1

P2

The spindle is basically designed for bending stiffness which l

ymax £ [y]

(5.17)

Fig. 5.4

c

Schematic diagram of a spindle

y1 of the spindle axis due to bending forces P1 and P2, and y2 of the spindle axis due to compliance of the spindle supports.

y3 of the centre or cutting tool due to compliance of the tapered joint.

5.4.1 Deflection of Spindle Axis due to Bending diagram. In this context, the following guidelines may be employed with success: 1. If the spindle is supported on a single anti-friction bearing at each end, it may be represented as a simply supported beam. 2. If the spindle is supported in a sleeve bearing, the supported journal is analysed as a beam on an elastic foundation; for purposes of the design diagram the sleeve bearing is replaced by a simple hinged support and a reactive moment Mr acting at the middle of the sleeve bearing. The reactive moment is given as

P2

Mr = k ◊ M where

M = bending moment at the support, k k = 0 at small loads to k = 0.3–0.35

Consider, for example, the spindle shown schematically in Fig. 5.5a. By replacing the rear ball bearing by a hinge, and the front sleeve bearing by a hinge and reactive moment Mr, the spindle can be reduced to the design diagram of Fig. 5.5b.

a

P1

b (a) P2

c P1

Mr (b)

determined by Macaulay’s method and is found to be y1 = where

˘ 1 È 2 Ê aˆ Í p1c (l + c) - 0.5 P2 abc ÁË1 - ˜¯ + M r lc ˙ 3EI Î l ˚

y1

(5.18)

(c)

Fig. 5.5 (a) Schematic diagram of the E = modulus of elasticity of the spindle material spindle (b) Design diagram of I = average moment of inertia of the spindle section the spindle (c) Deflected axis of the spindle

296

Machine Tool Design and Numerical Control

5.4.2 Deflection of Spindle Axis due to Compliance of Spindle Supports Let dA and dB represent the displacement of the rear and front supports, respectively. Owing to the compliance of sign purposes we are considering the most unfavourable case when bearing displacements are oppositely directed). From similar triangles OCC ¢ and OBB¢,

wherefrom,

d y2 = B x c+x

dA

A¢

B B¢

dB

C C¢

Fig. 5.6

y2 = ÊÁ1 + Ë

O

y2

Deflection of spindle due to compliance of the supports

cˆ ˜ dB x¯

(5.19)

Again, from similar triangles OAA¢ and OBB¢, dB d = A x l-x

wherefrom,

x=

l ◊ dB d A + dB

(5.20)

Substituting this value of x in Eq. (5.19), we get Cˆ c Ê y2 = dB Á1 + ˜ + d A Ë l¯ l

(5.21)

It is evident from Eq. (5.21) that displacement dB y2 of spindle nose than displacement dA of the rear bearing. Displacements dA and dB can be determined from the following expressions: RA KA R dB = B KB

dA =

(5.22) (5.23)

where RA and RB are reactions at supports A and B, respectively, while KA and KB is their respective stiffness. Reactions RA and RB can be found from the following equilibrium conditions applied to the design diagram of Fig. 5.5b: 1. Moment of All Forces about Point A = 0, i.e., SMA = 0

or

wherefrom, RB =

RB ◊ l – P2 ◊ a + Mr – P1(c + l) = 0 P2 a - M r + P1 (l + c) l

(5.24)

Design of Spindles and Spindle Supports 297

2. Moment of All Forces about Point B = 0, i.e., SMB = 0

or

RA ◊ l – P2 ◊ b – Mr + P1c = 0

wherefrom, RA =

P2 b + M r - P1c l

(5.25) y2 may be written as follows:

y2 =

P2 a - M r + P1 (l + c) Ê c ˆ P2 b + M r - P1c c x ÁË1 + ˜¯ + l ◊ KB l l ◊ KA l

(5.26)

The total deflection of the spindle nose can be determined as the sum of y1 (Eq. (5.18)) and y2 (Eq. (5.26)), i.e., y = y1 + y2 (5.26a)

y2

y

y1

Fig. 5.7.

Fig. 5.7 Total deflection of the spindle axis

5.4.3 Optimum Spacing between Spindle Supports An important parameter in spindle design is the ratio l = l/c. The optimum value of this ratio is the one that ensures minimum total y; it can be determined from the condition dy/dl = 0. The qualitative variations of y1, y2 and y1 + y2, for constant values of forces P1 and P2 are depicted in Fig. 5.8 as functions of ratio l/c. The point of minimum of the y1 + y2 curve yields the optimum value of ratio l/c which generally lies between 3 and 5. The value of lopt depends upon

y1 y2,y

y=y1 ,+y2

y1

1. ratio KB y2 KA of the stiffness of the front and rear bearings, and l 2. factor lopt K I F= B c Kc It Fig. 5.8 Effect of l/c ratio on y1, y2 and y1 + y2 3EI c where Kc = 3 = bending stiffness of the cantilever, c Ic = average moment of inertia of the spindle over the cantilever length, and Il = average moment of inertia of the spindle over the supported length. a=

298

Machine Tool Design and Numerical Control

l

The variation of l opt as a function of F is depicted in Fig. 5.9 for a = 1 and a = 10. When the spindle is mounted on anti-friction bearings, an additional check is necessary to ensure that the constraint lopt ≥ l min = 2.5 is not violated because for l < 2.5 the bearing play leads to considerable radial run out of the spindle nose. An opposite constraint on maximum span stems from the requirement that for normal functioning of the spindle driving gear, the stiffness of the span should not be less than 25–50 kgf/micron. This constraint is expressed through the following relationship: l£ where

Dl4/3 1/3

3 a =10 2 a =1 1

0 0

Fig. 5.9

(5.27)

k

1

2

3

F

Effect of spindle and supports stiffness on lopt

Dl = average diameter of the supported length of the spindle k = 0.05 in the case of normal accuracy machine tools and 0.1 for precision machine tools

the span should satisfy the condition, y lmax £ 10–4l

(5.28)

and the maximum span length lmax should be limited by the above constraint. This constraint is based upon of the bearing gap.

5.4.4 Deflection due to Compliance of the Tapered Joint The spindle ends of most machine tools have a tapered hole for accommodating a centre (as in lathes) or d from the spindle axis (Fig. 5.10), where a force P is acting, may be represented as1 y3 = d + qd

(5.29)

a q

P0

d

t

d

Fig. 5.10 Schematic diagram of a tapered joint

Design of Spindles and Spindle Supports 299

where

d = displacement of the shank or centre at the edge of taper due to contact compliance and q = angle of slope of the shank or centre at the edge of the taper

If the manufacturing errors of the taper are ignored, d and q can be determined from the following expressions: 4 Db C1 d= (5.30) (bdC2 + C3) micron pD q=

4 Pb 2 C1 (2bdC4 + C2) pD

(5.31)

In these expressions, C1 = C1 = 0.03–0.06, while for the 7 : 24 taper C1 = 0.02 micron cm2/kgf C2, C3, C4 tapers C2 = C3 = C4 = 1, while for the 7 : 24 taper C2 = C3 = 1.35 and C4 = 1 1/4

b=

1 Ê ˆ cm–1 ÁË 2.3C D 4 ˜¯ 1

D and d are expressed in cm Generally, displacement d due to contact compliance can be ignored in comparison with the displacement due to bending of the shank or centre. The expression for y3 can then be written as y3 =

4 Pb 2 C1 (2bdC4 + C2)d micron pD

If it is assumed that 2bdC4 is much greater than C2 (which is true), the for y3 is obtained: Pd 2 3 y3 = 2.55 b C1C4 micron D

(5.32)

(5.33)

Keeping in mind the expression for b and substituting C4 = 1 and C1 = 0.02 micron◊cm2/kgf, the expression for displacement of a shank or centre mounted in a 7 : 24 taper can be written as follows: y3 =

Pd 2 2D4

micron

(5.34)

Consequently, the stiffness of the 7 : 24 taper can be written as K7 : 24 =

P 2 D 4 kgf = 2 y3 d micron

(5.35)

As the d/D ratio increases, the stiffness of the tapered joint decreases. Equation (5.34) has been derived on the assumption that manufacturing errors of the taper are negligible. As a matter of fact, any difference in the taper angle of the hole and shank severely affects the stiffness of the tapered joint. A difference of 30–40¢ in taper angles can easily reduce the stiffness of a 7 : 24 taper by 10–15 times. The difference in taper angles should normally not exceed 1¢ and the shank should have the larger taper angle.

300

Machine Tool Design and Numerical Control

raised by applying an axial tightening force on the centre or shank. The tightening force should be such as to produce a pressure of p = 15–25 kgf/cm2 on the tapered surfaces. The required magnitude of the tightening force can be determined from the expression, P0 = p ◊ p(D – t tan a) ◊ t tan (a + r)

(5.36)

where r = angle of friction, generally r = 12–14° D and t are expressed in cm In large machine tools mechanical tightening devices must be employed to 2. ensure that the larger tightening force is not transmitted to the thrust bearings of the spindle. The principle of operation of such a device may be explained with the help of Fig. 5.11.2 The device consists of bolt 2 screwed into tapered shank 1, spring-loaded levers 3, pulling rod 4, strong spring-loaded helical spring 5 and hydraulic cylinder 6 (Fig. 5.11a). When the hydraulic cylinder applies pressure, spring 5 is compressed, rod 4 occupies the lowered position and levers 3 are tilted inwards due to the pressure of spring 7 (Fig. 5.11b). The cutting tool (e.g., face milling cutter) assembled on the adaptor with a tapered shank is of springs 7 and open the levers, which engulf the bolt head. The shank now hangs freely, supported by the levers. For tightening the shank, the pressure of the hydraulic cylinder is released. Consequently, spring 5 begins to expand; pulling rod 4 upwards. The tapered shank is also pulled upwards and tightened inside the tapered hole. This position is shown in Fig. 5.11c. 7

6

5

4

3 2 1

(a)

(b)

Fig. 5.11

(c)

(d)

Schematic diagram of a quick acting mechanical tightening device

Design of Spindles and Spindle Supports 301

For unloosening the shank, rod 3 is pushed downwards by means of the hydraulic cylinder. In their downward movement the levers push at the face of the tapered shank and unloosen it. The downward movement of rod 3 is stopped immediately as the shank becomes free. The shank along with the cutting tool now again hangs freely, supported only by spring-loaded levers. It can be easily removed from the spindle by a slight downward pull of the hand. This mechanised device not only serves the two functions listed above but has the added advantage of bein numerically controlled milling machines with automatic tool change provision. In some designs, a collet is used instead of spring-loaded levers and a packet of leaf springs in place of spring 5.

Permissible Deflection and Design for Stiffness: 5.4.5 Additional Check for Strength y1, y2 and y3 from Eqs (5.18), (5.26) from (5.32), respectively, the y] depends upon the machining accuracy required of the machine tool being designed. In general, it should be less than one-third of the maximum permissible tolerance on radial run out of the spindle nose. Machining accuracy depends not only upon the radial stiffness of the spindle, but also upon its axial and torsional stiffness. The axial displacement of the spindle unit consists of the axial deformation of the spindle plus the deformation of the spindle thrust bearings. For a majority of machine tools, the latter component is predominant. Therefore, the axial run out of the spindle can be kept within reasonable limits only by a proper selection of the thrust bearings. removal operations, such as gear and thread cutting in which the feed and primary cutting motions are kinematically linked. The torsional deformation of the spindle unit consists of the deformation of the spindle design calculations. As stated earlier in Sec. 5.4, spindles are designed for stiffness, primarily radial. However, in heavily loaded spindles the stiffness design must be substantiated by a strength check against fatigue failure. The strength check requires that n ≥ nmin (5.37) where

n = factor of safety against fatigue failure nmin = minimum value of safety factor, generally equal to 1.3 to 1.5

For spindles subjected to combined bending and torsion, the factor of safety n is calculated from the expression, n = where

(1 - a 4 )d e3s -1 10 (aM b ) 2 + (bM t ) 2

a = di /de = ratio of the internal diameter to the external; de is in cm s–1 = endurance limit of the spindle material, kgf/cm2

(5.38)

302

Machine Tool Design and Numerical Control

Mb = mean value of the bending moment acting on the spindle, kgf ◊ cm Mt = mean value of the torque acting on the spindle, kgf ◊ cm a b a can be determined from the following expression: a = K s (1 + C) where

(5.39)

Ks Ks = 1.7–2 C = Mba /Mb = ratio of amplitude of the bending moment to its average value b can be calculated from the following formula: b =

where

s -1 + Kt Ct sy

(5.40)

sy = yield stress of the spindle material, kgf/cm2 kt = Ct = Mta /Mt = ratio of amplitude of torque to its average value

Kt = 1.7–2

The values of C and Ct in Eqs (5.39) and (5.40), respectively depend upon the machining conditions. For Mba = Mta = 0, and therefore, C = Ct = 0. In simple turning and drilling operations, distinguished by formation of continuous chips, the characteristic values are C = Ct= 0.1 – 0.2. The variation of bending moment and torque is greater in intermittent cutting operations such as milling; for such operations C = Ct= 0.3–0.5. The most important parameter in spindle design is the diameter of the front bearing journal. Typical values of this diameter are given in Table 5.3.3 Table 5.3

Diameter (in mm) of front bearing journal3 Machine Tool

Power rating kW Lathe

Milling

Cylindrical Grinding

1.5–2.5

60–80

50–90

—

2.5–3.5

70–90

60–90

50–60

3.5–5.5

70–105

60–95

55–70

5.5–7.0

95–130

75–100

70–80

7.5–11.0

110–145

90–105

75–90

11.0–14.5

140–165

100–115

75–100

14.5–18.0

150–190

—

90–100

18.0–22.0

220

—

105

22.0–30.0

230

—

105

Design of Spindles and Spindle Supports 303

1. Additional Supports The radial stiffness of a spindle can be improved by supporting it at more than two points. If an intermediate support is provided, the spindle constraints change from free supports to clamped supports, thereby increasing its stiffness. However, in multiple bearing spindles the support journals should be machined in one setting, as otherwise skewing and jamming of the spindle can occur due to large misalignment. 2. Location of Bearings and Drive Elements The bearings should be located as near as possible to the between the spindle nose and the point where the drive element transmits torque is subjected to bending and torsion. It is, therefore, desirable that the drive element transmitting maximum torque to the spindle should be located as near as possible to the front bearing. 3. Balancing The satisfactory performance of high-speed spindles is possible only if the spindle unit (after mounting the gears, clutches, etc.) is dynamically balanced before assembly in the spindle head or headstock. For general-purpose machine tools the permissible value of disbalance is 25 g ◊ cm at 2000 rpm.

5.5 compliance of the front and rear spindle supports. The rotational accuracy, which is one of the basic funcoperating conditions of spindles, anti-friction, hydrodynamic, hydrostatic, and lately, air-lubricated bearings are used as spindle supports in different machine tools. Irrespective of the type of bearing, the common re1. 2. 3. 4. 5.

guiding accuracy, ability to perform satisfactorily under various conditions of spindle operation, high stiffness, minimum heating, as this can lead to additional spindle deformation, and vibration stability, which is governed mainly by the damping.

Anti-friction bearings are one of the most widely standardised elements in industry and are manufactured on a mass scale throughout the world. The distinguishing features of anti-friction bearings as compared to sliding bearings are: 1. 2. 3. 4.

Low frictional moments and heat generation. Low starting resistance. High load capacity per unit width of the bearing. Easy maintenance and less consumption of lubricants.

bearings. Ball bearings are less prone to heating, and therefore, permit larger rotational speeds. They are also cheaper than roller bearings and less sensitive to small alignment errors. However, roller bearings have higher load capacity.

304

Machine Tool Design and Numerical Control

A machine tool spindle experiences both axial and radial loads. These loads can be balanced either by bearings that take up radial and axial loads separately or by bearings that take up both. In this regard, it should be recalled that 1. cylindrical roller bearings can take up only radial load, 2. simple radial ball bearings are basically meant for taking up radial loads, but are generally able to support axial loads too, 3. angular contact ball bearings can take up radial loads as well as relatively large axial loads in one direction, 4. taper roller bearings can take up large radial and axial loads with equal ease, 5. ball thrust bearings are useful for supporting purely axial loads only, their maximum rotational speed is just about 60% of a radial ball bearing of equal size, and 6. cylindrical roller thrust bearings are not recommended for general use on account of sliding between rollers and races. The number of possible combinations of various anti-friction bearings that can be employed in machine ever, the viability of each combination must be assessed vis-a-vis the following parameters: 1. 2. 3. 4. 5. 6. 7. 8.

Radial stiffness of the spindle unit. Axial stiffness of the spindle unit. Radial run out of the spindle. Axial run out of the spindle. Heat generation. Maximum permissible rotational speed, restricted by bearing wear and its heating. Thermal deformation of the spindle. Ease of manufacture and assembly of the spindle unit.

The relative performance and technological indices for eight combinations are given in Table 5.4. 4 It may be pointed that these combinations are typical of a majority of spindle units in small- and medium-size machine tools. Based upon Table 5.4, some important operational features of different types of bearings can be summed up as under: 1. Use of taper roller bearings considerably increases both radial as well as axial run outs. 2. Use of only cylindrical roller bearings at the front support greatly enhances axial thermal deformation of the spindle nose. 3. Use of angular contact ball bearings at the front support results in low heat generation. This coupled with a fairly satisfactory axial stiffness permits high rotational speeds. It is evident from Table 5.4 that no single combination of bearings is ideal for all performance indices. In on the functional accuracy of the spindle unit. Table 5.55 contains this information for some of the major groups of machine tools. The procedure for bearing selection is not described here and the reader should refer to a standard text on Machine Design or a manufacturer’s catalogue for this purpose. However, some important features related to the performance of bearings are discussed as follows.

Design of Spindles and Spindle Supports 305

Table 5.4

Relative comparison of performance and technological indices of spindle supports Heat general

S. No.

Sketch

Radial Axial Radial Axial Front Common Permissible stiffness stiffness run out run out bearing rpm

Axial Ease of thermal manufacture deformation and assembly

1.

1.0

1.0

1.0

1.0

1.0

1.0

1.0

1.0

A

2.

0.95

0.71

1.0

1.0

0.5

0.6

1.0

3.0

B

3.

1.0

3.0

1.0

1.0

1.2

1.15

0.75

1.0

A

4.

0.78

3.0

1.0

1.0

1.2

1.15

0.75

0.6

A

5.

1.0

2.5

1.0

1.0

0.5

0.75

0.8

3.0

B

6.

0.93

1.0

2.0

1.5

0.75

0.8

0.6

0.8

A

7.

1.05

1.0

2.0

1.5

1.4

1.4

0.6

0.8

A

8.

0.7

1.0

1.0

1.0

0.5

0.7

1.2

0.8

C

Notation: A—Very complicated

Cylindrical roller bearing

B—Complicated

Angular contact ball bearing

Taper roller bearing

Ball thrust bearing

C—Simple Table 5.5 S. No. Machine tool 1.

Radial Axial Radial Axial stiffness stiffness run out run out

Heat generation

Permissible Thermal rpm deformation

Lathes (a) small-sized

I

I

D

D

D

D

I

(b) medium-sized

D

I

D

D

I

I

I

(c) single spindle automatic

D

I

D

D

I

I

D

Contd.

306

Machine Tool Design and Numerical Control

Table 5.5 2.

3.

Contd.

Work head spindle of universal and internal grinding machine (a) small-sized

D

U

D

D

U

U

U

(b) medium-sized and heavy-duty

D

I

D

D

U

U

U

Universal milling machine

D

D

D

D

I

I

D

Notation: D—of decisive importance; I—of average importance; U—of relatively minor importance

5.5.1

Preloading of Anti-friction Bearings

as well as the stiffness (compliance) of the spindle supports. The discussion which follows will reveal the The variation of spindle deformation d due to a radial force P is depicted in Fig. 5.12. If the bearing is assembled with a clearance, a reversal of the direction of the applied force results in an abrupt change of deformation (Fig. 5.12a). This is highly undesirable from the point of view of machining accuracy. Bearings assembled with interference are free of this shortcoming and are distinguished by a smooth d-P curve (Fig 5.12b). It can be seen from Fig. 5.12 that the rate of deformation is initially high but later on decreases and at large loads becomes virtually constant. This is due to the fact that as the load increases, its distribution between the rolling members becomes more uniform. Since a larger number of rolling members support the stiffness of the spindle unit. d

d

P

(a)

Fig. 5.12

P

(b)

Effect of radial force on spindle deflection when (a) bearing is assembled with a clearance (b) bearing is assembled with an interference

Design of Spindles and Spindle Supports 307

Table 5.6 S. No.

Effect of bearing play on spindle unit stiffness6

Play of front bearing, microns

spindle nose, microns bending, microns

compliance of support, microns

1.

15 (clearance)

14

16

30

2.

– 5 (interference)

13

6

19

3.

– 15 (interference)

11

5

16

It is evident from Table 5.6 that the assembly of a bearing with a small interference sharply reduces supnot so perceptibly. The reason for such behaviour is that as the interference increases, the end constraints change from a simply supported beam to those of a clamped beam. This is equivalent to providing an addinose. A glance at Table 5.6 revels that increase of interference from 5 to 15 microns barely yields a 1-micron reduction of . However, a large interference is accompanied by excessive heating and also reduces the bearing life on account of large contact deformation. Obviously, optimum interference is one which precludes clearance but does not result in excessive heating of the bearing. Interference in the assembly of rolling elements is achieved by preloading them.

D

Preloading of a bearing involves relative axial displacement of the inner and outer races by a small amount (Fig. 5.13). The methods of applying preloading in radial and angular contact ball bearings that are generally mounted in pairs are shown in Fig. 5.14. A constant preloading is achieved either by grinding Fig. 5.13 Schematic diagram depicting off the faces of the inner races (Fig. 5.14a) or by inserting spacpreloading by relative axial ing rings of different widths between the inner and outer races displacement of the bearing races (Fig. 5.14b). If the bearing rotates at high rpm, the initial preload has a tendency to weaken. In such cases, especially when bearings are small, the preloading can be applied by means of springs which ensure a constant preload that can be accurately adjusted (Fig. 5.14c). This method is adopted in precision bearings.

308

Machine Tool Design and Numerical Control

a

a

a (a)

a (b)

(c)

Fig. 5.14 Methods of preloading ball bearings

Cylindrical double roller bearings are generally mounted on tapered journals. The preloading is obtained through axial displacement of the inner ring with an adjusting nut (Fig. 5.15). The utility of this arrangement can be considerably improved by using a split nut. As the bearing wears, the preset value of preloading decreases. When this occurs, the split nut is removed, slightly ground and mounted again. The initial preloading is then reinstated by additional axial displacement of the adjusting nut by a distance equal to the ground layer removed from the split nut. However, this procedure should be discontinued after a few regrindings of the split nut, because when the bearing wear becomes large, the roller length is partially in contact with the worn surface of the race way and partially with the unworn surface. This results in excessive heat generation and non uniform rotation. Axial displacement of the bearing race by the threaded nut does not provide uniform contact of the face and can result in deformation of the spindle. The non-uniformity of face contact may be somewhat reduced by inserting a sleeve between the nut and bearing as shown in Fig. 5.15b. The spherical inner race is seriously impeded by large static friction between the inner race bore and the spindle. It requires an axial force of 2000–3000 kg to shift the race of a small bearing about 100 mm in diameter. Application of ness) of the bearing race, resulting in poor rotational accuracy of the spindle. To reduce friction at the time of adjustment of bearing play, oil is force fed into the interface between the race bore and spindle. Split nut

(a)

Sleeve

Split nut

(b)

Fig. 5.15 Methods of preloading cylindrical, double roller bearings

Design of Spindles and Spindle Supports 309

The oil is supplied by an injector through a small oil hole in the spindle and an annular groove (Fig. 5.16). In this way, the required axial force for bearing adjustment can be reduced ten-fold. After completing the Annular grooves

Oil hole

Injector

Fig. 5.16

Schematic diagram describing the oiling arrangement to reduce friction at the bearing-spindle interface during bearing play adjustment

Taper roller bearings are preloaded by the methods shown in Fig. 5.17. In the method shown in Fig. 5.17a, the inner and outer races are axially displaced with the help of nuts. This method is applied only in nonprecision bearings because the axes of inner and outer races get skewed. In the Gamet bearing arrangement (Fig. 5.17b), the outer race is axially displaced by means of springs, whereas in the Timken bearing (Fig. 5.17c), this is achieved by supplying oil or air under controlled pressure.

(a)

(b)

Fig. 5.17 Methods of preloading taper roller bearings

(c)

310

Machine Tool Design and Numerical Control

5.6

SLIDING BEARINGS

by using needle roller bearings. However, wide application of needle bearings in machine tool spindles is an eccentric load. When there is a constraint of space, generally sliding bearings are preferred. Sliding bearings are used when 1. rotational speeds are so high that anti-friction bearings become uneconomical due to their short service life, 2. accuracy of spindle rotation is required to be very high, and 3. the bearings are subjected to shocks and vibrations; the inherent damping of sliding bearings is Sliding bearing is a general term that covers all bearings that do not use rollers or balls. These bearings operate under conditions of sliding friction between the bearing bore and spindle journal, which are separated

to-metal contact; friction conditions at the interface of mating surfaces are of semi-liquid type and these bearings are known as sleeve bearings,

or hydrostatic journal bearings. The Hersi–Shtribek diagram shown in Fig. 5.18 depicts f as a function of the quantity mw l= p where

m = absolute viscosity of the lubricant w = angular velocity of rotation of the journal p = average pressure per unit area of the supporting surface

At very low rotational speed when l < l 1, the lubricant film is extremely thin—of the order of 0.1

f 1

3 2

fmin l1

l2

l3

mw l= p

Fig. 5.18 Hersi-Shtribek diagram

not change with l. This region which lies to the left of point For higher values of l lying between l1 and l2, the friction conditions are of semi-liquid type and the l > l2, liquid friction conditions prevail and they represent the operating conditions of hydrodynamic bearings.

Design of Spindles and Spindle Supports 311

The sketch of a sliding journal bearing is shown in Fig. 5.19. Diameter d of the journal is always less than diameter D of the bearing. At zero rotational speed, the journal rests on the bearing and metal-to-metal contact takes place at point A (Fig. 5.19a). As the journal begins to rotate in the anti-clockwise direction, it tends to roll up the bearing surface due to the friction force and moves to a position B consists of two parts—a converging wedge above line BE and a diverging wedge below it. Owing to the hydrodynamic effect a positive pressure builds up in the converging wedge. This hydrodynamic force increases with increase of rotational speed and overcomes the frictional force. As a result, the point of contact moves to point C (Fig. 5.19c). As long as l < l2, the metal-to-metal contact at point C persists. However, when the rotational speed is such that l > l 2, the hydrodynamic force becomes large enough to lift the journal and a D (Fig. 5.19d). F

x

x

x

x x

e

x

d

x

D

x

E

D

C

B

A hmin (a)

(b)

(c)

(d)

Fig. 5.19 Schematic diagram describing the working principle of a sliding journal bearing

bearings have less friction than a full journal bearing, but can be used only where the load always acts in one direction.

+

+

+

+

120°

120° (a)

+ +

(b)

(c)

Fig. 5.20 Types of sliding journal bearings: (a) Full (b) 120° partial (c) 120° fitted

5.6.1 Sleeve Bearings Sleeve bearings are designed for wear resistance. The design conditions are p=

P £ [p] d ◊l

(5.41)

312

Machine Tool Design and Numerical Control

where

p P d l v [p] [pv]

pv £ [pv] = bearing pressure = load on the journal = diameter of the journal = length of the bearing = peripheral speed = permissible value of bearing pressure = permissible value of the product of bearing pressure and peripheral speed

(5.42)

The permissible values of [p] and [pv] vary in a wide range depending upon factors such as bearing material, sliding velocity, cooling and lubrication conditions, etc. These are given for some important sliding bearing materials in Table 5.7. Table 5.7

Permissible values of [p] and [pv] for some bearing materials

S. No.

Material

v, m/s

[p], kgf/cm2

1.

Grey iron

0.5

40

1.0

20

2.0

1.0

0.2

90

18

2.0

0.5

1.0

1.0

120

120

5.0

5.0

25

2.

Anti-friction cast iron

3.

[pv] kgf, m/cm2 ◊ s

4.

Bronze

3

50

100

5.

Aluminium bronze

4

150

150

6.

Tin bronze

10

150

150

7.

Graphite bronze (20–25% porosity)

0.2

60

0.4

10

8.

White metal

12

250

300

9.

Aluminium alloy

12

250

300

10.

Zinc alloy

10

120

120

The selection of the sliding bearing material is based upon the following considerations: 1. High wear resistance. 2. High fatigue strength.

Design of Spindles and Spindle Supports 313

3. 4. 5. 6. 7.

High compressive strength. High thermal conductivity. High conformability to accommodate spindle deformations and reduce edge pressures. High corrosion resistance. Low modulus of elasticity.

Typical values of p and v that occur in machine tool spindle units and the recommended sliding bearing materials are given in Table 5.8. A sliding bearing made from anti-friction cast iron has poor conformability, therefore, the spindle should have high stiffness to avoid large pressures. Copper alloy compositions are used in the form of bimetallic sleeves. A layer of approximately 1.0 mm thickness is deposited on a steel or cast iron sleeve by the centrifugal casting method. Porous graphite bronze bearings are employed at low sliding speeds under conditions of variable loading. Aluminium alloy is employed as a replacement for white metal and zinc alloy for babbits. Table 5.8

Recommended compositions of sliding bearing materials

S. No.

v, m/s

p, kgf/m2
hcr, the bearing operates under liquid-friction conditions and the design may be accepted. If, however, hmin < hcr, then corrective measures must be taken. These may be 1. reducing hcr 2. increasing hmin either by increasing c or by increasing the l/d ratio as this yields a smaller value of e. Design Problem 2 having l/d, y and m values as found above will be able to take the operating load. The recommended values of hmin for different conditions are given in Table 5.12. These values satisfy the condition hmin > hcr. Table 5.12 Recommended value of hmin for different conditions S. No.

Operating condition

1.

In order to pass dirt particles and prevent scoring

0.0026

2.

Finely bored bronze bearing

0.0026

3.

hmin, mm

0.0026 and aircraft engines

4.

Babbit bearings running at high speeds

0.02

5.

General recommendation

0.0015–0.0002 mm per mm of the bearing diameter

For the known value of c and the selected value of hmin, the eccentricity ratio is determined from Eq. (5.50). Next, for the particular values of l/d ratio and , CL is found from Table 5.11. ing capacity is less than the load acting on it, the former must be improved by taking appropriate corrective measures, which include 1. reduction of y, C L, 2. increase of m, i.e., application of a more viscous lubricant, considerations, and therefore, only l may be increased, thus providing a greater l/d ratio. The l/d ratio the l/d ratio.

Check for Thermal Equilibrium

The rotation of a journal in a hydrodynamic bearing is resisted by the lubricant. The viscous friction offered by the lubricant results in a frictional force which must be check for hydrodynamic bearings is tb £ [tb]

(5.51)

Design of Spindles and Spindle Supports 319

where

tb = temperature of the bearing [tb] = permissible value of bearing temperature; generally [tb] = 60–75°C

The heat is dissipated through the housing, bearing body and journal. If the thermal equilibrium between heat generated and heat dissipated is established such that the bearing temperature is less than the permissible, then the design is accepted. However, if the reverse takes place, then either the design parameters should

The heat generated can be determined from the expression, wd ¸ kgf ◊ m/s ÔÔ 2 ˝ 3600 f ◊ P ◊ w d = ◊ k cal/h Ô Ô˛ 427 2

W = f ◊P◊

where

(5.52)

P = load on the bearing w = rotational speed of the journal

f =

f=

for short bearing having for bearings having

p mw y p

p mw + 0.55 y h y p

l Êdˆ < 1, h = Á ˜ Ël¯ d

(5.53)

(5.54)

1.5

l > 1, h = 1 d

The heat dissipation through the body and shaft is assumed to be proportional to the free surface area of the bearing assembly. It is given by the expression, W1 = kA(tb – ta)kcal/h where

k A ta tb

(5.55)

2

= ◊ h ◊ °C = free surface area of the bearing assembly, m2 = ambient temperature, °C = bearing temperature, °C k can be found from the following expression: k = 6 + 10 v

(5.56)

320

where

Machine Tool Design and Numerical Control

v

rotational speed of the journal and has a minimum value of 1 m/s Generally, k, lies between 25–35 kcal/m2 ◊ h ◊ °C The free surface area of the bearing assembly depends upon the design and size of the bearing. On an average, it may be taken equal to 25d2 or 20dl. However, for very simple assemblies, it may reduce to 12dl, whereas for bearings mounted in high housings it may go up to 40dl. An additional area of (5–8)d2 per journal should be added to the above to account for heat dissipation through the journal. The lower value of 5d2 is recommended for small bearings (d < 100 mm), while the higher value of 8d 2 is for large bearings (d > 100 mm). For thermal equilibrium, Eqs (5.52) and (5.55) are equated and the bearing temperature tb is calculated. If tb is found to be less than the permissible value, there is no need for forced circulation of the lubricant. However, if tb exceeds the permissible value, forced lubricant circulation is essential to carry away the extra expression: W2 = 60CQg (t0 – ti), kcal/h where

(5.57)

C ◊ °C Q g = density of the lubricant, g/cm3 to = oil temperature after circulation through the bearing, °C ti = oil temperature when it enters the bearing, °C

is calculated from the following thermal equilibrium equation after substituting t b¢ = [tb] the expression for W1. W = W1 + W2 (5.58) ticles more than 2–3 microns in size. The oil is supplied at a pressure of 0.1–0.2 kgf/cm2 which is enough to

Finite Bearing

Design considerations have been discussed till now with reference to an idealized

leakage. In an actual bearing, there is always some leakage of the lubricant at the ends. This reduces the load lubricant must be supplied to compensate the leakage loss. An important question that arises in this context is where to introduce the lubricant. In Fig. 5.19d, the wedge below line DF is diverging, and therefore,

in this region. np.

l/d

Design of Spindles and Spindle Supports 321

ratio and e. It is plotted in Fig. 5.21 as a function of pd/l for different values of e. Also plotted on the same dianF). f can be determined as n nf = p (5.59) nF Thus the load capacity of a finite hydrodynamic journal bearing can be expressed as P =

mw y

2

l ◊ d ◊ CL ◊ nP kgf

hP

e = 0.2

hF

0.9

0.4

0.95

0.8

0.6

0.90

0.8

0.82

0.7 0.6

0.80

0.5 0.4

(5.60)

e = 0.6 0.3

0.8

0.2

bearing to determine the bearing temperature, heat generated in the lubricant can be determined as follows:

0.1

e = 0.2 0.4

wd W = nf ◊ f ◊ P ◊ 2

(5.61)

0 0

1.0

2.0

3.0

pd l

nP Fig. 5.21 Leakage factors for load and friction nF have been presented force in 120° partial journal bearings in Fig. 5.21 only for a 120° partial bearing, these values can be used with satisfactory approximation for journal bearings having different angles.

