Engineering mechanics : statics 9781783323562, 1783323566

Table of contents :
Preface / Introduction / Co-planer System of Forces / Equilibrium of System of Forces / Analysis of Structure / Centroid and Moment of Inertia / Friction / Simple Stress & Strain / Objectives type Questions / Index.

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ENGINEERING MECHANICS STATICS

Special Thanks to Mr. Vibhor Sharma, Assistant Professor, Transportation, MIT Institute of Design, Pune for drawing artful illustrations and figures using Photoshop software which has enhanced the effectiveness of the text.

ENGINEERING MECHANICS STATICS

Vikrant Sharma Atul Kumar N S Baruaole† Mukesh Kumar

α Alpha Science International Ltd. Oxford, U.K.

Engineering Mechanics Statics 292 pgs. | 478 figs. | 11 tbls.

Vikrant Sharma Atul Kumar N S Baruaole† Mukesh Kumar Mechanical and Engineering Department Mody University Lakshmangarh Copyright © 2018 ALPHA SCIENCE INTERNATIONAL LTD. 7200 The Quorum, Oxford Business Park North Garsington Road, Oxford OX4 2JZ, U.K. www.alphasci.com ISBN 978-1-78332-356-2 E-ISBN 978-1-78332-432-3 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior written permission of the publisher.

Sincerely Dedicated to The Lord Ganesha & to our loving parents, family and friends

In the loving memory of our Co-Author & former Head of Department (Late) Sh. Nitin S. Baruaole

Preface Engineering Mechanics: Statics, deals with the behavior of bodies under the action of forces. It is the foundation of engineering disciplines hence, necessary for the students of engineering. The objective of this book is to develop a fundamental knowledge of ‘Statics’ and ability to analyze problems in real life with a simple and logical manner for carrying out substantial engineering design. Based on our classroom experiences, we have designed this textbook in a clear and concise language. The purpose is to present the concepts and principles of ‘Statics’ in a systematic approach to help the students understand the subject. We have invested a great deal of attention to learn the fundamental principles of statics and the development of problem-solving skills. The book would be useful for engineering students and the professionals as well. Large numbers of solved problems are selected to help students understand the intricacies of engineering problems and to analysis. Practice problems are provided for each unit, based on principles of statics and real-life engineering problems that are applicable to engineering designs. Each unit also lists a summary of key concepts to facilitate further learning. We wish to express our thanks to all colleagues and friends who have provided valuable inputs during preparation of this text book. Our special thanks to Dr. V. K. Jain, Dean, Mr. Satyajeet Anand and Mr. Mukul Kant Paliwal, Assistant Professors, CET Mody University for their valuable guidance. We also thank all of our students at CET, Mody University. Finally, our sincere thanks to Narosa Publishing House, for taking interest and publishing this book in the best possible form. Vikrant Sharma Atul Kumar N S Baruaole† Mukesh Kumar

Contents

Preface

0. Introduction

vii

0.1—0.5

0.1 Mechanics 0.2 Basic Concept 0.3 Fundamental Laws 0.4 Units

1. Co-planer System of Forces

0.3 0.3 0.4 0.5

1.1—1.76

1.1 Introduction to Force 1.2 Classification of Force 1.2.1 System of Forces 1.3 Resultant Force 1.3.1 Resultant of Collinear Forces 1.3.2 Resultant of Concurrent Forces 1.4 Resolution of a Force 1.5 Moment 1.5.1 Resultant Moment 1.5.2 Varignon’s Theorem (Principle of Moment) 1.5.3 Resultant of Parallel Force System 1.6 Couple 1.6.1 Moment of a Couple 1.6.2 Force–Couple System 1.7 Equivalent Systems of Forces Solved Examples Based on Parallelogram Law of Forces Solved Examples Based on Resolution of Forces Solved Examples Based on Co-planer Concurrent Forces Solved Examples Based on Moment & Non-Concurrent Forces Solved Examples Based on Parallel Forces Solved Examples Based on Force-Couple System and Equivalent System Summary Problems

2. Equilibrium of System of Forces

1.3 1.4 1.4 1.7 1.7 1.7 1.10 1.13 1.14 1.14 1.15 1.16 1.17 1.17 1.18 1.18 1.26 1.35 1.43 1.55 1.59 1.70 1.71

2.1—2.38

2.1 Equilibrium 2.2 Free Body Diagram (F.B.D.) 2.2.1 General Procedure for Constructing a Free Body Diagram 2.2.2 Examples on Free Body Diagram (FBD) 2.3 Equilibrium for Different Force System (Based on Number of Forces)

2.3 2.3 2.6 2.6 2.8

x  Engineering Mechanics Statics Solved Examples Based on Equilibrium of Forces Summary Problems

3. Analysis of Structure

2.10 2.34 2.35

3.1—3.49

3.1 Trusses 3.2 Types of Truss 3.3 Analysis of Truss 3.3.1 Assumptions in Truss Analysis 3.3.2 Methods of Truss Analysis Solved Examples Based on Method of Joints Solved Examples Based on Method of Sections Summary 3.4 Beam 3.5 Types of Beams 3.6 Types of Loading 3.7 Shear Force (V) and Bending Moment (M) 3.7.1 Sign Convention 3.8 Shear Force and Bending Moment Diagrams Solved Examples Based on Simply Supported Beam Solved Examples Based on Cantilever Beam Solved Examples Based on Overhang Beam Summary Problem

4. Centroid and Moment of Inertia

3.3 3.3 3.4 3.4 3.4 3.5 3.13 3.19 3.19 3.19 3.20 3.21 3.22 3.22 3.22 3.34 3.44 3.48 3.48

4.1—4.34

4.1 Introduction 4.2 Determination of Centroid 4.3 Centroid of Wire 4.4 Procedure to Determine Coordinates of Centroid of Composite Areas 4.5 Centroid of Common Areas Using Method of Integration 4.6 Centroid and Area of Some Common Plane Areas Solved Examples Based on Centroid Summary Problems 4.7 Area Moment of Inertia 4.8 Perpendicular Axis Theorem 4.9 Parallel Axis Theorem 4.10 Radius of Gyration 4.11 Area Moment of Inertia of Common Areas Solved Examples Based on Area Moment of Inertia Summary Problems

4.3 4.3 4.5 4.5 4.5 4.7 4.8 4.18 4.19 4.19 4.20 4.21 4.22 4.22 4.26 4.32 4.33

Contents  xi

5. Friction 5.1 Introduction 5.2 Coulomb’s Theory of Dry Friction 5.3 Coefficient of Friction 5.4 Angle of Friction 5.5 Angle of Repose 5.6 Applications of Friction 5.6.1 Ladder Friction 5.6.2 Wedge Friction 5.6.3 Belt Friction Solved Examples Based on Block Friction Solved Examples Based on Ladder Friction Solved Examples Based on Wedge Friction Solved Examples Based on Belt Friction Summary Problem

6. Simple Stress & Strain 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10

Stress 6.1.1 Types of Stress Strain 6.2.1 Shear Strain 6.2.2 Longitudinal and Lateral Strains Poisson’s Ratio Volumetric Strain Hooke’s Law and Elastic Moduli Bar of Varying Cross-Section Stress-Strain Diagram for Mild Steel Temperature Stress and Strain Composite System Factor of Safety Solved Examples Based on Stress and Strain Summary Problems



Objective Type Questions Review Problems Index

5.1—5.40 5.3 5.3 5.4 5.5 5.5 5.6 5.6 5.7 5.7 5.9 5.20 5.26 5.34 5.38 5.38

6.1—6.20 6.3 6.3 6.4 6.5 6.5 6.6 6.6 6.6 6.6 6.7 6.8 6.9 6.9 6.9 6.18 6.19 OTQ.1—OTQ.10 R.1—R.3 I.1–I.2

Unit 0

Introduction 0.1 Mechanics echanics is that branch of physics which deals with the state of rest or motion of M bodies under the action of forces. No other subject plays a bigger role in engineering analysis than mechanics. Though the principles of mechanics are few, they have varied application Mechanics is the study of in engineering. A detailed understanding of this forces and their effects. subject is an essential prerequisite for work in many fields of engineering. Mechanics is divided into three parts: 1. Mechanics of Rigid Bodies 2. Mechanics of Deformable Bodies and 3. Mechanics of Fluid. In this book, we will concentrate on mechanics of rigid bodies. Mechanics of rigid body deal with perfectly rigid bodies. It may be classified as follow:

Statics deals with the equilibrium of bodies under action of forces and Dynamics deals with the motion of bodies. Dynamics is further classified as, Kinematics and Kinetics. Kinematics is the study of motion of bodies without any reference to the cause of motion. (i.e. without any reference to effects of forces and masses). Kinetics is the study of bodies with reference to effects of forces and masses. In this book, we will concentrate on Statics.

0.2  Basic Concept The following concepts are important to understand before studying mechanics. 1. Space: It is the boundless three-dimensional region occupied by bodies which have relative position and direction. The position of the bodies

0.4  Engineering Mechanics Statics

can be defined by two lengths (in two-dimensional problems) or three lengths (in three dimensional problems). These lengths are known as the coordinates of bodies. 2. Time: Time provides a measure of when an event or sequence of events occurs. 3. Mass: Mass is the amount of matter in an object. It is a measure of inertia of a body, which is resistance of matter to a change of velocity. 4. Particle: A particle is a body of negligible mass and dimensions. Objects that are small compared to other objects can be idealized as particles. 5. Rigid body: A rigid body is non-deformable body i.e. the distance between any points in the body remains fixed.

0.3  Fundamental Laws The study of mechanics is based on following fundamental laws: 1. Newton’s Laws of Motion: Sir Isaac Newton in seventeenth century stated three laws of motion as follow, Law I: A particle remains at rest, or continuous to move in a straight line with uniform velocity, if there is no unbalanced force acting on it. Law II: The acceleration of a particle is proportional to the resultant force acting on the particle and is in the direction of this force. Law III: The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear. 2. Newton’s Law of Gravitation: This law states that two particles of mass m1 and m2 are mutually attracted with equal and opposite forces P and – P of magnitude P given by the formula,

P = G

m1m2 r2

Where, m1 and m2 = masses of two particles r = distance between two particles G = constant of gravitation 3. Law of Transmissibility: It state that, the point of application of a force can be transmitted to any other point along its line of action within the body. Force P acting at point O can be transmitted to point O’ along its line of action as shown in Figure 1.

Introduction  0.5

Fig. 1

0.4 Units In mechanics we use four fundamental measures. These are length, mass, force, and time. Although there are a number of different systems of units, only the two systems are most commonly used in engineering. They are shown in Table 1. Table 1. Fundamental Units Fundamental Measures

Symbol

SI Units

U.S. Customary Units

Length

L

Meter (m)

Foot (ft)

Mass

M

Kilogram (kg)

Slug

Force

P

Newton (N)

Pound (lb)

Time

T

Second (s)

Second (s)

Unit 1

In this unit, we discuss forces and effect of forces on particle and rigid bodies in detail. Particularly, we learn how to determine resultant force of different force systems. The moment of force and couple and concept of equivalent system are also introduced in this unit. Systematically understanding of these concepts is important to the design engineers.

Co-planer System of Forces 1.1  Introduction to Force Study of Statics and the whole study of Mechanics is actually the study about the actions of forces or force systems and the effect of these actions on bodies. So it is important to understand the action of forces, characteristics of force systems, and particular methods to analyze them.

What is force? Force is an action that Force is the action of a body about another body, has the ability to change and it changes or tends to changes the state of rest motion or motion of a body. Forces exist because of an interaction of one object with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. Force is measured using standard metric unit known as the Newton; Newton is abbreviated as ‘N’, so 15 N means 15 Newton of force. One Newton equals 1 kilogram multiplied by 1 meter per second squared. This means that a force of one Newton causes a 1-kilogram mass to have an acceleration of 1 m/sec2. 1 N = (1 kg) (1 m/s2) = 1 kg-m/s2. The force is a vector quantity as its effect depends on the direction as well as on the Force is a vector quantity magnitude of the action. Thus, for example 15 N is not a full description of the force acting on an object, in contrast 15 N acting in downward direction (or any direction) is a complete description of the force acting on an object. Force acting on an object may cause the object to change its shape, to start moving, to stop moving, to accelerate or decelerate. As shown in figure 1.1, the effect of the force applied to the bracket depends on P, the angle θ and the location of the point of application. If any of these specifications will change, it alters the Fig. 1.1 effect on the bracket. So magnitude, direction, sense and point of application are characteristics of the force. Magnitude represents the value of force, like, 15 N, 10 KN etc. It is also expressed in Pounds (lbs). The Magnitude can be represented graphically by drawing a vector to scale.

1.4  Engineering Mechanics Statics

Direction of the force is represented by line of action and angle it forms with some reference axis. The line of action is indefinitely long line on which the force is laying. Sense of the force is represented by an arrowhead; it specifies the direction in which the force moves along the line of action. The direction relates to the line of action of the force, and the sense is the way in which the force moves along that line. Point of application is the exact contact point (location) at which a force is applied to a body. Graphically it is represented by the tip of the arrowhead (it may be situated in the opposite end as the arrowhead) and it is unique to each force. It may happen that different forces share same point of application.

1.2 Classification of Force Forces are classified as either contact or body forces. Contact force is produced by direct physical contact of two objects, and they are distributed over a surface area of the body; examples of contact forces include applied force, frictional forces, normal forces etc. Body force is produced when body (object) is located in a force field such as a gravitational, electric or magnetic field. These types of forces results when the two interacting objects are not in physical contact with each other and they are distributed over the volume of the body. Forces are also considered as an external force or internal force. External force is applied externally to an object; they are either applied or reactive forces. Internal force is developed inside the body to resist deformation of body. Forces may be further classified as concentrated or distributed force. When the area over which contact force is applied is very small like a point, the force may be considered as concentrated on a point. The force which is distributed over an area is considered as distributed force.

1.2.1 System of Forces When two or more forces act on a body they are System of forces is simply called to form a system of forces. Most practical a particular set of forces problems of statics involve system of force. System of forces is simply a term used to describe a group of forces. System of forces may be classified as follows,

Co-planer System of Forces  1.5

Coplanar forces are those forces whose lines of action lies on the same plane. Coplanar forces may be collinear, concurrent, non-concurrent, and parallel. NonCoplanar forces are those forces whose lines of action does not lie on the same plane. Non-Coplanar forces may be concurrent, non-concurrent, and parallel. Non-coplanar forces are also known as space forces or spatial force system, as forces passes through space when they change their plane. Table 1.1 shows types and characteristics of different system of forces. Table 1.1.  Characteristics of Different System of Forces Collinear Forces are those forces whose lines of action lies on the same line. As shown in figure, forces P1 and P2 acting in same plane and having a common line of action, this system is coplanar collinear force system. Similarly forces P3 and P4 are collinear forces.

Coplanar Concurrent Forces are those forces whose lines of action passes through the same point and also lie on the same plane. The concurrent forces may or may not be collinear. As shown in figure, line of action of forces P1, P2 and P3 meet at common point O and also these forces lie on the same plane, this system is coplanar concurrent force system. Coplanar Non-Concurrent Forces are those forces whose lines of action does not pass through the same point but lie on the same plane. As shown in figure, forces P1, P2, P3 and P4 lie in the same plane but their lines of action do not meet at same point, this system is coplanar non-concurrent force system.

1.6  Engineering Mechanics Statics Coplanar Parallel Forces are those forces whose lines of action are parallel to each other and also lie on the same plane. As shown in figure, forces P1, P2, and P3 lie in the same plane and their lines of action are parallel to each other, this system is coplanar parallel force system. Parallel forces can be in the same or in opposite directions. Parallel forces which are having same direction are known as like parallel forces and parallel forces which are having different directions are known as unlike parallel forces. Non-Coplanar Concurrent Forces are those forces whose lines of action passes through the same point but do not lie on the same plane. As shown in figure, line of action of forces P1, P2 and P3 meet at common point O but do not lie on the same plane, this system is non-coplanar concurrent force system.

Non-Coplanar Non-Concurrent Forces are those forces whose lines of action does not pass through the same point and also not lie on the same plane. As shown in figure, line of action of forces P1, P2 and P3 do not pass through the same point and also not lie on the same plane, this system is non-coplanar non-concurrent force system.

Non-Coplanar Parallel Forces are those forces whose lines of action are parallel to each other but do not lie in the same plane. As shown in figure, forces P1, P2, and P3 are parallel to each other but do not lie in the same plane.

Co-planer System of Forces  1.7

1.3 Resultant Force

Resultant is a single force When number of forces are acting on an object then which produces the same it is possible to replace them with a single force effect as produced by which produces the same effect as produced by number of forces when all the given forces, this single force is known as acting together resultant force. Composition is the Resultant force is denoted by R. The procedure procedure to find out to find out single resultant force of a system of single resultant force of a forces is known as composition of forces. system of forces

1.3.1 Resultant of Collinear Forces The resultant of two or more collinear forces is simply their algebraic sum and acts on the same line on which lines of action of all collinear forces lie. If two forces P1 and P2 act in same direction as shown in figure 1.2, their resultant R = P1 + P2. If two forces P1 and P2 act in opposite direction as shown in figure 1.3, their resultant R = P1 – P2. As a matter of convenience, force acting towards right is considered as positive and force acting toward left is considered as negative.



Fig. 1.2

Fig. 1.3

1.3.2 Resultant of Concurrent Forces Here we may consider two cases of concurrent forces. Case I, resultant of two concurrent forces and Case II, resultant of more than two concurrent forces. CASE I. Resultant of two concurrent forces: Resultant of two concurrent forces can be found out by means of the parallelogram law or the triangle law. These methods are discussed in following section: Parallelogram Law of Forces (Analytical method) It state, “If two forces acting simultaneously at a point be represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant may be represented in magnitude and direction by the diagonal of the parallelogram passing through that point.” In figure 1.4, two forces P1 and P2 acting at point O are represented

Fig. 1.4

1.8  Engineering Mechanics Statics

in magnitude and direction by OA and OB respectively. By parallelogram law the resultant R is represented in magnitude and direction by OC. We now find magnitude and direction of resultant as below: Let, θ be the angle between two forces P1 and P2 and α is the angle between force P1 and resultant R. Now from C draw a perpendicular CD to OA produced. Now < DAC = < AOB = θ …(Corresponding angle) In ∆DAC, CD = P2 sin θ and AD = P2 cos θ In ∆DOC, OC2 = OD2 + CD2 = (OA +AD)2 + CD2 \ R2 = (P1 + P2 cos θ)2 + (P2 sin θ)2 \ R2 = P12 + 2P1P2 cos θ + P22 cos2 θ + P22 sin2 θ \ R2 = P12 + 2P1P2 cos θ + P22 (cos2 θ + sin2 θ) \ R =

P12 + P22 + 2 P1P2 cos θ ...  Magnitude of resultant

...(1.1)

Now in ∆DOC,

tan α =

CD CD = OD OA + AD

\

tan α =

P2 sin θ ...  Direction of resultant P1 + P2 cos θ

...(1.2)

Particular Cases

1. If θ = 90°, i.e. when the two forces act at right angle, then P R = P12 + P22 and tan α = 2 (since cos 90° = 0 and sin 90° = 1) P1



2. If the two forces are equal, i.e. P1 = P2



R =

P12 + P12 + 2 P12cos θ =

=

2 P12 × 2cos 2

=

4 P12 cos 2

θ 2

2 P12 (1 + cos θ) 2 (since 1 + cos q = 2 cos

θ ) 2

θ 2

θ 2 Triangle Law of Forces (Graphical method) ∴

R = 2 P1 cos

It state, “If two forces acting simultaneously at a point be represented in magnitude and direction by two sides of triangle taken in order, their resultant may be represented in magnitude and direction by the third side of triangle taken in opposite order”

Co-planer System of Forces  1.9

Triangle law is the equivalent statement of parallelogram law, which is shown in figure 1.5; this figure is one-half of the parallelograms. Here the tail of P2 is placed at the tip of P1 and R (resultant) is the vector that completes the triangle, drawn from the tail of P1 to tip the tip of P2. The result is identical if the tail of P1 is placed at the tip of P2 and R is drawn from the tail of P2 to the tip of P1.

Fig. 1.5

CASE II. Resultant of more than two concurrent forces: If three or more forces are acting at a point then their resultant is found analytically by a method known as rectangular components method. In this method all the forces acting at a point are resolved (resolution of force is discussed in detail in next section) into horizontal and vertical components and then algebraic sum of horizontal components (written as SH) and vertical components (written as SV) found separately. Then the resultant R is calculated as, (ΣH )2 + (ΣV )2 ...  Magnitude of resultant

...(1.3)

Let θ be the angle of resultant R with x-axis, then ΣV ...  Direction of resultant tan θ = ΣH Polygon Law of forces (Graphical method)

...(1.4)

R=

The resultant of three or more forces acting at a point is found graphically by polygon law of forces, it state, “If a number of forces acting simultaneously at a point be represented in magnitude and direction by the sides of a polygon taken in order, their resultant may be represented in magnitude and direction by the closing side of the polygon taken in opposite order” If four forces P1, P2, P3 and P4 are acting at a point as shown in figure 1.6, then the resultant of these forces is obtained by drawing a polygon of forces as explained below and shown in figure 1.7.

1.10  Engineering Mechanics Statics



Fig. 1.6

Fig. 1.7

With suitable scale from any convenient point A, draw AB parallel and equal to P1. From B draw BC parallel and equal to P2. From C draw CD parallel and equal to P3. From D draw DE parallel and equal to P4. Draw AE from the tail of P1 to the tip of P4, then AE represents resultant of four forces P1, P2, P3 and P4. Also AC represents resultant of P1 and P2 and AD represents resultant of AC and CD. ABCDEA is the force polygon and it can be extended to any number of forces.

1.4 Resolution of a Force Any force can be resolved into individual component forces in the same way as individual component forces can be composed Resolution is the process together. It is often convenient to decompose of splitting up the given a single force into two distinct forces; they are force into components, known as components. These forces, when without changing its acting together, have the same external effect on effect on the body a body as the original force. There are mainly two methods of resolving a force. 1. Perpendicular resolution 2. Non-perpendicular resolution 1. Perpendicular Resolution: Perpendicular resolution is the general method of splitting up a single force into two mutually perpendicular directions. Two perpendicular components are acting along x-axis and y-axis or any two perpendicular axes. Table 1.2 illustrates the resolution of a force into perpendicular components. Following sign conventions are used in drawing components, (a) Components acting horizontally towards right are positive and components acting horizontally towards left are negative. (b) Components acting vertically upward are positive and components acting vertically downwards are negative.

Co-planer System of Forces  1.11 Table 1.2

1.12  Engineering Mechanics Statics

When body is placed on an inclined plane, its weight ‘mg’ can be resolved into two components i.e. parallel to incline plane and perpendicular to incline plane. Here we can select x-axis along the plane and y-axis perpendicular to plane as shown in figure 1.8.

Fig. 1.8



2. Non-Perpendicular Resolution: In non-perpendicular resolution a force is resolved into two directions which are not perpendicular to each other. If force P is to be resolved into two non-perpendicular directions let say along OA and OB as shown in figure 1.9, where OA and OB are not perpendicular to each other. Now construct a parallelogram by keeping two components (i.e. components along OA and OB) along two adjacent sides of parallelogram and original given force along the diagonal of parallelogram. Fig. 1.9 Now apply sine rule in triangle OAC, P P P OB = OA = sin α sin β sin[180 – (α + β)]

Co-planer System of Forces  1.13



\  POA =

P sin β ... This is component of force P along OA sin[180 - (α + β)]



\  POB =

P sin α ... This is component of force P along OB sin[180 - (α + β)]

1.5 Moment

The rotational effect When a force causes an object to turn, this turning producead by force is effect is called moments. So the moment of a known as moment of the force is a measure of its tendency to cause a body force to rotate around a specific point or axis. Moment is the product of the force multiplied by the perpendicular distance from the line of action of the force to the pivot or point where the object will turn. The point about which moment is taken is known as moment center and the perpendicular distance is known as moment arm, as shown in figure 1.10. If moment arm d is zero, i.e. the line of action of the force P is passing through the moment center then moment due to this force will be zero.

Fig. 1.10

Moment = Force × Distance M = F × d ...(1.5) Since moment is product of force and perpendicular distance, its S.I. unit is N-m or KN-m. The direction of the rotation resulting from a moment is either clockwise or anticlockwise. When force causes an object to turn in clockwise direction, it is called clockwise moment and when force causes an object to turn in anticlockwise direction, it is called anticlockwise moment. Anticlockwise moments are regarded as positive, while clockwise moments are negative. (We shall assume this sign convention throughout this book) Anticlockwise moment → Positive

Clockwise moment → Negative

1.14  Engineering Mechanics Statics

1.5.1 Resultant Moment When two or more than two forces are acting about a point their combined effect is represented by the “Resultant Moment”. To find out the resultant moment about a point, first find the sum of the anticlockwise moments and clockwise moments about the point, than take the lesser of these two moments from the greater and the difference is the magnitude of the resultant moment. The direction of the resultant moment will be that of the greater of the two component moments.

1.5.2  Varignon’s Theorem (Principle of Moment) The Varignon theorem is a theorem given by French mathematician Pierre Varignon in 1687. It states, “The moment of resultant of a force system about any point is equal to the algebraic sum of moments of all other forces about the same point”. Let R be the resultant of two concurrent forces P1 and P2. Let R, P1 and P2 make angles θ, α and β with axis respectively as shown in figure 1.11. The moment of R about an arbitrary point A is, R × d, where d is the perpendicular distance from line of action of R to point A. Similarly if we consider d1 and d2 are perpendicular distances from line of action of P1 and P2 to point A, then the sum of moments of P1 and P2 about point A are P1 × d1 + P2 × d2.

Fig. 1.11

Now, the sum of vertical component of P1 and P2 is equal to the vertical component of their resultant R. (As R is the resultant of P1 and P2) i.e., Ry = P1 + P2 y

y

But, from figure, Ry = R cos θ, P1 = P1 cos α and P2 = P2 cos β y

\ R cos θ = P1 cos α + P2 cos β

y

Co-planer System of Forces  1.15

Now multiply on both sides by OA R × OA cos θ = P1 × OA cos α + P2 × OA cos β But, from figure, OA cos θ = d, OA cos α = d1 and OA cos β = d2 \ R × d = P1 × d1 + P2 × d2 ...(1.6) Hence, it is proved that, the moment of resultant of a force system about any point is equal to the algebraic sum of moments of all other forces about the same point.

1.5.3 Resultant of Parallel Force System Case I: Like Parallel Forces When two parallel forces are acting in same direction they are known as like parallel forces. Their resultant can be find out using following steps, consider figure 1.12. 1. Find resultant R = P1 + P2 2. Take the algebraic sum moment of forces about point O, (clockwise moment) Moment of P1 about O = P1 × AO (anticlockwise moment) Moment of P2 about O = P2 × BO 3. Position of resultant R can be obtained by applying Varignon’s theorem about point O. – P1 × AO + P2 × BO = R × CO In above equation, CO is the perpendicular distance between line of action of resultant R and reference point O.

Fig. 1.12

Case II: Unlike Parallel Forces When two parallel forces are acting in opposite direction they are known as unlike parallel forces. Their resultant can be find out using following steps, consider figure 1.13. 1. Find resultant R = P1 – P2 2. Take the algebraic sum moment of forces about point O, (clockwise moment) Moment of P1 about O = P1 × AO (clockwise moment) Moment of P2 about O = P2 × BO

1.16  Engineering Mechanics Statics

3. Position of resultant R can be obtained by applying Varignon’s theorem about point O. –P1 × AO – P2 × BO = –R × CO In above equation, CO is the perpendicular distance between line of action of resultant R and reference point O.



Fig. 1.13

1.6 Couple A couple consists of a pair of two forces which Two equal, unlike parallel, has the following properties: non-collinear forces form • Equal magnitude and opposite in a couple direction • Act along parallel lines of action • Separated by a perpendicular distance d, known as arm of couple. A couple is shown in figure 1.14. Since the two forces cancels out each other giving zero resultant, a couple produces purely rotational effect in the body without translation motion. A couple cannot be balanced by a single force but it can be balanced only by another couple of opposite nature.

Fig. 1.14

Co-planer System of Forces  1.17

1.6.1  Moment of a Couple Consider a couple as shown in figure 1.15. The moment of couple about point A is, MA = P × d2 – P × d1 MA = P (d2 – d1) But from figure 1.15, d2 – d1 = d MA = P × d The moment of a couple is the product of magnitude of one of the forces and arm Fig. 1.15 of couple. d from force P, the moment of couple Now consider point B at distance 2 about point B is, d d + P´ 2 2 æ d d ÷ö MB = P ççç + ÷÷ è 2 2ø MB = P × d ...(1.7) From above we see that, the moment of couple about any pivot point is P × d. It is important to note here that a couple does not have moment center, like moment of force. A couple has the same moment about all points on a body. MB = P ´

1.6.2  Force–Couple System The concept of couple is useful in application of parallel transfer of a force. Refer to figure 1.16, consider a force P is acting at point A and it is required to transfer from point A to point B without changing its magnitude and direction. Now apply two forces of equal magnitude (their magnitude are same as that of force P acting at point A) and opposite in direction at point B, as shown in figure 1.17.



Fig. 1.16

Fig. 1.17

Fig. 1.18

Observing figure 1.17, we see that two forces are acting in opposite direction at point A and B form a couple. Moment of this couple is P × d, anticlockwise. Thus, to shift a force to point B, a couple is required to be added to the system.

1.18  Engineering Mechanics Statics

This system which consisting a force and a couple at point B, as shown in figure 1.18 is known as force-couple system. So we can transfer a force parallel to itself at any point with force-couple system. It is also possible to replace a force-couple system into a single force by following exactly the reverse procedure as discuss above.

1.7 Equivalent Systems of Forces Two force systems that produce the same external effects on a rigid body are said to be equivalent, i.e. the sums of the forces and sums of the moments about a point are equal. Equivalent system consists of a single resultant force R through the given point and single resultant moment. Two systems are equivalent only when, 1. The sum of the forces in system 1 is equal to the sum of the forces in system 2 2. The sum of the moments about any point O in system 1 is equal to the sum of the moments about same point O in system 2.

Solved Examples Based on Parallelogram Law of Forces Example 1. Two forces 20 N and 25 N are acting at a point with an angle of 60° between them. Find the magnitude and direction of resultant. Solution: Here P1 = 20 N, P2 = 25 N and θ = 60° As we know from parallelogram law of forces,

R =

P12 + P22 + 2 P1P2 cos θ

\ R = 202 + 252 + 2 × 20 × 25cos60 R = 39.05 N   ...Magnitude of resultant  (Ans.) Now we know that P2 sin θ tan α = P1 + P2 cos θ \

tan α =

25sin 60 20 + 25cos60

α = 33.66°   ...Direction of resultant  (Ans.) \ Example 2. Force P1 of magnitude 10 N is acting along horizontal direction and force P2 acting along vertical direction. The resultant of P1 and P2 has magnitude of 20 N. Determine the magnitude of force P2 and direction of resultant. Solution: Here P1 = 10 N, R = 20 N and θ = 90° As we know that when θ = 90° in parallelogram then, R =

P12 + P22

Co-planer System of Forces  1.19

∴

20 =

102 + P22

\ 202 = 102 + P22 \ P2 = 17.32 N  (Ans.) Now we also know that when θ = 90° in parallelogram then P tan α = 2 P1 17.32 10 \ α = 59.5°  (Ans.) Example 3. Two forces acting at point and angle between them is 120°. The larger force is of magnitude 80 N and the resultant of these two forces is perpendicular to the smaller force. Find magnitude of smaller force and resultant. Solution: Here, P1 = 80 N, θ = 120°, and angle between smaller force i.e. P2 and resultant is 90°. As we know from parallelogram law of forces, P2 sin θ tan α = P1 + P2 cos θ \

tan α =

In above equation α = 30°. P2 sin120 \ tan 30° = 80 + P2 cos120 \

0.5773 =

P2 0.8660 P 0.8660 = 2 80 + P2 (–0.50) 80 – 0.5 P2

\ 80 – 0.5P2 = 1.5P2 \ 2P2 = 80 \ P2 = 40 N  (Ans.) Now from parallelogram law of forces, R =

P12 + P22 + 2 P1P2 cos θ

802 + 402 + 2 × 80 × 40 cos120

\

R =

\

R = 69.28 N  (Ans.)

Example 4. The maximum resultant of two forces P1 and P2 is 1000 N and

minimum magnitude is 400 N. Find values of P1 and P2.

Solution: The resultant of two forces will be maximum only when the two forces will act in same direction as shown in Figure Ex. 4 (a) and resultant is minimum only when the two forces will act in opposite direction as shown in Figure Ex. 4(b).

1.20  Engineering Mechanics Statics 1000 N O



400 N

1000 N O

400 N

Fig. Ex. 4(a)

Fig. Ex. 4(b)

Now from figure Ex. 4(a), angle between two forces θ = 0 From parallelogram law of forces, R =



P12 + P22 + 2 P1P2 cos θ

Substituting θ = 0

R =

P12 + P22 + 2 P1P2

\

R =

( P1 + P2 ) 2

\ R = P1 + P2 \ P1 + P2 = 1000   ...(Maximum resultant) Now from figure Ex. 4(b), angle between two forces θ = 180° From parallelogram law of forces, R =

...(Eq. 1)

P12 + P22 + 2 P1P2 cos θ

Substituting θ = 180° R = R =

\

P12 + P22 – 2 P1P2 ( P1 – P2 ) 2

\ R = P1 – P2 \ P1 – P2 = 400   ...(Minimum resultant) ...(Eq. 2) Solving Eq. 1 and Eq. 2, we get P1 = 700 N and P2 = 300 N  (Ans.) Example 5. Two forces of 375 N and 250 N are acting on a hook and passing through point O as shown in figure Ex. 5. Determine magnitude and direction of resultant of these two forces. 375 N

250 N

45°

30° O

Fig. Ex. 5

Co-planer System of Forces  1.21

Solution: Here, P1 = 375 N, P2 = 250 N and θ = 105°. As we know from parallelogram law of forces, R =

P12 + P22 + 2 P1P2 cos θ

\ R =

3752 + 2502 + 2 × 375× 250 cos105°

\ R = 393.18 N  ...Magnitude of resultant  (Ans.) Now, as we know from parallelogram law of forces, P2 sin θ P1 + P2 cos θ



tan α =

\

250 sin105° tan α = 375 + 250 cos105°

R 250 N

375 N

105°

\ α = 37.9°  ...Direction of resultant  (Ans.) Example 6. Find the magnitude of the two forces, such that if they act at right angles, their resultant is 5 N but when they act at 60°, their resultant is 37 N. Solution: First consider two forces P1 and P2 acting at right angle, \ angle between two forces θ = 90°. As we know that when θ = 90° in parallelogram then R =

P12 + P2 2

5 =

P12 + P2 2

\ \

25 = P12 + P22

...(Eq. 1)

Similarly, when the angle between the two forces is 60°, then resultant force from parallelogram law is, R = \

37 =

P12 + P22 + 2 P1P2 cos θ

P12 + P22 + 2 P1P2 cos 60°

\ 37 = P12 + P22 + 2P1P2 × 0.5 Substituting P12 + P22 = 25 from Eq. 1 \ 37 = 25 + 2P1P2 × 0.5 \ P1P2 = 37 – 25   \ P1P2 = 12 Now we know that, (P1 + P2)2 = P12 + P22 + 2P1P2 Substituting values from Eq. 1 and Eq. 2 (P1 + P2)2 = 25 + 2 × 12 = 49 \

P1 + P2 =

49 = 7

...(Eq. 2)

...(Eq. 3)

1.22  Engineering Mechanics Statics

Similarly, we know that, (P1 – P2)2 = P12 + P22 – 2P1P2 Substituting values from Eq. 1 and Eq. 2 (P1 – P2)2 = 25 – 2 × 12 = 1 \ P1 – P2 = 1 = 1 ... (Eq. 4) Solving Eq. 3 and Eq. 4 we get, P1 = 4 N and P2 = 3 N  (Ans.) Example 7. A disabled automobile is pulled using ropes subjected to the two forces as shown in figure Ex. 7. Determine the magnitude and direction of their resultant. 8 kN

30° 25°

4 kN

Fig. Ex. 7

Solution: Here, P1= 4 kN, P2 = 8 kN and θ = 55° 8 KN

R

55°

4 KN

As we know from parallelogram law of forces, R = \

R =

P12 + P22 + 2 P1P2 cos θ

42 + 82 + 2 × 4 × 8cos55°

\ R = 10.80 kN   ...Magnitude of resultant  (Ans.) Similarly we know from parallelogram law of forces, P2 sin θ tan α = P1 + P2 cos θ

8sin 55° 8 + 4cos55° α = 32.49°  ...Direction of resultant  (Ans.)

tan α =

Co-planer System of Forces  1.23

Example 8. Two forces of 240 N and 200 N are acting at point O as shown in figure Ex. 8. Determine magnitude of resultant and direction of resultant from positive x-axis. Y 200 N

240 N

29° 21°

X

O

Fig. Ex. 8

Solution: Here, P1 = 240 N, P2 = 200 N and θ = 90° – 21° + 29° = 98°. As we know from parallelogram law of forces, R =

P12 + P22 + 2 P1P2 cos θ

\ R =

2402 + 2002 + 2 × 240 × 200 cos98°

\

R = 290.24 N  ...Magnitude of resultant  (Ans.)

200 N R 240 N 98°

Now, as we know from parallelogram law of forces, P2 sin θ tan α = P1 + P2 cos θ 200sin 98° 240 + 200cos98° \ α = 43.02° Now direction of resultant from positive x axis is 21° + 43.02° = 64.02°  ...Direction of resultant  (Ans.) Example 9. Two forces of 600 N and 700 N are acting at point O as shown in figure Ex. 9. Determine magnitude of resultant and direction of resultant from x-axis. \

tan α =

1.24  Engineering Mechanics Statics 600 N

40°

60°

700 N

Fig. Ex. 9

Solution: Here, P1 = 700 N, P2 = 600 N and θ = 100°. As we know from parallelogram law of forces,

R =

700 N

P12 + P22 + 2 P1P2 cos θ

7002 + 6002 + 2 × 700 × 600 cos100° \ R = 839.12 N ...Magnitude of resultant  (Ans.) Now, as we know from parallelogram law of forces, P2 sin θ tan α = P1 + P2 cos θ \ R =

\

tan α =

600 sin 100° 700 + 600 cos100°

R 100°

600 N

\ α = 44.76° Now direction of resultant from x axis is 60° – 44.76° = 15.24°  (Ans.) Example 10. Two forces of 600 N and 500 N are acting at point O as shown in figure Ex. 10. Determine magnitude of resultant and direction of resultant from x axis.

Fig. Ex. 10

Co-planer System of Forces  1.25

Solution: Here, P1 = 600 N, P2 = 500 N and θ = tan –1 As we know from parallelogram law of forces,

4 2 – tan –1 = 54.16° 1 5

R =

P12 + P22 + 2 P1P2 cos θ

\ R =

6002 + 5002 + 2 × 600 × 500 cos 54.16°

\ R = 980.47 N   ...Magnitude of resultant  (Ans.) Now, as we know from parallelogram law of forces, P2 sin θ tan α = P1 + P2 cos θ \

tan α =

500 N 500 sin 54.16° 600 + 500 cos 54.16°

\ α = 24.41° Now direction of resultant from x-axis is –1 2 = 46.22°  (Ans.) 24.41° + tan 5

R 600 N

Example 11. Two forces of 82.5 kN and 74 kN are acting on a bracket as shown in figure Ex. 11. Determine magnitude of resultant and direction of resultant from x-axis.

