Strength of Materials [4 ed.] 0060453133, 9780060453138

Table of contents :
CHAPTER 1 Simple Stress
Introduction
Analysis of Internal Forces
Simple Stress
Shearing Stress
Bearing Stress
Thin-Walled Pressure Vessels
CHAPTER 2 Simple Strain
Introduction
Stress-Strain Diagram
Hooke's Law: Axial and Shearing Deformations
Poisson's Ratio: Biaxial and Triaxial Deformations
Statically Indeterminate Members
Thermal Stresses
CHAPTER 3 Torsion
Introduction and Assumptions
Derivation of Torsion Formulas
Flanged Bolt Couplings
Longitudinal Shearing Stress
Torsion of Thin-Walled Tubes, Shear Flow
Helical Springs
CHAPTER 4 Shear and Moment in Beams
Introduction
Shear and Moment
Interpretation of Vertical Shear and Bending Moment
Relations among Load, Shear, and Moment
Moving Loads
CHAPTER 5 Stresses in Beams
Introduction
Derivation of Flexure Formula
Economic Sections
Floor Framing
Unsymmetrical Beams
Analysis of Flexure Action
Derivation of Formula for Horizontal Shearing Stress
Design for Flexure and Shear
Spacing of Rivets or Bolts in Built-up Beams
CHAPTER 6 Beam Deflections
Introduction
Double Integration Method
Theorems of Area-Moment Method
Moment Diagrams by Parts
Deflection of Cantilever Beams
Deflections in Simply Supported Beams
Midspan Deflections
Conjugate-Beam Method
Deflections by the Method of Superposition
CHAPTER 7 Restrained Beams
Introduction
Redundant Supports in Propped and Restrained Beams
Application of Double-Integration and Superposition Methods
Application of Area Moment Method
Restrained Beam Equivalent to Simple Beam with End Moments
Design of Restrained Beams
CHAPTER 8 Continuous Beams
Introduction
Generalized Form of the Three-Moment Equation
Factors for the Three-Moment Equation
Application of the Three-Moment Equation
Reactions of Continuous Beams, Shear Diagrams
Continuous Beams with Fixed Ends
Deflections Determined by the Three-Moment Equation
Moment Distribution
CHAPTER 9 Combined Stresses
Introduction
Combined Axial and Flexural Loads
Kern of a Section; Loads Applied old Axes of Symmetry
Variation of Stress with Inclination of Element
Stress at a Point
Variation of Stress at a Point: Analytical Derivation
Mohr's Circle
Absolute Maximum Shearing Stress
Applications of Mohr's Circle to Combined Loadings
Transformation of Strain Components
The Strain Rosette
Relation Between Modulus of Rigidity and Modulus of Elasticity
CHAPTER 10 Reinforced Beams
Introduction
Beams of Different Materials
Shearing Stress and Deflection in Composite Beams
Reinforced Concrete Beams
Design of Reinforced Concrete Beams
Tee Beams of Reinforced Concrete
Shearing Stress and Bond Stress
CHAPTER 11 Columns
Introduction
Critical Load
Long Columns by Euler's Formula
Limitations of Euler's Formula
Intermediate Columns, Empirical Formulas
Eccentrically Loaded Columns
CHAPTER 12 Riveted, Bolted, and Welded Connections
Introduction
Types of Riveted and Bolted Joints: Definitions
Strength of a Simple Lap Joint: Bearing-Type Connection
Strength of a Complex Butt Joint: Bearing-Type Connection
Stresses in Bearing-Type Connections
Structural Bearing Type Joints
Eccentrically Loaded Bearing-Type Connections
Further Discussion of Bearing- and Friction-Type Connections
Welded Connections
Eccentrically Loaded Welded Connections
CHAPTER 13 Special Topics
Introduction
Repeated Loading, Fatigue
Stress Concentration
Theories of Failure
Energy Methods
Impact or Dynamic Loading
Shearing Stresses in Thin-Walled Members Subjected to Bending, Shear Flow
Shear Center
Unsymmetrical Bending
Curved Beams
Thick-Walled Cylinders
CHAPTER 14 inelastic Action
Introduction
Limit Torque
Limit Moment
Residual Stresses
Limit Analysis
APPENDIX A Moments of Inertia
Definition of Moment of Inertia
Polar Moment of inertia
Radius of Gyration
Transfer Formula for Moment of Inertia
Moments of Inertia by Integration
Moments of Inertia for Composite Areas
Product of Inertia
Product of Inertia Is Zero with Respect to Axes of Symmetry
Transfer Formula for Product of' Inertia
Moments of inertia with Respect to Inclined Axes
Mohr's Circle for Moments of Inertia
Principal Moments of inertia: Principal Axes
APPENDIX B Tables
Average Physical Properties of Common Metals: SI and U.S. Customary Units
Properties of Wide-Flange Sections (W Shapes): SI Units
Properties off-Beam Sections (S Shapes): SI Units
Properties of Channel Sections: SI Units
Properties of Equal Angle Sections: SI Units
Properties of Unequal Angle Sections: SI Units
Properties of Wide-Flange Sections (W Shapes): U.S. Customary Units
Properties of l-Beam Sections (S Shapes): U.S. Customary units
Properties of Channel Sections: U.S. Customary Units
Properties of Equal and Unequal Angle Sections: U.S. Customary Units
Index

Citation preview

Contents

Preface ix List of Symbols and abbreviations xi

1

CHAPTER 1 Simple Stress 1

1-1 1-2 1-3 1-4 1-5 1-6

Introduction 1 Analysis of Internal Forces 1 Simple Stress 4 Shearing Stress 15 Bearing Stress 19 Thin-Walled Pressure Vessels 22

2

CHAPTER 2 Simple Strain 30

2-1 2-2 2-3 2-4 2-5 2-6

Introduction 30 Stress-Strain Diagram 30 Hooke's Law: Axial and Shearing Deformations 34 Poisson's Ratio: Biaxial and Triaxial Deformations 42 Statically Indeterminate Members 45 Thermal Stresses 56

3

CHAPTER 3 Torsion 65

3-1 3-2 3-3 3-4 3-5 3-6

Introduction and Assumptions 65 Derivation of Torsion Formulas 66 Flanged Bolt Couplings 76 Longitudinal Shearing Stress 80 Torsion of Thin-Walled Tubes, Shear Flow 81 Helical Springs 84

4

CHAPTER 4 Shear and Moment in Beams 93

4-1 4-2 4-3 4-4 4-5

Introduction 93 Shear and Moment 95 Interpretation of Vertical Shear and Bending Moment 107 Relations among Load, Shear, and Moment 109 Moving Loads 125

5

CHAPTER 5 Stresses in Beams 131

5-1 5-2 5-3 5-4 5-5 5-6 5-7 5-8 5-9

Introduction 13 Derivation of Flexure Formula 131 Economic Sections 143 Floor Framing 148 Unsymmetrical Beams 152 Analysis of Flexure Action 158 Derivation of Formula for Horizontal Shearing Stress 161 Design for Flexure and Shear 171 Spacing of Rivets or Bolts in Built-up Beams 176

6

CHAPTER 6 Beam Deflections 182

6-1 6-2 6-3 6-4 6-5 6-6 6-7 6-8 6-9

Introduction 182 Double Integration Method 182 Theorems of Area-Moment Method 194 Moment Diagrams by Parts 198 Deflection of Cantilever Beams 207 Deflections in Simply Supported Beams 213 Midspan Deflections 224 Conjugate-Beam Method 228 Deflections by the Method of Superposition 232

7

CHAPTER 7 Restrained Beams 242

7-1 7-2 7-3 7-4 7-5 7-6

Introduction 242 Redundant Supports in Propped and Restrained Beams 242 Application of Double-Integration and Superposition Methods 243 Application of Area Moment Method 251 Restrained Beam Equivalent to Simple Beam with End Moments 259 Design of Restrained Beams 262

8

CHAPTER 8 Continuous Beams 267

8-1 8-2 8-3 8-4 8-5 8-6 8-7 8-8

Introduction 267 Generalized Form of the Three-Moment Equation 267 Factors for the Three-Moment Equation 271 Application of the Three-Moment Equation 275 Reactions of Continuous Beams, Shear Diagrams 28 I Continuous Beams with Fixed Ends 286 Deflections Determined by the Three-Moment Equation 292 Moment Distribution 297

9

CHAPTER 9 Combined Stresses 308

9-1 9-2 9-3 9-4 9-5 9-6 9-7 9-8 9-9 9-10 9-11 9-12

Introduction 308 Combined Axial and Flexural Loads 308 Kern of a Section; Loads Applied old Axes of Symmetry 316 Variation of Stress with Inclination of Element 320 Stress at a Point 322 Variation of Stress at a Point: Analytical Derivation 323 Mohr's Circle 326 Absolute Maximum Shearing Stress 336 Applications of Mohr's Circle to Combined Loadings 340 Transformation of Strain Components 352 The Strain Rosette 359 Relation Between Modulus of Rigidity and Modulus of Elasticity 362

10

CHAPTER 10 Reinforced Beams 366

10-1 10-2 10-3 10-4

Introduction 366 Beams of Different Materials 366 Shearing Stress and Deflection in Composite Beams 372 Reinforced Concrete Beams 372

10-5 10-6 10-7

Design of Reinforced Concrete Beams 377 Tee Beams of Reinforced Concrete 380 Shearing Stress and Bond Stress 382

11

CHAPTER 11 Columns 386

11-1 11-2 11-3 11-4 11-5 11-6

Introduction 386 Critical Load 387 Long Columns by Euler's Formula 388 Limitations of Euler's Formula 393 Intermediate Columns, Empirical Formulas 397 Eccentrically Loaded Columns 405

12 CHAPTER 12 Riveted, Bolted, and Welded Connections 411 12-1 12-2 12-3 12-4 12-5 12-6 12-7 12-8 12-9 12-10

Introduction 411 Types of Riveted and Bolted Joints: Definitions 411 Strength of a Simple Lap Joint: Bearing-Type Connection 413 Strength of a Complex Butt Joint: Bearing-Type Connection 415 Stresses in Bearing-Type Connections 420 Structural Bearing Type Joints 422 Eccentrically Loaded Bearing-Type Connections 425 Further Discussion of Bearing- and Friction-Type Connections 430 Welded Connections 431 Eccentrically Loaded Welded Connections 436

13

CHAPTER 13 Special Topics 443

13-1 13-2 13-3 13-4 13-5 13-6 13-7

Introduction 443 Repeated Loading, Fatigue 443 Stress Concentration 445 Theories of Failure 448 Energy Methods 450 Impact or Dynamic Loading 458 Shearing Stresses in Thin-Walled Members Subjected to Bending, Shear Flow 463 13-8 Shear Center 465 13-9 Unsymmetrical Bending 473 13-10 Curved Beams 481 13-11 Thick-Walled Cylinders 489

14

CHAPTER 14 inelastic Action 495

14-1 14-2 14-3 14-4 14-5

Introduction 495 Limit Torque 496 Limit Moment 497 Residual Stresses 501 Limit Analysis 508

15

APPENDIX A Moments of Inertia

15-1 15-2 15-3 15-4 15-5 15-6 15-7 15-8 15-9 15-10 15-11 15-12

Definition of Moment of Inertia 519 Polar Moment of inertia 520 Radius of Gyration 521 Transfer Formula for Moment of Inertia 522 Moments of Inertia by Integration 523 Moments of Inertia for Composite Areas 528 Product of Inertia 537 Product of Inertia Is Zero with Respect to Axes of Symmetry 538 Transfer Formula for Product of' Inertia 538 Moments of inertia with Respect to Inclined Axes 543 Mohr's Circle for Moments of Inertia 546 Principal Moments of inertia: Principal Axes 548

16

APPENDIX B Tables 551

16-1 16-2 16-3 16-4 16-5 16-6 16-7 16-8 16-9 16-10

Average Physical Properties of Common Metals: SI and U.S. Customary Units 552 Properties of Wide-Flange Sections (W Shapes): SI Units 554 Properties off-Beam Sections (S Shapes): SI Units 562 Properties of Channel Sections: SI Units 564 Properties of Equal Angle Sections: SI Units 566 Properties of Unequal Angle Sections: SI Units 568 Properties of Wide-Flange Sections (W Shapes): U.S. Customary Units 571 Properties of l-Beam Sections (S Shapes): U.S. Customary units 579 Properties of Channel Sections: U.S. Customary Units 581 Properties of Equal and Unequal Angle Sections: U.S. Customary Units 583

Index 589

Preface

Today, more than ever, engineering applications are often interdisciplinary, involving the interrelationship of several of the basic engineering sciences (mechanical, electrical, chemical, etc.). Therefore the modem engineer must have a fundamental knowledge in each of these areas. An understanding of how bodies re.spond to applied loads, the main area of emphasis in Strength of /Materials, is apart of this knowledge. Furthermore, for successful machine or structural design, a thorough mastery of strength of materials is a must. The unique feature of this fourth edition, as compared with previous editions, is that it uses both SI* and U.S. Customary Units. Since the United States has yet to adopt the SI system as its standard, there remains a need for engineers here to be trained in both sets of units. In this edition, the problems to be solved are divided almost evenly between SI and U.S. Customary Units, thus allowing the instructor to determine the proper balance for his or her students. This edition retains the general plan and features of the earlier editions, with the major emphasis still on elastic analysis, and, in addition, a chapter devoted to inelastic response. The importance of beam deflections in structural design warranted keeping the fairly complete treatment of this topic which includes energy methods, double-integration, area-moment, and moment distribution. However, since each of these topics is discussed separately, the instructor can easily choose only those methods that are relevant to his or her own presentation. Other features that are pertinent to this edition include an expanded discussion of plane stress, with a more thorough consideration of absolute maximum shearing stress, a revision of the chapter on connections to explain more fully

'

Sl is the official abbreviation for the intemationai system of units, Le Systémc International d`Unités.

ix

x

PREFACE

the distirictions between bearing-type and friction-type connections, and an updating of several topics due to changes in various design codes. Keeping in mind the special problems of students, we have, as in previous editions, endeavored to explain the fundamental concepts by using clear and concise language. The relatively large number of illustrative problems is intended to help the student bridge the gap between theory and application. The equations or principles used in the solutions of these problems are usually first stated in brackets, then the numerical values are substituted in the order in which the symbols appear in the equation. This technique enables the reader to follow the analysis more easily. The almost 1000 problems in this text have been carefully chosen to illustrate the fundamental concepts without overburdening the student with tedious numerical computation, wherever possible. The importance of free-body diagrams in strength of materials continues to be emphasized. The problems have been arranged largely in their order of difiiculty, and answers to about two-thirds of them accompany the appropriate problem statements. We continue to use a numbering plan that enables the reader to locate any cross reference quickly. In this scheme, all articles, figures, equations, tables, and problem statements, which are preceded by the number of the chapter in which they appear, are numbered consecutively throughout each chapter. The scheme is further simplified by having the numbers of the problems coincide with the numbers of the appropriate problem figures. The valuable suggestions and advice received from colleagues all over the world are sincerely acknowledged. To identify each contributor here would result in too lengthy a list, with the possibility of an inadvertent omission: each of these people has been thanked individually. However, a special debt is owed to Dr. Jean Landa Pytel, whose assistance in the preparation of this manuscript is greatly appreciated.

Andrew Pytel

Q

/

List of Symbols and Abbreviations

A A' a, b

b (`

C D, d E

e f

L L G g

h 1 lb.' A

I J J K k L L M

area partial area of beam section coordinates of centroid of moment diagram caused by simply ` supported loads breadth, width distance from neutral axis to extreme Tiber centroid of area diameter

modulus of elasticity in tension or compression eccentricity, base of natural logarithms frequency unit compressive stress in concrete unit tensile stress in reinforcing steel modulus of rigidity (i.e., modulus of elasticity in shear) gravitational acceleration (32.2 ft/s2; 9.81 m/s2) height, depth of beam moment of inertia of area moment of inertia with respect to neutral axis

centroidal moment of inertia polar moment of inertia centroidal polar moment of inertia stress concentration factor spring constant, radius of gyration length effective length for columns bending moment

xii

LIST OF SYMBOLS AND ABBREVIATIONS

m N n p p p pw e P by

mass normal force, factor of safety .

ratio of moduli of elasticity

p

Q q

R r S O' U 1, U 2

O' 3

O' b

Ue U Cl' '

Pr Ur U as

force, concentrated load, hoop tension power critical load for columns products of inertia pressure per unit area first moment of area shear flow reaction, resultant force, radius radius, radius ot' gyration section modulus (I/c) unit SUBSS, normal stress principal stresses unit bearing stress unit compressive stress critical unit stress in column formula unit flexural stress unit radial stress . allowable or working stress unit tensile stress, unit tangential stress

U

o up

unit normal stress in x. y, and z directions, respectively stress at yield point

T

torque, temperature

O' X ) o'y

91

0'

thickness, tangential deviation

f r

unit shearing stress uNit shearing stress in x-y plane

1' by

rectangular coordinates

u v, w V I

vertical shearing force velocity

v

total weight or load weight or load per unit of length

W Wo

x x y

|

}'»

D

z z

rectangular coordinates

coordinates of centroid or center of gravity deflection of beam . temperature coefficient of linear expansion angles unit shearing strain total elongation or contraction, deflection of beam, maximum deflection of column static deflection unit tensile or compressive strain principal strains unit tensile or compressive strain in the x, y, and z direction, respectively

q

a y,9 5 'Y

'Y6

69 t

el)

Hz

ex,

b y , 6:

I

LIST OF SYMBOLS AND ABBREVIATIONS

0 p

total angle of twist, slope angle for elastic curve radius of curvature, variable radius, mass density

v

Poisson's ratio

cu

angular velocity

CG

center of gravity degrees distribution factor factor of safety fixed end moment neutral axis proportional limit

deg

DF FS FEM NA

'PL YP

xiii

yield point

I

Chapter 1

1

Simple Stress

1-1 INTRODUCTION .Three fundamental areas of engineering mechanics are statics, dynamics, and strength of materials. Statics and dynamics are devoted primarily to the study of the external effects of forces on rigid bodies, that is, bodies for which the change in shape (deformation) can be neglected. In contrast, strength of materials deals with the relations between externally applied loads and their internal efilects on bodies. Moreover, the bodies are no longer assumed to be rigid, the deformations, however small, are of major interest. In mechanical design, the engineer must consider both dimensions and material properties to satisfy requirements of strength and rigidity. When loaded, a machine part or structure should neither break, nor deform excessively. The differences between rigid-body mechanics and strength of materials can be further emphasized by considering the following example. For the bar in Fig. l - l , it is a simple problem in statics to determine the force required to support the load W. A moment summation about the pin support determines P. This statics solution assumes the bar to be both rigid and strong enough to support the load. In strength of materials, however, the solution must extend further. We must investigate the bar itself to be sure that it will neither break nor be so flexible that it bends without supporting the load. P l I

Figure l-I Bar must neither break nor bend excessively.

L

W

Throughout this text we study the principles that govern the two fundamental concepts, strength and rigidity. In this first chapter we start with simple axial loadings, later, we consider twisting loads and bending loads, and finally, we discuss simultaneous combinations of these three basic types of loadings.

1-2 ANALYSIS OF INTERNAL FORCES Consider a body of arbitrary shape acted upon by the forces shown in Fig. 1-2. In statics, we would start by determining the resultant of the applied forces to determine whether or not the body remains at rest. If the resultant is zero, we

1/SIMPLE STRESS

2 a

II

I

F~.

•

a»

=~:'~l~.

° -

_ °'~._

\~~ s

. 1'

F.

Figure 1-2 Exploratory section a-a through loaded member.

a

have static equilibrium-a condition generally prevailing in structures. If the resultant is not zero, we may apply inertia forces to bring about dynamic equilibriurn. Such cases are discussed later under dynamic loading. For the present, we consider only cases involving static equilibrium. In strength of materials, we make an additional investigation of the internal distribution of the forces. This is done by passing an exploratory section a_a through the body and exposing the internal forces acting on the exploratory section that are necessary to maintain the equilibrium of either segment. In general, the internal forces reduce to a force and a couple that, for convenience, are resolved into components that are normal and tangent to the section, as

shown in Fig. 1-3. |

s.

y

FE

is

1

t

J-

\

t,

My: \

3

r

/z

m

x

Figure 1-3 Components of internal effects on exploratory section a-a.

The origin of the reference axes is always taken at the centroid which is the key reference point of the.section. Although we are not yet ready to show why this is so, we shall prove it as we progress, in particular, we shall prove it for normal forces in the next article. If the x axis is Normal to the section, the

section is known as the x surface or, more briefly, the x face. The notation used in Fig. 1-3 identifies both the exploratory section and the direction of the force .or moment component. The first subscript denotes the face on which the component acts, the second subscript indicates the direction of the particular component. Thus P is the force on the x face acting in the y direction.

1-2 ANALYSIS OF INTEFINAL FORCES

3

Each component reflects a different effect of the applied loads on the member and is given a special name, as follows;

Pa

J * x i a l force. This component measures the pulling (or pushing) action perpendicular to the section. A pull represents a tensile force that tends to elongate the member, whereas a push is a compressive force that tends to shorten

it. It is often denoted by P.

p,,, p

as iearforces. These are components of the total resistance to sliding the portion to one side of the exploratory section past the other. The resultant shear force is usually designated by V, and its components by V, and V, to identify their directions.

My(

Q0/Torlue. This component measures the resistance to

twisting the member and-is commonly given -the symbol T.

m,,, M . \J(8

riding moments. These components measure the resistance to bending the member about the you z axes and are often denoted merely by My or M, .