Multiple-wedge Bearings

The hydrodynamic journal bearing discussed till now has only one converging wedge and is called single-wedge bearing. One of the major shortcomings of these bearings is that the position of the journal inside the bearing b varies with the load and rotational speed. This positional change leads to unstable running of the journal, which is impermissible in spindles of high accuracy and precision machine tools. These requirements are met in bearings having several converging wedges B which uniformly surround the journal on all sides and ensure that during rotation the position of the journal changes very little or not at all. These bearings are known as multiple-wedge bearings. Figure 5.22 shows a self-adjusting tilting-pad journal bearing in which the wedges are formed due to tilt of the pads. These bearings generally have three to eight wedges. The recommended values and expressions for calculating the Fig. 5.22 Schematic diagram of a multiple-wedge bearing design parameters of multiple-wedge bearings are discussed be7 low.

1. Ratio of pad length l to journal diameter d l = 0.75d for grinding machines l = (0.85–0.9) for high precision lathes and boring machines

322

Machine Tool Design and Numerical Control

2. Pad width B The pad width should be such that it subtends an angle of b = 50–60° at the centre. It should generally satisfy the condition B = (0.6 – 0.8 )l 3. Radial clearance c

The minimum clearance between the pad and journal depends upon their surface Rz = 0.8–1.6 microns), then For d = 30–50 mm, For d = 50–100 mm, For d = 100 mm,

c = 3–5 micron c = 5–10 micron c = 10–15 micron

4. Viscosity of lubricant In multiple-wedge journal bearings, low viscosity oils are used. The recommended kinematic viscosity of the oil is 4–8 cS at 50°C, which corresponds to an absolute viscosity of 4–8 cP. It because the resultant mixture has poor lubricating properties and gradually becomes more viscous due to 5. Bearing capacity The bearing capacity is calculated approximately, assuming each wedge to be a slider bearing. The load capacity per wedge of an unloaded bearing is given by the expression, P0 = 10–2 where

mn dB 2 ◊ l (2c) 2

◊ CL2, kgf

(5.62)

m = absolute viscosity of the lubricant, cP n = rpm of the journal CL2 =

1.25 1 + ( B/l ) 2

d, B, l are in cm; c is in microns It is evident from Eq. (5.72) that considerable hydrodynamic force is developed in each wedge even if there is no load on the bearing. These forces balance each other and tend to retain the rotating journal eccentricity e. External load P and eccentricity e are related through the expression, 1 1 ˘ È P = P0 Í ˙ 2 (1 + e ) 2 ˚ Î (1 - 0.5e ) where

(5.63)

e = e/c is the eccentricity ratio.

Equations (5.62) and (5.63) can be used for solving the two general design problems which have been earlier discussed with reference to single-wedge bearings. As a rule, multiple-wedge hydrodynamic journal bearings are provided with forced circulation of the lubricant. Assuming that heat dissipation through the bearing assembly and housing is negligible as compared to the heat carried away by the circulating lubricant, the increase in temperature of the lubricant can be calculated from the expression, Dt =

860N F °C Cg Q

(5.64)

Design of Spindles and Spindle Supports 323

where

C g Q NF

◊ °C.

= = density of the lubricant, g/cm3 = = power loss due to friction, kW

The frictional power loss can be found from the following relationship: NF = 7.5 ¥ 10–10 m n2d3z kW where

(5.65)

m = absolute viscosity of the lubricant, cP z = number of wedges d = diameter of the journal, cm

5.6.3 Hydrostatic Journal Bearings of the order of 0.1–0.2 microns. Besides, the stiffness of hydrodynamic bearings changes with lubricant viscosity, temperature and rotational speed of the journal. These shortcomings are absent in hydrostatic journal bearings. The basic features and principle of operation of hydrostatic journal bearings are essentially similar to those of hydrostatic pad bearings discussed in Sec. 4.4.1. Hydrostatic journal bearings may be single-pad (Fig. 5.23a), multiple-pad (Fig. 5.23b) and multiple-recess (Fig. 5.23c) type. Constant pressure manifold Restrictor 2

1

3

4 Pad domain (a)

(b)

(c)

Fig. 5.23 Types of hydrostatic bearings: (a) Single-pad (b) Multiple-pad (c) Multiple-recess

The single-pad journal bearing has less than 180° included angle and normally supports only unidirectional load akin to partial hydrodynamic bearings. Multiple-pad bearings are employed when the load to be

324

Machine Tool Design and Numerical Control

supported is not unidirectional and varies as much as ± 180°, e.g., an oscillating load, a reversing load, etc. In multiple-pad bearings, the number and location of pads are dictated by the angular variation of the applied radial load. For instance, a single-pad having the included angle > 60° is adequate to support a purely unidirectional load. Two pads at an angle of 120° can be employed when the load varies within a maximum of 80°. If a purely reversing radial load acts on the journal, a two-pad bearing having pads 180° apart can be used. For the general case in which load variation can assume any arbitrary value, we can use multiple-pad bearings having three or more pads distributed round the journal. Single-pad and multiple-pad bearings are rarely used in machine tool spindles due to the fact that there is considerable variation in direction as well as the magnitude of the cutting force depending upon cutting conditions, type of machining operation and other factors. Multiple-recess bearings are akin to full (360°) hydrodynamic journal bearings and they can support all types of radial loads, including reversing and rotating loads. In common practise, the term hydrostatic journal bearing is understood to stand for a multiple-recess journal bearing and, therefore, the simpler term will be used in all subsequent discussion. The main difference between a multiple-pad and multiple-recess bearing is the absence of pressure-relieving grooves between the pads in the latter. Due to this, the pressure between

Figure 5.23c shows a full hydrostatic journal bearing having four oil pockets located at 90° to each other. The lubricant is supplied to each pocket through an individual restrictor. When there is no load on the journal, the latter occupies a concentric position. The clearances between the journal and bearing are equal and so are the pressures in the recesses. Now assume that an external load acts on the journal, moving it towards recess

p across the restrictor, resulting in higher recess pressure which tends to restore the journal to the initial

hydrostatic bearing pad when the journal is in a concentric position is known as pad preload force and it is given by the expression, where

Ppr = CL ◊ A ◊ p0

(5.66)

p0 h3 3 m /s mCF¢

(5.67)

CL = A = pad area, m2 p0 = pocket pressure, kgf/m2

Flow through the pocket is Qpr = where

h= m = absolute viscosity of the lubricant, kgf ◊ s/m2

Design of Spindles and Spindle Supports 325

1 CF¢ 8

Rectangular

pads with rectangular pockets are of two types: 1. Pads with equal sill lengths. 2. Pads with sill lengths in axial and circumferential directions proportional to the pad length in the same directions.

Strictly speaking, the multiple-recess hydrostatic journal bearings do not have individual pads. Pad coefcircumferential direction. Thus, in Eq. (5.66), A is taken as the area of the domain of each pocket. When a ra-

the practical hydrostatic journal bearing given in Figs 5.24–5.28 at supply pressure, pp = 2p0

(5.68)

1/CF¢

CL R

6.0

1.0 r

d

CL

0.8

5.0

0.6

4.0

0.4

3.0 1/C¢F

0.2 0.0 0.0

0.2

0.4

2.0 0.6

0.8

r 1.0 R

Fig. 5.24 Load and flow coefficients for a cylindrical pad with circular pocket

326

Machine Tool Design and Numerical Control

l/L

1/C¢F 20.0 I

L

⎫ ⎬ 0.9 ⎭

10.0 8.0

⎫ ⎬ 0.7 ⎭ ⎫ ⎬ 0.5 ⎭ ⎫ ⎬ 0.1 ⎭

6.0 4.0

d

q1

2.0 2q2h1

kgf mm

Re,

h=h3 >h2 >h1

kgf mm

Re,

kgf mm

(b)

(a)

Fig. 6.26 Polar plot of cutting process at the tool flank

the resulting transfer function is obtained as WCP(p) = WF (p) + WFL(p)

(6.40)

The resulting dynamic characteristic of the cutting process is obtained as the vector sum of the two characteristics represented by Eqs (6.35) and (6.39). The polar plots of the cutting process in accordance with In the frequency range 0–w 01, (curve h1), 0–w 02 (curve h2) and 0–w 03 (curve h3) the force lags behind the displacement whereas, beyond frequencies w 01, w 02 and w 03 of the respective curves, the force leads the displacement. Let us again reiterate that a phase lag is conducive to aggravation of vibrations, whereas a phase lead depicts damping, which in effect suppresses the vibrations. It is evident from Fig. 6.27 that at small values

lm, kgf mm

h1

h2 h3

h4 h=h1 t 02, the system remains stable till the rotation of point w by 2p brings it back to the starting position and the time-delay period becomes equal to t l2, i.e., for t 02 < t < t12, the system is stable. It may be thus seen that this is a case of conditional stability which means that for some values of t, i.e., nz, the system is dynamically stable, while for others it is unstable. The situation is similar to the one involving machining with b > blim min which was discussed in Sec. 6.6.3 with reference to primary chatter (see Fig. 6.37). The difference between conditional stability in primary and regenerative chatter is that in the latter case there is a succession of lobes such that the rpm ranges of relatively high dynamic stability alternate with those of low dynamic stability (Fig. 6.48). The width of the rpm range in which the system is stable can be determined approximately from the expression, Dnz = where

30 (w 1 – w 2) pq

(6.86) y

w 1, w 2 = circular frequencies at points where the W CL characteristic intersects the circle of radius unity q = whole number, i.e., 0, 1, 2, …

Dynamics of Machine Tools 377

Unstable zones blim

Stable zones

n, rpm

Fig. 6.48 Variation of limiting width of cut with rpm in case of regenerative chatter

It is evident from Eq. (6.86) that as nz decreases, i.e., q increases, the width of the stable rpm range is reduced. This is also obvious from Fig. 6.48. It may be pointed out that the asymptote to the lobes represents the boundary of unconditional stability. It should be interesting to compare the dynamic stability of the EES-cutting process system under conditions of regenerative and primary chatter. For unconditional stability against primary chatter, the WOL characteristic should intersect the –Re axis to the right of (–1, 0). However, for unconditional stability against regenerative chatter, the WOL characteristic must pass to the right of point (– 0.5, 0) (Fig. 6.45b). It may, therefore be concluded that all other factors remaining the same, the blim min in regenerative chatter is at least two times less than blim min in primary chatter. t From the procedure of constructing the W CL characteristic, it is obvious that chatter occurs only when the time-delay period is such that the point of the maximum radius vector lying close to frequency w n after rotation through f = tw n passes through point (–1, 0). In other words, this implies that the chatter frequency is t the frequency of that point of the W CL characteristic where it intersects the –Re axis. As the W tCL characteristic can cut the –Re axis a number of times for successive values of q = 0, 1, 2, …, the system is prone to chatter at a number of frequencies. The variation of the chatter frequency with the change of rpm can be established

nz =

30 wn pq

w sec–1

wherefrom, pq w n = nz ◊ 30

Unstable zones

w2

(6.87)

It is evident that in each unstable zone the chatter frequency increases with increase of rpm. This result is depicted in Fig. 6.49. It may be added in conclusion that chatter amplitudes also change with rpm and are

wn w1 Stable zones

n,rpm

Fig. 6.49 Variation of chatter frequency with rpm

378

Machine Tool Design and Numerical Control

maximum when the chatter frequency lies at the middle of its range of variation. The magnitude of the chatter amplitudes is determined by the non-linearity of EES.

6.7 Forced vibrations are caused by perturbances acting on EES and the cutting process (Fig. 6.2). These vibrations affect the machining accuracy, productivity and life of the cutting tool. The effect on productivity is dictated by the interaction of forced vibrations with chatter and the resulting change in the value of blim. In this section, we will mainly deal with this interaction as it is essential for proper assessment of the dynamic stability of the EES-cutting process system. The effect of forced vibrations on machining accuracy will also be discussed. The effect of forced vibrations on tool life is basically important to the production engineer and cutting tool designer. As such, it does not fall in the purview of this book. However, some light will be thrown on this aspect to help the machine tool designer get a better insight into the problem of forced vibrations in its totality. The forced vibrations affect the tool life through a change of the effective cutting speed. Let the amplitude of the vibratory displacement be A. The amplitude of the velocity of vibration will, therefore, be A ◊ w, where w is the circular frequency of vibrations. If the nominal cutting speed is v, the effective cutting speed will vary between v – Aw and v + Aw during each half cycle of vibration. It is obvious from the preceding discussion that high-frequency vibrations will have greater effect on tool life than low-frequency vibrations. It is often held, though somewhat erroneously in respect of machine tools, that frequencies of forced vibrations are equal to excitation frequencies and their harmonics. This is not really true. The complex nature of various perturbances in a running machine tool and their distribution over a fairly wide frequency range results in a wide-band type of excitation. The vibration spectrum, therefore, includes not only vibrations of the dominant excitation frequency and its harmonics, but also vibrations occurring with natural frequencies of the elements of EES. Thus, although the dominant excitation frequency may be quite low, the forced frequencies are determined from the vibration-spectrum record of the particular machine tool in operation. For instance, Fig 6.507 shows the vibration record of a face-milling operation. It may be pointed out that the frequencies 400 Hz, 10 kHz, and 15 kHz are natural frequencies of the spindle unit, milling cutter blade and cutting bit, respectively. A

0

0.4

6.0

10

11.4

Fig. 6.50 Vibration spectrum of the face-milling operation

15.5

f, kHz

Dynamics of Machine Tools 379

6.7.1 The sources of perturbance acting on the cutting process include the following: 1. Variable undeformed chip thickness due to inherent characteristics of certain machining operations, such as milling and broaching. 2. Variable undeformed chip thickness due to machining of a wave formed during a preceding pass or cut. 3. Periodic variation of the undeformed chip thickness while machining an eccentric workpiece. 4. Variable undeformed chip thickness at the time when the cutting tool enters the workpiece or comes out of it. 5. Change in the cutting speed during operations such as facing. 6. Change in kinematic cutting angles with change in the direction of feed. Most of the sources of perturbance of the cutting force are an unavoidable corollary of machining processes and cannot be eliminated. Their effect must, therefore, be taken into account. Of the large number of perturbances of the cutting process, the following two cases are of particular interest: 1. The dynamic perturbances are absent, i.e., y(t) = const, f (t) = const 2. The perturbance on EES is absent, i.e., f (t) = const but a dynamic perturbance y(t) involving periodic variation of the undeformed chip thickness is acting on the cutting process. Case I y(t) = const, f (t) = const Consider a parting operation on a lathe machine (Fig. 6.51). Suppose the machine is set for removing constant undeformed chip thickness a0, y of EES. In accordance with the objectives described above, let us analyse the machining accuracy in this case. The question of interaction between the static change y and blim does not arise in this particular case because a static change of the cutting force is incapable of causing chatter vibrations. From Eq. (6.7), y

W CL (p) =

a0 y

y WOL ( p ) = y (t ) 1 + WOL ( p )

at p = iw = 0, WOL = CEES ◊ ks, and the above expression may be y

W CL =

y CEES ◊ k s = a0 1 + CEES ◊ k s

Fig. 6.51

Schematic diagram of a parting operation

(6.88)

Machine Tool Design and Numerical Control

The ratio y/a0 is an index of machining accuracy as it shows to what extent an error of the workpiece surface is reproduced on the machined surface. The relationship between yla0 and CEES ◊ ks is shown in Fig. 6.52. The curve consists of two distinct portions:

y/a0 3

Undercutting

380

2

1. When CEES > 0; in this region any increase of CEES leads to an increase of deformation of the 1 system resulting in poorer machining accuracy. 2. When CEES < 0–; in this case the deformation changes sign and the cutting tool has a tendency CEES◊ks –1 1 0 to dig into the workpiece, thus removing an undeformed chip thickness greater than Fig. 6.52 Variation of static machining accuracy as a function of ks ◊ CEES intended. Qualitatively, the surface accuracy worsens as CEES increases. If CEES ◊ ks = –1, i.e., the curve approaches the asymptote passing through point (–1, 0) and the system can be considered statically unstable. In such a system, a random disturbance leads to a continuous penetration of the cutting tool into the workpiece until the tool breaks. This phenomenon is known as undercutting. The stability of a statically unstable system can be improved either by reducing CEES, i.e., improving the stiffness of EES, or by reducing ks, i.e., by reducing the undeformed chip width. The lathe operator often takes recourse to the latter measure in parting and grooving operations. For instance, in making a wide groove (of say 8 mm), undercutting can be avoided by using a thin tool (of say 4 mm) and completing the operation in two passes instead of doing the same operation in a single pass using an 8-mm wide tool. Case II f (t) = const, y(t) π const Let us assume that the dynamic perturbance on the cutting process is of the type y(t) = Ay sin wt

(6.89)

Such an expression is characteristic of the perturbance while machining an eccentric workpiece or a workpiece on which a vibratory wave has been left by the preceding cut. written as A yy = Ay where

AOL 1 + AOL

(6.90)

AOL = radius vector of the WOL characteristic Ay = amplitude of the harmonic perturbance acting on the cutting process

Once again, we would be interested in the ratio A yy /Ay which is an index of the machining accuracy. By substituting 1 + AOL = Ad, Eq. (6.90) can be written as A yy =

Ay Ad / AOL

(6.91)

Let us analyse the machining accuracy under forced vibrations of this type. Three distinct situations are possible:

Dynamics of Machine Tools 381

lm lm5 1. Ad /AOL < 1: This occurs when the point of frequency w of CEES◊ks the WOL characteristic lies to the left of line Im5 (see Fig. 6.53). In this situation, the amplitude of forced vibrations (–1,io) is greater than the amplitude of the vibratory wave being Re removed. Hence, the accuracy of the machined workpiece A d AOL worsens with each successive pass. If the undeformed chip width is increased, the WOL characteristic will expand, the w Ad /AOL value will further decrease, and the machining accuracy will worsen. wi 2. Ad /AOL > 1: This occurs when the point of frequency w of the WOL characteristic lies to the right of Im5 line. In this case, the amplitude of the vibratory wave reduces with Fig. 6.53 Polar plot of open-loop system each pass. When the undeformed chip width b is increased, transfer function WOL the ratio Ad /AOL increases, thus contributing to a reduction of the amplitude of the vibratory wave. It may be thus seen that in case of forced vibrations due to a low-frequency perturbance (w < w i 3. Ad /AOL = 1: This occurs at the point of intersection of the WOL characteristic with the Im5 line. In this situation, the amplitude of the vibratory wave on the machined surface remains unchanged.

If the above results are interpreted in terms of the dynamic stability of the EES-cutting process system, it may be concluded that 1. the system is unstable when Ad/AOL < 1, 2. the system is stable when Ad /AOL > 1, and 3. the system is on the threshold of stability when Ad /AOL = 1. It may be seen that the stability of the system is governed by the Im5, line as was the case with regenerative chatter. This is not surprising as the effect of harmonic perturbance is the same as that of machining a vibratory wave. However, the fact that identical conclusions have been drawn from two separate analysis points to the soundness of the approach. The reader by now would be able to fully appreciate that an increase in stability of the open-loop system will result in a contraction of the WOL characteristic. In the ultimate limit, the WOL characteristic can be reduced to a point, in which case the ratio Ad /AOL . The implication of this conclusion is that forced vibrations due to perturbance of the cutting process can be reduced to any desired extent and virtually eliminated by improving the dynamic stability of the EES-cutting process system. Thus, if a system is dynamically sound, the machine continues to run smoothly even in the case of resonance, e.g., with the impacting frequency in milling operations.

6.7.2 The sources of perturbance acting on EES include the following: 1. Periodic forces due to errors in gears, splines and keys, belt joints, inaccuracy of bearing racings, etc., in short, these perturbances arise due to deviation of the geometry and kinematic motion from the 2. Periodic forces due to disbalanced rotating parts.

382

Machine Tool Design and Numerical Control

3. Pulsating forces from the hydraulic circuit and coolant pumps. 4. Periodic forces and shocks transmitted to the machine tool through the foundation. 5. Variable inertial force at the time of reversal of the cutting tool or workpiece in machine tools with a translatory primary cutting motion. All the above mentioned perturbances have a harmful effect on the quality of machining and the life of the cutting tool. The designer should therefore constantly strive to suppress, and if possible, completely eliminate these perturbances. In this case, y(t) = const, f(t) π const Let the perturbance f(t) be of the type f (t) = Af sin wt Such a perturbance is a typical source for forced vibrations caused by manufacturing defects of the machine tool, vibrations transmitted through the foundation, etc. The response of the closed-loop system to perturbance f (t) f

WCL =

f

WEES y = f (t ) 1 + WOL

f

where WEES represents the transfer function of EES when P = 0 (Fig. 6.8). The above equation yields the following relationship between amplitudes f

Ay = Af where

Xin

f

f A ◊ Af AEES = EES 1 + AOL Ad

(6.92)

Afsinwt

f

A y = amplitude of the wave formed on the machined surface

Xout

EES

The perturbance and EES can be considered as on open-loop system representing the machine tool during its idle run (Fig. Fig. 6.54 Block diagram of machining 6.54). Consequently, f system during idle run Af ◊ A EES = A ir (6.93) where

Air is the relative amplitude of forced vibrations between the workpiece and cutting tool during idle run, measured in the direction of the normal to the surface of cut.

Equation (6.92) can now be written as follows: A A yf = ir Ad

lm CEES ◊ks (–1, io)

(6.94)

In analysing the machining accuracy under forced vibrations of this type we may come across the following three situations:

Re Ad

AOL

w

wi 1. Ad < 1: This occurs when the point of frequency w of the WOL characteristic lies inside the circle of unit radius drawn with its centre at point (–1, 0) (Fig. 6.55). In this Fig. 6.55 Polar plot of open-loop system case, the amplitude of the vibratory wave formed during transfer function WOL

Dynamics of Machine Tools 383

cutting is greater than Air. An increase in the undeformed chip width b, thus not only reduces the dynamic stability of the system, but also adversely affects the machining accuracy. 2. Ad > 1: This occurs when the point of frequency w of the WOL characteristic lies outside the circle. In this case, the amplitude of the vibratory wave formed during cutting is less than Air and the cutting process provides a damping effect. An increase of b in this case leads to a higher value of Ad. Thus in the case of forced vibrations due to force perturbance of frequency w < w i, an increase of b leads to better machining accuracy, although the stability of the system is reduced. 3. Ad = 1: This occurs at the point of intersection of the WOL characteristic with the circle of radius unity. Obviously, the amplitude of the vibratory wave during cutting is, in this case, the same as the relative amplitude of machine vibration during its idle run. A glance at Case 2 (Ad /AOL > 1 when the y(t) perturbance is considered) provides a useful, though somewhat paradoxical conclusion that machining accuracy can be improved by increasing b, i.e., the depth of cut in the case of forced vibrations occurring with a low frequency, provided, of course, that the increased value b is less than blim and does not result in chatter. As already explained, the increase in stability of the open-loop system of Fig. 6.3 reduces the size of the WOL characteristic, which in the limit reduces to a point. In this extreme situation 1 Ad = =1 1 + AOL Consequently from Eq. (6.79), A fy = A ir (6.95) A narrow WOL characteristic is typical of the working of precision machine tools as the depth of cut in precision operations is, as a rule, very small. It, therefore, transpires that in precision machine tool operation, forced vibrations cannot be eliminated completely even if the depth of cut is reduced to zero because even in this limiting case (Ad = 1), idle-run vibrations will be faithfully transferred onto the machined surface. The only possible method of improving the machining accuracy of precision machine tools is to reduce the level of idle-run vibrations by improving the quality of manufacture and assembly of components, such as gears, splines, keys, etc.

6.1 Determine the natural frequency of the system shown in Fig. 6.56. Given: K1 = 10 kgf/cm, K2 = 8 kgf/cm, K3 = K4 = 15 kgf/cm and m introduced under mass m? 6.2 Write the equations of motion and determine the natural frequencies of the two-degree of freedom vibration system shown in Fig. 6.57. (Hint: See the chapter on the free vibration of two-degree of freedom system in Refs 8 and 9.) 6.3 Figure 6.58 represents a two-degree of freedom system in normal coordinates. Draw the polar plot of the system in the frequency range 5–150 Hz, assuming that the resultant transfer function of the system is the vector sum of the transfer functions in normal coordinates. Given:

K1

K2

m

K3

K4

Fig. 6.56 Spring-mass system

384

Machine Tool Design and Numerical Control

K = 100 kgf/cm, m = 20 kg, l = 5 cm, l0 = 2 cm, d = 5 cm, P = 5 kgf. (Hint: See the section on coordinate coupling in Refs 8 and 9.)

Fig. 6.57 Two-degree of freedom system

Fig. 6.58

Two-degree of freedom system in normal coordinates

6.4 For the system shown in Fig. 6.57, transform the equations of motion into equations in principal or normal coordinates. If an exciting force P1 acts on mass m1, determine the transfer function of the system. (Hint: See the chapter on many degrees of freedom systems in Refs 8 and 9.) 6.5 Determine the transfer function of the system shown in Fig. 6.57 without decoupling the equations of motion. Assume that mass m1 is being acted upon by a force P1. 6.6 An orthogonal turning operation is being carried out by a sharp tool under the following conditions.

m/n r = 180 kgf/mm. 6.7 For the data given in the previous problem, draw the polar plot of the cutting process in the frequency KFL = 100 kgf/mm (see Eq. (6.37).) 6.8 A machine tool elastic system can be represented as a single degree of freedom system having stiffness = 9 ¥ 104 ◊ s/cm and mass = 20 ¥ 103 kgf ◊ s2/cm. The cutting above. Determine blim for the case of primary chatter. If the machining is done with an undeformed chip width b which exceeds blim by 25%, determine the range of cutting speed in which the cutting process will become unstable. 6.9 For the data of problem 6.8, plot the stability chart (i.e., chart showing variation of blim as a function of rpm) for the case of regenerative chatter during a single point turning operation. (Hint: 1. Select a value of b. y 2. Obtain the W CL plot as in Fig. 5.47a. 3. Determine the range of f = tw in which the system is unstable; from this determine the corresponding range of t and n = 60t. 4. Repeat step 3 for tw + p, 2w + 2p, etc. 5. Repeat steps 1 and 4 for another value of b.)

Dynamics of Machine Tools 385

1. Kudinov, VA, Theory of Chatter During Cutting (Friction). Peredovaya Tekhnologiya Mashinostroeniya, Academy of Sciences, USSR, 1955. 2. Kudinov, VA, et al., Vibration Behaviour Calculation of a Centre Lathe on Computer, ONTI, Moscow 1963. 3. Tlusty, J and M Polacek, “The stability of the machine tool against self-excited vibration machining,” Proceedings Int. Res. in Prod. Engg., Pittsburg, 1963. 4. Kudinov, VA, Dynamics of Machine Tools, Mashinostroenie Publishers, Moscow, 1967. 5. Raven, FH Automatic Control Engineering, McGraw-Hill Kogakusha, Tokyo, 1968, p. 57. 6. Tobias, FA and W Fishwick, “Theory of regenerative machine tool chatter”, The Engineer, Vol. 205, London, 1958, p. 199. 7. Mehta, NK, et al., “An investigation of tool wear and the vibration spectrum in milling”, Wear, Vol. 91, 1983, p. 219. 8. Grover, GK, Mechanical Vibrations, Nem Chand Bros., Roorkee, 1977. 9. Thomson, WT, Theory of Vibration, Prentice-Hall, New Delhi, 1975.

7 7.1

CONTROL SYSTEMS IN MACHINE TOOLS

FUNCTIONS, REQUIREMENTS AND CLASSIFICATION

The control systems of a machine tool are meant to generate controlling movements which are essential for A control system, in general, consists of

is generally a set of elements and mechanisms which appropriately modify the initial controlling

The two main functions of machine tool control systems are: C P

The critical importance of control systems in the overall machine tool performance dictates the following requirements in their design: R

A

7.2

CONTROL SYSTEMS FOR CHANGING SPEEDS AND FEEDS

Mechanical devices for stepped regulation of speeds and feeds in machine tools were discussed in Secs

Control Systems in Machine Tools 387

C

C

S S S

7.2.1

Speed and Feed Changing Systems with Simple Centralised Control

a and b

388

Machine Tool Design and Numerical Control

5

6

7

4 a b 8 3 2 1

9 10

11

Fig. 7.1 Speed-changing mechanism of a jig boring machine

a and b,

c

Control Systems in Machine Tools 389

4

2

3

1

z75

5 (z=75) (z=25) 7 6

20

b

8

9

10

13

12

11

a z14 14 45 31.5

8

.4 10

22

c z42

19

z14

5 12

250

15

5

16

6

18 17

z14 35

z14

180

Fig. 7.2 Feed-changing mechanism of a horizontal milling machine

a, b, c and d. x, y and z are

390

Machine Tool Design and Numerical Control

Drive I a b II

III

c

IV

View facing arrow A d (a)

(b)

1

3

(c)

(d) 2

4

Fig. 7.3(a) Single-lever control system of a milling machine speed box

Control Systems in Machine Tools 391

Cluster gear position a

b

c

d

L.H.

0

R.H.

R.H.

R.H.

0

’’

’’

a

Development of drum grooves controlling the cluster gears c b d

K 0

L.H.

’’

’’

0

R.H.

’’

’’

L.H.

0

L.H.

’’

R.H.

0

’’

’’

0

L.H.

’’

’’

0

R.H.

’’

’’

L.H.

0

R.H.

L.H.

R.H.

0

’’

’’

0

L.H.

’’

’’

0

R.H.

’’

’’

L.H.

0

L.H.

’’

R.H.

0

’’

’’

0

L.H.

’’

’’

0

R.H.

’’

’’

pD

z

x

y

Fig. 7.3(b) Development of groove profiles on drum cams 1 and 2

392

Machine Tool Design and Numerical Control

6

7 8 5

4

3

1

2

9

10

Fig. 7.4 Four-step joystick lever for changing speeds

7.2.2 Speed and Feed Changing Systems with Preselective Control

a preselective control system of speed and feed regulation on the machine

Then when the operation is over,

Control Systems in Machine Tools 393

F1 and F2

Lines for lubrication

V2

M2 V0 R2 Speed selector

F3

F4

F2

R1

M1

F1

Fig. 7.5 Schematic diagram of hydraulic preselective control system

F1 and F2 F3 and F occupy

394

Machine Tool Design and Numerical Control

Stop valve Direction– control valve

Discharge line

Fig. 7.6 Schematic diagram of a two-position shifting device

F3 and F which

Control Systems in Machine Tools 395

2

1

Plungers

Fig. 7.7 Schematic diagram of a three-position shifting device 2 2

and through valve V2 V

2

M1,

M2 R1 R2

7.2.3 Speed and Feed Changing Systems with Selective Control

L E

systems which permit changing of feed and speed from one value to another without having to pass through selective control systems.

396

Machine Tool Design and Numerical Control

x

x

x

2

4

3

1 x

Fig. 7.8 Single-lever selective speed-changing mechanism for four-speed horizontal boring machine

7.3

CONTROL SYSTEMS FOR EXECUTING FORMING AND AUXILIARY MOTIONS

Control Systems in Machine Tools 397

7.4

MANUAL CONTROL SYSTEMS

Fig. 7.9

main functions: S E

Schematic diagram depicting the closed-loop nature of manmachine interaction

398

Machine Tool Design and Numerical Control

ergonomics ergos

nomos

divided into the following two categories: C anthropometry functional anatomy or more popularly, biomechanics A

7.4.1 Anthropometric and Functional Anatomy Data

Fig. 7.10

H H/1.06

H/3.15

M - 144.8 cm F - 131.9 cm M 102.6 cm Hip Height F 100.1 cm

.08

H/ 7

5

H/ 3.92

H/1.90

H/ 4.63

F - 158.8 m

7 H/ 5.9

H/ 9.8

Shoulder height

Body Height

M - 169.6 m

Control Systems in Machine Tools 399

(a) Body dimensions of an average male and female (b) Dimensions of human limbs in proportion to body height 1

2

600

3

1680 1300 800

1500 1000

Fig. 7.11

Zone of working without excessive strain Zone of maximum comfort during working

Work-area data for an average-size operator

400

Machine Tool Design and Numerical Control

1 0.4 1.2 0.6 0.2 0.25 0.1

0.12 0.15 0.3 1 0.3 1.2

Fig. 7.12

The electric motor of the main drive of machine tools is

Table 7.1

Maximum pushing and pulling force as a fraction of body weight at different positions of the forearm

Maximum safe lifting force, kgf Adults Men

Youths

Women

Boys

Girls 15

12

Ergonomic Considerations Applied to the 7.4.2 Design of Control Members The anthropometric and functional anatomy data when considered along with factors such as ease and

Control Systems in Machine Tools 401

Design of Push Buttons mm and

Design of Toggles

The toggle is a miniature lever which is used as

a > 40°

d

I l=12–25mm d=3–12mm

Fig. 7.13 Toggle switch

Design of Knobs

b h l

motor of the main drive and other step switching operations requiring a relatively large switching Fig. 7.14 Bar knob

402

Machine Tool Design and Numerical Control

Table 7.2 S. No.

Design parameters of continuous function knobs

Type of knob

Sketch

Diameter

Depth

hand

Design of Cranks in which the handle is offset from and parallel to the

required,

mechanism for manual feeding and rapid manual

Fig. 7.15 Crank

Control Systems in Machine Tools 403

l

Table 7.3 S. No.

Load, kgf

Recommended values of crank size rpm

15 25

Design of Levers

ª

Crank lever arm length, mm

404

Machine Tool Design and Numerical Control

Design of Hand Wheels

A

Table 7.4 S. No.

Recommended diameters of hand wheels

Torque resistance, kgf ◊ cm

Wheel diameter, mm

should have a wavy inside surface of the rim to ensure a good

Design of Rotary Levers and Star Wheels A rotary lever

position varies during rotation and they frequently pass through

Fig. 7.16 Hand wheels

Control Systems in Machine Tools 405

that it provides the operator the choice of using a lever which is most convenient

Fig. 7.17 Star wheel

Ergonomic Considerations Applied to the Location of 7.4.3 Displays and Control Members

Quantitative Displays

Analogue

Fixed pointer and moving scale or dial

Digital

Fixed scale or dial and a moving pinter

+

1 4 7 3

406

Machine Tool Design and Numerical Control

0

5

10

0

5

10

and are preferred when precise readings at regular intervals

15

20

25

20

25

Poor 15

Good

Fig. 7.18 Example of scale subdivisions corresponding to 1 0

T

4

8

12

16

20

Poor

greater accuracy than required hinders rapid and 5

0

10

15

20

T Good

A/125

A/200

does not have to mentally carry out conversions and

A/90

Fig. 7.19 Example showing markings that should be numbered

T

T A/600

T

A/50

Fig. 7.20

T

A = Expected maximum reading distance

Recommended dimensions of scale markings

30± 2.5°

H= where

A 200

38± 2.1°

H = height of letter, mm A=

T

Fig. 7.21

Recommended range of location of displays

Control Systems in Machine Tools 407

Height of letter or numeral, mm

20

1

0

and functional anatomy considerations discussed in

see also

10

0

Fig. 7.22

are vertical rotary levers and star wheels which are

2 4 6 Distance between display and operator’s eye, m

Nomogram for determining height of letters or numerals depending upon the distance of display: — for instruments with moving scale and illumination intensity = 100 lx

of vertically outstretched forearm with the upper

- - - for instruments with fixed scale and illumination intensity = 100–200 lx – - – for instruments with fixed scale and illumination intensity = 300 lx

Table 7.5 S. No.