Fig. Ex. 11

Solution: Here, P1 = 82.5 kN, P2 = 74 kN and θ = 180° – tan –1

65 70 – tan –1 = 121.30° . 120 120

1.26  Engineering Mechanics Statics

As we know from parallelogram law of forces, R = \

R =

P12 + P22 + 2 P1P2 cos θ

82.52 + 742 + 2 × 82.5× 74 cos121.30°

R = 77.06 kN  ...Magnitude of resultant  (Ans.) \ Now, as we know from parallelogram law of forces, 82.5 N P2 sin θ tan α = P1 + P2 cos θ \

tan α =

74 sin 121.30° 82.5 + 74 cos121.30°

R

\ α = 55.13° Now direction of resultant from x-axis is 65 55.13° + tan –1 = 83.6°  (Ans.) 120

74 N

Solved Examples Based on Resolution of Forces Example 1. Resolve 100 N force into components along (a) x and y directions and (b) u and v directions. Refer figure Ex. 1 Y u Solution: (a) Components along x and y: Component along x = 100 cos 40° = 76.60 N  (Ans.) 100 N Component along y = 100 sin 40° = 64.28 N  (Ans.) 30° (b) Component along u and v: 40° X As angle between u and v is 100°, so this is non30° perpendicular resolution. Construct parallelogram by considering 100 N force v as a resultant and its component along u and v as shown Fig. Ex. 1 in figure Ex. 1(a) B

C

Pu 30° 30° O

100N

70° 80° Pv A

Fig. Ex. 1(a)

Co-planer System of Forces  1.27

Now in triangle OAC, apply sine rule 100 Pu Pv = = sin80° sin 70° sin 30° Pu =

100 sin 70° = 95.42 N   (Ans) sin 80°

Pv =

100 sin 30° = 50.77 N   (Ans) sin 80°

Example 2. If the force P as shown in figure Ex. 2 is resolved into components

parallel to the bars AB and BC, the magnitude of the component parallel to bar BC is 4 kN. What is the magnitudes of P and its component along AB? P

B 1.5m A



1m



C

4m

Fig. Ex. 2

Solution: Here taking a force P as a resultant, construct a parallelogram taking two side of parallelogram as two component of force P along AB and BC as shown in figure Ex. 2(a). Now, from figure Ex. 2 1.5 \ α = 56.31° tan α = 1.0

tan β =

From figure Ex. 2(a)

1.5 4.0

\ β = 20.56° Fig. Ex. 2(a)

θ = 90 – α \ θ = 90 – 56.31° = 33.69° Φ = 90 – β \ Φ = 90 – 20.56° = 69.54° Now apply sine rule to figure Ex. 2(a) P P P = AB = BC sin (180 – θ – Φ ) sin Φ sin θ

1.28  Engineering Mechanics Statics

P 4 = sin (180 – 33.69 - 69.54) sin 33.69 P =

\ \

4sin 76.87 sin 33.69

\ P = 7 kN  (Ans)

PAB 4 = sin 69.54 sin 33.69 PAB =

4sin 69.54 sin 33.69

\ PAB = 6.75 kN  (Ans)

Example 3. A force P is inclined at 60° to the horizontal. If the horizontal

component of force is 50 N find out vertical component of force. Solution: Here the force P is inclined to the horizontal at an angle θ = 60°. Horizontal component of force P = P cos θ 50 = 100 N P cos 60° = 50  \ P = cos60° Now vertical component of P = P sin θ = 100 sin 60° = 86.60 N  (Ans.) Example 4. A box is to be moved in the direction of OO’ as shown in figure Ex.4. It is identified that a 20 kN force in this direction is required but obstruction prevent direct application of such a force. Accordingly, the forces P1 and P2 are applied as shown. Determine the magnitude of P1 and P2. Solution: Here it is desired that 20 kN force in the direction of OO’ is the resultant of P1 and P2. Now the parallelogram may be constructed taking a 20 kN force as a resultant and by drawing lines parallel to the specified direction of P1 and P2 as shown in figure Ex. 4(a). Now in triangle OAC, apply sine rule Fig. Ex. 4 Figure Ex. 4(a) P1 20 P1 P2 20 kN = = sin 45° sin 60° sin 75° C \ P1 =

20sin 60° = 24.5 kN  (Ans) sin 45°

20 sin 75° = 27.32 kN   (Ans) \ P2 = sin 45°

75° 60° 45° A

P2

O

Fig. Ex. 4(a)

Example 5. Crank ABC is subjected to force P as shown in figure Ex. 5.

Knowing that P must have a 200 N component perpendicular to arm AB of the crank, determine the magnitude of force P and its component along AB.

Co-planer System of Forces  1.29 P 35° 40°

B

A

C

Fig. Ex. 5

Solution: Here component of force P along AB PAB and perpendicular to AB are as shown in figure Ex. 5(a). 75° Now we know that component of force P perpendicular to AB is 200 N 20 \ P sin 75° = 200 N 0N P 200 = 207 N   (Ans) \ P = sin 75° Fig. Ex. 5(a) Now component of P along AB = P cos 75° = 207 cos 75° = 53.58 N (Ans.) Example 6. A vertical force of 50 N acts downward at A as shown in figure Ex. 6. Determine the magnitudes of the two components of 50 N force along AB and AC.

Fig. Ex. 6

Solution: Here taking a 50 N force as a resultant, construct a parallelogram taking two adjacent sides of parallelogram as two component of force P along AB and AC as shown in figure Ex. 6(a). Now in triangle OAC, apply sine rule 50 P PAB = AC = sin 75° sin 60° sin 45°

1.30  Engineering Mechanics Statics

\ PAB = \ PAC =

50 sin 60° = 44.82 N   (Ans) sin 75°

O 45°

30° PAC

45° PAB

50 sin 45° = 36.6 N   (Ans) sin 75°

50N 75°

A

60°

Example 7. Resolve 800 N force into components along u and v directions as shown in figure Ex. 7.

C

Fig. Ex. 6(a)

Fig. Ex. 7

Solution: Construct parallelogram by considering 800 N force as a resultant and Pu its component along u and v as shown in figure Ex. 7(a). Now in triangle OAC, apply sine rule Pu Pv 800 O = = C sin 75° sin 30° sin 75° 75° 800N \

800 sin 75° = 800 N   (Ans) Pu = sin 75°

\

Pv =

800 sin 30° = 414.11 N   (Ans) sin 75°

30°

Pv

75° A

Fig. Ex. 7(a)

Example 8. Two forces P1 and P2 are applied to a hook as shown in figure Ex. 8. The resultant of the two forces has a magnitude of 300 N and makes an angle of 30° with positive x-axis (anticlockwise direction). Determine the magnitudes of P1 and P2. y

P2 45°

P1 30°

x

Fig. Ex. 8

Co-planer System of Forces  1.31

Solution: Here taking a 300 N force as a resultant, construct a parallelogram taking P1 and P2 as two adjacent sides of parallelogram as shown in figure Ex. 8(a). Now in triangle OAC, apply sine rule 300 P1 P2 = = sin 75° sin 75° sin 30° \ P1 = \

P2 =

300 sin 75° = 300 N   (Ans) sin 75° 300 sin 30° = 155.3 N   (Ans) sin 75°

Fig. Ex. 8(a)

Example 9. A bracket is subjected to two forces P1 and P2 as shown in figure Ex. 9. If the resultant of the two forces has a magnitude of 525 N and direction as shown in the figure, determine magnitudes of P1 and P2. P1

525N 2

2 3

1

4

1

P2

O

Fig. Ex. 9

Solution: Here first consider angle of P2, P1 and resultant force are respectively θ1, θ2, and θ3 from x-axis. 1 2 2 Now θ1 = tan –1 = 14.03° θ2 = tan –1 = 63.43° θ3 = tan –1 = 33.70° 4 1 3 49.4° C 47.73° 525 N

A P1

82.87° O

P2

Fig. Ex. 9(a)

Now taking a 525 N force as a resultant, construct a parallelogram taking P1 and P2 as two adjacent sides of parallelogram as shown in figure Ex.9 (a).

1.32  Engineering Mechanics Statics

Now in triangle OAC, apply sine rule

P1 P2 525 = = sin 49.4° sin 47.73° sin 82.87°

\

P2 =

525 sin 82.87° = 686.11 N sin 49.4°

\

P1 =

525 sin 47.73° = 511.66 N sin 49.4°

Example 10. A frame is subjected to a horizontal force of 200 N as shown in figure Ex. 10. If the component of force along AC is 250 N, directed from A towards C, determine the component of force along member AB and angle θ (00 ≤ θ ≤ 900) of member AB.

Fig. Ex. 10

Solution: Here taking a force 200 N as a resultant, construct a parallelogram taking two adjacent sides of parallelogram as two component of force along AC and AB as shown in Figure Ex. 10(a) PAB B Now in triangle OAC, apply sine rule \ \ \

250 PAB 200 = = sin θ sin (140 – θ)° sin 40° sin θ =

250 sin 40° = 0.8034 200

A



40° O

PAC C

200N

(140–)

Fig. Ex. 10(a)

θ = 53.46°  (Ans.) PAB =

250 sin (140 - 53.46)° = 310.6 N   (Ans.) sin 53.46°

Example 11. A wooden log is to be hoisted using two chains as shown in figure Ex. 11. If the resultant force is to be 1200 N in vertical upward direction determine the magnitudes of P1 and P2 acting on each chain and angle θ of P2 so that magnitude of P2 is to be minimum.

Co-planer System of Forces  1.33 P2

P1 

30°

Fig. Ex. 11

Solution: Here taking a force 1200 N as a resultant, construct a parallelogram taking two adjacent sides of parallelogram as two component of force along AC and AB as shown in figure Ex.11 (a). 1200N Now in triangle OAC, apply sine rule C 1200 P P2 = 1 =  (150–) sin (150 – θ)° sin θ sin 30° A \

P1

1200 sin 30° P2 = sin (150 – θ)°

P2

Now, for P2 to be minimum the denominator sin (150 – θ) must be maximum. \ (150 – θ) = 90°  \ θ = 60° 1200 sin 30° = 600 N   (Ans.) \ P2 = sin (150 – 60)° Now

P1 =

30°

O

Fig. Ex. 11(a)

P2 sin 60° 600 sin 60° = = 1039.23 N   (Ans.) sin 30° sin 30°

Example 12. The line of action of the 1000 N force is passing through the point A (–6, –2) and B (7,5) as shown in Figure Ex. 12. Determine the x and y components of 1000 N force. Y 1000N B(7,5)

X A(–6,–2)

Fig. Ex. 12 –1 Solution: Here angle made by 1000 N force with x axis is θ = tan

7 \ θ = 28.3° 13

Now x component of 1000 N force = 1000 cos 28.3° = 880.48 N  (Ans.) y component of 1000 N force = 1000 sin 28.3° = 474 N  (Ans.)

1.34  Engineering Mechanics Statics

Example 13. Resolve 500 N force acting on bar into components along OA and perpendicular to OA. Refer figure Ex. 13. A 20° 500N 30° O

Fig. Ex. 13

Solution: The component of 500 N force along OA and perpendicular to OA are, Component along OA = 500 cos 50° = 321.4 N  (Ans.) Component perpendicular to OA = 500 sin 50° = 383 N  (Ans.) Example 14. A frame is subjected to two forces as shown in figure Ex. 14 .

Replace them by equivalent forces along x and a direction.

400N 105° R

   450N Fig. Ex. 14       

  Fig. Ex. 14(a)

Solution: Here first find out resultant of given two forces using parallelogram law. Refer figure Ex. 14 (a) Now take P1 = 450 N, P2 = 400 N and angle θ between P1 and P2 = 105° As we know from parallelogram law of forces,

R =

P12 + P22 + 2 P1P2 cos θ

\

R =

4502 + 4002 + 2 × 450 × 400 cos105°

\ R = 519 N Now, as we know from parallelogram law of forces, P2 sin θ tan α = P1 + P2 cos θ \

tan α =

400 sin105° 450 + 400 cos105°

\ α = 48.11°

Co-planer System of Forces  1.35

Now taking a 519 N force as a resultant, construct a parallelogram taking two adjacent sides of parallelogram as two component of force along x and a direction as shown in figure Ex. 14(b). A

PX

O

45°113.1°

21.9° PA C 519N

Fig. Ex.14 (b)

Now in triangle OAC, apply sine rule Pa Px 519 = = sin113.1° sin 21.9° sin 45° \

Pa =

519 sin113.1° = 675.13 N   (Ans.) sin 45°

\

Px =

519 sin 21.9° = 273.76 N   (Ans.) sin 45°

Solved Examples Based on Co-planer Concurrent Forces Example 1. Determine resultant of the force system shown in the figure Ex. 1

Fig. Ex. 1

Solution: Resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components. +

→ΣH = 5 cos 30° + 10 cos 60° + 12 cos 40° = 18.52 kN

1.36  Engineering Mechanics Statics

+ ↑ ΣV = 5 sin 30° + 10 sin 60° –12 sin 40° = 3.45 kN Now we know that resultant R is calculated as,

R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(18.52) 2 + (3.45) 2

\ R = 18.84 kN  (Ans.) Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH tan θ =

\

3.45 18.52

θ = 10.55°  (Ans.)

Example 2. A hook is subjected to three forces as shown in the figure Ex. 2. Determine the magnitude of the resultant force and its direction from x-axis. 200N 2 500N

300N

1 1

1 2

2

Fig. Ex. 2

Solution: Consider θ1, θ2 and θ3 are the angles of forces 300 N, 200 N and 500 N respectively. –1 1 = 26.57° Now, θ1 = tan 2

–1 2 = 63.43° θ2 = tan 1



–1 1 = 26.57° θ3 = tan 2

Now resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components.

+

→ΣH = 300 cos 26.57° – 200 cos 63.43° – 500 cos 26.57°

Co-planer System of Forces  1.37

SH = – 268.33 N (here ‘–’ sign indicate that SH is acting toward negative x direction) +↑SV = 300 sin 26.57° + 200 sin 63.43° + 500 sin 26.57° \ SV = 536.71 N Now we know that resultant R is calculated as,

\



R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(268.33) 2 + (536.71) 2

R = 600 N \ Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH 536.71 \ tan θ = 268.33

\ q = 63.43°  (Ans.)

Example 3. Determine resultant of the force system shown in the figure Ex. 3.

Fig. Ex. 3

Solution: Resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components. +

→ΣH = –300 sin 20° + 350 cos 20° + 250 cos 30° – 600 sin 30° = 142.79 N +↑SV = –300 cos 20° + 350 sin 20° – 250 sin 30° + 600 cos 30° = 232.41 N Now we know that resultant R is calculated as,

R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(142.79) 2 + (232.41)2

\

R = 272.77 N  (Ans.)

1.38  Engineering Mechanics Statics

Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH tan θ =

\

232.41 142.79

\θ = 58.43°  (Ans.)

Example 4. The following forces act at a point: 30 N towards East, 20 N towards North, 35 N towards North West, 25 N inclines at 40° towards South of West, 40 N towards South. Determine the magnitude of the resultant force and its direction from East.

Fig. Ex. 4

Solution: The system of given forces is shown in figure Ex. 4. Resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components: +

→ΣH = 30 – 35 cos 45° – 25 cos 40° = –13.90 N (here ‘–’ sign indicate that SH is acting toward negative x direction i.e. towards West) +↑SV = 20 + 35 sin 45° – 25 sin 40° – 40 = –11.32 N (here ‘–’ sign indicate that SV is acting toward negative y direction i.e. towards South) Now we know that resultant R is calculated as,

R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(13.90) 2 + (11.32) 2

\ R = 17.92 N  (Ans.) Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH

Co-planer System of Forces  1.39

\

tan θ =

11.32 13.90

\ θ = 39.15°

Now angle of resultant with East = 180° + 39.15° = 219.15°  (Ans.) Example 5. Three forces act as shown in figure Ex. 5. Determine the magnitude

of the force P if the resultant is along x-axis. Also determine magnitude of resultant. Y 75N 45N 25° 25° X

O

35°

P

Fig. Ex. 5

Solution: As resultant is acting along x-axis, its vertical component become zero i.e. SV = 0. \ Resolving the forces along y-axes, and then equating sum of vertical components to zero, \ 75 sin 50° + 45 sin 25° – P sin 35° = 0 \ P = 133.32 N  (Ans.) Now as resultant is acting along x-axis, its magnitude is equal to SH R = SH = 75 cos 50° + 45 cos 25° + 133.32 cos 35° \ R = 198.20 N  (Ans.) Example 6. Three forces are acting on a bracket as shown in figure Ex. 6. Determine the magnitude and direction of resultant force. 200N

150N 3 4

30° 200mm

400mm

100N

Fig. Ex. 6

Solution: Consider θ1, and θ2 are the angles of forces 100 N, and 200 N respectively from horizontal.

1.40  Engineering Mechanics Statics

400 = 63.43° 200

Now

–1 θ1 = tan



–1 3 = 36.86° θ2 = tan 4

Resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components: +

→ΣH = 100 cos 63.46° + 150 cos 30° – 200 cos 36.86° = 14.56 N +↑SV = –100 sin 63.43° + 150 sin 30° + 200 sin 36.86° = 105.53 N Now we know that resultant R is calculated as,

R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(14.56) 2 + (105.53) 2

\ R = 106.52 N  (Ans.) Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH tan θ =

\

105.53   \ q = 82.14°  (Ans.) 14.56

Example 7. Four concurrent forces acts on the centre of mass during landing of airplane as shown in figure Ex. 7. Determine the magnitude of the resultant force and its direction from x-axis 2kN

8.9kN 45°

4.5 kN

30° 8.9 kN

Fig. Ex. 7

Solution: Resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components.

+

→ΣH = 2 sin 45° – 4.5 – 8.9 sin 30°

Co-planer System of Forces  1.41

= –7.53 kN (here ‘–’ sign indicate that SH is acting toward negative x direction) +↑SV = –2 cos 45° – 8.9 + 8.9 cos 30° = –2.60 kN (here ‘–’ sign indicate that SV is acting toward negative y direction) Now we know that resultant R is calculated as,

R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(7.53) 2 + (2.60) 2

\ R = 7.97 kN  (Ans.) Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH tan θ =

\

2.60 7.53

\ θ = 19°  (Ans.)

Example 8. Bracket AB is supported by cable BC as shown in figure Ex. 8. If the tension in cable BC is 150 N, determine magnitude of the resultant of the three forces and direction of resultant from x-axis. C T = 150N

45°

B

A 40° 40° 100N

125N

Fig. Ex. 8

Solution: Resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components: +

→ΣH = –150 cos 45° – 100 sin 40° + 125 cos 40° = –74.58 N (here ‘–’ sign indicate that SH is acting toward negative x direction) +↑SV = 150 sin 45° – 100 cos 40° – 125 sin 40° = –50.88 N (here ‘–’ sign indicate that SV is acting toward negative y direction) Now we know that resultant R is calculated as,

R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(74.58) 2 + (50.88) 2

1.42  Engineering Mechanics Statics

\ R = 90.28 N Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH \

tan θ =

50.88 \ 74.58

(Ans.)

θ = 34.30°  (Ans.)

Example 9. Three forces are acting on a hook as shown in figure Ex. 9. If the magnitude of the resultant force is 400 N and acting along positive y-axis, determine the magnitude of force P and the angle α.

Fig. Ex. 9

Solution: Consider θ is the angle of 200 N force with x axis, –1 3 = 36.86° Now θ = tan 4 As resultant is acting along y-axis, its horizontal component become zero i.e. SH = 0. \ Resolving the forces along x-axes, and then equating sum of horizontal components to zero, \ – P sin α + 250 cos 30° + 200 cos 36.86° = 0 P sin α = 376.52 (Eq. 1) \ Now resultant is acting along y axis, its magnitude is equal to SV \ 400 = SV \ 400 = P cos α + 250 sin 30° – 200 sin 36.86° (Eq. 2) P cos α = 394.98 Solving Eq. 1 and Eq. 2, \ a = 43.62°  (Ans.) Now from (Eq. 1), P sin 43.63° = 376.52 P = 545.68 N  (Ans.) Example 10. Three forces are acting on a bracket as shown in figure Ex. 10. If the magnitude of the resultant force is to be 250 N directed along positive u axis, determine force P and angle α.

Co-planer System of Forces  1.43 y

P u

 30° 65°

x 100N

225N

Fig. Ex. 10

Solution: Here find SH and equate with horizontal component of the resultant force, +

→ΣH = 250 cos 30° \ \ P sin α + 100 + 225 cos 65° = 250 cos 30° \ P sin α = 21.41 (Eq.1) Now find SV and equate with vertical component of the resultant force, \ +↑SV = 250 sin 30° \ P cos α – 225 sin 65° = 250 sin 30° \ P cos α = 328.91 (Eq.2) Solving Eq. 1 and Eq. 2, α = 3.72°  (Ans.) \ Now from (Eq. 1), P sin 3.72° = 21.41 P = 330 N  (Ans.)

Solved Examples Based on Moment & Non-Concurrent Forces Example 1. Determine the moment of the 100 N force about point O as shown in the figure Ex. 1. 100N 1.5m

0

Fig. Ex. 1

Solution: Since the perpendicular distance d from the force to the point O is 1.5 m, we use MO = P × d Treating anticlockwise moment as positive MO = –100 × 1.5 (Negative sign is used from our assumed sign convention of moment, as 100 N force tends to rotate the object in clockwise direction)

1.44  Engineering Mechanics Statics

\ MO = – 150 N.m MO = 150 N.m (Clockwise)  (Ans.) Example 2. Determine the moment of the 100 N force about point O as shown in the figure Ex. 2 100N 1.5m 60° 0

Fig. Ex. 2

Solution: Here first determine perpendicular distance from line of action of 100 N force to point O as shown in figure Ex. 2(a). MO = P × d \ MO = –100 × 1.5 sin 60° 100N 1.5m 60° O 60° d

Fig. Ex. 2(a)

MO = –129.9 N.m = 129.9 N.m (clockwise) Alternate Approach: Many times it is convenient to resolve the given force into its components and then take algebraic sum of moment of these components about given point. In above example given 100 N force can be resolve into horizontal and vertical components as shown in figure Ex. 2(b).

Fig. Ex. 2(b)

Treating anticlockwise moment as positive, sum of moment due to these components about point O is find out as below. (Note that, the line of action of horizontal component is passing through point O, so it produces zero moment) \ MO = –100 sin 60° × 1.5 MO = –129.9 N.m = 129.9 N.m (clockwise)  (Ans.)

Co-planer System of Forces  1.45

Example 3. Determine the moment of the 100 N force about point O as shown in the figure Ex. 3. 200 N

5 3 4

1m O 3m

Fig. Ex. 3 200 N

200 sin36.86° 5 3 4

200 cos 36.86° 1m O 3m

Fig. Ex. 3(a)

Solution: Here resolved 200 N force into horizontal and vertical components as shown in figure Ex. 3(a). Treating anticlockwise moment as positive, sum of moment due to these components about point O is, MO = –200 cos 36.86° × 1 – 200 sin 36.86° × 3 \ MO = –520 N.m = 520 N.m (clockwise)  (Ans.) Example 4. Three forces are acting in a plane as shown in figure Ex. 4. Determine moment of each forces about point A and B. B

40 50N

70N

30 A 20

100N

10

0

10

20

30

40

Fig. Ex. 4

Solution: Treating anticlockwise moment as positive, moment of 50 N force about point A and B are,

1.46  Engineering Mechanics Statics

MA= 50 × 10 = 500 N.mm (anticlockwise)  (Ans.) MB= –50 × 30 = –1500 N.mm = 1500 N.mm (clockwise)  (Ans.) Moment of 70 N force about point A and B are, MA = –70 × 10 = –700 N.mm = 700 N.mm (clockwise)  (Ans.) MB = 70 × 10 = 700 N.mm (anticlockwise)  (Ans.) Now resolved 100 N force into horizontal and vertical components and then find out moment due to these components about point A and B as below, Horizontal component of 100 N force = 100 cos 45° = 70.71 N Vertical component of 100 N force = 100 sin 45° = 70.71 N MA = (70.71 × 20) + (70.71 × 20) = 2828.4 N.mm (anticlockwise)  (Ans.) MB = (70.71 × 40) – (70.71 × 20) = 1414.2 N.mm (anticlockwise)  (Ans.) Example 5. A bracket is subjected to two forces as shown in figure Ex. 5. Determine (a) moment of 160 N force about point A, (b) moment of 210 N force about point A and B.

Fig. Ex. 5

Solution: Treating anticlockwise moment as positive, moment of 160 N force about point A is, MA = –160 × 0.7 = –112 N.m = 112 N.m (clockwise)  (Ans.) Moment of 210 N force about point A is, MA = –210 × 0.6 = –126 N.m = 126 N.m (clockwise)  (Ans.) MB = 210 × 0.4 = 84 N.m (anticlockwise)  (Ans.) Example 6. Force P is acting in a plane as shown in figure Ex. 6. If the magnitude of the moment due to the force P about O is 300 N-m, determine magnitude of force P. Y P

50° (5,1) X

O

Fig. Ex. 6

Co-planer System of Forces  1.47

Solution: Here first find out horizontal and vertical components of force P as shown in figure Ex. 6(a). Y P

P sin 50°

50° (5,1) P cos 50° X

O

Fig. Ex. 6(a)

Treating anticlockwise moment as positive, sum of moment due to these components about point O is, MO = (P sin 50° × 5) + (P cos 50° × 1) Now it is given that moment due to the force P about O is 300 N-m \ 300 = (P sin 50° × 5) + (P cos 50° × 1) \ 300 = P [(0.7660 × 5) + ( 0.6427 × 1)] \ P = 67 N  (Ans.) Example 7. Two equal forces of 150 N are applied as shown in figure Ex. 7. If the length of the bar AB is 500 mm and radius of pulley is 100 mm, determine sum of the moments of the forces (a) about point A and (b) about point B.

Fig. Ex. 7

Solution: Treating anticlockwise moment as positive, sum of moment due to these forces about point A is, SMA = (150 × 100) – (150 × 100) \ MA = 0  (Ans.) Now sum of moment about B is SMB = (150 × 100) – (150 cos 35°) (500 sin 45° + 100 cos 35°) – (150 sin 35°) (500 cos 45° + 100 sin 35°) \ MB = – 73860.57 N.mm = –73.86 N.m = 73.86 N.m (clockwise)  (Ans.)

1.48  Engineering Mechanics Statics

Example 8. A 200 N force is acting on a bracket as shown in figure Ex. 8. Determine the value of θ for which the moment about O is (a) zero and (b) a maximum. 3m A  5m 2000N

O

Fig. Ex. 8

Solution: For moment about O to be zero, the line of action of the given force must pass through point O as shown in figure Ex. 8 (a). –1 5 = 59°   (Ans.) \ θ = tan 3 Now for moment about O to be a maximum, the line of action of the given force must perpendicular to distance OA as shown in figure Ex. 8 (b).  –1 5  \ q =  tan  + (90°) = 149°   (Ans.) 3 

Fig. Ex. 8(a)

Example 9. Two forces acting on a bracket as shown in figure Ex. 9. Determine the sum of moment of these two forces about point O. 400N

600N

Fig. Ex. 8(b)

60° 1m

1m 2m

O

Fig. Ex. 9

Solution: Treating anticlockwise moment as positive, sum of moment due to these forces about point O is,

Co-planer System of Forces  1.49

\

SMO = – (600 × 1) + (400 sin 60° × 1) + (400 cos 60° × 2) MO = 146.41 (anticlockwise)  (Ans.)

Example 10. Cable AB and AC are used to support a pole as shown in figure Ex. 10. If the tension in cable AC is 250 N and the sum of the moments about O by the two cables is zero, determine the tension in cable AB. A

9m

5m B

4m O

C

Fig. Ex. 10

Solution: Let θ1 and θ2 are the angles of cable AB and AC with horizontal respectively. –1 9 –1 9 = 66° = 61° and θ2 = tan θ1 = tan 4 5 Now find out components of the cable tensions at point A. As sum of the moments about point O is zero, SMO = 0 Treating anticlockwise moment as positive and T be the tension in cable AB, T cos 61° × 9 – 250 cos 66° × 9 = 0 \ T = 209.74 N (Ans.) Example 11. Square ABCD is subjected to four forces and clockwise moment as shown in figure Ex. 11. Determine (a) magnitude and direction of the resultant force and (b) position of the resultant force from point A.

Fig. Ex. 11

1.50  Engineering Mechanics Statics

Solution: Here first resolve the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components, +

→ΣH = 10 – 30 = – 20 N (here ‘–’ sign indicate that SH is acting toward negative x Direction) +↑SV = 20 – 40 = – 20 N (here ‘–’ sign indicate that SV is acting toward negative y Direction) Now we know that resultant R is calculated as,

R =

( ΣH ) 2 + ( ΣV ) 2

\

R =

(20) 2 + (20) 2

R = 28.28 N (Ans.) \ Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH \

tan θ =

20 20

\ θ = 45°  (Ans.)

Now position of resultant force is determine by applying Varignon’s theorem about point A. Let x be the perpendicular distance between line of action of the resultant force and point A. \ Rx = SMA, treating anticlockwise moment as positive \ 28.28 x = 30 × 0.05 + 20 × 0.05 – 1.5 (note that the moment of 10 N and 40 N forces about point A is zero, as the line of action of these forces are passing through point A) D

C

A

B

45°

mm

.36

35 28.28N

Fig. Ex. 11(a)

\

x = 0.0353 m = 35.36 mm from point A , refer figure Ex. 11(a)  (Ans.)

Example 12. A rectangular block ABCD is subjected to four forces as shown in figure Ex. 12. Determine (a) magnitude and direction of the resultant force and (b) position of the resultant force from point A.

Co-planer System of Forces  1.51

Fig. Ex. 12

Solution: Here first resolve the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components, +

→ΣH = 100 – 400 – 100 cos 30° + 200 cos 30° = – 213.39 N (here ‘–’ sign indicate that SH is acting toward negative x Direction) +↑SV = –200 sin 30° – 100 sin 30° = –150 N (here ‘–’ sign indicate that SV is acting toward negative y Direction) Now we know that resultant R is calculated as,





R =

(ΣH )2 + (ΣV )2

\

R =

(213.39) 2 + (150) 2

\ R = 260.83 N (Ans.) Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH \

tan θ =

150 213.39

\ θ = 35.10°  (Ans.)

Now position of resultant force is determine by applying Varignon’s theorem about point A. Let x be the perpendicular distance between line of action of the resultant force and point A. D

C 35.1°

260.83N

m

0m

15 A

Fig. Ex. 12(a)

B

1.52  Engineering Mechanics Statics

\ Rx = SMA, treating anticlockwise moment as positive \ 260.83 x = (400 × 200) – (100 × 300) – (100 sin 30° × 100) – (200 cos 30° × 400) – (200 sin 30° × 150) \ x = 0.150 m = 150 mm from point A, refer figure Ex. 12(a)  (Ans.) Example 13. Three forces are acting on member AB as shown in figure Ex. 13. Determine (a) magnitude and direction of the resultant force, (b) point where line of action of resultant intersect member AB. Solution: Here first resolve the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components, +

→ΣH = – 25 cos 45° + 20 sin 25° = –9.22 N (here ‘–’ sign indicate that SH is acting toward negative x Direction) +↑SV = – 30 – 25 sin 45° – 20 cos 25° = – 65.80 N (here ‘–’ sign indicate that SV is acting toward negative y Direction) 30N 20N25° Now we know that resultant R is calculated as,



\

R = R =

( ΣH ) 2 + ( Σ V ) 2 2

(9.22) + (65.80)

25N 45° 2

\ R = 66.44 N  (Ans.) Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH \

65.80 tan θ = 9.22

A

B 1m 1m

2m

Fig. Ex. 13 82° A

B

2.07m

C

\ θ = 82°  (Ans.)

Now position of resultant force is determine by 66.44N applying Varignon’s theorem about point A. Let x be Fig. Ex. 13(a) the perpendicular distance between line of action of the resultant force and point A. \ R x = SMA, treating anticlockwise moment as positive \ 66.44 x = (–30 × 1) – (25 sin 45° × 2) – (20 cos 25° × 4) \ x = 2.07 m from point A, refer figure Ex.13(a) Now point where line of action of resultant intersect member AB from A as shown in figure Ex.13(a) is determine by again applying Varignon’s theorem as below, SV × horizontal distance AC = SMA 65.80 × distance AC = (–30 × 1) – (25 sin 45° × 2) – (20 cos 25° × 4) \ distance AC = 2.09 m  (Ans.)

Co-planer System of Forces  1.53

Example 14. A bracket is subjected to four forces as shown in figure Ex. 14.

Determine (a) magnitude and direction of the resultant force and (b) position of the resultant force from point O. 0.5 kN 2 kN 3 4

1.5 kN

1m O 1m

1m

1m 1.5m

45° 1kN

Fig. Ex. 14

Solution: Let θ be the angle of 2 kN force with horizontal. –1 3 = 36.86° θ = tan 4 Now resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components, +

→ΣH = 2 cos 36.86° – 1 sin 45° = 0.8931 kN +↑SV = 1.5 + 0.5 + 2 sin 36.86° – 1 cos 45° = 2.49 kN Now we know that resultant R is calculated as,

R =

(ΣH )2 + (ΣV )2

\

R =

(0.8931)2 + (2.49)2

R = 2.65 kN  (Ans.) \ Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH \

tan θ =

2.49 0.8931

\ q = 70.26°  (Ans.)

Now position of resultant force is determine by applying Varignon’s theorem about point O. Let x be the perpendicular distance between line of action of the resultant force and point O. \ R x = SMA, treating anticlockwise moment as positive

1.54  Engineering Mechanics Statics



\ 2.65 x = (1.5 × 1) + (0.5 × 2) + (2 sin 36.86° × 2) – (2 cos 36.86° × 1) – (1 cos 45° × 3) – (1 sin 45° × 1.5) \ x = 0.0442 m from point O  (Ans.)

Example 15. A bracket is subjected to four forces and a moment as shown in figure Ex. 15. Determine (a) magnitude and direction of the resultant force, (b) point where line of action of resultant intersect member OA measured from O.

Fig. Ex. 15

Solution: Resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components, +

→ΣH = 300 N +↑SV = –200 – 100 – 150 = –450 N (here ‘–’ sign indicate that is acting toward negative y direction) Now we know that resultant R is calculated as,

R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(300) 2 + (450) 2

\ R = 541 N  (Ans.) Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH

Fig. Ex. 15(a)

450 \ θ = 56.30°  (Ans.) 300 Now let x be the distance where line of action of resultant intersect member OA from O as shown in figure Ex.15(a) is determine by applying Varignon’s theorem as below, SH x = ∑ MO treating anticlockwise moment as positive 300 x = – (300 × 2) – (100 × 2) – (150 × 4) + 200 \ x = 4 m from O  (Ans.) \

tan θ =

Co-planer System of Forces  1.55

Solved Examples Based on Parallel Forces Example 1. Determine the magnitude of the resultant of parallel force system

shown in figure Ex. 1. Also determine the position of the resultant force from point O. 30N

60N

20N

40N

O 2m

3m

2m

4m

Fig. Ex. 1

Solution: Here magnitude of resultant is algebraic sum of all parallel forces. R = +↑SV = – 30 – 60 + 20 – 40 \ R = –110 N = 110 N downward  (Ans.) Now position of resultant is finding out by applying Varignon’s theorem at point O. \ Moment due to resultant R about point O = Sum of moment due to all forces about point O R d O

Fig. Ex. 1(a)

\ –R × d = – (30 × 2) – (60 × 5) + (20 × 7) – (40 × 11) (Treating anticlockwise moment as positive) –110 × d = –660 \ d = 6 m Refer figure Ex.1(a) Example 2. A pole is subjected to parallel forces as shown in figure Ex. 2. Determine (a) magnitude and direction of the resultant force and (b) position of the resultant force from point O.

Fig. Ex. 2

1.56  Engineering Mechanics Statics

Solution: Here magnitude of resultant is algebraic sum of all parallel forces. +

R = →ΣH = 200 – 800 – 100 + 150 \ R = –550 N = 550 N acting towards left  (Ans.) Now position of resultant is finding out by applying Varignon’s theorem at point O.

Fig. Ex. 2(a)

\ Moment due to resultant R about point O = Sum of moment due to all forces about point O 550 × d = –150 × 2 + 100 × 3 + 800 × 4 – 200 × 5 \ (Treating anticlockwise moment as positive) \ d = 4 m from O Refer figure Ex.2(a)  (Ans.) Example 3. Three parallel forces are acting on the lever as shown in figure Ex.3. Determine (a) magnitude and direction of the resultant force and (b) position of the resultant force from point O.

Fig. Ex. 3

Solution: Here magnitude of resultant is algebraic sum of all parallel forces. R = +↑SV = – 180 – 100 + 40 \ R = –240 N = 240 N downward  (Ans.) Now position of resultant is finding out by applying Varignon’s theorem at point O. 240N \ Moment due to resultant R about point O 2.125m = Sum of moment due to all forces about point O \ –R × d = –(180 × 2) – (100 × 3.5) + (40 × 5) O (Treating anticlockwise moment as positive) Fig. Ex. 3(a) –240 × d = –510 \ d = 2.125 m Refer figure Ex.3(a)  (Ans.)

Co-planer System of Forces  1.57

Example 4. Four parallel forces are acting on the lever as shown in figure Ex.4. Determine (a) magnitude and direction of the resultant force and (b) position of the resultant force from point O. 25N

25N

15N

20N

O

3m

2m

2m

3m

Fig. Ex. 4

Solution: Here magnitude of resultant is algebraic sum of all parallel forces. R = +↑SV = –25 + 25 + 15 – 20 \ R = – 5 N = 5 N downward  (Ans.) Now position of resultant is finding out by applying Varignon’s theorem at point O. \ Moment due to resultant R about point O = Sum of moment due to all forces about point O 5N

1m O

Fig. Ex. 4(a)

\ \

R × d = (25 × 5) – (25 × 2) + (15 × 2) – (20 × 5) (Treating anticlockwise moment as positive) 5 × d = 5 d = 1 m to the left of point O Refer figure Ex.4(a)  (Ans.)

Example 5. Four parallel forces are acting on lever as shown in figure Ex.5. Determine magnitude of P1 and P2 so that the four forces produce downward resultant of 150 N acting at 4 m from O

Fig. Ex. 5

1.58  Engineering Mechanics Statics

Solution: Here magnitude of resultant is algebraic sum of all parallel forces. R = +↑SV = 100 – P2 + P1 – 50 Now it is given that magnitude of resultant force is 150 N \ – 150 = – 50 + P1 – P2 + 100 \ P1 – P2 = – 200  \ P1 = P2 – 200 (Eq. 1) Now by applying Varignon’s theorem at point O, \ Moment due to resultant R about point O = Sum of moment due to all forces about point O \ 150 × 4 = – 2P1 + 5P2 – 700 (Treating anticlockwise moment as positive) 2P1 – 5P2 + 1300 = 0 2(P2 – 200) – 5P2 + 1300 = 0 Substituting value of P1 from (Eq. 1) 2P2 – 400 – 5P2 + 1300 = 0 900 = 5P2 – 2P2 900 = 3P2 \ P2 = 300 N  (Ans.) Substituting value of P2 in (Eq. 1) P1 = 300 – 200 P1 = 100 N  (Ans.) Example 6. Four parallel forces are acting on circle of diameter 2 m as shown in figure Ex.6. Determine (a) magnitude and direction of the resultant force and (b) position of the resultant force from point O. 30N

50N

45°

30°

60° 30°

80N

40N

Fig. Ex. 6

Solution: Here magnitude of resultant is algebraic sum of all parallel forces. R = +↑SV = 50 + 30 + 40 – 80 \ R = 40 N upward  (Ans.) Now position of resultant is finding out by applying Varignon’s theorem at point O. \ Moment due to resultant R about point O = Sum of moment due to all forces about point O \ 40 × d = (50 × 1 cos 30°) – (30 × 1 cos 45°) + (80 × 1 cos 60°) + (40 × 1 sin 30°) (Treating anticlockwise moment as positive) 40 × d = 82 d = 2.05 m from point O  (Ans.) \

Co-planer System of Forces  1.59

Solved Examples Based on Force-Coule System and Equivalent System Example 1. Replace 100 N force shown in figure Ex. 1 by equivalent force

couple at point A.

Fig. Ex. 1

Solution: Here transfer the force parallel to itself along with its moment at A as below: Introduce a system of 100 N forces at point A as shown in figure Ex. 1(a) Now force 100 N acting at point O is replaced at point A by keeping 100 N force as it is and a couple moment as shown in figure Ex. 1(b) M = 100 × 5 = 500 N.m anticlockwise.  (Ans.)

Fig. Ex. 1(a)

Fig. Ex. 1(b)

Example 2. Two systems of forces are shown in figure Ex. 2. Are they equivalent? 50N

2m

25N

4m

2m

System 1

System 2

25N

Fig. Ex. 2

1.60  Engineering Mechanics Statics

Solution: As we know that two systems are equivalent only when the sum of the forces and sum of the moments about an arbitrary point are equal. Now check for sum of the forces of two systems. +↑SFSystem 1 = 25 – 50 = –25 N + ↑SFSystem 2 = –25 N Now check for sums of the moments of two systems about left end. Treating anticlockwise moment as positive SMSystem 1 = –50 × 2 = –100 N.m = 100 N.m clockwise SMSystem 2 = –25 × 4 = –100 N.m = 100 N.m clockwise As sums of the forces of two systems are equal and sum of the moments of two systems about an arbitrary point are equal. Two systems are equivalent. Example 3. Replace the 200 N force shown in figure Ex. 3 by equivalent force couple at point A. 200N 50°

B

A 2m

Fig. Ex. 3

Solution: Here transfer the force parallel to itself along with its moment at A as below: Introduce a system of 200 N forces at point A as shown in figure Ex. 3(a) 200N

200N B

200 N A

200N     

Fig. Ex. 3(a)

460 Nm

Fig. Ex. (b)

Now force 200 N acting at point B is replaced at point A by keeping 200 N force as it is and a couple moment as shown in figure Ex. 3(b) M = 200 sin 50° × 3 = 460 N.m clockwise.  (Ans.) Example 4. The 50 N force is acting on lever as shown in figure Ex. 4. Replace the force by force and couple moment at point O that will have an equivalent effect.

Co-planer System of Forces  1.61

2m

50N

60°

O

Fig. Ex. 4

Solution: Here transfer the force parallel to itself along with its moment at O as below: Introduce a system of 50 N forces at point O as shown in figure Ex. 4(a) 50N

50N

50N

    Fig. Ex. 4(a)              Fig. Ex. 4(b)

Now force 200 N acting at point A is replaced at point O by keeping 50 N force as it is and a couple moment as shown in figure Ex. 4(b) M = 50 × 2 sin 60° = 86.60 N.m anticlockwise.  (Ans.) Example 5. A plate is subjected to two couples as shown in figure Ex.5. Determine the sum of the moments exerted on the plate by the two couples.