. From the preceding discussion, it is evident that the internal eHlect of a given loading depends on the selection and orientation of the exploratory section. In particular, if the loading acts in one plane, say, the by plane as is frequently the case, the six components in Fig. 1-3 reduce to only three, namely, the axial force P , (or P), the shear force P,,. (or V), and the bending moment My, (or M). Then, as shown in Fig. l-4a, these components are equivalent to the single

.%

b

'7

I

J

*s

I

Normal

I'

*I

component

I

»

P 'R

F2

/

Shear component

V

/

I

'ii

(a) Normal and shear components on arbitrary section a-a.

g.

\

\

\

,

*R FE

I 'Q b

lb) When exploratory section b _ b is perpendicular to resultant R of applied loads, only normal forces are produced.

Figure 1-4 (a) Normal and shear components on arbitrary section a=a.° (b) when exploratory section b-b is perpendicular to resulranx R of applied loads, only normal forces . are produced.

1/SIMPLE STRESS

4

resultant force R. A little reflection will show that if the exploratory section had been oriented differently, like b-b in Fig. l-4b where it is perpendicular to R, the shearing effect on the section would reduce to zero and the tensile effect would be at a maximum. The purpose of studying strength of materials is to ensure that the structures used will be safe against the maximum internal effects that may be produced by any combination of loading. We shall learn as our study proceeds that it is not always possible or convenient to select an exploratory section that is perpendicular to the resultant load, instead, we may have to start by analyzing the effects acting on a section like a-a in Figs. 1-2 and l-4a, and then learn how these effects combine to produce maximum internal effects like those on section b-b in Fig. l-4b. We shall study this procedure later in Chapter 9, which deals with combined stresses. For the present, we restrict our study to conditions of loading in which the section of maximum internal effect is evident by inspection.

1-3 SIMPLE STRESS I

I l

One of the basic problems facing the engineer is to select the proper material and propanion it to enable a structure or machine to perform its function efiiciently. For this purpose, it is essential to determine the strength, stiffness, and other properties of materials. A tabulation of the average properties of common metals is given in Appendix B, Table B-l, on page 552. Let us consider two bars of equal length but different materials, suspended from a common support as shown in Fig. 1-5. If we knew nothing about the bars except that they could support the indicated maximum axial loads [500 N (newtons) for bar l and 5000 N for bar 2], we could not tell which material is stronger. Of course, bar 2 supports a greater load, but we cannot compare

F I

'///...//////////////. .//{/,

Bar 1

Bar 2

¢

I

U

500 N

I

r

Ur 5000 N

Figure 1-5 loads.

Bars supporting maximum

strengths without having a common basis of comparison. In this instance, the cross-sectional areas are needed. So let us further specify that bar I has a cross~ sectional area of 10 mm2 and bar 2 has an area of 1000 mm2. Now it is simple to compare their strengths by reducing the data to load capacity per unit area. Here we note that the unit strength of bar I is

I

1-a SIMPLE STRESS

01

5

au-

500 N 10 I7llTl2

500 N 10 X 10'° m2

50 X 10° n/m2

and bar 2 has a unit strength 02 =:

5000 N 2 1000 mm

5000 N =.

1000

x

10'° mz

5 x 10° n/m2

Thus the material of bar l is ten times as strong as the material of bar 2.. The unit strength of a material is usually deNned as the stress* in the material. Stress is expressed symbolically as U

P A

(1-l)

where 0 (Greek lowercase letter sigma) is the stress or force per unit area, P is the applied load, and A is the cross-sectional area. Observe that maximum stress in tension or compression occurs over a section normal to the load, as indicated in Fig. l-4b. Shearing stress is discussed in the next section. From Eq. ( l - l ) it can be seen that the units for stress are the units of force divided by the units of area. In SI (which is the official abbreviation for the international system of units, Le Syszéme International d'Unizés), force is measured in newtons (N) and area is measured in square meters (m2). Thus the units for stress are newtons per square meter (N/m2). Frequently, one newton per

square meter is referred to as one pascal (Pa). Since the prefix M (read as "mega") refers to multiples of 106 in Sl, in the preceding example, the stress in bar I may be expressed as 50 MN/m2 (or 50 MPa) and that in bar 2 as 5 MN/mz (or 5 MPa). In U.S. Customary Units, force is measured in pounds (lb). With area measured in square inches, the units for stress are pounds per square inch (lb/in.2), frequently abbreviated as psi. Since "kip" is often used to represent kilopound ( l kip = 1000 lb, 2 kips = 2000 lb, ctc.). "ksi" is used as an abbreviation for l kip per square inch (1000 lb./in.2), for example, 8 ksi =: 8000 psi. in certain applications, such as soil mechanics, it is also common to measure stress in units of pounds per square foot (lb/ft2), abbreviated as psf. Although the expression in Eq. ( I - l ) is fairly simple; it requires careful discussion. Dividing load by area does not give the stress at all points in the cross-sectional area, it merely determines the average stress. A more precise definition of stress is obtained by dividing the differential load dP by the differential area over which it acts: . dp

(1-la) dA Next, let us see under what conditions o = P/A will accurately define the stress at all points of the cross section. The condition under which the stress is O'

constant or uniform is known as simple sf rc's.s. We shall show now that 8 uniform

. . • us¢ the arms am .M or la up/ .urvsx as synonymous with * Some ¢ng]nckB . load or force, and url!! .slrc's.s or .$l1'(.\.\ nll¢mlr.v when refcmng to the nntcnslty of load per unit area. In this book, stress wall always denote .Lora' /or um! urea. \

\

I

1/SIMPILE STRESS

6

\

stress distribution can exist only if the resultant of the applied loads passes through the centroid of the cross section.° Suppose that a cutting plane isolates the lower half of one of the bars in Fig. 1-5. Then, as shown in Fig. 1-6, the resisting forces over the cut section must balance the applied load P. A typical resisting force is dp. Applying the force and moment conditions of equilibrium, we obtain

m

(Z

*v

AF

/' o/ .

xdP=

X(o'

do)

dP 1-

J

TC I

I 9

I

II

1I

Q

\

l

I

2

Z

Pb =

0]

My:

AdA

II

[ Z 1c',=01

L

'I

l

Figure 1-6 For uniform stress, P must ` pass through the centroid C.

If we specify that the .stress distribution is to be constant over the cut section, a may be written outside the integrals in the preceding equations to ` . obtain P=a

f dA

=o'A

and therefore

Pb=(oA)b=a

I

xdA

Then, canceling the common factor a, we obtain b

:=

f x dA A

I

='.

J?

from which the coordinate b of the point C is recognized as being the x coordinate of the centroid of the section. By taking a moment summation about the x axis, 'There are certain exceptions to this male; they are caused by stress concentration (see p. 445), by abrupt changes in the cross section, and at points in the vicinity of the applied loads ' -(see p. 7).

L

7

1-3 SIMPLE STRESS

we could similarly show that 9' defines the y coordinate of' C. We conclude that a uniform stress distribution is obtained only when the resultant of the applied . loads passes through the centroid of that surface. It does not follow, however, that positioning the load through the centroid of the section always results in a uniform stress distribution. For example, Fig. 1-7 shows the profile of a flat bar of constant thickness.,The load P is applied at distribution is uniform the centerline of the bar. At sections b-b andf-f the stress ittdicated sections other the at but earlier; ed and illustrates the principle discuss the stresses are not uniform. r

»

'///.

/

."f'E'»111'"'{""' I 1

I

I I

I

I

I

e

I

Figure 1-7 Exceptions to uniform stress distribution occur at sections a-a, c-c, d-d,. and e-e.

a

f

I//

: IIIIII

I I

I

I I I I l;*l

I-.

l I

I

I I I l

I

I

I

I \

\

I I I I

I I I I I I/

I

I

i

I

HgI 2 I / / /

I

I

b

l

I I

I I I I 11114/

I c

I

I I |

I I II d

I

L;-LJ I

I

I I I

II

|

I

¢

e

|,» 'I

II

IIII

1\\\ \

\

\\\\I I

I I I I I I\ \ I I l \I I I I I I I I L I I I i I I I I I I I I I I I I I I I I I \ i \ \ \ \ I I II / ; f I / \ \l I \

\

4

d e

b

/

P

At section e-e, the stress distribution is not uniform because the line of action of P obviously does not pass through the centroid of the section. Nor are the stresses uniformly distributed across sections c-e and d-dbecause, although the action line of P does pass through the centroids of these sections, here there are abrupt changes in section. At such sections the stresses are usually highly localized and can be determined only by the mathematical theory of elasticity or SOIDC experimental method, such as photoelasticity. Also, the stress is not uniform across section a-a because here the section is too close to the point where the load is applied. Unless a section is located at a distance from the lend of the rod at least equal to the minimum width of the rod, we will not obtain a uniform stress distribution.* In order to visualize why sections c-c. d-d, and a-a do not have uniform stress, imagine that the applied force P produces stress lines that radiate out from the load and distribute themselves throughout the body as shown by the York,

(9335.~ '."'°*°"*° and J. n. Goodicr, Theory of Blasticily. 2nd

¢d., McGraw-Hill, New

1/SIMPLE STRESS

8

1

I

dashed lines an the figure. Although this concept is not actually correct, it does indicate the existence of stress concentration wherever the shape of the body interferes with the "free flow" of the stress lines. The bunching of these lines about the hole in section c-c, and around the sharp comer of section d-d. which indicates stress concentration, contrasts with the relatively smooth flow of stress . around the radius between sections e-e and f-f.

ILLUSTRATIVE PROBLEMS ~lol. A composite bar oonsists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in Fig. I-8a. Axial loads are applied at the positions indicated. Determine the stress in each section. Aluminum

Bronze

A

= 1.2 in.2

l

-~=n-

4000 lb-4~"l

|_

L 90001b

»|-

1.3 ft

Steel

A = 1.8 i1'1.2

A = 1.6 in.2

-nu-

7000 lb

2 I I c lb

v

1.6 to

~1.7 ft

•I

(a)

4000 lb

4000 lb

.

.

R+ Pb: L, 190001b *

4000 lb

9005 lb

I

Par

\

2000

•

E

Put I

a

(b)

Figure 1-8 Solution: To calculate the stresses, we must first determine the axial load in each section. The appropriate free-body diagrams are shown in°Fig. I -8b. from which we determine P., == 4000 lb (tension), P., = 5000 lb (compression). and PSl = 7000 lb (compression). The free~body diagrams have been drawn by 'isolating the portion of the bar- lying to the left of imaginary cutting planes. Identical results would be obtained if portions lying to the right of the cutting planes had been considered. ' The stresses in each section are

i

db, "

4000 lb = 3330 psi L2 in.2

(tension)

Ans.

0'.l=

5000 lb : 2780 psi L8 ill.:

(compression)

A is.

. 7000 lb Use = l 6 i n ' -=380ps1

(compression)

un

Ans.

0

ILLUSTRATIVE PROBLEMS

9

. Note that neither the lengths of the sections nor the materials from which the sections are made affect the calculation of the stressa. As you can see from this example, the first step in calculating the stress in a member is to determine the internal force carried by the member. This determination is accomplished by the analysis of correctly drawn free-body diagrams. Note that in this example, it would have been easier to determine the load in the steel portion section in the steel. by taldng the section lying to the might of the exploratory

F I

®i

of

H

A

C

I Ay _

I 1-

H,

I

é

G

r

r

30 kN

70 kN

A

I.

IH, -I

4 panels at 4 m = 16 m (a)

B AB

3 3m

4

i

BO'

as \\

` \

l

AS

AC

A

\

_-_-*E

A

C

h

1

A

I'

4m

(b)

CE

'30kN

4.

4 m-

(c)

Fleur# 1-9 102. For the .truss shown in Fig. I-9a. determine the stress in members .4 C`and BD. The cross-sectional area of' each member is 900 mmz.

Solution: The three assumptions used in the elementary analysis of trusses are as follows: l. Weights of the members are neglected. 2. All connections are smooth- pins. 3. All external loads are applied directly to the pins.

Using these three assumptions. the members of the truss may be analyzed as Iwo./bra' members-the internal force system carried by any member reduces to simply a single force (tension or eompressionl acting along the line of the member. The free-body diagram of the entire truss is shown in Fig, I-9a. An equilibrium analysis of this free-body diagram results in the following values for the external reactions: .4.,. = 40 kn. H.\. = 60 kn. and Ht 0.

-

..

10

1 /s\mpu2 STRESS s..

To determine the force in member AC, we pass an imaginary cutting plane which isolates joint A (sections), Fig, I-9a). The free-body diagram of_lolnt A IS shown in Fig, I-9b. Here AB and AC represent the forces in members AB and AC, respectively. Note that both members have been assumed to be in tendon. Analyzing the free-body diagram in Fig. I-9b, we have

I2F,=01

®A,+§/48=0 AB = -§,4, = -§(40) = -66.7 kN

@

1Ep,-=01

AC + §,48 = 0

Ac = -tAB = -§(-661) = 53.4 in The minus sign indicates that the 66.7 kN force in member AB is compressive. The force in member AC is 53.4 kn, tension. To determine the force in member BD, we pass an imaginary cutting plane that expires the force in member BD (section , Fig. 1-9a). The free-body diagram of the portion ofthetruss to the Is of section is shown in Fig. l-9c. (The portion of the truss to the right of station®wuld do have -been used.) The forces on members BD, HE, and CE are assumed to be tensile. To calculate the force BD, we eliminate the forces BE and CE by taking a moment summation about their point of intersection, E, and write

® -A,(8) + 30(4) - BDU) = 0

[ Z Ms = 0]

380 = -8A,, + 120 := -8(40)

+ 120

= _200

8D = -66.7 kN Therefore the force in member BD is 66.7 kN,°compression. The stresses in members AC and BD are °'.4c° =

53.4 kN 900 mm'

=

53.4 X 103 N 900 X l0°° m2

59,3 X 10° n/m2 ` '

__ 66.7 kN so

900 mm'

74.1

_

= 59.3 MPa

Ans.

(tension)

66.7 X 103 N 900 X lo* m2

x 10° n/m2 = 74.1 MPa

Ans.

(compression) In truss analysis, the method of analyzing a single joint, as shown in Fig. l-9b, is referred to as the method ofjobxts. The .analysis of a section of the truss composed of two or more joints, as shown in Fig. I-9c, is called the method of sections. lt must be reemphasized that the force internal to a member of a truss lies along the line of' the member only because sufficient assumptions arc made that reduce all members to two-force members. As discussed in Art. 1-2, the internal forces for an arbitrarily loaded member are considerably more complicated than \

simply an did force.

'

.

'

ILLUSTRATIVE PROBLEMS

11

103. The block of weight Win Fig. l-lOa hangs from the pin at A. The bars AB and AC are pinned to the support at B and C. The areas are 800 mm' for AB and 400 mm' for AC. Neglecting the weights of the bars, determine the maximum safe value of . W if the stress in A8 is limited to I 10 MPa and that in AC to 120 MPa. ///K/0//////////////////////////,//

C

B

845

PAC

60° I

/\

60° W

I

w

(b)

(3)

l'~'iglIre 1-10

Solution: The pin-connected bars AB and AC are two-force members dmc their weights are negligible and W is applied at a pin. Thus the internal force system in each bar is a single force acting along the line of the bar. Analyzing the free-body diagram of joint A in Fig. I-lob, we obtain

= 01

@

PAC

cos 60°

12 F, = 01

®

PAC

sin 60° + P , sin 400 -

[E F,

°°

PA, COS 40° = 0

Solving simultaneously, we have p,,,.= 0.508W

and

PAC

= 0.'r18

w

The value of Wthai would cause the stress in each bar to equal ins maximum sale magnitude is determined as follows. For AB: 0.508 = ( n o X 10° [P = UA] x 10'° mz)

w

H/

n/m2X8o0

173

x 103 n

2

173 kn

For AC:

[P = GA]

0.778 W = (120

x

w = 61.7 x

10° n/m2X400 x I0* m')

103 N = 61.7 kn

The maximum safe value of W is the smaller of the preceding two values:

W = 61.7 kN

Ans.

with the stress in AC' being the limiting condition. The value W == 173 kN must be discarded because the stress in .4C would exceed its limiting value of 120 MPa for W = 173 kn.

I

12

1/SIMPLE STRESS

PROBLEMS 104. A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kn. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2_ Ans. l l 9 m m 105. A homogeneous 800-kg bar AB is supported at either end by a cable as shown in Fig, P-l05. Calculate the smallest area of each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel. Ans. As, = 43.6 mm2. .4,, 32.7 mm2

.

Y//////////H I;/////{/K/271

Bronze = 4 in

Steel

L = 3 rn

go

Q

A

B

lOm

Figure P-105

106. The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C and a cable that runs from A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 in. and the bar weighs 6000 lb.

3 ft A

C

5 fs

q

Sn

Figure P-106

107. A rod is oorriposed of an aluminum section rigidly attached between steel and bronze sections, as shown in Fig, P-l07. Axial loads are applied at the positions indicated. lip = 3000 lb and the cross-sectional area of the rod is 0.5 in.2, determine the stress

in each section.

Ans. Steel

Aluminum

4p-4-

Bronze

_ .-4->-p

2 f¢

I

n o

3 f¢

~~|-

Us;

=

al

= "4.0 ksi:

18.0 ksi

Uhr

N

§

m

2.5 fc-»1

Figure P-l07

108. An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. P-I08. Axial loads are applied at the positions indicated. Find the maximum value ofPthat will not exceed a stress in steel of 140 MPa. in aluminum of90 MPa. or in bronze of 100 MPa. .-Ms. ¢

L _ , "`*\

"I

P

:

§".0

kN

13

PROBLEMS Aluminum A = 4.0 1111112

Steel

1 A = 500 mm 2

\ \ \

Bronze

N

I.

.I

.2.5 m

q

A = 200 mm'

1•

4P

`r

/

R2

/ /

4

Figure P-243 I I

r

244. A homogeneous bar with a cross-sectional area of 500 mm' is attached to rigid supports. It carries the axial loads P, = 25 kN and P2 = 50 kn, applied as shown in Fig. P-244. Determine the stress in segment BC. (Hint: Use the results of Prob. 243, and compute the reactions caused by P, and Pa acting separately. Then use the principle ofsuperpodtion to compute the reactions when both loads are applied.)

a I

i I

ix.

1

53

PROBLEMS I

I

/ /

8

A

t

I I

I I I

-»0.60

I

Figure P-244

m»l»

.1

C

l

»l-

1.20 m

F

D

I...

*

¢

0.90 m->

I

attached to unyielding supports. Compute 245. The composite bar in Fig. P-245 is firmly the stress in each material caused by the application of the axial load P = 50 kips. Au m=wwm

q H

l

l

§l~

Figure P-245 and P-246

Steel A = 2.0 an? 5 = 29 X 106 psi 1\

Aluminum A = 1.25 in.2 5 :: 10 X 106 psi

-->-P

4.

15 in.

10

if-_--l

246. Referring to the composite bar in Prob, 245, what maximum axial load P can be applied if the allowable stresses are 10 ksi for aluminum and 18 psi for steel? 247. The composite rod in Fig. P-247 is stress-free before the aid loads P, and P2 are applied. Assuming that the walls are rigid, calculate the stress in each material if p. = 150 kN and P2 = 90 kn. Ans, 011 = 86.2 MPa (C) Aluminum /

ZN

A

:

E

:;

900 mm?

Steel A =-. 2000 mm? Bronze E = 200 GPa A = 1200mm2\

I

Figure P-247 and P-248

k

E = 83 GPa

70 GPa

»l»2s0 mm -l-

500 mm

350 mm

*\\

-

248. Solve Prob. 247 if the right wall yields 0.80 mm. 249. There is a radial clearance of0.05 mm when a steel tube is placed over an aluminum tube. The inside diameter of the aluminum tube is 120 mm, and the wall thickness of each tube is 2.5 mm. Compute the contact pressure and tangential stress in each tube when the aluminum tube is subjected to an internal pressure of 5.0 MP; Ans. p = 2.80 MPa 250. In the assembly of the bronze tube and steel bolt shown in Fig. P-250, the pitch of the bolt thread is p = 3,5 in., the cross-sectional area of the bronze tube is L5 in." and of the steel bolt is g in." The nut is turned until there is a compressive stress of 4000 psi in the bronze tube. Find the stresses if the nut is then given one additional turn. How many turns of the nut will reduce these stresses to zero? Use Et, :: 12 x 106 psi and E,, = 29 X 10° psi. Ans. "be = 9130 psi, 1.78 turns

I

I

'/

up

//

of/ K

I/(///

_.

lin-

-

it

Figure P-250

_

¢

11

-

# I - 1 _

-+

al

l

.Q

-_.1"'

ww#//#ww/w/m//._{_F

I.

L = 40 in.

n

I

i

[as

2/SIMPLE STRAIN

54

are identical except 251. The two vertical rods attached to the light rigid bar in Fig. P-25 I for length. Before the load W was attached. the bar was horizontal and the rods were stress-free. Determine the load in each rod if II' = 6600 lb. 6000 lb Arts. Pa

.A

B

L=6f1

L=4N •

-4 to

n o

4 ft-

Figure P-251

2f:

252. The light rigid bar ABCD shown in Fig. P-252 is pinned .at B and connected to two vertical rods. Assuming that the bar was initially horizontal and the rods stress-free, determine the stress in each rod after the load P : 20 kips is applied. Ans.