Range of location of control members

Operator position

Vertical distance from mm

Horizontal distance between extreme control members, mm

Standing

(a)

Fig. 7.23

(b)

95 cm

133 cm

110 cm

135 cm

137 cm

160 cm

Sitting

(c)

Location of various control members while (a) operating a lever (b) grinding a tool (c) working on a bench

408

Machine Tool Design and Numerical Control

7.4.4 Compatibility in the Design of Control Members

population stereotypes

2

Fig. 7.24 Compatibility considerations in push-button control for rectilinear movement

OFF

ON

ON

OFF

OFF

ON

ON

OFF

Fig. 7.25

Compatibility considerations in on-off type push button and lever controls

Fig. 7.26

Compatibility considerations applied to continuous-function type levers

Control Systems in Machine Tools 409

wheels is that when an operator facing the wheel turns it in the

Fig. 7.27

Table 7.6 S. No.

Symbol

Compatibility considerations applied to hand wheels

Function symbols for use on control panels Function

S. No.

Symbol

Function

Limited rotation

rotation

rpm or cutting speed)

Stepless regulation of speed or feed

of rpm or cutting speed)

rotation

Accelerated travel

Manual Control

Cross feed

Local illumination

Contd.

410

Machine Tool Design and Numerical Control

Table 7.6

Contd.

Electric motor

rpm

(red in colour)

Coolant sump

7.5

AUTOMATIC CONTROL SYSTEMS

7.5.1

Mechanical Automatic Control Systems

!

Attention: Be cautious (yellow in colour)

Control Systems in Machine Tools 411

Cam-controlled Systems

s or

3 1

2

4

7 5

10 11

8

6

9

12

Fig. 7.28 Three-dimensional diagram of a cam-controlled automatic lathe

412

Machine Tool Design and Numerical Control

Template-controlled Systems

tracer controlled systems, the program

Z1, Z2 and Z3

4

5

6

3

Z1 2

Z2

Z3 7

1

Fig. 7.29 Working principle of a direct action, template controlled copy milling machine

Control Systems in Machine Tools 413

4

3 2

5

1 6

Fig. 7.30 Working principle of an indirect action, template controlled copy milling machine

7.5.2 Electrical Automatic Control Systems

414

Machine Tool Design and Numerical Control

Z

Z

Y

X

2

4

3

1 5

Fig. 7.31(a)

An electrical automatic control vertical milling machine

X

Contact board

X

plug is inserted, a particular contact on the contact to the drive or actuating mechanism of a particular

Indexing mechanism

Fig. 7.31(b) Drum-type plug board

X in Y and Z

Control Systems in Machine Tools 415

is initiated, it is necessary to actuate the limit switch which will determine the length of travel for the given

equipped with conventional machine tools does not presage radical changes in production planning and

7.6

ADAPTIVE CONTROL SYSTEMS

dictates resetting of new values of cutting speed and feed whenever a change occurs in the dimensions, depth

Adaptive control provides automatic control of the operating parameters, so that cutting speed and feed

An adaptive control system carries out three functions: Monitoring ptimisation E

416

Machine Tool Design and Numerical Control

Monitoring involves on line measurement of one or more parameters of the cutting process to evaluate

and in case of a discrepancy uses an optimising algorithm to carry out on line optimisation and calculate

of the cutting conditions are:

T A

Adaptive control systems are used in machine tools in the following two modes: T To achieve process optimisation from consideration of performance effectiveness criterion such as Commensurate with the function of limit constraint or process optimisation, adaptive control systems may (a) limit constraint adaptive control systems or adaptive control constraint (ACC) systems, and

Control Systems in Machine Tools 417

7.6.1 Limit Constraint Adaptive Control Systems

The torque and rpm of the machine spindle are measured and the actual consumed power (Nact) is compared with the limiting power (N N Nact smax

Amplifier and controller

Nmax

Feed drive

Carriage

Nact Torque transducer rpm Transducer

Spindle drive

Fig. 7.32 Limit constraint adaptive control system

7.6.2 Process Optimisation Adaptive Control Systems As already stated, the function of these systems is to control machining parameters in the course of an operation in such a manner that the machining process is carried out under optimum conditions for the

which calculates the optimum values of the controlling parameters in accordance with the algorithm and An adaptive control system designed from the viewpoint of minimum machining cost as the optimising

418

Machine Tool Design and Numerical Control

Temperature transducer Insulation Moment

Transducer

transducer Algorithm for wear rate calculation rpm

Spindle drive

Carriage

Drive control unit

Algorithm for opt. feed & speed calculation

Feed drive

Fig. 7.33 Process optimisation adaptive control system

References , Proc. II MTDR Conf.,

Numerical Control of Machine Tools 419

8 8.1

NUMERICAL CONTROL OF MACHINE TOOLS

FUNDAMENTAL CONCEPTS, CLASSIFICATION AND STRUCTURE OF NUMERICAL CONTROL SYSTEMS

A machine tool is said to be numerically controlled if it operates in a semi-automatic or automatic cycle as per instructions transmitted to it in a coded form. Strictly speaking, the term ‘numerical control’ is a misnomer, because the coded instructions are expressed not only through numerals, but also through letters, punctuation marks and other symbols. However, although ‘symbolic control’ would have, perhaps, been a more appropriate name, the term ‘numerical control’ (NC) has come to be so closely associated with control through symbols that it is now universally accepted and applied in the latter sense. It is obvious that numbers by themselves cannot do any work, leave aside operating a machine. A comprehensive electrical, electronic and mechanical processing and transmission system is required to affect the movement of a slide or cutting tool from information coded on a program medium, such as a punch card, punch tape, magnetic tape, etc. fundamental to a proper understanding of the functioning of numerical control systems. It, therefore, appears logical that this aspect should be dealt with in the very beginning. The elements of numerical control systems and their operating principle will now be described, starting with the example of turning (Fig. 8.1). Z Program X 5 20

5

+ +

Stepping motor

Amplifier

Program reader

Stepping motor

Decoder

Amplifier

Fig. 8.1 Schematic diagram of a numerical control system for simple turning operation

420

Machine Tool Design and Numerical Control

The cylindrical workpiece clamped in the lathe chuck is to be machined to produce a stepped shaft by removing the material which has been shown hatched. For the starting position of the tool as shown in the 1. 2. 3. 4.

Travel of the tool post in – X direction through 5 mm. Travel of the tool post in – Z direction through 25 mm. Travel of the tool post in + X direction through 5 mm. Travel of the tool post in + Z direction through 25 mm.

1. Switching instructions, e.g., feed direction in the given example; other switching instructions may be for spindle speeds, coolant on/off, tool change, etc. 2. Path magnitude instructions. For the time being let us forget switching instructions and see how path magnitude instructions are implemented. Also, for the sake of simplicity let us assume that the path magnitude is programmed in the unit pulse code. In this code, as many holes are punched on the paper tape (which is the program medium) as the number of impulses required. The number of impulses depends upon the least count of the system. For instance, if the least count = 1.0 mm, i.e., one impulse produces 1.0 mm displacement of the tool post, then 25 holes must be punched for obtaining a tool post travel of 25 mm. If the least count of the system were 0.1 mm, 250 impulses and hence 250 punched holes would be required to achieve the same displacement. hole and sends the corresponding information to the decoder. The decoder generates a pulse every time the lead screw of the lathe. The input of an impulse to the stepping motor results in a turning of its rotor through a particular angle. This rotary motion is transmitted to the lead screw and is converted through the screw-and-nut mechanism into translatory displacement of the tool post by a distance equal to the least count of the system. If we assume that the least count of the system shown in Fig. 8.1 is 1.0 mm, then the program would appear as shown in Fig. 8.2. It may be emphasised again that the switching instructions are, in this example, not coded. They may, as a hypothetical situation, be assumed to be initiated at the appropriate moment manually or electrically (see Sec. 7.5.2).

+Z

25

Stop

Stop

–Z

Stop

+X

Stop

–X

25

Fig. 8.2 Program in unit pulse code for the component of Fig. 8.1

The punched tape would appear as shown in Fig. 8.2. The set of instructions between two stop commands contains the necessary information for executing a particular motion. This is known as a sentence, e.g., STOP—Z 25STOP constitutes a sentence. The sentence consists of words (STOP), symbols (– sign), letters (Z) and numerals (25). While machining instructions are being carried out in accordance with a particular sentence, the information contained in the next sentence is read and stored temporarily in a memory device,

Numerical Control of Machine Tools 421

known as buffer storage. When the operation has been completed, the information from buffer storage is instantaneously transferred to the active storage, thus ensuring continuous running of the program. In the light of above discussions, the numerical control system of Fig. 8.1 may be represented in a general form by the block diagram of Fig. 8.3. Program medium Program reader

Amplifier

Buffer storage

Drive in X direction Operative member

Decoder

Amplifier

Drive in Z direction

Fig. 8.3 Block diagram of an NC lathe for simple turning

Let us now consider the example of a taper-turning operation (Fig. 8.4). The motions involved in the 1. 2. 3. 4. 5.

Travel of the tool post in the – X direction by 5 mm. Travel of the tool post in the – Z direction by 5 mm. Simultaneous travel of the tool post by 20 mm in the – Z direction and 4 mm in the + X direction. Travel of the tool post in the + X direction by 1.0 mm. Travel of the tool post in the + Z direction by 25 min. Program

Z Program reader

4 X 5 20

5 + +

Stepping motor

Amplifier

Decoder

Stepping motor

Interpolator

Amplifier

Fig. 8.4 Schematic diagram of a numerical control system for taper-turning operation

422

Machine Tool Design and Numerical Control

tapered surface. The generation of the tapered surface involves simultaneous movement of the tool in the – Z and + X directions such that 1. The movements in both directions start simultaneously, and 2. The displacement through 4 mm in the + X direction takes exactly the same time as the displacement through 20 mm in the – Z direction. The second requirement presages that stepping motor pulse rates in the two directions (and they control The task of calculating the feed rates of simultaneous movements and their coordination is done by a control element known as the interpolator. The interpolator is basically a microprocessor and is an essential part of tape for the machining cycle of the taper-turning operation is shown in Fig. 8.5 while the block diagram of the NC system is shown in Fig. 8.6.

+Z

Stop

Stop

Stop

Stop

–Z

Stop

–X +X

25

20

Fig. 8.5 Program in unit pulse code for me component of Fig. 8.4

Open-loop and Closed-loop Systems In the two systems-described above, the operative member of the machine tool is directed to move to a certain position. However, whether it precisely arrives at the desired position or not is not ascertained. In other words, there is no feedback on the accuracy of execution of the path magnitude commands. Such systems are known as open-loop systems. The NC systems depicted in Figs 8.3 and 8.6 are both open-loop systems. It may be useful to summarise at this stage that the elements

Drive in X direction

Interpolator

Amplifier

Decoder

Program reader

Buffer storage

Program medium

Operative member Amplifier

Drive in Z direction

Fig. 8.6 Block diagram of an NC lathe for taper turning

1. Program medium 2. Program reader 3. Buffer storage

Numerical Control of Machine Tools 423

4. Decoder 5. A 6. Drive Numerically controlled systems with a feedback device are known as closed-loop systems. The feedback arrangement consists of a transducer which monitors the actual displacement of the operative member. The actual displacement is compared with the programmed displacement and the difference signal is employed to actuate the drive motor until the command signal and actual position coincide. It may be thus seen that 1. A displacement measuring device 2. A comparator. The block diagram of closed-loop systems, based upon the open-loop systems of Figs 8.3 and 8.6 are shown in Figs 8.7 and 8.8, respectively.

Drive in X direction

?1 Transducer X-direction

Operative member

Drive in Z direction

?2

Amplifier

Program reader

Comparator Z-direction

Buffer storage

Decoder

Program medium

Amplifier

Comparator X-direction

Tachometer

Transducer Z-direction ?1

Tachometer

Fig. 8.7 Block diagram of a closed-loop NC system for simple turning

Drive in X direction

Amplifier

?1 Transducer X-direction

Operative member

Drive in Z direction

Amplifier

?2

Comparator Z-direction

Intrepolator

Decoder

Program reader

Buffer storage

Program medium

Comparator X-direction

Tachometer

Transducer Z-direction

Techomoter

Fig. 8.8 Block diagram of a closed-loop NC system for taper turning

?1

424

Machine Tool Design and Numerical Control

The displacement measuring device (transducer) and the comparator may be analog or digital. For compatibility between the transducer signal and comparator and between the transducer output and drive, it is necessary to use appropriate analog-to-digital (AD) and digital-to-analog (DA) converters. These are shown in Figs 8.7 and 8.8 by ? 1 and ? 2. In general, four combinations of the transducers and comparators

Combination

Converter ? 1

Converter ? 2

Analog comparator and analog transducer

Nil

Nil

Analog comparator and digital transducer

DA

Nil

Digital comparator and digital transducer

Nil

DA

Digital comparator and analog transducer

AD

DA

It may be added that a deceleration circuit is an integral element of all NC machines and it is placed before the motor control. The input devices, reading circuits, decoding circuits, interpolator, comparator or controller, position control circuit, velocity control circuit, deceleration circuit and auxiliary functions constitute the machine control unit (MCU) of the numerically controlled machine tool. The MCU is housed in a separate cabinet and together with the machine tool constitutes the NC machining system. Closed-loop systems are more reliable but also more expensive. Numerically controlled machine tools have traditionally been closed-loop systems, but in recent years, the concept of open-loop control is fast catching up. The successful operation of open-loop systems is governed by the reliability of stepping motors, i.e., their ability to respond reliably to a pulse command. If a stepping motor were to respond unfailingly to command signals, but not respond to spurious signals, then there would be no need for the feedback arrangement and the control systems would become considerably simpler and cheaper. At present, reliable stepping motors are available in a low horsepower range only. Therefore, open-loop control is generally employed only in positioning systems in which no material removal takes place during the movement controlled by the NC system.

Basic Length Unit (BLU)

The accuracy of travel in an axis of motion depends on the position resolution of the given axis and is called the basic length unit (BLU). It is also known as increment size and bit weight. In open loop systems, the BLU is the tool travel corresponding to one pulse of the stepper motor. For instance, if a stepper motor of 200 steps per revolution is moving the table through a lead screw of pitch 5 mm, the BLU of the system will be 5/200 = 0.025 mm. In a closed-loop system, it is necessary to distinguish between position resolution which is the smallest position increment in the part program and control resolution which is the smallest change in position that the feedback device can sense. For instance, if the optical encoder used as feed back transducer in a NC machine tool emits 1000 pulses per revolution of the shaft attached directly to a 5 mm pitch lead screw, i.e., it emits one pulse for every 5/1000 = 0.005 mm of linear displacement of the operative member, the control resolution of this system will be 0.005 mm. For best the system resolution or BLU. In both open loop as well as closed loop systems, the part dimensions are expressed as number of pulses corresponding to the distance of move. For instance, for the operative member to move by 5 mm in a NC

Numerical Control of Machine Tools 425

machine tool with BLU = 0.001 mm, the number of pulses to be generated by the stepper motor or servo motor = 5/0.001 = 5000. This quantity is entered as the reference value in the comparator. These pulses may actuate a stepper motor in an open-loop system or a dc servo motor in a closed-loop system. The rate at which these pulses are generated (pulses/min or pulses/sec) determines the feed rate at which the move is executed in the given axis of motion. Example: A 100 mm long workpiece is to be machined on a NC lathe at 500 rpm and feed rate of 0.1 mm/ rev. A stepper motor having step of 1.8° is used to provide the feed motion through a lead screw having a pitch of 5 mm. Determine BLU and the required total number of pulses and pulse rate. (i) As the step angle of the motor is 1.8°, 200 pulses will be required for one revolution of the lead screw. Hence BLU = 5.0/200 = 0.025 mm (ii) Pulses required to program for the move of 100 mm = 100/0.025 = 4000 (iii) The move feed rate per minute = 0.1 ¥ 500 = 50 mm/min For a feed rate of 50 mm/min the required pulse rate = 50/0.025 = 2000 pulses/min.

Point-to-point (Positioning) and Continuous-path (Contouring) Systems Depending upon 1. Point-to-point or positioning control systems. 2. Positioning-cum-straight-cut systems. 3. Continuous-path or contouring systems. In point-to-point systems, the table (or spindle) is moved by nu-

Y

is being drilled at position A (Fig. 8.9). After this hole has been drilled A and the drill are tracked out of contact with the workpiece, it is necessary to move the table so that point B occupies the working posiy tion under the spindle presently occupied by point A. Since the aim is B to position the table at the desired location, it is immaterial whether it X-axis and then along the Y-axis or vice versa. x As a matter of fact, to minimise the positioning time, the two motions should preferably be executed simultaneously, with the constraint that the displacement along each coordinate axis should cease when X 0 the displacement equals the assigned value. It may be thus seen that in point-to-point control systems, there need not be any coordination Fig. 8.9 Component drawing between the movements in the two directions. The only requirement is that the table should move to the desired location, irrespective of the path. An important feature of these systems is that the cutting tool is not in contact with the workpiece when the positioning movement is being executed. Point-to-point control is generally applied in drilling and boring machines. Positioning-cum-straight-cut systems are basically positioning systems with the additional capability of

426

Machine Tool Design and Numerical Control

1. The table speed during the machining operation should be equal to the feed rate and not that of the accelerated travel. 2. During the machining operation the simultaneous movements along the coordinate axes can not be coordinated. However, in some machines, it is possible to use both feeds simultaneously, thereby making it possible to take cuts at 45° to the coordinate axes. 3. The feed motor should have a higher capacity than the feed motor of a positioning system. This type of control is generally applied in milling machines and simple lathes. It permits in the former case machining of rectangular blocks and straight grooves, and in the latter case only cylindrical workpieces with perpendicular shoulders. The NC system shown in Fig. 8.1 is representative of this group. In the contouring systems the simultaneous movements of the tool and workpiece are coordinated to

segments of parabola to describe the locus of the tool cutting edge. This task is done by an appropriate interpolator, which may correspondingly be linear, circular or parabolic. To ensure reliability in maintaining the functional relationship between the movements, almost all contouring NC machine tools are equipped with a feedback arrangement, i.e., they function as closed-loop systems. Contouring NC machine tools are consequently more expensive than those operating on point-to-point control principle. Naturally, a contouring NC machine tool can carry out all the functions of point-to-point and positioning-cum-straight-cut control systems. Figures 8.6 and 8.8 represent contouring NC systems.

8.1.1

Coding System—Programming Mediums—Tape Format and Codes

Coding Systems In the decimal system, 970 = 9 ¥ 102 + 7 ¥ 101 + 0 ¥ 100 and in the binary system, 970 = 1 ¥ 29 + 1 ¥ 28 + 1 ¥ 27 + 1 ¥ 26 0 ¥ 25 + 0 ¥ 24 + 1 ¥ 23 + 0 ¥ 22 + 1 ¥ 21 + 0 ¥ 20 which is represented as 1 1 1 1 0 0 1 0 1 0. Now if 970 is encoded, it would appear in decimal and binary systems as depicted in Fig. 8.10a and b, respectively. It may be seen that for encoding any number less than 1000 in the decimal system, we need four rows, each containing 10 positions. In the binary system, we need only one row with an equal number of positions. Thus, the binary system of coding is much more economical than the decimal system. The biggest advantage of the binary system is that since any number is represented as a combination of 1 and 0, it can be easily realised in an on/off type electrical circuit – 1, i.e.,

0

1

2

3

4

5

6

7

8

9

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

0

10

1

10

2

10

3

10

+

(a) 9

2

8

2

7

2

6

2

5

4

2

2

+

+

3

2

2

2

+

1

2

0

2

+

(b)

Fig. 8.10 Decimal and binary encoding for 970

Numerical Control of Machine Tools 427

a punched hole corresponding to the on state of the circuit and 0, i.e., the absence of a punched hole corresponding to the off state of the electrical circuit.

binary number 1 1 1 1 0 0 1 0 1 0 represents 970 in the decimal system. Therefore, in the numerical control of machine tools the binary coded decimal (BCD) system is employed. The binary representation of numerals 1–9 being as given below. 1=0

0

0

1

2=0

0

1

0

3=0

0

1

1

4=0

1

0

0

5=0

1

0

1

6=0

1

1

0

7=0

1

1

1

8=1

0

0

0

9=1

0

0

1

the number 970 can be represented in BCD system as 9=1

0

0

1

7=0

1

1

1

0=0

0

0

0

i.e., the number is written in vertical columns, each row of the binary equivalent representing a number less than nine. As compared to the pure binary system, the BCD system is slightly more expensive in terms of consumption of the programming medium material, but it combines the advantages of binary representation with the ease of reading and writing, characteristic of the decimal system. Programming medium used in NC machine tools is punched tape. These tapes are available in 200 mm diameter rolls, containing 300–600 m length of tape, depending upon its thickness, and are made of paper, plastic or a plastic-aluminium sandwich. In early NC machine tools, punched cards and magnetic tape were also used as programming mediums, but subsequently punched tape became widely accepted by industry 1. 2. 3. 4. 5.

Low cost of perforator and program reader. Low cost of the tape proper. Ease of detecting damage to the tape. Possibility of punching the program manually. No hazard of accidental distortion of the coded information.

428

Machine Tool Design and Numerical Control

Tape Format and Codes tapes in use by various manufacturers of NC machine tools. In 1959, a standard format for tape size and accepted. A standard 25.4 mm (1≤) tape is shown in Fig. 8.11. When tape is edge guided use this edge Direction Hole numbering tape feed sequence “track” + –.009 Cumulative error in 6”

12

Round code holes .072+0.001 Dia –.002

3

.392+–.003 To feed hole

1,000+ –.003 Tape width

¸ 6 HolesÔPattern for 7 Holes ˝codes having 8 Holes Ô ˛

45

67

8 .100+–.002 Spacing of all hole centre lines

Fig. 8.11

Feed hole 0.46+.002 –.001 Thickness .004+–.0003

A standard eight track punched tape

The tape has eight tracks or channels of coding holes and a row of smaller holes for feeding the tape. The cross lines are known as rows. Each combination of holes in a row is regarded as a character. The switching and path information required for an operation constitutes a block of characters. The lengths of blocks obviously differ depending upon the quantum of information required for different operations. A block of characters (or simply block), thus corresponds to a program sentence. In a tape having eight coding tracks a total of 28 = 256 combinations are possible in each row. However, one of the tracks is reserved for the purpose of checking, and therefore, the number of possible characters is only 27 = 128. If the combination of ‘no holes’ is ignored as being ambiguous, the actual number of available characters becomes 27 – 1 = 127. The widely used EIA code effectively uses only six tracks of the tape for coding, because track No. 8 is used for coding only one particular statement (The END OF BLOCK statement), while track No. 5 is used for parity check (the purpose and method of parity check will be explained a little later). Thus, the number of available characters in the EIA code is only 26 – 1 = 63. Till not so long ago it was considered that 63 characters were adequate for controlling the operation of NC machine tools. However, with increasing sophistication of NC systems and the larger number of machine tool functions that are numerically controlled, the need was felt for a code that provided for more characters. This led to the development of a seven-bit code which has been accepted by the ISO. Most manufacturers of NC machine tools are gradually changing over to this code. A code similar to ISO is used in the USA where it is known as ASCII (American Standard Code for Information Interchange). The ASCII code is increasingly being referred as ANSCII (American National Standards Code for Information Interchange) since the name of the sponsoring organisation was changed from American Standards Institute to American National Standards Institute. In the ISO code track No. 8 is utilised for parity check, whereas the remaining seven tracks are used for coding. Thus, the ISO code has provision for 27 – 1 = 127 characters, which should be adequate for some considerable time to come, even if the current

Numerical Control of Machine Tools 429

The function of parity check is to keep a check on the mechanical condition of the punched tape. The EIA code adopts the odd parity check. This means that every character on the punched tape should consist of an odd number of holes. If the binary code requires an even number of holes for a particular numeral (e.g., 3 = 0 0 0 0 0 0 1 1) or symbol, then an extra hole is punched in track No. 5 to make the number of holes odd. Thus, 3 = 0 0 0 1 0 0 1 1 is the correct equivalent in the EIA code. If the tape ever gets mechanically damaged by means of a tear or rip, an extra hole will appear and the total number of holes in a row will become even. As soon as the parity check monitoring circuit detects an even number of holes in a row, a command is initiated to stop the tape reader and alert the operator. The ISO code employs an even parity check. The number of holes in a character must be even. If the number of required holes for a symbol or numeral is odd (e.g., 4 = 0 0 1 1 0 1 0 0), an extra hole is punched in track No. 8 to make the total number of holes even. In this case also, the tape is continuously monitored by the parity check circuit and the tape reader immediately stopped whenever an odd number of holes are detected in a row. The coded representation of some of the numerals and symbols in both EIA and ASCII codes is shown in Fig. 8.12. ASCII 8 7 6 5 4

EIA 244A 3 2 1

Tape Punch

8 7 6 5 4 EL X O CH 8

3 2 1 4 2 1

0 1 2 3 4 5 6 7 8 9 a b c d e f g h I j k l m n o p q r s t u v w x y z

.(period) ,(comma) / +(plus) –(minus) space delete carr.ret. or end of block back space tab end of record

Fig. 8.12 EIA and ASCII codes for numerals, symbol and letters

1. Digit zero is punched as the decimal equivalent of 16. The code is obtained as the binary representation of 16, i.e., 2, which implies a hole in track N05. However, since track N05 is reserved for odd parity check, the hole is actually punched in track N06.

430

Machine Tool Design and Numerical Control

2. Alphabets a to z a–i (9 alphabets), j–r (9 alphabets) and s–z (8 alphabets). Within each group, the alphabets are represented by their numerical equivalent, e.g., a = 1, b = 2, and so on, j = 1, k = 20 and so on. The punch code is obtained as the binary representation of the numeral (e.g. a = b = 2 = 20, and so on) and punching of additional holes in tracks 6 and 7 for alphabets a–i, track 7 only for alphabets j–r and tack 6 only for alphabets s–z.

1. Digits 0–9 are represented by their binary equivalent with two extra holes punched in tracks 5 and 6. 2. Alphabets a–z are represented by their binary equivalents 1–26 with one extra hole punched in track 7. The program is punched on the tape on a perforator. The perforator is a keyboard instrument like a typewriter or computer card punching machine. When a particular perforator key is pressed, the punch unit automatically cuts the correct holes, including the hole required for parity check. The operator of such a device need not have knowledge of the codes, etc., or any programming skill. As a matter of fact, the job can be easily done by a typist with a little amount of practise.

8.1.2 N C Hardware Program reader When the punched tape is passed through a punched tape reader, electrical connections are made in accordance with the punched holes. Thus the coded instructions on the tape are transformed into their electrical analogues which are utilised for controlling the various machine tool functions. Punched tape 1. Electromechanical 2. Photoelectric

passed. When there is no hole on the punched tape, the movable contact is aligned centrally inside the groove

character are read simultaneously by the eight sets of contacts. After a character has been read, the tape is moved forward to bring the next character under the reading position. These readers have a reading speed of 20 to 120 characters per second, depending upon the mass of the movable contact. The operation of a photoelectric tape reader is based upon the principle that if a beam of light falls on a photodiode, the latter generates an electrical signal. Under the tape, nine photodiodes are placed in a row, eight opposite the coded tracks and one opposite the feeding hole. If there is a hole on the tape, light escapes through this hole and the corresponding photodiode generates an electrical signal. Photoelectric tape readers do not have a drive sprocket, and, therefore, the signal from the photodiode placed opposite the feeding hole is used as a timer to trigger the simultaneous reading of all the eight coded holes. Photoelectric readers are capable of reading from 100 to 1000 characters per second.

Numerical Control of Machine Tools 431

speed of reading. The reading speed of electromechanical readers is adequate for positioning systems on which they are frequently used. It is in general true that all electromechanical tape reading devices damage the tape to some extent, especially at high-reading speeds. Therefore, from considerations of reliability, an increasing number of positioning systems are equipped with the slower variety of photoelectric readers. Almost without exception, continuous-path NC machine tools are equipped with high-speed photoelectric readers.

Decoder

The combination of holes read in a character must be changed into an unambiguous signal, which is utilised for switching a particular machine tool function or is the physical equivalent of a numerical quantity. This function is carried out by a decoder. A simple pyramid-type decoder is shown in Fig. 8.13. The signal from each track of the tape is fed to the relays. If there is no hole on the punched tape, the corresponding relay remains closed and occupies position 0 (dotted lines). In the case of a punched hole, the

is shown getting signals from four tracks. It, therefore, provides 24 = 16 electrical circuits. A larger decoder having relays for receiving signals from seven coding tracks would obviously provide 27 = 128 circuits. Thus, even after each of the coded symbol, letter and numeral of Fig. 8.12 is transformed by the decoder into a unique electrical circuit, there still remain 128 – 50 = 78 unutilised electrical circuits. These extra circuits are added to the redundancy to increase reliability. Track 1

Track 2

Track 3

Track 4 0

1

2

3

4

5

6

7

8

9

10

11 12

13

14

15

Fig. 8.13 A decoder for 16 electrical circuits

While decoding the numerals, it is desirable from the point of view of controlling the feed motor that the physical analogues of the numerals should be in the same proportion. For instance, the signal from the simple arrangement shown in Fig. 8.18, this is achieved by resistances Ri(i = 0, 1, 2, …, 7), such that R2 = 2R1 , R3 = 3R1, …, R7 = 7R1. Improved decoder designs employ logic elements or a diode matrix instead of relays. However, these designs are too complex to be discussed in an elementary exposition of the principles of numerical control.

Buffer Storage

The buffer storage is a memory device which temporarily retains data received from the decoder. The buffer storage is required in a NC system if the speed of tape reading is not adequate to ensure uninterrupted execution of programmed movements. A delay in transmitting information from the decoder

432

Machine Tool Design and Numerical Control

due to inadequate reading speed will momentarily stop the operative member of the machine tool. Such momentary stoppages are of little consequence in a point-to-point system because the workpiece is out of contact with the cutting tool during the controlled movement. However, in positioning-cum-straight-cut and continuous-path NC machine tools, the momentary stoppages will leave dwell marks on the machined surface, which is unacceptable. In positioning-cum-straight-cut NC machine tools, the length of programmed cutting movements next command. Therefore, tape reading delay is not a serious handicap for both point-to-point and positioning-cum-straight-cut NC machines tools. These machines are, therefore, seldom provided with buffer storage. In a

approximating lines or arcs, the greater is the accuracy. In order to achieve the accuracy generally required of or arcs. Let us consider that a rate of, say 300 mm/min the machining of this straight line segment will take 0.02 ¥ 60 = 0.004 s 300 An electromechanical reader with a reading speed of 100 characters per second will fail to provide continuous movement as the reading time per character, which is 1/100 = 0.01 s is greater than the time required for executing the elementary movement. It is, therefore, customary to equip continuous-path NC machine tools with a photoelectric tape reader and buffer storage. The buffer unit enables the control system to operate on an active storage while the data for the next movement is kept in the buffer storage. As soon as the movement controlled by the active storage is completed, data from the buffer storage is transferred to the active storage. The time required for data transfer from the buffer storage to the active storage is measured in millionths of a second (i.e., it is virtually instantaneous), and therefore, the cutting process is continuous for all practical purposes.

Interpolator In mathematics, interpolation is understood as the process of drawing as smooth a curve as possible through a number of given points. In application to a numerically controlled machining system, the aim of interpolation is to calculate the intermediate points of a curve, given its starting and end coordinates. Interpolation can be done both by analog and digital interpolators. However, only digital interpolation will be discussed here, as analog interpolation is seldom employed nowadays. A digital interpolator is basically a microprocessor or microcomputer which calculates the intermediate coordinate points of a curve in 1. The digital differential analyser (DDA) method. 2. The direct function estimation (DFE) method. for machining straight surfaces that are not parallel to either of the coordinate axes. Interpolation may be

Numerical Control of Machine Tools 433

straight lines, arcs of circles or segments of parabola. Interpolation of a higher order, e.g., by third-order of the interpolator becomes prohibitive. For a similar reason, three-dimensional interpolation is rarely used. Linear interpolator is primarily meant for straight lines but can be employed to approximate curved

required tolerances. It is obvious that the smaller the size of linear increments, the greater will be the degree is generally 0.005 mm. radius of curvature of the trajectory is denoted by R and the tolerance by t, the length l of the straight line increment can be found from the expression, l = 2.8 Rt

(8.3)

l = 2.8 Rt

Inside tolerance (IN TOL)

Outside tolerance (OUT TOL)

IN TOL OUT TOL

Fig. 8.14 Different methods of approximating a curved profile by straight lines

The coordinates of the ends of each straight-line increment must be found. This can either be done manually, or more practically, on a general-purpose computer. It must be appreciated that for machining a involving vast lengths of punched tape. reasonably small programs on machine tools that have provision of circular interpolation. For machining an arc using circular interpolation, all that needs to be programmed are the coordinates of the starting and end points of the arc, radius of the circle, coordinates of the centre and direction of movement. Thus, only required for machining the same arc by using linear interpolation. From a purely mathematical viewpoint, linear and circular interpolation techniques are basically similar because the circular interpolator also

interpolator which also generates the controlling signals that move the cutting tool along the programmed arc. It may be mentioned that from considerations of design simplicity of the circular interpolator and hence its cost, interpolation is sometimes restricted to one quadrant for a single set of commands. If a curve lies in

434

Machine Tool Design and Numerical Control

two or more quadrants, then a separate set of command statements must be prepared and punched for each portion of the curve lying in a different quadrant. Parabolic interpolation is particularly suitable for approximating complicated profiles which are

interpolation may be employed in two modes—tangent and three point. The tangent parabola is tangent to the curve at the beginning and end of the span (Fig. 8.15a). The three-point parabola joins three non-straight-line points on the curve in a trajectory which is either a complete parabola or a part of it case, it occurs half way between the middle and starting or end points. For identical span lengths, the and parabolic interpolations may be required simultaneously. In such cases, the transition from a line, arc or parabola to another arc or parabola is smoothened by means of function routines which are provided as software with circular and parabolic interpolators.

Tolerance Parabola Tolerance Circle Parabola

Circle

(a)

(b)

Fig. 8.15 Approximation of a curve by (a) tangent parabola (b) three-point parabola

By parabolic interpolation a curve can be approximated with about 50 times less points than with linear interpolation. However, widespread use of parabolic interpolators is restricted by their high cost and a majority of continuous-path NC machine tools are equipped with linear and circular interpolators only.

Drives The main drives of NC machine tools for control of spindle rotation hardly differ from those used on conventional machine tools. These may be simple stepped drives, stepped drives with electromagnetic clutches and stepless drives. All these have been described in Chap. 2 and the reader may refer to Secs are commonly used in spindle drive. However, with the advances in microprocessor controlled frequency converters, AC motors are being preferred in the newer generation of CNC machine tools. The feed motion drives are somewhat different. These drives will, therefore, be discussed here at some length.