Fig. Ex. 5

Solution: Here, 60 N couple acts in clockwise direction and 40 N couple acts in anticlockwise direction. Now treating anticlockwise as positive, \ SM = –60 × 100 + 40 × 180 = – 6000 + 7200 \ SM = 1200 N.mm anticlockwise  (Ans.)

1.62  Engineering Mechanics Statics

Example 6. A plate is subjected to two couples as shown in figure Ex.6. Determine the magnitude of force P if the resultant couple moment is 300 N.m anticlockwise.

Fig. Ex. 6

Solution: Let θ be the angle of force P with horizontal. 1 \ tan θ =   \ θ = 26.57° 2 Now resultant couple moment = sum of the moments exerted on the plate by the two couples. Treating anticlockwise moment as positive, \ 300 = (P cos 26.57° × 2) + (P sin 26.57° × 1) – (150 × 0.5) \ P = 168.16 N  (Ans.) Example 7. A rectangular plate is subjected to a force – couple moment system

as shown in figure Ex.7. Replace this system by a single force and determine its position from x-axis.

Fig. Ex. 7

Solution: Here first replace clockwise couple moment by two equal and opposite forces of magnitude 1500 N as shown in left hand part of figure Ex.7(a). The 1500 N force acting towards left at point O and another 1500 N force acting towards right at point A, forms a couple. Here arm of couple is d and moment of couple is given by M = P × d, where M =180 N.m clockwise, P = 1500 N and d is unknown.

Co-planer System of Forces  1.63

Fig. Ex. 7(a)

M 150 = = 0.1 m = 100 mm from x-axis as shown in the right hand P 1500 part of figure Ex.7(a).  (Ans.) \d=

Example 8. Replace the 100 N force actin at A on a square plate as shown in figure Ex. 8 by equivalent force and couple moment at point O.

Fig. Ex. 8

Solution: Here transfer the force parallel to itself along with its moment at O as below: Introduce 100 N forces at point O as shown in the left hand part of figure Ex. 8(a). Now calculate the moment of a 100 N force about point O, (treating anticlockwise as positive) 50°

100N

100N 281.75Nm O 50°

50° 100N

100N

Fig. Ex. 8(a)

M = – 100 cos 50° × 2 – 100 sin 50° × 2 M = –281.75 = 281.75 N.m clockwise.  (Ans.) equivalent force and couple moment at point O is shown in the right hand part of figure Ex.8(a).

1.64  Engineering Mechanics Statics

Example 9. Two couples act on a pole as shown in figure Ex. 9. Determine the magnitude of P so that resultant couple moment is 300 N.mm clockwise.

Fig. Ex. 9

Solution: The resultant couple moment is 300 N.mm clockwise, \ – 300 = sum of moments of two couples (treating anticlockwise as positive) \ – 300 = 150 × 500 – P × 800 sin 30° ( here 800 sin 30° is the arm of couple P) \ P = 188.25 N  (Ans.) Example 10. Replace the force system as shown in figure Ex. 10 by resultant force and couple moment at point O that will have an equivalent effect. 300N 0.75m O 300N

1m

1m

900N

Fig. Ex. 10

Solution: Here first determine the resultant force as below, +

→ΣH = –300 + 300 = 0 +↑SV = –900 N (here ‘–’ sign indicate that SV is acting toward negative y direction) Now we know that resultant R is calculated as, ( ΣH ) 2 + ( Σ V ) 2



R =

\

R =

\

R = 900 N

2

(0) + (900)

O

2

1125Nm

900N

Now determine sum of moments of all forces Fig. Ex. 10(a) about point O, (treating anticlockwise as positive) SMO = 300 × 0.75 + 900 × 1 = 1125 N.m anticlockwise

Co-planer System of Forces  1.65

Resultant force and couple moment at point O of given system is shown in figure Ex.10(a). Example 11. Replace the force system acting on a bracket as shown in figure Ex. 11 by resultant force and couple moment at point A that will have an equivalent effect.

Fig. Ex. 11

Solution: Here first resolve the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components, +

→ΣH = 200 cos 20° – 100 sin 30° = 138 N +↑SH = 200 sin 20° – 100 cos 30° = –18.19 N (here ‘–’ sign indicate that SV is acting toward negative y direction) Now we know that resultant R is calculated as, \

R = R =

( ΣH ) 2 + ( Σ V ) 2 (138) 2 + (18.19) 2

7.50° 1036.34 Nm

139.20N

\ R = 139.19 N Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH \

tan θ =

18.19 \ θ = 7.50° 138

Fig. Ex. 11(a)

Now determine sum of moments of all forces about point A, (treating anticlockwise as positive) SMA = 350 + 100 cos 30° × 2.5 + 200 cos 20° × 2.5 = 1036.34 N.m anticlockwise Resultant force and couple moment at point A of given system is shown in figure Ex.11 (a).

1.66  Engineering Mechanics Statics

Example 12. A square plate of 800 mm side is subjected to a couple and force as shown in figure Ex.12. Replace the force system acting on plate with an equivalent force – couple system at O. (O is center of square plate)

Fig. Ex. 12

Solution: It is identifying from given figure that two 60 N forces form a couple, therefore resultant force of the given force system is the remaining force i.e. 50 N. \ R = 50 N and its inclination θ with horizontal axis is 20°. For moment purpose first determine distance ab and cd. Refer figure Ex.12. In triangle oab oa cos 40° = 400 \ oa = 522.16 mm ab = 522.16 sin 40° \ ab = 335.64 mm In triangle ocd oc cos 30° = 400 \ oc = 461.89 mm cd = 461.89 sin 30° \ cd = 230.94 mm Now determine sum of moments of all forces about O, (treating anticlockwise as positive), 50N

866.73 Nm

Fig. Ex. 12(a)

SMO = – (50 cos 20° × 335.64) + (50 sin 20° × 400) – (60 sin 45° × 400) + (60 cos 45° × 230.90) + (60 cos 45° × 400) = 866.73 N.m anticlockwise Resultant force and couple moment at point O of given system is shown in figure Ex.12 (a).

Co-planer System of Forces  1.67

Example 13. A bracket is subjected to three forces as shown in figure Ex.13. Replace the force system with an equivalent force – couple system at B. 6m

A

200N 1m

30° 400N 2m 45° 300N B

3m

Fig. Ex. 13

Solution: Here first resolve the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components, +

→ΣH = 400 cos 30° + 300 sin 45° = 558.54 N +↑SV = 200 – 400 sin 30° – 300 cos 45° = – 212.13 N (here ‘–’ sign indicate that SV is acting toward negative y direction) Now we know that resultant R is calculated as, A



R =

( ΣH ) 2 + ( Σ V ) 2

\

R =

(558.54) 2 + (212.13) 2

R = 597.47 N \ Let θ be the angle of resultant R with x-axis, then ΣV tan θ = ΣH 212.13 \ θ = 20.80 558.54 Now determine sum of moments of all forces about B, (treating anticlockwise as positive) \

tan θ =

1541.34 N.M B 597.47N

Fig. Ex. 13(a)

SMB = (200 × 6) – (400 cos 30° × 2)

3 – (400 sin 30° × 6) – (300 sin 45° ×  2 – × 2 – (300 cos 45° × 3)   6

= –1541.34 N.m = 1541.34 N.m clockwise Resultant force and couple moment at point A of given system is shown in figure Ex.13 (a).

1.68  Engineering Mechanics Statics

Example 14. Two systems of forces are shown in figure Ex. 14. Determine the value of R and P if these two systems are equivalent. A 200N

3m

1m

C

A

150 3m

C

B

B

D D

R

2m

2m

250N

1m

E

E 1m

1m System 1

System 2 P

P

Fig. Ex. 14

Solution: As we know that two systems are equivalent only when the sum of the forces and sum of the moments about an arbitrary point are equal. SHSystem 1 = SHSystem 2 \ –150 + 200 – 250 = – R \ R = 200 N (Ans.) Now determine sums of the moments of two systems about B and equate them. (Treating anticlockwise moment as positive) SMSystem 1B = SMSystem 2B 150 × 1 – 250 × 2 = – P × 1 \ P = 350 N  (Ans.) Example 15. Two systems of forces are shown in figure Ex. 15. Are they equivalent?

Fig. Ex. 15

Solution: As we know that two systems are equivalent only when the sum of the forces and sum of the moments about an arbitrary point are equal.

Co-planer System of Forces  1.69

Now check for sum of the forces of two systems. +



→ ΣFSystem1 = 100 – 200 = –100 N



→ ΣFSystem 2 = –200 + 100 = –100 N

+

Now check for sums of the moments of two systems about O. (Treating anticlockwise moment as positive) SMSystem 1 = –100 × 3 + 500 = 200 N.m anticlockwise SMSystem 2 = 200 × 3 – 100 × 1.5 – 300 = 150 N.m anticlockwise As sums of the forces of two systems are equal but sum of the moments of two systems about an arbitrary point are not equal. Two systems are not equivalent. Example 16. Two equivalent system of forces and moments act on the plate are shown in figure Ex. 16. Determine the force P and couple M.

Fig. Ex. 16

Solution: As we know that two systems are equivalent only when the sum of the forces and sum of the moments about an arbitrary point are equal. +

+



→ ΣH System1 = → ΣH System 2



60 N = 60 N

+↑SVSystem 1 = +↑SVSystem 2 \ 100 – 20 = P – 60 \ P = 140 N  (Ans.) Now determine sums of the moments of two systems about A and equate them. (Treating Anticlockwise moment as positive) +

+

→ΣM System1A = →ΣM System 2 A

–60 × 3 – 20 × 4 + M = 200

\ M = 460 N. m  (Ans.)

1.70  Engineering Mechanics Statics

SUMMARY • Force is the action of a body about another body. ♦ Unit of the force is N, kN ♦ 1 N = (1 kg) (1 m/s2) = 1 kg-m/s2 • Magnitude, Direction, Sense and Point of application are characteristics of the force. • When two or more forces act on a body they are called to form a system of forces. ♦ Coplanar forces are those forces whose lines of action lie on the same plane. ♦ Non-Coplanar forces are those forces whose lines of action do not lie on the same plane • Resultant is a single force which produces the same effect as produced by number of forces when acting together. • The resultant of two or more collinear forces is simply their algebraic sum. • The resultant of two concurrent forces can be found out by means of the parallelogram law or the triangle law.



R =



tan α =





P2 sin θ .......... Direction of resultant P1 + P2 cos θ

• The resultant of more than two concurrent forces can be found out by means of rectangular components method or the polygon law.





P12 + P22 + 2 P1P2 cos θ .......... Magnitude of resultant



R =



tan θ =

(ΣH )2 + (ΣV )2 .......... Magnitude of resultant ΣV .......... Direction of resultant ΣH

• Resolution is the process of splitting up the given force into components, without changing its effect on the body. • The rotational effect produced by force is known as moment of the force. Moment = Force × Distance M = F × d • The location of the resultant of parallel force system is found out by means of Varignon’s Theorem (Principle of Moment) • Two equal, unlike parallel, non-collinear forces form a couple. • Force can be transfer parallel to itself by means of force couple system. • Two force systems that produce the same external effects on a rigid body are said to be equivalent.

Co-planer System of Forces  1.71

PROBLEMS 1. Two forces of 400 N and 150 N are acting on a hook as shown in figure Prob.1.Determine magnitude resultant of these two forces.

Fig. Prob. 1 2. Two forces of 375 N and P are acting on a hook as shown in figure Prob.2. If the magnitude of the resultant force is to be 200 N, directed along positive y-axis, determine the magnitude of force P.

Fig. Prob. 2 3. Two forces of 150 N and 200 N are acting on a plate as shown in figure Prob.3. Determine magnitude and direction of resultant of these two forces.

y

150 N

50°

200 N 30°

x

Fig. Prob. 3 4. Two forces of 160 N and 200 N are acting on a bracket as shown in figure Prob.4. Determine magnitude and direction of resultant of these two forces.

160 N 1m O 0.5m 200 N

2m

Fig. Prob. 4

1.72  Engineering Mechanics Statics 5. Resolve 300 N force in to components along u and v directions. Refer figure Prob.5.

v 300 N

70° 30°

u

Fig. Prob. 5 6. Resolve 300 N force in to components along u and v directions. Refer figure Prob.6.

60°

30°

u

O v

Fig. Prob. 6 7. Two forces of 150 N and 110 N are acting on a hook as shown in figure Prob.7. Determine (a) magnitude and direction of resultant of these two forces. (b) The magnitudes of two other forces Pu and Pv 110 N that would have the same resultant.

V

Y 20° U

40°

150 N 20°

20° X

Fig. Prob. 7 8. Determine magnitude of the resultant force and its direction from x-axis of three concurrent forces as shown in figure Prob.8.

Fig. Prob. 8 9. A rectangular plate is subjected to three concurrent forces as shown in figure Prob.9. Determine magnitude of the resultant force and its direction from x-axis.

50 N

60 N 35° 45°

55° 50 N

Fig. Prob. 9

Co-planer System of Forces  1.73 10. Determine magnitude of the resultant force and its direction from x-axis of three concurrent forces as shown in figure Prob.10.

150 N 4

250 N

3 1 O

3 1 1 400 N

Fig. Prob. 10 11. Determine magnitude of the resultant force and its direction from x-axis of four concurrent forces as shown in figure Prob.11.

Fig. Prob. 11 12. The magnitude of the resultant of three concurrent forces is 45 N and directed along positive y axis. Determine magnitude of force P. Refer figure Prob.12

Y 60 N 30°

P 30° 30°

X 50 N

Fig. Prob. 12 13. Determine the moments of the 120 N force shown in figure Prob.13 about points A, B and C.

A

B

70mm

C

80mm

100mm

Fig. Prob. 13 14. Determine moment of each forces shown in figure Prob.14 about points O.

Fig. Prob. 14

120 N

1.74  Engineering Mechanics Statics 1.5 m 150 N

1.5 m

15. Two forces are applied to a bracket as shown in figure Prob.15. Determine the sum of the moments of the two forces about point A and point B.

100 N

1.5 m

A

B

Fig. Prob. 15 16. Three forces are applied to a bracket as shown in figure Prob.16. Determine the sum of the moments of the three forces about point O.

200 N

30°

400 N 0.5 m O 0.5 m 0.5 m

1m 200 N

Fig. Prob. 16 17. The moment exerted by the weight W about point B is 300 N.m. Determine the moment by the weight W about point A. Refer figure Prob.17.

Fig. Prob. 17 18. Two forces are applied to a circular plate as shown in figure Prob.18. determine moment of each forces about point O.

30°

R = 0.8 m

50 N

O

45° 30 N

Fig. Prob. 18 19. Determine the resultant of the three forces and the couple C acting on a rectangular plate as shown in figure Prob.19.

30 N

50 N

1.4 m

50°

65°

1,2 m 40 N 45 Nm

Fig. Prob. 19

Co-planer System of Forces  1.75 20. Determine the resultant of the three forces acting on a beam as shown in figure Prob.20.

Fig. Prob. 20 21. Three parallel forces are acting on a beam as shown in figure Prob.21. Determine the resultant force and its location with respect to support A.

3.5 kN

4 kN

4.5 kN

A

B

1.4 m

1.5 m

1 m 1.2 m

Fig. Prob. 21 22. The resultant of the force system shown in figure Prob.22 is a 120 N.mm anticlockwise couple. Determine magnitude of P1, P2 and C.

300 mm 40° 200 mm

P1

C 50 N 200 mm

40°

P2

Fig. Prob. 22 23. Determine the moment of the couple shown in figure Prob.23. Also determine perpendicular distance between the two forces.

y 150 N

30°

150 N

30°

100 m m

Fig. Prob. 23

x

1.76  Engineering Mechanics Statics 24. If P = 80 N determine the couple moment produced. Refer figure Prob.24.

0.1 m

0.1 m P

P

Fig. Prob. 24 25. In Prob. No.24, If the couple moment produced by two forces P is 30 N.m. Determine magnitude of P. 26. A rectangular plate is subjected to two forces and the couple as shown in figure Prob. 26. Replace the force system with an equivalent force – couple system at point O

Fig. Prob. 26 27. A bracket is subjected to three forces as shown in figure Prob. 27. Replace the force system with an equivalent force – couple system at point A.

6.5 kN 30° 1m

60° 1m 4.5 kN A 1m

1m

Fig. Prob. 27

3.5 kN

Unit 2

In this unit, we discuss criteria for equilibrium of different force system and concept of free body diagram, which is perhaps the most important tool of mechanics. We then discuss analysis procedure for finding out unknown forces and moments using free body diagram and equilibrium equations.

Equilibrium of System of Forces 2.1 Equilibrium Equilibrium is the state of rest of a body i.e. the body does not move and also does not rotate about any point under the action of forces. For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body; Static equilibrium is the however it could be translating or rotating at a state in which the sum of constant velocity. the forces and moments When object is in equilibrium under the on each particle of the action of system of forces, following conditions system is zero are satisfied: 1. The algebraic sum of the components of the forces in any direction must equal to zero. 2. The algebraic sum of the moments of the forces about any point must equal zero. Equilibrium conditions for different force system are shown in Table 2.1 Table 2.1 Equilibrium Conditions for Different Force Systems Force System

Equilibrium Condition

Collinear force system Concurrent force system

Parallel force system

Non-concurrent force system

SP = 0 \ Resultant, R = 0 SH = 0 (sum of horizontal components) and SV = 0 (sum of vertical components) \ Resultant, R = 0 SP = 0 and SMO = 0 (O is any point in the plane of the forces) SH = 0 (sum of horizontal components) SV = 0 (sum of vertical components) and SMO = 0 (O is any point in the plane of the forces)

2.2  Free Body Diagram (F.B.D.) Free body diagram is one of the most important tools of mechanics. Free-body diagrams are fundamental to all engineering disciplines that are concerned with the effects that forces have on bodies. A free-body diagram of a body is a diagrammatic representation or a sketch of a body in which the body is shown completely separated from all surrounding bodies, including supports. The term free implies that all supports have been

2.4  Engineering Mechanics Statics

removed and replaced by the forces (reactions) that they exert on the body. Forces that we show on a free body diagram can be divided into two categories: (a) Reactive Forces: Reactive forces known as reactions are those forces that are exerted on a body by the supports to which it is attached. Reactions at supports and connections are of only two types: 1. If a support or connection prevents translation in some direction, then a force may be developed in that direction. 2. If a support or connection prevents rotation about some axis, then a moment may be developed around that axis (b) Applied forces: Forces acting on a body that are not provided by the supports are called applied forces. Different supports and their reactions are shown in Table 2.2 Table 2.2 Supports and their Reactions Support

Example

Normal contact on horizontal smooth surface

Free Body Diagram (F.B.D.)

Normal force oriented perpendicular to surface. Direction of normal force is towards the body.

W

W

N

Normal contact on inclined smooth surface

Rope, cable, wire

Normal force oriented perpendicular to inclined surface. Direction of normal force is towards the body.

W 

Cable 

Description

Tension oriented along cable. Direction of tension is away from the body.

T 

Spring Spring P

Spring force oriented along long axis of spring. Force is pull if spring is in tension, and force is push if spring is in compression.

Equilibrium of System of Forces  2.5 Normal contact on horizontal surface with friction

W P

P

W Rough Surface

F N

Normal contact on inclined surface with friction

W

W Rough Surface



F

Link

N

Normal force oriented perpendicular to surface. Friction force parallel to the surface. Normal force oriented perpendicular to inclined surface. Friction force parallel to the inclined surface. Force oriented along the link. Force can be push or pull

P Link

Roller

N 

Normal force oriented perpendicular to the surface supporting the roller.

 N

Pin/Hinge

Force in terms of components oriented along x and y-axes.

A

Fixed A

Ax M Ay

Force in terms of components oriented along x and y-axes. Moment about z-axis.

Sliding in guide Normal force oriented perpendicular to guide. N

2.6  Engineering Mechanics Statics

2.2.1  General Procedure for Constructing a Free Body Diagram

1. Choose the free body to be used, isolate it from any other body and sketch its outline. 2. Locate all external forces on the free body and clearly mark their magnitude and direction. This should include the weight of the free body, which is applied at the centre of gravity. 3. Locate and mark unknown external forces and reactions on the free-body diagram. 4. Include all dimensions that indicate the location and direction of forces. The free-body diagram of a rigid body can be reduced to that of a particle. The free-body of a particle is used to represent a point and all forces acting on it.

2.2.2  Examples on Free Body Diagram (FBD) Some examples on free body diagram are shown in Table 2.3 Table 2.3 Examples on Free Body Diagrams Different Systems

Free Body Diagram W

W 2

1

N2

2

1

N1

T W  W

N

W

W

W

W N5

W

W

N6

N1

N4

N2

N3

Equilibrium of System of Forces  2.7

W

B W 30° A



30° NB

Ax



Ay

10 N 10 N

20 Nm 20 Nm

A

A

B

B

Ax 3m

3m

3m

4m

3m

4m

Ay

By T 

A Ax

1m

2m Ay

A W (Smooth Surface) 

2.8  Engineering Mechanics Statics 10 N

15 N

1m

2m

10 N

15 N

C A

B 30°

1m

Ax 30° PCD

D

Ay

10 N 1.5m

N 1m

1m

Ax M 25 Nm Ay W

B

A 2m

1m

10 kN

2.3  Equilibrium for Different Force System    (Based on number of forces)

1. Equilibrium under Two Force System: If a body is in equilibrium under the action of two forces, then two forces must be collinear, of equal magnitude and act in opposite direction as shown in figure 1. P

P

Fig. 1

A body will not be in equilibrium under the action of two equal and opposite parallel forces, because these forces produce a couple.

Equilibrium of System of Forces  2.9

2. Equilibrium under Three Force System: If a body is in equilibrium under the action of three forces, then three forces acting on a body must be concurrent or parallel. To satisfy the condition SM = 0 the line of action of three non-parallel forces P1 P1, P2 and P3 must pass through the same point i.e. the three force must be concurrent as shown in figure 2. When the three forces P1, P2 and P3 acting on a body are parallel and acts in same direction then body will not be in equilibrium, as their resultant R = P1 + P2 + P3. If the three forces are acting P3 P2 in opposite direction and their magnitude is so adjusted that the resultant force is zero and sum of Fig. 2 moment of three forces about any point is zero then body will be in equilibrium.



Lami’s Theorem: It states, “If a body is in equilibrium under the action of three concurrent forces, then each force is proportional to the sine of angle between the other two forces” P1

  P3  P2

Let P1, P2 and P3 are three forces acting at point O. Let the angle between P1 and P2 is α, angle between P2 and P3 is γ and P3 and P1 is β, then according to Lami’s theorem, P P P1 = 2 = 3 (Eq 2.1) sin β sin ∝ sin γ

Below steps are following in equilibrium analysis of a body, 1. Draw a free-body diagram of the body which shows all of the forces and moment that act on the body. 2. Write the equilibrium equations for all the forces and moments that appear on the free-body diagram. 3. Solve the equilibrium equations for the unknowns.

2.10  Engineering Mechanics Statics

Solved Examples Based on Equilibrium of Forces Example 1. The system of forces shown in figure Ex. 1 is in equilibrium. String

BC is horizontal. Determine tension in string AB (T1), tension in string BC (T2) and tension in string CD (T3). A

B 35°

30°

B

C

40 N

50 N

Fig. Ex. 1

Solution: Consider the free body diagram at point B and C as shown in figure Ex. 1(a). Since the system is in equilibrium and three concurrent forces are acting at point B and C, we can use Lami’s theorem at point B and C.

Figure Ex. 1(a)

Consider point B from F.B.D. and apply Lami’s theorem. 40 T1 T2 = = sin 125° sin 90° sin 145° \ T1 =

40 sin 90°   \ T1 = 48.83 N  (Ans.) sin 125°

T2 =

40 sin 145°   \ T2 = 28 N  (Ans.) sin 125°

Consider point C from F.B.D. and apply Lami’s theorem. T3 50 28 = = sin 90° sin120° sin150° \ T3 =

28sin 90°   \ T3= 56 N  (Ans.) sin150°

Equilibrium of System of Forces  2.11

Example 2. Three concurrent forces acting at point O as shown in figure Ex 2 are in equilibrium. Find out the magnitude of force P and angle θ.

Fig. Ex. 2

Solution: Since the system is in equilibrium and three concurrent forces are acting at point O we can use Lami’s theorem at point O. P 15 20 = = ...(Eq. 1) sin 90° sin (180 − θ) sin (90 + θ) Now from Trigonometry formulas we know, sin (180 – θ) = sin θ and sin (90 + θ) = cos θ, putting these in Eq. 1, 15 20 P = = ...(Eq. 2) \ sin θ cos θ sin 90° 15 15 sin θ   \ tan θ = =   \ θ = 37°  (Ans.) 20 20 sin θ

\

Now from equation (2), P 15 = sin 90° sin 37° 15sin 90°   \ P = 25 N  (Ans.) sin 37° Alternate Approach: The above problem can also be solved by applying conditions of equilibrium, P =

\

+

i.e. →ΣH = 0 and +↑SV = 0 SH = 0: P cos θ – 20 = 0 20 \ P = cos θ \

SV = 0: P sin θ – 15 = 0 15 P = sin θ

...(Eq. 1)

...(Eq. 2)

2.12  Engineering Mechanics Statics

Now from equation Eq.1 and Eq.2 20 15 = cosθ sin θ

\

sin θ 15 15 \ tan θ = = \ θ = 37°  (Ans.) cos θ 20 20

Now from Eq. 1 P =

20   \ P = 25 N  (Ans.) cos37°

Example 3. Check whether two systems of forces shown in figure Ex. 3 are in equilibrium or not.

Fig. Ex. 3

Solution: A system is in equilibrium only when it satisfies all equations of equilibrium, i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) Now consider system 1 and apply equation of equilibrium,  4  3 SH =   60 –   80  5  5 = 48 – 48  \ SH = 0  4  3 SV = 100 –   80 –   60  5  5 = 100 – 64 – 36  \ SV = 0 As system 1 is concurrent force system, the sums of moment about point O is zero, i. e. SM = 0. So system 1 satisfies all equations of equilibrium, \ System 1 is in equilibrium  (Ans.) Now consider system 2 and apply equation of equilibrium, 4 3 SH =   60 –   80 5 5 = 48 – 48  \ SH = 0

Equilibrium of System of Forces  2.13

4 3 SV = 100 –   80 –   60 5 5 = 100 – 64 – 36  \ SV = 0 As system 2 is non-concurrent force system, taking the sum of moments about point O. Here since line of action of forces 60 N and 80 N passing through point O, they produces zero moment, but the 100 N force will cause a clockwise moment of 200 N.m. SMO = 200 N.m clockwise. \ So system 2 satisfies sum of forces equations of equilibrium but it does not satisfy sum of moment equation of equilibrium, \ System 2 is not in equilibrium  (Ans.)

Example 4. A 250 N block rest on an incline smooth surface as shown in figure

Ex.4. Determine normal reaction at contact and force P to maintain equilibrium.

Fig. Ex. 4

Solution: Applying conditions of equilibrium i.e. +SH = 0 and + SV = 0 to the free body diagram shown in figure Ex. 4 (a), 250 N P SH = 0: P – 250 sin 30° = 0 30° \ P = 125 N  (Ans.) SV = 0: N – 250 cos 30° = 0 \ N = 216.5 N  (Ans.) Alternate Approach: Since the system is in 30° equilibrium and three concurrent forces are acting at point N O as shown in figure Ex. 4 (b), we can use Lami’s theorem Fig. Ex. 4(a) at point O. N 30°

P 30°

O

250 N

Fig. Ex. 4(b)

2.14  Engineering Mechanics Statics



P 250 N = = sin150° sin 90° sin120°



250 P = sin 90° sin150°

\ \

P = 125 N  (Ans.) N 250 = sin120° sin 90° N = 216.5 N  (Ans.)

Example 5. A 100 N block rest on an incline smooth surface as shown in figure Ex.5. If θ = 40°, determine normal reaction at contact and force P to maintain equilibrium.

100 N

P 

20°

Fig. Ex. 5

Solution: Applying conditions of equilibrium i.e. +SH = 0 and +SV = 0 to the free body diagram shown in figure Ex. 5(a), SH = 0: P cos 40° – 100 sin 20° = 0 \ P = 44.65 N  (Ans.) SV = 0: N – P sin 40° – 100 cos 20° = 0 N = 44.65 sin 40° + 100 cos 20° \ N = 122.66 N  (Ans.) Alternate Approach: Since the system is in equilibrium and three concurrent forces are acting Fig. Ex. 5(a) at point O as shown in figure Ex. 5 (b), we can use Lami’s theorem at point O. P 100 N = = sin160° sin130° sin 70° \ \

P 100 = sin160° sin130° P = 44.65 N  (Ans.) N 100 = sin 70° sin130° N = 122.67 N  (Ans.)

Fig. Ex. 5(b)

Equilibrium of System of Forces  2.15

Example 6. In Example 5 if P = 60 N, determine the angle θ to maintain equilibrium. Solution: Applying conditions of equilibrium i.e. +SH = 0 and +SV = 0 to the free body diagram shown in figure Ex. 5(a), SH = 0: 60 cos θ – 100 sin 20° = 0  \ θ = 55.24°  (Ans.) Example 7. Determine the tension in each cord for equilibrium of 1000 N and 3 m long wooden log as shown in figure Ex.7.

Fig. Ex. 7

Solution: In order to maintain equilibrium the force developed in AB must be equal to weight of wooden log, i.e. 1000 N. Consider the free body diagram at point B figure Ex. 7(a). Since the system is in equilibrium and three concurrent forces are acting at point B, we can use Lami’s theorem, 1000 N P P 1000 BC BD = = sin150° sin135° sin 75° \ \

PBC 1000 = sin150° sin 75°

45°

PBC = 517.64 N  (Ans.) 1000 PBD = sin 75° sin135° PBD = 732.05 N  (Ans.)

60°

PBC

PBD

Fig. Ex. 7(a)

Example 8. Four forces are acting at point O as shown in figure Ex.8. Determine magnitude of P1 and P2 for equilibrium.

Fig. Ex. 8

2.16  Engineering Mechanics Statics +

Solution: Applying conditions of equilibrium i.e. →ΣH = 0 and +↑SV = 0 SH = 0: 80 + 60 cos 40° – P2 cos 30° = 0  \ P2 = 145.44 N  (Ans.) SV = 0: P1 – P2 sin 30° – 60 sin 40° = 0 P1 = 145.44 sin 30° + 60 sin 40°  \ P1 = 111.29 N  (Ans.) Example 9. A circular plate is subjected to three concurrent forces as shown in figure Ex.9. Determine magnitude of force P1 and P2 for equilibrium. 100 N

O P1

2 3 P2

Fig. Ex. 9

Solution: Since the system is in equilibrium and three concurrent forces are acting at point O we can use Lami’s theorem at point O. Refer figure Ex.9(a) 2 Let θ be the angle of 90 N force with horizontal, tan θ = 3 \ θ = 33.69° 100 N P1 P2 100 = = sin146.3° sin123.69° sin 90° \

100 P1 = sin 90° sin146.3°

33.69 °

56.3 °

P2 P1 P1 = 55.48 N  (Ans.) Fig. Ex. 9(a) P2 100   \ P2 = 83.20 N  (Ans.) = sin123.69 ° sin 90°

Alternate Approach: The above problem can also be solved by applying conditions of equilibrium, +

→ΣH = 0 and +↑SV = 0 i.e. SH = 0: P2 cos 56.30° – P1 cos 33.69° = 0 P2 cos 56.30° = P1 cos 33.69° \ P2 = 1.499 P1 ...(Eq.1) SV = 0: 100 – P2 sin 56.30° – P1 sin 33.69° = 0 100 = P2 sin 56.30° + P1 sin 33.69° Substitute value of P2 from Eq. 1

Equilibrium of System of Forces  2.17

\ 100 = 1.499 P1 sin 56.30° + P1 sin 33.69° 100 = 1.8023 P1  \ P1 = 55.48 N  (Ans.) Now from equation Eq. 1 P2 = 1.499 × 55.48  \ P2 = 83.17 N  (Ans.) Example 10. A 100 kg mass is supported by cables as shown in figure Ex.10.

determine tension in each cables.

Fig. Ex. 10

Solution: Consider the free body diagram as shown in figure Ex. 10 (a)

Fig. Ex. 10(a)

Applying conditions of equilibrium +↑SV = 0 to 100 kg block of free body diagram, SV = 0 : TAB – 100 × 9.81 = 0  \ TAB = 981 N  (Ans.) Now since the system is in equilibrium and three concurrent forces are acting at point B and D, we can use Lami’s theorem at point B and D. First applying Lami’s theorem at point B, TAB TBC T = BD = sin120° sin150° sin 90°

2.18  Engineering Mechanics Statics



TBC 981   \ TBC = 566.38 N  (Ans.) = sin150° sin120°



981 TBD =   \ TBD = 1132.76 N  (Ans.) sin120° sin 90°

Now applying Lami’s theorem at point D, TBD TDE T = DF = sin105° sin165° sin 90°

1132.76 TDE =   \ TDE = 303.52 N  (Ans.) sin105° sin165°



1132.76 TDF   \ TDF = 1172.71 N  (Ans.) = sin 90° sin105°

Example 11. Block A is supported by two weights of 100 N and 50 N as shown in figure Ex.11. Determine the weight of block A and angle θ for equilibrium.

50N



50°

100N

A

Fig. Ex. 11 +

Solution: Applying conditions of equilibrium i.e. →ΣH = 0 and +↑SV = 0 to free body diagram as shown in figure Ex.11(a) 50N 100N

50°



W

Fig. Ex. 11(a)



SH = 0: 100 cos θ – 50 cos 50° = 0  \ θ = 71.25°  (Ans.) SV = 0: 50 sin 50° + 100 sin 71.25° – W = 0  \ W = 133 N  (Ans.)

Equilibrium of System of Forces  2.19

Example 12. A roller 400 N rests against vertical and inclined smooth surfaces as shown in figure Ex.12. Determine the normal reactions at A and B.

B A 30°

Fig. Ex. 12

Solution: Consider free body diagram as shown in figure Ex.12 (a). Since the system is in equilibrium and three concurrent forces are acting at the centre of roller, we can use Lami’s theorem at point O. 400 N 30° NA NB NB 400N NA

30°

Fig. Ex. 12(a)



NA NB 400 = = sin 90° sin150° sin120°



NA 400   \ NA = 461.88 N  (Ans.) = sin 90° sin120°



400 NB   \ NB = 230.94 N  (Ans.) = sin150° sin120°

Alternate Approach: The above problem can also be solved by applying conditions of equilibrium, +

i.e. →ΣH = 0 and +↑SV = 0 SH = 0: NA sin 30° – NB = 0 ...(Eq. 1) NB = NA sin 30° SV = 0: NA cos 30° – 400 = 0  \ NA = 461.88 N  (Ans.) Now from Eq. 1 NB = 461.88 sin 30°  \ NB = 230.94 N  (Ans.)

2.20  Engineering Mechanics Statics

Example 13. Two smooth sphere of weight 150 N and radius 200 mm each rest in a channel as shown in figure Ex.13. Assuming all contact surfaces as smooth, determine reactions at points of contacts.

Fig. Ex. 13

Solution: First determine angle α. Refer figure Ex.13 (a), O1O2 = 200 + 200 = 400 mm,

m

0m

40

O2

O1E = 750 – 200 – 200 = 350 mm O1

O E 350   \ a = 29° cos α = 1 = O1O2 400 +



E 350mm

Now applying conditions of equilibrium i.e.

Fig. Ex. 13(a)

→ΣH = 0 and +↑SV = 0 to the right hand side roller of free body diagram shown in figure Ex. 13(b),

Fig. Ex. 13(b)

From Eq. 1,

SH = 0: NC cos 29° – ND = 0  \ ND = NC cos 29° ..(Eq.1) SV = 0: NC sin 29° – 150 = 0  \ NC = 309.39 N  (Ans.) ND = 309.39 cos 29°  \ ND = 270.60 N  (Ans.) +

Applying conditions of equilibrium i.e. →ΣH = 0 and +↑SV = 0 to the left hand side roller of free body diagram shown in figure Ex. 13(b),

Equilibrium of System of Forces  2.21

SH = 0: NA – NC cos 29° = 0 NA = 309.39 cos 29°  \ NA = 270.6 N  (Ans.) SV = 0: NB – 150 – NC sin 29° = 0 NB = 150 – 309.39 sin 29°  \ NB = 0 N  (Ans.)

\ \

Example 14. Two smooth sphere of weight 400 N and 200 N respectively rest

on an inclined surface as shown in figure Ex.14. Assuming all contact surfaces as smooth, determine reactions at points of contacts.

Fig. Ex. 14 +

Solution: Applying conditions of equilibrium i.e. →ΣH = 0 and +↑SV = 0 to the left hand side roller of free body diagram shown in figure Ex. 14(a), 200 N

400 N NB 



NB 40° NC

NA

10°

Fig. Ex. 14(a)



SH = 0: NA sin 10° – NB cos α = 0 SV = 0: NA cos 10° – 400 – NB sin α = 0 +

...(Eq.1) ...(Eq.2)

Applying conditions of equilibrium i.e. →ΣH = 0 and +↑SV = 0 to the right hand side roller of free body diagram shown in figure Ex. 14(a), ...(Eq.3) SH = 0: NB cos α – NC sin 40° = 0 ...(Eq.4) SV = 0: NB sin α – 200 + NC cos 40° = 0 Now from Eq. 1 and Eq. 3 NA sin 10° – NC sin 40° = 0 NA sin 10° = NC sin 40°  \ NA = 3.7016 NC ...(Eq.5)

2.22  Engineering Mechanics Statics

From Eq. 2 and Eq. 4 NA cos 10° + NC cos 40° – 200 – 400 = 0 NA cos 10° + NC cos 40° = 600 Substitute value of NA from Eq. 5, 3.7016 NC × cos 10° + NC cos 40° = 600  \ NC = 136 N  (Ans.) From Eq. 5, NA = 3.7016 × 136  \ NA = 503.41 N  (Ans.) Now from Eq. 3, ...(Eq. 6) NB cos α = 136 sin 40° = 87.419 From Eq. 4, ...(Eq.7) NB sin α = 200 – 136 cos 40° = 95.817 Dividing Eq. 7 by Eq. 6, 95.817 N B sin α = 87.419 N B cos α \ tan α = 1.096  \ α = 47.62° Now from Eq. 6, NB cos 47.62° = 87.419  \ NB = 129.69 N  (Ans.) Example 15. A roller of weight 800 N and radius 240 mm is pushed by force P over a step at A as shown in figure Ex.15. Determine magnitude of force P to just start the roller over the step. P 240mm 25° O 30mm

Fig. Ex. 15

Solution: Consider free body diagram of roller as shown in figure Ex.15(a). Consider the equilibrium state when roller just start over the step. Now since the system is in equilibrium, three forces acting on the roller must be concurrent at the centre of roller, and we can use Lami’s theorem.

Fig. Ex. 15(a)

Equilibrium of System of Forces  2.23

Referring to free body diagram, first find out angle α as below, OA = 240 mm and OB = 240 – 30 = 210 mm Now in ∆OAB, AB = Now

tan α =

OA2 − OB 2 = 2402 − 2102 = 116.18 N AB 116.18   \ tan α =   \ α = 28.950 OB 210

Now using Lami’s theorem, 800 P NA = = sin143.95° sin151.05° sin 65°

NA 800 =   \ NA = 1232 N sin 65° sin143.95°



P 800 =   \ P = 658 N  (Ans.) sin151.05° sin143.95°

Example 16. The boom OA and cable AC support a load of 300 N as shown in figure Ex.16. Knowing that boom AO exerts on pin A a force along OA, determine magnitude of force along OA and tension in cable AC.

Fig. Ex. 16

Solution: As three concurrent forces are acting at pin A, we can apply Lami’s theorem at A. Consider free body diagram at A as shown in figure Ex.16 (a) PAO TAC 300 PAO TAC = = sin110° sin 120° sin 130°

PAO 300 = sin 110° sin 120°

\ PAO = 276.48 N  (Ans.) \

TAC 300 = sin 110° sin130°

40°

30°

300 N

Fig. Ex. 16(a)

TAC = 244.56 N  (Ans.)

Example 17. A man is holding up the 600 N bar AB by applying perpendicular force of 400 N as shown in figure Ex.17. Determine angle θ at which he can support the bar.

2.24  Engineering Mechanics Statics

B

3.5

m

1.5

m

 A

Fig. Ex. 17

Solution: Consider free body diagram of bar as shown in figure Ex.17 (a). B 600 N

2.5

m

1.5

m

Ax

 400 N

A Ay

Fig. Ex. 17(a)

Applying conditions of equilibrium i.e. SMA = 0, treating anticlockwise moment as positive, SMA = 0: 600 × 2.5 cos θ – 400 × 1.5 = 0 600 × 2.5 cos θ = 400 × 1.5  \ θ = 66.42°  (Ans.) Example 18. The homogeneous bar AB of 300 N is resting as shown in figure

Ex.18. Assuming contact surface as smooth, determine the forces acting at A and B.