4

B 7

Aluminum

L = 4 fr

4 ft

A = 0.75 in.2

Un

31.4 ksi

Steel L : 3 ft A = 0.5 in.2 c 5 = 29 x 106 psi ID 'p

.l»2t¢ ~I-2fz --I

E = 10 X 106 psi

Figure P-252

'77

~_.,253/shown in Fig. P-253, a rigid beam°with negligible weight is pinned at one end and attached to two vertical rods. The beam was initially horizontal befol'e the load W = 50 kips was applied. Find the vertical movement of W Bronze A = 2 in.2 E = 12 X 106 psi

l 1

1..3 ft-l I

Steel

7/1

A=05m? E = 29 X 106 psi =3fz

5

I.

8 fr

L==10ftl

4aFigure P-253

PROBLEMS

55

254. As shown in Fig. P-254, a rigid bar with negligible mass is pinned at O and attached to two vertical rods. Assuming that the rods were initially stressfree, what maximum

load P can be applied without exceeding stresses of 150 MPa in the steel rod and 70 MPa in the bronze rod. Am. P :: 107 kN 2m

I

• l.5

m~¢I-1

.5 in

O I

I

P

75,7

Bronze A = 300 mrn2 E z 83 GP: L = 2 in

Steel

A = 900 mm2 E = 200 GPa L :. 1.5 m

,\'
u.

Max. -r = 14.3 ksi, 0 = l.23°

I

73

PROBLEMS

305. What is the minimum diameter of a solid steel shaft that will not twist through m0r¢ than 3° in a 6-m length when subjected to a torque of 12 kN~m? What maximum shearing sums is developed? Use G == 83 GPa. Ans. d = 114 mm, f = 41.3 MPa and 18 R long is used to transmit diameter in in. 14 shalt 306. A steel marine propeller 5000 hp at 189 rpm. lfG = 12 X l0' psi, determine the maximum shearing stress. 307. A solid steel shaft 5 m long is stressed to 80 MPa when twisted through 4°. Using G = 83 GPa, compute the shaft diameter. What power can be transmitted by the ' shaft at 20 Hz? Ans. d = 138 mm, P = 5.19 MW 308. A 2-in.-diameter steel shaft rotates at 240 rpm. If the shearing stress is limited to 12 ksi, determine the maximum horsepower that can be transmitted. 309. A steel propeller shaft is to transmit 4.5 MW at 3 Hz without exceeding a shearing stress of 50 MPa or twisting through more than 1° in a length of 26 diameters. Compute the proper diameter if G = 83 GPa. 310. Show that a hollow circular shaft whose inner diameter-fis half the outer diameter has a torsional strength equal to }2 of that of a solid shaft of the same outside diameter. 311. An aluminum shaft with a constant diameter of 50 mm is loaded by torques applied to gears attached to it as shown in Fig. P-311. Using G = 28 GPa, determine the relative angle of twist of g ~ar D relative to gear A. Ans.

0n/4 = 634°

600 N-m 900 N-m

1100 N

800 N-m

ax 3m

2 m'

Figure P-31 l 312. A flexible shaft consists of a 0.20-in.-diamet¢r steel wire encased in a stationary tube that flts closely enough to impose a frictional torque of 0.50 lb -in./in. Determine the maximum length of the shaft it` the shearing stress is not to exceed 20 ksi. What wll be the angular deformation of one end relativelto the other end? G = 12 X 10 psi. Ans. L = 62.8 iN., 0 = 29.9° 313. Determine the maximum torque that can be applied to a hollow circular steel shaft of l00-mm outside diameter and an 80-mm inside diameter without exceeding a shearing stress of 60 MPa or a twist of 0.5 deg/m. Use G = 83 GPa.

I'

3/TORSIQn

74

314. The steelshafl shown in Fig. P-314 rotates at 4 Hz with 35 kW taken off' at A, 20 kW removed at B, and 55 kW applied at C. Using G = 83 GPa, find the maximum shearing stress and the angle of rotation of gear A relative to gear C. Ans. Max. -r = 42.6 MPa, 0 = 6.0l° ~

Figure P-314

l

-A

'L

315. A 5-rn steel shaft rotating at 2 Hz has 70 kW applied at a gear that is 2 m from the left end where 20 kW are removed. At the right end, 30 kW are removed 'and another 20 kW leaves the shalt at 1.5 m from the right end. (a) Find the uniform shaft diameter so that the shearing stress will not exceed 60 MPa. (b) If a Uniform shaft diameter of 100 mm is specified, determine the angle by which one end of the shah lags behind the other end. Use G 83 GPa. Ans. d = 69.6 mm, 0 = 0.448°

-

.1

316. A compound shaft consisting of a steel segment and an aluminum segment is acted upon by two torques as shown in Fig. P-316. Determine the maximum permissible . value of Tsubjcct to the following conditions: 7sa s 83 MPa, To s 55 MPa, and the angle of rotation of the free end.is limited to 6°. For steel, G = 83 GPa and for . aluminum, G = 28.GPa. 2T

50 mm die

A

E

n40 mm die. T r Aluminum

Steel

900 mm

r"4.

600

(

mm"-.l

Figure P-316

317. A hollow bronze shaft of 3 in. outer diameter and 2 in; inner diameter is slipped over a solid steel shaft 2 in. in diameter and of the same length as the hollow shaft. The two shafts are then fastened rigidly together at their ends. For bronze, G = 6 X 10° psi, and for steel, G = 12 X 106 psi. What.torque can be applied to the composite shaft without exceeding a shearing stress of 8000 psi in the bronze or 12 ksi in the steel? Ans. T = 4230 ib- ft

318. A solid aluminum shaft 2 in. in diameter is subjected to two torques as shown in Fig. P-318. Determine the maximum shearing stress in each segment and the angle of rotation at the free end. Use G = 4 X l0° psi.

/

800lb-fz

200 lb.-fm

2 fz

I

3 re

0I l

Figure P-318 o

75

PROBLEMS

319. The compound shah shown in Fig, P-3l9 is attached to rigid suppose. For the 75 mm, 1 s 60 MPa, and G = 35 GPa. For bronze segment $8, the diameter is 50 mm, 1 s 80 MPa, and G == 83 GPa. If is eter diam the BC, the steel segment can be applied. a = 2 m and b -'= 1.5 m, compute the maximum torque Tthat

a

a

QA 4

i s

l,

q

E

ul

Z

.

Fugure P-319 and P-320

T

will be stressed 319, determine the ratio of lengths b/a so that each material ed? requir Tis torque What to its permissible limit; Aus. b/a = 1.19, T = 6.93 kN-rn P-321, to a solid shaft with built-in ends. 321. A torque T is applied, as shown in Fig. are T , = Tb/L and T; = TalI.. How walls the at s Prove that the resisting torque would these values be changed if the shaft were hollow?

320.

Iii Prob.

c

2

T1

T

r

I

r v

r

r

I

i

a

1

5:

Figure P-321

c

322. A solid steel shaft is loaded as shown in Fig. P-322. Using G = 83 GPA. determine the required diameter of the shaft if the shearing stress is limited to 60 MPa and the angle of rotation at the free end is not to exceed 4 deg. 750 N°m

I200N~m a

-2.5 m

Figure P-322

4.

-4

2.5 m

323. A shaft composed of segments AC. CD, and DB is fastened to rigid supports and loaded as shown m Fig. P-323. For bronze, G = 35 Gpa. for aluminum, G = 28 GPa, and for steel, G = 83 GPa. Determine the maximu m shearing stress developed in each segment.

TC=300N°M TA

TD=700 NIIN

_A

A Bronze

25 mm die.

-1.5 M

'c

If

Aluminum D

l

B Steel

25 mm die.

50 mm die. 2m

\

l

2.5 m

G

Ta

'k Figure P-323

1

76

3/ToRsloqfE' i.

324. The compound shaft shown in Fig. P-324 is attached to rigid supports. For the bronze segment AB, the maximum shearing stress is limited to 8000 psi and for the steel segment BC, it is limited to 12 ksi. Determine the diameters of eacb . s¢8m¢nt . so that each material will be simultaneously stressed to its permissible lunt When

a torque T = 12 kip-ft is applied. For bronze, G = 6 X 10° psi and for steel, G a 12 x 10° psi. Ans. db, = 4.26 in., d., = 2. 13 in. T

6 fr

A I

1

-

4 ft

Bronze

teel

g

"§

*A

Figure P-324

325. The two steel shaRes shown in Fig P-325, each with one end built into a rigid support, have flanges rigidly attached to their free ends. The shahs are to be bolted together at their flanges. However, initially there is a 6° mismatch in the location of the bolt hold, a shown in the figure. Determine the maximum shearing stress in each shaft after the shafts are bolted together. Use G = 12 X 10° psi and neglect deformations

-

of the bolts and flanges.

6°

Q

»

I

¢

s

L

•

b

I

I

\

I

I

\

I

I

\

II

n

-

*

g

Q

1

¥

. b

0 Q

o f

*

l

I L3.25 ft

J

D \I

I

I

2 in. die.

I 6.5 fm Figure P-325

\

#1 \

I

:ma

Q

I

co

A

• .-

\\

I I

A -

I

l n

E12

1

Q

1.5 in. die. .

-

n

A

3-3 FLANGED BOLT COUPLINGS O

A commonly used connection between two shafts is a flanged bolt coupling. It consists of flanges rigidly attached to the ends of the shafts and bolted together, as in Fig. 3-6. The torque is transmitted by the shearing force P created in the bolts. Assuming that the stress is uniformly distributed, the load in any' bolt is given by the simple stress equation P = A1 and equals (1rd2/4)1. It acts through the center of the bolt and tangent to the bolt circle. The torque resistance of one bolt is PR, where R is the radius of the bolt circle. Therefore, for any number of bolts, n, the torque capacity of the coupling is expressed by d2 T = PRo = '4 1Rn (3-4)

II

5

77

3-3 FLANGED BOLT COUPLINGS

P"AT'

no." 1' 4

7'

a -se s

\

`-N \

I

Q

-In.

I

I

*

I

\\

I

Ey

I

an

I / I N

l

T

Br

l \

I

I

i

J

J

_

/

¢

l

J

`\ Figure 3-6 Flanged bolt coupling.

of bolts, as in Fig. 3-7. Occasionally a coupling has two concentric rows subscript 2 refer to and circle outer the on bolts to refer l ipt Letting the subscr bolts on the inner circle, the torque capacity of the coupling is

T

=

PI RICI

+ P2 R2H2

(3-5)

's

f /

H

D

\.

Figure 3-7 Coupling with two concentric bolt circles.

. from the fact that the . between PI and P2 can be determined The.re1at10". comparatively rngxd flanges cause shear deformations in the bolts that are proportnonal to their radnal distances from the shaft axis. Thus the shearing strains are related by

ZN L

(0)

R. Re Using Hooke's law for shear, G = 1/7. we have t1

G.R.

12

G2R2

or P1/AI

G.R.

Pz/Az

G2 R2

(b)

3/TORSlON

78

g

i I

If the bolts .on the two circles have the same area. A, = .42, and if the bolts al-¢ made of the same material. G, = Go, the relation between P, and Pz reduces to

5 : LL R.

(3-6)

R:

This is the case shown in Fig. 3-7. Using the relation between P, and P2, Eq_ (3-5) will determine the torque capacity of the coupling. A similar procedure may be used for three or more concentric bolt circles, As we shall see in Chapter 12, this situation occurs in eccentrically loaded riveted or bolted connections.

PROBLEMS 326. A flanged bolt coupling consists of ten steel 20-mm-diameter bolts spaced evenly around a bolt circle 400 mm in diameter. Determine the torque capacity of the coupling if the allowable shearing stress in the bolts is 40 MPa. Ans. T = 25.1 k N - m 327. A flanged bolt coupling consists often steel I-in.-diameter bolts spaced evenly around a bolt circle 14 in. in diameter. Determine the torque capacity of the coupling if the allowable shearing stress in the bolts is 6000 psi. A ns. T = 6870 lb.-ft

328. A flanged bolt coupling consists of eight 10-mm-diameter steel bolts on a bolt circle 400 mm in diameter, and six I 0-mm-diameter steel bolts on a concentric bolt circle 300 mm in diameter, as shown in Fig. 3-7. What torque can be applied without exceeding a shearing stress of 60 MPa in the bolts? Ans. T = 10.7 k N - r n

329. A torque of 7000 lb- ft is to be carried by a Banged bolt coupling that consists of eight I-in.-diameter steel bolts on a circle of diameter 12 in. and six liin.-diameter steel bolts on a circle of diameter 9 in. Determine the shearing stress in the bolts. 330. Determine the number of l 0-mm-diameter steel bolts that must be used on the 400mm bolt circle of the coupling described in Prob. 328 to increase the torque capacity to 14 kN in.

-

»

Ans.

12 bolts

331. A flanged bolt coupling consists of six I-in. steel bolts evenly spaced around a bolt circle 12 in. in diameter, and four I-in. aluminum bolts on a concentric bolt circle . 8 in. in diannetcr. What torque can be applied without exceeding 9000 psi in the steel or 6000 pd in the aluminum? Assume G, = 12 X 10° psi and Gd = 4 X 10° psi. Ans. T = 77.8 kip.in. 332. In a rivet group subjected to a twisting couple I show that the torsion formula 1 = Tp/J can be used to find the shearing stress T at the center of any rivet. Let J = Z Ap2, where A is the area of a rivet at the radial distance p from the centroid of the rivet group. 333. A plate is fastened to a fixed member by four 20-mm diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed. (Hint: Use the results of Prob. 332.)

I

79

PROBLEMS

14 kN r

®`l®" " 8-0-.+~.mm 'V mm '*'- mm 1~y

'y

1

14 kN

Figure P-333

Udng 334. Six §-in.-diameter rivets fasten the plate in Fig. P~334 to the fixed member. the results of Prob. 332, determine the average shearing stress caused in each rivet by the 14 kip loads. What additional loads P can be applied before the average shearing stress in any rivet exceeds 8000 psi? Ans. Max. 1 = 14.0 ksi, P-'== 36.7 kips

- .

...bu-

Q.

1

.9

of

FE

.an ~s~

\

I

" § " \ \ \ \ ` \ \ \ \ \ \

ccrion. ¢

V 4/SHEAR AND MOMENT IN BEAMS

96

l I

section. This definition of shearing force (also called vertical shear or just shear) -

may be expressed mathematically as

V= (Z

i

Fy)L

I

(4-1)

the subscript L emphasizing that the vertical summation includes only the externai loads actingon the beam segment to the let of the section being considered. The resisting shear V, set up by the fibers in any section is always equal but oppositely directed to the shearing force V. In computing V, upward acting fonoes or loads are considered as positive. This rule of sign produces the effect shown in Fig. 4-4, in which a positive shearing force tends to move the left segment upward with respect to the right, and vice versa.

I

l

l

I

Positive shear

i

I I

Negative shear

Flew 4-4 Relative movements corresponding to signs of shearing force.

For complete equilibrium of the free-body diagram in Fig. 4-3b, the summation of moments must also balance. In this discussion, R, and I/ r are equal, thereby producing a couple M that is equal to R , x and is called the bending moment because it tends to bend the beam. The fiberSin the exploratory section must create a numerically equal resisting moment, M,, that acts as Shown.* In on beams, the free-body diagram carries a number of loads, as shown in Fig. 4-5, hence a more complete definition of bending moment is necessary.

1

Definition of Bending Moment

Bending moment is defined as the summation of moments about the centroidal axis of any selected section of all the loads acting either to the left or to the right side of the section, and is expressed mathematically as

I I

I

m = ( Z m».=(z M)R

(4-2)

the subscript L indicating that the bending moment is computed in terms of the loads acting to the left of the section, and the subscript R referring to loads to the tight of the section. Why the centroidal axis of the exploratory section must be chosen as the axis of bending moment may not be clear at this point, however, the reason is explained in Art. 5-2, Actually, in Fig. 4-5, where the loads are perpendicular to the beam, the axis of bending moment may be at point A, or B, or anywhere in the exploratory section, without changing the moment arms of the applied loads. But if the applied loads are inclined to the beam as shown in Fig. 4-6, the moment arms of the applied loads are unspecified unless the moment axis is at a definite

' An.

i

I I

: r

l I I I

I

4-3 shows that the bending moment, and hence the resisting moment, is always a

coupfl. I

1

4-2 SHEAFI AND MOMENT

1 1 11

Pa

18 v

97 P

Mr

m,

B

I

r K 1

v,

RE

Figure; 4-5

Figure 4-6

location in the exploratory section. Such inclined loads cause combined axial and bending effects which are discussed in Art. 9-2.

Sign of Bending Moment

0

To many engineers, bending moment is positive *f it produces bending of the beam concave upward, as in Fig. 4-7. We prefer to use an equivalent convention which states that upward aclirzg c.x'tcri1alhrc¢'s cwusc' positive bending moments with r

signs of bending moment.

I

r 4

Figure 4-7 Curvatures corresponding to Fositivc Bending

Negative Bending

Insofar as the left segment of a beam is concerned (Fig. 4-3b), this is equivalent to taking clockwise moments about the bending axis as positive, as indicated by the moment sense of R1. With respect to the right segment of a beam (Fig 4-Bc), this convention means that the moment sense of the upward reaction R: is positive in a counterclockwise direction. This convention has the advantage of permitting a bending moment to be computed, without any confusion in sign, in terms of the forces to either the left or the right of a section, depending on which requires the least arithmetical work. We never need think about whether a moment is clockwise or counterclockwise, upward acting forces always cause positive bending moments regardless of whether they act to the left or the right of the exploratory section. The definition ofshcaring force and bending moment may be summarized

mathematically as

V

::

M

II

Fy)l. ( y M)L =

.(4-1)

(E MM

(4-2)

IN which positive effects are produced by upward forces and negative effects by

downward forces. This rule of sign* will be used exclusively hereafter, and it will be further extended to give II positive sign to any quantity or expression in which * To avoid eonlhet with this rule, we mutt compute verttual shear in terms of' the forces lying lu the left of' the exploratory section. If the forces to the right of the section were used. would be it DCCCSSJFY to take downward lbreea as positive so as to agree with the sign convention shown in

l'lg. 4-4.

V

4/sHeAn AND MOMENT IN BEAMS

98

negative signs. such adjectives as "up" or "above" are used, and vice versa for t lying respecmen seg beam the to refer R and L subscripts the Remember that tively to the left and right of the exploratory section.

ILLUSTRATIVE PROBLEM 401. Write shear and moment equations for the beam loaded as shown in Fig. 4- 10a and sketch the shear and moment diagrams. I

Solutions: Begin bY computing the reactions. Applying E MR, = 0 gives RI = 63 in, and m,., = 0 yields R, = et in. A check of these values is given by E F, = 0. The sections in the beam at which the loading conditions change are called change of load points and are designated by the letters A, 8, C, and D. If a section a-a is taken through the beam anywhere between A and B. the external loads on it appear as in Fig. 4-8. Applying the definitions of venieal shear and bending moment, and noting that they apply only to external loads, we obtain

z

[ v = (21",)L_.

= (63 - 20.r) in

i

(0).

*...

[M = (2 M)Ll

#UAB

= 63x

- '

(63x

'i

(20x)

- 10x2) k N - m

(b)

These equations are valid only for values of x between 0 and 5 m, that is, between points A and B. To obtain shear and moment equations between B and (`. lake another exploratory section b-b anywhere between B and C. Note that the location of section b-b is still defined in terms of x as measured from the left end of the beam, although x now ranges between the limits of 5 m and 10 m. The effects of the external forces on this section are determined by applying the definitions of I

shear and moment to Fig. 4-9.

I

L V = ( E -Fy)!.]

Van = 63 - 100 = -37 kN

[M-=(ZM)Ll

Afg€' = 63x - l 00(x = (-37x

"°

(c) I

2.5)

+ 250) kN - m

(d)

The shear and moment equations for segment CD are obtained similarly by passing a section c-c anywhere between C and D. The external loads acting on the beam to the left of this section are shown in Fig. 4-1 l , from which we obtain IV:(ZFy)l.l

VCD=63"l00+67=

+30 k N

( e)

0

IM=

QX mill

m¢,,=63x- l00(x- 2.5)

+ 67(x

I0) I

=330x-420)kn-m

(fl

A simpler method of obtaining Mm is to consider the forces lying to the right of section C-C as shown in Fig. 4-IZ. Noting that downward forces produce negative bending moment, we also obtain. [M

z

(Z

M)xl

MCD

= -30(l4

= (30x

.»~

°'°

I

Q

x)

420) kN m .e

.J

i

20x kN

100 kN I

a

3.

$~ 2

zsm- I

l

-

unit

1

1

l

x

a

R1 = 6 3 kN

x

l

x

4




1 2 ft

1

Figure P-583

Q

584. A wide-flange section having the dimensions shown in Fig. P-584 supports a distributed load of Wo lb/ft on a simple span of length L ft. Determine the ratio of the maximum flexure stress to the maximum shear stress.

/ /

_Ll in.

/

/

/

/

/

10 in.

/ /

-85-1in (777-/,z7777l__

F-8 in.--4 t

1 in.

Figure P-584

PROBLEMS

175

585. A simply supported beam of length L carrier a uniformly distributed load of 6000 N/m and has the cross section shown in Fig. P~585. Find L to cause a maximum flexural stress of 16 MPa. What maximum shearing stress is then developed?