Numerical Control of Machine Tools 435

1. Feed Motion Drives of Point-to-point Systems The feed drives of point-to-point NC machine tools are generally powered by stepper motors. In most commercially available stepper motors, each pulse turns the rotor of the stepper motor through 1.8°. It is thus obvious that it would take 200 pulses to complete one revolution of the rotor. Motors generally handle pulse frequencies ranging from 80–1600 Hz. The corresponding minimum rpm of the motor is, therefore, 80 ¥ 60/200 = 24 rpm and the maximum 1600 ¥ 60/200 = 480 rpm. Improved models of stepper motors can handle pulse frequencies of up to 4.5 kHz, but these are rarely used in NC machine tools because the torque developed is extremely low. Even normal frequency-range stepping motors can develop a torque of the magnitude of 200 kg ◊ cm. This is which metal removal must take place during the control movement. 2. Feed Motion Drives of Continuous-path Systems Hydraulic and electrical stepless drives are commonly used in continuous-path systems. These drives have been described in Sec. 2.9. The electrical drive packages are economically viable in systems having relatively low power requirements, e.g., NC lathes. For power requirements greater than 7 hp, hydraulic drives perform better. These drives are, therefore, more commonly to be found on heavy-duty machine tools.

see Fig. 1.27) and a hydraulic motor (generally radial-piston type; Sec. 1.4). The output of the stepping motor is utilised Electric servo motors are now the most widely used drives for feed motion control. The motor is directly coupled to the lead screw or connected to it through a toothed belt drive. The feed rate is changed by varying the rpm of the servo motor, which is controlled electrically via the machine control unit. DC servomotors, brushless DC servomotors, AC servomotors and linear motors are being increasingly used in the current generation of CNC machine tools.

Displacement-measurement Devices A device for measuring the actual travel of a moving member is an essential element of all closed-loop NC systems. Of course, it would be ideal to measure the workpiece progress. It is, therefore, tacitly assumed that by measuring the travel of the moving member, we obtain fairly accurate information about the correct workpiece dimension. Before displacement-measuring devices are described, it is necessary to differentiate between two basic modes of listing the coordinate positions. These are absolute and incremental. In the absolute measurement while in the incremental system the position of a point is determined by its departure from the preceding point. Figure 8.16a and b depict how the position of a point (point B) systems. One apparent drawback of the incremental system is that, it being an additive system an error during any stage of measurement will be carried over to all subsequent measurements, thus giving wrong positional readings. On the other hand, the relative simplicity and lower cost of incremental-measurement devices enjoins that they should be given serious consideration. In the early stage of development of NC machine tools there existed a popular, though somewhat misplaced belief that the absolute dimensioning system was inherently superior than the incremental. As a matter of fact, on any part drawing there are some

436

Machine Tool Design and Numerical Control

incremental-measuring systems have a place in NC machine tools. Some guidelines for selecting one or the 1. Incremental-measuring systems are viable for small and medium displacement positioning NC machines and continuous-path machines with digital interpolators. 2. Absolute measuring systems are preferable for positioning, straight-line and continuous-path control on large machine tools. Y

Y B

yBA increm

B

yB

A

A

xA

yA

yA

xB O

(a)

X

xA O

xBA increm

(b)

X

Fig. 8.16 Methods of listing the coordinate positions: (a) Absolute (b) Incremental

1. Digital 2. Analogue 1. Digital Displacement-measuring Devices In digital displacement-measuring devices, the position of the moving member is monitored by a feedback transducer which converts minimum increments of the displacement into discrete numerical values. Digital-measuring devices are used both for absolute and incremental systems. A simple digital measuring system for incremental method is shown in Fig. 8.17. It consists of a transparent calibrated scale and a scanning device. One of them is stationary while the other is mounted on the moving generate a pulse every time a mark of the scale is passed. The pulses are added in the electronic counter whose reading serves as an accurate indication of the distance travelled. Counter 5 4 8 2

Scanning device

Transparent scale

Fig. 8.17 Schematic diagram of a digital displacement-measuring system

Numerical Control of Machine Tools 437

The cost of simple digital scales, such as the one shown in Fig. 8.17 becomes prohibitive if they are

s1

0111=7

two scales—a main scale and an auxiliary scale which is inclined at an angle d to the former. The optical interaction of the gratings of the main and auxiliary scales produces a Moire pattern whose width depends upon d reducing the accuracy requirement for the graduation spacing. The device for measuring linear displacement, described above, is based on direct measurement of the distance travelled by the moving member. The linear displacement can however be measured indirectly in terms of the lead-screw rotation. The reduction involved in converting rotary motion into translatory motion offers the advantage of increasing the spacing of graduations. A simple graduated disc is mounted on the lead screw and rotates with the latter. The discrete signals are generated again by a photoelectric scanning device. If extremely small displacement increments have to be discretised, then an arrangement consisting of a main scale and an auxiliary grating is employed. The rotary feedback devices are less expensive. However, they are less accurate as mechanical factors such as inaccuracy of the screw, screw compression, backlash, etc., affect the readings. It may be thus seen that the accuracy and stiffness of the lead screw and nut improve the machining accuracy directly as well as indirectly by contributing to the accuracy of the feedback device reading. Digital systems for absolute displacement measurement make use of coded binary scales or discs. The coded binary scale consists of a number of parallel uncoded scales (Fig. 8.18). The top-most scale has the least pitch, while the pitch of all successive scales increases by two times.

s2

0

2

1

2

2

2

3

2 s3 s4

Fig. 8.18 Binary coded scale

Thus, the topmost scale gives the decimal readout of the 2° position of the binary coded number, while the successive scales give the readouts of 21, 22, 23 … positions. The accuracy of measurement depends upon the pitch of the topmost scale; the smaller the pitch the higher the accuracy. The total length that can be measured scales, it can measure a maximum displacement of 24 + 23 + 22 + 21 + 20 scale has 12–20 parallel uncoded scales. Coded binary scales have very close resolution (about 2.5 microns).

438

Machine Tool Design and Numerical Control

Table 8.1 Decimal code

Decimal code

0

0000

5

1110

1

0001

6

1010

2

0011

7

1011

3

0010

8

1001

4

0110

9

1000

In spite of their high accuracy and reliability, NC machine tools due to their very high cost. The system which is widely used in practice consists of a series of coded discs, each with a separate scanning device. The discs (Fig. 8.192) consisting of four circular binary scales employ a special binary . This code is shown in Table 8.1 and enables the four scales to represent digits between 0 and 9 only. In comparison, four scales in the pure binary code would represent digits between 0 and 15 (23 + 22 + 21 + 20 = 15). A number of such discs are coupled in succession through a gearing having a the readout of units, the next one gives the readout of tens, and so on.

0010=3

Fig. 8.19

Binary coded ring using modified Grey code

2. Analogue Displacement-measuring Devices In these systems, the displacement is converted into another physical analogue which can be easily measured. The possible physical analogues may be extremely varied in nature, such as mechanical, magnetic, electrical, photo electrical, etc. Most of the devices work with an electrical analogue, which may be monitored through voltage, frequency or phase. In actual practise, cost, accuracy and reliability considerations overwhelmingly favour phase-sensitive transducers which, not A typical transducer consists of a single-phase wound rotor and a two-phase wound stator, in which windings are displaced by 90° with respect to each other. If an alternating voltage is applied to the rotor winding, then the sine and cosine components of the amplitude can be tapped from the two stator coils. If the rotor winding is coupled to the lead screw of the machine tool, the voltages tapped from the stator coils will indicate the angle of rotation of the lead screw, and hence the displacement of the moving member (slide, table, etc.). This type of measuring device is known as resolver, as it resolves the angle of rotation into sine and cosine components. A device with three stator windings displaced through 120° is known as a selsyn transducer.

Numerical Control of Machine Tools 439

A system which is just the reverse of the one described above can also be employed for displacement measurement. In this arrangement a symmetrical, two-phase voltage is applied to the two stator coils. The the rotor winding. Besides the permanent factors like ratio of windings, the emf induced in the rotor winding depends upon its relative position with respect to the stator coils. Hence, if the rotor winding is coupled to the machine tool lead screw, the voltage induced in it is an indication of its angular position, and hence the displacement of the moving member. This measuring device is known as synchro owing to the fact that it is widely used in synchronising or follow-on control circuits. Synchro devices with a three-phase stator winding are more common. Their working principle is the same as that of two-phase synchros. In application to NC machine tools, synchros are used on a much wider scale than resolvers. A single synchro can represent an angular rotation unambiguously in a very narrow range, corresponding to a rectilinear displacement of about 4–6 mm. This is obviously not enough for most machine tools. Therefore,

switching system successively switches from one transducer to another as the actual and desired displacement In a synchro- or resolver-based feedback system, the cost of precision gears, racks, pinions, etc., that are free of backlash can be quite high. This restricts their application to small-sized NC machine tools in which the displacements are small and can be measured by one or at the most, two synchros, and to machine tools having not too stringent accuracy requirements (so that the accuracy requirements of the gearing system are correspondingly low). An analogue device which can measure large displacements is the linear inductosyn. The linear inductosyn consists of a scale and a slider made of a nonmagnetic material. The scale carries a printed rectangular winding, whereas the slider carries two windings which are 90° out of phase. The scale and slider windings interact in the same manner as the stator and rotor windings of a rotary resolver. As the slider moves with respect to the scale, the emf induced in its windings changes. From the emf values, the equivalent angle of rotation may be determined and subsequently the displacement may be calculated. Linear inductosyns are generally designed for a scale spacing of 2 mm. The scales, which have a standard length of 200 mm, may be placed end to end for obtaining longer scales. Nowadays, linear inductosyns often form part of a measuring system which includes rotary resolvers and an inductosyn. Resolvers do measurement of the absolute displacement only. In comparison with digital displacement-measuring devices, analogue devices are generally simpler, more robust and less expensive. However, the measuring process displacement-measuring devices are suitable for small and medium size NC machine tools of positioning as well as continuous path types.

Comparators

Depending upon the type of feedback transducer employed for measuring displacement, the comparator may be 1. discrete-function type, and 2. continuous-function type. The discrete-function comparator operates in conjunction with digital feedback transducers. The distinguishing feature of these comparators is that they produce an output signal only when the programmed

440

Machine Tool Design and Numerical Control

and actual displacement values coincide. On the other hand, a continuous function comparator emits a signal all the time till the programmed and actual displacements are different. These comparators operate in conjunction with analog feedback transducers. 1. Discrete-function Comparator The simplest discrete-function comparator is a preselection counter. As the name implies, it consists of a preselector and a pulse counter. The counter is a standard electronic instrument. The preselector is a set of decade switches which may be set at a desired value either manually or through instructions generated by the control unit in accordance with the punched program. The setting on the preselector depends upon the elementary displacement of the moving member realised from one pulse (least count of the system). The output signal can be generated by the preselector in three different ways using a subtraction circuit, addition circuit and detection circuit, which also determine the setting on the preselector. 2. Continuous-function Comparator

Before describing a continuous-function comparator, it is necessary

to the programmed displacement is constant, while the other coming from the feedback transducer is corresponding to the programmed displacement is fed to the primary winding of transformer I, while the variable voltage (due to change of the relative phase angle) from the analog feedback transducer is supplied to the primary winding of transformer II. If the rotors of the programmed transducer and feedback transducer are perpendicular, the comparator output is zero. If the two are not perpendicular, the comparator output is a dc signal whose polarity depends upon the sign of the phase difference. Continuous-function comparators and analog transducers form a simple, robust, reliable and relatively inexpensive feedback system, which explains their wider application.

8.2

MANUAL PART PROGRAMMING

The program for machining a particular workpiece depends not only upon the workpiece shape and dimensions but also upon the NC machine tool on which it is machined. The program for machining a workpiece on one machine tool. The cause of variation lies in the difference between formats of the machine tools. Machine tool formats differ not only from one type of machine tool to another (say the lathe and milling machine), formats of individual machine tools will be of little use to the reader of a general text of his nature. It is felt that a general description of machine tool formats will serve a more useful purpose and equip the reader with

8.2.1 Machine Tool Formats The information which the programmer must have about the concerned machine tool is as follows. 1. Does the machine tool have a positioning, positioning-cum-straight-cut or a continuous-path control system? 2. Does the machine tool have an absolute or incremental mode of listing the coordinate positions?

Numerical Control of Machine Tools 441

5. How many digits are used for specifying dimensions? Where is the decimal point located? Does the machine tool control have the provision for suppressing leading zeros or trailing zeros or none?

Mode of Listing Coordinates The two modes of listing coordinate positions-absolute and incremental were described in Sec. 8.1.2 (Fig. 8.16). Although both systems can be used with equal ease, there are some points that need to be highlighted for better appreciation of the distinguishing features of the two. One apparent drawback of the incremental system is that, it being additive, an error during any stage of system is observed in case of interruptions that force stoppage of the machine, e.g., after tool breakage. In case of a machine with absolute system, the operator changes the tool and rewinds the program to the beginning of the block in which the interruption occured and takes the tool to the starting position by invoking the absolute coordinates. In an incremental system, it is impossible to return the tool to the starting position of the segment in which the interruption occured. Therefore, in an incremental system, every time an interruption occurs, the part program has to be rewound fully and restarted from the very beginning. Another advantage of the absolute system lies in the ease with which dimensional data in the program can In the incremental system the part program would have to be corrected from that point onwards where the In the incremental system, for the segment of part program between starting of tool from home position and returning to it after completion of machining, the sum of increments of the individual motions is zero. measuring devices used in incremental systems are simpler and cheaper. The recommendations for selection of the mode of listing of coordinates, based on these and other factors have already been discussed in Sec. 8.1.2

Zero System (i) Fixed (ii) Full zero shift (iii) F

(a) Zero point or machine zero is the point where the X, Y and Z axes of the machine tool intersect. It is a datum point characteristic of the machine tool which may be a physical point on the machine tool or an imaginary point programmed in the machine control unit. (b) S characteristic of the workpiece and must be located accurately with respect to the machine zero to

442

Machine Tool Design and Numerical Control

coordinates with respect to the machine zero are (Xsp, Ysp) as assigned by the part programmer. When the coordinates of the set-up point are entered in the MCU console manually and the start button is pressed, the spindle moves to the set-up point of the job (lower left-hand corner in the given case) so that it comes exactly under the spindle and the sides of the job are strictly parallel to the X and Y axes. This setting calls for considerable skill of the operator. In this zero system, the coordinates of the part are taken from the machine

+Y

FIXED MACHINE ZERO

YSP +X 0, 0

SET POINT

XSP

Fig. 8.20 Fixed-zero system

In the system with full zero shift (Fig. 8.21) the operator clamps the workpiece in a convenient position on the machine table. He then enters the set point coordinates in the MCU console and presses the start button. As a result the centre of the spindle will move to the position (Xsp, Ysp) with respect to the machine zero. However, this location of the spindle will not coincide with the set-up point of the workpiece. To make the two coincident, the operator manipulates the zero shift dials of the machine tool to shift the machine zero to the position from where the set-up point of the part will have the coordinates (Xsp, Ysp). In principle, the machine zero can be shifted to any location by the procedure described above. However, for convenience, it is generally shifted to the lower left-hand corner to obviate the necessity of recalculating the dimensions.

high skill. select any point as the machine zero without any constraint on the sign of the programmed dimensions, i.e., the dimensions in the part program may be positive as well as negative. Suppose for the part shown in Fig. 8.22, it is desired to set the machine zero at the centre of the job. The set-up point is the lower lefthand corner of the job having coordinates (– Xsp, – Ysp) with respect to the planned machine zero. To carry out zero setting of the machine, the operator takes the spindle to the set-up point and enters the coordinates (– Xsp, – Ysp). This will set the machine zero at the centre of the job.

Numerical Control of Machine Tools 443

+Y

Y

SET POINT

XSP

XSP

YSP YSP

+X

ZERO AFTER FULL SHIFT X FIXED MACHINE ZERO

Fig. 8.21 Full-zero shift system

FLOATING ZERO +Y

+X –YSP

–XSP

SET POINT

Fig. 8.22 Full-floating zero system

Programming Formats X dimension, followed by Y dimension followed by end of block statement. The positions of direction words X and Y are

444

Machine Tool Design and Numerical Control

or full-zero shift as all the dimensions are positive. It should be kept in mind that even if X or Y location does not change, it must be repeated in the next block. In a machine tool that has a tab-sequential format, the tab character is utilised to separate the words in a

a repeated word (dimension) need not be punched and may be passed over by pressing the tab character, because except the tool change function all other numbers are retained in the MCU memory. In the word-address format, the words in a block information is preceded by an identifying word (X or Y) which acts as the address. Thus the major distinguishing feature of the word-address format is that X and Y characters must be punched. This makes the program and M 50–59 (cam setting) commands remain in effect till they are replaced by a new one. However, commands M02 and M06 are cancelled automatically after they are implemented once, and must, therefore, be programmed in every block in which they are required. The meaning of these commands will be explained shortly in Sec. 8.2.2. The variable-block format combines word-address and tab-sequential formats for both flexibility word-address pair is separated by a tab character. The repeated words need not be punched.

Specification of Dimensions X 3.3, it the X remaining three digits represent numbers to the right of the decimal point. Thus, for instance, on a machine having the format X 3.3 the dimension X have provision for suppression of either leading zeros or trailing zeros. In the word 0 0 1 5 0 0, the two zeros before 15 are leading zeros, while the two zeros after 15 are trailing zeros. If the control unit of the machine tool reads from right to left, it is provided with leading zero suppression while if the words are read from left to right the trailing zeros are suppressed. On a machine tool having leading zero suppression, the word 0 0 1 5 0 0 will be written as 1500, whereas on a machine tool with trailing zero suppression the same word will be written as 0015. The provision of suppressing leading or trailing zeros considerably shortens the length of the punched tape required for a program. It should, however, be noted that zero suppression is applicable only to have an X 2.4, Y 2.4 dimension format, while those working in the metric system have an X 3.3, Y 3.3 format. The machine tool format details discussed above will now be elaborated with the help of a few simple programming problems for point-to-point NC systems. The workpiece is a rectangular block and the numerical control is to be applied to position the spindle at points 1, 2 and 3 (see Fig. 8.23a). The dimensions d, respectively. The X and Y coordinates of the points in the three systems are given in Table 8.2.

Numerical Control of Machine Tools 445

2

2

50 45

3

3 25

65

1

5

1

5

25 20

55 60

45

20 20 75

(a)

Y

X

(b)

2

2

20 3

3

45

20

25

1 5

1 X

5

25 55 (c)

Fig. 8.23

25 (d)

(a) Component drawing (b) Programming dimensions in fixed-zero system (c) Programming dimensions in full-zero shift system (d) Programming dimensions in full-floating zero system

Table 8.2

Coordinates of programmed points on machine tools with different zero systems Point

Fixed zero

Full-zero shift

F

X

Y

X

Y

X

Y

1

25

25

5

5

– 25

–20

2

25

65

5

45

– 25

20

3

75

45

55

25

25

0

The possible formats are numerous, but the examples described below will, it is hoped, not only help the reader to grasp differences between formats but also impress upon him that programming fundamentals are basically similar and one can easily adapt to programming in any format. X 3.3, Y 3.3, and it has no provision for suppressing zeros. format is X 2.4, Y 2.4 and it has provision for suppressing trailing zeros. format is X 2.4, Y 3.3 and it has provision for suppressing trailing zeros.

446

Machine Tool Design and Numerical Control

X 3.3, Y 3.3 and it has provision for suppressing leading zeros. The program manuscripts and characters punched on the tape are shown below for all the four programming problems. It may be noted that every block starts with a sequence number and ends with an end of block statement. An EOB statement is also required at the beginning to initiate the program. Problem 1 Seq.

X 3.3

Y 3.3

001

025000

025000

EOB

002

025000

065000

EOB

003

075000

045000

EOB

EOB 0 0 1 0 2 5 0 0 0 0 2 5 0 0 0 EOB 0 0 2 0 2 5 0 0 0 0 6 5 0 0 0 EOB 0 0 3 0 7 5 0 0 0 0 4 5 0 0 0 EOB

Program manuscript

Characters on tape Problem 2 Seq.

X 2.4

Y 2.4

001

TAB

– 25

TAB

– 20

EOB

002

TAB

…

TAB

20

EOB

003

TAB

25

TAB

0

EOB

0 0 1 TAB – 2 5 TAB – 2 0 EOB 0 0 2 TAB TAB 2 0 EOB 0 0 3 TAB 2 5 TAB 0 EOB

EOB

Program manuscript

Characters on tape Problem 3 Seq.

X 2.4

Y 3.3

N 001

X 05

Y 005

EOB

N 002

—

Y 045

EOB

N 003

X 55

Y 025

EOB

Numerical Control of Machine Tools 447

N 0 0 1 X 0 5 Y 0 0 5 EOB N 0 0 2 Y 0 4 5 EOB N 0 0 3 X 5 5 Y 0 2 5 EOB

EOB

Program manuscript

Characters on tape Problem 4 Seq.

X 3.3

Y 3.3

N 001

TAB

X 25000

TAB

Y 20000

EOB

N 002

TAB

…

TAB

Y 20000

EOB

N 003

TAB

X 25000

TAB

Y0

EOB

N 0 0 1 TAB X 2 5 0 0 0 TAB Y 2 0 0 0 0 EOB N 0 0 2 TAB TAB Y 2 0 0 0 0 EOB N 0 0 3 TAB X 2 5 0 0 0 TAB Y 0 EOB

EOB

Program manuscript

Characters on tape

8.2.2 Function Codes Up to this point, we have discussed numerical control principles for programming of positioning movements only. In addition to the control of machine table or spindle movements, there are a number of functions which are essential to the process of material removal for making a part. Examples of these functions are tool changing, switching the spindle on, turning the coolant on, setting the proper feed rate and rpm, etc. These functions may be executed manually by the operator or may be numerically controlled. Depending upon the complexity of the NC system, we may have, at one end, a machine tool on which none of the functions are program-controlled and at the other extreme a machine tool on which all functions are program-controlled. On most NC machine tools, a few functions such as loading and unloading are done manually while a majority of the functions are programmed. The important functions and their codes are described as follows.

Sequence-number Function represented by N plus three digits.

Tool-selection Function This function is coded on machine tools that have programmable turrets and automatic tool changers. It is represented by T Speed Function

This function is coded on machine tools which have provision for automatic speed

machine tools have, in most of the cases, automatic control of both or none. The speed function is represented by S plus three digits. On machine tools having stepped regulation of spindle revolutions, the speed steps may be assigned three digit numbers. For instance, if a machine tool speed box has 24 speed steps, then numbers then the three digits may directly represent the rpm or the magic-three code equivalent. The manner in which

Machine Tool Design and Numerical Control

448

number of digits before the decimal point. Consider, for instance, the rpm value 325. The second and third digit of the code would be 3 + 3 = 6. Consequently, the magic three-code equivalent of 325 will be 632. A few more converted code values are shown for illustration. Speed, rpm

Magic three-code quivalent

10

510

100

610

730

673

3250

732

Feed-rate Function

The feed rate is expressed through F plus three digits. The digits may represent the feed rate in mm/min, mm/rev, or the magic-three code equivalent. The manner in which the machine control

min or mm/rev values into the magic three-code equivalent is done in the same way as explained above. It is obtained by subtracting from 3 the number of zeros immediately to the right of the decimal point. A few converted code values are shown to illustrate the point. Feed rate, mm/rev

Magic three-code equivalent

0.195

319

0.0258 0.0079

226 179

Dimensional-data Functions

The dimensional data is represented by a symbol plus 5–8 digits. The

X Y

primary X dimension primary Y dimension

Z A B

primary Z dimension angular dimension about the X-axis angular dimension about the Y-axis

C

angular dimension about the Z-axis

U V W

secondary dimension parallel to the X-axis secondary dimension parallel to the Y-axis secondary dimension parallel to the Z-axis

I J K

interpolation parameter parallel to the X-axis interpolation parameter parallel to the Y-axis interpolation parameter parallel to the Z-axis

Numerical Control of Machine Tools 449

ly controlled by them. For, instance a NC milling machine with control of X, Y and Z axis would be referred as a 3-axis machine, while one with additional control of angular dimension A would be termed as 4-axis machine. Some early NC machine tools had provision of 1/2 axis control, wherein the control of dimension was implemented by means of preset cams and templates but their actuation was done through a command in the part program. For instance, a drilling machine with X-Y control for positioning, but cam control of the 1 depth of hole would be referred as 2 axis machine. 2

Preparatory Functions

The preparatory function instructs the machine tool to get prepared for the operation to follow. In a block of the program manuscript, the preparatory function comes immediately after the sequence number. Preparatory functions are represented by plus two digits. Thus there is provision for to a function. However, most manufacturers of NC machine tool follow the standard codes that have come to be accepted over the years. Some of the important preparatory function codes are given in Table 8.3. Table 8.3 Code

Codes of preparatory functions

Function

XY plane selection for interpolation XZ plane selection for interpolation YZ plane selection for interpolation

Contd.

450

Machine Tool Design and Numerical Control

Table 8.3

Contd.

Numerical Control of Machine Tools 451

Miscellaneous Functions Miscellaneous functions are used to describe instructions other than dimensions. Miscellaneous functions are not a part of the cutting process but are nonetheless important from the point of view of greater productivity. These functions are denoted by M plus two digits. The codes of some miscellaneous functions are given in Table 8.4. Table 8.4 Codes of Miscellaneous Functions Code M00 M01 M02 M03 M04 M05 M06 M08 M09 M10 M1l M13 M14 M19 M21 M22 M23 M30 M41–45 M51–59 M98 M99

Function

452

Machine Tool Design and Numerical Control

8.2.3 Complete Program—Point-to-point Problem The complete program including the preparatory and miscellaneous function codes will be written for the part shown in Fig. 8.24. 2

3

50 45

25 1 5 5 55 60

Fig. 8.24 Component drawing

1. Drilling a hole f 6 at point 1. 2. Drilling a hole f 6 at point 2. 3. Making a threaded hole M12 at point 3; machining hole M12 involves drilling a hole f 10.3 and then tapping (the reader may be reminded that 12 mm taps come in sets of 2). The machine tool format is

In this format ( . ) represents tab-sequential format, N3 represents sequence number, X 3.3, Y 3.3 represent

1. The part is to be machined on a machine tool which has provision for absolute and incremental dimensioning, input of dimensions in mm or inches, and point-to-point control. 2. The machine tool has full-range zero shift. 3. The machine zero is shifted to the set-up point, which happens to be the lower left-hand corner of the part. 4. The machine control unit has provision for suppressing trailing zeros. 5. To avoid encumbrance in loading and unloading of the workpiece and tool changing, the table must be positioned at point 4 (X = 100, Y = 100) during these operations. The coordinates of the programmed points are shown in Table 8.5.

Numerical Control of Machine Tools 453

Table 8.5

Coordinates of programmed points of part shown in Fig. 8.48 X

Y

Point

On drawing

On tape

On drawing

On tape

1

5

005

5

005

2

5

005

45

045

3

55

055

25

025

4

100

1

100

1

The complete program manuscript of the part is given in Table 8.6. Table 8.6 N

T/E

Program manuscript of part shown in Fig. 8.48 T/E

000 EOR

X

T/E

0

Y

T/E

M

T/E

0

Set machine zero

005

T

90

E

010

T

71

E

015

T

00

E

020

T

80

T

1

T

1

T

06

E

025

T

81

T

005

T

005

T

51

E

030

T

T

T

T

03

E

035

T

T

T

040

T

T

T

045

T

80

T

1

T

050

T

81

T

055

T

055

T

T

060

T

T

065

T

80

T

1

T

070

T

84

T

055

T

075

T

T

080

T

T

045

Remarks

T

Clamp f6 drill

E

T

05

E

1

T

06

E

025

T

52

E

T

T

03

E

T

T

05

E

1

T

06

E

025

T

53

E

T

T

03

E

T

T

05

E

Clamp f10.3 drill

Clamp Tap No. 1 of M12

Contd.

454

Machine Tool Design and Numerical Control

Table 8.6 Contd. 085

T

80

T

1

T

1

T

06

E

090

T

84

T

055

T

025

T

54

E

095

T

T

T

T

03

E

100

T

T

T

T

05

E

105

T

T

02

E

80

T

1

T

1

Clamp tap No. 2 of M12

Note: 1. The T/E column may contain a TAB word (T) or End of Block word EOB (E) command. In some older models of NC machines, the % symbol is used instead of EOR command. 3. N000 block instructs the operator to set the machine zero at the lower left-hand corner of the part.

the programmed position in X and Y coordinates, rapid quill movement to clearance height (feed engagement point), feed to the required depth and retracting of the quill at rapid rate. The depths of clearance plane and feed motion are controlled by the cams and switches that are suitably preset. The appropriate set of cams and switches is activated by the M51–59 command in the program. at rapid rate to the programmed position in X and Y coordinates with the spindle in up position. drilling cycle except that retracting of the quill to the clearance plane is carried out at threading feed rate with the spindle rotation reversed. Further retracting is carried out at rapid rate.

M03 M05

Spindle start clockwise direction Spindle stop

M06 M02 M51–54

Tool change End of program, machine stop, tape rewind Cam and switch settings for drilling and tapping operations

Numerical Control of Machine Tools 455

8.2.4 Complete Program—Positioning-cum-straight-cut Problem Consider the part shown in Fig. 8.25. The program is to be prepared for drilling two 10-mm holes 1 and 2 and milling 10-mm wide slot. The machine tool format is

10 2

10

3

4

60 7

7

format has two M2 statements. In this format, two miscellaneous functions may be coded in a single block. The program has to be written within the constraints of the fol-

30

1

10 10 70

1. The part is to be made on a point-to-point type

Fig. 8.25 Component drawing

absolute dimensioning in mm only. 2. The machine zero is selected at the centre of the workpiece. 3. The groove is machined with a 10-mm diameter end mill cutter. 4. A numerical information must be preceded by the proper sign. 5. During loading and unloading of the workpiece and tool changing, the spindle is stationed at point 5, whose coordinates are given in Table 8.7. Table 8.7

Coordinates of programmed points of part shown in Fig. 8.49 X

Y

Point

On drawing

On tape

On drawing

On tape

1

– 25

– 025000

– 20

– 020000

2

+ 25

+ 025000

+ 20

+ 020000

3

– 42

– 042000

0

+ 000000

4

+ 42

+ 042000

0

+ 000000

5

+ 65

+ 065000

+ 70

+ 070000

For milling the groove, the cutter must start from point 3 (providing for an approach of 7 mm) and stop at point 4. With the machine zero located at the workpiece centre, the coordinates of the programmed points will be as shown in Table 8.7. The complete program manuscript of the part is given in Table 8.8.

456

Machine Tool Design and Numerical Control

Table 8.8

EOR

Program manuscript of part shown in Fig. 8.49

N

X

Y

000

– 035000

– 030000

M

M

E

Remarks Set machine zero

005

80

+ 065000

+ 070000

06

06

E

010

81

– 025000

– 020000

51

13

E

015

81

+ 025000

+ 020000

51

13

E

020

80

+ 065000

+ 070000

05

09

E

025

80

+ 065000

+ 070000

06

06

E

Change f10 end mill

030

78

– 042000

+ 000000

52

52

E

Clamp quill, start

035

79

+ 042000

+ 000000

52

13

E

040

78

+ 042000

+ 000000

05

09

E

045

81

+ 042000

+ 000000

05

09

E

050

80

+ 065000

+ 070000

02

02

E

Clamp f10 drill

Unclamp quill

Note: 1. The number of digits in each column is maintained constant even if this means giving the same command twice in a single block, e.g., M06 is repeated twice in sequence number 005. 2. The set-up information in N000 instructs the operator to set machine zero at the centre of the part.

necessary to manually clamp and unclamp the quill. In this cycle, the table or spindle moves at rapid rate to the programmed portion in X and Y coordinates. The quill moves at rapid to the clearance plane and further at cutting feed rate to the programmed depth. The machine then stops. The operator clamps the quill and presses the start button to recommence the program execution. The process is repeated in reverse after the milling operation. tion to the programmed position in X and Y coordinates at cutting feed rate. table/spindle to the desired location with the spindle up.

Numerical Control of Machine Tools 457

8.2.5 Complete Program—Continuous-path (Contouring) Problem It may be recalled that in continuous-path control, the resulting motion is obtained from two or more a circle (or its part) and a straight line are particular cases. A free-formed surface can be approximated by circular or parabolic arcs and straight lines. Thus, before taking up a continuous-path programming problem, parabolic arcs is of relevance only to machine tools that have provision of parabolic interpolation. Since such machine tools are very few, programming for a parabolic arc is not included in the discussion below. We shall describe here the programming procedure to be adopted for straight lines and circular arcs and some additional features like tool offset which are essential for continuous-path programming. Though continuous-path programming is described below for a lathe, it is equally applicable to other machine

1. Most NC lathes follow the incremental system –X of the position dimensioning. 2. The movement in the direction of axial feed +Z –Z is denoted by Z, while the movement in the direction of radial feed is denoted by X. Thus, the movement of the cutting edge takes place in the XZ-plane. The notation of axial +X movement by Z is on account of a universally Fig. 8.26 Notation of coordinate axes of a lathe accepted convention which requires that the axis of machine tool spindle be designated as the Z direction (see Fig. 8.). It may be added that most continuous-path NC machine tools use the word-address format.

Programming for a Straight Line Parallel to X- or Z-axis

A straight-line motion parallel to the X- or Z-axis is programmed on a machine tool with continuous path control through I and K characters. For instance, a movement of 100 mm parallel to the X-axis will be programmed as X 100 I 99999. Here I represents the sine function of the angle between the direction of motion and the Z-axis. For a motion parallel to the X-axis, this angle is 90°, and therefore, sin 90° = 1 ª 0.99999. The number 99999 thus represents sin 90°. The decimal point is not written because the machine tool control knows that the value of a sine function is always less than one. A 10 mm travel parallel to the Z-axis will be programmed Z10K 99999; the character

458

Machine Tool Design and Numerical Control

K represents the cosine function of the angle between the direction of motion and the Z-axis. This angle being 0°, the K value is cos 0° = 1 ª 0.99999.

Programming for a Straight Line Inclined to the X- and Z-axis

An inclined straight line may

Case I The departures in the X and Z directions are known, as also the angle (Fig. 8.27). The I and K values are determined as explained above. For the given case the I value will be sin 36.87° = 0.6, while the K value will be cos 36.87° = 0.8. If the control has provision for suppression of trailing zeros, the complete statements will be I 6 and K 8 respectively. Case II The departures in the X and Z directions are known as also the length of the tool path, but the angle is not known (Fig. 8.28). In this case, the I and K values are not provided in the program and the tool motion along the desired path is achieved by specifying a feed-rate number (FRN). The FRN function code is F plus formula, FRN =

10 ¥ desired feed rate in mm/min length of tool path in mm

q

30 50

40

30

q = 36.37° sin q = 0.6 cos q = 0.8

Fig. 8.27

40

Programming for a straight line with given inclination angle

Fig. 8.28

Programming for a straight line with given departures

When the machine control unit receives instructions for simultaneous X and Z motions but no accompanying values of I and K characters, it knows that the F function is to be treated as the feed-rate number to control the tool motion along the desired path.