Fig. Ex. 18

Equilibrium of System of Forces  2.25 +

Solution: Applying conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 to free body diagram shown in figure Ex. 18(a), NB B

2.5 m

300 N 60°

A

Ax

Ay

Fig. Ex. 18(a) +

→ΣH = 0: NB + Ax = 0 ...(Eq.1) +↑ SV = 0: Ay – 300 = 0  \ Ay = 300 N  (Ans.) SMA = 0: 300 × 2.5 cos 60° – NB × 5 sin 60° = 0 300 × 2.5 cos 60° = NB × 5 sin 60°  \ NB = 86.60 N  (Ans.) \ AX = –86.60 N  (Ans.) Now from Eq. 1, 86.60 + AX = 0  Here, ‘–ve’ sign indicate that force AX will act in negative x direction. Example 19. The homogeneous bar AB of 1000 N is resting as shown in figure Ex.19. Assuming contact surface as smooth, determine the tension in the rope CD.

Fig. Ex. 19

+

Solution: Applying conditions of equilibrium i.e. →ΣH = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 19(a), T

C

B NB

25° 25°

A 1000 N NA

Fig. Ex. 19(a)

2.26  Engineering Mechanics Statics +

→ΣH = 0: NB sin 25° – T = 0 ...(Eq.1) T = NB sin 25° SMA = 0: T × 1225 sin 25° – NB × 800 + 1000 × 612.5 cos 25° = 0 Substituting value of T from Eq. 1 × 800 + 1000 × 612.5 cos 25° = 0 NB sin 25° × 1225 sin 25° – NB = 955 N \ Now from Eq. 1, T = 955 sin 25°  \ T = 403.60 N  (Ans.) Example 20. The homogeneous bar AB of 600 N is resting as shown in figure Ex.20. Assuming contact surface as smooth, determine the reactions at A and B.

Fig. Ex. 20

Applying +

conditions

of

equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 20(a), SH = 0: NA – NB = 0 \ NA = NB ...(Eq.1) SV = 0: T – 600 = 0 ...(Eq.2) \ T = 600 N

NA

360mm

Solution:

T

30

0m

20

0m

m

m

 600 N 480 mm

Fig. Ex. 20(a)

360   \ θ = 36.86° Now from free body diagram tan θ = 480 SMB = 0: –T × 200 cos 36.86° + 600 × 600 cos 36.86° – NA × 360 = 0 Substituting T = 600 N from Eq. 2, –600 × 200 cos 36.86° + 600 × 600 cos 36.86° = NA × 360 \ NA = 533.40 N  (Ans.) Now from Eq. 1, NA = NB  \ NB = 533.40 N  (Ans.)

NB

Equilibrium of System of Forces  2.27

Example 21. The homogeneous bar ABC of 20 kg is supported as shown in figure Ex.21. If weight of the bar is concentrate at G, determine the reactions at A and tension T. A T 1m

0.5m G C 2m

Fig. Ex. 21

Solution: Applying conditions of equilibrium

Ay

+

i.e. →ΣH = 0, +↑SV= 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 21(a), SH = 0: Ax = 0  (Ans.)

SV = 0: Ay + T – 20 × 9.81 = 0

A

1m G C 1.5m

\ Ay + T = 196.2 N  ...(Eq.1) \

Ax

T

0.5m 20×9.81

SMA = 0: 20 × 9.81 × 0.5 – T × 2 = 0

Fig. Ex. 21(a)

T = 49 N  (Ans.)

Now from Eq. 1, Ay + 49 = 196.2  \ Ay = 147.2 N  (Ans.) Example 22. The homogeneous bar ABC of is attached to a pin at A and rest on roller support at B as shown in figure Ex.22. Neglecting the weight of the bar, determine the reactions at A and B. 150N

B

A

3m

1m

1m

Fig. Ex. 22 +

Solution: Applying conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 22(a),

2.28  Engineering Mechanics Statics 150N

A

Ax

1m

B

3m

45° NB

Ay

Fig. Ex. 22(a)

SH = 0: Ax – NB sin 45° = 0 ...(Eq.1) \ Ax = NB sin 45° ...(Eq.2) SV = 0: Ay + NB cos 45° – 150 = 0 SMB = 0: –Ay × 3 – 150 × 1 = 0 \ Ay = –50 N  (Ans.) Here, ‘–ve’ sign indicate that force Ay will act in negative y direction. Now from Eq. 2, –50 + NB cos 45° – 150 = 0 \ NB = 282.84 N  (Ans.) NB cos 45° = 200 Now from Eq. 1, Ax = 282.82 sin 45°  \ Ax = 200 N  (Ans.) Example 23. The 100 N load is attached to one and of a rope and the load is

held at rest by the force T applied to the other end of the rope as shown in figure Ex.23. Assuming pulley as frictionless, determine the reactions at A and force T.

Fig. Ex. 23

+

Solution: Applying conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 23(a),

Equilibrium of System of Forces  2.29 Ay 30° A

Ax T

100N

Fig. Ex. 23(a)

SH = 0: Ax + T cos 30° = 0 ...(Eq.1) \ Ax = –T cos 30° SV = 0: –100 – T sin 30° + Ay = 0 ...(Eq.2) Ay = 100 + T sin 30° SMA = 0: –T × 100 + 100 × 100 = 0  \ T = 100 N  (Ans.) Now from Eq. 1, Ax = – 100 cos 30°  \ Ax = –86.60 N  (Ans.) Here, ‘–ve’ sign indicate that force AX will act in negative x direction. Now from Eq. 2, Ay = 100 + 100 sin 30°  \ Ay = 150 N  (Ans.) Example 24. A 2 m long cantilever of negligible weight is used to support 200 N load as shown in figure Ex. 24. Determine the reactions at A. A

B 2m 200N

Fig. Ex. 24

+

Solution: Applying conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 24(a), Ax SH = 0: Ax = 0  (Ans.) B \

CA

SH = 0: Ay – 200 = 0 Ay = 200 N  (Ans.)

SMA = 0: CA – 200 × 2 = 0 \ CA = 400 N.m (anticlockwise)  (Ans.)

2m 200N

Ay

Fig. Ex. 24(a)

2.30  Engineering Mechanics Statics

Example 25. A 2 m long cantilever of negligible weight is used to support 50 N/m distributed load as shown in figure Ex.25. Determine the reactions at A.

Fig. Ex .25

+

Solution: Applying conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 25(a),

Fig. Ex. 25(a)

Here, load is distributed as 50 N/m over 2 m length total load acting in downward direction is 50 × 2 = 100 N and for moment purpose the total load (as a point load) is assume to act at the centre of distribution as shown by dash line in free body diagram. SH = 0:  Ax = 0  (Ans.) SV = 0: Ay – 50 × 2 = 0  \ Ay = 100 N  (Ans.) SMA = 0: CA – 50 × 2 × 1 = 0  \ CA = 100 N.m (anticlockwise)  (Ans.) Example 26. The bar AB is pinned at A and rest on roller at B. The bar AB is loaded as shown in figure Ex.26. Neglecting the weight of the bar, determine the reactions at A and B. 10N 30° 15Nm B

A 1m

1m

1m

Fig. Ex. 26 +

Solution: Applying conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 26(a),

Equilibrium of System of Forces  2.31



SH = 0: Ax – 10 cos 30° = 0

\

Ax = 8.66 N  (Ans.)



SV = 0: Ay + By – 10 sin 30° = 0 ...(Eq.1)



SMA = 0: By × 3 – 15 – 10 sin 30° × 1 = 0

Fig. Ex. 26(a)

\ By = 6.67 N  (Ans.)

Now from Eq. 1, Ay + 6.67 – 10 sin 30° = 0  \ Ay = –1.67 N  (Ans.) Here, ‘–ve’ sign indicate that force Ay will act in negative y direction. Example 27. A bracket is pinned at A and rest on roller at B. It is subjected to a 100 N force as shown in figure Ex.27. Neglecting the weight of the bracket, determine the reactions at A and B.

Fig. Ex. 27 +

Solution: Applying conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 27(a), 100N SV = 0: Ax + 100 = 0 \

Ax = –100 N  (Ans.)

Here, ‘–ve’ sign indicate that force AX will act in negative x direction. \

SV = 0: Ay + By = 0

...(Eq.1)

SMA = 0: By × 50 – 100 × 170 = 0 By = 340 N  (Ans.)

Now from Eq. 1,

A Ax Ay

Fig. Ex. 27(a)

Ay + 340 = 0 \

Ay = –340 N  (Ans.)

Here, ‘–ve’ sign indicate that force Ay will act in negative y direction.

By

2.32  Engineering Mechanics Statics

Example 28. A bracket is fixed at A and loaded as shown in figure Ex.28. Neglecting the weight of the bracket, determine the reactions at A. 400N 1.5m 300N 200Nm

1.5m

200Nm 2.5m A

Fig. Ex. 28 +

Solution: Applying conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 28(a), SH = 0: Ax – 300 – 200 = 0 \

Ax = 500 N  (Ans.)



SV = 0: Ay – 400 = 0

\ Ay = 400 N  (Ans.)

SMA = 0:

CA – 200 + 200 × 2.5 + 300 × 4 – 400 × 1.5 = 0 \ CA = – 900 N.m = 900 N. m (Clockwise)  (Ans.) Here, ‘–ve’ sign indicate that moment reaction CA will act in opposite direction.

Fig. Ex. 28(a)

Example 29. The weight of the pickup is 3600 N and acting at G as shown in figure Ex.29. Determine the load W on the pickup for which normal reactions at front and rear wheels are equal.

Fig. Ex. 29

Equilibrium of System of Forces  2.33

Solution: Applying conditions of equilibrium i.e. +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 29(a), Here given that normal reactions at front and rear wheels are equal, NA = NB = N \ SV = 0: 2N – 3600 – W = 0

Fig. Ex. 29(a)

\ W = 2N – 3600 SMA = 0: 3600 × 450 – N × 1120 + W × 1280 = 0 Substituting value of W from Eq. 1 3600 × 450 – N × 1120 + (2N – 3600) × 1280 = 0 \ N = 2075 N Now from Eq. 1 W = 2 × 2075 – 3600 \ W = 550  (Ans.)

...(Eq.1)

Example 30. The motorcycle shown in figure Ex.30, has a mass of 150 kg and distance between front and rear wheel is 1.6 m. If the rear wheel exerts a 1000 N force on the ground, determine the location of motorcycle’s centre of gravity from front wheel also determine reaction at front wheel.

A

1.6m

B

Fig. Ex. 30

Solution: Applying conditions of equilibrium i.e. +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to free body diagram shown in figure Ex. 30(a),

2.34  Engineering Mechanics Statics

Fig. Ex. 30(a)

SV = 0: NA + 1000 – 150 × 9.81 = 0  \ NA = 471.5 N  (Ans.) SMA = 0: 1000 × 1.6 – 150 × 9.81 × x = 0  \ x = 1.08 m  (Ans.)

Summary

• Equilibrium is the state of rest of a body i.e. the body does not move and also does not rotate about any point under the action of forces. • For collinear force system equilibrium conditions are, SP = 0  \ Resultant, R = 0 • For concurrent force system equilibrium conditions are, SH = 0 and SV = 0  \ Resultant, R = 0 • For parallel force system equilibrium conditions are, SP = 0 and SMO = 0 • For non-concurrent force system equilibrium conditions are, SH = 0, SV = 0 and SMO = 0 • A free body diagram is a sketch of the body showing all the forces acting on it. • If a body is in equilibrium under the action of two forces then two forces must be collinear, of equal magnitude and act in opposite direction. • If a body is in equilibrium under the action of three forces, then three forces acting on a body must be concurrent or parallel. • Lami’s theorem is useful for three concurrent forces.

Equilibrium of System of Forces  2.35

Problems 1. The system of forces shown in figure Prob. 1 is in equilibrium. Determine P1 and P2.

Y 50 N P1 40°

30° 60°

50° 70 N

P2

Fig. Prob. 1 2. The system of forces shown in figure Prob. 2 is in equilibrium. Determine P1 and P2.

15 N 20° P2

P1 40° 80N

40N

O

Fig. Prob. 2 3. Two cables are tied together and loaded at B as shown in figure Prob. 3. Determine tension in cable AB and cable CB.

A

C 30°

40° B 50 Kg

Fig. Prob. 3 4. A 500 N block is supported by a strut and two cables as shown in figure Prob. 4. Determine the force in strut CB and cable AB.

A

B

20° 60°

500 N

C

Fig. Prob. 4 5. The force P is applied to ring at B as shown in figure Prob. 5. If the tension in cable AB and BC is 150 N, determine magnitude of force P.

Fig. Prob. 5

2.36  Engineering Mechanics Statics 6. Determine the tension developed in each cable to maintain equilibrium of the 800 N load. Refer figure Prob. 6

E A 4 30°

C 3

B D 800 N

45°

Fig. Prob. 6 7. The 500 N roller is supported by the cable AB rests against a smooth wall as shown in figure Prob. 7. Determine force in cable and reaction on roller from the wall. Radius of roller is 150 mm.

A 250mm

B

Fig. Prob. 7 8. Neglecting the weight of bracket, determine all reactions at O due to 160 N force acting on bracket as shown in figure Prob. 8.

2m

3m

60°

160 N

O

Fig. Prob. 8 9. Three cylinders, each of diameter 300 mm and weight 500 N are stacked as shown in figure Prob. 9. Determine reactions exerted on cylinders by the inclined surfaces. 30°

30°

Fig. Prob. 9

Equilibrium of System of Forces  2.37 10. A cantilever is subjected to loading as shown in figure Prob. 10. Determine reactions at the fixed support.

Fig. Prob. 10 11. A bracket is subjected to the loading as shown in figure Prob. 11. Determine force P and reactions at pin support O.

15 N 30° 2m P 3m O 1m

Fig. Prob. 11 12. A beam is subjected to loading as shown in figure Prob. 12. Determine reactions at A and B. Take AC = 1.5 m and CB = 2 m.

20°

80 N

100 N A

C

Fig. Prob. 12 13. A bar of 200 N with roller end is supported by wire CB and two surfaces as shown in figure Prob. 13. Determine reactions at A and B and tension in wire CB.

Fig. Prob. 13

B

2.38  Engineering Mechanics Statics 14 A bracket is subjected to the loading as shown in figure Prob. 14. Determine reactions at A and B.

30 N 400 mm

40 N 400 mm

400 mm

A 300 mm 25 N 300 mm

B 500 mm

Fig. Prob. 14 15. A uniform bar AB of length 1.5 m and weight 200 N resting on two inclined smooth surfaces as shown in figure Prob. 15. Determine angle θ for equilibrium of bar.

A

 45°

Fig. Prob. 15

B 30°

Unit 3

In this unit, we discuss analysis of structures. In first part of this unit, we focus on analysis of truss where we determine forces acting on interconnected members of truss. In second part of this unit, we focus on analysis of beams where we determine internal forces induced in a beam because of external transverse loading. Analysis of these forces is important before design of any members.

Analysis of Structure 3.1 trusses Truss is one of the important engineering structures. Truss is used to support the load. A frame which is composed of members A truss is a structure joined at their ends to form a structure, called a composed of truss. Members of a truss are two-force members interconnected members that are subjected to two equal and opposite joined at their ends. forces directed along member. Truss structures are used in buildings, bridges, roofs, transmission towers etc. Truss structure in two dimensions, is known as plane truss. Basic triangular truss is as shown in figure 1, where members of a truss are connected by a pin joints at their ends. For keeping the truss structure stationary all degrees of freedom must be restricted. A truss which does not collapse under loading is known as rigid or stable truss. For a stable truss, the following relation exists, n = 2 j – 3 Where,  n  is the number of members and j  is the Fig. 1 number of pin joints. A truss for which above relationship does not satisfy is known as unstable truss.

3.2  Types of Truss Some standard types of truss structure are shown in figure 2. Warren truss consists of longitudinal members joined by angled crossmembers that forms alternate inverted equilateral triangle-shaped spaces along its length. Howe truss includes vertical members and diagonals that slope upward towards the centre of its length. Pratt truss includes vertical members and diagonals that slope downward towards the centre of its length. K truss includes vertical members and two tilted members forming K.

3.4  Engineering Mechanics Statics

    Fig. 2

3.3 Analysis of Truss When a truss is loaded its members are subjected to either tensile or compressive forces. In truss analysis we determine these forces induced in members. In figure 3, a member of a truss is shown with tensile and compressive forces.

Fig. 3.

3.3.1 Assumptions in Truss Analysis Following assumptions are made in analysis of a truss, 1. All members of a truss are connected by frictionless pins. 2. Load or forces may be applied at joints only. 3. Each member may have two joints only. 4. Weights of members are assumed to be negligible. 5. The truss is a stable truss.

3.3.2 Methods of Truss Analysis There are commonly two methods that can be used for the analysis of a truss, 1. Method of joints 2. Method of sections 1. Method of Joints: In method of joints each joint of the members are treated as a particle. Forces induced in each members of a truss are determine as follow, • First, if necessary, determine reactions at the supports considering entire truss as a single object.

Analysis of Structure  3.5





• Isolate an individual joint and draw free body diagram by assuming forces in members as a tensile force. • Apply conditions of equilibrium i.e. SH = 0, and SV = 0 to the free body diagram of the joint. • If the calculated value from equilibrium equations is positive then our assumption of tensile force in member is valid. If the calculated value from equilibrium equations is negative then our assumption of tensile force in member is not valid and nature of the force in member is compressive. • Repeat this process for all joints and determine forces in all the members of a truss. 2. Method of Sections: In method of sections forces induced only in certain members of a truss are determine as follow, • First, if necessary, determine reactions at the supports considering entire truss as a single object. • Select a section of a truss by cutting maximum three members. • Draw the free body diagram of the section by showing axial forces in the members (away from joints). • Apply conditions of equilibrium i.e. SH = 0, SV = 0 and SM = 0 to the free body diagram of the section.

Solved Examples Based on Method of Joints Example 1. Determine the forces in all the members of a truss with the loading and support system shown in figure Ex. 1.

Fig. Ex. 1

Solution: To determine reactions, consider the free body diagram of the entire truss as shown in figure Ex. 1(a). The triangle PQR is a right angled triangle with angle PRQ = 90°

3.6  Engineering Mechanics Statics

PR = PQ cos 60° = 6 × 0.5 = 3 m Distance of line of action of 50 kN force from P, PS = PR cos 60° = 3 × 0.5 = 1.5 m Now apply conditions of equilibrium i.e. SM = 0 about P, (Treating anticlockwise moment as positive) Fig. Ex. 1(a) SMP = 0:  RQ × 6 – 50 × 1.5 = 0 \ RQ = 12.5 kN Now apply conditions of equilibrium i.e. +↑SV = 0 to free body diagram SV = 0 : RP + RQ = 50 kN Substituting RQ = 12.5 kN, RP = 37.5 kN +

Joint P: Now apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint P as shown in figure Ex. 1(b). ...(Eq. 1) SH = 0: FPQ + FPR cos 60° = 0 SV = 0: FPR sin 60° + Rp = 0 FPR = −

Rp sin 60°

= – 43.3 kN (Compression)  (Ans.)

Substituting in Eq. 1 FPQ + (–43.3) cos 60° = 0 Fig. Ex. 1(b) FPQ = 21.65 kN (Tensile)  (Ans.) Joint Q: Now apply conditions of equilibrium i.e. +↑SV = 0 to free body diagram of joint Q as shown in figure Ex. 1(c). SV = 0: FRQ sin 30° + 12.5 = 0 12.5 = – 25 kN (Compression)  (Ans.) FQR = – sin 30° Example 2. Use the method of joints to determine the force in each member of the truss shown in figure Ex. 2.

Fig. Ex. 2

Fig. Ex. 1(c)

Analysis of Structure  3.7

Solution: For this Simple truss, the member forces can be determined without solving the support reactions. +

Joint B: Apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint B as shown in figure Ex. 2(a). SH = 0: FBC cos 40° – 1500 – FAB cos 50° = 0 ...(Eq. 1) SV = 0: – FAB sin 50° – FBC sin 40° = 0 FAB sin 50° = –FBC sin 40° Fig. Ex. 2(a) ....(Eq. 2) FAB = –0.84 FBC Substituting the value of FAB in Eq. 1 FBC cos 40° – 1500 – (–0.84 FBC) cos 50° = 0 Solving, FBC = 1145.03 kN (Tensile)  (Ans.) From Eq. 2, FAB = –0.84 × 1145.03 = –961.83 kN (Compressive)  (Ans.) +

Joint C: Apply conditions of equilibrium i.e. →ΣH = 0 to free body diagram of joint C as shown in figure Ex. 2(b). SH = 0: – FBC cos 40° – FAC = 0 –1145.03 cos 40° – FAC = 0 FAC = –877.14 kN (Compression)  (Ans.) Example 3. Use the method of joints to determine the force in each member of the truss shown in figure Ex. 3.

Fig. Ex. 2(b)

Fig. Ex. 3

Solution: For this simple truss the member forces can be determined without solving the support reactions.

3.8  Engineering Mechanics Statics +

Joint B: Apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint B as shown in figure Ex. 3(a). 4 3 SH = 0 : – FAB + FBC – 20 cos 45° = 0   ...(Eq. 1) 5 5 SV = 0 : –

3 4 FAB – FBC – 20 sin 45° = 0   ...(Eq. 2) 5 5

Solving Eq. 1 and Eq. 2 yields FAB = –28.28 kN (compression)  (Ans.) FBC = 20.8 kN (Tensile)  (Ans.)

Fig. Ex. 3(a)

+

Joint C: Apply conditions of equilibrium i.e. →ΣH = 0 to free body diagram of joint C as shown in figure Ex. 3(b). 3 SH = 0 : – FBC – FAC = 0 5 FAC = –

3 3 FBC = – × 20.8 = – 12.48kN 5 5

(Compression)  (Ans.) Example 4. Use the method of joints to determine the force in each member of the truss shown in figure Ex. 4

Fig. Ex. 3(b)

Fig. Ex. 4

Solution: For this simple truss the member forces can be determined without solving the support reactions. +

Joint D: Apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint D as shown in figure Ex. 4(a). SH = 0: –FAD + FCD = 0 ...(Eq. 1) SV = 0: FBD – 2000 = 0 FBD = 2000 kN (Tensile)  (Ans.)

Fig. Ex. 4(a)

Analysis of Structure  3.9 +

Joint B: Apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint B as shown in figure Ex. 4(b). 3 4 ...(Eq. 2) SH = 0 : FBC – FAB – 1500 = 0 5 5

SV = 0 : –

Fig. Ex. 4(b)

4 3 FBC – FAB – 2000 = 0 ...(Eq. 3) 5 5

Solving Eq. 2 and Eq. 3 yields, FAB = –2400 kN (Compressive) FBC = –700 kN (Compressive) +

Joint C: Apply conditions of equilibrium i.e. →ΣH = 0 to free body diagram of joint D as shown in figure Ex. 4(c). 3 SH = 0 : – FCD – FBC = 0 5 3 – FCD – (–700) = 0 5 FCD = 420 kN (Tensile)  (Ans.) Now from Eq. 1, – FAD + 420 = 0 FAD = 420 kN (Tensile)  (Ans.)

Fig. Ex. 4(c)

Example 5. Use the method of joints to determine the force in each member of the truss shown in figure Ex. 5.

Fig. Ex. 5

Solution: To determine reactions, consider the free body diagram of the entire +

truss as shown in figure Ex. 5(a) and apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 about A, (Treating anticlockwise moment as positive)

3.10  Engineering Mechanics Statics

Fig. Ex. 5(a)

SH = 0: Ax = 0 SV = 0: Ay + Dy – 5 – 7 = 0 Ay + Dy = 12 SMA = 0: Dy × 8 – 7 × 6 – 5 × 2 = 0 Dy = 6.5 kN From Eq. 1, Ay + 6.5 = 12  Ay = 5.5 kN

...(Eq. 1)

+

Joint D: Apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint D, as shown in figure Ex. 5(b). SH = 0: –FDE – FCD cos 60° = 0 ...(Eq. 2)

SV = 0: FCD sin 60° + 6.5 = 0 FCD = –7.50 kN (Compressive)  (Ans.)

From Eq. 2, –FDE – (–7.50) cos 60° = 0

Fig. Ex. 5(b)

FDE = 3.752 kN (Tensile)  (Ans.) +

Joint A: Apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint A as shown in figure Ex. 5(c). SH = 0: FAE + FAB cos 60° = 0  ...(Eq. 3)

SV = 0: FAB sin 60o + 5.5 = 0

FAB = –6.35 kN (Compressive)  (Ans.) Substituting the value in Eq. 3, FAE + (–6.35) cos 60° = 0 FAE = 3.175 kN (Tensile)  (Ans.)

Fig. Ex. 5(c)

Analysis of Structure  3.11 +

Joint B: Apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint B as shown in figure Ex. 5(d). SH = 0: FBC + FBE cos 60° – FAB cos 60° = 0  ...(Eq. 4) SV = 0: –FAB sin 60° – FBE sin 60° – 5 = 0 Substituting value of FAB, –(-6.35) sin 60° – FBE sin 60° – 5 = 0 FBE = 0.576 kN (Tensile)  (Ans.)

Fig. Ex. 5(d)

Substituting value of FAB and FBE in Eq. 4, FBC + (0.576) cos 60° – (–6.35) cos 60° = 0 FBC = –3.4763 kN (Compressive)  (Ans.) +

Joint C: Apply conditions of equilibrium i.e. →ΣH = 0 to free body diagram of joint C as shown in figure Ex. 5(e). SH = 0: FCD cos 60° – FCE cos 60° – FBC = 0 Substituting value of FCS and FBC (–7.5) cos 60° – FCE cos 60° – (–3.476) = 0 FCE = – 0.548 kN (Compressive)  (Ans.)

Fig. Ex. 5(e)

Example 6. Use the method of joints to determine the force in each member of the truss shown in figure Ex. 6.

Fig. Ex. 6

Solution: For this simple truss the member forces can be determined without solving the support reactions. +

Joint A: Apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 to free body diagram of joint A as shown in figure Ex. 6(a). SH = 0: FAE + FAB cos 30° = 0 ...(Eq. 1) SV = 0: FAB sin 30° – 800 = 0 FAB = 1600 kN (Tensile)  (Ans.)

3.12  Engineering Mechanics Statics

Substituting the value in Eq. 1, FAE + 1600 cos 30° = 0 FAE = –1385.64 kN (Compressive)  (Ans.) Joint B: Apply conditions of equilibrium i.e., +SH = 0 and +SV = 0 to free body diagram of joint B as shown in figure Ex. 6(b). SH = 0: FBC – FAB = 0 FBC = FAB = 1600 kN (Tensile)  (Ans.) SV = 0: –FBE – 650 = 0 FBE = –650 kN (Compressive)  (Ans.)

Fig. Ex. 6(a)

+

Joint E: Apply conditions of equilibrium i.e. →ΣH = 0 and +↑SV = 0 to free body diagram of joint E as shown in figure Ex. 6(c). SH = 0: FEF – FEA – FBE cos 60° + FCE cos 60° = 0 SV = 0: FBE sin 60° + FCE sin 60° – 1000 = 0 Substituting value of FBE, –650 sin 60° + FCE sin 60° = 1000 FCE = 1804.70 kN (Tensile)  (Ans.) Substituting values of FEA, FBE and FCE in Eq. 2 FEF – (–1385.64) – (–650) cos 60° + (1804.70) cos 60° = 0 FEF = –2613 kN (Compressive)  (Ans.)

Fig. Ex. 6(b)

...(Eq. 2)

Fig. Ex. 6(c)

+

Joint C: Apply conditions of equilibrium i.e. →ΣH = 0 and +↑SV = 0 to free body diagram of joint C as shown in figure Ex. 6(d).

Fig. Ex. 6(d)

SH = 0: FCD + FCF cos 60° – FCE cos 60° – FCB cos 30° = 0 SV = 0: –FCF sin 60° – FCE sin 60° – FCB sin 30° = 0 Substituting value of FCE and FCB, –FCF sin 60° – 1804.70 sin 60° – 1600 sin 30° = 0 FCF = –2728.45 kN (Compressive)  (Ans.) Substituting values of FCF, FCE and FCB in Eq. 3 FCD + (–2728.45) cos 60° – 1804.70 cos 60° – 1600 cos 30° = 0 FCD = 3652.21 kN (Tensile)  (Ans.)

...(Eq. 3)

Analysis of Structure  3.13

Solved Examples Based on Method of Sections Example 1. The length of each truss member in figure Ex. 1 is 7 m. Find the

forces in members CD, DE & EF using method of sections.

Fig. Ex. 1

Solution: To determine reactions, consider the free body diagram of the entire +

truss as shown in figure Ex. 1(a) and apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 about A, (Treating anticlockwise moment as positive)

Fig. Ex. 1(a)

SH = 0: Ax = 0 SV = 0: Ay + By –1200 – 1000 – 800 – 1000 = 0 ...(Eq.1) Ay + By = 4000 SMA = 0: By × 35 – 1200 × 7 – 1000 × 14 – 800 × 21 – 1000 × 28 = 0 By = 1920 kN From Eq. 1, Ay + 1920 = 4000  Ay = 2080 kN Now consider free body diagram of the cut part of the truss to the left section of member DF as shown in figure Ex. 1(b) and apply conditions of equilibrium i.e. +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive)

3.14  Engineering Mechanics Statics

Fig. Ex. 1(b)

SME = 0: 1200 × 7 – 2080 × 14 – FCD × (7 Sin 60°) = 0 FCD = –3418 kN (Compressive)  (Ans.) SMD = 0: 1000 × 3.5 + 1200 × 10.5 – 2080 × 17.5 + FEF × (7 Sin 60°) = 0 FEF = 3348.6 kN (Tensile)  (Ans.) SV = 0: FDE sin 60° – 1000 – 1200 + 2080 = 0 FDE = 138.56 kN (Tensile)  (Ans.) Example 2. Find the forces in members CD, CF, and FG of the truss shown in figure Ex. 2 using method of sections.

Fig. Ex. 2

Solution: To determine reactions, consider the free body diagram of the entire truss as shown in figure Ex. 2(a) and apply conditions of equilibrium SM = 0 about A, (Treating anticlockwise moment as positive)

Analysis of Structure  3.15

SMA = 0: Ey × 30 – 1000 × 10 – 1500 × 20 = 0 Ey = 1333.33 kN Now consider free body diagram of the cut part of the truss to the right section of member CG as shown in figure Ex. 2(b) and apply conditions of equilibrium i.e. SM = 0 (Treating anticlockwise moment as positive),

Fig. Ex. 2(b)

SME = 0: –FCF sin 60° × 10 + 1500 × 10 = 0 FCF = 1732.10 kN (Tensile)  (Ans.) SMF = 0: FCD × (10 sin 30°) + 1333.33 × 10 = 0 FCD = –2666.66 kN (Compressive)  (Ans.) SMC = 0: –FFG × (15 tan 30°) – 1500 × 5 + 1333.33 × 15 = 0 FFG = 1443.49 kN (Tensile)  (Ans.) Example 3. Find the forces in members BC, BF and EF of the truss shown in figure Ex. 3 using the member of sections.

3.16  Engineering Mechanics Statics

Fig. Ex. 3

Solution: For this truss, the forces in members can be determined without solving for the support reactions. Now consider free body diagram of the cut part of the truss to the right section of member BE as shown in figure Ex. 3(a) and apply conditions of equilibrium i.e. +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive), −1 7 = 54.46° q = tan 5 SMB = 0: –FEF × 7 – 5 × 10 = 0 FEF = –7.14 kN (Compressive)  (Ans.) SMF = 0: FBC × 7 – 5 × 5 = 0 FBC = 3.57 kN (Tensile)  (Ans.) SV = 0: FBF sin 54.46° – 5 = 0 FBF = 6.14 kN (Tensile)  (Ans.)

Fig. Ex. 3(a)

Example 4. Find the forces in members AB and FG of the truss shown in figure Ex. 4, using method of sections.

Fig. Ex. 4

Analysis of Structure  3.17

Solution: For this truss, the forces in members can be determined without solving for the support reactions. Now consider free body diagram of the cut part of the truss to the right section as shown in figure Ex. 4(a) and apply conditions of equilibrium i.e. SM = 0 (Treating anticlockwise moment as positive),

Fig. Ex. 4(a)



SMF = 0: FAB × 5 – 6 × 3 – 4 × 6 = 0 FAB = 8.4 kN (Tensile)  (Ans.)



SMB = 0: – FFG × 5 – 6 × 3 – 4 × 6 = 0



FFG = –8.4kN (Compression)  (Ans.)

Example 5. Find the forces in members BC, CF and EF of the truss shown in figure Ex. 5. using method of sections.

Fig. Ex. 5

Solution: To determine reactions, consider the free body diagram of the entire truss as shown in figure Ex. 5(a) and apply conditions of equilibrium SM = 0 about A, (Treating anticlockwise moment as positive)

3.18  Engineering Mechanics Statics

Fig. Ex. 5(a)

SM = 0: Dy × 48 – 50 × 12 – 35 × 36 = 0 Dy = 38.75 kN Now consider free body diagram of the cut part of the truss to the right of joint F as shown in figure Ex. 5(b) and apply conditions of equilibrium i.e. SM = 0 (Treat ing anticlockwise moment as positive),

Fig. Ex. 5(b)



−1 q = tan

6 = 26.57° 12

SMF = 0: FBC × 6 – 35 × 12 + 38.75 × 24 = 0 FBC = –85 kN (Compressive)  (Ans.) SMC = 0: –FEF × (12 cos 26.57°) + 38.75 × 12 = 0 FEF = 43.33 kN (Tensile)  (Ans.) SMB = 0: –FCF × (24 sin 26.57°) – 35 × 24 + 38.75 × 36 = 0 FCF = 51.73 kN (Tensile)  (Ans.)

Analysis of Structure  3.19

SUMMARY

• A truss is a structure composed of interconnected members joined at their ends. • Truss structure in two dimensions, is known as plane truss. • A truss which does not collapse under loading is known as rigid or stable truss. • For a stable truss, the following relation exists, n = 2 j – 3 Where, n is the number of members and j is the number of pin joints. • A truss for which above relationship does not satisfy is known as unstable truss. • In truss analysis we determine forces induced in members due to loading. • When a truss is loaded its members are subjected to either tensile or compressive forces. • There are commonly two methods that can be used for the analysis of a truss. Method of joints: This method is used to determine forces induced in all members of a truss Method of sections: This method is used to determine forces induced only in certain members of a truss. • A truss member that supports no forces is known as zero force member.

3.4 Beam Beam is a slender* structural member subjected to lateral loads that is, line of action of forces perpendicular to the axis. Internal forces are developed inside structural member when they are subjected to external forces. Knowledge of internal forces that structural member must support is important before the members are designed. * A structural member is said to be slender when the dimensions of its cross section are small compared to its length.

3.5 Types of Beams Beams are generally designated by the way in which they are supported. Table 3.1 describe different types of beam and their free body diagram, which shows reactions at support.

3.20  Engineering Mechanics Statics Table 3.1 Types of Beam Types of Beam Simply Supported Beam

Description

Example and Free Body Diagram

A beam with a pin support at one end (A) and a roller support at the other (B). Horizontal and vertical reactions at pin support and vertical reaction at roller support.

Cantilever Beam A beam which is fixed at one end (A) and free at the other. Horizontal, vertical and moment reactions at fixed support.

Beam with an Overhang

A beam which is simply supported at one end (A) intermediate support/supports (B) and also projects beyond the support.

3.6 Types of loading All types of beam discussed in previous section are subjected to one or more than one types of loading. Table 3.2 describe different types of loading on the beam. Table 3.2 Types of Loading on Beam Types of Loading Concentrated Load or Point Load

Description Concentrated load is acts over a very small area. Its intensity is N or kN.

Uniformly Uniformly distributed load Distributed Load acts uniformly over a finite area of the beam. Its intensity is expressed in per unit length i.e. N/m, kN/m etc. For finding reactions, this load is assumed as total load (product of intensity of distributed load and spreading distance) acting at the centre of gravity of distribution of load.

Example

Analysis of Structure  3.21 Uniformly Varying Load

Uniformly varying load varies linearly over a finite area of the beam. Its intensity is zero at one end, and maximum at other end of distribution which is expressed in per unit length i.e. N/m, kN/m etc. For finding reactions, this load is assumed as total load (total area of distribution i.e. area of the triangle) acting at the centre of gravity of distribution of load i.e. centroid of the triangle.

Concentrated Moment

Concentrated moment is the external moment acts at certain points on the beam.

3.7 Shear Force (V) and Bending Moment (M) Shear force and bending moment are internal forces, and internal moment induced within beam because of loading. To determine the forces and moments within the beam, we ‘cut’ the beam by plane at an arbitrary cross section and isolate left-hand part of the beam as a free Fig. 3.1(a) body from right-hand part of the beam. As an illustration, consider a cantilever beam AB loaded by a force P at its free end as shown in Figure 3.1 (a). We cut the beam at cross section C at distance x from fixed end, and isolate left-hand part as free Fig. 3.1(b) body as shown in Figure 3.1 (b). To maintain equilibrium of the free body, it is subjected to some system of forces and moments at the cut cross section as shown. The component V perpendicular to the beam’s axis is called the shear force and the couple M is called the bending moment. (Because the load P is perpendicular to the axis of Fig. 3.1(c) the beam, no axial force i.e. force parallel to the beam’s axis exists at the cross section). Both V and M act in the plane of the beam. The shear force and bending moment on the part of the beam to the right of the cross section C are shown in Figure 3.1 (c). Note that V and M are equal in magnitude but opposite in direction to the internal forces and moment on the free body diagram as shown in Figure 3.1 (b).

3.22  Engineering Mechanics Statics

3.7.1 Sign Convention For the sake of consistency, it is required to adopt sign conventions for shear force and bending moment. The direction of the shear force and bending moment shown in Figure 3.1 (b) and Figure 3.1 (c) are the established positive directions of these quantities. A positive shear force V tends to rotate beam element clockwise. A positive bending moment M tends to bend beam element concave upward (smiling of the beam).

3.8 Shear Force and Bending Moment Diagrams Shear force and bending moment diagrams are plots of V and M as functions of position x. These diagrams help us understand how the shear force and bending moment change throughout a beam and show locations where these have maximum and minimum values.

Solved Examples Based on Simply Supported Beam Example 1. The simply supported beam AB is subjected to concentrated load as shown in figure Ex. 1. Determine, (a) reactions at supports A and B, and (b) value of shear force and bending moment at C and D.

Fig. Ex. 1 +

Solution: To determine reactions, apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 1(a). SH = 0: Ax = 0 SV = 0: Ay + By – 20 = 0  ...(1) SMA = 0: By × 6 – 20 × 3 = 0 \ By = 10 kN  (Ans.) From Eq. (1), Ay + 10 = 20 Fig. Ex. 1(a) \ Ay = 10 kN  (Ans.) Now to determine shear force and bending moment at C, first cut the beam by a plane at C and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 1(b).

Analysis of Structure  3.23

SH = 0: Ax = 0 SV = 0: Ay – V = 0 10 – V = 0 V = 10 kN  (Ans.) \ Fig. Ex. 1(b) SMC = 0: M – Ay × 2 = 0 M = Ay × 2 = 10 × 2 \ M = 20 kN.m  (Ans.) Now to determine shear force and bending moment at D, first cut the beam by a plane at D and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 1(c). SH = 0: Ax = 0 SV = 0: Ay – 20 – V = 0 V = Ay – 20 = 10 – 20 \ V = –10 kN  (Ans.) SMD = 0: M + 20 × 2 – Ay × 5 = 0 M = Ay × 5 – 20 × 2 Fig. Ex. 1(c) = 10 × 5 – 20 × 2  \ M = 10 kN.m  (Ans.) Alternate Approach: Shear force and bending moment at C, also determine by considering the free body diagram of the part of the beam to the right of the plane. Refer figure Ex. 1(d). SV = 0: V – 20 + By = 0 V = 20 – By = 20 – 10 \ V = 10 kN  (Ans.) SMC = 0: –M – 20 × 1 + By × 4 = 0 M = –20 × 1 + By × 4 = –20 × 1 + 10 × 4 Fig. Ex. 1(d) \ M = 20 kN.m  (Ans.) Shear force and bending moment at D, also determine by considering the free body diagram of the part of the beam to the right of the plane. Refer figure Ex. 1(e). SV = 0: V + By = 0 V = – By  \ V = –10 kN  (Ans.) SMD = 0: –M + By × 1 = 0 Fig. Ex 1(e) M = By × 1 = 10 × 1  \ M = 10 kN.m  (Ans.) Example 2. The simply supported beam AB is subjected to two concentrated load as shown in figure Ex. 2. Determine, (a) reactions at supports A and B, and (b) draw the shear force and bending moment diagrams for this beam.

3.24  Engineering Mechanics Statics

Fig. Ex. 2

Solution: Part 1: To determine reactions, apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV= 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 2(a). SH = 0: Ax = 0 SV = 0: Ay + By – 20 – 30 = 0  ...(1) SMA = 0: By × 5 – 30 × 3.5 – 20 × 1.5 = 0 \ By = 27 N  (Ans.) From Eq. (1), Ay + By = 50  \ Ay + 27 = 50 Fig. Ex. 2(a) Ay = 23 N  (Ans.) Part 2: First cut the beam by a plane at point E between A and C (0 < x < 1.5) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 2(b). SV = 0: 23 – V = 0 \ V = 23 N  (Ans.) SME = 0: M – 23 x = 0 \ M = 23 x Fig. Ex. 2(b) Substituting x = 0 m, \ M = 0 N.m  (Ans.) Substituting x = 1.5 m, \ M = 34.5 N.m  (Ans.) Now cut the beam by a plane at point F between C and D (1.5 < x < 3.5) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 2(c). SV = 0: 23 – 20 – V = 0 \ V = 3 N  (Ans.) SMF = 0: M + 20 (x – 1.5) – 23 x = 0 \ M = 23 x - 20 (x – 1.5) Fig. Ex. 2(c) Substituting x = 3.5 m, M = 23 × 3.5 – 20 (3.5 – 1.5) \ M = 40.5 N.m  (Ans.)