200 mm

1250 mm

150 mm

Figure P-585

300 mm

586. The distributed load shown in Fig. P-586 is supported by a box beam having the same cross section as that in Prob. 585. Determine the maximum value of Wo that will not exceed a flexural stress of 10 MPa or a shearing stress of 1.0 MPa. Wo

+

1111111llI11II111&1li11l1i! A

\~

on

I Figure P-586

RE

R1

587. A beam carries two concentrated loads P and a triangular load of 3P as shown in Fig. P-587. The beam section is the same as that in Fig. P-577-on page 17 l- Determine the safe value of P if a/s 1200 psi and -r s 200 psi. Ans.

p=742Ib

3P P

P

l

•

_¢l&lll 4 ft

Figure P-587

1 2 ft

J

4 ft

Rx

588. The distributed load shown in Fig. P-588 is supported So a wide-flange section of

the given dimensions. Determine the maximum value of we that wall not exceed a flexural stress of 10 MPa or a shearing stress of L0 MPa.

Lu -v-2 m

R1

I

I.

mm'

We

I

°l-*-25 mm

l

I I

1

) l

250 mm

A

4m

l

25 mm--I \

RE

my

5

`25 mm I"-200 mm-°-I-F

\

-

-

:

u

p

Figure P-588

5/STRESSES IN BEAMS

176

load of es two concentrated loads W and a total distributed 589. A channel section carri and bottom the above in. 2.17 is NA the that Verify P-589. 4Was shown in Fig. that W of value um maxim the that I~A = 62 in.'. Use these values to determine compression of IO 000 will not exceed allowable stresses in tension of 6000 psi, in psi. or in shear of 8000 psi.

I L

LE , _;

*--6 in.-l 1-1 in.

lin.--l

HHHH

r

UP a

l

»

GN

\

I

z*A/M L" r

-2 f1

1

RE

RE

T 6 in.

1

1

Fisvrfr P-S89 and a concentrated load P as 590. A box beam carries a distributed load of 200 lb./ft P if Uf s 1200 psi and shown in Fig. P-590. Determine the maximum value of 7 s 150 psi. A ns. P = 3480 lb

-*D

3" /

200 lb./ft

5h

l

1

1 2 in.

-HL

/

L

81 Few P-590

|

I

r

1

10 fa

rrfrrf

8 in.

l

l

I)//)//

r

I,'/

10 in.

R2

5-9 SPACING OF RIVETS OR BOLTS IN BUILT-UP BEAMS ,-

ts In our analysis of flexure action (An. 5-6) we showed that the various elemen consider now We another. the one past slide to tend composing a built-up beam action. size and spacing of rivets or bolts in a built-up beam to resist this sliding bolts. or The first step is to calculate the force to be resisted by such rivets Figure 5-31 shows a beam composed of three planks bolted together by ing stress at the two rows of bolts spaced e apart. Equation (5-4) gives the shear as planks upper two contact surface between the T

"

V 1bQ

where Q is the static moment about the NA of the shaded area in the end view» Multiplying this shearing stress by the shaded area eb in the top view gives the force F to be resisted in a length e: or

I

5.9 SPACING OF RNETS OR

sous IN BUILT-UP BEAMS

4

y

O O

.

177

O

/

O

Top view f

I*

c

:in

4

f

I

I

C

FFT1

»

I I I I' l I I 1

I

II l

I

g

I l

I I l I

I

L Front view

I

I I

I 1 I I I I I l

I

Lw

J

run

0'

I I

mm 41

Q

s

U so

Endvxew

Figlnre 5-31

F =

~t(

eb = )

V Ve Q(eb)=I Q lb 1

The same result can be obtained more directly by using the concept of shear flow, which is the longitudinal shearing force developed per unit length. Thus in the length e, Eq. (5-4a) determines the shear force to be

F

vQ

as before. Friction being neglected, this force is resisted by the shearing or bearing strength R of the bolts, whichever is smaller. Equating R to F gives

Ve R = -, Q

(5-7)

If the vertical shear varies in a beam, V is the average vertical shear in the interval e; but it is usually taken as the maximum Vin this interval, especially in built-up steel girders where the length of the interval is taken as a panel length equal to the depth of the girder. In this case, Eq. (5-7) gives the bolt pitch in each panel length. -

ILLUSTRATIVE

PROBLEM

S9l. A plate and angle girder is fabricated by attaching the short legs of four 125 X 90 x 13 mm angles to a web plate I 100 mm by 10 mm to form a section $20 mm deep, as shown in Fig. 5-32. The moment of inertia' about the NA. is I = 4140 X l0° mm'°. A t a section where If' = 450 in. determine the spacing between I9-mm rivets that fasten the angles to the web plate. Use 1 =. 100 MPa, in bearing, use 00 = 220 MPa for rivets in single shear and 00 = 280 MPa for rivets in double shear.

5/STRESSES IN BEAMS

178 23.7

""'vL'Z'"°°

mm*

Q V' l

*

560 -25.7 = 536.3 mm

NA

U

l

Q

1120 mm *Web plate 1100 mmx 10 mm 1.

qi

.I

L

Figure 5-32

Solution: The rivets must resist the longitudinal force tending to slide the two .Ranges past the web. Hence it is the static moment of the area of these two flange . antics that must be used in Eq. (5-7). Referring to Fig. 5-32, we obtain `

Q = 2(2630X536.3) = 2820

X

10' mm' = 2820 X l0'° m'

The shearing distance of a 19-mm rivet in double shear is R,

:

wxz) :

(0.019)2(100 X 106X2)

:

56.7 in

The bearing resistance against the web plate is R, = (dl)0b =` (0.019)(0.010X.280 X 106) = .53.2 kN

Using the lower of these values in Eq. (5-7), we get the required rivet pitch e

z

II *f Q

= (53.2 X 10'X4140 x 10'°) = O.l74 m (450 X I0'X2820 X 10-6)

174 mm

Ans.

PROQLEMS 592. A wide-flange section is formed ay bolting together three planks, each 80 mm by 2(X) mm, arranged as shown in Fig. P-592. If each bolt can withstand a sheadng fomc of 8 kn, determine the pitch if the beam is loaded so as to cause a maximum . siwafivs sums of 1.4 MPa, Ans. e = 84.2 mm .GL l 1

II

I

I

I

II

II II II

I II l

1l

I

I

l

80 mm x200 mm

/

H

Figure P-592

'

Current specihcatiom of AlSC call for no deduction for rivet holes in computing 1, pro vided that the rivet hole area docs not exceed 15% of the gross flange area. If it does, only the area in excess of 15% need be considered in modifying I to deduct for rivet holes. I

PROBLEMS

179

593. A box beam, built up as shown in Fig. P-593, is secured by screws spaced 5 in. apart The beam supports a concentrated load P at the third point of a simply supported span 12 ft long. Determine.the maximum value of Pthat will not exceed r = 120 psi in the beam or a shearing force.of 300 lb in the screws. What is the maximum flexuraL stress in the beam?

_

[-81°.--I iv J

` 10 in

•

Figure P-593 594. A distributed load of W0 lb/Ft is applied over the midge 6 lt of a simply supported span 12 ft long. The beam section is that in Prob. 593, but used here so that the 8-in. dimension is vertical. Determine the maximum value of we if 0/S 1200 psi, -r s 120 psi, and the screws have a shear strength of 200 lb and a itch of 2 in. Aus. We 514 Ib/Ft

-

.-

595. A concentrated load P is carried at midspan of a simply supported I 2-R span. The beam is made of 2-in. by 6-in. pieces screwed together, as shown in Fig P-595. If the maximum flexural stress developed is 1400 psi. find the maximum shearing stress and the pitch of the screws if each st new can resist 200 lb.

as

'/

nu- /' I I I

/

4-

/ /

/7

I ¢

/

f

. in. n

\`

=_

Ag

a

,1

I

¢

I

.unw

fig.

1

1 21 in.

I x

Figure P-595

41I \

\

y

I

pam

P-$96

S96. Three planks 4 in. by 6 in., arranged as shown in Fug. P496 and secured by bolts spaced I ft apart, are used to support a concentrated load Pat the center of a simply supported span 12 R long. If P causes a maximum tlcxural stress of 1200 psi, de-

termine the bolt diameters, assuming that the shear between the plaamks is transmitted by friction only. The bolts are tightened to a tension of 20 ksi and the coeElcient of friction between the planks is 0.40. Ares.

d = 0.713 in.

597. A plate and angle girder similar to that 'shown in Fig. 5-32 is fabricated by riveting the short legs of four 125 x 75 X 13 mm awes to a web plate 1000 mm by 10 mm to form a section 1020 mm deep. Cover plates. each 300 mm by 10 mm, are then

5/STRESSES IN BEAMS

180

riveted to the flange angles making the overall height 1040 mm. The moment of inertia of the entire section about the NA is I = 4770 x 10° mm'. Using the allowable stresses specified in Illustrative Problem 591, determine the rivet pitch t`or~22-mm rivets, attaching the angles to the web plate at a section where V = 450 kn. 598. As shown in Fig. P-598, two C380 X 60 channels are riveted together by pairs of 19-mm rivets spaced 200 mm apart along the length of the beam. What maximum vertical shear V can be applied to the section without exceeding the stresses given v in illustrative Problem 59l?

25.9 kN

.4 n5.

fi

n

H* I

Figure P-598 599. A beam is formed by bolting together two WZOO X 100 sections as shown in Fig. P-599. It is used to support a uniformly distributed load of 30 kN/m (including-the weight of the beam) on a simply supported span of 10 m. Compute the maximum flexural stress and the pitch between bolts that have a shearing strength of 30 kn.

I

NAJVB I

Figure P-599

SUMMARY

For homogeneous beams. originally straight, conying transverse loads i n the plane of symmetry, the bending moment creates flexural stresses expressed by »

U

v

Ally 1

\` v

(5-2)

The flexural stresses vary directly with their distance y from the neutral axis, which coincides with thecentroidal axis of the cross section. For beams that arc symmetrical with respect to the NA, the maximum flexural stresses occur at the section of maximum bending moment at the extreme fibers of the section. The distance from the NA to the extreme fibers being denoted by C, the flexure formula becomes Max. a =

_Mg I

91 S

(5-Za, b)

SUMMARY

181

in which S = I/c represents the section modulus of the beam. For geometric shapes, values of S are tabulated in Table 5-1 (page I 36); for structural shapes, the values are given in Appendix B. The vertical shear sets up numerically equal shearing stresses on longitudinal and transverse sections (Eq. 5-5, page 164), which are determined from *r

V V = - A' v = -lb J lb Q

(5.4)

in which A' is the partial area of the cross section above a line drawn through the point at which the shearing stress is desired. Q = AT is the static moment about the NA of this area (or of the area below this line). Maximum shearing stresses occur at the section of maximum Vand usually at the NA. For rectangular beams, the maximum shearing stress is

31

Max. -r

(5-6)

2 bh

In wide-flange beams, a very close approximation is Max. 1

V

-|

Awcb

where '.4,,,,b is the web area between the flanges. ` The rivet or bolt pitch in built-up beams is given by e=

-RI vQ

(5-7)

where R is the rivet or bolt resistance in the pitch length e, I is the moment of inertia of the gross section about the NA, V is the maximum vertical shear in the interval e, and Q is the moment of area about the NA of the elements whose sliding is resisted by the rivets or bolts.

l.

\

u

Chapter 6

Beam Deflections

m.

6-1 INTRODUCTION In this chapter we consider the rigidity of beams. Frequently the design of: beam is determined by its rigidity rather than by its strength. For example, in designing metalworking equipment for precision work, such as lathes, milling machines, or grinders, the deformations must be kept below the Permissible tolerances of the work being machined. One of the most important applications of beam deflections is to obtain equations with which, in combination with the conditions of static equilibrium, statically indeterminate beams can be analyzed. (See Chapters 7 and 8.) Several methods are available for determining beam deflections. Althougjt based on the same principles, they differ in technique and in their immediate objective. We consider first a variation of the double-integration method that srwly broadens and simplifies its application. Another method, the area-moment method, is thought to be the most direct of any, especially when the deflection at a specific location is desired. After a preliminary discussion (Art. 6»4), it will be found to be not only simple, but extremely rapid to apply. A variation of it which we shall consider in Art. 8-7, is also rapid and easy to use. . Other methods are the conjugate-beam method and the method of superposition. The conjugate-beam method is a variation of the areaemornent method but differs from it in technique. The method of superposition is not an independent method; it uses the deflection formulas for certain fundamental type of loadings to obtain results for loadings that consist of combinations of these . fundanaentd types.

6-2 DOUBLE-JNTEGRATION METHOD The edge view of the neutral surface of a deflected beam is called the elastic curve of the beam. lt is shown greatly exaggerated in Fig. 6-1. This article show$ how to determine the equation of this curve, that is, how to determine the vertical displacement y of any point in terms of its x coordi rate. Select the left end of the beam as the origin of an x axis directed along the original undeflected position of the beam, and a y axis directed positive upward. The deflections are assumed to be so small that there is no appreciable difference

182

I

183

6-2 DOUBLE-INTEGRATION METHOD

A

y

`.l.l,ae O

al

Elalstic curve x

I

I I

>

x

al

1

(do

_.

0

Segnnenf. of beam .

Figlllre 6-1 Elastic curve.

between the original length of the beam and the projection ofits deflected length. Consequently, the elastic curve is very flat and its slope at any point is very small. The value of the slope, tan 0 = dy/dx, may therefore with only a small error be set equal to 0, hence

Q

(a)

dx

d6 dx

and

(be

If we now consider the variation in 6 in a diH'erential length ds caused by bending in the beam, it is evident that (C) ds = p dO where p is the radius of curvature over the arc length ds. Because the elastic curve is very flat, ds is practically equivalent to do; so from Eqs. (c) and (b) we

obtain

-1 = be ds p

";:

Q dx

or

-1

dl'

y

p

(d)

In deriving the flexure formula in Art. 5-2, we obtained on page 135 the relation

1 p

M EI

(5-1)

Equating the values of l/p from Eqs. (d) and (5-1), we have dy

E1

is - M

(6-1)

This is known as the differential equation of the elastic curve of a beam. The product El, called thejlexural rigidity of the beam, is usually constant along the beam.

6/BEAM oEFLEcnb;

184

The approximations we have made do not seriously invalidate Eq_ (6-I). for if we replace l/p by its exact value as found in any calculus text, we have from Eq. (5-l)

d`yldx2

.r

Ii + ldy/dv)']3/2

.01 EI

1

Since do/dr is very small. its square is negligible compared with unity, and hence we obtain

.

d2.1. do2

M EI

which is the same as Eq. (6»I ). IfEq. (6-1) is now integrated. assuming El constant. we obtain

d.

f

E1 -l

do

M dx + C1

(6.2)

This is the slope equation specifying the slope or value of dy/dx at any point. Note that here M represents the moment equation expressed in terms of x, and C, is a constant to be evaluated from the given conditions of loading. We now integrate Eq. (6-2) to obtain

.

Ely =

H

mdxdx+ C l x +

c,

(6-3) I n

This is the required deflection equation of the elastic curve specifying the value ofy for any value o f f ; C2 is another constant of integration that must be evaluated from the given conditions of the beam and its loading. If the loading conditions change along the beam, there is a corresponding change in the moment equation. This requires that a separate moment equation be written between each change of load point and that two integrations of Eq. (6-1) be made for each such moment equation. Evaluation of the constants introduced by each integration can become very involved. Fortunately, these complications can be avoided by writing a single moment equation in 'such a way that it becomes continuous for the entire length of the beam in spite of the

discontinuity of loading. For example, consider the beam shown in Fig. 6-2. Using the definition M = (Z M)!_ discussed in Art. 4-2. we find that the moment equations between the change of load points are . Mas = 480x N ' IH Mac = [480x

500(x - 2)1 N. m

Mc, = [480x - 500(x - 2)

- i?lx - 3)21 N - m

Observe that the equation for MCU will also be valid for both

MAY

and MBC

provided that the terms (x - 2) and (x - 3>2 are neglected for values of x less than 2 m and 3 in, respectively. In other words, the terms (x - 2) and (x - 3)2 are nonexistent for values ofx for which the terms in parentheses are negative.

¢

185

6-2 DOUBLE-INTEGRATION METHOD ¢

y

I

500 N

4

!

--4

A

I

1

B

1. I

450 N/m C

4

1

1

1

1

D

x

7 - - - » ¢

1

-Mm -L-2 m

2m

P

R2=920N

RI ==480N

Figure 6-2

As a reminder of` these restrictions, we adopt a notation in which the usual . With this form of parentheses is replaced by pointed brackets, namely, ( change in notation, we obtain a single moment equation M = (480x - 500(x

- 2) - %Q(x - 3>2) N ~ m

which is valid for the entire beam if we postulate that the terms between the pointed brackets do not exist for negative values, otherwise the term is to be treated like any ordinary expression. As another example, consider the beam in Fig. 6-3a. Here the distributed load extends only over the segment BC. We can create continuity. however, by assuming that the distributed load extends beyond C and adding an equal upwarddistiibuted load to cancel its effect beyond C, as shown in Fig 6-3b. The general 600 x 600 N

y

I

4oonm

400 N/m

I

| B.

I

I

1

I

|

I

1 r

1

~C

A

-

J'

I

li

R1

£1

D

run, 500 N

X

A

B

I

|

I

1

|

n

1

Ill

12 ma »2 m

R2 =1a0o x

lm R1=500N

C`HHn l l l l l l l l

|

I

i

u

T`T'TTT" I

,

I c

~lLLLi>8_.x

3m

2m

R,

2m

i300 N

(b)

(a)

Figure 6-3 Technique of establishing continuity of loading.

moment equation, written for the lat segment DE in our new notation using pointed brackets, is x - l " + 1°§x + l 300{x 6))N.rn M = (500.r As before, we Specify that the terms between the pointed brackets do not exist for negative values. Notice that all loadings are automatically included in the general moment equation by writing it for the last segment of the beam. Q

- 4>*

_

* The justification for ignoring negative values of the terms in the pointed brackens depends on the fact that the general moment equation is written using the definition M = (E u . which means that we consider the effects of loads lying only no the left of an exploratory section. A negative value of the terms in a pointed bracket Indie-ates a loading that is to the right of an exploratory section. whereas a zero value merely indicates the start of a loading.

186

6/BEAM DEFLECTIONS

ILLUSTRATIVE PROBLEMS 601. A concentrated load of 300 N is supported as shown in Fig. 6-4. Determine the equations of the elastic curve between each change of load point and the maximum

deflection in the beam.

y

I A

I

300 N 2m l

x

4

R1-ICD N

8

c

lm

_in I

D

/

x

I

9

R,-200 N

Figure 6-4

Solution: Writing the general moment equation for the last segment BC of the beam, applying the dflbrentid equation of the elastic curve, and integrating twice, we obtain the following slope and detection equations:

-

d2y

£ / 5 = M = (l00x - 300(x - 2)) N - m

(0)

~dy E12-1 = (50.r 2 - l 50(x - 2) 2 + CI) N - m 2

Ely =

(%°x' - 50(x - 2>3 + CIX + co) N

(b)

- m3

I

(c)

To evaluate the two constants of integration that are physically equivalent to slop¢ and deflection at the origin, we apply the following boundary conditions:

I . At A where x = 0, the deflection y =0. Substituting these values in Eq. (c), we is to be ignored for negative values. find that C2 = 0. Remember that ( x 2. At the other support where x = 3. the deflection y is also zero. Knowing that C2 = 0 and substituting these values in tbe deflection equation (c), we obtain

2>3

0 ==§(3)3-500-2):+3c1

or C l = - l 3 3 N - m 2

Having thus evaluated the constants of integration, we return to Eqs. (b) and (c) to rewrite the slope and detection equations in the conventional form shown in the tabulation below. Continuing the solution, we assume that the maximum deflection will occur in the segment AB. Its location may be found by differentiating Eq. (e) with respect to x and setting the derivative equal to zero, or, what amounts to the same thing, setting the slope equation (d) equal to zero and solving for the point of zero slope. We obtain

Sox'

- £33 = 0

or x = 1.63 m

Since this value of x is valid for the segment AB, our assumption that the maximum deflection occurs in this region is confirmed. Hence, to determine the maximum deflection, we substitute x = 1.63 in Eq. (e), which yields

187

ILLUSTRATNE PROBLEMS SegmentAB(0 s x s 2 m )

d 51-i'= (50x2 - l33)N-m' dx

(d)

l33/) N. ml

Ely =

(f)

A

Segm entBC (2rns xs3m )

_ _

= 150X I50 _ 1331 n.m 2? 50 _ 2)-'- l33x]N- my Ely = [»_ do

1

BIZ

(f)

1

50(x

(8)

Max. Ely = -145

un

n~m'

The negative value obtained indicates that the deflection y is downward from the ' regard to do, is x axis. Frequently only the magnitude of the deflection without desired; this is denoted by 6, the use ofy being reserved to indicate a directed value of deflection. In Eq. (c), the unit of the product Ely is N - m3. This follows from integrating Eq. (a) twice. With M in units of N - m, the first integration gives N • m2 as the unit of the slope equation. A second integration results in Nun' as the unit of the 9

deflection equation. For consistent units, E must be in units of N/m2 and I in units of m'. Then the deflection y will be determined in meters. For example, if E = in GPa and 1 = 1.5 X 106 mm' = 1.5 x 10" m', Eq. In) becomes (10 X I0°)'

(0)

In@lUing*&. (a)$=we obtain the slope equation: '~'¢

J

_

1. - a x ef-'LMI ) 2 " do: 3* \

2

w

go (x - a)3 +c.

o L2

2 (

(b)

However, the s i p = .dy/dr is zero at x = 0, so C, = 0. We may now integrate the slope equation (with CO = 0) and obtain the deflection equation:

Ely = \q,(l1 ._

I3

__.

al 6

Wo _»_(13

4

._

a2)x2 -

52 ( x - a)4 + Co

(c)

Since g = 0 at x = 0, we End that Co = 0 also. Observe that selecting the origin of axes at the perfectly restrained wall where the slope and deflection arc zero makes the constants of integration do zero. The value of the maximum deflection, which occurs at the free end, is denoted by 6. Evidently 6 = -or so On substituting .r = L and simplifying, we obtain £16

_ ub(L - 0) (U + L34 + La' - as 8

3

On: important variation of this result occurs when a = 0. Then the entire length of the beam is uniformly loaded and the maximum deflection is given by q

LE up' = £16 = *o 8 8 1

PROBLEMS 605. Determine the maximum deflection 45 in a simply supported beam of length L carrying a concentrated load P at midspan. Ans. 6 = pL3/4851 . 606. Determine the maximum deflection 6 in a simply supported beam of length L carrying a uniformly distributed load of intensity Wo applied over its entire length. Anr. 6 =- (5/384X(0,'/£l1 -= ( 5 / 3 8 4 x w L ' / E l 1 of 607. Determine the maximum value Ely for the cantilever beam loaded as shown in ' at the wall, Fig. P-607. Take the origin 4

191

PROBLEMS

P

>1

a

I

nl

-I

L

Figmrc P-607

beam shown in Fig, P-608; 608. Find the equation of the elastic curve for the cantilever at the free end. Take the We to wall the at zero it carries a load that varies from origin at the wall.

za- INN lll |

Figure P-608

/

•

carries two symmetrically placed 609. As shown in Fig. P-609, a simply supported beam of concentrated loads. Compute the maximum deflection 5 and compare one-half answer your Check 233. page 6-2, Table 7, case of this result with the midspan 6 by letting a = L/2 and comparing it with the answer to Problem 605. Aus. s =- Ir0/24£I1313 - 4az>

a J

P

I

I

a F

L

1

Fi8\It¢ P-609

I:

If:

R.