Programming for a Circular Arc 1. 2. 3. 4.

While programming for a circular arc, it is necessary to determine

The direction of the arc, The X and Z incremental departures, The arc centre offset, and The feed-rate number.

The direction of motion along a circular arc (clockwise or anticlockwise) is determined by looking in the direction of the Y-axis. The EIA standard recommends the – Y direction for this purpose, i.e., the arc motion

Numerical Control of Machine Tools 459

along the arc is given a clockwise or anticlockwise direction as seen from underneath the lathe looking up. However, some machine tool manufacturers employ the + Y direction for this purpose. The programmer must along the arc. The arc centre offset is programmed through I and K characters. The I character represents the distance of the starting point of the arc from its centre in the X direction, while the K character represents the corresponding distance in the Z direction. The method of determining the incremental departures in X and Z directions as also the values of I and K characters is illustrated in the examples shown in Fig. 8.29. While programming for a circular arc which passes through more than one quadrant, a separate block of instructions must be punched for each quadrant. The values of I and K characters are absolute (i.e., positive) and the provision for suppression of trailing or leading zeros is applicable to them too. In the block of instructions for motion along a circular arc the F function is represented through the feed-rate number (FRN) which is represented as

FRN =

10 ¥ desired feed rate (mm/min) Radius of arc (mm) –X

–X I = R, K = 0 X = –R, Z = – R

I = 0, K = R X = R, Z = –R Z

Z I = R, K = 0 X = R, Z = R

–Z

–Z I = R, K = 0 X = –R, Z = R

X X –X

–X

I = A, K = B X=R–A Z=–B

I = 0, K = R– A, X=B Z=R–A

B –Z

A

A B

I = A, K = B X = –(R– A) Z=B

Fig. 8.29

B I = R, K = 0 X=RA Z=B

Z

A

Z

–Z B

X

A X

Method of determination of I and K values and the X and Z departures for programming circular arcs

Tool Offset Suppose the part shown in Fig. 8.30 has to be produced on an NC lathe. If the lathe tool were to have an absolutely sharp nose, the tool departures would be the same as those calculated from the part drawing. However, if the tool nose radius is equal to r, then the tool departures refer to the movement of

460

Machine Tool Design and Numerical Control

the centre of the nose radius. Tool departures are thus offset as compared to the part dimensions. The offset can be easily computed and the programmed departures determined beforehand by the programmer from the known values of part dimensions and tool radius. This procedure for tool-radius compensation is applicable to all machining processes in which material is removed by the periphery of the cutter of radius r, e.g., plain milling. Described below are methods of calculating tool departures of a few typical cases.

Z

X

Fig. 8.30 Component drawing and trajectory of a tool of nose radius r

Case I Rectangular bend (Fig. 8.31) Dimension

Tool departure Z

X

AB

– A¢ B¢ = – (AB + r)

0

BC

0

B¢C¢ = BC

CD

– C¢D¢ = – (CD – 2r)

0

DE

0

D¢E¢ = DE

EF

– E¢F¢ = – (EF + r)

0

D

F

E

F¢

E¢

C

D¢ C¢

B

B¢

A

A¢

Fig. 8.31 Calculation of tool offset for a rectangular bend

Numerical Control of Machine Tools 461

Case II Tapered bend (Fig. 8.32) Dimension

Tool departure Z

X

AB

A¢B¢ = AB – DZ = AB – r tan q/2

0

BC

FC¢ = – (BE + DZ – DZ ) = – BE

FB¢ = EC + DX – DX = EC

CD

C¢ D¢ = – (CD + DZ ) = – (CD + r tan q/2)

0

DZ E

A

B

B

q

C¢¢

X

B¢

C

D

R

A¢ D

A

B¢

X

A¢

C

F DZ

D¢

C¢ C¢

D¢

Z

Fig. 8.32 Calculation of tool offset for a tapered bend

Z

Fig. 8.33 Calculation of tool offset for an anticlockwise circular bend

Case III (a) Circular bend, anticlockwise (Fig. 8.33) Dimension

Tool departures

I

K

Z

X

AB

– A¢ B¢ = – AB

0

—

99999

BC B¢C≤

– (R – r)

R–r

R–r

0

C≤C¢

0

r

99999

—

CD

– C ¢D¢ = (CD + r)

0

—

99999

Case III (b) Circular bend, clockwise (Fig. 8.34) B¢¢¢ B

A

B¢¢¢

B

B¢¢ B¢¢ D

B¢ X A¢

C

D¢

C¢

Z

Fig. 8.34 Calculation of tool offset for a clockwise circular bend

462

Machine Tool Design and Numerical Control

Before compiling the table of tool departures, it is necessary to highlight that departure A¢ B¢ is less than the dimension AB by the radius of the tool r, but with additional departure D along the Z-axis corresponding to the segment BB≤ of the circular arc. This departure can be found as R2 - r2

D = BB≤¢ = OB – OB≤¢ = R –

Dimension

Tool departures

I

K

Z

X

AB

– (AB – r + D)

0

0

99999

BC

– (R + r – D)

R

r

(R + r – D)

CD

– CD

0

0

99999

Complete Program Manuscript

The program manuscript will be prepared for the part shown in

1. Machine tool has provision for absolute as well as incremental programming. 2. The machine control unit accepts the word-address format. 3. D

0

R1

45°

f 20

f 40

f 60

L3

20 C2 P6

L5

L2

P3

L4

50 42 10

20

50 Cutter zero postion (S P)

C1

P1

L1 P2

P4 P5

20

f 10

R5

10

20

70

Fig. 8.35 Component drawing

Numerical Control of Machine Tools 463

N

X

000

05

Z

I

K

F

S

T

M

M

E

Remarks

E

Zero setting of tools

EOR N005

E

N010

E

N015

E

N020

X002

N025

X05

F0004 S650 Z05

E T01 M06

N030 Z07

N040

X-042

N045

X002

K99999

X005

N060

N070

Z-007

I003

Z-01

I7071

Z021242

N075

X007

N080

X003

N085

Z-007

I007

X002

N095

X017

M03 M08 E

K99999

E

K007

E

K99999

E

K7071

E

K99999

E

K0

E

I99999 Z-023

N090

E F00002

Z-018757 X01

E

I99999

Z-013

N050

N065

Turning tool

E

N035

N055

E

E K99999

E F0004

I99999

M09

E

M05

E

N100

Z027

K99999

T02 M06

E

N110

Z-03

K99999

M03

E

N115

X-017

N120

X0002

I99999

E F00002

M08

E

Parting tool

464

Machine Tool Design and Numerical Control

N125

X-034

N130

X0002

N135

X05

N140

X05

I99999

E F0004

I99999

M09

E E

K99999

M05 M02 E

4. Control unit has provision for suppression of trailing zeros. 5. Two miscellaneous functions may be programmed in a block, provided they are not contradictory. 6. The spindle should make 500 rev/min; the machine control unit accepts the S function in the magic-three code; the code equivalent of 500 is 650. 7. The feed rate for cutting operations is 0.2 mm rev; the machine control unit accepts the F function as mm/rev or mm/min; the F function is represented as F 4.1 (0000.0). 8. The feed rate for accelerated travel is 4 mm/rev; the F function representation and format is the same as above. 9. An approach of 5 mm should be provided before the tool starts cutting. 10 A dwell of two seconds should be provided when the feed rate is changed; the dwell period in seconds (two digits) is punched after the X character; for instance, a two-second dwell is written as X02. 11. Nose radius of the turning tool (designated as T01) is 3 mm. 12. Job of total length 90 mm is cut off using a 5 mm wide parting tool (designated as T02). 13. Format of I and K codes during circular interpolation is I 3.3 and K 3.3.

M03

Spindle on, clockwise direction

M08

Coolant on

M05

Spindle stop

M09

Coolant off

M02

End of program, machine stop, program is rewound

Numerical Control of Machine Tools 465

The program manuscript has been prepared keeping the tool-radius compensation in view. The reader would do well to refer to the tool radius compensation problems discussed earlier. Special attention of the 1. 2. 3. 4. 5. 6. 7.

N040 N050 N055 N060 N070 N075 N080

8.3

tool departure of the nose radius centre in the X direction is – 42 mm. tool departure in the Z direction is – (5 + 10 – 3 + 1) = – 13 mm (see Fig. 8.34). see Fig. 8.34. tool departure = 20 – r tan 22.5° = 18.7574 mm (see Fig. 8.32). tool departure = 20 + r tan 22.5° = 212426 mm (see Fig. 8.32). see Fig. 8.33. tool departure = – (20 + r) = – 23 mm (see Fig. 8.33).

COMPUTER AIDED PART PROGRAMMING

8.3.1 Computer and the Numerical Control Programming Languages The programs for machining simple parts are small and do not require much calculation. Such programs may be prepared manually by the methods described in the preceding

X

End point

formed surfaces, a lot of calculations have to be done and the program also is often very long. Consider, for instance, that a cutting tool of radius r straight lines. It is then required to determine the length of

Z L

the coordinates of the starting and end points of each line and calculate the departures of the tool centre. This exercise is quite cumbersome even for one line. Considering the fact Starting that the number of approximating lines can run into hundreds, point one can easily imagine the volume of calculations involved and the time required for manually punching the program Fig. 8.36 Diagram depicting the quantities required to be calculated for each running into thousands of statements. Moreover, chances of elementary line errors creeping in during such voluminous calculations are considerable. A computer does calculations many times faster and also commits no errors, provided the routines for calculations are correct. Thus, the use of a computer in such cases not only helps save hundreds of hours of monotonous work, but also ensures an error-free program. C and C++

with instructions having typical machine tool application, such as LINE, POINT, CIRCLE, RADIUS,

466

Machine Tool Design and Numerical Control

TOLERANCE, etc. A general processor endows the computer with the capability to understand terms which a NC programmer would require to prepare machining instructions for a particular part. 2. A part program which the programmer writes for a particular workpiece. The script of this part program differs from the manually prepared program manuscript in that control statements are expressed through ‘words’ comprising the general processor. program preparation procedure. At this stage, the programmer can only receive a computer listing of coordinates and tool departures. For obtaining the part program, he must complete the control statements by which machining of the part is contemplated. For this, the computer must be provided with another set of compiler instructions through which format details of the machine tool control unit are taken into account. This set of instructions, known as a POST PROCESSOR, adapts the general information about coordinates Large general processors are of a universal nature and include programming and post-processing instructions for almost any machine tool. Many machine tool manufacturers develop their own programs and post processors for their equipment. In any case, it is the programmer’s responsibility to clearly understand the programming and post processing instructions of the machine tool for which he undertakes to prepare the program. Even by a conservative estimate, more than a hundred general processors have been developed by now. However, most of them have gained limited application only because they were developed by

APT (Automatically Programmed Tools) for production-engineering purposes and is still the most powerful of all existing general processors. A brainchild of MIT (USA), the APT is continuously under development and over the years four versions of APT have been produced. The APT-IV version contains approximately 400 words and can handle the most complex of three-dimensional programming problems. In fact, the APT is rather too powerful for handling the commonly encountered simple programming problems. Its versatility and volume which require a large computer with high storage capacity thus become a disadvantage. Attempts have been made to develop slimmed-down versions of APT that are simpler and can be handled by a medium- or small-size computer. The important ones are ADAPT, AUTOMAP and EXAPT. ADAPT is a subset of the APT language which consists of 160 words plus punctuation. It is particularly suitable for two-dimensional programming problems and requires only a medium-size computer. AUTOMAP is a still more limited version of APT which is suitable for two-dimensional programming of lines and circles. This program consists of 50 words and punctuation and is quite easy to learn. EXAPT (Extended subset of APT) consists of a group of APT-based processors developed in West 1. EXAPT I, which is applicable to positioning-type NC systems. 2. EXAPT II, which is applicable to lathe operations. 3. EXAPT III, which is applicable primarily to drilling and milling operations involving 2.5 dimension control, i.e., continuous-path control in the XY-plane and control of depth only along the Z-axis. A few general processors that are not compatible in syntax or vocabulary with APT have also found fairly wide application either due to their simplicity or by virtue of their link to NC machine tools of certain popular manufacturers. Examples of such general processors are SNAP, AUTOSPOT, SYMAP, etc.

Numerical Control of Machine Tools 467

SNAP computer. It does not require a postprocessor if used with Brown and Sharpe machines. This program is effective mainly for point-to-point work. AUTOSPOT (Automatic System for Positioning Tools) contains about 100 words plus punctuation and is used mainly for point-to-point programming problems. SYMAP 1. 2. 3. 4.

SYMAP (P) for positioning control systems. SYMAP (S) for straight-line control systems. SYMAP (B) for 2.5-dimension control systems. SYMAP (PS) which is a combination of SYMAP (P) and SYMAP (S) and is applicable to drilling and straight-line milling. 5. SYMAP (DB) for lathes with continuous-path control. 6. SYMAP (DS) for lathes with straight-line control.

The number of words in SYMAP languages is less than the corresponding EXAPT languages, and, therefore, they can be used with medium-size computers. UNIAPT provides a limited version of APT to be implemented on small computers, thus allowing many small shops to possess computer assisted part programming. It was developed by United Computing Company of Carson, California, USA. SPLIT (Sunstrand Processing Language Internally Translated) is a proprietary system intended for Sunstrand machine tools. It can handle upto 5-axis positioning and also has contouring ability. No separate post-processor is required.

8.3.2 APT Programming System The APT programming system has been selected for describing computer aided part programming because it is, despite its complexity, the most popular among the general processors on account of its application to 1. Motion statements, which describe the path of the cutting tool. 2. cutting tool. 3. P particular machine tool. 4. Auxiliary statements that are not covered by the above three, e.g., cutter diameter, tolerances, etc.

APT Motion Statements for Point-to-point Programming It should be abundantly clear to the reader by now that in a point-to-point control problem the cutter is directed to proceed from one point to another, the path being of no consequence. The APT words to direct a cutting tool from point 1 (APT notation P1) to point 2 (P2) and then on to point 3 (P3) would be FROM/PI

468

Machine Tool Design and Numerical Control

geometry statements for various surfaces, lines and coordinates will be described later. If the coordinates of the starting points are not known, the program must be provided with X, Y and Z incremental departures to the target point. Suppose the tool is stationed at a point A (coordinates unknown) and has to be directed to move to a point B to which the incremental departures are X = 20, Y = 10, Z = 30. The APT statement would be

If the points to which the cutting tool is successively directed form a certain pattern, the programming is considerably

P3

20 holes

on which 300 holes have to be drilled (15 holes along the X-axis and 20 holes along the Y-axis forming the grid of 300

PAT1 = PATERN/LINEAR, P1, P2, 15 PAT2 = PATERN/LINEAR, P1, P3, 20 The word PATERN should not be misconstrued as wrong the APT language to PATERN due to the restriction of a maximum of six letters in a word. The motion statements for the

P1

P2 15 Holes

Fig. 8.37 Component drawing

FROM/SETPT CYCLE/DRILL, CAM, 1 CYCLE/OFF n1, n2, n3 will omit points n1 n2 and n3.

APT Motion Statements for Continuous-path Programming

In a continuous-path motion the cutter may, in general, be considered to be guided in its path by three planes. One of the planes guides the bottom surface of the tool and is known as the part surface (PS or PSURF). The plane guiding the side of the cutting tool, i.e., the plane along which the tool side moves is known as the drive surface (DS or DSURF).

Numerical Control of Machine Tools 469

A third plane is used to stop the cutter or change its direction. This plane is known as the check surface (CS or CSURF). These planes are shown in Fig. 8.38a for a tool moving on a straight line and in Fig. 8.38b for a and may be real or imaginary. Depending upon the machining problem, the cutter may be required to stop at the check surface in one of

check surface after the/sign. 4°

RF

CSURF

Range of GOFWD command

U DS

PSURF

F DSURF

UR

CS (a)

GOON/

(d)

CSURF

DSURF

GOTO/

CSURF

DSURF

DSURF

CSURF

PSURF (b)

GOPAST/

GOBACK

GOFWD

(e)

(c)

Fig. 8.38

(a) Drive, part and check surfaces in straight line motion (b) Drive, part and check surfaces in curvilinar motion (c) Stopping statements

left of the drive surface, to the right of the drive surface or on the drive surface. The positioning statements the/sign. The axis of the cutter is aligned with the help of the direction vector components, I, J and K. The motion statement is TLAXIS/I, J, K. If the tool axis is to be aligned normal to the part surface, this is done by the command TLAXIS/NORMPS. After the cutter has been brought TO, ON or PAST the check surface, its further motion can be controlled by specifying the direction relative to the previous motion. The direction commands and their APT statements are given below.

470

Machine Tool Design and Numerical Control

In all these statements, the/sign should be followed by the drive, part and check surfaces of the particular operation in the given sequence. If the part surface is common for a number of motions, then it may be programmed only once. While assigning the direction at a turn, the programmer should imagine himself to be aligned with the drive surface and looking in the direction of the previous motion. Let us apply the motion statements described above to a simple programming problem shown in Fig. 8.39. The end mill cutter is stationed at set point (SP or SETPT). From the set point it should proceed to point 1 which is formed by the intersection of planes H and A. Then it should follow the path marked by points 1–2–3–4–5–6. The tool end should be guided by known as the start-up statement. For the given example this statement is PSURF

3

E

4 I

H B 1

2 A

F 5

6 G

SP

Fig. 8.39 Example describing the directions of turns

FROM/SP

used in point-to-point programming. Also, note that the three surfaces have been programmed in the sequence that the part surface does not change during the whole path, APT statements for subsequent tool movements

Numerical Control of Machine Tools 471

The start-up statement, as given above, is programmed in its most general form. Simpler start-up statements are possible with the help of two or even one surface. In the case of the statement with the help of two surfaces FROM/SP the cutter will move from the set point to the start-up point on the drive surface by the shortest path. If the part surface happens to be the XY-plane itself (i.e., Z coordinate = 0), the start-up statement can be written as FROM/SP

The motion in the direction of a vector V is programmed through the statement FROM/SP INDIRV/ the values of the X, Y and Z components of the vector following the/sign. The MACRO statement in the part program tells the APT system that the block of motion statement that follows is to be stored for later, generally repeated execution. The end of the block of statements is signalled by a TERMAC command. The MACRO statement is assigned a symbol, e.g., ABC123 = MACRO Later on when the block of motion statements bound between the MACRO and TERMAC words is CALL/ABC123 The MACRO command is a very powerful feature of APT and is similar to a subroutine in Fortran. The MACRO command helps in reducing the total number of statements required in a given APT program.

APT Geometry Statements the surface to be machined by the cutting tool. A typical geometry statement, as shown below, consists of P1 = POINT/4.000, 2.000, 3.000 The second part gives a general description of the geometrical element (POINT in the given case) and X, Y and Z coordinates of the point. The geometrical elements that constitute a trajectory may be obtained, and hence described in a number upon as a geometrical element that passes through a given point and is tangent to a given circle. The various described below.

472

Machine Tool Design and Numerical Control

POINT

P1 = POINT/X, Y, Z P2 = POINT/INTOF, L1, L2 3. Points obtained as the intersection of a line L1 and a circle C1 (Fig. 8.40). When a line intersects a P3 = POINT/YSMALL, INTOF, L1, C1 or P3 = POINT/XSMALL, INTOF, L1, C1 or

Y

Y P4

P5

L1

C2

C1

P3 C1 P6

0

X

0

X

Fig. 8.40 Points obtained by the intersection of a line and circle

Fig. 8.41 Points obtained by the intersection of two circles

Here the words XSMALL and YSMALL indicate that P3 is that point of intersection which has smaller values of the X and Y coordinates as compared to the other point of intersection P4. 4. Points, obtained by the intersection of two circles C1 and C2 (Fig. 8.41). P5 = POINT/XSMALL, INTOF, C1, C2 or Point P6 may be described in an identical manner. 5. Point, which represents the centre of a circle C1. P7 = POINT/CENTER, C1.

Numerical Control of Machine Tools 473

LINE 1. Line, obtained by joining two points P1 and P2 L1 = LINE/P1, P2 2. Line, passing through a point P1 and making a particular angle (say 45°) with the X-axis 3. Line, passing through a given point P1 and making a particular angle (say 45°) with another line (say L2) 4. Line passing through a given point P1 and parallel to another line (say L3) L4 = LINE/PI, PARLEL, L3

Y

10

10

5. Line, parallel to another line (say L3) and located at a given distance from it (say 10 mm). As is evident from Fig. 8.42, there may be two lines satisfying the above description. For instance, line L5 may be described as

L5

Similarly, line L6 may be described as

L3

L6 = LINE/PARLEL, L3, YSMALL, 10

L6

0

6. Line passing through a given point (P1) and tangent to a given circle (C1).

X

Fig. 8.42 Lines parallel to a given line

It is evident from Fig. 8.43 that two lines satisfy the above description. The APT statements of the two Y

L7 = LINE/PI, LEFT, TANTO, C1 L7

indicates that line L7 lies to the left of the circle when looking from that point towards the circle.

C1

P1

L8

PLANE 1. Plane, passing through three points P1, P2 and P3 which do not lie on a straight line PL1 = PLANE/P1, P2, P3 2. Plane, passing through a given point P1 and parallel to a given plane PL1

0

Fig. 8.43

X

Two lines tangent to a given circle and passing through a given point

PL2 = PLANE/PI, PARLEL, PL1

474

Machine Tool Design and Numerical Control

3. Plane, parallel to a given plane PL1 and located at a particular distance (say 10 mm) from it. There may be two planes satisfying the above description. For instance, the required plane may have aZ PL4 = PLANE/PARLEL, PL1, ZSMALL, 10 4. Plane, parallel to the XY-plane and located at a given distance (say 10 mm) from it. If the Z coordinate of the required plane is positive, the APT statement would be PL5 = PLANE/0, 0, 1, 10 If the Z coordinate of the required plane is negative, the APT statement would be PL6 = PLANE/0, 0, 1,–10 CIRCLE X, Y and Z) of the centre and radius R CI = CIRCLE/X, Y, Z, R R C2 = CIRCLE/CENTER, P1, RADIUS, R 3. Circle, having centre P1 and tangent to another circle C1. It is evident from Fig. 8.44, that there may be two such circles C3 and C4. Their respective APT C3 = CIRCLE/CENTER, P1 SMALL, TANTO, C1 4. Circle of a given radius R and tangent to two lines L1 and L2. It is clear from Fig. 8.45 that there are four circles satisfying the above description. The APT statement of circle C5 is Y Y

L1

C6

L2

C1 C7 C3 P1 C5

C4

C8 0

Fig. 8.44

X

Two circles having a given centre and tangent to a given circle

0

Fig. 8.45

X

Four circles tangent to two given lines

Numerical Control of Machine Tools 475

Y coordinate of a line parallel to L2 and passing through the centre of circle C5 is greater than that of line L2. Also, XSMALL, L1 indicates that the X coordinate of a line parallel to L1 and passing through the centre of circle C5 is less than the X coordinate of line L1. The APT statements of circles, C6, C7 and C8 can be written in an identical manner,

C8 = CIRCLE/YSMALL, L2 XSMALL, L1, R

APT Postprocessor Statements

Postprocessor statements can differ depending upon the type of machine tool. Even for a particular type of machine tool, they can vary depending upon the manufacturer. The APT post processor statements should, therefore, be used in conformity with the “postprocessor listing” of the particular machine tool. A few APT postprocessor statements that are more or less universally applicable on all machine tools are described below. 1. COOLNT/ON directs the control unit to switch on the coolant. If the machine tool has provision COOLNT/MIST or COOLNT/FLOOD, whichever is appropriate. 2. COOLNT/OFF directs the control unit to switch off the coolant. 3. SPINDL/is used for controlling the spindle speed. The spindle speed may be expressed in rpm (SPINDL/600 RPM), surface speed in ft/min (SPINDL/150, SFM) and surface speed in m/min

4. 5. 6.

7.

8. 9.

tool has only stepped speed regulation, the computer automatically selects the nearest rpm available on the speed box. SPINDL/ON. SPINDL/OFF. FEDRAT/is used for controlling the feed rate. If the feed rate is expressed in mm/min, then only the numerical value need be indicated. For instance, the APT statement FEDRAT/150 would automatically be interpreted as a feed rate of 150 mm/min. However, if the feed rate is expressed in mm/rev, then the word MMPR should follow the numerical value, e.g., a feed rate of 0.2 mm/rev would be programmed by the APT statement FEDRAT/0.2, MMPR. RAPID directs the operative member of the machine tool to move at an accelerated feed rate. A RAPID statement is valid only for a single execution. Therefore, it must be programmed before every motion statement in which a rapid traverse is desired. PITCH prepares a lathe machine for the threading mode. A thread of 8TPI would be programmed by the statement PITCH/8. THREAD/is used to start a threading cycle.

the postprocessor used on the Cincinnati milling machine with Bendix control unit. The APT statement for the BXCIN post processor would be MACHIN/BXCIN. 11. TURRET/is used to select a particular turret position or tool from an automatic tool-changing device. Supposing a machine tool has a front and a rear turret, then the APT statement for position No.1 of the front turret would be TURRET/1, 0 FRONT. the next tool change. LOADTL/is the command for loading of the selected tool.

476

Machine Tool Design and Numerical Control

12. CYCLE/is used to execute a particular cycle of operation, e.g., a drilling cycle, tapping cycle, etc. The APT statement for a drilling cycle would be CYCLE/DRILL. This word is generally employed on two-axis control systems, the depth being controlled by a cam. However, if the machine tool also has Z-axis control, then information about the movements in the Z-axis direction may also be programmed. For instance, the statement CYCLE/DRILL, RAPID, TO, – 10, FEDTO, – 15, MMPM, 150 would call the drilling cycle, direct the tool to move rapidly to the Z coordinate – 10 mm, and then move further at a feed rate of 150 mm/min to the Z coordinate – 15 mm.

14. STOP commands the machine tool, coolant and spindle to stop but does not shut off the MCU. rewinding. 16. SEQ No/ is a command for giving instructions regarding the automatic allocation of sequence numbers, e.g., SEQ No/ON, INCR, 5 instructs commencement of sequence numbering and their incrementing by 5. SEQ No./l00, INCR, 1 instructs sequence number to begin from 100 and incrementing by 1. SEQ No./OFF instructs termination of sequence numbering. 17. DELAY/T provides a dwell of T seconds.

APT Auxiliary Statements

2. CLPRNT directs the computer to print out a listing of the coordinates of the points that determine various cutter motions in its trajectory. 3. REMARK statement enables the part programmer to make a note in the program. It does not affect the part program and can be employed to direct the operator or simply to elaborate a particular statement. between the words. Examples of the two types of REMARK statements are given below. REMARK/T15 would direct the operator to clamp tool No. 15 REMARK/INTOL means that the curve is approximated by chord 4. CUTTER/is used to specify the diameter of the cutting tool. The dimension follows the/sign, e.g., the APT statement for a 50 mm diameter cutter would be CUTTER/50. When this statement is given in the

Numerical Control of Machine Tools 477

program, the computer automatically calculates the tool centre offset and prepares tape instructions to describe the movement of the centre of the cutter. tolerance value follows the/sign, e.g., a tolerance of 0.005 mm would be expressed through the APT statement INTOL/0.005. For a tolerance of 0.05 mm, the APT statement would be OUTTOL/0.05. The concept of INTOL and

9. The $ symbol is used at the end of a line when the statement has to continue in the next line. 10. The $$ symbol is used to separate the APT statement to be punched on the tape from some additional remarks regarding the statement which the programmer desires in the print out. For instance, in the a part of the APT program, whereas the portion to the right of the $$ symbol is only printed out in the part program listing.

APT Computation Statements

The computation operators used in APT for algebraic and trigonometric calculations are the same as those used in FORTRAN. A few of them are listed alongside. surfaces required to describe the geometry of the part—the points as P1, P2, …, etc., the lines as L1, L2, …, etc., the planes as PL1, PL2, …, etc., and the circles as C1, C2, …, etc. There are no hard and fast rules concerning the sequence of various statements in an APT program.

Arithmetic operation or trigonometric function

APT operator

1. 2.

+, X + Y –, X – Y

+, X + Y – X–Y

3.

x, XxY

*, X*Y

4.

÷, X ÷ Y

/, X/Y

5.

X

2

X**2 SQRTF(X)

6.

X

7.

e

x

8.

sin X

SINF(X)

9.

cos X

COSF(X)

10.

In X

11.

arc tan X

12. 13.

tan X log10 X

EXPF(X)

ATANF(X) TANF(X)

478

Machine Tool Design and Numerical Control

1. Identify the part and postprocessor required. 2. Write the auxiliary statements for coordinate listing (CLPRNT), tolerances (INTOL/OUTTOL/), cutter diameter (CUTTER/), unit system (UNITS/MM), etc. 3. Write the postprocessor statements for feed rate (FEDRAT/) and spindle speed (SPINDL/). 4. W 5. Write the motion statements. Often, auxiliary and postprocessors statements are interspersed with the motion statements for convenience. This pertains particularly to the statements COOLNT/ON, COOLNT/OFF, SPINDL/ON and SPINDL/OFF. Also, a few geometry statements may sometimes be given separate along with the motion statements. Thus, the procedure given above should be taken as a general guideline, but, wherever necessary, the sequence of statements can be changed if this provides for a program which is easier to understand. A word of caution is, statement in which it is required.

8.3.3 APT Program—Point-to-point Problem Problem 1

PD, 11. The holes are to be made with a 15 mm HSS drill in the plate of mild steel. The required machining Z-axis control, but the feed and spindle rpm are set manually. The APT program for the above part is given as follows. PARTNO FLAT PLATE NO 1 MACHIN/PD, 11 Y

4(25, 75)

Z

3(50, 75)

SP (75, 50) 5(30, 50) 5P (75, 50) 1(25, 25)

2(50, 25) X 10

0

X

Fig. 8.46 Component drawing demand special attention

Numerical Control of Machine Tools 479

UNITS/MM SETPT = POINT/75, 50, 50 P1 = POINT/25, 25, 5 P2 = POINT/50, 25, 5 P3 = POINT/50, 75, 5 P4 = POINT/25, 75, 5 P5 = POINT/30, 50, 5 REMARK/SET, FEDRAT, 0.5 MMPR REMARK/SET, SPINDL, 600 RPM REMARK/DRILL, 15 MM SPINDL/ON COOLNT/ON FROM/SETPT RAPID

COOLNT/OFF SPINDL/OFF RAPID REWIND FINI

1. A 5-mm approach is provided, which explains the Z-coordinate value equal to 5 mm for points 1 to 5,

480

Machine Tool Design and Numerical Control

2. An over travel of 5 mm is provided, which explains the fact that the drill tip is directed to move to Z = – 15 mm in all the CYCLE/statements, although the plate is only 10-mm thick, 3. T plate thickness (10 mm) and approach (5 mm). Problem 2 Write the APT program for drilling a grid of holes in the part shown in Fig. 8.37. PD, 01. The diameter of spindle speed = 5000 rpm. The feed and speed are programmed. Drilling cycle is used with control of depth by a cam (say CAM 1). The coordinates of P1 are (50, 50) and the holes in the grid are spaced 10 mm apart. PART NO FLAT PLATE NO 2 MACHIN/PD, 01 UNITS/MM CLPRNT SETPT = POINT/0, 0, 50 P1 = POINT/50, 50, 5 P2 = POINT/190, 50, 5 P3 = POINT/50, 240, 5 PAT 1 = PATERN/LINEAR, PI, P2, 15 PAT 2 = PATERN/LINEAR, P1, P3, 20 FEDRAT/0.2,MMPR SPINDL/5000, CLW REMARK/DRILL, 3 MM SPINDL/ON COOLNT/ON FROM/SETPT RAPID CYCLE/DRILL, CAM, 1 $$ SET CAM1 CYCLE/OFF SPINDL/OFF COOLNT/OFF RAPID REWIND FINI

Numerical Control of Machine Tools 481

8.3.4 APT Program—Continuous-path Problem Problem 1 (Milling Problem) Write the APT program for machining the part shown in Fig. 8.47.

Let us additionally assume that the particular machine tool has numerical control of feed-rate regulation, but the spindle speed is adjusted manually. Y C1

R

30

P5 (150, 100) (80, 100)

L4

L5 P4 (130, 50) P3 (150, 40) L2

L3

SP

P1 (50, 25)

L1

P2 (150,25) X

Z

X

8 Table

Fig. 8.47 Component drawing

The APT program for the part is given below. PARTNO BRACKET MACHIN/ABM, 8 UNITS/MM CLPRNT

482

Machine Tool Design and Numerical Control

SETPT = POINT/0, 0, 0 $ $ USE 8 MM SHIM TO LOCATE TOOL P1 = POINT/50, 25, 0 P2 = POINT/150, 25, 0 P3 = POINT/150, 40, 0 P4 = POINT/130, 50, 0 P5 = POINT/50, 100, 0 L1 = LINE/P1, P2 L2 = LINE/P2, P3 L3 = LINE/P3, P4 C1 = CIRCLE/80, 100, 30 L5 = LINE/PI, LEFT, TANTO, C1 PL1 = PLANE/0, 0, 1, 0 CUTTER/10 FEDRAT/0.1, MMPR REMARK/SET, SPINDL, 1000, RPM INTOL/0.01, $$ CUT CHORDS WITHIN 0.01 MM OF PERFECT CIRCLE FROM/SETPT SPINDL/ON COOLNT/ON

SPINDL/OFF COOLNT/OFF RAPID FINI Problem 2 Milling Problem Involving Repetitive Machining in a Particular Cycle. Write the APT program for milling a pocket in the part shown in Fig. 8.48. 1. From set point to point P1. 2. Vertical feed to a particular depth (plunge cut).

Numerical Control of Machine Tools 483

3. 4. 5. 6. 7. 8.

Move to point P2 (start up statement). Move to point P5. Move along the path P5-P6-P7-P8 shown by arrows. Move from point P8 to point P9. Repeat the path followed in (5) above till the pocket milling is completed. Repetition of the cycle from (2) to (7) above with a new depth of cut. Y

P3 (50, 120)

P7 L4 P6 P1 (50, 50)

P4 (120, 120)

P9 P8 P5 L2

L5 L3 L1

P2 (120, 50)

S.P. X Z

S.P. (0, 10) 0

5 10

X

Table

Fig. 8.48 Component drawing

postprocessor. The feed rate and spindle speeds are set manually, but the machine has Z-axis control. Let us rate = 0.1 mm/rev, spindle speed = 1000 rpm and maximum depth of cut = 5 mm. Obviously, the pocket of depth 5 mm can be obtained in this case in one cycle. However, if the maximum allowable depth of cut were to be less, the APT program would not have changed and all that would have been necessary is to repeat the program statement with a different Z coordinate of the cutter tip.