Analysis of Structure  3.25

Next cut the beam by a plane at point G between D and B (3.5 < x < 5) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 2(d). SV = 0: 23 – 20 – 30 – V = 0 \ V = –27 N  (Ans.) SMG = 0: M + 30 (x – 3.5) + 20 (x – 1.5) – 23 x = 0 \ M = 23 x – 20 (x – 1.5) – 30 (x – 3.5)

Fig. Ex. 2(d)

Substituting x = 5 m, M = 23 × 5 – 20 (5 – 1.5) – 30 (5 – 3.5)  \ M = 0 N.m

(Ans.)

Fig. Ex. 2(f)

The shear force and bending moment diagrams are shown in figure Ex. 2(e) and figure Ex. 2(f). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2. Note that for each concentrated force, the shear force diagram jumps by an amount equal to the force, and remain constant between two concentrated forces. The bending moment diagram is an inclined straight line between two concentrated forces.

3.26  Engineering Mechanics Statics

Example 3. The simply supported beam AB is subjected to a concentrated load and a moment as shown in figure Ex. 3. Determine, (a) reactions at supports A and B, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex. 3

Solution: Part 1 +

To determine reactions, apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 3(a). SH = 0: Ax = 0

SV = 0: Ay + By – 25 = 0  ...(1)



SMA = 0: By × 4 – 25 × 1 – 35 = 0

\

By = 15 N  (Ans.)

From Eq. (1),

Fig. Ex. 3(a)

Ay + By = 25 \ Ay + 15 = 25 \

Ay = 10 N  (Ans.)

Part 2: First cut the beam by a plane at point E between A and C (0 < x < 1) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 3(b). SV = 0: 10 – V = 0 \ V = 10 N  (Ans.) \

Fig. Ex. 3(b)

SME = 0: M – 10 x = 0 M =10 x

Substituting x = 0 m, \

M = 0 N.m  (Ans.)

Substituting x = 1 m, \

M = 10 N.m  (Ans.)

Fig. Ex. 3(c)

Analysis of Structure  3.27

Now cut the beam by a plane at point F between C and D (1 < x < 2) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 3(c). Fig. Ex. 3(d)

Fig. Ex. 3(f)

SV = 0: 10 – 25 – V = 0  \ V = –15 N  (Ans.) SMF = 0: M + 25 (x – 1) – 10 x = 0  \ M = 10 x – 25 (x – 1) Substituting x = 2 m, M = 10 × 2 – 25 (2 – 1)  \ M = –5 N.m  (Ans.) Next cut the beam by a plane at point G between D and B (2 < x < 4) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 3(d).

3.28  Engineering Mechanics Statics

SV = 0: 10 – 25 – V = 0  \ V = –15 N  (Ans.) SMG = 0: M – 35 + 25 (x – 1) – 10 x = 0 \ M = 35 – 25 (x – 1) + 10 x Substituting x = 2 m, M = 35 – 25 (2 – 1) + 10 × 2  \ M = 30 N.m  (Ans.) Substituting x = 4 m, M = 35 – 25 (4 – 1) + 10 × 4  \ M = 0 N.m  (Ans.) The shear force and bending moment diagrams are shown in figure Ex. 3(e) and figure Ex. 3(f). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2. Note that for a concentrated moment, the bending moment diagram jumps by an amount equal to the moment. Example 4. The simply supported beam AB is subjected to uniformly distributed load as shown in figure Ex. 4. Determine, (a) reactions at supports A and B, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex. 4

Solution: Part 1: To determine reactions, apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 4(a). As discussed in section 3.6, distributed load is assumed as total load (product of intensity of distributed load and spreading distance) acting at the centre of gravity of distribution of load. SH = 0: Ax = 0 SV = 0: Ay + By – 10 × 5 = 0  ...(1) Fig. Ex. 4(a) SMA = 0: By × 5 – 10 × 5 × 2.5 = 0 \ By = 25 N  (Ans.) From Eq. (1), Ay + By = 50 \ Ay + 25 = 50 \

Ay = 25 N  (Ans.)

Fig. Ex. 4(b)

Part 2: Now cut the beam by a plane at point C between A and B (0 < x < 5) and then apply equilibrium equations to the free body diagram of the part of the beam

Analysis of Structure  3.29

to the left of the plane. Refer figure Ex. 4(b).

Fig. Ex. 4(d)

SV = 0: 25 – 10x – V = 0 \ V = 25 – 10x ...(2) Substituting x = 0 m, \ V = 25 N  (Ans.) Substituting x = 5 m, \ V = –25 N  (Ans.) The location of the section on beam where shear force is zero is found by substituting V = 0 in Eq. (2), x = 2.5 m  x Now, SMC = 0 : M + 10 x   – 25 x = 0 2  x M = 25 x – 10 x   2 Substituting x = 0 m,  \ M = 0 N.m  (Ans.) Substituting x = 5 m,  \ M = 0 N.m  (Ans.) Substituting x = 2.5 m, \

3.30  Engineering Mechanics Statics

 2.5  M = 25 × 2.5 – 10 × 2.5     \ M = 18.75 N.m  (Ans.)  2  The shear force and bending moment diagrams are shown in figure Ex. 4(c) and figure Ex. 4(d). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2. Note that for distributed load, the shear force diagram is an inclined straight line between start and end of distribution. The bending moment diagram is a parabola between start and end of distribution. The maximum bending moment occurs where the shear force is zero. Example 5. The simply supported beam AB is subjected to uniformly varying load as shown in figure Ex. 5. Determine, (a) reactions at supports A and B, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex.5

Solution: Part 1 +

To determine reactions, apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 5(a). As discussed in section 3.6, uniformly varying load is assumed as total load (total area of distribution i.e. area of the triangle) acting at the centre of gravity of distribution of load i.e. centroid of the triangle. Fig. Ex. 5(a) SH = 0: Ax = 0 1 SV = 0 : Ay + By – × 3 × 24 = 0 ...(1) 2 SMA = 0 : By × 3 –

1 2 × 3 × 24 × × 3 = 0 2 3

\ By = 24 N  (Ans.) From Eq. (1), Ay + By = 36 \ Ay + 24 = 36 \ Ay = 12 N  (Ans.)

Fig. Ex. 5(b)

Analysis of Structure  3.31

Part 2: Now cut the beam by a plane at point C between A and B (0 < x < 3) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 5(b). w 24 Let w be the intensity at section C, we have from similar triangle, = or w = 8 x x 3 1 SV = 0 : 12 – × x × 8 x – V = 0 2 V = 12 – 4x2 \ ...(2) Substituting x = 0 m,  \ V = 12 N  (Ans.) Substituting x = 3 m,  \ V = –24 N  (Ans.) The location of the section on beam where shear force is zero is found by substituting V = 0 in Eq. (2),

Fig. Ex. 5(d)

\

x = 1.73 m

 x SMC = 0 : M + 4 x 2 ×   – 12 x = 0 3

2 x M = 12 x – 4 x   3 Substituting x = 0 m,  \ M = 0 N.m  (Ans.) Substituting x = 3 m,  \ M = 0 N.m  (Ans.) Substituting x = 1.73 m,

\

3.32  Engineering Mechanics Statics

 1.73  \  M = 12 × 1.73 – 4 × 1.732 ×    \ M = 13.86 N.m  (Ans.)  3  The shear force and bending moment diagrams are shown in figure Ex. 5(c) and figure Ex. 5(d). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2. Note that for uniformly varying load, the shear force diagram is a parabola between start and end of distribution. The bending moment diagram is a third-degree polynomial between start and end of distribution. The maximum bending moment occurs where the shear force is zero. Example 6. The simply supported beam AB is subjected to uniformly distributed load and concentrated load as shown in figure Ex. 6. Determine, (a) reactions at supports A and B, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex. 6

Solution: Part 1: To determine reactions, apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 6(a). As discussed in section 3.6, distributed load is assumed as total load (product of intensity of distributed load and spreading distance) acting at the centre of gravity of distribution of load. SH = 0: Ax = 0 ...(1) SV = 0: Ay + By – 30 – 15 × 2 = 0 SMA = 0: By × 5 – 30 × 2 – 15 × 2 × 4 = 0 \ By = 36 N  (Ans.) From Eq. (1), Ay + By = 60  \ Ay + 36 = 60 \ Ay = 24 N  (Ans.) Part 2: First cut the beam by a plane at point E between A and C (0 < x < 2) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 6(b).

Fig. Ex. 6(a)

Fig. Ex. 6(b)

Fig. Ex. 6(c)

Analysis of Structure  3.33

SV = 0: 24 – V = 0  \ V = 24 N  (Ans.) SME = 0: M – 24x = 0  \ M = 24x Substituting x = 0 m,  \ M = 0 N.m  (Ans.) Substituting x = 2 m,  \ M = 48 N.m  (Ans.) Now cut the beam by a plane at point F between C and D (2 < x < 3) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 6(c). SV = 0: 24 – 30 – V = 0 \ V = –6 N  (Ans.) SMF = 0: M + 30 (x – 2) – 24x = 0 \ M = 24x – 30 (x – 2) Substituting x = 3 m,  \ M = 42 N.m  (Ans.)

Fig. Ex. 6(e)

Fig. Ex. 6(f)

Fig. Ex. 6(d)

3.34  Engineering Mechanics Statics

Next cut the beam by a plane at point G between D and B (3 < x < 5) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 6(d). SV = 0: 24 – 30 – 15 (x – 3) – V = 0 V = 24 – 30 – 15 (x – 3) \ Substituting x = 5 m,  \ V = –36 N  (Ans.)  x – 3 SMG = 0 : M + 15( x – 3)   + 30( x – 2) – 24 x = 0  2   x – 3 M = 24 x – 30( x – 2) – 15( x – 3)    2  Substituting x = 5 m,  \ M = 0 N.m  (Ans.) The shear force and bending moment diagrams are shown in figure Ex. 6(e) and figure Ex. 6(f). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2. \

Solved Examples Based on Cantilever Beam Example 1. A cantilever beam AB is subjected to concentrated load at free end

B as shown in figure Ex. 1. Determine, (a) reactions at fixed support A, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex. 1

Solution: Part 1: To determine reactions, +

apply conditions of equilibrium i.e. →ΣH = 0, +↑SV= 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 1(a). SH = 0: Ax = 0 SV = 0: Ay – 10 = 0 \ Ay = 10 kN  (Ans.) SMA = 0: CA – 10 × 4 = 0 \ CA = 40 kN.m  (Ans.)

Fig. Ex. 1(a)

Fig. Ex. 1(b)

Analysis of Structure  3.35

Part 2: Now cut the beam by a plane at point C between A and B (0 < x < 4) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 1(b).

Fig. Ex. 1(d)

\

SV = 0: 10 – V = 0  \ V = 10 kN  (Ans.) SMC = 0: M + 40 – 10 x = 0 M =10 x – 40

Substituting x = 0 m,  \ M = – 40 kN.m  (Ans.) Substituting x = 4 m,  \ M = 0 N.m  (Ans.) The shear force and bending moment diagrams are shown in figure Ex. 1(c) and figure Ex. 1(d). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2. Example 2. A cantilever beam AB is subjected to two concentrated load as shown in figure Ex. 2. Determine, (a) reactions at fixed support A, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex. 2

3.36  Engineering Mechanics Statics

Solution: Part 1: To determine reactions, apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 2(a). SH = 0: Ax = 0 SV = 0: Ay – 5 – 8 = 0 \ Ay = 13 kN  (Ans.) SMA = 0: CA – 5 × 2 – 8 × 4 = 0 \ CA = 42 kN.m  (Ans.)

Fig. Ex. 2(a)

Part 2: First cut the beam by a plane at point E between A and C (0 < x < 2) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 2(b). SV = 0: 13 – V = 0  \ V = 13 kN  (Ans.) SME = 0: M + 42 – 13x = 0 \ M = 13x – 42 Substituting x = 0 m, \ M = – 42 kN.m  (Ans.) Substituting x = 2 m, \ M = – 16 kN.m  (Ans.)



Fig. Ex. 2(b)

Fig. Ex. 2(c)

Now cut the beam by a plane at point F between C and D (2 < x < 4) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 2(c).

Fig. Ex. 2(d)

Analysis of Structure  3.37

Fig. Ex. 2(f)



SV = 0: 13 – 5 – V = 0  \ V = 8 kN  (Ans.) SMF = 0: M + 42 + 5 (x – 2) – 13 x = 0

\ M =13 x – 42 - 5 (x – 2) Substituting x = 4 m, M = 0 kN.m  (Ans.) \ Next cut the beam by a plane at point G between D and B (4 < x < 5) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 2(d). SV = 0: 13 – 5 – 8 – V = 0  \ V = 0 kN  (Ans.) \

SMG = 0: M + 42 + 8 (x – 4) + 5 (x – 2) – 13 x = 0 M =13 x – 42 – 8 (x – 4) – 5 (x – 2)

Substituting x = 5 m, \ M = 0 kN.m  (Ans.) The shear force and bending moment diagrams are shown in figure Ex. 2(e) and figure Ex. 2(f). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2.

3.38  Engineering Mechanics Statics

Example 3. A cantilever beam AB is subjected to uniformly distributed load as shown in figure Ex. 3. Determine, (a) reactions at supports A, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex. 3

Solution: Part 1: To determine reactions, apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 3(a).

Fig. Ex. 3(a)

As discussed in section 3.6, distributed load is assumed as total load (product of intensity of distributed load and spreading distance) acting at the centre of gravity of distribution of load. SH = 0: Ax = 0 \ \

SV = 0: Ay – 20 × 5 = 0 Ay = 100 N  (Ans.) SMA = 0: CA – 20 × 5 × 2.5 = 0 CA = 250 N.m  (Ans.)

Part 2: Now cut the beam by a plane at point C between A and B (0 < x < 5) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 3(b). SV = 0: 100 – 20 x – V = 0 V = 100 – 20 x \ Substituting x = 0 m, \ V = 100 N  (Ans.) Fig. Ex. 3(b)

Analysis of Structure  3.39

Substituting x = 5 m, \ V = 0 N  (Ans.) \

 x SMC = 0 : M + 250 + 20 x   – 100 x = 0 2  x M = 100 x – 20 x   – 250  2

Substituting x = 0 m, M = –250 N.m  (Ans.) \ Substituting x = 5 m, \ M = 0 N.m  (Ans.)

Fig. Ex. 3(d)

The shear force and bending moment diagrams are shown in figure Ex. 3(c) and figure Ex. 3(d). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2.

3.40  Engineering Mechanics Statics

Example 4. A cantilever beam AB is subjected to uniformly varying load as shown in figure Ex. 4. Determine, (a) reactions at supports A, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex.4

Solution: Part 1 +

To determine reactions, apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 4(a).

Fig. Ex. 4(a)

As discussed in section 3.6, uniformly varying load is assumed as total load (total area of distribution i.e. area of the triangle) acting at the centre of gravity of distribution of load i.e. centroid of the triangle. SH = 0: Ax = 0 1 SV = 0: Ay – × 12 × 60 = 0 2 \ Ay = 360 N  (Ans.)

SMA = 0: C A –

1 1 × 12 × 60 × × 12 = 0 2 3

\ CA = 1440 N.m  (Ans.)

Fig. Ex. 4(b)

Analysis of Structure  3.41

Fig. Ex. 4(d)

Part 2: Now cut the beam by a plane at point C between A and B and then apply equilibrium equations to the free body diagram of the part of the beam to the right of the plane. Refer figure Ex. 4(b). Here we measure x from B towards C. 1 SV = 0: V – × 5 x × x = 0 2 \

V =

1 × 5x × x 2

Substituting x = 0 m, \ V = 0 N  (Ans.) Substituting x = 12 m, \ V = 360 N  (Ans.)

3.42  Engineering Mechanics Statics

\

SMC = 0: – M –

1  x × 5x × x   = 0 2 3

1  x M = – × 5 x × x   2 3

Substituting x = 0 m, M = 0  (Ans.) \ Substituting x = 12 m, \ M = –1440 N.m  (Ans.) The shear force and bending moment diagrams are shown in figure Ex. 4(c) and figure Ex. 4(d). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2. Example 5. A cantilever beam AB is subjected to uniformly distributed load and a concentrated load as shown in figure Ex. 5. Determine, (a) reactions at supports A, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex. 5

Solution: Part 1: To determine reactions, apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 5(a). As discussed in section 3.6, distributed load is assumed as total load (product of intensity of distributed load and spreading distance) acting at the centre of gravity of distribution of load. SH = 0: Ax = 0 SV = 0: Ay – 10 × 3 – 5 = 0 \ Ay = 35 kN  (Ans.) Fig. Ex. 5(a) SMA = 0: CA – 10 × 3 × 1.5 – 5 × 6 = 0 \ CA = 75 kN.m  (Ans.) Part 2: First cut the beam by a plane at point D between A and C (0 < x < 3) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 5(b). SV = 0: 35 – 10 x – V = 0 \ V = 35 – 10 x

Analysis of Structure  3.43

Substituting x = 0 m, V = 35 kN  (Ans.) \ Substituting x = 3 m, \ V = 5 kN  (Ans.)  x SMD = 0: M + 75 + 10 x   – 35 x = 0 2 Fig. Ex. 5(b) x   \ M = 35 x – 10 x   – 75 2 Substituting x = 0 m, M = –75 kN.m  (Ans.) \ Substituting x = 3 m, \ M = –15 kN.m  (Ans.) Now cut the beam by a plane at point E between A and B (3 < x < 6) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 5(c). SV = 0: 35 – 10 × 3 – V = 0  \ V = 5 kN (Ans.) SME = 0: M + 75 + 10 × 3 × (x – 1.5) – 35 x = 0 \ M = 35x – 75 – 10 × 3 × (x – 1.5) Substituting x = 6 m, \ M = 0 kN.m  (Ans.)

Fig. Ex. 5(c)

Fig. Ex. 5(e)

3.44  Engineering Mechanics Statics

The shear force and bending moment diagrams are shown in figure Ex. 5(d) and figure Ex. 5(e). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2.

Solved Examples Based on Overhang Beam Example 1. A overhanging beam AC is subjected to a concentrated load as shown in figure Ex. 1. Determine, (a) reactions at supports A and B, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex. 1

Solution: Part 1: To determine reactions, apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 1(a). SH = 0: Ax = 0 SV = 0: Ay + By – 10 = 0  ...(1) SMA = 0: By × 3 – 10 × 5 = 0 \ By = 16.67 kN  (Ans.) From Eq. (1), Fig. Ex. 1(a) Ay + By = 10 \ Ay + 16.67 = 10 \ Ay = –6.67 kN  (Ans.) Here ‘–ve’ sign indicate that, reaction at A will act in negative y direction. Part 2: First cut the beam by a plane at point D between A and B (0 < x < 3) and then apply equilibrium equations to the free body diagram of the part of the beam to the left Fig. Ex. 1(b) of the plane. Refer figure Ex. 1(b).

Fig. Ex. 1(c)

Analysis of Structure  3.45

SV = 0: –6.67 – V = 0  \ V = –6.67 kN  (Ans.) SMD = 0: M + 6.67 x = 0  \ M = –6.67 x Substituting x = 0 m, \ M = 0 kN.m  (Ans.) Substituting x = 3 m, \ M = –20 kN.m  (Ans.)

Fig. Ex. 1(e)

Now cut the beam by a plane at point E between B and C (3 < x < 6) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 1(c). SV = 0: –6.67 + 16.67 – V = 0 \ V = 10 kN  (Ans.) \

SME = 0: M + 6.67 x – 16.67 (x – 3) = 0 M = 16.67 (x – 3) – 6.67 x

Substituting x = 5 m,

M = 16.67 (5 – 3) – 6.67 × 5

\

M = 0 kN.m  (Ans.)

The shear force and bending moment diagrams are shown in figure Ex. 1(d) and figure Ex. 1(e). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2.

3.46  Engineering Mechanics Statics

Example 2. A overhanging beam AC is subjected to a concentrated load and uniformly distributed load as shown in figure Ex.2. Determine, (a) reactions at supports A and B, and (b) draw the shear force and bending moment diagrams for this beam.

Fig. Ex.2

Solution: Part 1: To determine reactions, +

apply conditions of equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the entire beam as shown in figure Ex. 2(a). SH = 0: Ax = 0

Fig. Ex. 2(a)

SV = 0: Ay + By – 20 × 4 – 10 = 0



...(1)

SMA = 0: By × 4 – 20 × 4 × 2 – 10 × 6 = 0



By = 55 N  (Ans.)

\ From Eq. (1),

Ay + By = 90

\

Ay + 55 = 90

\

Ay = 35 N  (Ans.)

Part 2: First cut the beam by a plane at point D between A and B (0 < x < 4) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 2(b). SV = 0: 35 – 20x – V = 0 V = 35 – 20x ...(2) \ Substituting x = 0 m, \ V = 35 N  (Ans.) Substituting x = 4 m, Fig. Ex. 2(b) \ V = –45 N  (Ans.) The location of the section on beam where shear force is zero is found by substituting V= 0 in Eq. (2), \ x = 1.75 m

Analysis of Structure  3.47

 x SMD = 0: M + 20 x   – 35 x = 0 2  x \ M = 35 x – 20 x   2 Substituting x = 0 m, \ M = 0 N.m  (Ans.) Substituting x = 4 m, \ M = –20 N.m  (Ans.) Fig. Ex. 2(c) Substituting x = 1.75 m, \ M = 30.625 N.m  (Ans.) Now cut the beam by a plane at point E between B and C (4 < x < 6) and then apply equilibrium equations to the free body diagram of the part of the beam to the left of the plane. Refer figure Ex. 2(c).

Fig. Ex. 2(e)

3.48  Engineering Mechanics Statics

SV = 0: 35 – 20 × 4 + 55 – V = 0  \ V = 10 N  (Ans.) SME = 0: M – 55 (x – 4) + 20 × 4 (x – 2) – 35 x = 0 \ M = 35 x – 20 × 4 (x – 2) + 55 (x – 4) Substituting x = 6 m, \ M = 0 kN.m  (Ans.) The shear force and bending moment diagrams are shown in figure Ex. 2(d) and figure Ex. 2(e). These are the plots of V and M as a function of position x for different segments of beam derived in Part 2.

SUMMARY

• Beam is a slender structural member subjected to lateral loads. • Simply supported beam, Cantilever, beam with overhang, Fixed beam etc are different types of beam. • Shear Force (V) and Bending Moment (M) are internal forces and moment induced in a beam. • Internal forces are forces and moments develop within beam due to external loading. • A positive shear force V tends to rotate beam element clockwise. A positive bending moment M tends to bend beam element concave upward. • Shear force and bending moment diagrams are plots of V and M as functions of position x.

Problem Determine shear force and bending moment equations for different sections of the beam and draw complete shear force and bending moment diagrams for the beam shown in figure Prob. 1 to figure Prob. 8.

Fig. Prob. 1

Fig. Prob. 2

Fig. Prob. 3

Fig. Prob. 4

Analysis of Structure  3.49

Fig. Prob. 5

Fig. Prob. 6

Fig. Prob. 7 Fig. Prob. 8

Unit 4

In this unit, we discuss concept of centroid and moment of inertia. In first part of this unit, we focus on center of gravity and centroid and show how to determine location of centroid for composite areas. In second part of this unit, we focus on area moment of inertia that measures how an area is distributed about particular axes. These concepts are useful in distributed force system. We show how to determine area moment of inertia of composite areas.

Centroid and Moment of Inertia 4.1 Introduction The force of attraction exerted by the earth on body is known as force of gravity or weight of the body, denoted by W. This force on free body diagram is represented at the center The weight of the body of gravity of the body. Actually a rigid body is is distributed force. collection of particles and earth exerts a force on each of the particles. Thus weight of the body is distributed over the entire volume of the object, but it is represented by single equivalent force. It is convenient to find resultant of distributed force system so that all the forces may be assumed to be concentrated on a point. Distributed quantities like length, area, mass and weight may be assumed to be concentrated on a point for the ease in the analysis. This point for length and area is called as Centroid, for mass, is called as Center of Mass and for weight, is called as Center of Gravity. Center of Gravity Center of gravity of a body is a point through which the resultant of the gravity forces act C.G. is the average irrespective of the orientation of the body. For position of a distribution symmetrical object C.G. would be in the exact of weight. center of object. However for non-symmetrical objects C. G. would be in any number of positions, depending on weight distribution. The location of the C.G. remains fixed as long as the body does not change shape. If an object’s shape changes, the location of the C.G. also changes. Center of Mass Center of mass of a body is a point where the entire mass of a body is concentrated. In a uniform gravitational field the center of gravity coincides with center of mass. The center of mass may lie outside the object. Centroid Centroid is geometrical center of area or shape. It is also defined as a point where the whole area of the shape is assumed to be concentrated.

4.2 Determination of Centroid To determine mathematically the coordinates of centroid, we apply principle of moment. Consider the area A shown in figure 1. Let area ‘A’ is composed of a number of small areas a1, a2, a3, a4, .... etc.

4.4  Engineering Mechanics Statics

\ A = a1 + a2 + a3 + a4 + ... + an Let, x1 = the distance of the centroid of the area a1 from y axis x2 = the distance of the centroid of the area a2 from y axis x3 = the distance of the centroid of the area a3 from y axis x4 = the distance of the centroid of the area a4 from y axis The moments of all small areas about y axis is = a1 x1 + a2 x2 + a3 x3 + a4 x4 + ... Let G be the centroid of the total area whose distance from the y axis is x Y a1

a a4 a2 3

AA G

x1 x2 x3 x4 x X

Fig. 1

Then the moment of total area about y axis = Ax The moment of all small areas about the y axis must be equal to the moment of total area about the same axis. a1 x1 + a2 x2+ a3 x3 + a4 x4 + ... = Ax

x =

a1 x1 + a2 x2 + a3 x3 + a4 x4 + .... A

In the same way if we apply principle of moment about the x axis, then y = a1 y1 + a2 y2 + a3 y3 + a4 y4 + .... A \ Using following expressions centroid of any given irregular shape may be found. n



x =



y =

∑ i=1 xi Ai n ∑ i=1 Ai n ∑ i=1 yi Ai n ∑ i=1 Ai

...(Eq. 4.1)

...(Eq. 4.2)

By taking the limits of Eq. 4.1 and Eq. 4.2, as Ai → 0, the summations become integrals ∫ x dA and y = ∫ y dA x = ...(Eq. 4.3) ∫ dA ∫ dA

Centroid and Moment of Inertia  4.5

Here x and y are the location of the centroid of area element dA. is known as first moment of area about the y axis and moment of area about the x axis.

∫ y dA

∫ x dA

is known as first

4.3 Centroid of Wire Centroid of a linear element like when wire bent into some shape may be found using expression; l x + l x + .... + ln xn ∑ lx = X = 1 1 2 2 ...(Eq. 4.4) l1 + l2 + ..... + ln ∑l

Y =

l1 y1 + l2 y2 + ..... + ln yn ∑ ly = l1 + l2 + ..... + ln ∑l

...(Eq. 4.5)

where l1, l2, ..., ln are the lengths of such regular line elements for which centroid coordinates (x1, y1), (x2, y2),..., (xn, yn) are known.

4.4  Procedure to Determine Coordinates of Centroid of Composite Areas Many engineering objects can be considered as composite bodies made up of connected simple shapes like a rectangle, triangle, circle and semicircle. Knowing the location of the centroid of the simple shapes, we can easily determine the location of the centroid of complex composite areas. 1. Divide the given area into different simple shapes having known centrodial distance. 2. Establish the coordinate axes and determine the coordinates x , y of the centroid of each part. 3. Determine x , y by applying the center of gravity equations. 4. If section is symmetric about its own x axis we can find y coordinate of centroid directly without any calculations as it lie on x axis. If section is symmetric about its own y axis we can find x coordinate directly without any calculations as it lie on y axis.

4.5 Centroid of Common Areas Using Method of Integration

1. Centroid of a triangular area: Consider a triangle shown in figure 2. We have to determine the x coordinate of centroid of this triangle. Now consider area dA in the form of vertical strip of width dx as shown h in figure 3. Here height of the strip is x. b

4.6  Engineering Mechanics Statics y

y

dA h

h

x b

      

x x

dx

Fig. 2                 Fig. 3

Now using Eq. 4.3, Solving,

x = x =

∫ x dA ∫ dA

b

=

∫o

h x dx b ∫ dA

2 b 3

1 h. 3 Here it is important to understand the orientation of the area. If the orientation of triangular area will change, the coordinates of centroid will also change. 2. Centroid of Semicircular Area: Consider a semicircular lamina with radius r as shown in figure 4. Here semicircular area is symmetric about y-axis, therefore x = 0. Select an elementary area in form of strip as shown in figure 4. The strip can be considered as a triangle whose base is rdq and altitude is r. The location of the centroid of the elementary sector is A.

Similarly considering a horizontal strip we can determine y =

Fig. 4

2r (Centroid of triangle) \ Distance O A = 3 \ y coordinate the centroid of element is y =

2 r sin θ 3

Centroid and Moment of Inertia  4.7

1 1 Area of the strip, dA = × r × rd θ = r 2 d θ 2 2 Using Eq. 4.3, π

y =





∫0 π ∫0 dA

y dA

π

=

2

1

∫0 3 r sin θ ⋅ 2 r π

1

∫0 2 r

2

2

dθ

dθ

r3 π ∫0 sin θd θ = 3 2 r π dθ 2 ∫0

r3 r3 [ − cos θ]0π ×2 3 3 = = r2 π r2 [θ]0 π 3 2 4r y = \ 3π

4.6 Centroid and Area of Some Common Plane Areas Centroid and area of some common plane areas are shown in table 4.1. Table 4.1 Centroid and Area Shape

Area

h

Centroid

x =

b 2

y =

h 2

y

b×d

x b

1 bh 2

y x

pr2

b 3 y = h 3

x =

x =0 y =0

4.8  Engineering Mechanics Statics

π⋅ r2 2

r y

x = r, y =

4r 3π

x d

π⋅ r2 4

x =

4r 4r y = 3π 3π

Solved Examples Based on Centroid Example 1. Determine coordinates of centroid of the shaded area shown in figure Ex. 1. All dimensions are in mm. Y 200

80

40 40

20 X 300

Fig. Ex. 1

Solution: First divide the given area into three parts i.e. rectangle, semicircle and triangle. Rectangle: a1 = 300 × 140 = 42000 mm2 300 140 = 150 mm, y1 = = 70 mm x1 = 2 2 Semicircle: a2 =

πr 2 π × 202 = = 628.31 mm 2 2 2 4r = 291.5 mm 3π 40 = 40 mm y2 = 20 + 2

x2 = 300 –

Centroid and Moment of Inertia  4.9

Triangle:

a2 =

1 1 b × h = × 100 × 100 = 5000 mm 2 2 2



x3 =

b 100 = = 33.33 mm 3 3



y3 = 140 –

h 100 = 140 – = 106.67 mm 3 3

Now coordinates of the centroid are a x − a2 x2 – a3 x3 x = 1 1 a1 – a2 – a3 =

y =

=

42000 ×150 − 628.31× 291.5 − 5000 × 33.33 = 163.59 mm   (Ans.) 42000 − 628.31 − 5000 a1 y1 − a2 y2 – a3 y3 a1 – a2 – a3 42000 × 70 − 628.31× 40 − 5000 ×106.67 = 65.47 mm 42000 − 628.31 − 5000

Example 2. Determine coordinates of centroid of the shaded area shown in figure Ex. 2. All dimensions are in mm.

Fig. Ex. 2

Solution: First divide the given area into two parts i.e. rectangle 1 and rectangle 2. Rectangle 1: a1 = 30 × 160 = 4800 mm2 30 = 15 mm x1 = 2

y1 =

160 = 80 mm 2

4.10  Engineering Mechanics Statics

a2 = 90 × 40 = 3600 mm2 90 = 30 + 45 = 75 mm x2 = 30 + 2

Rectangle 2:

y2 =



40 = 20 mm 2

Now coordinates of the centroid are a x +a x 4800 × 15 + 3600 × 75 x = 11 2 2 = = 40.71 mm   (Ans.) a1 + a2 4800 + 3600 y =

a1 y1 + a2 y2 4800 × 80 + 3600 × 20 = = 54.28 mm   (Ans.) a1 + a2 4800 + 3600

Example 3. Determine coordinates of centroid of the shaded area shown in

figure Ex. 3.

25 mm

30

m

m

50 mm

25 mm

X

O 50 mm

50 mm

Fig. Ex. 3

Solution: Here it is important to note that given area is symmetric about its own vertical axis therefore x coordinate of centroid of area will lie on its own vertical axis and we have to determine only y coordinate of centroid. Now divide the given area into four parts i.e. rectangle, triangle 1, triangle 2 and semicircle. Rectangle: a1 = 50 × 100 = 5000 mm2 50 = 25 mm y1 = 2 Triangle 1: a2 =

1 × 25 × 50 = 625 mm 2



y2 =

2 × 50 = 33.33 mm 3

Triangle 2:

a3 =

1 × 25 × 50 = 625 mm 2



y3 =

2 × 50 = 33.33 mm 3

Centroid and Moment of Inertia  4.11

Semicircle:

a4 =

πr 2 3.14 × 302 = = 1413.71 mm 2 2 2



y4 =

4r 4 × 30 = = 12.73 mm 3π 3 × 3.14

Now y coordinate of the centroid is a y − a y − a3 y3 − a4 y4 y = 1 1 2 2 a1 − a2 − a3 − a4 =

5000 × 25 – 625× 33.33 − 625× 33.33 − 1413.71×12.73 = 27.96 mm (Ans.) 5000 − 625 − 625 − 1413.71

Example 4. figure Ex. 4.

Determine coordinates of centroid of the shaded area shown in Y R 2 cm

O R 1 cm X

3 cm 10 cm

Fig. Ex. 4

Solution: First divide the given area into four parts i.e. rectangle, triangle, semicircle and circle. a1 = 7 × 4 = 28 cm2 Rectangle: 7 x1 = 3 + = 6.5 cm 2

y1 =

4 = 2 cm 2

Triangle:

a2 =

1 × 3 × 4 = 6 cm 2 2



x2 =

2 × 3 = 2 cm 3



y2 =

1 × 4 = 1.33 cm 3

Semicircle:

a3 =

πr 2 π × 22 = = 6.28 cm 2 2 2

4.12  Engineering Mechanics Statics

x3 = 10 +

4r = 10.84 cm 3π

y3 = 2 cm a4 = pr2 = 3.14 cm2 Circle: x4 = 10 cm y4 = 2 cm Now coordinates of the centroid are a x + a x + a3 x3 − a4 x4 x = 1 1 2 2 a1 + a2 + a3 − a4 =

y =

=

28× 6.5 + 6 × 2 + 6.28×10.84 − 3.14 ×10 = 6.21cm   (Ans.) 28 + 6 + 6.28 − 3.14

a1 y1 + a2 y2 + a3 y3 − a4 y4 a1 + a2 + a3 − a4 28× 2 + 6 ×1.33 + 6.28× 2 − 3.14 × 2 = 1.89 cm   (Ans.) 28 + 6 + 6.28 − 3.14

Example 5. Determine coordinates of centroid of the shaded area shown in figure Ex. 5. All dimensions are in mm.

Fig. Ex. 5

Solution: First divide the given area into two parts i.e. quarter circle and semicircle Quarter circle: a1 =

πr 2 π × 302 = = 706.5 mm 2 4 4

x1 =

4r 4 × 30 = = 12.73 mm 3π 3 × 3.14



y1 =

4r 4 × 30 = = 12.73 mm 3π 3 × 3.14

Semicircle: a2 =

πr 2 3.14 × 152 = = 353.25 mm 2 2 2

Centroid and Moment of Inertia  4.13

x2 = 15 mm 4r 4 ×15 = = 6.37 mm y2 = 3π 3 × 3.14 Now coordinates of the centroid are a x −a x x = 1 1 2 2 a1 − a2 =

y =

=

706.5 × 12.73 − 353.25 × 15 = 10.46 mm   (Ans.) 706.5 − 353.25 a1 y1 − a2 y2 a1 − a2 706.5 × 12.73 − 353.25 × 6.37 = 19.1 mm   (Ans.) 706.5 − 353.25

Example 6. Determine coordinates of centroid of the shaded area shown in figure Ex. 6.

18 cm

3 cm 2 cm

2 cm

9 cm

Fig. Ex. 6

Solution: Here it is important to note that given area is symmetric about its own vertical axis therefore x coordinate of centroid of area will lie on its own vertical axis and we have to determine only y coordinate of centroid. Now divide the given area into two parts i.e. rectangle 1 and rectangle 2 Rectangle 1: a1 = 18 × 9 = 162 cm2 18 = 9 cm y1 = 2 Rectangle 2:

a2 = 2 × 3 = 6 cm2 3 y2 = 2 + = 3.5 cm 2

4.14  Engineering Mechanics Statics

Now y coordinate of the centroid is a y −a y y = 1 1 2 2 a1 − a2 =

162 × 9 − 6 × 3.5 = 9.21cm   (Ans.) 162 − 6

Example 7. Determine coordinates of centroid of the shaded area shown in

figure Ex. 7. All dimensions are in mm. Y

800 200

1200

200

200 X 600

Fig. Ex. 7

Solution: First divide the given area into two parts i.e. rectangle 1, rectangle 2 and rectangle 3 Rectangle 1: a1 = 600 × 200 = 120000 mm2 600 = 300 mm x1 = 2 200 = 100 mm 2



y1 =

Rectangle 2:

a2 = 1200 × 200 = 240000 mm2 200 = 300 mm x2 = 200 + 2



1200 = 800 mm 2



y2 = 200 +

Rectangle 3:

a3 = 8000 × 200 = 160000 mm2 800 = 400 mm x3 = 2 200 = 1500 mm y3 = 1400 + 2



Centroid and Moment of Inertia  4.15

Now coordinates of the centroid are a x + a x + a3 x3 x = 1 1 2 2 a1 + a2 + a3 =

y =

=

120000 × 300 + 240000 × 300 + 160000 × 400 = 330.76 mm   (Ans.) 120000 + 240000 + 160000 a1 y1 + a2 y2 + a3 y3 a1 + a2 + a3 120000 ×100 + 240000 × 800 + 160000 ×1500 = 853.84 mm   (Ans.) 120000 + 240000 + 160000

Example 8. Determine coordinates of centroid of the shaded area shown in

figure Ex. 8.

10 cm

10 cm

4 cm

5 cm

20 cm

Fig. Ex. 8

Solution: Here it is important to note that given area is symmetric about its own vertical axis therefore x coordinate of centroid of area will lie on its own vertical axis and we have to determine only y coordinate of centroid. Now divide the given area into two parts i.e. triangle and rectangle 1 2 Triangle: a1 = × 20 × 25 = 250 cm 2 25 = 8.33 cm 3



y1 =

Rectangle:

a2 = 4 × 10 = 40 cm2 10 y2 = 5 + = 10 cm 2



Now y coordinate of the centroid is a y −a y y = 1 1 2 2 a1 − a2

4.16  Engineering Mechanics Statics

=

250 × 8.33 − 40 ×10 = 8.01cm   (Ans.) 250 − 40

Example 9. Determine coordinates of centroid of the shaded area shown in figure Ex. 9.