610. The simply supported beam shown in Fig. P-6l0 carries a unilbrm load of intensity deWo symmetrically distributed over part of its length. Determine the maximum flection 6 and check your result by letting a = 0 and comparing with the answer to Problem 606.

We

11111.11 *l 2b .1.

J-6

-

lrsli¢~~~ p-sno

I

G-4?

I

pa

152

611. Compute the value of E16 at midspan for the beam loaded as shown in Fig. P-6l I . If E = 10 GPa, what value of I is required to limit the midspan dethwtion to I /360 of the span? Au. E18 = SOON-m1, l= 4,50 X 10° mm' 300 N

uuI"i 1 Ji

0... all

Figure P-6\ I

2,

...g

_I

'|

n

' -

4 I . .

v '

12.

6/BEAM DEFLECTIONS

192

612. Compute the midspan value of E16 for the beam loaded as shown in Fig. P-6 I 2. Ans. E16 = 6960 N m'

-

600 N'm 1

l

1

l

1

-1

,,,_1.

_|.

3m

2m

R)

U

Figure P-612

1

613. IFE = 29 X 10° psi, what value of I is required to limit the midspan de flection to 1/360 of the span for the beam in Fig. P-6l3? Aus. 66.4 in.'

2400 :be _

4

>

1

.1.

l 4.

1.

1r

oh

_ 1

F

I

Iy

I

,;_2 ft.. I

Gft

£1 614

R2

Figure P-613

For the beam loaded as shown in Fug. P-614, calculate the slope of the elastic cu we over the right support

J: i ' i i

I

411

240 lb

IN

.1.

1

411

RE

I

-2 fr

J

R2

Figure P-614

615. Comping the szlue of Ely at the right end of the overhanging beam shown in Fig. P-615. Ans.

Ely = -2850 lb- fm'

1000 lb

I

1

I

•

R1

6 f1

.1.

400 Ib/It

.

1

41;

R2

I

a nFigure P-615

616. For the beam loaded as shown in Fig. P-616, determine (a) the deflection and slope under the load PaW (b) the maximum deflection between the supports. Ans. (b) Max. Ely = Pa 1b/9 VS?

PROBLEMS

193 'P a

'y

,A L

Figure P-616 and P-6l7

9

617. Replace the load P in Prob. 616 by a clockwise couple M applied at the right end and determine the slope and deletion ax the right end. Ans. EI dy/dr = ' ( M / 3 X L + 210; E15 = (Mb/6)(2 I, + b) 618. A simply supported beam can'ies a couple M applied as shown in Fig, P-618. &termine the equation of the elastic curve and the deflection at the point of appliantion

of the couple. Then, letting a = L and a = 0. compare your solution of the elastic curve with cases I I and 12 in Table 6-2 on page 233. Ana. Ely = 1un/3I.xL' - 3La + Za?) ~\

M

""n

'g-»

I

Figure P-618

'-

L

l

I

619. Determine the value of Ely midway between the supports for the beam loandeé as shown in Fig. P~6l9. 20ONrn M

1800 N°m I

r

1

r

l

1

r

r

1

I

I

k it

Figure P-6l9

us

a

, 1

J I

2

L. l u

12,

m-I

R;

620. Find the midspan deflection 6 for the beam shown in Fig, P-620. carrying two triangularly distributed loads. (Hint: For convenience, select the organ of the ax¢s at the midspan position of the elastic curve.)

.-ins.

wo

5

(9/' nwox

L';£1>

HUI UP 4* HUH .J

J. B

L

"*

2

Rx

We

J

i

L

a

J L

)

Ra Figure P-620 \

6/BEAM DEFLECTIONS

194

621. Determine the value of E16 midway between the supports for the beam shown in Fig. P-62 l. Check your result by letting a = 0 and comparing with Prob. 606. (Apply the hint given in Prob. 620.)

wo 1

I

a

L

a

I I

Figure P-62 l

6-3 THEOREMS OF AREA-MOMENT METHOD A useful arid simple method of determining slopes and deflections in beams involves the area of the moment diagram and also the moment of that areathe area-moment method. We discuss first the two basic theorems of the method: then, after showing how to compute the area and moment of area of the moment diagram, we shall apply the method to several types of problems. The method is especially useful in directly determining the slope or deflection at a specified position. Depending as it does oh the geometry of the elastic curve, the areamoment method emphasizes the physical significance of slope and deflection. The area-moment method is subject to the same limitations as the doubleintegration method, but in order to present it in its entirety as a completely independent alternative method, we repeat a small portion of the preceding article. Figure 6-8a shows a simple beam that supports an arbitrary loading. The elastic curve is the edge view of the neutral surface and is shown, with greatly exaggerated deflections, in Fig. 6-8b. the moment diagram is assumed to be as

in Fig. 6-8c. As we saw in the derivation of the flexure formula, Art. 5-2, two adjacentplane sections of an originally straight beam will rotate through the angle d' relative to each other. This is demonstrated in the enlarged detail of Fig. 6-8b, in which it is also apparent that the arc distance ds measured along the elastic curve between these two sections equals p dt), where p is the radius of curvature of the elastic curve at the given position. From Eq. (5-l) we have

_

H

1 p

EI

and since ds = p d6, we now write

or

l

M

p

EI

dO

M E1 ds

go

ds

(a)

l . .

6-3 THEOREMS OF AREA-MOMENT METHOD

195

lo Arbitrary loading

rI

I

(a)

RE

L

R: .-

A

I I

7

I

.

B,

g,

\

f

JB

1

>

Ida

/{

I

d6

(b)-

I I

I I I I I I -I I

I

I

I

I I

Enlargement of segment CD

x

(C)

I

I

}d ' / A

>

I

>

pa

\\m \

\

I

do

I

Figure 6-8 Area-moment theorems.

In most practical cases the elastic curve is so flat that no serious error is made in assuming the length ds to equal its projection do. With this assumption, we obtain . »

dB

M dx EI

(b)

It is evident that tangents drawn to the elastic curve at C and D in Fig 6-8b are separated by the same angle d6 by which sections OC and OD (in the enlarged deta.il) rotate relative to each other. Hence the change in slope between tangents drawn to the elastic curve at any two points A and B will equal the sum of such small angles: '

.

6,48

all;

=f

Us

84

dO =

of

is

As! dx

-me

(c)

in Fig. 6-8b that the distance from B on the elastic curve (mea- . Note sured perpend icular. to the original position' of the beam) that will intersect a tangent drawn to this curve at any other point A is the sum of the intercepts dt

created by tangents to the cu rve at adjacent points. Each of these intercepts may

be considered as the arc of a circle of radius X subtended by the angle do: dt == x dB

6/BEAM DEFLECTIONS

196

Hence x do

is/A

Replacing dO by the value in Eq. (b), we obtain l IS/A

i

L.

Ra

n

El

(d)

x(M dr)

The length IB/,4 is known as the deviation of B from a tangent drawn at A. or as the tangential deviation of B with respect to A. The subscript indicates that the deviation is measured from B relative to a reference tangent drawn at A. Figure 6-9 illustrates the difference between the deviation Is/,4 of B from a reference tangent at A, and the deviation Lua of A from a reference tangent at B. In general, such deviatioNs are unequal. ,

"i"I

J 1... I

'I

Reference tangent at A Reference tangent at

.

Figure 6-9 Inequality of

B

L4/1l

and Ia/.4

.

The geometric significance of Eqs. (c) and (d) gives rise to the two basic theorems of the area-moment method. From themoment diagram in Fig 6-8c we see that M dx is the area of the shaded element located a distance x from the ordinate through B. Since j' M dx means a summation of such elements, Eq. . (c) may be expressed as

l

M = E (areas

(6-4)

This is the *algebraic expression of Theorem I, which is stated as follows: Theorem I. The change in slope between tangents drawn to the elastic curve at any two points A and B is mud to the product of l / E I multiplied by the area of the moment diagram between these two points.

1

Figure 6-8c shows that the expression x(M dx) that appears under the integral sign in Eq. (d) is the moment of area of the shaded element about the ordinate at B. Hence the geometric significance of the integral x(M do) is that the integral is equivalent to the moment of area about the ordinate at B of that part of the moment diagram between A and B. Thus we obtain the algebraic form of Theorem II:

f

t8/A =

1

E; (area)m 'la

This is stated more formally as: J

(6-5)

6-3 THEOREMS OF AREA-MOMENT METHOD

197

Theorem II. The deviation of any point B relative to a tangent drawn to the elastic curve at any other point A, in a direction perpendicular to the original position of the beam, is equal to the product of l/EI multiplied by the moment of area about B of that part of the moment diagram between points A and B.

The product El is called flexural rigidity. Note that we have tacitly assumed that E and I remain constant throughout the length of the beam; this is usually the case. If they vary, however, they cannot be written outside the integral sign, and the manner of their variation with x must be known. Such variations are usually taken care of by dividing each moment ordinate by EI to obtain an M/EI diagram that is used in place of the moment diagram in 'the foregoing theorems. In the two theorems, (areas is the area of the moment diagram between points A and B, and x`8 is the moment arm of this area measured from B. When the area of the moment diagram is composed of several parts (this is explained in .Art. 6-4), the expression (31'63).4B pa includes the moment of area of all such parts. The moment of area is always taken about an ordinate through the point at which the deviation is being computed. An automatic method of using the correct axis for moments is to give X' the same subscript, for example, B (meaning that moment arms are to be measured from this point), as appears in the numerator of the subscript to t (i.e., B/A). One rule of sign is very important: The deviation at any point is positive if the point lies above the reference tangent from which the deviation is measured, and negative if the point lies below the reference tangent. Positive and negative deviations are shown in Fig. 6-10. Conversely, a computed positive value for deviation means that the point must lie above the reference tangent.

-

A

'i

I

I

4 A

/A

'as B

£8/A

Y

(a) Positive deviation; B located above reference tangent

(b) Negative deviation; B located below reference tangent

Figlure 6-10 Signs of deviations.

A \

`\49

51 I-'\. A

tangent

l

8 9.48

(a) Positive chaunge of slope; 943 is counterclockwise from left

a..

(b) Negative change of slope; is clockwise from left tangent

Figure 6-1 l Signs of change of slope.

Another rule of sign that concerns slopes is shown in Fig. 6-1 l. A positive value for the change in slope 6,8 means that the tangent at the rightmost point

6/BEAM DEFLECtOrS

198

the leftmost 8 is measured in a counterclockwise direction from the tangent at versa. point, and vice

6-4 MOMENT DIAGRAMS BY PARTS In order to apply the theorems of the area~moment method, we should be able to compute easily and accurately the area under any part of a moment diagram, and also the moment of such an area about any axis. A method of doing this from calculus is to integrate the two expressions l' M dx and x(M do) between proper limits, noting that the bending moment M must be expressed as a function

f

of x. Our purpose here, however, is to discuss a method of dividing moment diagrams into parts whose areas and centroids are known; this permits simple numerical calculations to replace integrations. The first step is to learn how to draw moment effects of each separate loading (hereafter called moment diagram by parts) instead of a conventional moment diagram. The construction of moment diagrams by parts depends on two basic principles:

1. The resultant bending moment at any section caused by any load system is the algebraic sum of the bending moments at that section caused by each load acting separately. This statement is expressed algebraically `

by

(4-2)

M=(2M)L=-.(2M)R

where (Z M); indicates the sum of the moments caused by all the forces to the left of the section, and (Z AI)R is the sum of the moments caused by all the forces to the right of the section. 2. The moment effect of any single specified loading is always some variation of the general equation '

J'

(0)

Kr"

The graph of this equation is shown in Fig. 6-12. The shaded area and the . location of its centroid are easily shown by calculus to be q

=

a

I n+l

(b)

l -b n+2

(c)

r e a ~ b h x I I

where b is the base and h is the height.

.

When the area under the curve between positions like A and B in Fig. is 6-13 computed, Eqs. (b) arid (c) refer to the shaded area between the curve. the ordinate at B and the tangent at .4. To this area must be added, of course. the shaded trapezoidal area between the tangent and the x axis.

6-4 MOMENT DIAGRAMS BY PARTS

199 y

VI

l

y=hx"

\


-

b

4

in

II

Figure 6-12

y

1

|

h I

e

A

I

4

.

/ (1

I/{

O

b

x

Figure 6-13

Table 6-1 demonstrates the truth of the second basic principle stated earlier, namely, that the moment effect of any load is some variation of the equation y = ex". This table gives data on four cantilever beams, each loaded differently with increasingly complex loads. ' Note that a cantilever loaded by a couple C has a moment equation of the type y = kx" in which k = -C and n is zero, namely, M = -Co. In other words, a couple type of loading produces a moment equation of zero degree. Similarly, a concentrated load produces a moment equation of the first degree, a uniform load produces a moment equation of the second degree, and so on. In the column headed "Area," the area of the moment diagram is expressed in terms of a factor multiplied by the general base distance b and the maximum height h of the moment diagram. The position of the centroid of each moment diagram from the maximum ordinate of the diagram is defined as a factor of the base distance. These factors or coefficients, which increase very simply, are obtained from Eqs. (b) and (c) by assigning to n the value of the degree of each equation, that is, n = 0, n = l, n = 2, and so on. An example will illustrate how Table 6-1 is used to draw moment diagrams by parts. The simple beam in Fig. 6-14 is 3 m long and supports a uniformly distributed load of 300 N/m over the right 2 m of the span. At any section a-a between A and B, the moment effect defined by M = ( Z M)1_ is caused only by RI. Also, at any section b-b between B and C, the moment effect will be due to Re and to the portion of the uniformly distributed

c

qI

9

qI

QI

QI _g

i

v

|

ws

wI I

we

I/4% sum

l

»

-=

*IZ

'f

~1 z

9 -= z701" 'I

Z

z.7°m"

~iz

l--s

7J-==v

o--w

'1=Q

7=9

7-=Q

'I q WBJDBIP IUBWOW

n

l

l

pig

PUZ

is I

•

(0x9- = W '°a'I)

01:2

uonanbe lU8lJJOl.U IO eeJ5eQ

'79 to 5

z

=:

=W

W

z X 5 ' = W

rd»

J-==w

w xuewow) uon9nbe IUBUIOW

(x uon°°6 Au!

I

I

Z

1

I

-l

1

. I

i

L

_

I

-l

I

I

I

7

-l

I

n

'I I

V-=-1l L-p'

H*

. ,-,

-w

to l l t t l

.2

-1--r-u-

owe 1°A°lnu¥O

8u!.&.1¢A &1u1.lo_;!u(]

pmyluaouog

v°\nQ!11$!1> Klwqnvn

¢

ogdnog)

VI

a°\av.\.

6uln¢°la°°dAJ.

S5)NlQYQ"l H3A3'1lJ.NVO

\

coz

a

201

6-4 MOMENT DIAGRAMS BY PARTS 'a

'L.=,1

A*

'

8

1b _ _ i _300 N/m~ I

I

r

1

I

r

r

Ib

1

I

1

r

2m

C

In, I SR I

-

of: an,-

+600 N- m Line of

( ) ,,_,5\\

(d)

w// ; I

3m

aoolwm

-(300x21x 1 =-600N»m

(b)

`

I'

lm

mu

M: -mwxmxm -Moment caused

by UDL

Moment

caused by R

81

/

Moment diagram by parts

1

(a)1

zero-moment

_

-600 N-m Equivalent cantilever

loading.

Resultant moment

Conventional moment diagram

Fg\lre 6-I4 Moment diagzaaum byparm

load included between B and b-b. Note that defining the bending moment in terms of the forces to the let of the section means that the uniformly distributed load has no moment eHlect on segment AB. Actually, the moment elect of R, at any section of the beam is equivalent to the cantilever loading at (a), whereas the moment effect of the uniform loading on any section of the beam is equivalent to. the cantilever loading at (b). . By referring to Table 6- I , we can plot the moment diagrams of beams (a) and (b) on a Common baseline (the line of zero Moment), as shown at (c) in the figure. That the algebraic sum of the shaded areas of (c) will yield the resultant or conventional moment diagram is evident from the fact that the moment at any section of the original beam is mud to the sum of the moments at that section caused by the individual loads (basic principle l). Hence, if the triangular area is revolved about the line of zero moment as an axis, we obtain diagram (d). The shaded area of diagram (d) is evidently equal to the area of the conventional moment diagram (e) obtained by plotting the resultant moment at any section as ordinates to a horizontal baseline Hence the conventional moment diagram may be replaced by an equiyalcnt moment diagram constructed of

parts, as in Fig. 6~l4c, whose areas and centroids can be easily computed from

the data in Table 6-1.

6/eemm DEFLECTIONS

202

To compute the moment of area of the moment diagram, we observe that the moment of the area of the conventional moment diagram (Fig. 6-l4e) is equivalent to the sum°of the moments of area of its parts, as drawn in Fig. 6-l4c. Hence, noting that each such area iS the product of a coefficient listed . in Table 6-1 multiplied by the dimensions of the circumscribing rectangle, and that its oentroidal location is the product of a similar coefficient times the~ba$¢~ length of this rectangle, we obtain as the moment of area with respect to the right end C II

»>?

\

i-4

5. 3

Earl

n

.q

II

u 'H

2

f-.

P

x 600 2

Xi

II

700 N~m3

X

;°°)(..l 2 1 4

2X

)

.

X

Ans.

The symbol ac means that moment arms are to be measured from C. From Theorem II of the area-moment method, this result represents the product Elf(/,4 , where FC/A is the deviation Of point C from a tangent drawn to the elastic curve at point A. . . . One .final observation: It is usually unnecessary to compute the beam re-. actions. Since the conventional bending moment at C must be zero (beCcuse no loads act to the right of C to cause bending moment), the moment of the reaction R, at C (equal to 3R1) must balance the bending 'moment of --600 N m at C caused by the loading. . . By now it may be evident that the technique of drawing moment diagrams by parts is really a graphical interpretation of the method of writing a general moment equation described on page 185. Thus for the beam shown in Fig. 6~l5, the general moment equation is .

-

M=

RIX+M1(X

al)0'

(x

W2

'

02>2

s

1-

I'

43

.l

"

6b (x

Us

3

. b-l

.Ill ill |`°'*n->ull' Q-

2

1

RE

r

Ra

Figure 6-15

If each of the terms in pointed brackets is replaced by a new variable that starts at the position where each loading begins, we see that each term in the general moment equation is the moment equation of a cantilever type given in

I

203

ILLUSTRATNE PROBLEMS

Table 6-1. Thus replacing (X

- ui) bY u. (x

" 61) by

obtain

v, and ( x

- 03> by

2.

we

w

M = R | x + M I u 0 - ; -' U 2 - 5 2 2 3

ILLUSTRATIVE PROBLEMS moment 622. For the beam shown in Fig. 6-l6a, compute the moment of area of the diagram about the left end.

8,.uIlllH?

A I

2m

I

3m

I

I

R1 = 300 N I I, I I I I

I.

!

I

(a)

I I

I I

I l

I

I I I I

I

M..300X 5 , R1=300N =1500N-m

600 N

H2 I I I I I I

W

,axeoo. 900 N

(I1)

I I I

1500 N m

lm

600 N/m

2

I 2m

3m

=-900N-m

SOON-m

I

I

.

ii!-'lsdox 1

I I

I

I I I I I I

l

I I I

I

I,

5m

D

m. I 11 I

|-.

I I

600 N / m

I5m.