484

Machine Tool Design and Numerical Control

PARTNO BLOCK UNITS/MM CLPRNT SETPT = POINT/0, 0, 10 P1 = POINT/50, 50, 0 P2 = POINT/120, 50, 0 P3 = POINT/50, 120, 0 P4 = POINT/120, 120, 0 P10 = POINT/50, 50, 10 LI = LINE/PI, P2 L2 = LINE/P2, P4 L4 = LINE/P1, P3 L5 = PL1 = PLANE/0, 0, 1, – 5 CUTTER/10 REMARK/SET, FEDRAT, 0.1, MMPR REMARK/SET, SPINDL, 1000, RPM FROM/SETPT RAPID SPINDL/ON COOLNT/ON A = 10 $$ VARIABLE A DEFINED EQUAL TO 10 MM B = 20 $$ VARIABLE B DEFINED EQUAL TO 20 MM ABC123 = MACRO $$ MACRO STATEMENT IDENTIFIED AS ABC123

TERMAC $$ THE MACRO LOOP ENDS HERE A = A + 20 $$ A IS NOW EQUAL TO 30 B = B + 20 $ B IS NOW EQUAL TO 40

Numerical Control of Machine Tools 485

CALL/ABC123 $$ EXECUTE MACRO CYCLE A = A + 20 $$ A IS NOW EQUAL TO 50 B = B + 20 $$ B IS NOW EQUAL TO 60 CALL/ABC123 ………… ………… ETC.

SPINDL/OFF COOLNT/OFF RAPID FINI Problem 3 (Lathe Problem) Write the APT program for the part shown in Fig. 8.35. tool used for the machining operations has nose radius r = rate = 0.3 mm/rev and spindle speed = 500 rpm. Both the speed and feed are regulated through numerical control. The designation of lathe axes for APT programming is the same as that for milling machines, i.e., XY. The XY coordinates are changed to ZX coordinates by the post processor when the control tape is punched. The APT program for the part is given as follows. PARTNO SHAFT NO 1 MACHIN/PLATH, 10 UNITS/MM CLPRNT SETPT = POINT/50, – 50 P1 = POINT/120, – 5 P2 = POINT/105, – 5 P3 = POINT/80, – 10 P4 = POINT/70, – 20 P5 = POINT/50, – 30 P6 = POINT/20, – 30 P7 = POINT/115, – 10 LI = LINE/PI, P2 CI = CIRCLE/100, – 5, 5 , C1 L3 = LINE/P3, P4 C2 = CIRCLE/50, – 30, 10

486

Machine Tool Design and Numerical Control

L5 = LINE/P5, P6 L6 = LINE/P6, PERPTO, L5 CUTTER/6 FEDRAT/0.3, MMPR SPINDL/500, RPM INTOL/0.01 OUTTOL/0.01 FROM/SETPT RAPID COOLNT/MIST

SPINDL/OFF COOLNT/OFF RAPID REWIND FINI

Review Questions 8.1 In manual part programming and tape preparation for a NC drilling machine, the spindle speed was coded as S684 in the magic 3 code. Determine the spindle rpm. 8.2 (i) C1 = CIRCLE/CENTER, P1, TANTO, L1 (ii) FEDRAT/4.0 8.3 The work table of a CNC machine is driven by a stepper motor having 200 steps in one rotation and a lead screw of pitch 4 mm. Determine the BLU of the machine tool. 8.4 In a NC machine tool, a stepper motor with step angle of 1.8° drives the work table by means of a lead screw of pitch 2 mm. Determine the basic length unit value of the machine tool.

Numerical Control of Machine Tools 487

8.5 The vertical travel of a CNC machine is powered by a DC motor by means of a lead screw of pitch machine tool in the vertical direction is 0.005 mm, determine the pulses generated by the encoder for a move of 9 mm in the vertical direction. 8.6 Y

R

=3

5

2 Zero in the full-floating zero system 4

3

50

70

35 1 10 10

20

25

25 70 X

40 Fixed machine zero

Zero in the full-zero shift system

Fig. 8.49 Component drawing

8.7 For the part shown in Fig. 8.49, prepare the program manuscripts for positioning the machine tool

(ii) Full zero shift, tab sequential, X 2.4, Y 2.4, zeros not suppressed. (iv) Fixed zero, variable-block format, X 3.3, Y 2.4, trailing zeros suppressed.

488

Machine Tool Design and Numerical Control

8.8 Depict the characters on the punched tape for the program manuscripts of the previous problem. Compare the punched tapes and determine the most economical programming format. 8.9 Prepare the complete program manuscript, including the preparatory, miscellaneous and tool change functions for drilling 10 mm holes at points 1 and 2, and 15 mm holes at points 3 and 4 in the 20-mm thick steel plate shown in Fig. 8.49. The details of machine tool format and the programming and

(ii) Point-to-point control (iii) Optional absolute or incremental dimensioning in mm or inches (iv) Tool change position is (150, 150) (v) Fixed-zero system with provision for suppression of trailing zeros (vi) Z-axis control with the help of cams. 8.10 Prepare the complete program manuscript for the component shown in Fig. 8.35, except that all the dimensions are increased two times. The cutter zero position is the same as in Fig. 8.35. The format (i) Word address format and provision for suppression of trailing zeros (ii) Optional absolute or incremental dimensioning (iii) Dimensions only in mm (iv) Spindle speed = 350 rpm (v) Feed = 0.08 mm/rev (vi) Feed for accelerated travel = 5 mm/rev (vii) Feed and spindle speed values accepted in magic three code only (viii) Tool nose radius = 5 mm (ix) Interpolation codes are I 3.3, K 3.3. 8.11 Write the APT program for the hole drilling problem described in Q. 8.9 above. The set point and tool-changing position is (150, 150). The format details and machining and programming instructions (i) (ii) (iii) (iv) (v) (vi)

Three-axis control Postprocessor DRIL 5X Spindle speed and feed settings are numerically controlled Cutting speed = 30 m/min Feed = 0.1 mm/rev Through holes to be drilled in the 20-mm thick plate.

8.12 Write the APT program for making a grid of 625 (25 ¥ 25) holes of diameter 2 mm spaced at 5 mm from each other in a 10-mm thick steel plate. The set point is (0, 0) and the nearest corner hole is located at (40, 40). The required cutting speed = 40 m/min and the feed = 0.08 mm/rev. Both the spindle speed and feed are set manually. The Z-axis movement is cam-controlled. No post processor is required. 8.13 Write the APT program for machining the periphery of the 20-mm thick steel plate shown in Fig. 8.49 by a 6-mm HSS end mill cutter. The required cutting speed and feed are 30 m/min and 0.1 mm/rev, respectively and both are set manually. The rapid travel feed is 3 mm/rev. Assume a suitable set point and appropriate values of the tolerances for circular interpolation. The machine requires a postprocessor MILL XYZ.

Numerical Control of Machine Tools 489

1.0

1.0

0.5

8.14 Write the APT program for machining the lathe component described in Q. 8.9 above. Both the speed and feed settings are numerically controlled and a postprocessor LATHE 7 is required. 8.15 chining the holes among the following two on 3 4 a machine tool with absolute dimensioning, 2 tion (8,4) 5 0.5 (a) 1 – 2 – 3 – 4 – 5 – 6 6 1 (b) 1 – 3 – 4 – 6 – 2 – 5 points 1 and 2, assuming that the NC ma-

Fixed zero

1.5

4.0

1.5

FIg. 8.50

X2.3,Y2.3 suppression of trailing zeros and absolute dimensioning. The sequence of machining the holes is 3 – 4 – 6 – 2 – 5 – 1 (iii) Write a NC program for drilling of holes in the sequence 1 – 2 – 3 – 4 – 5 – 6. The machine tool 8.16 machine using f EOB, full zero shift, trailing zeros suppressed. Speed is 500 rpm (S3) and feed is 4.3 in/min (F2.4). Tool proceeds from machine zero to A in absolute mode. Subsequent movements are incremental in the direction of the arrow. Use M51 for Z-axis control. 2.0 1.0

R 1.0

5.0

1.5

A(3.0, 3.0) 3.0 + M/c Zero

1.5

5.0 All dimensions in inches

Fig. 8.51

8.17 Write the complete NC part program for contour milling and drilling of hole in the part shown in Fig. 8.52 on a 2.5 axis vertical milling machine. RPM = 2000 in magic- 3 code, feed = 50 mm/min in 4.1 format. Machine tool format is full range zero shift, absolute dimensions X3.3, Y3.3, tab sequential,

490

Machine Tool Design and Numerical Control

suppression of trailing zeros. Contour milling is done with f 20-mm end mill cutter and proceeds in the direction of the arrow starting from point 0. Y

5

f1

R3

0

30 X

50 100

Fig. 8.52

8.18 Write the NC part program for making two f12 mm holes at 1 and 2 and milling a 10 mm wide and 5-mm deep slot in the 10-mm thick plate shown in Fig. 8.53 on a 2.5 axis machine. The machine tool zeroes. Speed and feed are set manually and need not be programmed. Home position for tool change is (150, 150). Milling is initiated at point 3. 100 15 15

1 2

10

3 100 50

2 Holes, M12

10

2

2

15 Ym

15 50 Ym

Fig. 8.53

Numerical Control of Machine Tools 491

8.19 Prepare a NC part program for the part shown in Fig. 8.54. (i) Word address format (iii) Absolute and incremental dimensioning (iv) Fixed zero in Z and X (v) Trailing zeroes suppressed (i) (ii) (iii) (iv) (v)

Feed in mm/min; s = 0.2 mm/rev, accelerated feed = 3 mm/rev Speed in magic-3 code, rpm = 500 Home position = (150, 40) 6 mm groove is for parting off Tool nose radius = 2 mm Machine Zero

2.5 ¥ 45°

f 35

f 40

f 50

f 40

Z f 20

X 6

20

5

25

30

15

120

Fig. 8.54 y

L5

R1.0

1.0

BRACKET, postprocessor – TRDL, RPM – 500, feed = 4.0 in/min, inside tolerance = 0.001 inch, outside tolerance = 0, set point is (–2,–2,0) and the X-Y plane coincides with the top surface of the component. Contour cutting starts from P1 in the direction of L1. 8.21 Write the complete APT program for contour machining of 5 mm thick part shown in Fig. 8.56 on a 3-axis milling machine using f10-mm HSS end mill cutter. The feed rate is 125 mm/min and the cutting speed is 500 rev/min. Inside and outside tolerance for approximation of circular arc is 0.0125 and 0.0025 mm, respectively. The

P5

2.0 L6

P6

P4 C1

5.0

8.20 Write the APT part program for contour milling of the component shown in Fig. 8.55 using f

L4

L7

P3 L3

L2 P1

L1

P2

3.0 SPe

6.0

All dimensions in inches

Fig. 8.55

35° X

492

Machine Tool Design and Numerical Control

postprocessor is MACHINE/QRS and the part name is BRACKET. Machining should commence from point A and proceed along the Y-axis. All the geometrical entities required to describe the part are market with symbols on Fig. 8.56. Y L2 C1

R15 P4

P7 30

L6

L3

L1

40

5

25

P6

P2

P3

P5 L5

10

L8

15°

L4

P1 A

L7

X 25

125

S.P. 25

Fig. 8.56

8.22 Write the APT part program for external turning, facing and boring operations on the given part ‘SHAFT’ n = 500 rpm, s = 0.1 mm/rev. Hole diameter 15 is predrilled. Tool No. 1 is used for turning and Tool No. 2 for facing and boring. Facing is done from the centre outwards to the operator. Boring cut is taken on the generatix closer to the operator.

f 80

X

20

30

f 30

f 50

f 20

f 15

M/C and Zero Part Zero

Z 15 20

Tool Home (80, – 40)

75

FIg. 8.57

8.23 Write the complete APT program for the part shown in Fig. 8.58. The f 0.75 holes have already been outside and inside tolerances on circular interpolation are 0.001 inch. Speed and feed should be 400 rpm and 3.0 in/min, respectively. Part name is Plate and post processor is MACHIN/MILL.

Numerical Control of Machine Tools 493

2.0 Part thickness 1 = inch 2

.75

R 1.0

0.875

4.25

3.0

6.0

3.0

1.875

1.0

1.875

3.75

f0

SP (0, 0)

Fig. 8.58

References 1. 2. 3. 4.

Simon, W, The Numerical Control of Machine Tools, Edward-Arnold London, 1973, p. 307. Ibid., p. 101. Ibid., p. 78. Ibid., p. 139.

494

Machine Tool Design and Numerical Control

9

EXTENSIONS OF NUMERICAL CONTROL—CNC, DNC, MACHINING CENTRES

It was explained in Sec. 8.1 that a NC machine tool system consists of the machine control unit (MCU) and the machine tool itself. The MCU which is housed in a separate cabinet consists of the following elements: 1. 2. 3. 4. 5. 6. 7. 8.

Input device such as punch tape reader Reading and parity logic check circuits Decoders Interpolators and comparators Position control circuit Velocity control circuit Deceleration circuit Auxiliary function controls such as spindle on/off, coolant on/off, etc.

Elements 1–4 receive, decode and process the input information to generate data in a format suitable for the machine tool control. These elements thus constitute a data processing unit (DPU). On the other hand, elements 5–8 use the data generated by the DPU to generate proper control signals for the machine and thus constitute the control loop unit (CLU) of the MCU. In the NC machine tools, both the DPU and CLU functions are implemented in hardware, i.e., by electronic elements built into the control unit. As the data input from punched tape forms part of the control, the tape has to be read every time for machining a component, moving block-by-block between execution of program sentences. In view of the above features, NC machine tools suffer from the following drawbacks: key followed by punching of the correct data on the tape. When the duplicate tape is prepared, the on a separate length of tape which is then spliced into the original tape. It is thus seen that program 2. No provision for changing speed and feed during the cutting process. Hence, the part programmer must assign the speed and feed for worst case conditions, thereby reducing productivity. 3. Punched tape, being made of paper or thin plastic sheet is susceptible to wear, tear and damage, resulting in frequent stoppages of the machine tool. 4. Tape readers contain mechanical moving parts which are not suited for high frequency operations. This makes the tape reader the least reliable hardware component of the NC machine system, resulting in frequent stoppages of the machine tool and large down time. 5. No provision of feedback on operational performance to provide timely information to the management. 6. As the control is hardwired, the control features can not be easily altered or upgraded to keep pace with the developments in electronics and computer science. These drawbacks were the cause of widespread dissatisfaction among users of NC machines and held back the wide acceptance of NC technology for at least two decades till the midsixties. By this time, it

Extensions of Numerical Control—CNC, DNC, Machining Centres 495

became clear that despite the all too obvious advantages and tremendous potential, NC technology will not be able to deliver the promised results unless solutions were found to the problems listed above. The various developments towards achieving this goal constitute extensions of numerical control and form the subject matter of the discussion that follows.

9.1 The development of large scale integration (LSI) in electronics in the sixties gave a massive thrust to miniaturisation of electronics circuitry and saw the emergence of mainframe computers that packed much greater computation power in relatively small physical size, compared with the existing computers of the time that were built with SSI (Small Scale Integration) and MSI (Medium Scale Integration) circuits. These developments in computer technology led to the logical attempt at using a computer to take over the data processing functions of the MCU of the NC machine tools. The part programs were stored in the bulk memory of the computer and down loaded to the NC machine tool through direct connection in real time. This eliminated the punched tape and tape reader from the NC machining system, thus relieving it of the components that were the cause of most of the drawbacks listed above. As the mainframe computers were expensive, a single computer was used to store part programs for several NC machine tools and distribute them on demand. This control came to be known as distributive numerical control (DNC-1). As the computer was used mainly as a substitute of the tape reader, this mode of control was also referred to as behind the tape reader (BTR) system, implying thereby that the computer was directly linked to the NC controller, bypassing the tape reader. The distributive numerical control had its own drawbacks. Firstly, breakdown of the computer meant shut down of all NC machines connected to it. To meet this eventuality, a standby computer was kept in readiness. Keeping in mind that mainframe computers were fairly expensive at that time, the cost of maintaining two computers made the actual implementation of distributive numerical control prohibitively expensive. This

9.2 in the emergence of very large scale integration (VLSI) and microchips. This, together with advances in computer technology led to the development and rapid growth of mini and later micro computers that packed enormous computing power in small physical size. Furthermore, dramatic reductions in cost made it attractive to equip each NC machine with a dedicated computer, to take over the functions of the DPU. This came to be known as computer numerical control (CNC). Unlike the hard wired conventional NC controller, the physical components of CNC units are virtually the same, irrespective of whether they control a lathe or a milling machine. It is not the control unit elements, but the control (executive) program that makes a control unit think as a lathe or a milling machine. The control program is loaded into the CNC computer memory by the control manufacturer. Normally, the user does not tamper with the control program, but like all computer in the early years of its existence CNC was often referred to as soft wired control and the machine tools as

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soft wired numerically controlled machine tools. In the early CNC machine tools, some of the DPU functions such as feed rate generation and interpolation were left to be performed by hard wired components as they were more cost effective and allowed the use of smaller, less expensive computers. These machines are known as hybrid CNC machines. Developments in computer technology and their continuous cost reduction and easy availability have overtaken events at a pace that was hard to visualise even a decade ago and almost all CNC machine tools manufactured today are fully computer controlled. In CNC machine tools, the part program punched on tape is run only once and then stored in the computer memory. The punched tape is retained only as a back up medium. This rids the CNC controller of the tape reader during regular operation. In this sense, computer numerical control carries out the same functions as the distributive numerical control described above, with the additional advantage that failure of the control in this case affects only one machine. In recent CNC systems, the tape reader has been eliminated altogether by incorporating manual data input (MDI) consoles as off line programming modules. The MDI consoles are elaborate alphanumeric keyboards which allow writing of fairly complex part programs directly into the computer memory. In off line programming, the part program is written on a personal computer (PC) using the appropriate programming software and is then downloaded into the CNC system through a data communication line. We thus see that CNC was the next logical extension of the concept of numerical control which made NC technology more reliable and cost effective. The promise held forth by the concept of numerical control machine tools worldwide. The availability of a dedicated computer with each CNC machine tool helped in equipping these machines with advanced control and programming features, thereby further enhancing their utility. Some of these features are: 1. Improved diagnostics The diagnostics system monitors one or several parameters and sends warning signals in case of imminent breakdown. In case back-up components are incorporated in the CNC system, the faulty component is automatically disconnected and the back-up component is activated. The diagnostic system maintains a record of the downtime and of the reasons behind it. 2. In-process compensation This feature allows adjustments for errors sensed by in-process inspection probes and gauges and offset compensations for tool length and radius. 3. Advanced operating features These features include ease of editing of part programs, graphic display program storage facility. 4. Advanced programming facilities These include (i) Provision of various interpolation schemes, viz. circular, parabolic, cubic, etc. (ii) Use of macros: Under this facility frequently utilized subroutines such as bolt hole patterns, pocket milling, drilling and tapping cycles can be permanently stored. (iii) Axis inversion or mirror image: This facility allows a single program to be used for symmetrical leftand right-hand parts by inverting the sign of the X or Y axis around the program zero. Symmetrical machining in all four quadrants is possible by inversion of a quadrant about one axis and then the inversion of two quadrants about the other axis.

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9.2.1 CNC EFFICIENCY Concepts of Sampling Rate and Motor Pulsing Rate

With the emergence of multi axis, multifunction CNC machine tools known as machining centres, the axis positioning requirements have drastically increased, while tool management and other functions have become vastly more complex. In order to implement the enhanced control and programming features, today’s CNC controllers must have far greater power as regards their control and computational capabilities. Closing the control loop is a function that a control system must repeat hundreds of times each second as it executes motion commands. Essentially, this function is performed by software routines or algorithms adjusting the servo signals until the commanded and actual positions match. The time it takes to actually process and execute a motion determines the sampling rate of a CNC machine. The higher the accuracy of the tool path and the density of points in a part program, the more important is the sampling rate. The sampling rate is the interval at which the CNC controller receives position feedback from servo loops. The CNC factors this information into both the position target and velocity of each new servo command. Usually, a more complex part has a higher point density that requires more axis moves, thereby increasing the information the control must process. If the control cannot process the information fast enough, then accuracy may be compromised. Whether talking about processing of blocks of the part program or their execution (sampling), the critical issue is for the CNC to perform these tasks as quickly as necessary to keep up with a programmed feed rate. Let us say that currently a given CNC machine processes 100 moves per second or 6000 moves per minute. If a part with a high tolerance has an average move of 0.001 inch, then 6000 ¥ 0.001 = 6 inches/minute is the maximum possible feed rate. Most materials can be machined at much higher rates. Therefore, to improve second, otherwise, we would be machining at lower feed rates than permitted, thereby lowering productivity. The highest priority for the CNC is managing the axis motors. The motors are driven as they receive electrical pulses at a set rate. The larger the pulse, the faster the motor is driven. Two problems can occur due to mismatch between motor response (motor pulsing rate) and updating of the moves (sampling rate of CNC controller). Overshooting when the motor is quick to respond to changes in the distance to go, but the distance to go is updated slowly. In other words, the motor has already reached the position it was commanded to reach, but the distance to go has not been updated, so the motor keeps on going. Undershooting occurs when the motor is slow to respond to the distance to go. In other words, the distance to go is updated before the motor has a chance to move to position. The best system is one in which the motor pulsing rate matches the sampling rate of the CNC controller. In case of mismatch between the two, the compromise will involve either loss of production by (lower feed rate) or accuracy (large move size). Early CNC machines were sluggish with the motor updated every 20 milliseconds (50 times a second). Some new multiprocessor machines can update the motors every millisecond (1000 times per second). CNC machines with slow update rate should be used only for machining of parts with low resolution (large approximating segments and hence large INTOL and OUTTOL). If a part requiring high resolution is machined on a CNC machine with slow update rate, several moves will be bunched in one update and details will be lost, thus compromising the accuracy of the machined surfaces.

CNC Processing Efficiency Processing is the conversion of moves from the part program into a form the CNC motor understands (axis motor commands). For example, even a simple move such as X10 may

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these moves are processed. In general, the CNC must consider several factors when processing moves: 1. 2. 3. 4. 5. If processing requires too many calculations, it may become a bottleneck. The fewer calculations the processor has to make to convert a move into motor pulses, the more moves can be processed in the same time. Thus, inadequacy of processing speed of CNC controller can be compensated to some extent by reducing the processing content of the CNC part program blocks. Cutter radius compensation requires the radius compensation facility should be used sparingly, especially while running a program which requires

the machining moves at a faster rate.

values. Which mode should be used depends on how the control has been set up to make calculations. Regardless of the mode adopted in the program, the CNC calculates in the mode for which it is set. For example, if a given CNC is set for calculations in the absolute mode, then all the programmed incremental moves will be converted to absolute values before the calculations begin. If the programmed motions were in absolute mode, then this conversion would have been avoided. The same applies to inch and millimeter modes. For a given CNC, all table moves are either in inch or millimeter units, regardless of the units adopted in the program. If a given CNC performs table motions in inch units, then any program written in millimeters will have to be converted into inch units. If the program had been written in inch units to begin with, then the millimeter to inch conversion would have been avoided. It is easy to check whether a given CNC runs in incremental/absolute mode by running two point-to-point part programs that machine the same part in both

9.3

MACHINING CENTRES

It was explained in Sec. 1.7 that the impetus for development of numerical control largely came from the need to implement automation in batch production with reduction of non-productive time as one of the main objectives. As a consequence of this, the cutting tool spent a higher percentage of its time in actual operation, unlike the manufacturing on conventional machine tools on which the time spent in actual cutting often constituted as little as 5% of the total manufacturing time. The implication arising out of this factor was that the drives, gears, guideways, etc., of NC machine tools experienced wear at a much faster rate than the corresponding parts of conventional machine tools. For these reasons, NC machine tools soon got to be equipped with special low friction elements such as ball recirculating lead screws, anti-friction guideways, hydrodynamic and hydrostatic bearings, etc. sure that the control members were ergonomically designed and placed at a convenient height within easy

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reach. In NC machine tools, the role of the operator was drastically altered from one carrying out a machining operation to one monitoring it. This called for a fresh look at some of the fundamental postulates of machine Traditionally, machine tools were kept as simple as possible with emphasis on operator skill to accomplish complex machining tasks. With wages of skilled workers rising faster than the cost of equipment, there appeared objective conditions for taking measures to enhance equipment capability to reduce the reliance on operator skill. One of the measures that found favour was to combine several operations on a single machine. For example, if drilling, boring and milling operations could be accomplished on one machine by one operator in a single set-up instead of on three machines by three operators in three separate set-ups, then the savings in set-up time and total operator wages are too obvious to require special comments. Machining in one set-up provides constancy of setting datum, which makes for higher accuracy of machining of complex parts. Execution of several operations on a single machine tool required large number of tools. Consequently, provision of automatic tool changers coupled with a facility for storing large number of tools right on the machine became almost an obligatory feature for machine tools combining several operations. CNC machine tools with the above features are known as machining centres. They usually have two (or more) work tables so that while machining is being carried at one work table, the other one is in a embodiment of the concept of taking the machine to the work, so that the work does not have to travel from one machine tool to another for undergoing a succession of operations. In view of the features discussed above, machining centres are best suited for machining of heavy jobs requiring complex machining, e.g., housings. Machining centres are generally designed to carry out operations such as milling, drilling, reaming, machining centre for producing simple parts that can be easily machined on CNC drilling or milling machines is not cost effective.

top, whereas a horizontal machining centre is more suitable for box like, cubic and prismatic parts in which moving column type, depending on whether the table with the workpiece travels longitudinally with respect to the column or vice versa. Apart from longitudinal travel, complex machining centres are capable of other programmable movements such as tilt and swivel of spindle head, etc. Turning centres are analogous to machining centres and are essentially CNC lathes, in which the multiple operation and automatic tool changing capabilities are incorporated. In turning centres, the tool post is replaced by an indexable turret carrying usually up to 12 tools. Unlike lathes, the tools are generally placed above the workpiece to make the chips fall directly down. The type of tool changing mechanism employed in machining centres depends on the number of tools. from those used in turning centres, because in the former, the turret actually represents multiple-spindle indexable head, wherein only the tool in the machining position rotates, while the rest are stationary. Some machining centres carry two turrets, of which the idle one is used for loading and presetting of tools, thereby considerably reducing the machine down time. Tool magazines are used when the number of tools is more than 12. Tool magazines are of two types: 1. Rotary drum type 2. Chain type

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A drum-type tool magazine is shown in Fig. 9.1. These magazines are employed on machining centres carrying between 12 to 50 tools. When the number of tools lies between 12–25, the magazine is mounted on top of the spindle head. This arrangement minimises the distance between the spindle axis and the tool changing position of the magazine, thereby providing for a simple tool loading mechanism. When the number of tools lies between 25 and 50, the drum type tool magazine is mounted either on the column of the machining centre or an independent column. In the latter case, the independent column possesses an additional linear movement by which it is moved to the loading position for executing the tool change.

Arm

Arm movements

Fig. 9.1 Drum-type tool magazine

Chain type magazines (Fig. 9.2) are used when the number of tools on the machining centre is greater than 50. These magazines too are mounted overhead, on the side of the column of the machining centre and on an independent column.

Fig. 9.2 Chain-type tool magazine

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In drum as well as chain type magazines, tool transfer from the magazine to the spindle and vice versa is executed by a changer arm (Fig. 9.3). The arm 1 occupies vertical position when not in 3 operation. For executing tool change, it is shifted to the horizontal position in which it simultaneously grips two tools, one in the magazine 2 and the other mounted in the spindle 3. Both the tools are 2 released by applying pneumatic pressure and the 1 arm is pulled outwards. It is now rotated through 180° for the tools to interchange their positions. By pushing the arm inwards the new tool is inserted in the spindle and the used tool is put back in the Fig. 9.3 Changer arm mechanism used with tool magazine. The arm is now rotated through 90° to magazines occupy the idle position. It was mentioned earlier that machining centres are particularly well suited for machining of complex heavy parts. These parts constitute the so called main parts of engineering equipments. They hardly comprise 10% of the total number of parts of equipment by number, but their share of the total value of the equipment may be 50% and more. In view of the high penalty for producing a reject, the premium on accuracy of machining of such parts is very high and this constitutes the major area of application of machining centres.

9.4 a central computer. In case the number of CNC machines is large the link may be two-stage, whereby groups of CNC machines are controlled by satellite computers, which in turn, are linked to the central computer. There are three components required for DNC—the CNC, a computer, a serial line connecting them, generally a standard RS-232-C communications cable and the necessary communications software. In fact one external computer may be connected to several CNC machines. The MCU of each of the CNC machine tools is under the control of the DNC computer, but the individual CNC machine is controlled by its MCU. All modern CNC machines are provided with a serial port or a RS-232-C port so they can be directly connected to the DNC computer. Older NC and CNC controllers that have little or no internal memory or communicaspecial interface unit which is plugged into the tape reader port. Such systems are known as behind the tape reader (BTR) systems. Once this system is installed, even these NC controls can be integrated in the DNC network. They can then run long programs that would have otherwise required miles of paper tape. The central computer stores the part programs required by all the CNC machines of the system. These programs are downloaded to the computers of individual CNC machines on demand. Unlike distributive numerical control discussed in Sec. 9.1, the central computer does not directly control the working of CNC machines. The real time control rests with the computers of the individual CNC machines. Generally, enough part programs are transferred from the central computer to the computers of CNC machines to keep the machines busy for a shift or day, as the case may be. Thus, routine operation of CNC machine tools is controlled by their individual dedicated computers, independent of the central computer. This has two advantages. Firstly, the CNC machines can be kept running for several hours from programs stored

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in their memory in case of malfunctions of the central computer, thereby preventing immediate shut down of the CNC machine tools and at the same time providing valuable buffer period during which the fault of the central computer can be looked into. Secondly, being free of real time control of individual machines merely for the purpose of downloading of part program, the central computer can be gainfully employed for data collection and supervisory functions. As each CNC machine tool has a dedicated computer, it becomes possible to have a two-way communication between the central computer and the CNC machine tools. This helps in collecting valuable data on crucial performance characteristics such as production piece count, tool usage, machine utilisation and real time status and down time record of individual CNC machine tools. The central computer prepares performance reports based on the collected data, thereby providing valuable feedback to the shop and plant management. With proper software support, the central computer of DNC-2 system can handle other manufacturing planning and control functions such as process planning, materials requirement planning, scheduling, etc. Decision regarding the additional functions to be handled depends on the capacity of the central computer. Conversely, knowing the functions to be handled, the central computer could be selected accordingly. It is thus seen that direct numerical control is a major step in the direction of closed-loop computer aided factory operation and management and a forerunner of computer integrated manufacturing. The functions of a DNC may thus be summed up as follows: 1. 2. 3. 4.

Numerical control without punched tape. NC part program storage, including writing and editing of new programs. Sending part programs to CNC machines in real time on demand, known as program downloading. Receiving part programs from CNC machine tools, if such programs are prepared or edited on the MCU, known as program uploading. 5. P information. 6. Collecting performance data and preparing performance reports. 7. Drip feeding of long part programs. Drip feeding is resorted to when long part programs cannot be accomodated fully in the MCU of CNC machines. Recent CNCs have controllers that allow them to load upto 16 megabytes of programs. However, huge memory options still have limitations. Being new technology, they are expensive. They also require long download times. For instance, it may take an hour to download a program with one million moves. On the other hand, it is not uncommon for CAM systems to generate CNC programs with a million or so moves. Most CNC controllers can not store such large programs, which therefore have to be split into parts. At each split, there is possibility of discontinuity in machined surface or a burr which affects the quality of the product. In such cases, the program is fed to the MCU from the central computer and the CNC must execute each move as it is received. This process of sending a move at a time from the central computer to the CNC is known as drip feeding block of program already executed to vacate part program memory. The DNC computer sends the next segment of the part program before the completion of the previous segment. The MCU is thus not The use of DNC involves additional cost in terms of network, computer, terminals and the software. 1. when the interconnected CNC machines are large in number, 2. when drip feeding is essential for execution of long part programs,

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3. when part program sizes are very large and can not be held in the part program memory of the MCU, 4. when part program variety is large and batch sizes are small. In such cases, each CNC machine has to be fed a large number of part programs every day, thus justifying the use of DNC, and 5. engage the CNC machine for this purpose and keep it idle.

1. Eliminates local input device such as tape reader or MDI, thus helping in entry of error free programs. 2. Large storage capacity of DNC computer makes it possible to store several part programs. 3. If a part is to be machined on different machine tools, then separate program entry into the individual downloading. 4. Better management of part programs which allows the supervisor to organise the operators work and

9.4.1 DNC Efficiency As mentioned before, there are three components in a DNC, namely the CNC, a computer and a serial line connecting them. The program to be machined is sent by the computer a move at a time to the CNC via the serial line. While the CNC is executing the current move and processing the next move, the serial line is downloading the moves after that. The CNC is thus doing three things: loading moves, processing moves and turn, implies increasing the speed at which the CNC loads, processes and executes moves. Further, there must be compatibility between the speeds of loading, processing and execution. Any one of the three which is less

Loading a program for DNC involves transfer of part program from an external computer to the CNC a move at a time over a serial line (RS – 232 line). The speed at which the moves are sent over a serial line is called the baud rate. The baud rate corresponds to the number of bits per second that can be transferred between the computer and CNC. In general, for each character (letter or number in a move), 10 bits are sent over the serial line eight bit character plus one start bit and one stop bit. Typical industry baud rates are 9600, 19200 and 38400. So, at 9600 baud, 960 characters are transferred per second, while at 38400 baud rate, 3840 characters are transferred. If an average move is 10 characters long, then at 9600 baud we will be transferring 960 characters per second for a maximum of 96 moves per second. So, the obvious conclusion is that the The communication (talk) between the computer and CNC over a serial line is governed by a protocol, also known as hand shaking. The reason for a protocol is to make sure that the computer and the CNC are not trying to send information to each other at the same time. When CNC is sending information, it is not listening to the information that is being sent to it. This is how information is lost. To avoid losing information, the computer and CNC follow a protocol to make sure that they do not communicate at the same time. There are two protocols in use today X ON/X OFF and X Modem.