Fig. Ex. 9

Solution: First divide the given area into two parts i.e. triangle 1 and triangle 2 1 2 a1 = × 40 × 50 = 1000 cm Triangle 1: 2

x1 =

50 = 16.67 cm 3



y1 =

40 = 13.33 cm 3

Triangle 2:

a2 =

1 × 40 × 30 = 600 cm 2 2



x2 =

30 = 10 cm 3



y2 =

40 = 13.33 cm 3

Now coordinates of the centroid are a x −a x x = 1 1 2 2 a1 − a2 =

y =

=

1000 × 16.67 − 600 × 10 = 26.67 cm   (Ans.) 1000 − 600 a1 y1 − a2 y2 a1 − a2 1000 × 13.33 − 600 × 13.33 = 13.33 cm   (Ans.) 1000 − 600

Centroid and Moment of Inertia  4.17

Example 10. Determine coordinates of centroid

Y

of the shaded area shown in figure Ex. 10. All dimensions are in mm. Solution: Here it is important to note that given 100 200 area is symmetric about its own vertical axis X therefore x coordinate of centroid of area will lie Fig. Ex. 10 on its own vertical axis and we have to determine only y coordinate of centroid. Now divide the given area into two parts i.e. semicircle 1 and semicircle 2 Semicircle 1: a1 =

y1 =

Semicircle 2: a2 =

y2 =

πr 2 π × 2002 = = 62831.85 mm 2 2 2 4r 4 × 200 = = 84.88 mm 3π 3π πr 2 π ×1002 = = 15707.96 mm 2 2 2

4r 4 ×100 = = 42.44 mm 3π 3π

Now y coordinate of the centroid is y = a1 y1 − a2 y2 a1 − a2 62831.85× 84.88 − 15707.96 × 42.44 = 99.02 mm (Ans.) 62831.85 − 15707.96 Example 11. Determine location of centroid of homogeneous bend wire shown in figure Ex. 11. All dimensions are in mm. =

Fig. Ex. 11

Solution: Here we consider three segments of wire, AB, BC and CA. Segment AB: l1 = 120 mm x1 = 0

4.18  Engineering Mechanics Statics

y1 =



120 = 60 mm 2

l2 = 130 mm 50 = 25 mm x1 = 2

Segment BC:

y1 =



120 = 60 mm 2

Segment CA:

l3 = 50 mm 50 = 25 mm x3 = 2

y3 = 0 Now coordinates of the centroid are 120 × 0 + 130 × 25 + 50 × 25 l x +l x +l x = 15 mm   (Ans.) x = 1 1 2 2 3 3 = 120 + 130 + 50 l1 + l2 + l3 l y +l y +l y 120 × 60 + 130 × 60 + 50 × 0 = 50 mm   (Ans.) y = 1 1 2 2 3 3 = 120 + 130 + 50 l1 + l2 + l3

SUMMARY

Weight of the body is distributed force system. Center of gravity is the average position of a distribution of weight. Center of mass is the average position of a distribution of mass. Centroid is the average position of a distribution of area or shape. If an area has symmetry about one axis its centroid lies on the axis. If an area has symmetry about two axes its centroid lie at the intersection of axes. • Coordinates of centroid of composite areas are determined using following equations

• • • • • •

n



x =

∑ i=1xi Ai n ∑ i=1 Ai



y =

∑ i=1yi Ai n ∑ i=1 Ai

n

Centroid and Moment of Inertia  4.19

PROBLEMS Determine coordinates of centroid of the following shaded area. y

3 cm

12 cm

6 cm x 18 cm

y

6 cm

12 cm

4 cm 4 cm 3 cm

2 cm

x 5 cm 2 cm 2 cm

3 cm

y

y

60 mm 40 mm 20 mm

75 mm

60 mm

x x

75 mm 20 mm

4.7 Area Moment of Inertia Moment of inertia is a measure of an object’s resistance to changes to its rotation. It is also defined as the capacity of a cross section to resist bending. Moment of inertia depends Moment of Inertia is the upon how the mass is distributed relative to the second moment of area. axis of rotation. The moment of inertia of an area about any axis is the second moment of area about that axis. Consider an area A in the X – Y plane as shown in figure 5.

4.20  Engineering Mechanics Statics Y

A x

dA

y

X

O

Fig. 5

Moment of inertia of area A about the x axis is defined as, Ix =

∫y

2

dA ,

...(Eq. 4.6)

here y is the y coordinate of differential element of area dA. Moment of inertia of area A about the y axis is defined as, Iy = ∫ x dA , here x is the x coordinate of differential element of area dA. The moment of inertia is denoted by I and its unit is mm4, m4. 2

...(Eq. 4.7)

4.8 Perpendicular Axis Theorem Consider a plane area ‘A’ lying in plane XY as shown in figure 6. X and Y are two mutually perpendicular axes and Z is perpendicular axis. Y

A x

dA

y

r

X

O

Z

Fig. 6

Centroid and Moment of Inertia  4.21

Consider differential element of area dA. Let x be the distance of dA from Y axis, y be the distance of dA from X axis and r be the distance of dA from Z axis. Now moment of inertia of area A about Z axis which is also known as Polar moment of inertia is defines as, Iz = 2

∫r

2

dA

2

...(Eq. 4.8)

2

Now from figure, r = x + y Therefore Eq. 4.8 can be written as, Iz = ∫ ( x 2 + y 2 )dA =

∫y

2

dA + ∫ x 2 dA ,

From Eq. 4.6 and Eq. 4.7 we can write, ...(Eq. 4.9) Iz = Ix + Iy Theorem of perpendicular axis states that if Ix and Iy be the moment of inertia of a plane area about two mutually perpendicular axis X and Y in the plane of the area then the moment of inertia of the area Iz about the axis Z perpendicular to the plane and passing through the intersection of X and Y is given by Eq. 4.9.

4.9  Parallel Axis Theorem Sometime the moments of inertia of an area are known for a particular coordinate system but we need their values in terms of some other coordinate system. When the coordinate systems are parallel, the desired moments of inertia can be obtained using parallel axis theorem. Consider a plane area ‘A’ lying in plane XY as shown in figure 7. Centroidal axis y dA A

y Centroidal axis x

G

h

A

Parallel axis

B

Fig. 7

Let, Centroidal axis x in the plane of area A and passing through the centroid of the area. AB is the axis in the plane of area ‘A’ and parallel to the centroidal axis x h be the distance between axis AB and centroidal axis x.

4.22  Engineering Mechanics Statics

Now consider differential element of area dA at distance y from centroidal axis. Moment of inertia of the area dA about AB,

2 IAB = ∫ ( y + h) dA

Expanding above integration, IAB =

∫y

2

dA + ∫ 2 y h dA + ∫ h 2 dA

=

∫y

2

dA + 2h ∫ y dA + h 2 ∫ dA



From above we see that the first integral A about centroidal axis x = Ix Second integral

∫ y dA = Ay = 0, as

∫y

2

dA is moment of inertia of area

y =0

And third integral is h2A ...(Eq. 4.10) IAB = Ix + h2 A This is a parallel-axis theorem which relates the moment of inertia of A about the centroidal axis through the centroid to the moment of inertia about the parallel axis AB.

4.10 Radius of Gyration Consider an area A which has a moment of inertia IAB with respect to the AB axis as shown in figure 8. Let us imagine that we concentrate this area into a thin strip parallel to the AB axis as shown in right hand part of figure 8. If the area A is to have the same moment of inertia with respect to the AB axis, the strip should be placed at a distance k from the AB axis, where k is defined by the radius of gyration.

k =

I AB A

A

A A A k

B

B

Fig. 8

4.11 Area Moment of Inertia of Common Areas

1. Area M.I. of rectangle about base: Consider a rectangle of base b and height h. X is the reference axis passing through base.

Centroid and Moment of Inertia  4.23

Now consider an elemental strip of thickness dy located at a distance y from the reference axis X as shown in figure 9. Using Eq. 4.6 Ix =

∫y

2

dA

Area of elemental strip dA = b dy Ix = Ix =

h

h

∫0 y b dy = b ∫0 2

dy

h h

 y3  y dy = b    3 0

y

2

X

3

bh    ...(Eq. 4.11) 3

b

Fig. 9.

2. Area M.I. of rectangle about centrodial axis: Area moment of inertia of rectangle about its centrodial x’ axis (horizontal axis passing through centroid of rectangle) is determine by applying parallel axis theorem. By parallel axis theorem, Ix = Ix′ + Ah2 Where h = Distance between the reference axis and parallel centroidal axis. Ix′ = Ix – Ah2

bh3 3 2 3 bh3 bh3  h  bh – b×h×  = – Ix = 3 3 4 2

From Eq. 4.11, Ix =

Ix′ =

bh3 12

...(Eq. 4.12)

3. Area M.I. of triangle about base: Consider a triangle of base b and height h. X is the reference axis passing through base. Now consider an elemental strip of thickness dy located at a distance y from the reference axis as shown in figure 10.

h

dy l y X b

Fig. 10

4.24  Engineering Mechanics Statics

Using Eq. 4.6 Ix =



∫y

2

dA

b Area of elemental strip dA = (h – y )dy h 2b Ix = ∫ y (h − y ) dy h h

b 2 b  y 3h y 4  − Ix = ∫ y (h − y ) dy =   h 4 0 h 3 bh3 Ix = 12



...(Eq. 4.13)

4. Area M.I. of triangle about centrodial axis: Area moment of inertia of triangle about its centrodial x’ axis (horizontal axis passing through centroid of rectangle) is determine by applying parallel axis theorem. By parallel axis theorem, Ix = Ix′ + Ah2 Where h = Distance between the reference axis and parallel centroidal axis. Ix′ = Ix – Ah2



From Eq. 4.13, Ix =

bh3 12 2



Ix′ =

3 bh3 1 bh3  h  bh – b×h×  = – 12 2 12 18 3



Ix′ =

bh3 36

Area moment of inertia of some common plane areas are shown in table 4.2. Table 4.2 Area Moment of Inertia of Plane Area. Shape y

h

Moment of Inertia

Rectangle b 2

y

C

x h 2 x b

Ix′ =

1 1 3 bh Ix = bh3 3 12

Iy′ =

1 1 3 b h Iy = b 3 h 3 12

Centroid and Moment of Inertia  4.25 Right Triangle y

b 3

y

Ix′ =

1 3 1 3 bh Ix = bh 36 12

Ix′ =

1 3 1 3 hb hb Iy = 12 36

Ix′ =

1 3 1 3 bh Ix = bh 36 12

Iy′ =

1 3 hb 48

h x h 3

C

x

b

Isosceles Triangle

Circle y

r x

C

Ix′ = I y ′ =

1 4 πr 4

Semicircle

y

C

x 4r 3 x

r

1 Ix = I y = πr 4 8 Ix′ = 0.1098r4

Quarter Circle y

Ix = I y = r

1 4 πr 16

x

C

x r

Ix′ = Iy′ = 0.05488r4

4.26  Engineering Mechanics Statics

Solved Examples Based on Area Moment of Inertia Example 1. Determine Ix and Iy of the shaded area shown in figure Ex. 1. All

dimensions are in mm.

y

60

20

40

x

Fig. Ex.1

Solution: Here the x axis is passing through the base of rectangle so we use following expression to determine Ix  Ix =

bh3 40 × 603 = = 2880000 mm 4   (Ans.) 3 3

Now to determine Iy we use parallel axis theorem. Iy = Iy’ + Ah2 Here h is the distance between y axis and y’ axis (vertical axis passing through centroid of rectangle) =

60 × 403 + 60 × 40 × 402 = 4160000 mm 4   (Ans.) 12

Example 2. Determine IxC of the shaded area shown in figure Ex. 2. All dimensions are in mm. 20

80

Xc 35 20 80

Fig. Ex. 2

Centroid and Moment of Inertia  4.27

Solution: Here first we divide the given area into two parts i.e. rectangle 1 and rectangle 2 then we apply parallel axis theorem to both rectangle about Rectangle 1:

IX = Ix′ + Ah2 C

=

20 × 803 + 20 × 80 × (60 – 35) 2 12

IX = 1853333.33 mm4 C

Rectangle 2:

IX = Ix′ + Ah2 C

=

80 × 203 + 20 × 80 × (35 – 10) 2 12

IX = 1053333.33 mm4 C

Total IX = (IX of rectangle 1) + (IX of rectangle 2) C

C

C

Total IX = 1853333.33 + 1053333.33 C

\ Moment of inertia of given area about x = 2906666.66 mm4  (Ans.) C

Example 3. A 20 × 20 mm square is removed from a 40 × 40 mm square as

shown in figure Ex. 3. Determine IxC and IyC. Solution: Here first we divide the given area into two parts i.e. square 1 and square 2. As xC is centroidal axis which is passing through centroid of given area we can apply Eq. 4.12 to both rectangle. Square 1:

IX = I x , = C

=

bh3 12

40 × 403 12

IX = 213333.33 mm C

Rectangle 2:

IX = I x , = C

=

Yc

Xc 4

bh3 12

Fig. Ex. 3

3

20 × 20 12

IX = 13333.33 mm4 C

Total

IX = (IX of square 1) – (IX of square 2)

Total

IX = 213333.33 – 13333.33

C

C

C

C

\ Moment of inertia of given area about xC = 200000 mm4  (Ans.)

4.28  Engineering Mechanics Statics

Example 4. Determine IxC and IyC of the shaded area shown in figure Ex. 4. yc 0.165 m

0.8 m C

xc 0.1m

0.1m

0.265 m

0.5 m

Fig. Ex. 4

Solution: Here first we divide the given area into two parts i.e. rectangle 1 and rectangle 2. Now we apply parallel axis theorem to both rectangle about xC. Rectangle 1: IX = Ix’ + Ah2 C

=

0.1× 0.83 + 0.1 × 0.8 × (0.4 – 0.265) 2 12

IX = 5.72 × 10–3 m4 C Rectangle 2: IX = Ix’ + Ah2 C

=

0.5× 0.13 + 0.5 × 0.1 × (0.265 – 0.05) 2 12

IX = 2.35 × 10–3 m4 Total Total

C

IX = (IX of rectangle 1) + (IX of rectangle 2) C

C

C

–3

IXC = 5.72 × 10 + 2.35 × 10

–3

\ Moment of inertia of given area about xC = 8.07 × 10–3 m4  (Ans.) Now we apply parallel axis theorem to both rectangle about yC Rectangle 1:

Iy = Iy’ + Ah2 = C

0.8× 0.13 + 0.1 × 0.8 × (0.165 – 0.05) 2 12

Iy = 1.12 × 10–3 m4 C

Rectangle 2:

Iy = Iy’ + Ah2 = C

0.1× 0.53 + 0.5 × 0.1 × (0.35 – 0.165) 2 12

Iy = 2.75 × 10–3 m4 Total

C

Iy = (Iy of rectangle 1) + (Iy of rectangle 2) C

C

C

Total Iy = 1.12 × 10–3 + 2.75 × 10–3 C \ Moment of inertia of given area about yC = 3.87 × 10–3 m4  (Ans.)

Centroid and Moment of Inertia  4.29

Example 5. Determine Ix of the shaded area shown in figure Ex. 5.

Fig. Ex. 5

Solution: Here first we divide the given area into two parts i.e. triangle and rectangle. Triangle: Here the x axis is passing through the base of triangle so we use following expression to determine Ix bh3 20 × 253 = = 26041.7 cm 4 12 12 Rectangle: Here we apply parallel axis theorem about x Ix = Ix’ + Ah2 Ix =

4 × 103 + (4 × 10) × 52 = 1333.3 cm4 12 Total Ix = (Ix of triangle) – (Ix of rectangle) = 26041.7 – 1333.3 \ Moment of inertia of given area about x = 24708.4 cm4  (Ans.) =

Example 6. Determine area moment of inertia about centroidal axes of shaded area shown in figure Ex. 6.

Fig. Ex. 6

4.30  Engineering Mechanics Statics

Solution: Centroidal axes means the x’ and y’ axis which are passing through centroid of given area. Therefore, first we determine centroid of given area. It is important to note that given area is symmetric about its own vertical axis therefore x coordinate of centroid of area will lie on its own vertical axis and we have to determine only y coordinate of centroid. a y + a y + a3 y3 y = 1 1 2 2 (y distance is measured from base of given area) a1 + a2 + a3 =

(18 × 2) × 15 + (2 × 12) × 8 + (2 × 5) × 1 (18 × 2) + (2 × 12) + (2 × 5)

y = 10.6 cm Now we divide the given area into three parts i.e. rectangle 1, rectangle 2 and rectangle 3. Apply parallel axis theorem to all rectangles about xC. Rectangle 1: IXC = Ix’ + Ah2 =

1 (18 × 23 ) + (18 × 2) (15 – 10.6) 2 12

IX = 708.96 cm4 C Rectangle 2: IX = Ix’ + Ah2 C

=

1 (2 × 123 ) + (2 × 2) (10.6 – 8) 2 12

IX = 315.04 cm4 C IX = Ix’ + Ah2



Rectangle 3:

C

=

1 (5 × 23 ) + (2 × 5) (10.6 – 1) 2 12

IX = 924.93 cm4 C

Total IX = (IX of rectangle 1) + (IX of rectangle 2) + (IX of rectangle 3) C

Total

C

C

C

IX = 708.96 + 315.04 + 924.93 C

\ Moment of inertia of given area about xC = 1948.93 cm4  (Ans.) Now we apply parallel axis theorem to all rectangles about yC. In following calculation, distance h is zero because the vertical axis of individual areas coincide with vertical axis of composite area. Rectangle 1: Iy = Iy’ + Ah2 C

=

1 (2 × 183 ) + 0 12

Iy = 972 cm4 C

Centroid and Moment of Inertia  4.31

Iy = Iy’ + Ah2

Rectangle 2:

C

1 (12 × 23 ) + 0 12 Iy = 8 cm4 C Rectangle 3: Iy = Iy’ + Ah2 =

C

=

1 (2 × 53 ) + 0 12

Iy = 20.83 cm4 C

Total Iy = (Iy of rectangle 1) + (Iy of rectangle 2) + (Iy of rectangle 3) C

C

C

C

Total Iy = 972 + 8 + 20.83 C \ Moment of inertia of given area about yC = 1000.83 cm4  (Ans.) Example 7. Determine area moment of inertia about centroidal x’ axes of shaded area shown in figure Ex. 7.

40 cm

14 cm

20 cm

18 mm

Fig. Ex. 7

Solution: Centroidal x’ means the axis which are passing through centroid of given area. Therefore first we determine centroid of given area. It is important to note that given area is symmetric about its own vertical axis therefore x coordinate of centroid of area will lie on its own vertical axis and we have to determine only y coordinate of centroid. a y − 2[a2 y2 ] y = 1 1 (y distance is measured from base of given area) a1 − 2a2 =

(18 × 74) × 37 − 2 × [(3.14 × 7 × 7) × 27] (18 × 74) − 2 × (3.14 × 7 × 7)

y = 40 mm

4.32  Engineering Mechanics Statics

Now we divide the given area into three parts i.e. rectangle and two semicircles. Apply parallel axis theorem to all areas about xC. Rectangle: IX = Ix’ + Ah2 C

=

18 × 743 + (18 × 74) × (40 – 37) 2 12

IX = 619824 mm4 C

Two semicircle: IX = 2[Ix’ + Ah2] C

 3.14 × 7 4  3.14 × 7 2  2 (40 27) + − = 2    8 2    4 IX = 27887.125 mm C Total IX = (IX of rectangle ) + (IX of two semicircle) C C C Total IX = 619824 + 27887.125 C \ Moment of inertia of given area about xC = 647711.125 mm4  (Ans.)

SUMMARY • Moment of inertia is a measure of an object’s resistance to changes to its rotation. • Moment of inertia measures how an area is distributed about particular axes. • Area moment of inertia is defined as, 2 Ix = ∫ y dA, ,





∫ x dA 2

• Radius of Gyration is defined as



Iy =

k =

I A

• The parallel-axis theorem is usually used to calculate the moment of inertia about a second axis which is parallel to first axis and moment of inertia about first axis is known.

Centroid and Moment of Inertia  4.33

PROBLEMS 1. Determine Ix , Iy, Kx and Ky of shaded area shown in figure Prob. 1.

Fig. Prob. 1

2. Determine Ix , and Iy, of shaded area shown in figure Prob. 2. y

60 mm 20 mm

60 mm

x 20 mm

Fig. Prob. 2

3. Determine Ix, and Iy, of shaded area shown in figure Prob. 3. All dimensions are in mm. y 800 200

1200

200

200 x 600

Fig. Prob. 3

4.34  Engineering Mechanics Statics 4. Determine Ix¢ and Iy¢ of shaded area shown in figure Prob. 4. All dimensions are in mm. y 20

160

40 x 90

Fig. Prob. 4

5. Determine area moment of inertia about centroidal axes of shaded area shown in figure Prob. 5. 2 cm

12 cm

2 cm

5 cm

18 cm

2 cm

Fig. Prob. 5

Unit 5

In most of the problems that we have analysed up to last unit, the contact surfaces assumed to be smooth or frictionless. Actually when two objects are come in contact, generally friction force develop between them. In this unit, we focus on friction force and models for calculating friction forces. We discuss methods of analysing problems that involve friction and sliding.

Friction 5.1 Introduction Friction is defined as the resistance force which Frictional force is tangent one body offers to the motion of a second body to the surfaces of contact when the second body tends to slide or slide over of the bodies. the former. In many situations, friction forces are helpful. On the other hand, friction can also be unfavourable. Friction may be classified into two types, Dry friction and Fluid friction. In this book, we will concentrated on Dry friction Friction

Dry Friction

Static Friction

Fluid Friction

Kinetic Friction

Dry friction refers to the friction force that exists between two un-lubricated solid surfaces. Fluid friction acts between moving surfaces that are separated by a layer of fluid. Static friction experienced by a body when it is at rest under the action of forces. Kinetic friction experienced by a body when it is moving.

5.2 Coulomb’s Theory of Dry Friction Consider the block of weight W rest on rough horizontal surface and subjected to horizontal force P as shown in left hand part of Figure 1. The free body diagram of the block is shown in right hand part of Figure 1. W

P

P

W

F N

Fig. 1

If applied force P is small, static friction force F balance P and the block will not move. If the force P is increased, the friction force F also increases to

5.4  Engineering Mechanics Statics

maintain equilibrium until its magnitude reaches Maximum frictional force a maximum value Fmax. When P exceeds Fmax Fmax is called as limiting the friction force cannot balance it and the block friction. start sliding and friction force F drops to a kinetic value. [There are irregularities of the surfaces in contact. When one surface comes in contact At limiting friction body with other, the actual area of contact is very less is said to have impending than the surface area of contact. Because of this, motion. pressure due to the reaction force is very high, hence irregularities deform a little and cold welds are formed at contact points. So to start relative sliding between surfaces enough force is required to break these cold welds. Once the welds break when one surface start sliding over other, a smaller force is enough to keeps it moving with uniform velocity. This is one of the reason that coefficient of static friction is more than coefficient of kinetic friction.] If P is further increased F remains constant. The plot of F verses P is shown in Figure 2. F No Motion Fs

Fmax Motion

Fk F=P

P

Fig. 2

5.3 Coefficient of Friction In order to determine the maximum frictional force corresponding to any normal force, a certain experimental constant known as the coefficient of friction (μ), is used. Coefficient of friction does not depend upon the area of the surface in contact but depend on nature of the surfaces in contact. The coefficient of static friction μs is defined as the ratio of limiting friction (Fmax) to corresponding normal force, F μs = max or Fmax = μs N ...(Eq. 5.1) N Typical values of μs for various materials are shown in Table 5.1.

Friction  5.5 Table 5.1 Values of μs for Various Materials Materials

Coefficient of Static Friction

Wood on wood

0.25 to 0.50

Wood on leather

0.25 to 0.50

Metal on metal

0.15 to 0.30

Metal on wood

0.20 to 0.60

Metal on stone

0.30 to 0.70

Metal on leather

0.30 to 0.60

Stone on stone

0.40 to 0.65

Rubber on concrete

0.60 to 0.90

If two surfaces move relative to each other, the ratio of the friction developed to the corresponding normal force is known as coefficient of kinetic friction μk. F μk = k ...(Eq. 5.2) N

5.4  Angle of Friction The angle of static friction (fs) is defined as the angle between normal reaction and resultant force when motion is impending. W It is evident from Figure 3, F tan ∅s = max P N But from Eq. 5.1, Fmax Fmax μs = s N R N ...(Eq. 5.3) \ μs = tan fs Fig. 3 Angle of static friction ∅s = tan–1 μs ...(Eq. 5.4) The angle of kinetic friction (∅k) is defined as the angle between normal reaction and resultant force when two surfaces move relative to each other. ...(Eq. 5.5) Angle of kinetic friction ∅k = tan–1 μk

5.5  Angle of Repose Angle of repose is the maximum angle of an inclined plane from horizontal at which body is at impending motion. Consider a block of weight W on an inclined plane as shown in left hand part of Figure 4. The free body diagram at impending motion (block is about to slide down the inclined plane) is shown in right hand part of Figure 4.

5.6  Engineering Mechanics Statics W 

F



N

Fig. 4

Applying conditions of equilibrium i.e. SH = 0, and SV = 0 along and normal to plane, SH = 0: F – W sin α = 0 Replacing friction force by μs N from (Eq. 5.1), ...Eq. (a) μs N – W sin α = 0  \ μs N = W sin α SV = 0: N – W cos α = 0  \ N = W cos α Substitute value of N in Eq. (a) μs W cos α = W sin α  \ μs = tan α But from (Eq. 5.3), μs = tan ∅s \ tan ∅s = tan α ...(Eq. 5.6) \ ∅s = α From (Eq. 5.6), we can say that, Angle of repose (α) = Angle of static friction (∅s).

5.6  Applications of Friction The various equations discussed in previous sections have a number of applications in engineering practices. Following are some of the important applications of friction.

5.6.1 Ladder Friction A ladder is a device used for climbing on the walls. A ladder only stands if there is friction between ladder and the wall, and ladder and the ground. The left hand part of Figure 5 shows a ladder AB of weight W with its end A resting on the ground and end B leaning against a wall. The free body diagram of the ladder shown in right hand part of Figure 5 is a non-concurrent force system. Now by applying equilibrium equations assuming ladder impending, we can solved problem of ladder friction.

Friction  5.7 B

B

NB FB

W

A

A

 FA

NA

Fig. 5

5.6.2  Wedge Friction A wedge is a simple device of triangular cross-section used for small adjustment in the position of a body or to apply large forces. Wedges mainly depend on friction during functioning. When a wedge is pushed forward, its faces exert large normal forces which can be used to lift the load. The left hand part of Figure 6 shows a wedge used to lift a large load W. The free body diagram of the wedge and load is shown in right hand part of Figure 6. The force P necessary to start raising the load can be determined by applying equations of equilibrium assuming weight and wedge are impending. W F3 N3  N2

W

F2 F2

P

P

N2 

 F1 N1

Fig. 6

5.6.3 Belt Friction Friction is useful in belt or rope and driving surfaces for the transmission of power. If a belt or rope passes over a rough cylinder or pulley, the tension on the two sides of the pulley will not be equal. The tension in the belt or rope on the two sides of a rough pulley is determined as follow.

5.8  Engineering Mechanics Statics

The left hand part of the Figure 7 shows a thin flexible flat belt passing over a cylinder. The angle of contact is β and belt tensions are T1 and T2. T1 is the force on low tension side and T2 is the force on high tension side. Here we assume that tension T1 is known, and we have to determine largest force T2 that is acting on another end of belt.

Fig. 7

The free body diagram of the differential element is shown in right hand part of the Figure 7. As tension in the belt varies with position, the belt tension on the right side of the element is T and tension on the left side of the element is T + ∆T. The force ∆N is the normal force exerted on the element by cylinder. As we have to determine largest value of T2 that will not cause the belt to slip, we assume that friction force is equal to its maximum value μs ∆N. Now applying equilibrium equations, +  ∆α   ∆α  →ΣH = 0: –(T + ∆T )cos  ...(Eq. 5.7)  + T cos   + µ s ∆N 2    2   ∆α   ∆α  +↑SV = 0: –(T + ∆T )sin   – T cos   + ∆N  2   2  Since

dα is small, cos 2

...(Eq. 5.8)

 dα   dα  dα  dα  . The term dT sin    ≈ 1 and sin  ≈   2   2  2  2 

is small quantity of the second order and may be neglected. Hence (Eq. 5.7) and (Eq. 5.8) become, μs dN – dT = 0 dN – T da = 0 Eliminating dN, dT = μsda ...(Eq. 5.9) T This expression is a differential equation that describes how the force T changes with position α. By integrating (Eq. 5.9) the relationship between T1 and T2 is determined as follow,

Friction  5.9 T2

∫T1

\ \

In

dT = T

β

∫0 µ s d α

T2 = msb T1 T2 = ems b T1

...(Eq. 5.10)

Where, e is the base of natural logarithms and b is angle of contact measured in radians.

Solved Examples Based on Block Friction Example 1. A 400 N block is resting on rough horizontal surface is subjected to 80 N force as shown in Figure Ex. 1. Determine the friction force between the block and the surface. Assume ms = 0.5. 80 N 400 N

Fig. Ex. 1

Solution: Assume equilibrium of the block and apply conditions of equilibrium i.e. +

→ΣH = 0, and +↑SV = 0 to the free body diagram of the block as shown in figure Ex.1(a). 400 N SH = 0: 80 – F = 0 \ \

F = 80 N N = 400 N

Now from (Eq. 5.1), Fmax = μs N, we can check our assumption of equilibrium of the block as below, Fmax = 0.5 × 400 \

80 N

SV = 0: N – 400 = 0 F N

Fig. Ex. 1(a)

Fmax = 200 N

Since calculated value of friction force from equilibrium analysis is less than Fmax (F < Fmax), we conclude that the block is in static equilibrium and the value of the friction force is, F = 80 N  (Ans.)

5.10  Engineering Mechanics Statics

Example 2. In Example 1, if block is subjected to 120 N force, determine the friction force between the block and the surface. Assume ms = 0.25 and mk = 0.20. Solution: Assume equilibrium of the block and apply conditions of equilibrium +

i.e. →ΣH = 0, and +↑SV = 0 to the free body diagram of the block as shown in figure Ex. 2. 400 N SH = 0: 120 – F = 0 \ \

F = 120 N

120 N

SV = 0: N – 400 = 0 N = 400 N

Now from (Eq. 5.1), Fmax = μs N, we can check our assumption of equilibrium of the block as below,

F N

Fig. Ex. 2(a)

Fmax = 0.25 × 400 Fmax = 100 N \ Since calculated value of friction force from equilibrium analysis is greater than Fmax (F > Fmax), we conclude that the block will not be in equilibrium and slide to the right and the value of the friction force is determined from (Eq. 5.2), F = Fk = μk N Fk = 0.20 × 400 = 80 N  (Ans.) Example 3. A 60 kg block is subjected to force P = 400 N as shown in Figure Ex. 3. Determine the friction force between the block and the surface. Assume ms = 0.6. 400 N 30°

Fig. Ex. 3

Solution: Assume equilibrium of the block and apply conditions of equilibrium +

i.e. →ΣH = 0, and +↑SV = 0 to the free body diagram of the block as shown in figure Ex. 3(a). SH = 0: 400 cos 30° – F = 0 \ \

F = 346.41 N SV = 0: N – 60 × 9.81 – 400 sin 30° N = 788.6 N

Fig. Ex. 3(a)

Friction  5.11

Now from (Eq. 5.1), Fmax = μs N, we can check our assumption of equilibrium of the block as below, Fmax = 0.6 × 788.6  \ Fmax = 473.16 N Since calculated value of friction force from equilibrium analysis is less than Fmax (F < Fmax), we conclude that the block is in static equilibrium and the value of the friction force is, F = 346.41 N  (Ans.) Example 4. Determine the force P required to start moving the 250 N block as

shown in Figure Ex. 4 up the inclined surface. Assume ms = 0.3.

P 30º

Fig. Ex. 4

Solution: Here we can consider impending motion of the block up the plane and then apply conditions of equilibrium i.e. SH = 0 250 N along the inclined plane, and SV = 0 perpendicular 30° to the inclined plane. Free body diagram of the block is shown in figure Ex 4(a). F SV = 0: N – 250 cos 30° = 0 P N = 216.50 N \ SH = 0: P – F – 250 sin 30° = 0 N Now for impending motion from (Eq. 5.1), we Fig. Ex. 4(a) know Fmax = μs N. \ P – μs N – 250 sin 30° = 0 P = 0.30 × 216.50 + 250 sin 30°  \ P = 190 N  (Ans.) Example 5. Determine the horizontal force P required to start moving the 500 N block as shown in Figure Ex. 5 up the inclined surface. Assume ms = 0.3. P

30º

Fig. Ex. 5

Solution: Here we can consider impending motion of the block up the plane and then apply conditions of equilibrium i.e. SH = 0 along the inclined plane, and

5.12  Engineering Mechanics Statics

SV = 0 perpendicular to the inclined plane. Free body diagram of the block is shown in figure Ex 5(a). 500 N SH = 0: P cos 30° – F – 500 sin 30° = 0 30° P cos 30° = F + 500 sin 30° Now for impending motion from (Eq. 5.1), we P F 30° know Fmax = μs N. \ P cos 30° = μs N + 500 sin 30° ...(Eq. 1) Now, SV = 0: N – P sin 30° – 500 cos 30° = 0 \

N = P sin 30° + 500 cos 30°

N

Fig. Ex. 5(a)

Substituting value of N in to Eq. 1, P cos 30° = 0.30 (P sin 30° + 500 cos 30°) + 500 sin 30° \ P = 530.58 N  (Ans.) Example 6. The 50 N force as shown in Figure Ex. 6 is required to produce impending motion of the block down the inclined surface. If the weight of the block is 250 N, determine the coefficient of static friction (ms) between the block and the inclined surface.

50 N

30°

Fig. Ex. 6

Solution: Here we can consider impending motion of the block down the plane and then apply conditions of equilibrium i.e. SH = 0 along the inclined plane, and SV = 0 perpendicular to the inclined plane. Free body diagram of the block is shown in figure Ex 6(a). SH = 0: 50 cos 30° + 250 sin 30° – F = 0 250 N 30° 50 N 30°

F N

Fig. Ex. 6(a)

Friction  5.13

F = 50 cos 30° + 250 sin 30° Now for impending motion from (Eq. 5.1), we know Fmax = μs N. ...(Eq. 1) \ μs N = 50 cos 30° + 250 sin 30° Now, SV = 0: N – 250 cos 30° + 50 sin 30° = 0 \ N = 250 cos 30° – 50 sin 30° Substituting value of N in to Eq. 1, μs (250 cos 30° – 50 sin 30°) = 50 cos 30° + 250 sin 30° \ μs = 0.8788  (Ans.) Example 7. A 150 N block is subjected to the horizontal force P as shown in Figure Ex. 7. Determine the range of horizontal force P that will keep the 150 N block in equilibrium. Assume ms = 0.25.

Fig. Ex. 7

Solution: Here it is clear that slipping of the block down or up the inclined plane will impend when the magnitude of force P reaches to 150 N its minimum value or maximum value respectively. 45° At minimum value of P, the block will impend down the plane and at maximum value of P, the block will P impend up the plane. 45° Now first consider impending motion of the block down the plane and then apply conditions of N F equilibrium i.e. SH = 0 along the inclined plane, and Fig. Ex. 7(a) SV = 0 perpendicular to the inclined plane. Free body diagram of the block is shown in figure Ex 7(a). SH = 0: P cos 45° + F – 150 sin 45° = 0 Now for impending motion from (Eq. 5.1), we know Fmax = μs N. \ P cos 45° + μs N – 150 sin 45° = 0 Now, \

SV = 0: N – 150 cos 45° – P sin 45° = 0 N = 150 cos 45° + P sin 45°

Substituting value of N in to Eq. 1, P cos 45° + 0.25 (150 cos 45° + P sin 45°) – 150 sin 45° = 0 \ P = Pmin = 90 N  (Ans.)

...(Eq. 1)

5.14  Engineering Mechanics Statics 150 N 45° P 45°

F

N

Fig. Ex. 7(b)

Now consider impending motion of the block up the plane and then apply conditions of equilibrium i.e. SH = 0 along the inclined plane, and SV = 0 perpendicular to the inclined plane. Free body diagram of the block is shown in figure Ex 7(b). SH = 0: P cos 45° – F – 150 sin 45° = 0 Now for impending motion from (Eq. 5.1), we know Fmax = μs N. \ P cos 45° – μs N – 150 sin 45° = 0 ...(Eq. 2) Now, SV = 0: N – 150 cos 45° – P sin 45° = 0 \ N = 150 cos 45° + P sin 45° Substituting value of N in to Eq. 2, P cos 45° – 0.25 (150 cos 45° + P sin 45°) – 150 sin 45° = 0 \ P = Pmax = 250 N  (Ans.) \ Range of horizontal force P is from 90 N to 250 N that can keep the block in equilibrium. Example 8. A 50 kg block is attached to mass M using string which is passing over a smooth pulley as shown in Figure Ex. 8. Determine the range of mass M that will keep the block in equilibrium. Assume ms = 0.30.

Fig. Ex. 8

Solution: Here it is clear that slipping of the block down or up the inclined plane will impend when the mass M reaches to its minimum value or maximum value respectively. At minimum value of mass M, the block will impend down the plane and at maximum value of mass M, the block will impend up the plane. Also as pulley is smooth, the tension in string (M × 9.81) will remain same throughout.

Friction  5.15

Now first consider impending motion of the block down the plane and then apply conditions of equilibrium i.e. SH = 0 along the inclined plane, and SV = 0 perpendicular to the inclined plane. Free body diagram of the block is shown in figure Ex 8(a). SH = 0: F + M × 9.81 – 50 × 9.81 × sin 30° = 0 Now for impending motion from (Eq. 5.1), we know Fmax = μs N. \ μs N + M × 9.81 – 50 × 9.81 × sin 30° = 0 M × 9.81 = 50 × 9.81 × sin 30° – 0.30 × N Now, \

Fig. Ex. 8(a)

...(Eq.1)

SV = 0: N – 50 × 9.81 × sin 30° = 0 N = 245.25

Substituting value of N in to Eq. 1, M × 9.81 = 50 × 9.81 × sin 30° – 0.30 × 245.25 \ M = Mmin = 17.5 kg  (Ans.)

Fig. Ex. 8(b)

Now consider impending motion of the block up the plane and then apply conditions of equilibrium i.e. SH = 0 along the inclined plane, and SV = 0 perpendicular to the inclined plane. Free body diagram of the block is shown in figure Ex 8(b). SH = 0: – F + M × 9.81 – 50 × 9.81 × sin 30° = 0 Now for impending motion from (Eq. 5.1), we know Fmax = μs N. \ – μs N + M × 9.81 – 50 × 9.81 × sin 30° = 0 M × 9.81 = 50 × 9.81 × sin 30° + 0.30 × N ...(Eq.2) Now, SV = 0: N – 50 × 9.81 × sin 30° = 0 \ N = 245.25 Substituting value of N in to Eq. 2, M × 9.81 = 50 × 9.81 × sin 30° + 0.30 × 245.25 \ M = Mmax = 32.5 kg  (Ans.) \ Range of mass M is from 17.5 kg to 32.5 kg that can keep the block in equilibrium.

5.16  Engineering Mechanics Statics

Example 9. Block A of 100 N and block B of 50 N are resting on a horizontal surface as shown in Figure Ex. 9. Determine magnitude of horizontal force P applied to block A to initiate motion. Assume coefficient of static friction between block A and horizontal surface as 0.20. B P

A

Fig. Ex. 9

Solution: Here we can consider impending motion of the block A towards right +

and then apply conditions of equilibrium i.e. →ΣH = 0, and +↑SV = 0. Free body diagram of the entire system is shown in figure Ex 9(a). SH = 0: P – F = 0 50 N \ P = F Now for impending motion from (Eq. 5.1), we P know Fmax = μs N. 100 N ...(Eq.1) \ P = μs N F SV = 0: N – 50 – 100 = 0 N \ N = 150 N Substituting value of N in to Eq. 1, Fig. Ex. 9(a) P = 0.2 × 150 \ P = 30 N  (Ans.) Example 10. In Ex. 9 if the horizontal force P is applied to Block B as shown in Figure Ex. 10. Determine magnitude of horizontal force P to initiate motion. Assume coefficient of static friction between block A and horizontal surface as 0.20 and between block A and block B as 0.30. B

P

A

Fig. Ex. 10

Solution: Here there are two possible ways in which motion can impend: impending motion of entire system on horizontal surface, or impending motion of block B on block A. First consider impending motion of the entire system towards right and then +

apply conditions of equilibrium i.e. →ΣH = 0, and +↑SV = 0. Free body diagram of the entire system is shown in figure Ex 10(a).

Friction  5.17

SH = 0: P – F = 0 50 N P \ P = F Now for impending motion from (Eq. 5.1), we 100 N know Fmax = μs N. ...(Eq.1) \ P = μs N F SV = 0: N – 50 – 100 = 0  \ N = 150 N N Substituting value of N in to Eq. 1, Fig. Ex. 10(a) P = 0.2 × 150 \ P = 30 N Now consider impending motion of the block B on block A and then apply +

conditions of equilibrium i.e. →ΣH = 0, and +↑SV = 0 to block A and block B separately. Free body diagram of block B and block A are shown in figure Ex 10(b).

Fig. Ex. 10(b)

Block B: SH = 0: P – F2 = 0 ...(Eq.2) \ P = F2 Now for impending motion from (Eq. 5.1), we know Fmax = μs N. ...(Eq. 3) P = μs N2 \ SV = 0: N2 – 50 = 0  \ N2 = 50 N Substituting value of N in to Eq. 3, P = 0.3 × 50  \ P = 15 N From Eq. 2, P = F2  \ F2 = 15 N Block A: SH = 0: F2 – F1 = 0 \ F2 = F1  \ F1 = 15 N SV = 0: N1 – 50 – 100 = 0  \ N1 = 150 N Now for impending motion from (Eq. 5.1), we know Fmax = μs N. \ (F1)max = μs N = 0.2 × 150  \ (F1)max = 30 N

5.18  Engineering Mechanics Statics

We have determined that for impending motion of entire system on horizontal surface, P = 30 N and for impending motion of block B on block A, P = 15 N. Therefore, the motion will initiate of block B on block A at P = 15 N  (Ans.) It is also clear from analysis of block A that calculated value of F1 = 15 N < (F1)max = 30 N. Therefore at P = 15 N motion will not initiate between block A and horizontal surface. Example 11. Two boxes A and B of weight 250 N and 150 N respectively are placed on an inclined surface as shown in Figure Ex. 11. Determine value of θ so that motion of boxes impend down the plane. Assume coefficient of static friction between block A and inclined surface as 0.30 and between block B and inclined surface as 0.20.