In

2m 3rd degree curve .(b)

m 900 N~m

lm

600x 1 600 N*m

R2 ='ego N

Equivalent cantilever loadings

Figure 6-16 Solution: From statics, the reactions are computed to be.R, = 300 N and RE = 600 N. The moment diagram by parts is constricted by applying the deNtition M = (Z M)!. to segments .48 and BC. and the definition M = (E M), to seginnent CD. Observe that it is simpler to compute the bending moment for an exploratory section in segment CD by taking moments of forces to the right of the section rather than to the left. Computing bending moment by applying either M = (E M); or M = (2 M), will generally indicate the simplest manner of drawing the moment diagram by parts. . . The equivalent cantilever loadings are shown in Fig. 6-l6c. The bending moment diagram in Fig. 6-l6b is checked by the fact that the moment at C in terms of the forces to the left of C is 1500 - 900 = 600 N in, which equals, as it should, the moment at C expressed in terms of the forces acting to the right of C.

-

6/BEAM DEFLECTIONS

204

A is now comThe moment of the area of the M diagram about the left end putod as equal to the sum of the moments of area of its parts. Referring to Table 6-1 for the coelbcients of area and of centroidal distances, we obtain: ltareaM • x] (3F¢8)AD • -'FA

=,`E. as]

1 )(-3 W2 ')(5 + 3 .)

1500x 5 2

=(

X

2X 5

X

900 X 3

-(

4

)(2 + 45 X 3) Ans.

I1.I3 kN.m3

What value of EI! dog this result represent? 623. For the overhanging beam shover in Fig. 6-17a, compute the moment of area about C d the moment diagram included between the supports at A and C.

60 lbfft

B;

Al 4 ft

i°' I

I

r

I

| _.

4 f¢

r

l

r

8 ft

|

I

I 1

I

I

I

r

"t I

r

1

I

RE

ca)

I

I

4ft

sf:

\

_

12 fc .

R1

11231 = 14401b-fr

12f¢

(¢).

\

5; M=

cure

/

/`

nu

8 ft 4 ft M=-(60x 8)x =-1920 lb.-ft

4/

K 2nd degree -480 lb.-ft

4

12R,.

7

curve 2nd degree

V

f

D

4 fI.

m= - (60x

4')x g

480 lb'ft/( un-

60 lb./ft

up 4 fu

-1920 lb.°ft Equivalent cantilever loadings

Figure 6-17

.r

Solution: At any section between A and C, the conventional bending moment is computed more easily by applying M = ( Z M),,, whereas between C and D.it is simpler to apply M = (2 Mln. The moment diagram by parts shown in Fig. 6-l7b is therefore oonstruaetd by combining the cantilever loadings in Fig. 6- l 7c. In this problem, the value of the reactions need not be computed. The moment at C caused by R. is found from the fact that the bending moment at C of all forces to the left of C must equal the bending moment of all forces to the right of C, which is in accord with the fundamental definition M == ( Z My. == ( Z M)8. In other words. I 2n, 1920 = -480; hence 12& B 1920 480 as 1440 lb°R.

I

r

PROBLEMS

205 We obtain the moment of area of the M diagram between A and C abo .it C

by applying [(area),4¢°J?e =

=(

(area),¢e' ic

Earl

l440 x 12 -.

2

34 560

:

ii

1920xs

xI_:H

10 240

z

3

AIR.

24 320 Ib-R'

To

F*

PROBLEMS For each of the beams in Problems 624 to 629, compute the moment 0/ area of the M diagram between the reactions about both the left and the right reaction.

624. Beam loaded as shown in Fig, P-624.

Ans.

(3r¢3),,,-2, = 11.7 in~In'

1000 N

M=400N°m 2m

'TI

2m

2m

8

L

0

R

Figure P-624

625. Beam loaded as shown in Fig. P-625. (Hint: Dralw the moment diasrw by pans from right to left.) / /

»;~

SOON

I

I

2m

1

nl-

Im

400 N:z1

l

A

Figure P-625

lm

V

8

RE

R!

626. Beam loaded as shown in Fig. P-626. 400 lb/tr.

HIHIHH HHIIHI

A k 1 .A a l a

Figure P-626

Rx

~!~ " | '

G D . u l I.

I 'I'

B r 1

Q .

.Las

R2

6/BEAM DEFLECTIONS

206

627. Beam loaded as shown in Fig. P-627. (Hint: Resolve the trapezoidal loading into a uniformly distributed load and a uniformly varying load.) 600 N/m

A

./

MTI"IIIIIII

* * l m .|. .

8 4 I

3m

n

RE

R2

Figure P-627

sis. Bealm loaded with 3 uniformly yalying load and a couple as shown in Fig. P-628. Ans. M=400lb-ft

.._lrHH{

A

2 f¢

-1-

2fz

ullllllll

lo' 1b.ft3

200 lb./ft

' B l

i

Sf:

D .

31

(area),48-jc] = 30.0 X

R2

Flss P-628 and P-629 629. Sdvc Prob. 628 if the sense of the couple is counterclockwise instead of clockwise as shown in Fig, P-628. °

630. For the beam loaded as shown in Fig. P-630, compute the value of (area),48~x',4 . From this result determine whether the tangent drawn to the elastic curve at B slopes up or down to the right. (Hint: Fefer to Eq. (6-5) and Fig. 6-10.) Ans. (areal,48°J?,4 = -463 N m3, slope is down to right

-

200 N 400 N/m

A

,.&11Illll»

I -1...L

J

I

I

3m

I

2m

R2

R1

Figure P-630

631. Dctcrnaine the vduc of the couple M for the beam loaded as shown in Fig. P-63 l so that the moment of area about A of the M diagram between A and B will be zero. What is the physic significance of this result? M

AH 2r¢

A

151

100 lb./fi I

r

r

I

I

|

I

r

B

4 f¢

152

Figure P-631

207

ILLUSTRATIVE PROBLEMS

632. For the beam loaded as shown in Fig. P-632, compute the value of (2\f¢3),4a'i¢» From this result, is the tangent drawn to the elastic curve at B directed up or down to the right? (Hint: Refer to Eq. (6-5) and Fig. 6-I0.)

Ans. (m3p,.x', = 1.z7 in.m', up IO right

zoon 800 Film I

1

r

4

I

` *

.L

L

2m

s

Figure P-632

1

r

>

A

. 1

lm

p

U

lm

F52

R1

8-5 DEFLECTION OF CANTILEVER BEAMS It will be recalled that the tangential deviation at any point is the distance from the point on the elastic curve to a tangent drawn to the curve at some other point (Art. 6-3 and Fig. 6-8). As a consequence, the tangential deviation is generally not equal to the deflection. In cantilever beams, however, the wall is usually assumed to be perfectly fixed, and hence the tangent drawn to the elastic curve at the Ivan will be horizontal, as in Fig. 6-18. Therefore if the tangential deviation at A is measured from a tangent drawn at 8, the magnitude of the deviation t,,1, will equal the deflection PA at A. l l l

5.4 =

l£A/8l

I

Ill

V

II

A

Figure 6-18

Several examples will illustrate how area-moment IMi11¢iPl¢S are used to determine slope and deflection in cantilever bums. Other umm of bealuns an considered later.

ILLUSTRATIVE PROBLEMS 633. For the cantilever beam in Fig, 6-19. it is assumed that E = L728 x 10' 15.6 in.'. What value ofP will cause a I-in. deflection at the flue end?

ad. / -

Solution: The moment diagram by parts is drawn as shown. Evidently the deflecdrawn at C. tion a at A is numerically equal to the deviation of A from a tangent Since the deviation at A is negative because A lies below the tangent, we have from Theorem II.

Q

6/BEAM DEFLECTIONS

208 P

90 lb 4 fc

A]

.LIB

6 ft

Cl

6 fr

I

61° I , / | /

A'1

4 fs

I

I - 6 P lb.»ft

10 f t

--900 lb.-ft 1

Luc

-6 from which

E (=1r¢a),.€

u

1

El

-

XA

-6P X 6

2

Figure 6-19

)(4+ _23 )+( X6

EIRE = (l44P + 30 000) lb

-900 x 10 2

2

)(~3

. fx'

X

iN

Substituting numerical values for E, I. and 6, and multiplying the right-hand member by 1728 to make the equation dimensionally homogeneous as discussed in Illustrative Problem 601, we obtain

`

L728 X 106 X 15.6 X 1 = (144P + 30 0O0)( 1728)

15 600 :: l44P+ 30 000

P = -100 lb

A ns.

The minus value of P indicates th the direction of P must be reversed to that originally assumed, that is, P must act upward. 634. Compute the maximum slope and deflection for the cantilever beam shown in Fig. 6-20a that carries a load varying uniformly from zero at the wall to we at the free . end. Solution: Although the given cantilever loading is not one of the types given in Table 6-1, it is easily transformed into them by replacing the given loading in Fig. 6-20a by those shown in Fi 6-;0c, that Q, superbosing a downward. uniforrnlv distributed load and an upward uniformly varying load. Thus we obtain the parts of the moment diagram shown in Fig. 6-20b. 'The elastic curve in Fig. 6-20a shows that the maximum slope and de flection occur at the free end A. The angle 04 is clearly equal to the change in slope aha measured between the tangents drawn to the elastic curve at A and B. Then by Theorem I we obtain ,

l @,.=v,,.=-(ar¢=),.=-L3

EI

I

-4.

l 6 .L_

WL2

3

WoLf

2

~L

W0 L3

El r

J

209

ILLUSTRATIVE PROBLEMS

Wo

jllm Ullr1--.V

M=-

s

T

V=.

(8)

+

L

\

2 3

L

4

t

2

w0L2 " 0

-_

3

wal. 2

w01.2

and degree curve

waz.

L

,¢.l

HUH .

(¢) l

2nd degree" curve (b)

2

I

I

I

4

1

q

t

f

I

[

2

We

I

Finn" 6-20

According to Fig. 6-1 I on page 197 the minus sign means that the angle is measured in a clockwise sense from the left tangent at A to the right tangent at B; hence the slope an A s u p to the right as shown. g. which is the The maximum deflection 6 at A is numerically equal to deviation of A from a tangent drawn at B. Since the deviation at A is negative because A lies below the tangent, we have from Theorem II

£4/a

l

- i n

_

'go (2I6a),.3 - XA 5

_ [(4 »)(» )1

1

]

EI

W°L2

6

4 5

We

I . L2 3 2

3

4 »x-11

which reduces to 11 be 120 El

Am:

Remember that the symbol 1] means that the moment arms of the areas are to be measured from A.

635. For the cantilever beam loaded as shown in Fig. 6-2la, compute the value of E16 at A and at B. Solution: The deflections at A and at B are ntnherieally equal to the deviations of A and B from the horizontal tangent drawn to the elastic cane at C. Since A and B are both below this tangent, the deviations to/c and have are both negative. Applying Theorem II, we have

`°'5»¢ =

and

in/c

`5a = in/c

=

5I

- 3, I

(&!'¢8).4C°X.4

(O)

(arca)1¢~xlI

(5)

Observe that in Eq. (a) we use the M diagram between A and €. whereas in EQ. (5) we need only the M diagram between 8 and C. Let us examine hrs; the various

ways in which the M diagram by pans can be drawn so that the simplest diagram is used.

6/BEAM DEFLECTIONS

210 150 N/m

c

1

I/,f A l'*-2 ru

6.4" ( .

l`°**',»..-'

1

68 >

':-

(

- =300 N

"/8/('

-7

m

M= -300 x J = -900 N - m

>

I

(H)

4m

I

dl % .m "'"°'1§'03'n go 4 m

300 \ -l= \"00 N-m

\4

-900 N-m -300 x 1 ::

-300 N-m

i

am

nu.

m

(b)

V = 300 N

1

-.M= -900 N»rn 7 150 N/m

4L 4Il

|

l

'I

2m -300N°m

'2 m M=-300x 1

I

150 N / m

H11 a~a@_M'

600 N~m 2m a

l

•

2m

1.

1

(c)

-300 N-m

I

-(150 X 2) X 1 -300 NF In

to:

/A

M=2X300-/ =600 N-m

7

/Q

-900 N~m

300 N

1M :

I

2 m.

-=900 N - m

M=-900 N - m

150 N/m

111111

\

2m n

2'")-

4m

2m

1300 N-m

I

2m

M=(150X2)X1=

(d)

150 N/In 300 N°m

IHHHIH 4m

-1200 N-m

.

W=\

M="(150X 4 l X 2 =

-.

-1200 N - m

Figure 6-21 Variations of moment diagram by parts.

First. the A1diagram in Fig 6-2lb is drawn by expressing the bending moment at any section in ten's of the loads to the right of the section in accordance with

the basic definition M = (Z Ml8- The equivalent cantilever loadings are also shown. Second, the Mdiagram can also bedrawn by expressing the bending moment between A and B by the definition M = ( Z M)/., and that between B and C by As! s ( Z MM. This results in the M diagram and equivalent cantilever loadings shown in Fig. 6-2 lc. . Third, still another method is to replace the given loading by superposing the loadings in Fig. 6-2 ld. This gives what is probably the simplest M diagram by parts.

211

PPOBLEMS

Although identical faults will be obtained So using any of the aforementioned means to diagrams, we apply Eq. (a) to the third one (Fig. 6-2ld). Noting that £11 obtain; wi: A, take the moment of area about

2 X 300 3

l

'

EI

-5

3

2

+4X

.

which reduces to EI6,4 = 4100 N-m'

)(~4 x *H 3

4 X 1200

) -(

3

2

Am.

M Since the deflection at B is expressed in terms of (area)8€, either of the segment diagrams in.part (b) or (c) of Fix- 6-21 may be used, for they are identical for BC. Taddng moments about 8 as indicated by in in Eq. (b), we obtain

-( =El l

~

--

5 Jr

..

from which we have

'

2 X 600 2

lx

IX

)(-3

oH

°°°(~2

2 ) - 2( x

)

.

Aus.

-E158 = 400- 1800 or £I6,=l4 0on-m '

by Although not shown in Fig. 6-21, a fourth way to draw the M diagram beam a 109: page on 4-3 Art. ~in parts from B to C is to use the concept discussed section may be cut at any section and the effect of the loads to one side of the a shear replaced by the shear and moment. at the cut section. At B, the load causes from defects whose fig N -300 of couple moment bending a force of -300 N and may B to C produce a triangular and a rectangular moment diagram. The student check. a as 58 compute to them use and well sketch these diagrams

PROBLEMS 636. The cantilever beam shown in Fig. ..P-636 has a rectangular cross section 50 mm wide by h mm high. Find the height h if the maximum deflection is not to exceed ' 10 m m . U s e E = lOG Pa.' Ans. 'lx = 619 mm

41 r

c

E

-.

I 8 t

x

J

R2

l

I

`

zC/A

I I I I

\

D Figure 6-22 Geometry of area-moment method applied to simple beams.

The problem is to determine the value of the deflection 6 at some position 8. If a tangent to the elastic curve is drawn at A, the deviation IBM at B from this tangent is evidently not the required deflection 6. However, the sum of 6 and IB/.4 constitutes the distance EF; and if both EF and :BM were known, 6 could easily be found. Hence the distance EF must also be found. This is done by noting that the triangle AEF is similar to the triangle ACD, of which the leg CD equals the deviation to of C from the reference tangent drawn at A. The proper procedure to apply is obtained by reversing the steps of the preceding analysis into the following order

1. Compute Ic'/,4 using the relation I

fc'/A :

.

EI ] (area) c,4 Xc

2. From the relations between similar triangles, determine EF in terms of /c/.4-

We obtain

-

EF =

x

Z°fC/A

3. Compute to from the relation Is/,4 =

EI(3-/¢3)zu • XI1

4. Since EF is the sum of a and I8/A. the` value of 6 is given by 5==EF-

18/,4

215

J

ILLUSTRATIVE PROBLEMS

As mentioned previously, this procedure may seem long, but actually it is rapid. Several examples will demonstrate the method. Only simple loadings are used, since this concentrates attention on basic ideas. For more complex loadings the basic method is unchanged, the more complicated M diagram by parts being constructed as indicated in Art. 6-4.

ILLUSTRATIVE PROBLEMS

1 l

649.. The simple beam in Fig. 6-23 supports a concentrated load of 300 N at 2 m from the left support. Compute the value of £16 at B, which is I m from the left support. y Q

l

1m

I'

A

. --

4

\

s

l

RI =100 N

lm

lm

B

|

6

DV

-la'/ I

'\

£8/A

x

/ 4

Elanstic CIIIVB

I

e

I

l20on I C/A

I I

I

I

I

E

3m


. .\. 4m

Figure P-657

.

I

2m

I

R1

-

6S8• F th ouple. be of the shown in Fug. . P 658, Iind the educ of ER at the point of applxcatnon Q ,

Aus.

£16 = (m0/31.x1.2

q

- up + 26)

-n a

Figure P-658

L

u

»&8>»

6/BEAM DEFLECTIONS

222 load placed anywhere on the soan, 659. A simple beam supports a wncentrated maximum d¢U¢¢ii0I1 an Fi_P-659. Measuring x from A, show that the x == (U - lf)/3.

h;wii

'1

\

P b

a

A

C

8

I

r

'Ra

Rll

Figure P-659

660. A imply supported beam is loaded by a couple M at its righ.t end, as shown in Fig. P-660. Show that the maximum deflection occurs at x = 0.577L.

L

Figure P-660

661. Compute the midspan deflection of the symmetrically loaded beam shown in Fig. P-66l. Check your answer by letting a = L/2 and comparing with the answer to case 6 .in Table 6-2 on page 233. Also compare one-half your answer with the midspan deflection of case 7 in Table 6-2.

P

P

a

1

lu"'Il l.._a

uwiu

I

G-1

a

L

A

.-

I

.

L

31

R2

152

RE Figure p-661

Figure P-662

0.

662. Determine its maximum deflection of the beam shown in Fig. P-662. Check your result by kiting a = L/2 and comparing with case 8 in Table 6-2. Also, use your result to check the answer to Prob. 653.

663. Detennnine the maximum deflection of the beam carrying a uniformly distributed load over the middle portion, as shown in Fig P-663. Check your answer by letting 2b = aL a d comparing with case 8 in Table 6-2. Ans. £16 = (wob/24)(L' 2Lb1 + be)

-

-¢»~l~2»~p~ HHH I

I

R1

I Ra

Figure P-663 *a

PROBLEMS

223

664- The middle half of the beam shown in Fig, P-664 had a moment ofinenia L5 times that of the rest of the beam. Find the midspan deflection. (Hint: Convcn the M diagram into an M/EI diagram.) Aus. 17 pa1/l851 P

*

I

I l5I I ¢+¢+¢+= I

IL

é,

Figure P-664 and P-665

665. Replace the concentrated load in Prob. 664 by a uniformly distributed load of intensity Wo acting over the middle half of the beam. Find the maximum deflection. 666.' Determine the value of E16 at the right end of the overhanging beam shown in Fig

P-666. 515 == (~J>"/24x4a + 3b)

Ans.

wo

I.

I

1

r

1

I

,Air

I

Figure P-666

I .al

a

I

b

'

667. Determine the value of E15 at the right end of the overhanging bum sbovwn in Fig. P-667. Is the deflection up or down?

100 lim 60 lb I l

AS Film P-667

»3 ftl

an .

l

15.

I

I

»aft*l

R2

668. For the beam shown in Fig. P~668, compute the value of Pthat will cause the tangent to the elastic curve over support $1 to be horizontal. What will then be the value

of E16 under the 100-lb load? Aus.

.p=561bx£!a=918lb 413

P

100 lb I

.4 few Figure P-668

RE

3 ft

6 ft F2

6/BEAM DEFLECTIONS

224

the bcam shown in Fig, 669. Compute the value of E16 midway between the supports of 9. P-66 s

120 lblft

120 lblfc

IH 6Hft H

Bft

i

HHIH RE

RE

Figure P-669

670. Determine the value of E16 at the left end of the overhanging beam shown in Fig.

P-670.

.~lfzs.

E16 = 428 N

- m~' down

900 N/m M=600N'm

,¢lll{llllll

4

m

I

\

3m

on

or.

R1

I

Figure P-670

6-7 MIDSPAN DEFLECTIONS In a symmetrically loaded simple beam, the tangent drawn to the elastic curve at midspan is horizontal and parallel to the unloaded beam. In such beams, the deviation at either support from the midspan tangent is equal to the midspan \

deflection. For simple beams that are unsymmetrically loaded, the midspan deflection can be found as easily as for a symmetrically loaded beam. All that needs to be done is to add a symmetrically placed load for each load actually acting on the beam. The effect of this transformation to symmetry is to double the actual midspan deflection. In other words, the actual midspan deflection is equal to one-half the midspan deflection of the transformed symmetrically loaded beam. Note that there is so little difference between midspan deflection and the actual maximum deflection that practically the two values may be considered equivalent.

ILLUSTRATIVE PROBLEMS 671. A simply supported beam of length L carries a load that varies uniformly from zero at the left end to We at the right end, as shown in Fig. 6-27a. Determine the midspan deflection.

Solution: Create symmetry. as shown in Fig. 6-27b, by adding a load that varies uniformly from zero at the right end to We at the left end. The result is a uniformly distributed load of intensity W0 over the entire span. The deviation at C' from the

ILLUSTRATIVE PROBLEMS

225

IHHI*b _` r "- " in&l{

i1 `*nH""'I

*-.-

L "" 2

""

A

L 2

7

R1

Hz .