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The X ON/X OFF is the faster of the two protocols. The DNC computer talks to the CNC as fast as it can, but if it talks faster than the CNC can listen, the CNC sends an XOFF to the computer, telling the computer to stop sending until it can catch up. When the CNC has caught up, it sends an XON to recommence the communication. The most frustrating part of using XON/XOFF is that if the serial line has a failure, one gets to know about it only after machining is over. Serial communications are not 100% reliable. When the computer is sending 38400 bits per second, it is possible that a few bits may be lost in transition if there is lot of electrical noise in the shop or the serial cable is very long. A cable is like a capacitor and its capacitance increases with increasing length, thus increasing the possibility of corrupting the data being transferred. Since the CNC has no way of detecting and preventing serials errors, the part is in most cases scrapped. It is therefore recommended to position the computer near the CNC to reduce the cable length and also take measures to minimise electrical noise in the shop from other electrical equipments. The X modem protocol allows the CNC to detect and prevent errors by breaking up the program into small packets. Each packet has a checksum, which is computed by totaling the bit value of every character in the packet. The DNC computer computes this sum while sending the characters and the CNC while receiving them. After an entire packet has been sent, the computer sends the checksum it computed to the CNC which compares it with its own checksum. If the checksums match, the CNC sends an ACK (Acknowledge) signal to the computer requesting to it send the next packet. If the checksum is bad, which means the packet has an error, the CNC sends a NAK (Not acknowledged) signal and requests it to resend the packet. In this way, most errors are detected and prevented thus protecting the part. With all the ACK and NAK communications, the X modem is slightly slower than the XON/XOFF protocol, but is still recommended for DNC systems in view of the greater reliability. may employ preprocessing and compression of the part programs before sending them over the serial line. Preprocessing means that the DNC computer does part of the processing job that the CNC normally does. This saves CNC time and speeds up the DNC. Compressing means that the computer shrinks the program before sending it over the serial line and the CNC then expands it once it has been received. Preprocessing by itself can make the program bigger and take longer to send, but since the DNC computer processes faster than the CNC, there is a net saving of time. Similarly, compressing makes the program smaller and takes less time to send, but it takes longer to get processed in the CNC. The decision of which of the two to use is based on the processing speed of the CNC. In most CNCs, the moves are processed as soon as they are received and the CNC is generally waiting for moves to arrive. In such cases, the CNC is said to be buffer starved. Compression is the best option in this case, because although it will take a little longer with each move in order to decompress it, the CNC is waiting anyway. The basic rule is that if the CNC is waiting and doing nothing part of the time then it should be given more to do (compressing). On the other hand, if the CNC is overloaded and has too much to do, then part of the burden should be passed on to the computer (prepreprocessing).

9.5

CNC PROGRAMMING

It was mentioned in Sec. 9.2 that CNC machine tools had advanced programming facilities such as macros, that make programming simpler, viz., 1. position data input in decimal system, 2. provision of programming more than one G code in a single block,

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3. automatic tool diameter compensation (offset), and 4. automatic tool length compensation (offset). The important programming features available on CNC machine tools will be initially described and later illustrated in application to example programming problems.

Tool Diameter Offset (Compensation) for Milling 9.5.1 Operation (G41, G42, G40) The concept of tool diameter compensation was explained in Sec. 8.2.5. As the part program refers to the trajectory followed by the cutter centre, a different part program is obtained for each cutter size. In NC cutter size and then writes the part program for the trajectory followed by the cutter centre with the help of appropriate trigonometric calculations. This has two disadvantages: 1. P 2. A new part program has to be written if the cutter size is changed. CNC machines have the facility of automatic cutter diameter compensation, which allows the cutter size to be ignored, i.e., allows the part program to be written as if the cutter diameter were zero. The size of the cutter is entered by the operator in a tool offset register, usually designated as D-register. The diameter of each cutter is entered under an independent sequence number as follows. Tool diameter offset register (D-register) Tool No.

Register No.

Diameter

01

01

15 mm

02

02

25 mm

…

…

…

…

…

…

08

11

12 mm

Though it is desirable, it is not necessary that the cutter of a particular No. should be entered under the same sequence number in the D-register. For instance, cutter No. 8 (f 12 mm) has been entered in Register D-11. In the part program, D11 is incorporated in the block pertaining to machining with tool No. 8. The part program is written imagining the cutting diameter to be zero. By invoking register D11, the CNC computer automatically calculates the offsets. If a different cutter is required at a later stage (say a cutter of diameter f follows:

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Tool No.

Register No.

Diameter

08

11

18 mm

It is thus seen that the cutter diameter offset facility in CNC machines is free of the aforementioned drawbacks of NC machines tools. Cutter diameter compensation can be to the left (G41) or right (G42). To determine which of the two is applicable the programmer imagines himself to be positioned ahead of the cutter, facing the direction in which the cutter is moving. Depending on whether the job lies to the left or right of the observer so positioned, the cutter diameter compensation is designated as left (Fig. 9.4a) or right (Fig. 9.4b). Both left- and right-cutter diameter offsets are cancelled by G40. Programmer

G41 (Left)

Starting point of cutter (a)

G42 (Right)

Programmer

Starting point (b)

Fig. 9.4 Tool diameter compensation left (G41) and right (G42) on machining centres

Tool Length Offset (Compensation) for Milling and Drilling 9.5.2 Operations (G45, G46) Tool length offset is a very useful CNC programming feature which controls the Z-axis motions in CNC machine tools and machining centres. This facility has two types of applications in CNC programming: 1. It allows tools of different lengths to be used in CNC milling and drilling machines without making any change in the part program. 2. It allows tools of different lengths to be used in various spindles of machining centers without presetting The difference of tool length is entered as an offset value in H-register by the operator. The stored value is then used as a positive or negative compensation with respect to the programmed motion in Z-axis, as the case may be. Tool length offset is thus essentially an addition or subtraction operation carried out by the CNC controller. Tool length offset is programmed by G45 and G46 codes. When the G45 code is used, the offset value entered in the H-register is added to the programmed motion in Z-axis, while in case of the G46 code it is subtracted. Usually G45 is used for convenience.

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Consider the example of CNC drilling shown in Fig. 9.5. The tip of the drill (see Tool No. 1) is positioned at 50 mm from the surface of the job and the spindle has to be programmed to move rapid in Z-axis so that the drill tip stops 5 mm short of the job surface. Consequently, the distance to be travelled by the drill is 45 mm in the minus Z direction. The programmer may choose to include 30 mm of this travel in the part program and the rest of 15 mm as compensation in the tool length offset register (say H 01). The statement (say 050) in the part program would read as: +z T01

–z T02

20 T03 20 50

Fig. 9.5 Tool length off set in drilling operation on a machining centre

N050 G00 G 45 Z – 30 H01 The entry in the H-register will be as follows: Tool No.

H-register No.

Tool length offset

01

01

15

The offset value (+ 15) is added to the programmed Z axis motion (+ 30) and the net motion (30 + 15 = 45) occurs in the direction associated with the sign of the programmed motion (minus Z). Hence, the total motion is 45 mm in the minus Z direction. To illustrate the foregoing, a few possible combinations of the programmed motion and offset are given below for the example under consideration to yield a net motion of 45 mm in the minus Z direction:

508

Machine Tool Design and Numerical Control

Programmed motion

Tool length offset

–20

+25

–25

+20

Now, suppose that the given drill (Tool No. 1) is replaced by another drill (T02) which is 20 mm shorter. In this case, the distance to be travelled by the tool tip to reach the plane 5 mm short of the job surface will be 65 mm. Retaining the same statement in the program, viz., N050 G00 G45 Z – 30 H01 the entry in the H-register will be +35 to yield a net motion of – (30 + 35) = – 65 mm. The entry in the

Tool No.

H-register No.

Tool length offset

01

H01

35

Let us consider now that the given drill (Tool No. 1) is replaced by another tool (T03) which is 20 mm longer. In this case, the distance from the tool tip to the plane 5 mm short of the job surface will be of 25 mm. Hence, the entry in the H-register will be –05 to yield a net motion of – (30 – 05) = – 25. The entry in the H

Tool No.

H-register No.

Tool length offset

01

H01

– 05

The approach described above is also valid for milling cutters of different lengths on CNC milling machines. Let us now discuss the application of tool length offset facility in reference to machining centres on which several cutting tools (milling cutters, drills) are mounted in the various spindles. Suppose the zero of the machining centre in the Z-axis is set to lie in plane A which is 5 mm above the top surface of the part (Fig. 9.6). The setting is done by making tool T04 (f 10 mm end mill cutter) touch the top of a 5 mm plate, placed on the table and pressing a button to set the Z-axis read out to zero. In this position, the spindle face occupies a position in plane B at a certain distance Zs from the zero. Henceforth, whenever the machining centre is programmed to move to zero position in Z-axis, the spindle face will occupy the set position in plane B. Now when tools T01 (face milling cutter f 30), T02 (drill f 10) and T03 (counter boring cutter f 12) lengths. The offsets can be determined by moving the tools from their respective zero position to plane A. The displacement from the zero position is noted on the Z-axis read out and provides the offset for the corresponding tool. The length offsets of the various tools are entered in the H-register with the radii. Thus, in

Extensions of Numerical Control—CNC, DNC, Machining Centres 509

machining centres both the tool diameter and tool length offsets are entered in a single register. For the tools shown in Fig. 9.6, the entries in the H-register will be as follows: Plane B T02

T04

T 03

T01 Zs

+Z Plane A 12 5 –12 –Z

Fig. 9.6 Tool length offset on machining centres

Tool No.

H-register No.

Tool length offset mm

Tool diameter mm

01

01

5.0

30.0

02

02

– 12.0

10.0

03

03

12.0

12.0

04

04

0.00

10.0

When the programmer writes a part program for operations in which these tools are used, he can ignore the variations of tool length and assume that for all the tools the zero in Z-axis occurs when the tip of the tool lies in plane A. The CNC controller automatically adds or subtracts the offsets given in the appropriate H-registers. For example, if the f 12-mm counter boring cutter is to be programmed to make a 10 mm deep slot and is positioned at zero prior to this operation, then the statement (say N010) in the program for the Z-axis motion will be N010 G01 G45 Z – 15 H03 The net motion of tool T03 in the direction of Z-axis will be – (15 + 12) = – 27 mm. Note that the motion will occur at feed rate (G01). To illustrate again, let us assume that a 15-mm deep hole is to be drilled in the job using the f 15 drill (T02). It is obvious that prior to these operations the tool must be positioned at a location of + Z7 or greater, so that the drill is clear of the surface to be machined. Suppose the tool T02 is positioned at 15 mm in the plus Z direction, such that its tip is at a distance of 3 mm from plane A. From this position, the statements (say N040 the job and then at feed rate to the desired depth will be as follows: N040 G00 G45 Z –15 H02 N050 G01 Z –20

510

Machine Tool Design and Numerical Control

In the statement N040, the rapid net motion of the tool will be – (15 – 12) = – 03 mm. Subsequently, in N050 the tool will move at feed rate by 20 mm (5 mm to the job surface from plane A and 15 mm further to the required depth of hole). As tool T02 is longer than the tool T04 that was used for zero setting of Z-axis, it becomes necessary to take the precaution of positioning it properly before executing a Z-axis movement, as otherwise it can lead to serious accident and damage to both tool and machine. This problem can be avoided by using the longest tool for setting the zero of the Z-axis. In this case, all the offsets entered in the H-register will be positive. In the example case shown in Fig. 9.6, it would be desirable to use tool T02 for zero setting of the Z-axis. In that case, the offsets of the other three tools will be as follows: Tool No.

H-register No.

Tool length offset mm

Tool diameter mm

01

01

17

30

02

02

0

10.0

03

03

24

12.0

04

04

12

10

Tool length offset is cancelled by G49 command.

9.5.3 Zero Setting in CNC Machining Centres (G55) The zero systems in NC machine tools and the manner in which they are set were discussed in Sec. 8.2.1. The

are different from those adopted for NC machine tools. For a CNC machine tool with full zero shift, the machine zero Y is shifted to the desired location by means of G55 command used outside the program. Suppose the machine zero is to be shifted from the absolute datum to a point A (Fig. 9.7) on the component V where a drilled hole exists from a previous operation. A probe A B (or may be even a drill) of proper size that allows it to slide into the hole with minimum clearance is mounted in the machine H X tool spindle and the spindle/table is moved in jog mode so that Machine zero the probe is positioned exactly above point A. This is checked by Fig. 9.7 Zero setting in machining inserting the probe (drill) in the hole. The readings of the X and Y centres with full zero shift coordinates are noted on the digital read out (say Ax, Ay). The operator now presses the following keys to shift the zero to point A: G55 ENTER Ax ENTER EOB G55 ENTER AY ENTER EOB

Extensions of Numerical Control—CNC, DNC, Machining Centres 511

The bars under the commands indicate that they are entered manually by the operator on the MCU console. If the zero were to be shifted to the lower left-hand corner of the part (Point B), then a round bar of known radius R will be mounted in the spindle and the spindle/table moved in the jog mode so that the side of the bar touches the edge V of the component. The digital read out of the X coordinate is noted (say Bx) and the operator now presses the following keys to shift the Y-axis to point B: G55 ENTER Y(Bx + R) ENTER EOB Similarly, the bar of radius R is aligned with the edge H of the component. If the digital out of the Y coordinate is By, then the X axis is shifted to point B by the operator bypressing the following keys in the given order: G55 ENTER X(By + R) ENTER EOB It is assumed that the edges of the component are aligned parallel to the coordinate axes and the component is properly clamped before commencing the zero shift operation. allocated for this function, therefore, an unallocated code G56 will be used for this purpose. Suppose a segment of the part program has already been writY

shift the zero to point C (Fig. 9.8) which has coordinates (Cx, CY) (noted from the digital readout). The statement in the program for this purpose (say N080) will be,

D C

N080 G00 XCX,YCY G56 In some CNC machining centres command G45 when used in conjuction with X, Y coordinates and D-register serves for setting

Machine zero

Fig. 9.8

same command when used in conjunction with the Z-direction coordinate serves for tool length offset (Sec. 9.5.2). Suppose the

X

Zero setting in machining centres with full floating zero

register (D01 for X-axis and D02 for Y-axis are zero) then the statements (say N040 and N050) for setting of N040 G00 X50 G45 D01 N050 G00 Y55 G45 D02 Suppose the part programmer makes provision for a movement of 30 mm each in the X and Y directions. The entries in the registers D01 and D02 will, in this case, be 50 – 30 = 20 mm and 55 – 30 = 25 mm, N040 G00 X30 G45 D01 N050 G00 Y30 G45 D02 shifted to another location without making any change in the part program. Suppose due to some reasons

512

Machine Tool Design and Numerical Control

point C, but may also be necessitated by change in the size of the component or a new zero location after the full zero shift. Suppose the coordinates of point D are XI00, Y115. The statements in the program will remain unchanged, but the operator will be instructed to change the entries in registers D01 and D02 to 100 – 30 = 70 mm and 115 – 30 = 85 mm, respectively.

Special Programming Features in CNC Milling and 9.5.4 Drilling Operations Tool Radius Vector Setting (G39)

Consider Fig. 9.9a. At point A where a change of direction takes place, the position of the cutter centre must be shifted from Ol to O2 without violating the contact at point A, so that the tool is aligned tangent to the new surface before machining commences. This is done by rotating the centre of the tool using G39 command, followed by the interpolation parameters I, J, K to indicate the vector direction of the new motion. The interpolation parameters represent the signed values of the X, Y and Z increments of the new motion. For the dimensions shown in Fig. 9.9a, the statement (say N 010) in the part program for directing the tool from surface P to surface Q will be N010 G39 I30 J-20 N040 G00 X50 G45 D01 N050 G00 Y55 G45 D02 Over travel

+

O1 O2

P

P

A

A 20 Q

Q 30 (a)

(b)

Fig. 9.9 Tool radius vector setting in machining centres (a) by rotation of tool centre (b) by overtravel of tool

In the recent models of CNC machine tools, the reorientation of the cutter is taken care of by the CNC controller automatically and it is not required to use the G39 command for this purpose. In these CNC systems, the cutter travels past the programmed point to a position where it becomes tangent to the new surface. It then changes direction to commence machining of the new surface (Fig. 9.9b). The required over travel is calculated by the CNC controller.

Extensions of Numerical Control—CNC, DNC, Machining Centres 513

Do Loop (G51) for points in which certain features are repeated. The general structure of a do loop statement is N sequence number G51 number of repetitions The operations to be repeated are placed between the above command and the do loop termination command G50. The departures in the DO LOOP are given in the incremental dimension mode.

Subroutine Programming (M98)

A subroutine is an independent program which is placed at the end

three digits). It is possible to use more than one sub routines in a program. Execution of the main program is transferred to the subroutine by means of a M98 command. The general structure of this statement is as follows:

Note that the subroutine commences with fresh sequence numbering (N010). The subsequent steps of the subroutine are programmed with sequence numbers N020, N030 … and so on. The last statement of the subroutine program returns the execution from the subroutine to the line of the main program immediately succeeding the line where the diversion to the subroutine was executed. This is done by a M99 command and the general structure of this statement is N sequence number M99 The move to the starting point of the subroutine is in absolute dimensioning mode, whereas the departures within the subroutine are given in incremental dimension mode.

Mirror Imaging (M21, M22) In a large number of engineering components, the features to be machined are placed symmetrically about one axis or both. Hence, they represent mirror images. For example, in Fig. 9.10a, feature 1 is a mirror image of feature 2 about the X-axis and in Fig. 9.10b feature 3 is a mirror image Y

Y

1

4

5

3

X

X

X 6

2 (a)

Y

(b)

©

Fig. 9.10 Mirror image (a) about X-axis (b) about Y-axis (c) about X- and Y-axes both

514

Machine Tool Design and Numerical Control

of feature 4 about the Y-axis. For the relative location of features shown in Fig. 9.10c, feature 6 can be treated

facility known as mirror imaging. It is a simple concept whereby the part program written for features lying in one quadrant is copied for similar features in other quadrants (as explained in Fig. 9.10) by reversing the sign of the coordinates or increments. To wit, the sign of X coordinate or X increment is reversed in mirror imaging about the Y-axis and vice versa. The commands for minor imaging about the X and Y-axes are M21 and M22, respectively and the general structure of the statement is N sequence number M21 N sequence number M22 The CNC program that is to be copied by mirror imaging is stored as a subroutine at the end of the main program. The execution of the main program is shifted to the subroutine by means of the M98 command as described above and this is followed by the command for mirror imaging (M21 or M22 as the case may be). Both M21 and M22 are cancelled by M23. Before issuing a mirror image command a second time, it is necessary to cancel the previous mirror image command. This may not be obligatory in some controllers, but is desirable from the standpoint of safety.

Canned Cycles memory with a particular G-code address. Several such cycles are by now standardised, e.g., G81-drilling cycle, G84-tapping cycle, G85-boring cycle, etc. Canned cycles are of two types: 1. Fixed 2. Variable canned cycle is repeated at several locations in a program with the help of a Do Loop command, then it is known as a repetitive canned cycle. Variable canned cycles are essentially subroutines that allow the

will be discussed here. In almost all canned cycles, there are three levels of tool positions: 1. I 2. R rapid feed rate. This level generally lies between 0.5–3.0 mm from the part surface. It is also referred to as clearance plane when not used in the context of canned cycles. 3. Z-level represents the plane up to which the tool travels in the Z direction at feed rate.

The general structure of a canned cycle is

Extensions of Numerical Control—CNC, DNC, Machining Centres 515

N

G

G

X

Y

Z

R

F

L

No. of repetitions Feed rate R level Z level Coordinates of the location to be machined 98/99 depending on the desired plane of withdrawal Cycle code (e.g., 81 for drilling) Sequence number

A canned cycle is cancelled by G80.

Programming Examples for CNC Milling and Drilling 9.5.5 Machines and Machining Centres Example 9.1 Write the CNC program for contour milling of the given part (Fig. 9.11) on a vertical spindle CNC milling machine. The machine has provision of full zero shift and cutter diameter compensation. The tool radius vector must be reset whenever there is a change in the direction of cutter path. Contour milling is done by a f 10 mm end mill cutter at a rpm of 1200 and feed of 50 mm/min. The operator is instructed to normally carry out the following operations: 1. Shift the machine zero to the lower left-hand corner of the part with the help of a bar using G55 command as explained in Sec. 9.5.3. Assuming that a bar of f 20 mm is used for this purpose, the G55 Enter X(50 + 10) ENTER EOB G55 Enter Y(40 + 10) ENTER EOB 2. Enter the diameter of the end mill cutter in the D-register (say D01) Before writing the program, it would be relevant to mention certain features that though not essential, are desirable from the standpoint of safe programming, namely (i) cancel all previous cycles (G80); tool diameter compensation (G40) and tool length compensation (G49) (ii) identify a clearance plane at a certain distance (say 2 mm) from the top surface of the part so that the tool may be programmed to move at rapid feed rate up to this plane.

516

Machine Tool Design and Numerical Control

(iii) provide a certain overlap (say 2 mm) for milling. 60

30

35 D

40

20

C

B

50

70

15 F

E

H

P

G R 7.5

Ym Q

50

A

Xm f10 end mill

2

20

10

–Zm 2

Fig. 9.11

Sample part. Contour milling problem illustrating the cutter diameter compensation facility

The CNC program is given below with an explanation of the instructions side by side: N010 G71 G40 G49 G80

:

Metric data, cancel cutter dia compensation, cutter length compensation and previously programmed cycles.

N020 G92 X0Y0Z0

:

Initialise to synchronise program zero with the zero of the digital display.

Extensions of Numerical Control—CNC, DNC, Machining Centres 517

N030 G94 S1200 M03

:

Set spindle rpm = 1200 and spindle on clockwise. Feed in mm/min.

N040 G90 G17 G00 G4l X60 Y50 J50 D01

:

Absolute dimensions, interpolation in X-Y plane, rapid rate, cutter diameter compensation left, current movement to point (60, 50), next movement in the vertical direction of dimension 50 mm, cutter dia available in register D01.

N050 G19 G00 Z-18

:

Interpolation in Y-Z plane to the clearance plane at rapid feed rate.

N060 G91 G19 G01 Z-14 F50

:

Incremental dimensions, move to overlap plane at feed rate of 50 mm/min.

N070 G17 G01 Y50 M08

:

Machine side AB, coolant on.

N080 G39 I30 J20

:

Reset radius vector of the tool.

N090 G01 X30 Y20

:

Machine side BC.

:

Machine side CD.

:

Machine side DE.

N150 G01 X-15

:

Machine EF.

N160 G03 Y-15 J7.5

:

Note that G39 is not required as the circle is in smooth continuation of the previous linear motion; semicircle FGH is machined.

N170 G01 X15

:

Machine HP.

:

Machine side PQ.

N210 G01 X-65

:

Machine side QA.

N220 G19 G00 Z14

:

Withdraw tool to clearance plane.

N230 G17 G40 X-10 Y-10 M09

:

Cancel cutter diameter compensation at a safe distance from the job (– 10, – 10) and switch coolant off.

N240 G28 Z0

:

Return tool to zero in Z-axis.

N100 G39 I35 N110 G01 X35 N120 G39 J-40 N130 G01 Y-40 N140 G39 I-15

N180 G39 J-15 N190 G01 Y-15 N200 G39 I-65

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Machine Tool Design and Numerical Control

N250 G28 X0 Y0 M05

:

Return tool to zero in X-Y plane and switch off the spindle.

N260 M30

:

End of program. Rewind.

The program could also have been written with tool diameter compensation right (G42). In this case, the milling of the contour would have proceeded from point A in the direction AQ and the tool would have taken the path AQPHGFEDCBA in completing the milling of the part contour.

Example 9.2 Write the CNC program for drilling three holes in the part shown (Fig. 9.12) on a CNC drilling machine. A G45 command program. 70

15

15

Hole f 8

T01(f 10) 70

T02(f 8) T03(f 6)

Hole f 10

60

+Zm

10

–Zm

50

Ym

10

Hole f 6

Xm

2

40

50

20

12

30

3

Fig. 9.12 Sample part. Drilling problem illustrating the tool length offset facility

Drilling of the f 6, f 8, and f10 mm holes is carried out at 2200, 1600 and 1200 rpm, respectively and a feed rate of 40 mm/min. Drills f10, f 8, and f 6 are designated as tools T01, T02 and T03 and their offsets are recorded in registers H01, H02 and H03, respectively. Certain additional points which deserve special mention are as follows:

Extensions of Numerical Control—CNC, DNC, Machining Centres 519

1. A 2. For machining the through hole f10 mm, the drill must overshoot the lower face of the job by at least R cot f, where R is radius of the drill and f the half lip angle. Assuming f = 60°, the over travel = 2.87 mm. Hence, a provision of 3 mm of over travel is made in the program. 3. The zero will be set at the centre of the job (centre of the hole f10). For the part clamped as shown in show the coordinates of the centre of the job as (95, 85).

Register No.

Offset

D01

70 mm (X offset)

D02

70 mm (Y offset)

Similarly, the length offsets (Z-axis offsets) stored in the H-registers are as under: Tool No.

Register No.

Tool length offset

Tool diameter

T01

H01

10 mm

10 mm

T02

H02

20 mm

8 mm

T03

H03

30 mm

6 mm

It may be mentioned that the given offset values entered in the D and H registers were selected arbitrarily. As explained earlier in Sec. 9.5.2 and 9.5.3, values other than the above may be used 5. Before changing the tools, the spindle of the drilling machine must be raised to the zero datum in the Z-axis. The CNC program is given below with an explanation of the instructions side by side. The statements which were explained earlier in example 1 have not been elaborated here. N010 G71 G40 G49 G80 N020 G92 X0Y0Z0 N030 M06 T01

: Clamp f10 mm drill in the spindle.

N040 G94 S1200 M03 N050 G45 G00 X25 D01

: Actual travel = 25 + value in register D01 = 25 + 70 = 95 mm.

N060 G45 G00 Y15 D02

: Actual travel = 15 + value in register D02 = 15 + 70 = 85 mm. Floating datum is set at job centre.

520

Machine Tool Design and Numerical Control

N070 G91 G19 G45 G00 Z-18 : Actual travel = – (18 + value in register H01) = – (18 + 10) = – 28 mm H01 which is the desired travel at rapid rate to the clearance plane. N080 G01 Z-25 F40 M08

: Z-axis motion at feed rate of 40 mm/min = – (2 + 20 + 3) = – 25 mm. Coolant on.

N090 G00 Z25 M09

: Retrace drill f 10 to clearance plane at rapid feed rate. Coolant off.

N100 G28 Z0

: Return tool to zero in Z-axis for tool change.

N110 G00 X20 Y20 M05

: Move to position above hole f8 at rapid feed rate. Stop the spindle.

N120 M06 T02

: Clamp f 8 mm drill.

N130 S1600 M03 N140 G19 G45 G00 Z-18 H02 : Actual travel = – (18 + value in register H02) = – (18 + 20) = – 38 mm. N150 G01 Z-14 F40 M08

: Z-axis motion at feed rate of 40 mm/min = – (2 + 12) = – 14 mm.

N160 G00 Z14 M09

: Withdraw drill f 8 to clearance plane at rapid feed rate. Coolant off.

N170 G28 Z0 N180 G00 X-45 Y-45 M05

: Move to position above hole f 6 at rapid feed rate. Stop the spindle. Note that the X and Y values show the increments with signs from hole f 8 to f 6 (note G91 in Seq. No. N070).

N190 M06 T03

: Clamp f 6 mm drill.

N200 S2200 M03 N210 G19 G45 G00 Z-18 H03 : Actual travel = – (18 + value in register H03) = – (18 + 30) = – 48 mm. N220 G01 Z-14 F40 M08

: See N150 for explanation.

N230 G00 Z14 M09

: See N160 for explanation.

N240 G28 Z0 M05 N250 G28 X0 Y0 N260 M30

Example 9.3 Write the CNC program for contour milling of the plate shown in Fig. 9.13 and drilling of 12 holes f6 on a vertical machining centre. Contour milling is carried out by a f10 mm end mill cutter at 1200 rpm and feed of 50 mm/min. The f6 mm holes are drilled at 2200 rpm. Drilling is carried out at a feed rate of 0.1 mm/rev. Command G39 for resetting tool radius vector is not required.

Extensions of Numerical Control—CNC, DNC, Machining Centres 521

12 holes f 6

D

30

2

I

90

3 II

10 10 10

C 1

T 02 Drill f 6 10 10 10

60 T 01 End mill f 10

IV

III

+Z Ym 10

30

40

30

10

B 50

A

120

Xm

–Z 40

30 2

10

20

2

Fig. 9.13 Sample part. Problem illustrating mirror image and subroutine programming facilities

As far as the location of the holes is concerned, they constitute four identical patterns of 3 holes in each quadrant (assuming the centre of the job as the centre of coordinates). The program for machining of the holes will, of the holes will be machined using the mirror imaging facility available in CNC programming. The important points are explained below: 1. The end mill cutter is designated as Tool T01, while the f 6 mm drill is designated as T02. Their offsets are recorded in registers H01 and H02, respectively. The values entered in the H registers are as under: Tool No.

Register No.

Tool length offset

Tool diameter

T01

H01

10 mm

10 mm

T02

H02

10 mm

6 mm

2. For contour milling of the plate, the machine zero will be shifted outside the program to the lower left-hand corner of the part with the help of G55 command using the same procedure as described in Example 9.1. Next, for machining of the holes, the zero will be shifted to the centre of the job. Having

522

Machine Tool Design and Numerical Control

shifted the machine zero to the lower left-hand corner of the job, the digital display of the CNC at the centre of the job the operator is instructed to enter the following values in the D-registers: Register No.

Offset

D01

20 mm (X offset)

D02

30 mm (Y offset)

3. An approach of 2 mm (clearance plane) and an over travel of 2 mm (overlap plane) are assumed. 4. The spindle of the machining centre should be directed to the home position in X, Y and Z for effecting the tool change. The CNC program for machining of the part is given as follows with an explanation of the instructions side-by-side. N010 G71 G40 G49 G80 N020 G92 X0 Y0 Z0 N030 M06 T01

: Clamp f10 mm end mill cutter.

N040 G94 S1200 M03 N050 G90 G17 G00 G42 X0 Y0 H01 : Note that cutter diameter. compensation has been employed. Cutter diameter is to be taken from register H01. N060 G91 G45 G19 G00 Z-28 H01

: Actual travel to clearance plane = – (28 + length offset in register H01). = – (28 + 10) = – 38 mm. Incremental dimensions henceforth.

N070 G0l Z-24 F50 N080 G17 G0l X120 M08

: Machine AB.

N090 G0l Y90

: Machine BC.

N100 G0l X-120

: Machine CD.

N110 G0l Y-90

: Machine DA.

N120 G19 G00 Z24

: Withdraw tool to clearance plane.

N130 G17 G40 X-10 Y-10 M09

: Cancel cutter diameter compensation at a safe distance from the job. Coolant off.

N140 G28 X0Y0Z0 M05

: Go to home.

Extensions of Numerical Control—CNC, DNC, Machining Centres 523

N150 M06 T02

: Clamp f6 mm drill.

N160 G45 G00 X40 D01

: Actual travel = 40 + value in register D01 = 40 + 20 = 60 mm.

N170 G45 G00 Y15 D02

: Actual travel = 15 + value in register D02 =15 + 30 = 45 mm.

N180 G95 S2200 F0.l

: RPM = 2200, feed rate 0.1 mm/rev.

N190 G91 G19 G45 G00 Z-10 H02

: Actual travel = – (10 + value in register H02) = – (10 + 10) = – 20 mm. Hence, initial plane of canned cycle is now 10 mm from part surface.

N200 G81 G99 Z-12 R-8 M08

: which is 8 mm from the initial plane and movement at feed rate cycle will not be initiated as the X, Y coordinates are not given.

N210 G90 N220 P330 M98

: Call subroutine. Drill pattern I.

N230 M21

: Mirror image in X-axis.

N240 P330 M98

: Drill pattern II.

N250 M22

: Mirror image of pattern II in Y-axis.

N260 P330 M98

: Drill pattern III.

N270 M23

: Cancel mirror image.

M280 M22

: Mirror image of pattern I in Y-axis.

N290 P330 M98

: Drill pattern IV.

N300 G80 G49 M09 N310 G28 X0 Y0 Z0 M05 N320 M30 : 330 N010 X20 Y15

: Soil routine starts. Drill hole 1 of pattern I.

N020 G91 X0 Y20

: Drill hole 2 of pattern I.

N030 X30 Y-10

: Drill hole 3 of pattern I.

N040 M99

: Return to main program.

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Machine Tool Design and Numerical Control

Example 9.4 Write the CNC program for drilling 10 holes f 6 mm in the part shown in Fig. 9.14 using canned cycle programming facility on a CNC vertical drilling machine. The machine has provision of full zero shift. Drilling of the holes is to be carried out at rpm of 2000 and feed of 0.1 mm/rev. 90

72

10

60

3 2

Ym

15

1

15 50

Xm

‘I’ level 30

2

‘R’ level 15 ‘Z’ level 2

Fig. 9.14 Sample part. Problem illustrating the canned cycle programming facility

From the given dimensions, it is clear that the holes are spaced at 10 mm in the X-direction and 8 mm the part following the procedure described in Example 9.1. The tool length offset (say 15 mm) is entered in register H01. The CNC program is given below with an explanation of the instructions side-by-side. N010 G71 G40 G49 G80 G91 N020 G92 X0 Y0 Z0

Extensions of Numerical Control—CNC, DNC, Machining Centres 525

N030 S2000 M03 N040 G95 F0.1 M08 N050 G19 G45 G00 Z-5 H01

:

N060 G81 G99 X15 Y15 Z-19 R-8 F0.1

:

N070 G91 X10 Y8 L9

:

Actual travel = – (5 + value in register H01) = – (5 + 15) = – 20 mm, i.e., 10 mm above the top

Repeat drilling cycle for the remaining 9 holes

N080 G80 M09 N090 G28 X0 Y0 Z0 M05 N100 M30

9.5.6 Zero Setting in CNC Turning Centres Turning centres have provision of two types of zero setting: 1. Full zero shift 2. F Xm

–Z¢A

ZA

Zm

0¢ –X¢A Xp

A XA Zp

Fig. 9.15 Zero setting on turning centre with full zero shift

In CNC turning centres with full zero shift facility, the machine zero generally lies in the space to the right of the head stock and above the spindle axis, but for convenience it is usually shifted to the centre on the right-hand face of the workpiece in clamped position. (Fig. 9.15). The procedure adopted for this purpose is as follows:

526

Machine Tool Design and Numerical Control

which the centre is desired to be shifted. This point A is referred to as program datum. 2. The coordinates of point A representing the centre of the tool tip (say – Z¢A – X A¢ ) in the coordinate system of the machine zero are noted on the digital display of the CNC controller. ZA, XA) are found by moving the tool from point A to the centre of the job 0 and noting down the corresponding Z and X increments. Alternately, the distance between 0 and A in the Z and X directions may be measured by means of suitable end standards (end bars, slip gauges, etc).

N Seq. number G50 X0Z0 N Seq. number G00 – Z¢A – X¢A N Seq. number G50 ZA XA motions will be executed in the coordinate system with its centre at point 0. At the end of the program the (machine zero) with the G50 command. usually it is set at the centre of the right-hand face of the clamped workpiece for convenience. In these turning centres, T00 is designated as the reference tool and is used only for setting the zero datum and is not used later in the program. The procedure for setting the zero datum at point 0 (Fig. 9.16) is as follows: 1. Touch the centre of the face of the workpiece with the tool T00 moving the tool post in the jog made (Fig. 9.l6a). 3. Move the tool post away from the workpiece and touch the job at a known diameter measured accurately with a micrometer or Vernier callipers (Fig. 9.16b). 4. Key in the measured diameter as an X value and press the datum setting button.