Fig. Ex. 11

Solution: Considering impending motion of the blocks down the plane and then apply conditions of equilibrium i.e. SH = 0 along the inclined plane, and SV = 0 perpendicular to the inclined plane to blocks A and B separately. Free body diagram of the blocks are shown in figure Ex. 11(a). 250 N

150 N



R



R F N N

F

Fig. Ex. 11(a)

Block A: SH = 0: R + 250 sin θ – F = 0 R = F – 250 sin θ Now for impending motion from (Eq. 5.1), we know Fmax = μs N. \ R = 0.30 N – 250 sin θ ...(Eq. 1) Now, SV = 0: N – 250 cos θ = 0 \ N = 250 cos θ

Friction  5.19

Substituting value of N in to Eq. 1, R = 0.30 × 250 cos θ – 250 sin θ ...(Eq. 2) SH = 0: – R + 150 sin θ – F = 0 Block B: R = 150 sin θ – F Now for impending motion from (Eq. 5.1), we know Fmax = μs N. R = 150 sin θ – 0.20 N ...(Eq. 3) \ Now, SV = 0: N – 150 cos θ = 0 \ N = 150 cos θ Substituting value of N in to Eq. 3, R = 150 sin θ – 0.20 × 150 cos θ ...(Eq. 4) From Eq. 2 and Eq. 4, 0.30 × 250 cos θ – 250 sin θ = 150 sin θ – 0.20 × 150 cos θ 75 cos θ – 250 sin θ = 150 sin θ – 30 cos θ \ 105 cos θ = 400 sin θ \ θ = 14.70°  (Ans.) Example 12. A 300 N crate is subjected to 80 N force as shown in Figure Ex. 12. Does the crate slide, tip or remains in equilibrium? Assume ms = 0.3. 1m

80 N 2m 1m

Fig. Ex. 12

Solution: Here we have to understand that there are two possibility of impending motion: impending sliding or impending tipping (In all previous examples possibility of tipping was neglected). Now assume equilibrium of the crate and apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anti-clockwise moment as positive) to the free body diagram of the crate as shown in figure Ex.12(a). Note that the normal reaction must act at a distance x from the crate’s centre line in order to counteract tipping effect caused by applied force.

Fig. Ex. 12(a)

5.20  Engineering Mechanics Statics

SH = 0: 80 – F = 0 \ F = 80 N SV = 0: N – 300 = 0 \ N = 300 N \ x = 0.27 m SMO = 0: 300x – 80 × 1 = 0 Now from (Eq. 5.1), Fmax = μs N, \ Fmax = 90 N Fmax = 0.3 × 300 Since calculated value of friction force from equilibrium analysis is less than Fmax (F < Fmax), we conclude that the crate will not slide. Now largest possible value for x is 0.5 i.e. half the width of the crate. Since calculated value of x from equilibrium analysis is less than largest possible value of x, we conclude that the crate will not tip. \ Crate remains in equilibrium when subjected to 80 N force  (Ans.)

Solved Examples Based on Ladder Friction Example 1. A 5 m ladder of weight 300 N is placed against a smooth vertical wall as shown in Figure Ex. 1. Does the ladder remains in equilibrium? Assume ms between ladder and horizontal surface as 0.3.

Fig. Ex. 1

Solution: Assume equilibrium of the ladder and apply conditions of equilibrium +

i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the ladder as shown in Figure Ex. 1(a). SH = 0: FA – NB = 0  \ FA = NB ...(Eq.1) SV = 0: NA – 300 = 0  \ NA = 300 N SMA = 0: NB × 5 sin 60° – 300 × 2.5 cos 60° = 0  \ NB = 86.60 N From Eq. 1, FA = 86. 60 N

Friction  5.21

Fig. Ex. 1(a)

Now from (Eq. 5.1), Fmax = μs N, we can check our assumption of equilibrium of the ladder as below, Fmax = 0.3 × NA = 0.3 × 300 \ Fmax = 90 N Since calculated value of friction force from equilibrium analysis is less than Fmax (86.60 < 90), we conclude that the ladder is in static equilibrium.  (Ans.) Example 2. A 800 N man starts climbing a ladder that placed against a wall as shown in Figure Ex. 2. Neglecting the weight of the ladder, determine how far up the ladder the man can climb before the ladder starts slipping. Assume ms at both surfaces as 0.25. B

30° 8m

x

A

Fig. Ex. 2

Solution: Considering impending motion of the ladder, apply conditions of +

equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the ladder as shown in figure Ex.2(a).

5.22  Engineering Mechanics Statics B NB 30° FB

x 800 N A FA NA

Fig. Ex. 2(a)

SH = 0: FA – NB = 0



FA = NB

\

Now for impending motion we know from (Eq. 5.1), Fmax = μs N \

0.25 NA = NB



...(Eq. 1)

SV = 0: NA – 800 + FB = 0

\ NA = 800 – FB = 800 – 0.25 NB Substituting the value of NA in Eq. 1, 0.25 (800 – 0.25 NB) = NB \

NB = 188.23 N

From Eq. (1), 0.25 NA = 188.23 \

NA = 752.94 N

SMA = 0: NB × 8 cos 30° + 0.25 NB × 8 sin 30° – 800 × x sin 30° = 0 Substituting the value NB , 188.23 × 8 cos 30° + 0.25 × 188.23 × 8 sin 30° = 800 × x sin 30° Solving,  \

x = 3.73 m  (Ans.)

Example 3. A 600 N man starts climbing a ladder of weight 100 N that placed against a wall as shown in Figure Ex. 3. The centre of mass of 10 m ladder is at its midpoint. What is the largest value of α for which a man can climb to the top of the ladder without slipping? Assume coefficient of static friction between ladder and floor as 0.4 and coefficient of static friction between ladder and the wall as 0.3.

Friction  5.23

B



A

Fig. Ex. 3

Solution: Considering impending motion of the ladder, apply conditions of +

equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the ladder as shown in figure Ex.3(a). 600 N

NB FB 5m 100 N  FA NA

Fig. Ex. 3(a)

SH = 0: NB – FA = 0 NB = FA \ Now for impending motion we know from (Eq. 5.1), Fmax = μs N \ NB = 0.4 NA SV = 0: NA – 600 – 100 + FB = 0 \ NA = 700 – FB = 700 – 0.3 NB Substituting the value of NA in Eq. 1, NB = 0.4 (700 – 0.3 NB)  \ NB = 250 N From Eq. 1, 250 = 0.4 NA  \ NA = 625 N

...(Eq. 1)

5.24  Engineering Mechanics Statics

SMA = 0: 100 × 5 cos α + 600 × 10 cos α – NB × 10 sin α – FB × 10 cos α = 0 Substituting the value NB, 100 × 5 cos α + 600 × 10 cos α – 250 × 10 sin α – 0.3 × 250 × 10 cos α = 0 Solving, \ α = 66.50°  (Ans.) Example 4. A 600 N man starts climbing a ladder of weight 100 N that placed

against a smooth wall as shown in Figure Ex. 4. The centre of mass of 10 m ladder is at its midpoint. Determine the minimum coefficient of static friction between the ladder and the floor for which a man can climb to the top of the ladder without slipping.

B

30°

A

Fig. Ex. 4

Solution: Considering impending motion of the ladder, apply conditions of +

equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the ladder as shown in figure Ex.4(a).

Fig. Ex. 4(a)

Friction  5.25

SH = 0: NB – FA = 0 \ NB = FA Now for impending motion we know from (Eq. 5.1), Fmax = μs N ...(Eq. 1) \ NB = μs NA SV = 0: NA – 600 – 100 = 0 NA = 700 \ Substituting the value of NA in Eq. 1, NB = μs × 700 SMA = 0: 100 × 5 sin 30° + 600 × 10 sin 30° – NB × 10 cos 30° = 0 Substituting the value NB, 100 × 5 sin 30° + 600 × 10 sin 30° – (μs × 700) × 10 cos 30° = 0 Solving, \ μs = 0.536  (Ans.) Example 5. A 600 N man starts climbing a ladder that placed against a wall as shown in Figure Ex. 5. The weight of the ladder is 80 N. Determine how far up the ladder the person can climb before the ladder starts slipping. Assume ms at both surfaces as 0.20. 1m B

3m

x

A

4m

Fig. Ex. 5

Solution: First determine inclination of ladder with horizontal as below, 3 tan α =   \ α = 36.87° 4 Now considering impending motion of the ladder, apply conditions of +

equilibrium i.e. →ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the free body diagram of the ladder as shown in figure Ex.5(a). SH = 0: FA – NB sin 36.87° + FB cos 36.87° = 0 \ FA = NB sin 36.87° – FB cos 36.87° Now for impending motion we know from (Eq. 5.1), Fmax = μs N

5.26  Engineering Mechanics Statics

\ 0.2 NA = NB sin 36.87° – 0.2 × NB cos 36.87° ...(Eq.1) NA = 2.20 NB SV = 0: NA – 600 – 80 + NB cos 36.87° + 0.2 × NB sin 36.87° = 0 \ NA + 0.919 NB = 680 Substituting the value of NA from Eq. 1, 2.20 NB + 0.919 NB = 680  \ NB = 218 N From Eq. (1), NA = 2.20 × 218  \ NA = 479.6 N SMA = 0: NB × 5 – 600 × x cos 36.87° – 80 × 3 cos 36.87° = 0 Substituting the value NB, 218 × 5 – 600 × x cos 36.87° – 80 × 3 cos 36.87° = 0 Solving \ x = 1.87 m  (Ans.)

Fig. Ex. 5(a)

Solved Examples Based on Wedge Friction Example 1. The weight of the block A is 500 N. Neglecting the weight of wedge B, determine the force P required to produce impending motion of block A. Assume coefficient of static friction at all contact surfaces as 0.3. Refer Figure Ex. 1.

Fig. Ex. 1

Friction  5.27

Solution: Considering impending motion, apply conditions of equilibrium i.e. +

→ΣH = 0, and +↑SV = 0 to the block A and wedge B separately. Free body diagrams of the block and the wedge are shown in figure Ex 1(a). Here we have to understand that the wedge is subjected to more number of unknowns compare to block, therefore we apply equations of equilibrium to block first. Block A: SH = 0: F2 cos 15° + N2 sin 15° – N3 = 0 Now for impending motion from (Eq. 5.1), we know Fmax = μs N. ...(Eq.1) 0.3 × N2 cos 15° + N2 sin 15° = N3 Now, SV = 0: – F2 sin 15° + N2 cos 15° – 500 – F3 = 0 – 0.3 × N2 sin 15° + N2 cos 15° – 500 – 0.3 N3 = 0 Substituting value of N3 from Eq. 1, – 0.3 × N2 sin 15° + N2 cos 15° – 0.3 (0.3 × N2 cos 15° + N2 sin 15°) = 500 Solving, N2 = 690.79 N Substituting value of N2 in Eq. (1), N3 = 378.95 N F2 = 0.3 × N2 = 0.3 × 690.79  \ F2 = 207.23 N Now, F3 = 0.3 × N3 = 0.3 × 378.95  \ F3 = 113.685 N W F3 N3 15° N

2

F2 F2

P

N2 15° F1 N1

Fig. Ex. 1(a)

Wedge B: SH = 0: P – F1 – F2 cos 15° – N2 sin 15° = 0 P = F1 + F2 cos 15° + N2 sin 15° Now for impending motion from (Eq. 5.1), we know Fmax = μs N. ...(Eq. 2) P = 0.3 N1 + F2 cos 15° + N2 sin 15° Now, SV = 0: N1 – N2 cos 15° + F2 sin 15° = 0 Substituting values of N2 and F2 obtained in previous part, \ N1 = 690.79 × cos 15° – 207.23 × sin 15° \ N1 = 613.62 N

5.28  Engineering Mechanics Statics

Substituting value of N1, F2 and N2 in to Eq. 2, P = 0.3 × 613.62 + 207.23 × cos 15° + 690.79 sin 15° \ P = 563 N (considering approximate value)  (Ans.) Alternate Approach: This example can also be solved by considering angle of static friction, ∅s = tan–1 μs from (Eq. 5.4). Here consider free body diagrams of the block and the wedge as shown in figure Ex 1(b). R1, R2 and R3 on free body diagram are the resultant forces of normal reaction and friction force at that surfaces. Angle of friction is measured in between normal reaction and resultant force. Here ∅s = tan–1 0.3 = 16.69°

Fig. Ex. 1(b)

Now considering equilibrium of block, apply Lami’s theorem to the free body diagram shown in figure Ex 1(c). 500 R3 R2 = = sin 138.38° sin 73.31° sin 148.31° R2 = 721.08 N and R3 = 395.46 N

\

15°+

R1

R2

 P

 R3 R2



500 N

Fig. Ex. 1(c)



15°+

Fig. Ex. 1(d)

Now considering equilibrium of wedge, apply Lami’s theorem to the free body diagram shown in figure Ex 1(d). R2 P = sin 106.69° sin 131.62° \

P = 563 N (considering approximate value)  (Ans.)

Friction  5.29

Example 2. Block A and B each of weight 150 N resting on horizontal surface are to be pushed apart by 20° wedge as shown in Figure Ex. 2. Neglecting the weight of the wedge determine the force P required to start movements of blocks. Assume coefficient of static friction at all contact surfaces as 0.25.

Fig. Ex. 2

Solution: Considering impending motion, apply conditions of equilibrium +

i.e. →ΣH = 0, and +↑SV = 0 to the blocks and wedge separately. Free body diagrams of the blocks and the wedge are shown in figure Ex 2(a). Here we have to understand that the wedge is subjected to more number of unknowns compare to block, therefore we apply equations of equilibrium to block first. As two blocks are identical we can consider only one block for analysis. P

F2

N2

10°

F2

10° N2

A N2 10°

N2 F2

F2

10°

F1

F1 N1

N1

Fig. Ex. 2(a)

Block A: SH = 0: –F1 + N2 cos 10° – F2 sin 10° = 0 Now for impending motion from (Eq. 5.1), we know Fmax = μs N. –0.25 N1+ N2 cos 10° – 0.25 N2 sin 10° = 0 ...(Eq.1) \ N1 = 3.76 N2 Now, SV = 0: N1 – N2 sin 10° – 0.25 N2 cos 10° – 150 = 0 Substituting value of N1 from Eq. 1, 3.76 N2 – N2 sin 10° – 0.25 N2 cos 10° = 150 Solving, N2 = 44.91 N

5.30  Engineering Mechanics Statics

Substituting value of N2 in Eq. (1), N1 = 168.86 N Now, F2 = 0.25 × N2 = 0.25 × 44.91 \ F2 = 11.22 N Wedge: SV = 0: 2 (F2 cos 10°) + 2 (N2 sin 10°) – P = 0 Substituting values of N2 and F2 obtained in previous part, P = 2 (11.22 × cos 10°) + 2 (44.91 × sin 10°) \ P = 37.68 N  (Ans.) Example 3. A block of 1000 N is to be raised by means of the wedges A and B as shown in Figure Ex. 3. Neglecting the weight of the wedges determine the force P required to produce impending motion of the block upwards. Assume coefficient of static friction at all contact surfaces as 0.25.

Fig. Ex. 3

Solution: Considering impending motion, apply conditions of equilibrium i.e. +

→ΣH = 0, and +↑SV = 0 to the block and wedges separately. Free body diagrams of the block and the wedge are shown in figure Ex 3(a). As two wedges are identical we can consider only one wedge for analysis.

Fig. Ex. 3(a)

Friction  5.31

Wedge A: SH = 0: F2 cos 15° + N2 sin 15° – F2 cos 15° – N2 sin 15° = 0 Now, SV = 0: 2 (N2 cos 15°) – 2 (F2 sin 15°) – 1000 = 0 Now for impending motion from (Eq. 5.1), we know Fmax = μs N 2(N2 cos 15°) – 2 (0.25 × N2 sin 15°) = 1000 Solving,

N2 = 555.55 N

Now,

F2 = 0.25 × N2 = 0.25 × 555.55  \ F2 = 138.88 N

Wedge B:

SH = 0: – P + 0.25 N1 + F2 cos 15° + N2 sin 15° = 0

Substituting values of N2 and F2 obtained in previous part, P = 0.25 N1 + 138.88 cos 15° + 555.55 sin 15°

...(Eq.1)



SV = 0: N1 + 138.88 sin 15° – 555.55 cos 15° = 0

Solving,

N1 = 500.68 N

Substituting values of N1 in Eq.1

P = 0.25 × 500.68 + 138.88 cos 15° + 555.55 sin 15°

\

P = 403.1 N  (Ans.)

Example 4. A block of 1000 N is rest on horizontal surface as shown in Figure Ex. 4. Neglecting the weight of the wedge determine the force P required to produce impending motion of the block. Assume coefficient of static friction at all contact surfaces as 0.3. P

1000 N

18°

Fig. Ex. 4

Solution: Considering impending motion, apply conditions of equilibrium +

i.e. →ΣH = 0, and +↑SV = 0 to the block and wedge separately. Free body diagrams of the block and the wedge are shown in figure Ex 4(a). Here we have to understand that the wedge is subjected to more number of unknowns compare to block, therefore we apply equations of equilibrium to block first.

5.32  Engineering Mechanics Statics P

N2

N3

1000 N F2

F3

F2 N2

F1 N1

Fig. Ex. 4(a)

Block: SH = 0: F1 – N2 = 0 \ F1 = N2 Now for impending motion from (Eq. 5.1), we know Fmax = μs N ...(Eq.1) \ 0.3 N1 = N2 Now, SV = 0: N1 – 1000 – F2 = 0 N1 – 1000 – 0.3 N2 = 0 Substituting value of N2 from Eq. 1 N1 – 0.3 (0.3 N1) = 1000 Solving, N1 = 1098.9 N From Eq. (1), N2 = 329.6 N Wedge: SH = 0: N2 – N3 cos 18° + F3 sin 18° = 0 N2 – N3 cos 18° + (0.3 N3) sin 18° = 0 Substituting values of N2 obtained in previous part, 329.6 – N3 cos 18° + (0.3 N3) sin 18° = 0 Solving, N3 = 384.46 N SV = 0: – P + F2 + F3 cos 18° + N3 sin 18° = 0 P = F2 + F3 cos 18° + N3 sin 18° P = 0.3 × N2 + 0.3 × N3 cos 18° + N3 sin 18° Substituting values of N2 and N3 P = 0.3 × 329.6 + 0.3 × 384.46 cos 18° + 384.46 sin 18° \ P = 327.37 N  (Ans.) Example 5. A beam AB is supported by wedge as shown in Figure Ex. 5. Neglecting the weight of the wedge and beam determine the force P required to produce impending motion of the wedge towards left. Assume coefficient of static friction at all contact surfaces as 0.3.

Friction  5.33 500 N

800 N

B

A

P

10° 2m

2m 5m

Fig. Ex. 5

Solution: Considering impending motion, apply conditions of equilibrium i.e. +

→ΣH = 0, +↑SV = 0 and SM = 0 (Treating anticlockwise moment as positive) to the beam and wedge separately. Free body diagrams of the beam and the wedge are shown in figure Ex 5(a). Here we have to understand that the wedge is subjected to more number of unknowns compare to beam, therefore we apply equations of equilibrium to beam first. Beam: SH = 0: Ax – F2 cos 10° – N2 sin 10° = 0

Fig. Ex. 5(a)

Now, SV = 0: Ay – 500 – 800 + N2 cos 10° – F2 sin 10° = 0 Now for impending motion from (Eq. 5.1), we know Fmax = μs N ...(Eq.1) Ay – 500 – 800 + N2 cos 10° – 0.3 N2 × sin 10° = 0 SMB = 0: – 6Ay + 800 × 2 + 500 × 4 = 0 Ay = 600 N Solving, Substituting value of Ay in Eq. 1 600 – 500 – 800 + N2 cos 10° – 0.3 N2 × sin 10° = 0 Solving, N2 = 750.42 N As F2 = 0.3 N2 = 0.3 × 750.42  \ F2 = 225.12 N Wedge: SH = 0: – P + F1 + F2 cos 10° + N2 sin 10° = 0 Substituting values of N2 and F2 obtained in previous part, P = 0.3 N1 + 225.126 cos 10° + 750.42 sin 10° ...(Eq.2)

5.34  Engineering Mechanics Statics

SV = 0: N1 + F2 sin 10° – N2 cos 10° = 0 N1 + 225.12 sin 10° – 750.42 cos 10° = 0 Solving, N1 = 699.92 N Substituting values of N1 in Eq. 2 P = 0.3 × 699.92 + 225.12 cos 10° + 750.42 sin 10° \ P = 561.98 N  (Ans.)

Solved Examples Based on Belt Friction Example 1. The 500 N block is attached to a rope that passes over a fixed cylinder as shown in Figure Ex. 1. Determine the range of the horizontal force P for which the system will be at rest. Assume coefficient of static friction between the rope and the cylinder as 0.3. P

500 N

Fig. Ex. 1

Solution: Considering impending motion, apply (Eq. 5.10)

T2 = ems b to figure T1

Ex 1(a). In this equation T2 is tension on higher side and β=

π rad. 2

The maximum value of P for equilibrium occurs when the block is on the verge of moving upward. For this case, T2 = Pmax and T1 = 500 N π 

\

0.3   Pmax 2  = e 500

Fig. Ex. 1(a)

Solving, Pmax = 801 N  (Ans.) The minimum value of P for equilibrium occurs when the block is on the verge of moving downward. For this case, T2 = 500 N and T1 = Pmax π 

\

0.3   500 2  = e Pmin

Friction  5.35

Solving, Pmin = 312.30 N  (Ans.) \ Range of horizontal force P is from 801 N to 312.30 N that can keep the block in equilibrium. Example 2. The 500 N block is attached to a rope that passed over a fixed

cylinder as shown in Figure Ex. 2. If P = 250 N, determine the minimum value of angle α at which block begins to slip. Assume coefficient of static friction between the rope and the cylinder as 0.3.

Fig. Ex. 2

T Solution: Considering impending motion, apply (Eq. 5.10) 2 = ems b to figure T1 Ex 2(a). Here T2 = 500 N and T1 = 250 N  500 \ = e0.3 b 250 N 250 In

500 = 0.3 β 250

Solving,

 180  β = 2.31 rad = 2.31    π 

Now,

β = 132.38° α = 132.38° – 90°  \ α = 42.4°  (Ans.)

500 N

Fig. Ex. 2(a)

Example 3. The 200 N block is suspended from a rope that passes over two fixed cylinder as shown in Figure Ex. 3. Determine the smallest vertical force P required to exert on the rope to support the block. Assume coefficient of static friction between the rope and the right cylinder as 0.2 and between the rope and the left cylinder as 0.3.

200 N

P

Fig. Ex. 3

5.36  Engineering Mechanics Statics

T2 = emsb to the right T1 cylinder of figure Ex. 3(a). In this equation T2 = 200 N, T1 = T (tension in the rope π between two cylinders) and β = rad. 2 Solution: Considering impending motion, apply (Eq. 5.10)

T

T

200 N

P

Fig. Ex. 3(a) π 

0.2   200 2  \ = e T Solving, T = 146.08 N T Now apply (Eq. 5.10) 2 = emsb to the left cylinder of figure Ex. 3(a). In this T1 π rad. equation T2 = T, T1 = P and β = 2

146.08 = e P

\ Solving,

π  0.3   2 

P = 91.18 N  (Ans.)

Example 4. Two blocks of 175 N and 350 N respectively are suspended from a rope that passes over a fixed cylinder as shown in Figure Ex. 4. Determine minimum coefficient of friction between rope and the cylinder that keep system in equilibrium. Solution: Considering impending motion, apply (Eq. 5.10)

175 N 350 N

Fig. Ex. 4

T2 = emsb to figure Ex. 4(a). T1



T2 = 350 N, T1 = 175 N and β = π rad.

Here \

350 = em(p) 175

Solving,

μ = 0.2206  (Ans.)

175 N 350 N

Fig. Ex. 4(a)

Example 5. Block A of weight 200 N connected to block B of weight w using a rope that passes over a fixed cylinder as shown in Figure Ex. 5. Determine minimum weight of block B for which no motion occurs. Assume coefficient of static friction between the rope and the cylinder as 0.2 and between the block A and horizontal surface as 0.3.

Friction  5.37 140 N A

B

Fig. Ex. 5

Solution: Consider impending motion of the block A towards right and apply +

conditions of equilibrium i.e. →ΣH = 0, and +↑SV = 0 to the free body diagram of the block A as shown in figure Ex. 5(a) SH = 0: 140 – F – T = 0 200 N Now for impending motion from (Eq. 5.1), we know Fmax = μs N. ...(Eq.1) T = 140 – 0.3 N 140 N T SV = 0: N – 200 = 0 \ N = 200 N F Substituting value of N in Eq. 1, N T = 140 – 0.3 × 200 Fig. Ex. 5(a) \ T = 80 N Now considering impending motion, apply T (Eq. 5.10) 2 = emsb to figure Ex. 5(b). T1

Fig. Ex. 5(b)

Here T2 = T (tension in the rope between block A and cylinder), T1 = w and π rad. β= 2 π  0.2   80 2 \ = e   w Solving, w = 58.43 N  (Ans.)

5.38  Engineering Mechanics Statics

SUMMARY

• Frictional force is tangent to the surfaces of contact of the two bodies. • Coulomb’s law of friction before sliding is, F < μs N • Coulomb’s law of friction at impending motion is, Fmax = μs N • Coulomb’s law of friction after sliding is F = μk N • Maximum frictional force Fmax is called as limiting friction. • At limiting friction body is said to have impending motion. • The angle of friction is defined as, ∅ = tan–1 μ • At impending motion, Angle of repose (α) = Angle of static friction (∅s). • At impending slip between cable/ rope and cylinder, the ratio of tension on high side and tension on low side are related as, T2 = eμsβ T1

PROBLEMS 1. A 300 N block is subjected to force of 400 N as shown in figure Prob. 1. Determine the friction force between the block and the surface. Assume ms = 0.3 and mk = 0.25. 2. The 10 N force as shown in figure Prob. 2 is required to produce impending motion of the block up the inclined surface. If the weight of the block is 5 kg, determine the coefficient of static friction (ms) between the block and the inclined surface.

400 N 25°

Fig. Prob. 1 10 N 20°

30°

Fig. Prob. 2

3. The 1 kg block is subjected to the load of 2 kg as shown in figure Prob. 3. Determine the friction force between the block and the inclined surface. Assume ms = 0.9.

2 Kg

1 Kg 2 4

Fig. Prob. 3

Friction  5.39 4. Block A of 100 N and block B of 400 N are connected with cable that passes over a frictionless pulley as shown in figure Prob. 4. At impending motion of the blocks, determine coefficient of static friction between block B and inclined surface. Assume ms = 0.25 between block A and B. 5. Block A of 200 N and block B of 100 N are connected with cable that passes over a frictionless pulley as shown in figure Prob. 5. Determine the range of force P that keep blocks in equilibrium. Assume ms = 0.20 between block A and inclined surface.

A B 30°

Fig. Prob. 4 P

A A 30°

B

Fig. Prob. 5

6. Block A weight 100 N and block B weight 60 N. Horizontal force P is acting to block A as shown in figure Prob. 6. Determine, (a) if the system is in equilibrium at P = 40 N, (b) maximum value of P that keep the system in equilibrium. Assume ms = 0.30 between block A and B and ms = 0.25 between block A and horizontal surface. 7. A crate of weight 500 N is subjected to horizontal force P as shown in figure Prob. 7. Determine the force P required to cause tipping of the crate also determine coefficient of static friction for tipping.

50 N 45° B P A

Fig. Prob. 6 1m

P

500 N

2m

1m

Fig. Prob. 7 m

P

0m

50°

50

8. A 300 N block is resting on an inclined surface and subjected to the force P as shown in figure Prob. 8. Determine maximum value of force P for equilibrium of block and also check whether the impending motion will be by tipping or by slipping. Assume ms = 0.50 between block and inclined surface.

80

20°

0m

m

Fig. Prob. 8

5.40  Engineering Mechanics Statics 9. The rod AC of mass 5 kg rest as shown in figure Prob. 9. Determine coefficient of statics friction for impending motion of the rod.

C

1.2

B

m

0.5 m 30° A

Fig. Prob. 9

10. The rod AB of 3 m and weight 5 kg is placed as shown in figure Prob. 10. Determine maximum value of angle θ at which the rod can be placed before it slip. Assume ms = 0.25 for all contact surfaces.

Fig. Prob. 10

11. The weight of block A is 500 N and block B is 50 N. Determine the force P required to produce impending motion of block A. Assume = 0.30 for all contact surfaces. Refer figure Prob. 11.

P

B A 70°

Fig. Prob. 11

12 The weight of block A is 800 N and wedge B is 400 N. Determine the force P required to produce impending motion of block A. Assume = 0.20 for all contact surfaces. Refer figure Prob. 12.

A

8° B 8°

P

Fig. Prob. 12

13. A rope is wrapped around a rod as shown in figure Prob. 13. If tension on one end of rope is 180 N and on other end is 20 N, determine the least number of turns of rope around rod in order to prevent slipping of the rope. Assume ms = 0.20 between rope and rod.

20 N 180 N

Fig. Prob. 13

14. Determine the range of weight W that may be applied without causing the 250 N block to slip. Assume ms = 0.25 between block and inclined surface and ms = 0.30 between the rope and the cylinder.

W 40°

Fig. Prob. 14

Unit 6

In this unit, we discuss concept of stress and strain. We study the stress induced in the member due to axial loading. We also discuss the deformation caused by the load. Systematically understanding of these concepts is important to the design engineers.

Simple Stress & Strain 6.1 stress All bodies offer an equal internal resistance to the externally applied forces. The magnitude of the resisting force per unit area is called stress. The stress is denoted by Greek letter sigma (σ). P s = A

Stress is internal resistance set up by a body when it is deformed. ...(Eq. 6.1)

In above equation P is external force expresses in Newton (N) and A is original cross section area of the specimen and expresses in square meter or millimetre (m2 or mm2). So, SI unit of stress is N/m2 or N/mm2. 1 N/m2 is called Pascal (Pa). 1 kPa = 103 Pa = 103 N/ m2 1 MPa = 106 Pa = 106 N/ m2 = 1 N/mm2 1 GPa = 109 Pa = 109 N/ m2 Figure 1 shows square bar subjected to pull force P and internal resistance at cutting section.

Fig. Ex. 1

6.1.1 Types of Stress Generally there are two types of stress, (a) Normal Stress (b) Shear Stress (a) Normal Stress: The stress developed on a plane normal to it is called normal stress. It is equal to the force acting on the body per unit normal area. A normal stress is a stress that occurs when a member is loaded by an axial force and member is placed in tension or compression. If the member is subjected to axial tension, the stress developed at a section is called tensile stress.

6.4  Engineering Mechanics Statics

If the member is subjected to axial compression, the stress developed at a section is called compressive stress.



(b) Shear Stress: The stress developed when a member is subjected to a force that is parallel or tangent to the surface is called shear stress. It is denoted by Greek letter tau (τ). It is obtained by dividing the magnitude of the resultant shear force (V) by the cross sectional area (A) V t = ...(Eq. 6.2) A

6.2 Strain

If Lo is the initial gauge length and L is the observed length under a given load, the gauge elongation, dL = L – Lo as shown in figure 2. The elongation (or contraction) per unit of the initial gauge length is given as: δL e = L0

Strain is defined as the change in length per unit length. ...(Eq. 6.3)

Simple Stress & Strain  6.5

Fig. 2

This expression defines the tension (or compression) strain. Since this is associated with the normal stress, it is usually called the normal strain. It is a dimensionless quantity (mm/mm). This is also called the nominal or engineering strain.

6.2.1 Shear Strain Consider a block OABC fixed on horizontal surface as shown in figure 3. It is subjected to force P acting on its upper face. The deformation of the block because of force P appears as OA’B’C. Now shear strain is defined as, BB′ = tan φ Shear strain = CB

P A

A

B

 O

C

Fig. 3

6.2.2  Longitudinal and Lateral Strains Consider a circular test specimen of diameter d and length L as shown in figure 4. If a tensile load P is applied to the specimen its length increases. The increase in length per unit length is called the longitudinal strain. Since the volume of the specimen remains constant, therefore, the increase in length is accompanied by a decrease in diameter. This decrease in diameter per unit diameter is called the lateral strain. The longitudinal and lateral strains are of opposite nature. δL eLongitudinal = L

eLateral =

δd d

B

...(Eq. 6.4) Fig. 4

6.6  Engineering Mechanics Statics

6.3  Poisson’s Ratio The ratio of the lateral strain to the longitudinal strain is called the Poisson’s ratio. It is denoted by v (Greek letter Nue). Poisson’s ratio for most Lateral strain of the materials varies Poisson’s ratio = Longitudinal strain from 0.25 to 0.40.

6.4  Volumetric Strain It is the ratio of the change in volume of the body to its original volume V0, when subjected to hydrostatic stress. ∆V ...(Eq. 6.5) Volumetric strain eV = V

6.5 Hooke’s Law And Elastic Moduli According to this Law, stress is directly proportional to strain, within the elastic limits. σ ∝ ε or σ = Eε (a) Modulus of Elasticity, E: It is defined as the ratio of normal stress to normal strain within the elastic limits. In above equation the constant of proportionality, E is called modulus of Elasticity. This equation can be used to find out change in length of the bar as below P σ A E = = ε δL L PL dL = ...(Eq. 6.6) AE

(b) Modules of Rigidity, G: It is defined as the ratio of shearing stress to shearing strain. τ ...(Eq. 6.7) G = γ



(c) Bulk Modulus, K: It is the ratio of hydrostatic stress to volumetric strain.

6.6  Bar of Varying Cross-Section Consider a bar of varying cross-section shown in figure 5. The free body diagrams for the various parts are shown in figure 6. The stresses in the various parts are: P P P −P sAB = 1 , σ BC = 1 2 , σCD = 3 A1 A2 A3

Simple Stress & Strain  6.7

Fig. 5

Fig. 6

The total elongation in the bar is find out as

δL =

n

PL

∑ Ai Ei i =1



...(Eq. 6.8)

i i

6.7  Stress-Strain Diagram For Mild Steel Stress-Starin diagram is a graphical representation of the relationship between stress (σ) and strain (ε). It shows behaviour of material under loading. The curve varies from material to material. The stress-strain diagram for a ductile material like mild steel is shown in figure 7. The curve starts from the origin, showing thereby that there is no initial stress of strain in the specimen.

Fig. 7

6.8  Engineering Mechanics Statics

Salient Point of the Diagram A: Upto point A, Hooke s law is obeyed and  stress is proportional to strain or elongation is proportional to the load. Therefore, OA is a straight line. Point A is called the limit of proportionality or proportional limit. B: Up to point B, the material remains elastic, i.e.  on removal of the load, deformations are completely recovered. AB is not a straight line. Point B is called the elastic limit point. Beyond point B, the material goes to the plastic stage until the upper yield point C is reached. C & D: Beyond point B plastic deformation occurs and material is not totally recoverable. Here permanent deformation takes place when load is removed. At this point the cross-sectional area of the material starts decreasing. Points C & D are termed as upper and lower yield points respectively. The stress at the yield point is called the yield strength. E: Between DE, the specimen elongates by a considerable amount without any increase in stress. Point E corresponds to ultimate strength of a material. The maximum load which a specimen can withstand without failure is known as load at ultimate strength. F: Beyond point E, the bar begins to form neck and its cross-sectional area decreases at a rapid rate. The apparent stress deceases but the actual or true stress goes on increasing until the specimen breaks at point F, called the point of fracture. The fracture of ductile material is of the cup and cone type.

6.8  Temperature Stress and Strain When the temperature of a material is changed, its dimensions also change. A stress is setup in the material for change in dimension due to temperature change is prevented. This is called the temperature stress. Let, L be the length of the member at temperature t, dt is the change in temperature, α is the coefficient of linear expansion for the material change in length Now, δL = αLδt Expanded length = L (1 + α dt) α L δt \ Temperature strain et = L(1 + α δt ) \ Temperature strain et = adt ...(Eq. 6.9) Temperature stress st = et E ...(Eq. 6.10) at = a dt E When the temperature increases, length increases. Since this increase in length is prevented, compressive stress is developed in the material. The reverse phenomena occur when the temperature is decreased and tensile stress is developed.

Simple Stress & Strain  6.9

6.9  Composite System Composite system consists of two or more bar of different materials in parallel. In such system the sharing of load by each material can be found using equilibrium. Consider a composite system shown in figure 8 P consisting of different materials and area of crosssection, subjected to load P. P = P1 + P2 + P3 1 2 3 Now using following compatibility equation we A1 A2 A3 L can determine load or stresses in each materials. P1L PL PL = 2 = 3 A1E1 A2 E2 A3 E3

6.10  Factor Of Safety

Fig. 8

It is the ratio of the maximum permissible stress to which a member can be subjected to the allowable or working stress. ultimate load Factor of safety = allowable load Factor of safety =

ultimate stress allowable stress

Solved Examples Based on Stress and Strain Example 1. A metal wire is 5 mm diameter and 2 m long is subjected to 24 N tensile force that stretches 0.3 mm. Determine stress and strain in the wire. Solution: Area of cross-section A = π r2 , where r is radius of wire \ A = π × (2.5)2 = 19.634 mm2 P 24 Now we know, s = = A 19.634

s = 1.22 N/mm2  (Ans.) δL 0.3 = = 0.00015   (Ans.) Now we know, e = L0 2000 \

Example 2. A circular rod of diameter 8 mm and 300 mm long is subjected to a tensile force 20 kN. Take E = 200 kN/mm2. Find stress, strain and elongation of the bar due to applied load. Solution: Area of cross-section A = πr2, where r is radius of wire \ A = π × (4)2 = 50.26 mm2 P 20 × 1000 Now we know, s = = A 50.26

6.10  Engineering Mechanics Statics

s = 397.93 N/mm2  (Ans.) σ 397.93 Now we know, e = = E 200 × 1000 \

\ \

e = 0.001989  (Ans.) dL = e × L = 0.001989 × 300 dL = 0.5968 mm  (Ans.)

Example 3. A hollow circular tube of 200 mm long is subjected to compressive load of 26 kN. The inner and outer diameters of the tube are 40 mm and 45 mm respectively. The shortening of the tube due to load is 0.12 mm. Determine stress and strain in the tube. Solution: Area of cross-section A = π (outer radius2 – inner radius2) \ A = π × (22.52 – 202) = 333.79 mm2

Now we know, s =

P 26 × 1000 = A 333.79

s = 77.89 N/ mm2  (Ans.) δL 0.12 = = 0.0006   (Ans.) Now we know, e = L0 200 \

Example 4. A rectangular bar of 100 mm × 40 mm and 2 m long is subjected to

an axial tensile load of 60 kN. If the extension in the length of the bar is 1 mm, determine stress, strain and modulus of elasticity of the bar. Solution: Area of cross-section A = 100 × 40 = 4000 mm2 P 60 × 1000 Now we know, s = = A 4000 s = 15 N/mm2  (Ans.) δL 1 = = 0.0005   (Ans.) Now we know, e = L0 2000 \

Now,

E =

σ 15 = = 30 × 103 N/mm 2   (Ans.) ε 0.0005

Example 5. A steel specimen of 10 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.50 mm under a load of 40 kN and the load at elastic limit is 55 kN. The maximum load is 70 kN. The total extension at fracture is 28 mm and diameter at neck is 7 mm. Find (i) The stress at elastic limit. (ii) Young’s modulus. (iii) Percentage elongation. (iv) Percentage reduction in area. (v) Ultimate tensile stress. Solution: Area of cross-section A = π × (5)2 = 78.53 mm2



(i) Stress at elastic limit, s = \

P 55 × 1000 = A 78.53

s = 700.36 N/ mm2  (Ans.)

Simple Stress & Strain  6.11

Now we know, e =



(ii)

δL 0.50 = = 0.0025 L0 200

P 40 × 1000 2 E = A = 78.53 = 203744 N/mm   (Ans.) ε 0.0025

(iii) Percentage elongation =

Total extension 28 = = 14%   (Ans.) L0 200

(iv) Percentage reduction in area = =

π(5) 2 − π(3.5) 2 π(5) 2

(v) Ultimate tensile stress = =

Initial area – Final area Initial area

= 51%   (Ans.)

Maximum load Area

70 × 1000 = 891.37 N/mm 2   (Ans.) 78.53

Example 6. A steel wire 1.5 m long and 5 mm in diameter is extended by 0.50 mm due to weight suspended from the wire. If the same weight is suspended from the brass wire, 3 m long and 3 mm in diameter, it is elongated by 5.35 mm. Determine the modulus of elasticity of brass if that of steel is 2 × 105 N/mm2 Solution: For the type of given loading, a change in the length of wire is given by, PL dl = AE

Now for the steel wire,

0.50 =

P ×1500 = 0.0003819 P π(2.5) 2 × (2 × 105 )

\ Weight suspended P = 1309.24 N Now for the brass wire, 1309.24 × 3000 5.35 = π(1.5) 2 × EB \

EB = 103861.7 N/mm2  (Ans.)