Lu,

co' was. 2

l

2

I I

M

(1

I I J

4

(cm I

I

I

I

I

2

V-0

R=

L2

M

I-

I

(d)

L

I

'T s

I

I I *I41

I

1001'2

I

I I

I

I

aoL f

I

I

2

Wo

I

"'¢1»°=°

I

rof I

I

.g

Q

i

(H)

woL

Wo

"'°'TTr*"

o8

,L B

wal. 2

z

Figure 6-27 Transformation to symmetry. midspan tangent drawn at B is mud to 26 or twice the actual midspan deflection in Fig. 6-27a. Since we are considering the deviation of C from the midspan tangent drawn at B, we need the moment diagram of only half the beam. This M diagram may be drawn by parts from C to B as shown in Fig. 6~27c or from B to C as shown in Fig. 6-27d. To facilitate understanding of Fig. 6-27d, the free-body qiagmn of the right half of the beam is also shown adjacent to it. Note that the midspan moment M is found from the fact that its effect at the right end must be WW and opposite to that of the load in order to give a resultant bending moment of zero at the end of the beam. We apply the area-moment theorem to the M diagram in pan (d), leaving the student to verify that using part (c) yields the same result. We obtain

[5Uc/8 =

from which

51a

(WoLf 8

2

5 L4 768 We

_*Fel

x'

1. 11.

2

3

8

8 15

II

ZEISS =

(3l'C3)(.l8 •

5 384 WU

Aus.

572. Determine the midspan value of E16 caused by the loads shown in Fig. 6-28a.

Solution: The transformation to symmetry is shown in Fig. 6-28b. Since we need the M diagram for only half the beam, this is shown fOr the left half in Fig. 6-28c, together with the corresponding free-body diagram. Note that the midspan moment

.is found as described in the preceding problem. Also note that the reaction in the

transformed beam is equal to one half the symmetrically applied loads; that is, it is equivalent to the sum of the original loads.

6/BEAM DEFLECTIONS

226

s

an

60 lblft

l

I 6 f-.L2

:.

g,

¢»-L 4 ft--*

;2 f u u u

60 lblft l60 lb 4 ft

ljllll

60 lblfz

4 ft

I

2 ft

(b) 'Transformation to symannetry

`~~~ __l2°-.-/"

$A/8

f

(Q) Original loading

1

J

W

320 lb

320 lb

I I

1)

I

Sf:

960 lb- ft

`240 lb I V 80 lb 60 lb./ft

)

L

-480 lb.°ft

'I'

4 or

_

2 ft

___6_ fn

-480 lb- fr

Q

ii'£"w .4 fm.

I

I

2 ft

M= 960 Ib~ft

320 lb

(c)

Figun 6-28 Midspan deflection. I \

The deviation Lua of A from the horizontal tangent drawn at B is equal to » twice the midspan deflection of the original loading. We obtain

IE//A/I

= (3f¢3),4s°3EA]

ZEISS = ( 9 6 0 x 6§l ( ) - ( 2

480 X 4 3

4

4 )()-(

480 X 6 2

pa)

which reduces to

E16 = 6880 1b~ ft'

Ans. I

\

PROBLEMS 673. For the beam shown in Fig. P-673, show that the' midspan deflection is 6 (pb/4851 you - 4b').

P A

Ag.

b

a

C

8

x 1

1 1

.J

A r

IR,

Figure P-673

PHOBLEMS

227

674. Fund the deflection midway between the supports for the overhanging beam shown in Fig P-674. ,p

\

P

Figure P-674

L

4 4.

a

675. Repeat Prob. 674 for the overhanging beam shown in Fig. P-675. Are.

6 = ~,a2L2/321:/

wo 1

r

r

'I

L

a

Figure P-675

676. Determine the midspan def'lection'for the simply supported beam loaded by the couple shown in Fig. P-676. .4n5. a = 3,w1.2/64£/ I

M

Q

L

4

Figure P-676

677. Determine the midspan deflection for the beam loaded as shown in Fig. P-6?7_

wo '\

\llllIrw k

L cu

u-

I

-q

J

L

-'v

I

13,

Figure P~677

678. Determine the midspan value of E16 for the beam shown in Fig P-678. ARE.

Elf = 2810 N-rn' 600 N 900 N

3m l

I l lm

lm

a

Figure P-678

R1

152

6/BEAM DEFLECTIONS

228

679. Determine the midspan value of E16 for the beam shown in Fig. P-679 that carries a uniformly varying load over part of the span.

E16 = 3904 lb- R]

Ans. 60 lb/fx

.

Mlm .|.

Ji

Sf:

B.

»l-2

41%

r

I

fe

R1

r

R2

Figure P-679

680. Determine the midspan value of E15 for the beam loaded as shown in Fig. P-680. 600 N

M = 600 N~m'

I

lm

2m

m

JJ fol

R2

Figure P-680

681. Show that the midspan value of Era is (wob/48xL' 2Lb2 + be) for the beam in pan (a) of Fig. P-68l. Then use this result to find the midspan E16 of the loading in pan (b) by assuming the loading to extend over two separate intervals that start from midspan and adding the results. Ans. E16 '= . 9280 N-m3 »-

£ 2

'I'

b-*1 W0

1r

r -

1 r

_

r

t

1

I

w

1r

_

-

l A

I'

»2 m

gr

151

I

¢

r

1 r

I

I

1

r

1

r

1

r

1

i

L

u

-800_ N/m

'I

R1

R2 (a)

..l

3m

r

I

-1 L

4

.-.RE

(b)

Figure P-681

6-8 CONJUGATE-BEAM METHOD Successive differentiation of the deflection equation discloses the following relations:

Ely

El

g=

d

.

= deflection b

slope

d2y E1&- = moment

=M

6-8 CONJUGATE-BEAM METHOD

229

dS

v = L4 dx __ any _ do

5 / % = shear = x

do;y _ load $4

EI

do .

dx'

It is evident that the relations among deflection, slope, and moment are the same as those among moment, shear, and load. This suggests that the areamoment method can be used to determine bending moment from the load diagram, just as deflection was obtained from the moment diagram. For example, in the load diagram in Fig. 6-29, the bending moment at B should equal (area of load diagram),¢B 558- That it does is seen from

-

¢

1

\

('Won)§ x

Wax 2 2

Thus we could apply area-moment principles to compute bending moment, although this is impractical because better methods are available.

I

A

Figure 6-29

I

lllw.illi *

Nevertheless, the similarity of relations among load, shear, and moment, and among moment, slope, and deflection suggests that the relations among moment, slope, and deflection can be found by using the methods developed in Chapter 4 for computing shear and moment from load diagrams. We need merely assume that a beam is loaded, not with the actual loads, but with the M/EI diagram corresponding to these loads. Treating this M/EI diagram as a fictitious loading, we compute the shear and moment at any point caused by this loading These fictitious shears and moments correspond to the actual slopes and detlections in the beam at corresponding points. This technique is known as the conjugate-beam method and sometimes as the method of elastic weights. Applying the principles of shear and moment to a beam loaded with an MLEI diagram, we conclude that

1. The actual slope

= the fictitious shear

2. The actual deflection = the fictitious moment

(6-6) (6-7)

The method is especially useful for simply supported beams. For other beams, such as cantilevers or overhanging beams, artificial constraints must be applied, they are discussed later. To evaluate the conjugate-beam method, let us compare it and the areamoment method when applied to a simple beam. Only in simply supported

12

6/BEAM DEFLECTIONS

230

beams can the conjugate-beam method be applied directly without using artificial . constraints. Figure 6-30a shows a simply supported beam carrying a distributed load of intensity W0~ The moment diagram for this loading (drawn by parts in Fig. 6-30b) is multiplied by l/E! and used as the conjugate-beam loading shown simply supported on the span L in Fig. 6-30c. The reaction R, of this conj ugate beam is found by setting moments of the fictitious loads about B equal to zero. We obtain L2 L L W0 L2 • L _L (0) [ E M = 0] R.L 4 251 2 -

or

lm x)

()

The right-hand member of Eq. (al will be recognized as (I/EI)(area)8.4 - in. that is, /s/,4- Obviouslyjsolving for R, is equivalent to I8/.4/L. which is the actual slope at A; this fact is evident from the geometry of the elastic curve in Fig. 6302. Nevertheless, this is confirmation of rule 1 of the conjugate-beam method: the fictitious shear equals the actual slope at the corresponding point in the actual beam. w0L2

r

'2

L

I I

V

luinlu 4.-f L

Ill

Al~ ..,8¢»

'i

l

w0L2

2

(b) Moment diagram (by parts)

won _ L

--72L,-f" C/4

-

I;

.I

45

wa-w=-.u»-ai»¢

RJ s 2El 1

g

2

2EI

£a/A

w0L2 2El

v L)l**_L 4 3

L

g

R:

(c) Conjugate-beam loading

Figure 6-30 Connpadson of conjugate-beam and area-moment methods.

To obtain the deflection at any point on the actual beam, definition of bending moment to the conjugate.loads:

D¢Bemlony= (Z M)L = R,x - A I on

=RIX"IA1

x x "Ag 3 4

I r

apply the

5x + A, x Z (b)

231

6_8 CONJUGATE-BEAM METHOD

However, in terms of the moment diagram in Fig, 6-30b, [A1(x/3) - I4z(x/4)] equals ( I/El)(area)¢,4 °x`€- which equals 1c/.4 on the elastic curve in Fig. 6-30a. ' Hence Eq. (b) may be rewritten as (C)

1c'/4

.V = RI

0x = (I8/,4 IL)x, is equivalent to the following area-moment

which, since R , x relation:

x

I8/A

(J

(d)

I(*/4

This is the result previously obtained in An. 6-6 for deflections in simple beams by the area-moment method. Thus the conjugate-beam method, which uses the fictitious shears and moments of an M/EI loading to determine actual slopes and deflections, involves precisely the same computations ZS the area»moment method but has the disadvantage of obscuring the physical significances of the computations. This disadvantage is even more pronounced when the method is used with cantilever and overhanging beams, where certain artificial constraints must be applied. Nevertheless, the conjugate-beam method oilers an occasional advantage in certain routine work, in that it permits direct application of the definitions of shear and moment to the fictitious loading to find slope and deflection without any

need of an elastic curve. Now for a word about the need for artificial constraints in certain cases. For the cantilever beam in Fig. 6-3la, the M/El diagram appears as in Fig. 6-3 lb. This diagram cannot be applied directly as a fictitious load to a cantilever with the wall at the right end C because the fictitious shear and moment at B would be zero whereas the actual slope and deflection at B are not zero. Therefore the diagram of fictitious loads must be modified, as in Fig. 6-3 Ic, so that the fictitious loading will correspond to the actual slope and deflection at the free end. Wo l r

1 r

1

B'~

I

1

1

I

I

I

_ I

1

r

r

1

\

g

I

1 1

.n l

q

-

B

C

L

.

'|

Arel\=A

0-

I

/

(a) Actual loading and elastic curve

m)§m»uv=»~

nola 251

I

M

L

l

Ar~8l\=A

Iv

°. mol.: ask

(c) Conjugate-beun loading

Figure 6-31 method.

Constraints required to solve cantilever beams by the conjugate-beam

6/BEAM DEFLECTIONS

232

The reason for supplying aniticial constraints when solving cantilever problems should now be clear. To produce an actual zero slope at C in the original cantilever requires that the fictitious shear must be zero at C; therefore

[V

2

Ft°)L]

0= V

A

From this we see that the fictitious shear of the conjugate beam at B must equal the area A of the M/EI diagram. Also. to produce a zero fictitious moment at C. we must calculate the fictitious restraint M from [Mc = (Z Mn]

0=m+vL-A

L 4

_

*

Only after the artificial constraints M and V have been found can the fldtious shearand moment (corresponding to actual slope and deflection) be computed. Hence cantilever problems can be solved more simply and more dizealy by the area-moment rNethod.

PROBLEMS Phoblcms 653 to 665 inclusive and C3S8S 6 through 12 in Table 6-2 (page 233) may be assigned for solution by the conjugate-beam method.

I

3-9 DEFLECTIONS BY THE METHOD OF SUPERPOSITION in a supplementary method of determining slopes and deflections, the results of a few simple loadings are used to obtain those for more complicated loadings. This procedure, called the method of superposition, determines the slope or defleaion at any point in a beam as the resultant of the slopes and deflections at that point caused by each of the loads acting separately. The only restriction on this nnahod is that the effect produced by each load must be independent of that pnoduccd by the other loads, that is, each separate load must not cause an excessive change in the original shape or length of the beam. The tcditinique of supetposition is advantageous primarily for loadings that combine the types in Table 6-2. For partially distributed loads, the method of superposition requires integration (see Illustrative Problem 683). In such cases, the double-integration :method is preferable; if the deflection at only one specific location is desired, the area-mornent method is generally best.

ILLUSTRATIVE PROBLEMS 682. Using the method of superposition, compute the midspan value ofEI6 for the beam carrying two concentrated loads, shown in Hz. 6-»32a.

(continues)

§ § 8

I

5

Q

J;

*Q

-J

Ii?-9

I

I -e

'9

9

-o

mf

[M

.:

I

11

1

.7°~*

ml'8 n

EIE

I.

- I'S

'

Fo

HS I

I

an

'S

~:

/38

.am

g s2

E

-Q

I

(0

~;l~

v

1

- ,WS + 15101

.._I

v

v

o

u |-

a

,9

" r

.is

N

N

>
2

(0)

i I

a

ILLUSTRATIVE PROBLEMS

247 I

I

do = m,,x + V,,x2 Era;

l0(x- 3)3 +

2.

6"

(b) -0

(c)

To evaluate MA and VA, we note that at the other restrained end B where x = 12 ft, the slope and deflection are also zero. Hence substituting x = 12 ft in Eqs. (b) and (c), we obtain

'.

12nd,, + (12)214; 10(9)' 2

(d)

: M A + ( 1 2 )V,4' - " 10 (9)' =0

(f)

(12)

s

6

2

Solving these equations simultaneously yields I

Ans. `

VA = 190 lb and MA = -531 lb-R

The negative sign for MA indicates that the direction of M, was incorrectly assumed, as was stated earlier. The moment is actually negative, and there should now be no confusion about sign in its subsequent use. Having determined V, and MA, we apply the conditions of static equilibrium to. the free-body diagram in Fig. 7-5 to cornpute_the shear and moment at B. A vertical summation of forces gives

[ Z F,,=-0]

V8+ 190

Va = 350 lb

60(9) = 0

Ans.

We can now determine MY from the condition E Ma = 0, but it is simpler (and less confusing with respect to sign) to apply the definition of bending moment: [M = (2 Mk]

Ma = 12v,,

+ MA

= l2(l90)

a n

(60 X 9x§)

531

n

2430 = -681 lb.~ft

ARE.

As was shown in the preceding problem, this solution may also be adapted to a beam propped at B instead of fixed, by changing the boundary conditions at B to zero moment and zero deflection. Then by substituting x = 12 ft in Eqs. (a) and (c), we could determine MA and VA if this beam were propped at B. To apply the method of superposition, we replace the given loading repeated in Fig. 7-6a by the equivalent cantilever loadings in (b), (c), and (co). We now apply the conditions that 01 + 01 - 03 = 0 and 61 + 62 - 63 - 363 = ii. Using Table 7-1, we then obtain M,,(l2) +

[ZEI6=0] and

[2516 = 0 ]

(l2V,,)(12)

MA(l2)2 + (12v,,X12)2

2

_ »

2430(9)

3

2 u

p

2430(9)2

3

0

3 x 24

4

3?

9)

0

The solution of these equations yields, as before

V = 190 lb

and MA = -531 lb.~ilt

Check

What would the answers be if the beam had been propped at A instead of being . Gxed?

I

N

'7/RESTRAINED BEAMS

248

I 61 _L

" \

-

*

-u

1 2 ft

MA

M-MA' (b)

t

MA

C

3ft~

Set

\

lbfft

ZI.-}u°i°1l1!E'"' b )

12 f t 3

M-12VA

VA VA

/

VB

(C)

ogftbln

(a)

i

4I

r

9 ft 60 lb/ft

»

I7-1 r

l

pa 39j_f_

M

-2430 lb.~ft

(Dr Figure '7-6 Solution by superposition.

703. Two cantilever beams, having the same cross section and made of the same material, jointly support a distributed load of intensity We as shown in Fig. 7-7. Determine the force P at the roller between them by superposition. . i 1

4*

A 'I

4

L0

luulllluuu a

'I

_'in

.

I

C

Figure 7~7

Solution: The force P at the roller may be determined by the condition that at B both cantilevers have the same deflection. The deflection at B for the lower cantilever is found from case I of Table 6-2 (page 233) to be

6 _ Pa' 35]

The upper carilevcr is loaded with a combination of cases 3 and 2 of Table 6-2.,the resultant deflection at BB¢ing

6

wma 2

::

6L' 245/(

+

a

2

- 411)

Pa 3

3EI

Equating these deflections gives, for P

.

p = l6a Wo- ( 6L2+ a z _ 4La)

Ans.

249

PROBLEMS

PROBLEMS In solving the following problems, use superposition or double-integration as directed by your instructor. Unless otherwise stated, the supports are assumed to remain at the same level. Additional problems may be selectedfrom those in Arr. 7-4.

704. For the propped beam shown in Fig. P-704, fznd R and draw the shear and moment diagrams. Ans.

R = (w0/1'/8UX4L

- be

HlHW4'lHl_/ .I* /

Q

a

b

>-

L

r

a

R

Figure P-704

705. Compute the reaction R and sketch the shear and moment diagrams for the propped . beam shown in Fig. P-705. Ans. R -= sumo \

VI

F 1

HUi

I

L

Figure P-705

I

in

706. For the propped beam shown in Fig. P-706, determine the reaction R and sketch the shear and moment diagrams. w

Figure P-'706

707. A couple M is applied at the propped end of the beam shown in Fig. P301 Compute R at the propped end and also the wall restraining moment. .4n.s'. R = 3A{/2L M

as.

O

Figure P-70'1

L

7/RESTRAINED BEAMS

250

708. Two identical cantilever beams in contact at their ends support a distributed load over one of them as shown in Fig. P-708. Determine the restrammg moment at each wall. Ans. m, = -3w0L2/16, m, = -5w0,2/I6 wo

n

*l.

L Figure P-708

709. The beam in Fig. P-709 is supported at the left end by a spring that deflects l in. for each 300 lb. For the beam E = 1.5 X 106 psi and I = 144 in.'. Compute the deflection of the spring. Ans.

Q

1_1

12 f t

6 = 1.74 in.

200 lblfz I 1

I

I

F

n

P

1

I

r

I

I

P

1

»

|

r

300 lblin.

Fiazure P-709 710. Two timber beams are mounted at right angles and in contact with each other at their midpoints. The upper beam .4 is 2 in. wide by 4 in. deep and simply supported on an 8-ft span, the lower beam B is 3 in. wide by 8 in. deep and simply supported on a 10-ft span. At their cross-over point, they jointly support a load P = 2000 lb. Determine the contact force between the beams. 711. A cantilever beam BD rests on a simple beam AC as shown in Fig. P-71 l. Both beams are of the same material and are 3 in. wide by 8 in. deep. If they jointly carry a load P = 1400 lb, compute the maximum flexural stress developed in the beams. Ans. a = 1200 psi

_

P=1400 lb

8 fm

12 f t

8 fm

*`

4 ft

'c

k,

Figure P-711

712. There is a small initial clearance A between the left end of the beam shown in Fig. P-712 and the roller support. Determine the reaction at the roller support after the Uniformly distributed load is applied.

251

7_4 APPLICATION OF AREA-MOMENT METHOD

'I' L. uluullll

Lf

•

2

2

U

We

{


>- _*

L

a

A

L

>}-4

J

8

wolf

m,

l

m,

_

l __

30 ` =

II

WL '

'T WL

was," 'y'.

`§5"`

f*

L a

m, = m. -

B

b

m

M

. -

w°°L'

WL'

384

384

WL

b

w

Midspan Ely =

L' 768

.-

WL' 384

WL' 384

lo

L

-§

t - - - - - - 1

M

M =

A Y

As,

5w¢L2 7wgL SQL ='-» M ,EI = 96 48 ax J' 3840

=i"3(3:='_ ») L L

B

B

f

::

L' Midspan Ely = 768

96

2

.0

E

y

U

§GFE> A

7

PL* 55

5

-- WL 96

S

f

f

. A

lr

We 6

_ WL . Max. EI 12

\

Wo

I

-_ ._

12

5 A = ' ' W 0 L 2 = k

B

L 2

4

1

»

2

§

Max. Ely =

»

us u

A

\

T

W0L2

m,=m,=-

We

5

PL "°

L

1

4

-

M,

B

. 4

3

a

_

- 6EIA LE 651A L2

4

.. Val3 v-»

1920

r

NRESTRAINED BEAMS

264 2

Q

Plaza L2

p, L 8

Mu.M=-°L

12

I

Max. M = -8040 N m The negative sign of moment indicates a tensile stress at the top Fibers. In an unsymmetrical section like a T beam, this would be important, but in a symmetrical section, only the numerical value of moment need be used. From the flexure formula, . the section modulus required is therefore

8040 1

80 X 10° n

_ 100.5 X 10" m3

100.5 X 103 mm'

Ans.

747. Select a suitable wide-flange beam to support the loads shown in Fig. 7-15 without exceeding a flexural stress of 18 000 psi. Compute the midspan de flection of this beam. E = 29 x 10° psi.

P1 =60u0 lb

P2 =9000 lb I

I

3 ft

L

I

7

5 ft

12 ft

Figure 7-15

A

Solution: Because of the unsymmetrical loading, the end at which the maximum moment occurs is not evident, and computations for moments at each end must therefore be made. At t'he left end, we obtain

M/.

Pgdh2 L2

_

P2Gb2

L2

6000(3)(9)2 (12)2

_.