0

(a)

Fig. 9.16

0

(b)

Zero setting on turning centre with full floating zero: (a) Setting of zero in Z (b) Setting of zero in X

Extensions of Numerical Control—CNC, DNC, Machining Centres 527

After these steps, the zero will be set at the point 0 for the reference tool T00.

N seq. No. G50 XA ZA

9.5.7 Tool Nose Radius and Tool Position Offsets in CNC Turning Centres The tool nose radius (TNR) compensation in lathe tools is similar to what was discussed with reference to milling cutters in Sec 9.5.1. The part program is written assuming the tool nose radius as zero and the actual value of the TNR is entered in the appropriate register. As in milling, the TNR offset may be left (G41) and right (G42). In right-hand coordinate system which is commonly employed in turning centres, the + X axis points upwards or away from the operator, hence the G41 command will be valid for facing from outer diameter towards the centre and the boring operations, while the G42 command will be valid for external turning operation and facing from the centre towards the outer diameter (Fig. 9.17). In this case, G02 will represent circular interpolation clockwise and G03 circular interpolation counterclockwise.

G 41

G 42

Fig. 9.17 Tool nose radius (TNR) offset left (G41) and right (G42) in turning centres

However, in CNC lathes, the left-hand coordinate system is sometimes encountered. In this system, the + X axis points towards the operator, therefore, the meaning of the commands discussed above is reversed, i.e., G41 is used for TNR compensation for external turning and facing from the centre towards the outer diameter, while G42 is used for TNR compensation for the boring operations and facing from the outer diameter towards the centre. Similarly, the commands used for programming the clockwise and counterclockwise circular arcs are also reversed.

528

Machine Tool Design and Numerical Control

making their tips touch the face and the known diameter of the workpiece and noting the corresponding Z and X values on the digital display. The tool nose radius and the Z and X offsets of each tool are entered in a register. For the tools shown in Fig. 9.18 in the right-hand coordinate system, the values in the registers will be as shown in the table below. Tool 01

Tool 02

Tool 03 Tool 00

Tool 00 Tool 00

X

8

8

–5 –12

10.0 Z

9

Tools 00 are identical in all THREE casses

Fig. 9.18 Tool length offset on turning centres

Tool No.

Offset register number

X offset, mm

Z offset mm

TNR mm

TNV No.

01

01

– 5.0

10.0

2.0

3

02

02

8.0

– 12.0

1.5

3

03

03

8.0

9.0

3.0

3

The TNR, though small, is similar to the radius of the cutter in milling, except that TNR compensation direction in which the tool may be oriented, there are eight standardised TNV numbers. These are shown in Fig. 9.19. The TNV number of the given tool is entered together with TNR and the X and Z offsets in the appropriate register. The tool offsets and tool nose radius are shown in the program as a T word followed by four digits, of T0202 refers to tool number 02 for which the offsets are entered in register 02.

Extensions of Numerical Control—CNC, DNC, Machining Centres 529

No.1

No.4

No.2

No.3

No.5

No.6

6 2

No.7

1

No.8 7 5 3

4

8

Fig. 9.19

Standard tool nose vector (TNV) numbers of tools used on turning centres. Arrows indicate the suggested directions of cutting motions

9.5.8 Special Programming Features in CNC Lathe Programming Spindle Speed lathes with a geared head stock, there are two or more gear ranges. In such machines, the proper range in with the help of an S command.

Diameter Programming In case of parts machined on lathes, one is generally used to referring to the radial dimensions in terms of diameters. Keeping this in view, most CNC turning centres and lathes have provision for what is known as diameter programming, whereby for every programmed movement in the X-axis, the actual displacement is half. For instance, if the tool is programmed to make an incremental movement in X-axis by 4 mm, the radial displacement of the tool is 2 mm which provides for a reduction of the original diameter by 4 mm. However, before writing the CNC program for a given part, it is necessary to check whether the particular machine on which the part is to be machined has radial or diametral programming. Circular Interpolation

Circular interpolation, as explained in Chapter 8 (Sec. 8.2.5) is carried out by the X, Z, I and K characters. It is important to note that even in the CNC turning centres and lathes with diameter programming for X-coordinate, the I value for circular interpolation is programmed in terms of

530

Machine Tool Design and Numerical Control

the radius. In some CNC turning centres and lathes, there is also provision for programming of circular arcs by specifying the location of the end of the arc (through its X and Z coordinates) and its radius through R followed by the particular value. The respective command for clock and anticlockwise circular interpolation in this case appears as follows: N – G02 X – Z – R – N – G03 X – Z – R –

Canned Cycles and drilling operations. For the same reasons, canned cycles are widely used in part programming for CNC codes between G81 and G89 are utilised. As in milling and drilling, canned cycles for turning centres may be

motion started. Thus, on completion of canned cycle the tool returns to the point from where the cycle had commenced. The commands for some of the simple canned cycles are given below: 1. Simple turning, boring and facing (Fig. 9.20) N – G86 X – Z – F –

2. Taper turning (Fig. 9.21) N – G87 X – Z – I – F – with respect to point A. The sign of I is governed by whether the direction of taper with respect to point A coincides with the positive or the negative direction of the X-axis. 3. Taper face turning (Fig. 9.22) N – G88 X – Z – K – F – with respect to point A. The sign of K is governed by whether the direction of taper with respect to point A coincides with the positive or negative direction of the Z-axis. In cases 2 and 3 above, the vectors I and K should be written with the proper sign, as indicated in the see Sec. 9.5.7).

Extensions of Numerical Control—CNC, DNC, Machining Centres 531

4

A 1

3

2

2

3

1

A

4

(b)

(a)

A

1

3 2

4

2 A

1

(c)

Fig. 9.20

4

3

(d)

Canned cycle for (a) turning (b) boring (c) facing (d) internal facing. Cross hatched portion represents material to be removed

4

A

3 1

–I

2 1

2 A

+/

x

3 4 z (a)

Fig. 9.21

(b)

Taper turning canned cycle (a) external (b) internal. Cross hatched portion represents material to be removed

532

Machine Tool Design and Numerical Control

+K

–K

A

1

3 4

2

4

2

X

3 1 A Z

(a)

Fig. 9.22

(b)

Taper face turning canned cycle: (a) External (b) Internal. Cross hatched portion represents material to be removed

9.5.9 Programming Examples for CNC Turning Centres A few problems are discussed below to illustrate the various features of programming on CNC turning to record that the essential fundamentals will remain unchanged for CNC lathes and turning centres with left-hand coordinate system, except that the application of TNR compensation left (G4l) and right (G42) and circular interpolation clockwise (G02) and counterclockwise (G03) will be reversed, as explained earlier in Sec. 9.5.7.

Example 9.5 Write the CNC part program for machining of the component shown in Fig. 9.23 on CNC turning centre with full zero shift facility. The outer diameter turning and facing operations are carried out by tool T01 while the ber are entered in Register 01, while for tool T02 the tool length offset is entered in register 02 (Note that for parting tool there is no TNR compensation). The offsets of the two tools were determined by the procedure explained in Sec. 9.5.7. The outer diameter turning and facing operations are carried out at an rpm of 1000 and feed of 0.2 mm/rev., while the parting operation is carried out at an rpm of 500 and feed of 0.1 mm/rev. It is assumed that a centre hole has been premachined on the job face to allow T01 to be fed to the desired depth for the facing operation.

Extensions of Numerical Control—CNC, DNC, Machining Centres 533

Xm 70

100 Zm

O¢ T0202 100 6 R5 4

+

1

f 25

+ +

A

R5

2

3

f 40

T0101

Xp

45° 5

0

f 15

7

B

30

Zp O¢ MACHINE ZERO O PART ZERO A PROGRAM DATUM

7.5

10

10 45

8

Fig. 9.23 Sample part. Problem illustrating TNR offset right (G42) and full zero shift facilities

The CNC program is given below with the explanation of the instructions side-by-side. N010 G71 G40 G49 G80

:

N020 G50 X0 Z0

:

N030 G00 X-100 Z-100

:

Rapid to point A.

N040 G50 X60 Z70

:

Shift centre of coordinate system to point 0.

N050 G95

:

Feed in mm/rev.

:

Rapid to point B before commencement of facing operation.

N090 G0l G42 X0 Z0F0.2

:

TNR compensation right at the centre of job

N100 X5 M08

:

Machine 0–1

N060 T0101 M06 N070 G90 G00 X0 Z3 N080 S1000 M03

534

Machine Tool Design and Numerical Control

Nll0 G03 X15 Z-5 K5 I0

:

Machine 1–2

N120 G0l Z-15

:

Machine 2–3

N130 G02 X25 Z-20 I5 K0

:

Machine 3–4

N140 G0l Z-30

:

Machine 4–5

N150 X40 Z-37.5

:

Machine 5–6

N160 Z-53

:

Machine 6–7

N170 X45

:

Move to safe point away from job surface

N180 G00G40 X60Z70 M09

:

Cancel TNR compensation. Move rapid to poet A

N190 G50 X-l00 Z-l00

:

N200 G28 X0Z0 M05

:

Move tool to home; spindle stop

N250 G00 X42 Z-53

:

Move 8 mm wide parting tool to the position for commencement of cutting off operation

N260 G0l X-2 F0.l M08

:

Carry out cutting off operation

N210 M06 T0202 N220 G00 X-l00 Z-l00 N230 G50 X60 Z70 N240 S500 M03

N270 G00 X60 Z70 M09 N280 G50 X-l00 Z-l00 N290 G28 X0Z0 M05 N300 M30

Example 9.6 Write the CNC part program for machining of the component shown in Fig. 9.24 on CNC turning centre with full zero shift facility. The facing operation (from outer diameter towards the centre) and the boring operation are both carried out by tool T01 for which the length offsets, TNR compensation and TNV number are entered in register 01. Machining is earned out at constant peripheral speed of 100 m/min and feed of 0.1 mm/rev.

Extensions of Numerical Control—CNC, DNC, Machining Centres 535

Xm 70

100

Xp A R5

B 1

T0101

2 4 45°

O 100

50

3

5

6

f70

f40

R5

f15

7

f50

8 0 10

7.5

10

C Zp

50

Fig. 9.24 Sample part. Problem illustrating TNR offset left (G41)

The CNC program is given below together with the explanation of the instructions side-by-side. N010 G71 G40 G49 G80

:

N020 G50 X0Z0 N030 G00 X-100 Z-100 N040 G50 X100 Z70

:

N050 T0101 M06 N060 G96 S100 M03

:

Constant surface speed of 100 m/min. Spindle on.

N080 G90 G00 X75 Z3

:

Rapid to point B.

N090 G0l G41 X75 Z0 F0.l

:

TNR compensation left.

N100 X50 M08

:

Machine 1–2.

N110 G02 X40 Z-5 R5

:

Machine 2–3.

N120 G0l Z-15

:

Machine 3–4.

N130 X25 Z-22.5

:

Machine 4–5.

N070 G95

Zm

536

Machine Tool Design and Numerical Control

N140 Z-32.5

:

Machine 5–6.

N150 G03 X15 Z-37.5 R5

:

Machine 6–7.

N160 G01 Z-53

:

Machine 7–8.

N170 X10

Move to safe point away from job surface.

N180 G00 G40 Z30 M09

Cancel TNR compensation. Move rapid to point C.

N190 G00 X100 Z70

Move to point A.

N200 G50 X-100 Z-100 N210 G28 X0Z0 M05 N220 M30

Example 9.7 Making use of canned cycles write the CNC program for machining the part shown by dark lines in Fig. 9.25

from the blank: Xp 2

1

C A 2¢ 3

B

15

f 45

f 32

O

f 20

4 E

f 10

f 30

f 40

D

Zp

5

35 40

Fig. 9.25 Sample part. Problem illustrating canned cycle programming facility

(a) Blocks A and B: by turning canned cycle (the allowance being large, it is removed in two cuts). These cycles commence from point 1 having coordinates X48Z2.

Extensions of Numerical Control—CNC, DNC, Machining Centres 537

(b) Block C: by taper turning canned cycle. This cycle commences from point 2 having coordinates (c) Block D: by taper face turning canned cycle. This cycle commences from point 3 having coordinates X22 Z2. (d) Block E: by facing canned cycle. This cycle commences from point 4 having coordinates X12Z2. The simple and taper turning operations are carried out by tool T01, while the taper face turning and facing compensation and TNV numbers of the tools T01 and T02 are entered in registers 01 and 02, respectively. The procedures to be followed for these operations are explained in Sec. 9.5.6 and 9.5.7, respectively. The program datum is selected at X50 Z30. The CNC program is given below together with an explanation of the instructions side-by-side. N010 G71 G40 G49 G80

:

N020 G50 X50 Z20

:

N030 G95

:

N040 M06 T0101 N050 S1200 M03

:

N060 G00 G90 X48 Z5

:

Move rapid to convenient point near point 1.

N070 G01 G42 X48 Z2 M08

:

TNR compensation right at point 1.

N080 G86 X32 Z-25 F0.1

:

Machine block A using turning canned cycle.

N090 X20 Z-25

:

Machine block B.

N100 G80 G00 X48 Z-20

:

Move rapid to a convenient point near point 2.

N110 G01 G42 X48 Z-22

:

TNR compensation right at point 2.

N120 G87 X40 Z-40 I-6 F0.1

:

Machine block C using taper turning canned cycle.

N130 G80 G00 X50 Z20 M09

:

Move to program datum.

N140 M05

:

N150 M06 T0202

:

N160 G00 M03 X22 Z5

:

Move rapid to convenient point near point 3.

N170 G01 G42X22Z2 M08

:

TNR compensation right at point 3.

N180 G88 X10 Z-5 K-6 F0.1

:

Machine block D using taper face turning canned cycle.

538

Machine Tool Design and Numerical Control

N190 G80 G00 X12 Z5

:

Move rapid to a convenient point near point 4.

N200 G01 G42 X12 Z2

:

TNR compensation right at point 4.

N210 G86 X-2 Z-5 F0.1

:

Machine block E using facing canned cycle.

N220 G80 G00 X50 Z20 M09 N230 M05 N240 M30

Review Questions 9.1 Write the CNC part program for contour milling of the part shown in Fig. 8.71 on a vertical spindle CNC milling, assuming the point SP to be concident with the machine zero. The 20-mm thick plate is to be machined using a f10-mm end mill cutter at an rpm of 1200 and feed of 0.2 mm/rev. The machine has provision of full zero shift and cutter diameter compensation and requires the use of G39 command for effecting change of direction of the cutter path. 9.2 For the part shown in Fig. 8.73, write the complete CNC part program for contour milling and drilling on a CNC machining centre. Contour milling of the 20-mm thick plate is done by a f15-mm end mill cutter at an rpm of 1000 and feed of 50 mm/min. Through holes f10 mm are made at positions 1, 2 and 4, while a 10 mm deep hole f15 mm is made at position 3. Drilling is done at a speed of 94.2 m/ min and feed of 0.1 mm/rev. The machining centre has provision for full zero shift, cutter diameter compensation and tool length offset. Show the various tools at certain assumed length offsets and describe the entries to be made in the H-registers. 9.3 Write the CNC part program for drilling a 12 ¥ 10 grid of f 4 mm holes 5 mm deep in a 100 ¥ 100 ¥ 10 mm plate. The holes are arranged in 12 rows and 10 columns and are spaced 6 mm apart. The grid is symmetrical about the centre of the plate. Drilling is to be carried out at an rpm of 2000 and feed of 0.1 mm/rev. 9.4 of the part shown in Fig. 8.59. system at a distance of (100, 100) from the centre of the job face. Facing and turning operations are carried out by Tool T01 at a constant peripheral speed of 120 m/min and feed of 0.15 mm/rev. The parting off operation is carried out by tool T02 at rpm of 500 and feed of 0.08 mm/rev. Assume suitable length offsets and tool nose radius of the tools. Depict the tool positions for the assumed offsets and describe the entries to be made in the tool registers. 9.5 Write the CNC program for machining of the part shown in Fig. 9.26. Turning of outer diameter is done with a tool having TNV number 3, while facing and boring operations are carried out by a tool carried out at cutting speed of 120 m/min and feed of 0.1 mm/rev. Assume suitable offsets of the tools and describe the entries in the tool registers.

Extensions of Numerical Control—CNC, DNC, Machining Centres 539

22 12

3

f45

f32

f36

f15 f55

f22

f35

30°

10

15

10

50

Fig. 9.26 Example part

9.6 Write the CNC part program for machining of the part shown in Fig. 9.26 from a stepped blank. Machining is to be done on the 55 mm long step of f60. The blank has a through hole of f12. Write the program illustrating the use of canned cycles for turning, taper turning, boring, taper boring and taper face turning. Assume suitable values of offsets and machining parameters. 9.7 Write the CNC part program for external turning, facing and boring operations on the part shown in Fig. 9.27. The rapid and cutting feed value are 4.0 and 0.1 mm/rev, respectively and spindle rotates at 500 rpm. The turning operation is done with a turning tool (Tool No. 1) and the facing and boring operations with a boring tool (Tool No. 2). Tool changing is done at machine zero. Hole dia. 15 has been drilled in a previous setting. Use the appropriate tool nose radius compensation commands assuming tool nose radius of 2 mm for both the tools. A- Point of commencement of the turning as well as the facing and boring operations. B- Point employed for zero setting in the part coordinate system. Xm 300

XP

Zm

A (5,30)

20

f 30

f 50

f 15

15 20

f 20

f 80

200

30

75

Fig. 9.27

ZP

B (50, 50)

540

Machine Tool Design and Numerical Control

9.8 Write the CNC part program for facing and external turning of the part and its parting as shown in Fig. 9.28. Given n = 500 rpm; s = 0.1 mm/rev, tool nose radius = 3.0 mm for the turning tool. For the parting operation s = 0.05 mm/rev and constant surface is 60 m/min. A- approach point for start of facing and turning B- point for tool setting C- approach point for start of parting. Xm 300 M/C zero

Zm 200

XP C (90, – 80) 5 20

B (90, 90)

60 10

30

f 40

f 60

f 80

R 10

A (15, 0) ZP

Fig. 9.28

9.9 Write the CNC part program for contour milling of the part shown in Fig. 9.29 using cutter diameter compensation left. All dimensions are in inches. Given n = 500 rpm, s = 0.1 mm/rev, diameter of end mill cutter = 10 mm. Use G39 command at the points involving change of direction. (overlap) 0.1 0.5 0.1 (approach)

yp 4.5

R 1.5 4.5 5.0

ym

A

2.0

y

4.5

5.0

M/c table

B xm

z

A = 5.9, B = 4.45

Fig. 9.29

1.5

xp

Extensions of Numerical Control—CNC, DNC, Machining Centres 541

9.10 Write the CNC part program for the part shown in Fig. 9.30 using tool length compensation. Use the tool length off set facility for zero setting the machine zero to point A. Hole f 0.6 is made in two cuts, f 0.4, then f 0.6. Length of f 0.2 inch drill is 3.0 inch. Yp 0.1 0.5

R-plane 1.5

0.3 1.0

A

3.0 f 0.2

4.0

Ym

f0.6

5.0

1.5

f 0.4 1.0

Xp

0.2 3.0 M/c table

Xm

4.0 Drill f 0.6 5.0

Drill f 0.4 Drill f 0.2

6.0

Fig. 9.30

All dimensions in inches

Index 543

INDEX

A Absolute Dimensioning System 435 AC Variable Speed Drives 133 ADAPT 466 Adaptive Control Systems 415 Additional Check for Strength of spindle 301 Aerodynamic Bearings 329 Aerostatic Bearings 330 Air-lubricated Bearings 328 Amplitude-frequency plot 338 Amplitude-phase-frequency Plot 337 Analogue Displacement-measuring Devices 438 Analogue Displays 405 Angular Indexing 161 Anthropometric and Functional Anatomy Data 398 Anti-friction Bearings 303 Anti-friction Guideway with Recirculation of Rolling Elements 269 Aperture in Box-type Structure 183 APT 466 APT Auxiliary Statements 476 APT Computation Statements 477 APT Geometry Statements 471 APT Motion Statements 467 APT Postprocessor Statements 475 APT Program for Drilling Holes 478 APT Program for Milling a Pocket 482 APT Programming Language 466 ASCII Code 428 Automap 466 Automatic Control Systems 396, 410 Auxiliary motions 2, 3, 59 Average Sideway Pressure 246

B Basic Length Unit 424 Baud Rate 503

Bearing Clearance 315 Bearing Materials 312 Bed Sections and Wall Arrangements 198 Behind the Tape Reader 501 Belt Transmission 37 Bolt Arrangement for columns 187 Box-type Structure 183 BSW Thread 147, 150, 164 Buffer Storage 431

C Cam-controlled Systems 411 Cam Mechanism 40 Canned Cycles 514, 530 for simple turning 530 for taper face turning 530 for taper turning 530 Capillary Compensated Pad Bearing 257 Capillary Restrictor 257 Cast Iron Slideways 238 Chain Transmission 38 Chatter Vibrations 348 Check Surface 469 Chip Thickness C Circular Interpolation 433

Clearances in Slideways 240 Closed Hydrostatic Slideways 263 Closed-type Anti-friction Ways 268 Clutches 48, 50 electromagnetic 51 friction 21, 31, 33, 42, 49, 50, 51 CNC Programming 504 Coding Systems 426 Column Sections 214

544

Index

Combination of Two Flat Slideways 245 Combination of V and Flat Slideways 243 Combinations of Various Anti-friction Bearings 304 Comparators 439 continuous-function type 439 discrete-function type 439 Compatibility in the Design of Control Members 408 Composite Slider Bearing 253 Computer Aided Part Programming 465 Computer Numerical Control 495 Cone Variators without Intermediate Member 135 Constant-DP-type Reducing Valve 129 Restrictor 261 C Constant-gap Restrictor 261 Continuous-function-type Levers 408 Continuous-path Systems 425 Control Loop Unit (CLU) 494 Control Systems 386 Control Systems for Changing Speeds and Feeds 386 Control Systems for Forming and Auxiliary Motions 396 Control Systems with Centralised Control 387 Control Systems with Individual Control 387 Couplings 48, 50 Crank-and-Rocker Mechanism 39 Cross Receptance 351 Cutting Process Closed-Loop System 336 Cutting Process Dynamic Characteristic 356 Cutting Speed 3, 4, 12, 13, 20, 55, 65, 66 Cutting Speed C

D Data Processing Unit (DPU) 494 Decoder 431

Supports 296 Design Criteria for Machine Tool Structures 167 Design Criteria for Slideways 243 Design for Stiffness 194 Design for Stiffness of Spindle 301

Design for Strength 191 Design of Speed Box: Selection of Range Ratio 76 Design of Aerostatic Slideways 265 Design of Anti-friction Guideways 267 Design of Anti-friction Ways for Stiffness 271 Design of Bases and Tables 219 Design of Beds 197 Design of Columns 213 Design of Cranks 402 Design of Cross Rails, Arms, Saddles and Carriages 221 Design of Hand Wheels 404 Design of Housings 216 Design of Hydrodynamic Slideways 252 Design of Hydrostatic Slideways 254 Design of Knobs 401 Design of Lathe Bed 203 Design of Levers 403 Design of Push Buttons 401 Design of Rams 222 Design of Rolling-friction Power Screws 281 Design of Rotary Levers and Star Wheels 404 Design of Slideways 234 Design of Slideways for Stiffness 249 Design of Slideways for Wear Resistance 243 Design of Sliding-friction Power Screws 276 Design of Toggles 401 Design Procedure of Machine Tool Structures 188 Design Process Applied to Machine Tools 57 Design recommendations for Displays 406 Design Requirements to Spindle Units 288 Determination of Gear Dimensions 112, 117 Determination of Shaft and Gear Dimensions 112 Determination of Shaft Dimensions 112, 115 Determining Number of Teeth of Gears 108 Diagonal Stiffeners 199 Differential Indexing 159 Differential Mechanism 46 Digital Displacement-measuring Devices 436 Digital Displays 406 Dimensional-data Functions 448

Index 545

D Format 444 Direction-control Valves 29 Direct Numerical Control 501 Direct Receptance 350 Displacement-measurement Devices 435 Display 405 Distributive Numerical Control 495 DNC E Drip Feeding 502 Drive Motion 1, 2, 3 Drives of NC Machine Tools 434 Drive Surface 468 Drives Using Multiple-speed Motors 95 Drives with Flow-control Valves 127 Drives with Fluid Delivery Control 126 Factors 179 Dynamic A Dynamic Characteristic of the Cutting Process 352 Dynamic Characteristic of the Equivalent Elastic System 340 Dynamic C Dynamic Compliance 179 Dynamic Coupling 343 Dynamic Stability of EES 336 Dynamic Stiffness 178

E Economic Effectiveness of Machine Tool 52 EIA code 428 Electrical Automatic Control Systems 413 Electric Motor 20 Selection of 20 Electrical Stepless Regulation 130 Electrodynamic Exciters 350 Electrohydraulic Exciters 350 Electromagnetic Exciters 350 Equivalent Elastic System 336 Ergonomics 398 Errors in Anti-friction Ways 272 Exapt 466

F Facing Operation 6

Faceplate Variators 134 Feed Box 90 design of 69, 90 Feed Boxes with Change Gears 121 Feed Boxes with Gear Cone and Sliding Key 121 Feed Boxes with Sliding Gears 121 Feed Boxes with Tumbler Gear (Norton’s Gear) 122 Feed Box with Meander’s Mechanism 123 Feed-changing Mechanism of a Horizontal Milling Machine 388 Feed Motion 1, 2, 3, 20, 43 Feed Motion Drives 435 Feed Per Minute 4, 9, 42, 66 Feed Per Revolution 4 Feed Per Tooth 4, 9, 66, 67 Feed-rate Function 448 Finite Hydrodynamic Bearing 320 Fixed Block Format 443 Forced Vibration due to Perturbance of the Cutting Process 379 Forced Vibrations due to Perturbance of EES 381 Forced Vibrations of Machine Tools 378 Force Required for Shifting a Toggle 400 Function Codes 447 Functions of Control Systems 386 Functions of Guideways 233 Functions of Machine Tool Structures 167 Functions of Spindle Unit 288 Function Symbols on Control Panels 409

G Gearing Diagram 105, 155 general recommendaitions 105 Gearing Diagram of Drilling Machine 151 Gearing Diagram of Horizontal Milling Machine 155 Gearing Diagram of Speed Box of a Lathe with Mechanical Variator 143 Gearing Diagram of Speed Box of a Thread Cutting Lathe 141 Gear Transmission 35 Generalised Model of Dynamic Cutting Force 362 General Processor 465

546

Index

Geneva Mechanism 43

H Hand Wheels 409 Hersi-Shtribek Diagram 310 Hole Combinations in ASCII Code 430 Hole Combinations in EIA Code 429 Horizontal Milling Machine 155 Hydraulic Cylinders 28 Hydraulic Drive 23, 24, 30 rotary 2, 3, 19, 21, 23, 24, 25, 29, 31, 34, 39, 60, 61 translatory 2, 19, 23, 24, 25, 28, 30, 34, 39, 40, 42, 43, 60, 61 Hydraulic Preselective Control System 393 Hydraulic Stepless Regulation 125 Hydraulic Transmission 23 Hydrodynamic Force 252

multiple-pad 323 multiple-recess 323, 324 single-pad 323, 324 Hydrostatic Slideway with a Restrictor 257 Hydrostatic Slideway without Restrictor 255

I Incremental dimensioning system 435 Indexing 162 Indexing For Machining of Helical Grooves 162 Indexing Head 157 Interpolator 432

K Kudinov’s Dynamic Cutting Force Expression 355

L Layout of Machine Tool 60 Limit Constraint Adaptive Control System 417 Limited Travel Anti-friction Guideway 268 Linear Inductosyn 439 Linear Interpolator 433 Cylindrical Pad 325 Load and Flow C

Hydrodynamic Bearing 316 Load Capacity of a Hydrostatic ournal 327 Location of Control Members 407

M Machine Control Unit 424, 494 Machine Tool Design 54 safety and convenience of controls 56 simplicity of design 56 Machine Tool Drives 19 Machine Tool Formats 440 Machine Tools 1, 52

Machining Centres 498 Machining Cost 68, 90 Machining Time 4, 8 turning operation 4 facing operation 6 boring operation 6 drilling operation 8 plane milling operation 10 symmetrical face milling operation 10 asymmetrical face milling operation 11 shaping operation 13 cylindrical grinding: external traverse cut 14 cylindrical grinding: internal 15 surface grinding 17 Magic-three Code 447 Male and Female Body Dimensions 398 Man-machine System 397 Manual Control System 396 Manual Data Input 496 Manual Part Programming 440 Many Degrees of Freedom System in Generalised Coordinates 343 Many Degrees of Freedom System in Normal Coordinates 342 Materials of Machine Tool Structures 170 Materials of Slideways 236

Index 547

Materials of Spindles 289 Maximum Force Required for Pressing a Button 400 Maximum Sideway Pressure 246 Maximum Turning Force for Operating Levers 400 Mechanical Automatic Control Systems 410 Mechanical Exciters 350 Mechanical Stepless Regulation 134 Mechanical Transmission 34 Metering-in Circuit 128, 129 Metering-out Circuit 128, 129 Milling Machine 389 Mirror Imaging 513 Miscellaneous Functions 451 Mode Coupling 346 Model Technique in Design of Structures 224 Modular Thread 147, 151 Moment of Inertia of Lathe Bed Section 208 Motor Pulsing Rate 497 Multiple Start Threads 149 Multiple-wedge Bearings 321

N N C Hardware 430 NC Systems 422 closed-loop 423 open-loop 422 Numerical Control 419 fundamental Concepts 419 Nut-and-Screw Transmission 41 Nyquist Criterion 337

O ON-OFF-type Push Buttons 408 Open-type Anti-friction Ways 267 Optimum Location of Displays 406 Optimum Spacing Between Spindle Supports 297 Compensated Pad Bearing 259 O O Restrictor 259 Overlap Factor 373

P Parabolic Interpolators 434

Parity Check 428 Part Surface 468 Penetration Rate 359 Perpendicular Stiffeners 199 Phase-frequency Plot 338 Piezoelectric Exciters 350 PIV Drive 140 Plain Indexing 158 Point-to-point Systems 425 Polar Plot 337 Polar Plot of the Cutting Process 357 Positioning-cum-straight-cut Systems 425 POST PROCESSOR 466 Power Screws 276 Preloading of Anti-friction Bearings 306 Preparatory Functions 449 Preselective Centralised Control 387 Preselective Control System 392 hydraulic 392 mechanical 392 Pressure Valves 31 Process Optimisation Adaptive Control Systems 417 Programming for a Circular Arc 458 Programming for a Straight Line 457 Programming Formats 443 Program Reader 430 Protecting Devices for Slideways 274 Pumps 25 constant delivery 25 radial piston 27, 28 variable delivery pumps 25, 27, 28 Push button 408

Q Quantitative Displays 405 Quick Acting Mechanical Tightening Device 300

R Rack-and-Pinion Transmission 42 Ratchet-Gear Mechanism 43 Reduction of Axial Dimensions of the Speed Box 106

548

Index

Reduction of Radial Dimensions of Speed Boxes 107 Regenerative Chatter 372 Resolver 438 Resultant Receptance 351 Reversing Mechanism 44 Ribs and Stiffeners 186 rpm Values Constitute 69, 70 Arithmetic progression 69 geometric progression 70 harmonic progression 71 logarithmic progression 73

S Sampling Rate 497 Selective Centralised Control 387 Selective Control Systems 395 Self-excited Vibrations 348 Sequence-number Function 447 Set-up Point 441 Shapes of Slideways 235 Simple Centralised Control 387 multiple lever 387 single lever 387 Single Degree of Freedom Elastic System 341 Single-lever Control System 389 Sleeve Bearings 311 Slider Crank Mechanism 39 Slideways with Hydraulic Relief 273 Slideways with Rolling Members 273 Sliding Bearings 310 Sliding ournal Bearings 311 Special Programming Features in CNC 512 Speed and Feed Changing Systems with Simple Centralised Control 387 Speed Boxes Built into the Spindle Head 118 Speed Boxes Designed as Separate Unit 119 Speed Boxes with Change Gears 119 Speed Boxes with Friction Clutches 120 Speed Boxes with Sliding Gears 120 Speed Box with a Combined Structure 102

Speed Box with Broken Geometrical Progression 101 Speed Box with Overlapping Speed Steps 99 Speed-changing Mechanism of a ig-boring Machine 387 Speed Chart 88 Speed Function 447 Speed Box using a Two-speed Motor 97 Spheroidal and Cone Variators with Intermediate Member 137 Spheroidal Cast Iron Slideways 238 Spindle Ends 289 Stability Analysis 366 Standard Spindle Speeds 81 Static Coupling 343 Static Stiffness 176 Static Stiffness with Respect to Dynamic Stability 176 Static Stiffness with Respect to the Workpiece Accuracy 176 Steel Slideways 238 Stepless Regulation 124, 125, 130, 134 Stepped Regulation of Speed 69 Stiffener Arrangement 186 Stiffness of Aerostatic Slideways 266 Stiffness of Hydrostatic ournal Bearing 328 Structural Diagrams 82 Subroutine Programming 513 Symap 467 Synchro Devices 439

T Tab-sequential Format 444 Tape Format and Codes 428 Template-controlled Systems 412 Thermal Equilibrium in Hydrodynamic Bearing 318 Thick Film Bearings 310 Thin Film Bearings 310 Thread Cutting on Lathe 146 Throttles 33 Time Constant of Chip Formation 356 Tlusty’s Dynamic Cutting Force Expression 354 Tobias’ Dynamic Cutting Force Expression 358 Tool Diameter Offset 505

Index 549

Tool Length Offset 506 Tool Magazines 499 chain type 499 rotary drum type 499 Tool Nose Radius (TNR) Compensation 527 Tool Nose Vector (TNV) 528 Tool Offset 459 Tool-radius Compensation 460 circular bend, anticlockwise 461 circular bend, clockwise 461 rectangular bend 460 tapered bend 461 Tool Radius Vector Setting 512 Tool-selection Function 447 Torsional Rigidity of Bed with Perpendicular Stiffeners 200 Torsional Rigidity of a Rectangular Box-type Section 202 Torsional Rigidity of Beds with Diagonal Stiffeners 202 Transfer Function 339 Transfer Function of EES 339 Transfer Function of the Cutting Process 355, 357 Turning Centres 499 Turning Moment for Switching Knobs 400 Types of Anti-friction Ways 234 Types of Slideways 233

U Unit Stiffness 171 Unit Stiffness Values of Engineering Materials 171 Unit Strength under Bending 173 Unit Strength under Tension 171

Unit Strength under Torsion 172

V Variable-block Format 444 Variators with Axially Displaceable Cones 138 Velocity Coupling 343

W Ward–Leonard System 132 Wave Cutting 364 Wave Removal 364 Wave Removal over Wave Cutting 366 Word-address Format 444 Work-area for an Average-size Operator 399 Working Motions 1, 2, 3

X X Modem Protocol 504 X ON/X OFF Protocols 504

Z Zero Film Bearings 310 Zero Point 441 Zero Setting in CNC Machining Centres 510 Zero Setting in CNC Turning Centres 525 Zero System 441

full zero shift 441