Example 7. A solid circular steel rod of 25 mm in diameter and 300 mm long is rigidly fastened to the end of a square brass bar of 10 × 10 mm and 250 mm long as shown in figure Ex. 7. An axial tensile force of 20 kN is applied at each of extreme ends. Determine the elongation of assembly. Take Es = 200 GPa and EB = 90 GPa.

6.12  Engineering Mechanics Statics

Fig. Ex. 7

Solution: Let dlb and dls be the elongations in brass and steel. Then total increase in length of the assembly is, PL PL dl = δlb + δls = b b + s s Ab Eb As Es =

20 ×1000 × 250 20 ×1000 × 300 + (10 ×10) × (90 ×1000) π(12.5)2 × (200 ×1000)

dl = 0.555 + 0.0611 = 0.616 mm  (Ans.) Example 8. A member PQRS of uniform diameter 250 mm has been subjected to point loads as shown in figure Ex. 8. Determine the net change in the length of the bar. Take E = 200 × 109 N/m2.

Fig. Ex. 8

Solution: Cross-sectional area A = p × (125)2 = 49087.38 mm2 The free body diagram of each segment of the member are shown in figure Ex. 8(a).

Fig. Ex. 8(a)

Segment PQ; dl1 =

(150 × 1000) × (150) 49087.38× (200 × 109 )

Simple Stress & Strain  6.13

Segment QR; dl2 = Segment RS;

dl3 =

(60 × 1000) × (400) 49087.38× (200 × 109 ) (180 × 1000) × (150) 49087.38× (200 × 109 )

Total contraction in length, dl = dl1 + dl2 + dl3 = 7.48 × 10–9 mm  (Ans.) Example 9. A straight bar of 900 mm long is 40 mm in diameter for the first 500 mm length and 20 mm diameter for the remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar. Take E = 2 × 105 N/mm2. Solution: Here extension of the bar dl = dlb + dls PL1 PL2 + = A1E A2 E

=

20 ×1000 × 500 20 ×1000 × 400 + 2 5 π(20) × (2 ×10 ) π(10) 2 × (2 ×105 )

dl = 0.0397 + 0.1273 = 0.524 mm  (Ans.)

Example 10. A bar made up of two square sections. A steel of 15 × 15 mm square cross-section and 300 mm length and aluminium bar of 30 × 30 mm and 400 mm long. The bar is subjected to a compressive force P. If total decrease in length of the bar is 0.15 mm, than determine the value of P. Take E for steel as 205 × 103 N/mm2 and for aluminium 75 × 103 N/mm2. PLS PLA + Solution: Here extension of the bar dl = AS ES AA E A

Solving,

300 400   + 0.15 = P  15×15× 205×1000 30 × 30 × 75×1000  p = 12 kN  (Ans.)

Example 11. A rectangular block of 20 mm wide and 10 mm deep and 100 mm long is subjected to 40 kN tensile load. Measurement show that the elongation of the block is 0.0700 mm and decrease in width is 0.00500 mm. Determine modulus of elasticity and Poisson’s ratio. Solution: Cross-sectional area A = 10 × 20 = 200 mm2 P 40 × 1000 Now we know, s = = A 200

s = 200 N/ mm2 δL 0.0700 = = 0.0007 Longitudinal strain e = L0 100

\

6.14  Engineering Mechanics Statics

E =

Now,

eLateral =

Poisson’s ratio =

σ 200 = = 285714 N/mm 2   (Ans.) ε 0.0007 δw 0.00500 = = 0.00025 w 20

Lateral strain Longitudinal strain

=

0.00025 = 0.357   (Ans.) 0.0007

Example 12. A steel bar of 100 mm wide, 25 mm thick and 500 long is subjected to an axial pull of 90 kN. Find the change in length, width and thickness of the bar. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.30. Solution: Cross-sectional area A = 100 × 25 = 2500 mm2 P 90 × 1000 Now we know, s = = A 2500

s = 36 N/ mm2 σ 36 = = 0.00018 Longitudinal strain e = E 2 × 105

\

Lateral strain = Poisson’s ratio × Longitudinal strain = 0.30 × 0.00018 = 0.000054 δL e = Now, L0 \ dL = e × Lo Change in length = 0.00018 × 500 = 0.09 mm  (Ans.) δw eLateral = w \ dw = eLateral × w Above equation shows that change in width = lateral strain × original width \ Change in width 0.00054 × 100 = 0.054 mm  (Ans.) Now change in thickness = lateral strain × original thickness = 0.00054 × 25 = 0.0135  (Ans.) Example 13. A rod of length 150 mm and diameter 10 mm is subjected to an

axial pull of 10 kN. Determine the change in the dimension of the rod. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.30. Solution: Cross-sectional area A = π × r2 = π × 52 = 78.53mm2 10 ×1000 ×150 PL Change in length of the rod dl = = AE 78.53× 2 ×105 \

dl = 0.095 mm  (Ans.)

Simple Stress & Strain  6.15

Longitudinal strain e =

δL 0.095 = = 0.000633 L0 150

Now, Poisson’s ratio =

Lateral strain Longitudinal strain

Lateral strain = Poisson’s ratio × Longitudinal strain = 0.30 × 0.000633 = 0.000189 δd Also, lateral strain eLateral = d dd = eLateral × d dd = 0.000189 × 10 = 0.0018 mm  (Ans.)



Example 14. An aluminium rod of 25 mm diameter and 150 mm long is subjected to tensile load of 50 kN. The elongation of the rod is 0.1095 mm and its diameter is reduced by 0.00607 mm. Determine Poisson’s ratio of the material. δL 0.1095 = = 0.00073 Solution: Longitudinal strain e = L0 150

Lateral strain eLateral = Now,

δd 0.00607 = = 0.000242 d 25

Poisson’s ratio =

Lateral strain Longitudinal strain

Poisson’s ratio =

0.000242 = 0.33   (Ans.) 0.00073

Example 15. A hard rubber bar is deformed by 0.5 mm as shown in figure Ex. 15 by dash line due to force P. Determine the P A A B B shear strain at point O. Solution: The shear strain is represented by the 150 angle between AOA1. mm

Shear strain –1 ∅ = tan

AA′ 0.5 = tan –1 = 0.19°   (Ans.) OA 150

O

C

Fig. Ex. 15

Example 16. A composite bar of a circular rod of copper of 50 mm diameter is fixed into a steel tube of internal diameter 50 mm and external diameter 80 mm. If the composite bar of length L is subjected to 150 kN load, determine the stress developed in the two materials. Take E for steel = 2 × 105 N/mm2 and E for copper = 1.2 × 105 N/mm2 Solution: Area of cross-section of copper rod AC = π × (25)2 = 1963.49 mm2 Area of cross-section of steel tube AS = π × (402 – 252) = 3063 mm2

6.16  Engineering Mechanics Statics

Now from equation of equilibrium, 150 kN load = Load on copper rod + Load on steel tube 150 × 1000 = PC + PS From compatibility condition, PL PC L = S AS ES AC EC

...(Eq. 1)

PC PS 5 = 1963.49 × 1.2 × 10 3063 × 2 × 105 PC = 0.3846 PS

...(Eq. 2)

Substituting in Eq. 1

150 × 1000 = 0.3846 PS + PS

PS = 108334.5 N And,

PC = 0.3846 × 108334.5 = 41665.46 N

Now, Stress in copper sC =

41665.46 = 21.22 N/mm 2   (Ans.) 1963.49

Stress in steel sS =

108334.5 = 35.36 N/mm 2   (Ans.) 3063



Example 17. A composite tube consist of a steel tube 300 mm internal diameter and 20 mm thickness and an outer brass tube 340 mm internal diameter and 20 mm thickness. The composite tube of length L is subjected to an axial load of 800 kN. Find the stress developed in two materials. Take E for steel = 2 × 105 N/mm2 and E for brass = 1 × 105 N/mm2. Solution: Area of cross-section of steel tube AS = π × (1702 – 1502) = 20106.19 mm2 Area of cross-section of brass tube AB = π × (1902 – 1702) = 22619.46 mm2 Now from equation of equilibrium, 800 kN load = Load on steel tube + Load on brass tube ...(Eq. 1) 800 × 1000 = PS + PB From compatibility condition, PB L PL = S AB EB AS ES

PS PB = 5 20106.19 × 2 × 105 22619.46 × 1 × 10 PB = 0.5625 PS Substituting in Eq. 1 800 × 1000 = 0.5625 PS + PS

...(Eq. 2)

Simple Stress & Strain  6.17

PS = 512000 N And, PB = 0.5625 × 512000 = 288000 N Now, Stress in brass 288000 = 12.73 N/mm 2   (Ans.) sB = 22619.46 Stress in steel sS =

512000 = 25.46 N/mm 2   (Ans.) 20106.19

Example 18. Two aluminium rod and one copper rod together support a load of 125 kN as shown in figure Ex. 18. If area of each aluminium rod is 1000 mm2 and that of copper is 1200 mm2. Determine stresses in each rod. Take E for aluminium = 1 × 105 N/mm2 and E for copper = 1.2 × 105 N/mm2. 125 KN

A

C

A

150 mm 200 mm

Fig. Ex. 18

Solution: Area of cross-section of aluminium rod AA = 1000 mm2 Area of cross-section of copper AC = 1200 mm2 Now from equation of equilibrium, 125 kN load = 2 (Load on aluminium rod) + Load on copper rod ...(Eq. 1) 125 × 1000 = 2PA + PC From compatibility condition, PA LA P L = C C AA E A AC EC

PC × 200 PA × 150 5 = 1000 × 1 × 10 1200 × 1.2 × 105

PA = 0.9259 PC Substituting in Eq. 1 125 × 1000 = 2 × 0.9259 PC + PC PC = 43831.16 N

...(Eq. 2)

6.18  Engineering Mechanics Statics

PA = 0.9259 × 43831.16 = 40583.27 N 43831.16 = 36.52 N/mm 2   (Ans.) Now, Stress in copper sC = 1200

and,

Stress in aluminium sA =

40583.27 = 40.58 N/mm 2   (Ans.) 1000

Example 19. A rod is 1 m long at 15°C. Determine the expansion of the rod when the temperature is raised to 90°. Take E = 1 × 105 N/mm2 and coefficient of thermal expansion α = 0.000014 per °C. If the expansion is prevented find the stress induced in the material. Solution: Here rise in temperature from initial temperature is, T = 75°C. Expansion of the rod = α T L = 0.000014 × 75 × 1 × 1000 = 1.05 mm  (Ans.) Stress induced in the rod st = α T E = 0.000014 × 75 × 1 × 105 = 105 N/ mm2  (Ans.) Example 20. A steel rod of length 2 m and diameter 15 mm is at temperature

of 10°C. Determine the force exerted by the rod when prevented for expansion at temperature 50°C. Take E = 2 × 105 N/mm2 and α = 0.000012 per °C. Solution: Here rise in temperature from initial temperature is T = 40°C. Expansion of the rod = α T L = 0.000012 × 40 × 2 × 1000 = 0.96 mm Stress induced in the rod st A = α T E = 0.000012 × 40 × 2 × 105 = 96 N/ mm2 Force exerted by the rod = st A = 96 × p (7.5)2 = 16964.6 N  (Ans.)

SUMMARY • Stress is internal resistance setup by a body when it is deformed. P and its unit is N/mm2 Mathematically σ = A



•

Normal stress and Shear stress are types of stress. The stress developed on a plane normal to it is called normal stress. Tensile stress and compressive stress are types of normal stress. The stress developed when a member is subjected to a force that is parallel or tangent to the surface is called shear stress. V Mathematically τ = A



• Strain is defined as the change in length per unit length. δL εLongitudinal = L δd δw δb = = εLateral = d w b

Simple Stress & Strain  6.19



• The ratio of the lateral strain to the longitudinal strain is called the Poisson’s ratio. • The ratio of the change in volume of the body to its original volume is known as volumetric strain. • The ratio of normal stress to normal strain is constant and this constant is known as modulus of elasticity. • The ratio of shearing stress to shearing strain is known as modulus of rigidity. • Elongation of the bar due to axial loading is given by, PL δL = AE



• A stress is setup in the material for change in dimension due to temperature change is prevented. This is called the temperature stress. Temperature stress σt = α δt E Temperature strain εt = α δt

PROBLEMS

1. A rectangular bar having a cross-sectional area of 100 mm2 has a tensile force of 20 kN applied to it. Determine the stress in the bar.



2. A circular wire has a tensile force of 30.0 N applied to it and this force produces a stress of 2.06 MPa in the wire. Determine the diameter of the wire.



3. A square-sectioned support of side 24 mm is loaded with a compressive force of 20 kN. Determine the compressive stress in the support.



4. A metal bar which is part of a frame is 30 mm diameter and 250 mm long. It has tensile force acting on it of 35 kN which tends to stretch it. The modulus of elasticity E is 205 × 109 N/m2. Calculate the stress and strain in the bar and the amount it stretches.



5. A bolt having a diameter of 10 mm is loaded so that the shear stress in it is 200 MPa. Determine the value of the shear force on the bolt.



6. A 2.5 m long and 15 mm diameter steel bar is stretched by 20 mm on application of an axial load of 800 N. Calculate the stress, strain and Young’s modulus.



7. A rectangular metal bar has a width of 20 mm and can support a maximum compressive stress of 25 MPa; (a) determine the minimum breadth of the bar when loaded with a force of 5 kN. (b) If the bar is 1.5 m long and decreases in length by 0.25 mm when the force is applied, determine the strain.



8. A mild steel specimen of cross-sectional area 200 mm2 and gauge length 100 mm is subjected to a tensile test and the following data is obtained: within the limit of proportionality, a load of 70 kN produced an extension of 0.150 mm, load at yield point is 90 kN, maximum load on specimen is 125 kN, final cross-sectional

6.20  Engineering Mechanics Statics area of waist at fracture is 79 mm2, and the gauge length had increased to 105 mm at fracture. Determine for the specimen: (a) Young’s modulus of elasticity, (b) the yield stress, (c) the tensile strength, (d) the percentage elongation, and (e) the percentage reduction in area.

9. A load of 40 kN is applied to a steel wire. If the unit stress in the wire must not exceed 90 N/mm2 what is the minimum diameter of the wire is required? What will be the extension of 3.80 metre length of wire? Take E = 2 × 105 N/mm2.



10. A rod of length 150 mm and diameter 10 mm is subjected to an axial force of 15 kN. The modulus of elasticity of the material is 2 × 105 N/mm2 and Poisson’s ratio is 0.30. Calculate longitudinal and lateral strain.



11. A brass bar having cross-sectional area of 750 mm2 is subjected to axial force as shown in figure prob. 11. Find the total elongation of the bar. Take E = 1 × 105 N/ mm2.

Fig. Prob. 11

12. A composite rod is 800 mm long, its two ends are 30 mm2 and 20 mm2 in area and Length are 300 mm and 200 mm respectively. The middle portion of the rod is 15 mm2 in area. If the rod is subjected to an axial tensile load of 20 kN, find its total elongation. Take E = 200 GPa



13. Determine the total strain in a bar made up of 50 mm diameter solid for a length of 200 mm and a hollow circular section of outer diameter 45 mm and inner diameter of 25 mm for a length of 120 mm. Take E = 200 kN/mm2. The axial load is 80 kN.



14. A tensile stress is to be applied along the axis of a cylindrical rod that has a diameter of 20 mm. Determine the magnitude of the load required to produce 0.0045 mm change in diameter. Take E = 97 × 103 N/mm2 and poisons ratio is 0.34.



15. A hollow steel rod of outer diameter 20 mm, inner diameter 12 mm and length 2 m is subjected to compressive load of 40 kN. Determine, (i) change in length of the rod (ii) lateral strain, and change in diameters of the rod. Take E = 200 × 103 N/mm2 and poisons ratio = 0.30.



16. A reinforced concrete column is 200 × 200 mm in cross-section and carries a load of 300 kN. The column is provided with 6 bars of 12 mm diameter. Find the stress in the concrete and the bar. Take E for bar = 2 × 105 N/mm2 and E for concrete = 1.4 × 104 N/mm2.

Objective Type Questions 1. Study of forces and the conditions of equilibrium of bodies subjected to the action of forces is known as (a) Statics (b) Dynamics (c) Kinematics (d) Kinetics 2. The action of one body on another body that changes or tends to change the state of the body is known as (a) Mass (b) Rigid body (c) Force (d) None of these 3. The intensity of the force is its (a) Magnitude (b) Direction (c) Sense (d) Arrow head 4. Force can be characterized by (a) Magnitude (b) Direction (c) Point of application (d) All of these 5. The forces whose lines of action are passing through one point, are known as (a) Concurrent force system (b) Non-concurrent force system (c) Parallel force system (d) None of these 6. The forces whose lines of actions are passing through one common line of action are known as (a) Concurrent force system (b) Non-concurrent force system (c) Collinear force system (d) Parallel force system 7. The forces whose lines of actions are parallel to each other and lie in the same plane are known as (a) Concurrent force system (b) Non-concurrent force system (c) Collinear force system (d) Parallel force system 8. The single force that can replace the original system of forces without changing its external effect is known as (a) Resultant (b) Couple (c) Moment (d) None of these 9. The resultant of the two forces by parallelogram law be represented in magnitude and direction by (a) Diagonal of the parallelogram which does not pass through point of intersection of two forces (b) Diagonal of the parallelogram which passes through point of intersection of two forces (c) Longer side of the other two sides (d) Shorter side of the other two sides 10. The resultant of two forces P1 and P2 acting at an angle θ is 2 2 (a) P1 + P2 + 2 P1 P2 cos q

2 2 (b) P1 - P2 - 2 P1 P2 cos q

OTQ.2  Engineering Mechanics Statics

P 2 + P22 + 2 P1 P2 sin q P12 + P22 - 2 P1 P2 sin q (d) (c) 1 11. The resultant is maximum and minimum respectively when the angles between two forces are (a) 90° and 0° (b) 180° and 90° (c) 0° and 180° (d) 90° and 180° 12. If two forces acting simultaneously at a point be represented in magnitude and direction by two sides of triangle taken in order, their resultant may be represented in magnitude and direction by the third side of triangle taken in opposite order, this is (a) Parallelogram law (b) Triangle law (c) Polygon law (d) None of these 13. The resultant of two equal concurrent forces of magnitude P is θ θ θ θ 2 P sin (c) P cos (d) 3P cos (a) 2 P cos (b) 2 2 2 2 14. The resultant of two equal concurrent forces of magnitude P and angle between them 90° is (a)

p 2

(b)

p 2



(c)

2P



(d)

2 p

15. The resultant of more than two concurrent forces is (a) (ΣV ) 2



(b)

(ΣH )2 − (ΣV ) 2

2 2 (d) (ΣH ) + (ΣV ) (c) (ΣH )2 16. Moment of a force produce (a) Rotation of a body about fixed point (b) Translation of a body about fixed point (c) Rotation and translation of a body about a fixed point (d) None of these 17. Varignon’s theorem can be applied to determine (a) Position of resultant (b) Location of centroid (c) Magnitude of resultant (d) None of these 18. A couple is a (a) Pair of parallel forces of different magnitude separated by distance acting in opposite direction. (b) Pair of parallel forces of same magnitude separated by distance acting in opposite direction. (c) Pair of parallel forces of different magnitude separated by distance acting in same direction. (d) Pair of parallel forces of same magnitude separated by distance acting in same direction.

and and and and

Objective Type Questions  OTQ.3

19. Consider following statements. The couple is unchanged if, I) it is shifted to any other position, II) it is rotated through any angle , III) it is replaced by another pair of forces with same rotational effect, of these statements (a) I alone is correct (b) II alone is correct (c) I and II are correct (d) I, II and III are correct 20. Moment of a couple is (a) Independent of a point (b) Depend upon point (c) Depend upon axis (d) None of these 21. If moment of force applied on a door is 16 N.m and force applied is 4 N then distance of handle from pivot is (a) 2 m (b) 6 m (c) 4 m (d) 8 m 22. If moment arm is zero, then moment produce will be (a) Doubled (b) 1 (c) Zero (d) None of these 23. Moment depends upon (a) Magnitude of force (b) Moment arm (c) both (a) and (b) (d) none of these 24. If the body is at rest or in uniform velocity, it is said to be in (a) Rest (b) Equilibrium (c) Uniform motion (d) None of these 25. Number of forces acting at a point will be in equilibrium if (a) Their sum in horizontal direction is zero S = 0 (b) Their sum in vertical direction is zero S = 0 (c) Both (a) and (b) (d) None of these 26. Consider following statement Two forces can be in equilibrium only if they are I) equal in magnitude II) Opposite in direction III) Collinear in action. Of these statements (a) I and II are correct (b) I and III are correct (c) II and III are correct (d) All are correct 27. For equilibrium of co-planer non-concurrent force system, the following conditions are to be satisfied (a) SH = 0, SV = 0 (b) SH = 0, SM = 0 (c) SV = 0, SM = 0 (d) SH = 0, SV = 0, SM = 0 28. A sketch of the body isolated from its surrounding is known as (a) Free body diagram (b) Equilibrium (c) Force system diagram (d) None of these

OTQ.4  Engineering Mechanics Statics

29. The resultant of a system of forces acting on a body is zero, this state is known as (a) Co-planar forces (b) Equilibrium (c) Free body diagram (d) None of these 30. “If a body is in equilibrium under the action of three co-planar and concurrent forces, each of the forces is proportional to the sine of the angle between the other two”, this is the statement of (a) Triangle law of forces (b) Lami’s theorem (c) Parallelogram law of forces (d) Polygon law of forces 31. If a body is acted upon by a number of co-planar non-concurrent forces, it may (a) Rotate about itself without moving (b) Move in any one direction rotating about itself (c) Be in equilibrium (d) All of these 32. If three co-planar forces acting upon a rigid body keep it in equilibrium, then they must (a) Meet at a point (b) Be all parallel (c) Both (a) and (b) (d) None of these 33. The concurrent force system is in equilibrium if their resultant is (a) Zero (b) Positive (c) Negative (d) None of these 34. Lami’s theorem is applicable for (a) Two concurrent forces in equilibrium (b) Three concurrent forces in equilibrium (c) More than three concurrent forces in equilibrium (d) None of these 35. The normal reaction on a smooth roller from horizontal surface will act (a) Horizontal to the plane of contact (b) Perpendicular to the plane of contact (c) Inclined to the plane of contact (d) None of these 36. Simplest form of a perfect frame is (a) Rectangle (b) Square (c) Triangle (d) Pentagon 37. Minimum number of members that form a simple truss are (a) 2 (b) 3 (c) 4 (d) 5 38. Redundant truss is a type of (a) Perfect truss (b) Imperfect truss (c) Stable truss (d) None of the above

Objective Type Questions  OTQ.5

39. Which axial force is determined while analyzing a truss?  (a) Compressive force (b) Tensile force (d) None of the above (c) Both (a) and (b) 40. In analysis of truss by method of joints, the number of unknowns at a joint should not be (a) > 2 (b) > 3 (c) > 4 (d) None of these 41. In analysis of truss by method of sections, the number of unknowns at a section should not be (a) > 3 (b) > 4 (c) > 5 (d) None of these 42. Method of joints is more suitable when (a) Forces in few of the members only is desired (b) Forces in all the members are desired (c) Reactions at supports are desired (d) None of these 43. Method of sections is more suitable when (a) Forces in few of the members only is desired (b) Forces in all the members are desired (c) Reactions at supports are desired (d) None of these 44. The force system in method of joints involves (a) Collinear forces (b) Concurrent forces (c) Parallel forces (d) Non concurrent forces 45. The force system in method of joints involves (a) Collinear forces (b) Concurrent forces (c) Parallel forces (d) Non concurrent forces 46. Beam is a structural member subjected to (a) Axial load (b) Transverse load (c) Twisting moment (d) No load 47. In cantilever beam (a) One end is fixed and other end is free (b) Both ends are fixed (c) Both ends are supported on roller (d) None of these 48. In simply supported beam (a) One end is fixed and other end is free (b) Both ends are fixed (c) One end is supported on roller and other end is hinged (d) None of these

OTQ.6  Engineering Mechanics Statics

49. The bending moment of cantilever carrying a point load at free end is (a) Parabola (b) Triangle with maximum height at fixed end (c) Triangle with maximum height at free end (d) None of these 50. A sudden increase or decrease in shear force diagram between two points indicate (a) No loading (b) Distributed load between two points (c) Concentrated load between two points (d) None of these 51. If the bending moment diagram is parabolic curve between two points then it indicate (a) No loading (b) Uniformly distributed load between two points (c) Concentrated load between two points (d) None of these 52. Bending moment diagram for any part of a simply supported beam between two concentrated load is (a) Horizontal straight line (b) Vertical straight line (c) Inclined line (d) Parabola 53. The point through which the whole weight of the body acts is known as (a) Center of gravity (b) Center of mass (c) Centroid (d) Moment of inertia 54. The centroid and centre of gravity coincide if (a) The body has uniform density (b) Centre of gravity and center of mass coincide (c) g is uniform throughout the Earth (d) None of these 55. Centre of gravity of a body is a (a) Point in the body at which g is constant. (b) Point in the body which changes with orientation of the body. (c) Point in the body at which the entire weight is assumed to be concentrated. (d) None of these. 56. The first moment of an area about the x-axis is (a) ∫ x dA



(b)

∫ y dA

(c)

∫x

2

dA

(d)

∫y

2

dA

Objective Type Questions  OTQ.7

57. If an area has an axis of symmetry then (a) Its first moment about that axis is zero (b) Its centroid lies on that axis (c) Both (a) and ((b) (d) None of these 58. For an area having two axes of symmetry, the centroid lies on (a) Horizontal axis (b) Vertical axis (c) Intersection point of two axes (d) Not on any axes 59. If an area is symmetric about x axis, its (b) y = 0 (a) x = 0 (c) Both (a) and (b) (d) None of these 60. If an area is symmetric about y axis, its (b) y = 0 (a) x = 0 (c) Both (a) and (b) (d) None of these 61. The centroid of a semi-circle lies at a distance of __________ from its base measured along the vertical radius. (a) 4r/ 3π (b) 3r/ 8 (c) 3r/4π (d) 4r/ 2π 62. The unit of coordinates of centroid from reference axis is in (a) kg (b) N.mm (d) gram (d) mm 63. Moment of inertia is denoted by (a) I (b) C (c) M (d) K 64. The unit of moment of inertia is (a) L (b) L2 (c) L3 (d) L4 65. Second moment are also termed as (a) Center of gravity (b) Moment of force (c) Moment of inertia (d) Couple 66. According to perpendicular axis theorem, Ix = Iz + Iy (d) Iy = Ix + Iz (a) Iz = Ix – Iy (b) Iz = Ix + Iy (c) 67. The moment of inertia of a rectangle base ‘b’ and depth ‘d’ about the base is bd 8 bd 8 bd 2 bd 8 (b) (c) (d) 3 12 6 12 68. The moments of inertia of an area about x and y axes are Ix and Iy respectively. Its polar moment of Inertia is (a) Ix Iy (b) Ix/Iy (c) Ix + Iy (d) Ix – Iy 69. The moment of inertia of a circular section of diameter d is π π π π 4 4 (d ) 4 (c) (d ) 4 (d) (d ) (b) (a) (d ) 48 32 64 16 (a)

OTQ.8  Engineering Mechanics Statics

70. Area moment of inertia relative to a non-centroidal axis is (a) Larger than centroidal axis values (b) Smaller than centroidal axis values (c) Can be larger or smaller than centroidal axis value (d) Equal to centroidal axis values 71. Moment of inertia can also be calculated by (a) Integration (b) Differentiation (c) Moments (d) None of these 72. The moment of inertia of a triangle base ‘b’ and depth ‘d’ about the centroidal axis is bd 8 bd 2 bd 3 bd 8 (c) (b) (d) 36 6 24 12 73. The moment of inertia of a triangle base ‘b’ and depth ‘d’ about the base is (a)

bd 8 bd 8 bd 2 db8 (b) d (d) (c) 12 36 6 24 74. The relation used by radius of gyration is _____ (c) I = Ak2 (d) I = A2k2 (a) I = Ak (b) I = A2k 75. If X-X is the centrodial axis of an area ‘A’ and A-B is another axis at a distance ‘d’ and parallel to X-X, then by parallel axis theorem (b) IAB = IXX + Ad2 (a) IXX = IAB + Ad2 (d) None of these (c) IAB + IXX = Ad2 76. Moment of inertia of any composite area about an axis passing through its centroid is (a) Minimum (b) Maximum (c) Depend upon dimensions of area (d) None of these 77. The polar moment of inertia is measured about (a) x axis (b) y axis (c) z axis (d) None of these 78. Moment of inertia of hollow circular section of outer diameter D and inner diameter d is π π π π ( D 4 – d 4 ) (d) (D4 – d 4 ) (a) ( D 4 − d 4 ) (b) ( D 4 − d 4 ) (c) 64 48 16 32 (a)

79. Friction force act in a direction ______ to the direction of motion of an object. (a) Same (b) Perpendicular ( c) Opposite (d) None of these 80. The maximum frictional force which comes into play when a body just begins to slide on a surface is known as

Objective Type Questions  OTQ.9

(a) Static friction (b) Kinetic friction (c) Limiting friction (d) None of these 81. The body will move only when (a) Force of friction = applied force (b) Force of friction < applied force (c) Force of friction > applied force (d) All of the above 82. Which friction of the following is the highest? (a) Static friction (b) Kinetic friction (c) Rolling friction (d) None of these 83. Friction force is a (a) Scalar quantity (b) Vector quantity (c) Can be scalar or vector (d) None of these 84. To start a crate moving requires ___________ to keep it moving at a constant speed on a horizontal rough surface. (a) More force than (b) The same force (c) Less force than (d) None of these 85. A body of weight W is placed on an inclined plane. The angle made by the inclined plane with the horizontal, when the body is on the point of moving down is called (a) Angle of inclination (b) Angle of friction (c) Angle of repose (d) None of these 86. The ratio of static friction to kinetic friction is (a) Equal to one (b) Greater than one (c) Less than one (d) None of these 87. Static friction is always (a) Equal to kinetic friction (b) Less than kinetic friction (c) Greater than kinetic friction (d) None of these 88. Coefficient of friction depends upon (a) Nature of surface (b) Area of contact (c) Both (a) and (b) (d) None of these 89. If the angle of friction is 25°, the coefficient of friction is (a) 0.404 (b) 0.466 (c) 0.927 (d) None of these 90. The ratio of the limiting force of friction (F) to the normal reaction (N) is known as (a) Force of friction (b) Angle of friction (c) Angle of repose (d) Coefficient of friction 91. The coefficient of friction (µ) is equal to (a) tan φ (b) cos φ (c) sin φ (d) None of these

OTQ.10  Engineering Mechanics Statics

92. Stress is (a) External force (b) Friction force (c) Internal resistive force (d) None of these 93. The ratio of linear stress to linear strain is known as (a) Poisson’s ratio (b) Bulk modulus (c) Modulus of rigidity (d) Modulus of elasticity 94. The ratio of lateral strain to longitudinal strain is called (a) Modulus of elasticity (b) Modulus of rigidity (b) Poisson’s ratio (c) None of these 95. Which of the following is a dimensionless quantity? (a) Poison’s ratio (b) Strain (c) Both (a) and (b) (d) None of these 96. The total extension in a bar, consists of 3 bars of same material, of varying sections is (a) P/E(L1/A1+L2/A2+L3/A3) (b) P/E(L1A1+L2A2+L3A3) (c) PE(L1/A1+L2/A2+L3/A3) (d) PE(L1/A1+L2/A2+L3/A3) Where P = Load applied, E = young’s modulus for the bar, L1,2,3 = Length of corresponding bars, A1,2,3 = Area of corresponding bars. 97. The change in length due to tensile or compressive force acting on a bar is given by AE PL PLA E (d) (b) (c) (a) PL AE E PLA Where, P = force acting on a bar; L = length of the bar; E = modulus of elasticity; A = cross section area 98. When compressive stress is applied axially on a circular rod its I. diameter increases II. length decreases III. volume decreases Of these statements, (a) Only I is correct (b) Only II is correct (c) Both I and II are correct (d) None of these 99. Tensile Strain is (a) Increase in length / original length (b) Decrease in length / original length (c) Change in volume / original volume (d) None of these 100. Hooke’s Law is truly valid up to (a) Elastic limit (b) Proportional limit (c) Plastic limit (d) None of these

Review Problems 1. The resultant of two forces P and Q is 1400 N vertical. Determine the force Q and the corresponding angle θ for the system of forces as shown in figure.

2. Determine the magnitude of force F and direction θ so that the resultant of the three forces as shown in figure is vertically downward with a magnitude of 6 kN.

3. Determine the support reactions for the beam loaded as shown in figure.

4. Two identical rollers, each of weights 1000 N are supported by an inclined plane as shown in figure. Assuming smooth surfaces find the reactions induced at the points of supports.

R.2  Engineering Mechanics Statics

5. Two spheres A and B of diameter 80 mm and 120 mm respectively are held in equilibrium by separate strings as shown in figure. Sphere B rests against vertical wall. If masses of sphere A and B are 10 kg and 20 kg respectively, determine the tension in the string and reactions at point of contact.

6. Locate the coordinates of centroid of shaded region shown in figure.

7. Determine the y coordinate of the centroid of shaded area shown in figure.

Review Problems  R.3

8. A block of weight W1 = 1290 N rests on a horizontal surface and supports another block of weight W2 = 570 N on top of it as shown in figure. Find the force applied to lower block that will be necessary to cause the slipping to impend. Assume coefficient of friction between blocks = 0.25 and between block 1 and horizontal surface = 0.40.

9. Determine the forces in each member of truss shown in figure.

10. A tensile load of 50 kN is acting on the rod of 50 mm diameter and length of 5 m. Determine length of a bore of 25 mm diameter that can be made central in the rod, if the total extension is not to exceed by 25% under the same tensile load. Take E = 2.05 × 105 N/mm2

Index A Analysis of truss 3.1, 3.4 Angle of friction 5.5, 5.28, 5.38 Angle of repose 5.5, 5.6, 5.38 Anticlockwise moment 1.13, 1.14, 1.15, 1.44, 1.45, 1.46, 1.47, 1.48, 1.49, 1.50, 1.52, 1.55, 1.56, 1.57, 1.58, 3.6, 3.9, 3.13, 3.13, 3.14, 3.15, 3.16, 3.17, 3.17, 3.18, 3.22, 3.24, 3.26, 3.28, 3.30, 3.32, 3.34, 3.36, 3.38, 3.40, 3.42, 3.44, 3.46, 5.20, 5.21, 5.23, 5.24, 5.33 Applied forces 2.4, 6.3 Area M.I. of rectangle 4.22 4.23 Area M.I. of triangle 4.23, 4.24 Area moment of inertia 4.1, 4.19, 4.23, 4.24, 4.26, 4.29, 4.31, 4.32 Area moment of inertia 4.1, 4.19, 4.23, 4.24, 4.26, 4.29, 4.31, 4.32 B Bar of varying cross-section 6.6 Beam 3.1, 3.19, 3.20, 3.21, 3.22, 3.34, 3.44, 3.46, 3.48 Belt friction 5.34 Bending moment 3.21, 3.22, 3.30 3.32, 3.34, 3.42 3.44, 3.48 4.19 Bending moment diagram 3.22, 3.25, 3.34, 3.40, 3.42, 3.44, 3.48 Block friction 5.9 Bulk modulus 6.6 C Cantilever beam 3.20, 3.34, 3.38, 3.42, 3.48 Center of gravity 4.1, 4.3, 4.5, 4.18, Center of mass 4.3, 4.18 Centroid 4.3, 4.5, 4.6, 4.7, 4.8, 4.10, 4.14, 4.18, 4.19, 4.21, 4.22, 4.26 Centroid of a triangular area 4.5 Centroid of semicircular area 4.6 Characteristics of the force 1.3, 1.70

Clockwise moment 1.13, 1.14, 1.15, 1.43, 1.48, 1.55, 2.24, 3.13, 3.17, 3.22, 3.34, 3.44 Coefficient of friction 5.4, 5.36 Collinear forces 1.5, 1.7, 1.70 Composite system 6.9 Coplanar concurrent forces 1.5, 1.6 Coplanar non-concurrent forces 1.5, Coplanar parallel forces 1.6 Co-planer concurrent forces 1.35, Coulomb’s theory of dry friction Couple 1.1, 1.16, 1.17, 1.59, 1.65, 2.8, 3.21 D Determination of centroid 4.1 E Equilibrium 0.1, 2.1, 2.3, 2.8, 2.9, 2.10, 2.14, 2.12, 2.19, 2.34, 2.5, 3.5, 3.13, 3.21, 3.22, 3.34, 3.44, 5.6, 5.7, 5.9, 5.20, 5.29 Equilibrium of forces 2.10 Equilibrium under three force system 2.9 Equilibrium under two force system 2.8 Equivalent systems of forces 1.18 F Factor of safety 6.9 Force 0.3, 0.4, 0.5, 1.1, 1.3, 1.4, 1.7, 1.8, 1.9, 1.10, 1.13, 1.14, 1.15, 1.16, 1.17, 1.18, 1.26, 1.35, 1.43, 1.55, 1.59, 1.70, 1.71, 2.3, 2.6, 2.8, 2.10, 2.34, 3.3, 3.4, 3.5, 3.13, 3.19, 3.21, 3.22, 3.34, 3.44, 3.48, 4.3, 4.18, 5.3, 5.4, 5.5, 5.6, 5.7, 5.9, 5.26, 5.34, 5.38, 6.3, 6.5, 6.9 Force–couple system 1.17 Force-couple system and equivalent system 1.59 Free body diagram 2.3, 2.6, 3.19, 3.22, 3.34, 3.44, 4.3, 5.3, 5.5, 5.6, 5.7, 5.9 Friction 1.4, 2.5, 3.4, 5.1, 5.3, 5.4, 5.5, 5.6, 5.7, 5.20, 5.26, 5.34, 5.38,

I.2  Index H Hooke’s law 6.6 L Ladder friction 5.6, 5.20, 5.50 Lami’s theorem 2.9, 5.28 Lateral strain 6.5, 6.6 Law of transmissibility 0.4 Limiting friction 5.4, 5.38 Longitudinal strain 6.5, 6.6, M Mechanics 0.3 0.4, 0.5, 1.3, 2.3, Method of joints 3.4, 3.5 Method of sections 3.4, 3.5, 3.13 Modules of rigidity 6.6, Modulus of elasticity 6.6, 6.19, Moment 1.13, 1.14, 1.15, 1.17, 1.18, 1.43, 1.59, 2.3, 3.21, 3.22, 3.34, 3.44, 3.48, 4.3, 4.19, 4.21, 4.22, 4.26, 4.32, Moment & non-concurrent forces 1.43 Moment of a couple 1.17 N Newton’s law of gravitation 0.4 Newton’s laws of motion 0.4 Non-coplanar concurrent forces 1.6 Non-coplanar non-concurrent forces 1.6 Non-coplanar parallel forces 1.6 Non-perpendicular resolution 1.10, 1.12 Normal stress 6.3, 6.6, 6.18 O Overhang beam 3.44 P Parallel axis theorem 4.21, 4.26, 4.32 Parallel forces 1.6, 1.15, 1.55, 2.8 Parallelogram law of forces 1.7, 1.18 Parallelogram law of forces 1.7, 1.18 Particle 0.4, 2.3, 2.6, 3.4, 4.3 Perpendicular axis theorem 4.20 Perpendicular resolution 1.10, 1.12, 1.26 Poisson’s ratio 6.6 Polygon law of forces 1.9

R Radius of gyration 4.22, 4.32 Reactive forces 1.4, 2.4 Resolution 1.9, 1.10, 1.26, 1.70 Resolution of forces 1.26 Resultant 0.4, 1.7, 1.14, 1.15, 1.16, 1.18, 1.26, 1.35, 1.55, 1.70, 2.34, 4.3, 5.5 Resultant moment 1.14, 1.18 Resultant of collinear forces Resultant of concurrent forces 1.7 Resultant of parallel force system 1.55, 1.70 Rigid body 0.3, 1.18, 1.70, 2.3, 2.6, 4.3, S Shear force 3.21, 3.22, 3.30, 3.34, 3.44, 3.48, 6.4, 6.19 Shear force diagram 3.25, 3.30, Shear stress 6.3, 6.4, 6.18, 6.19 Simply supported beam 3.20, 3.22, 3.48 Solved examples based on 1.18, 1.26, 1.35, 1.43, 1.55, 1.59, 2.10, 3.5, 3.13, 3.22, 3.34, 3.44, 4.8, 4.26, 5.9, 5.20, 5.26, 5.34, 6.9 Space 0.3, 3.3, Strain 6.3, 6.4, 6.5, 6.6, 6.7, 6.8, 6.9 Stress 6.3, 6.6, 6.7, 6.8, 6.9, 6.18 Stress and strain 6.1, 6.8, 6.9, 6.19 Stress-strain diagram 6.7 System of forces 1.3, 1.4, 1.7, 1.70, 2.3, 2.10 T Temperature strain 6.8, 6.19 Temperature stress 6.8 Triangle law of forces 1.8 Truss 3.3, 3.4, 3.5, 3.19 Types of beams 3.19 Types of loading 3.20 Types of truss 3.3 V Varignon’s theorem 1.14, 1.15, 1.50 Volumetric strain 6.6, 6.19 W Wedge friction 5.7, 5.26