9000(8)(4)2 = - 1 8 l20 lb.-ft (12)2

I

I

I

At the right end, the moment is

i I

MR

P102b L2

nu-

P202/> L2

,

6000(3)2(9)

9000(8)2(4)

(i2)1°

(I2)2

=~l93401b-ft

Substituting the larger numerical value ofend moment in the flexure formula. we find that the required section modulus for live loads is

A suitable beam is W8 X 18 with S

15.2 in.3 and 1 ::. 61.9 i i

Table 7-2

f

PROBLEMS

265

gives the midspan deflection,* including the dead weight of the beam, as

Pa* w0LE Ely= E - - ( 3 L - 4 b ) + 48 384 Ely

6000(3)2 [3(I2) 48

Ely = 27 000

-nu

+ 60000

4(3)l +

+ 918

r:

* I

9000(4>=

48

1302)

.u.

4(4)I

+

I7(I2)' 384

87 920 1b-a'

Substituting numerical values For E and I. we then convey the right-hand side to inches by multiplying by 1728: (29 X 10°X61.9)y = (87 920Xl728)

from which y = 0.0846 in.

A/1.5.

PROBLEMS 748. A restrained beam 6 m long supports a concentrated load of 30 kN at 2 m from the left end and another concentrated load of 50 kN at 1.5 rn from the right end. Select the lightest wide-flange beam that will support these loads without exceeding a flexural stress of 120 MPa. Neglect the mass of the beam. Compute the midspan

deflection of this beam if E = 200 GPa.

a

Ans. W360 X 33, 6 = 3.2! mm 749. A timber beam 6 in. wide by 12 in. deep and 20 ft long is perfectly restrained at both ends. It supports a uniformly distributed load of 240 lb/ft over its entire length, and a concentrated load P at 8 ft from the left end. Determine P so as not to exceed a flexural stress of 1500 psi or a midspan deflection of 1/360 of the span. Assume that E = 1.5 X 10° psi. ' 750. An S200 X 34 steel beam 6 m long is perfectly restrained at both ends. It cames a concentrated load of 30 kN at 2 in from the left end and another concentrated load of 45 kN at 1.0 m from die right end. Compute the maximum flexural st2ressa1nd the midspan deflection. Neglect the mass of the beam. Use E = 200 GP; .4 ns. 6 = 168 MPa 751 A timber beam with a square cross section supports the loads shown in Fig, P-75 L Determine the cross-sectional dimensions if the allowable flexural stress is 1-450 psi. What is the maximum shearing stress developed in the beam?

.

I n

I

450 lb

mu 1 q

>

in

I

4.

1 Sf:

Figure P-751

'

200 lb if

.

1

.

for a concentrated load the term b ` th In computing all of th .. . . midspan deflection IS two segments MIO which the load dwtdes the length of the beam, C sm Cr c

»

7/RESTRAINED BEAMS

266

752. Using Table 7-2, check the values of end moment and midspan deflection for the restrained beam in Prob. 7 I 3 (page 251). 753. An S130 X 22 steel beam 4 m long carries a load varying uniformly from zero at the left end to 15 iN/rn at the right end. The beam is perfectly restrained against rotation at its ends, but the right end settles 10 mm relative to the left end. Determine the ratio of the maximum flexural stress to the flexural stress if no settlement had occurred. Use E = 200 GPa. Ans. 1.06

SUMMARY The principles of beam deflections studied in Chapter 6 are applied here to obtain additional equations that can be combined with the equations of static equilibrium to solve problems involving statically indeterminate beams. In propped beams, we generally use the fact that the deflection under the redundant support is zero (if the support does not settle) or some known value (if the support does settle). For beams perfectly restrained at the ends, the elastic curve is such that there is no change in slope between the ends, and the deflection of one end relative to the other end is zero. The method of superposition is usually the easiest way to determine the redundant support in propped beams. lt is also the best method of determining the end moments in restrained beams subjected to loadings of the types listed in Table 7-2. With the double-integration method or the area-moment method, we generally take the shear and moment at one end as the redundant supports. Either method is as simple to apply as the other. The double-integration method is essentially mathematical and automatically determines deflections as well as redundant supports. The area-moment method, by emphasizing the geometric relations between the elastic curve and the moment diagram, is perhaps more direct in obtaining the equations that determine the redundant supports, but additional work is needed if deflections are required. In Art. 7-5, the reduction of a restrained beam to the combination of a simply supported beam carrying the given loading and another simply supported beam subjected to end couples is valuable in visualizing which restrained end carries the larger moment. This concept is particularly advantageous in applying the area-moment method to continuous beams, as will be explained in the next chapter. In addition, it provides the basis for a rapid method of drawing the shear diagram, also explained in the next chapter.

i*

1

n

1

1

Chapter 8

Continuous Beams

8-1 INTRODUCTION

s I

In this chapter we consider beams that are continuous over two or more spans, thereby having one or more redundant supports. It is possible to determine these redundancies by applying the deflection relations developed in Chapter 6, but 3 more convenient method is to consider the unknown bending moments at the supports of the beam as the redundancies. After these bending moments are found, it is comparatively simple to determine the reactions, as we show i n Art. 8-5. We present two methods of solving for the moments at the supports, either or both of which may be studied, depending on the available time. in the first method it is necessary to find a general relation between the bending moments at a n y three sections in a beam. This relation is known as the three-moment equation and is easily derived by applying the area-moment theorems. We shall show how this equation is used to determine deflections as well as redundancies in any type of beam. Actually, the three-moment equation can be used to solve all the problems in Chapters 6 and 7, however, in some instances, it is best used in combination with the area-moment or the double-integration method. Such combinations of techniques will also be discussed. An alternate method of solving continuous beams is moment distribution. which is described in Art. 8-8. This method is completely independent of the three-moment equation, but Art. 8-5 for determining shear diagrams is common to both. Before the moment-distribution method may be applied, however, each span of the continuous beam must be assumed to be perfectly restrained at the supports and the fixed end moments must be computed. Usually the span loadings are such that the flxed end moments are readily obtained by superposition of the general results tabulated in Table 7-2 on page 263. However, this table does not list results for distributed loads over part of a span, for these loadings the three-moment equation is preferable if a more general list is not available.

I 8-2 GENERALIZED FORM OF THE THREE-MOMENT EQUATION

A portion of a beam that is loaded and supported in an arbitrary manner is shown in Fig. 8-la. At any three points 1, 2, and 3, pass cutting sections and replace ,the effects of the loads to the left or right of these sections by the proper 267

268

8/CONTINUOUS BEAMS

I

I

1

I 1

I

I

I I

Arbitrary l loading | I

. 2

2

L,

:iv

.

uN

l

l

ii

3

L,

1

1

|.

(a) Loading diagram -l"

m,

Load on span 1 I I

VI

Load on span 2

I

J

I

H,

¢v

'I

A

i

v,

m.

i.

1

y..

(b) Free»body di agrams of beam aegrnents

Figure 8-1 General loading on any beam. values of vertical shear and bending moment. Thus the beam segre is between points 1 and 2 and between points 2 and 3 (hereafter referred to as spans l and 2, respectively) may be isolated by means of the free-body diagrams in Fig. 8-lb. The lengths of the spans (or segments) are L, and L2, and the bending moments at points l , 2, and 3 are M, , My, M3 , the vertical shears at these points are V,, V_2 (just to the left of point 2), V2 (just to the right of" point 2), and V., just to the left of point 3. _ . The technique discussed in Art. 7-5 enables us to resolve the free-body diagrams of the beam segments into simply supported spans that carry the actual beam loading, and spans loaded only by the bending moments and held in equilibrium by the -couple reactions R', on span l and by RQ on span 2. This equivalent loading is shown in Figs. 8-2a and 8-2b, respectively. When these loadings 3II€ superposed, they produce the free-body diagrams in Fig. 8-1b. Hence the vertical shears at points l , 2, and 3 are equal to the algebraic sum of the simple beam reaction and couple reaction at these respective points. In this manner, the moment diagram of each original beam segment is resolved into the moment diagram of the loads assumed to be carried on a simply supported span and the trapezoidal moment diagram caused by the bend ing moments in the original beam at the selected points l, 2, and 3. These diagrams are shown in Figs. 8-2c and 8-2d, respectively. For clarity, the elastic curve of the beam has beendrawn separately in Fig.

8-3. The deflection of the curve is greatly exaggerated in order to show the geometric relations, Note that points l, 2, and 3 lie on it. A tangent drawn to the elastic curve at point 2 determines the tangential deviations I1/2 at point l and I3/2 at point 3. Another line draw n throu gh point 2 parallel to the initial position of the unloaded beam (which has been assumed horizontal for convenience) determines the heights of points l and 3 above point 2 t o be h, and I13. There are formed the shaded similar triangles having the bases Li and L2 and the altitudes (hi " f an) and (13/2 - h3).

I

. I

I

s

|

3; 's

D

I I

I

269

8-2 GENERALIZEO FORM OF THE THREE-MOMENT EOUATION

(Q) md; on simply supported spam

Load

1

LE

L

(usu me M,

(assume M1 > M2) (b) Lo¢diD8 by end , moments and be-ncima maple

l

L.

r I

J

Re 1

n

R2

Figure P-802 802. See Fig. P-802. When b = L/2. how does your result compare with case 2 of Table 8-1° 803. See Fig. P-803.

Ans.

WL3

We

H |

I

Will A

k

L

L

I

I

2

Figure P-803

H1

804. See Fig. P-804. Check your result by subtracting the answer for Prob. 803 from 'case 2 otlTable 8-1. _.We

wo

/

MU7w.nllfH

I

1L

L 1

I

2

2

R2,

RE

Figure P-804 I

805. See Fig. P-805. The roller support may resist upward or downward reaction. U); 6,45/L = (Af/Lx3b2 Ans, 6Aa"/L = -(m/L X30*

-

a

a

*

b

I

*

L

Figure P~805

UI

275

ILLUSTRATNE PROBLEMS

806. Sec Fig. P-806.

Ans. 6,45/L = 1414 n- m2; 6,45/L = 987 N m'~

¢.J

r R

1

HUH 2m

2 m.

I

R2

R1

Figure P-806

400 N/m

807. See Fig. P-807. Solve by combining the results for Probs. 805 and 806. 490 Ba am

M=200N°m

mllllM

I

JJ

lM

Figure P-807 andP~808

4°

lm

)

.

2m

|»

I

R1

R2

808. .Solve Prob. 807 if the couple is applied in a counterclockwise sense. 809. See Fig. p-809.

'

Ans. 6A5/L

l

Figure P-809 'and P-810 I

1

1

;:

MU 6 ft

= 10.26 kip - 52 100 ism M : 240 Et~ft .

35; -4

r.;ac 3

810. Solve Prob. 809 if the couple' is.applied in a counterclockwise sense.

i

F F

i E

8-4 APPLICATION -OF THE THREE-MOMENT EQUATION

I !

We now see how the.three-moment equation may be applied to determine the moments over the supports in various types of continuous beams. Later articles

will show how these moments are used to determine the reactions of continuous beams. and will 'describe a speedy -method of drawing shear and moment . diagrams. •

ILLUSTRATIVE PRQBIIEMS 8 l L For the continuous beam in Fig. 8-6. determine the values of the moments over the supports. The supports are assumed to be rigid or. what amounts to the same thing. to have equal deformations. This assumption applies to all problems unless stated otherwise.

276

8/CONTINUOUS BEAMS I

60 lb

90 lb I

60 lb./fc

I

4 _ft

Y

~--1 f i

= 10 ft

!

I

llllMl

-Span 1

Is J

30 l b / f t

I (To H l'l"1

4 ft

J

u

R1

Span 2 = 12 f t -

*

R2

Ra

Figure 8-6 Solution:

l

Apply the three-moment equation to points over the supports. Since the

supports remain at the same level, heights h, and /13 are -zero, and the equation reduces IO

M1 L1 + 2!vI2(LI + Le) + MJL2 +

6,4,5, + 6/426,

L1

Le

0

I

(H)

We begin by noting that the bending moment at support 1, caused by the

-

load to the left of R, , is M, = -60(4) = -240 lb ft, whereas over support $3 the moment M3 is zero since no loads act to the right of R3. Observe that the negative sign of M, must be retained when substituting its value in the three-moment equation* The factors in Eq. (a) are found by using Table 8-1. Th.e load on span I is that of case 3, hence we have ' 6A5,

8 '66 we L3

L,

8

60

-

(60)(10)3 = 8000 lb fr?

(b)

i

|

For span 2, the factor is found by adding the results listed for cases 1 and 5, hence we obtain . 6A52 L2

W0d2

pa (L2 - IN2) +

AL

L

I

i

(2L2 -- ¢/2)

90(4) 30(8)2 (144-16)+ (2 12 4(12)

x

i

144 - 6 4 ) I

3840 + 8960 = 12 800 lb~f12

(c) I

* A common error involves applying the three-moment equation

l

between the overhang and span I as shown in Fig. 8-7, thus completely forgetting that the 60-lb load causes the overhang to deflect downward an unknown distance h. Under these conditions, the right-hand part of the general three mornent equation .contains the unknown h and is not zero. However, after M, and M! have been found, the general three-moment equation may be applied between spans 0 and I to solve for the detection h. This procedure will be discussed in Illustrative Problem 858 on page 293.

1

I

60 lb I

1'

/111

h

I/'

l

I1-span

0-»l.R

-~ Y I I 1

4

1L -

-

A

P

I

P

n

i.

R2

(b) Axial compression

(a) Axial tension

Figure 9-2

ILLUSTRATIVE PROBLEM l

901. A cantilever beam G»-uv-nitg. 9-3) ha the profile shown so that it will provide sufficient clearances for large pulleys mounted on the line shaft it supports. The reaction of the line shaft is a load P = 25 kn. Determine the resultant normal stresses at A and B at the wad).

?~ ¢/

F

150

L

7r é !

7/

an

I

AL50

I(.

450 mm. •

f

//

4/ 8

150 mm

I

I

p,=20 k N

a PJ/215 kN

Figure 9-3

P= 25 kN

I l

I

311

ILLUSTRATIVE PROBLEM

Solution: We begin by determining the bending moment of P. This is computed in terms of its components P, = 20 kN and P, := 15 kN by taking moments about the centroidal axis of section AB: III..

( 2 McM]

x 10'x0.4501 + (20 x

-(15

II

[M

103X0.150)

--3750 N~ m

=,

Since P, acts down, its moment effect is negative (Art. 4-2) and the opposite moment effect of P, must be positive. The negative sign of the bending moment at AB indicates that the beam curvature at section A-B is concave downward (An. 4~2), thereby causing tension at A and compression at B. Having thus interpreted the sign of the bending moment, we use only its absolute value in applying Eq_ (9-1). However, it may not as yet be obvious that the axial tensile effect is due solely to Pa. Use the Principles of mechanics to convert the given load into either of the equivalent loadings in. Fig. 9-4. It is evident from the principle of transmissibility that the entire moment effect is due to P, in Fla 9-4b and therefore the axial effect is caused by P, alone. Or we may add a pair of collinear forces each equal to P, as 111 Fig. 9-4c, thereby reducing the system to that shown in Fig. 9-4d. Once again we see that the axial effect is caused by P,, since the bending moment., which consists of -0.450P, plus the couple 0.l50P,, is equivalent to the bending moment as computed previously.

'

7

/

/.

I

C I

5.

\ ¢

\

/ / / / /

II

/ /

9/ I

I

/ / /

1;

\

/

P

pa)

(b)

/

/~

"Ye

/ // . / / // // / /

1;

I150 mmT- _L

/

1

450

)

p,

in in

/ / / / /

m= (0150 PJN-m

2

)

al

Pi I

21 -450 mm

\r

'Py (ii)

( )

>

or

*8 Figure 9-4

We are now ready to compute the resultant stresses by applying Eq. (9-1). At A we obtain U

074

P A

+

Mc

6M

T

he'

)]

0(3150) 20 x I0' + (0.050x0.I sol ($.050x0. I50)' (2.67 X 10°)

+ (20.00 X 10°) =:

22.67 MPa

Ans. I

'I Iu.

9/COMBlNED STRESSES

312 At B, where the flexural stress is compressive, we have

M

P

0

=

0'

8

6M biz

I

.4

_

20 X 103 (0.50X0.l50)

= (2.67 X 10°)

)] 6(3750) (0_050)(0_I50)2

(20.00

x

106) = -17.33 MPa

A115.

The signs indicate that the stress is tensile at A and compressive at 8.

PROBLEMS 902. Compare the maximum stress in a bent rod I in. square, where the load P is 1 in. off center as shown in Fig. P-902, with the maximum stress if the rod were straight and the load applied axially. This problem illustrates why lateral deflection in columns is so dangerous. Ans. 7 to I

P --.> \-;~

L

J

* 0

-

ll

v

Q

.in.

l

1

P Figure P-902

903. A cast iron link is 40 mm wide by 200 mm high by 500 mm long. The allowable stresses are 40 MPa in tension and 80 MPa in compression. Compute the largest compressive load that can be applied to the ends of the link along a longitudinal axis that is located 150 mm above the bottom of the link. 904. To avoid interference, a link in a machine is designed so that its cross-sectional area is reduced one half at section A-B as shown in Fig. P-904. If the thickness of the link is 50 mm, compute the maximum force P that can be applied if the maximumnormal stress on section A-B is limited to 80 MPa. Ans.

I

40 kN

A

1

*u

I

A

"P

40ifnm

P

e

1

>

l

I

I

B

1.

40 mm Y

; -|

.i

1

Figure P-904

I

1

I I

905. A wooden . beam 100 mm by 200 mm, supported as shown in Fig. P-905, carries a load P. What is the largest safe value of P if the maximum stress is not to exceed

10 MPa?

iI

I

313

PROBLEMS

p 40°

100 mm

it-ten

r

4;

L

1200 mm

i) I

n

4

u

-

I

Q

1 .L

Ill

I II

Figure P-905

906. For the 2-in. by 6-in. wooden beam shown in Fig. P-906; determine the normal stresses at A and 8. Are these the points of maximum normal stress IF not, where are they located and what are their values" Ans. OA = -921 psi, Us = 599 psi, max. i f . = 852 psi and - I 174 psi

P-=2000 1b

ft

IA

..|

2 in, 7

»

15°

•

I

1

|

1 Q

¢

;

6 in.

l

l

/

l I

q

I-I I

ALl.

' I

.l f £ '

B

, ~ 1

3 ft

I >-i Figure P306

4

I I

907. Determine the largest load P that can be supported by the circular steer bracket shown in Fig. P-907 if the normal stress on section A-B is limited to 80 MPa.

1

I

\

I

I I

R

250 mm A

.\\' Section A -B I

Figure P-967

I

die. = 100 mm

314

9/COMBINED STRESSES.

908. A punch press has the cast-steel frame shown in Fig. P-908. Determine the greatest force _P that_can be exerted at the jaws of the punch without exceeding a stress of 18 lest at section A-8. The properties of the area are as shown and 1-1 is the centroidal axis. -

•

120 in. I#v//n

r

Q

D

B 12

I

in

Section A -B I/.//7//./ I / / / / I / / / / / / ' / / / / / / / ' / / / / / / I / /

I1-1= 4000 in." Area = 120 in?

Figure P-908 h

•

909. The bent steel bar shown in Fig. P-909 is 200 mm square. Determine the normal stresses at A and B. Ans. O34 = 6.03 MPa,o'8 = -19.5 MPa

450 kN J

6*

x

0

I

'L _

/

K \ 4 .

4

\ \ \ \ \ l

R

uucp nun

Figure P-909

y

.

400 mm

-

-4----

100 mm

910. A timber beam AD, 6 in. thick by 10 in. high and loaded as shown in Fig. P~9l0, is pinned at its lower end and supported by a horizontal cable CE. Compute the maximum compressive stress developed in the beam. Ans.

0

=

1603 psi

315

PROBLEMS

3000 lb

v

I

6 ft .e0001b 4 fc

4

Figure P-910

of concrete is 911. A concrete dam has the profile shown in Fig. P-911. If the density sive the maximum compres 2400 kg/rn' and that ofwa.ter is 1000 kg/m3, determine dam ish = 15 m. the behind stress on section m-n if the depth of the water

"Ir

r

m

=--;'1'-

1

25 m

h

mf. I//////

!

.9

n

s

A

1

I / / / / / / 7 / / / , / .I / / / ' / ' »

s m *I

Figure P»~9l I

in Fig, P-912 if ,P 912. Compute the stresses at A and B on the link loaded as shown . lb. 3000 = F and 9000 lb Ans.

of,

= 2250 psi; Us = -?50 psi

P

n.

L

6 in.* *

T l

IA

in.

4

I .4.e.. 4

I I

• I

l b

Lu.

I

12 in.

I

2 in. y/

/ / .K

"1 I

",/x*

i

Gin

'A /A

f

i

n,

8

Cross~ section

1*

FE

Figure P-912 and P-9 13 913. Solve Prob. 912 i f p = 6 kips and F = 10 kips. i

316

9/COMBINED STRESSES

914. The structure shown in Fig. P-9 l 4 is hinged to 6.xed supports at A and C`. Assume that the pin connections at ,4. 8. and C are frictionless. The bars are each 4 in. by 4 in. in section. Compute the maximum compressive stress developed in bar AD. Ans.

o('

= V200 psi

r .|

I

I

S001b

500 lb

5 fn-

- 5 ft

T"§=\ .riff

-6 f t -

2 ft

I

so;

I

1 /

x

A

Figure P-914

915. For the structure in Prob. 914, compute the maximum tensile stress developed in bar CB. r \

916. The structure shown in Fig. P-9 16 is hinged to fixed supports at .4 and E. Compute the maximum compressive stress developed in bar BDE if its cross section is 200 mm square. Neglect the weights of the members. °

»20 kN

1 I I

i

1

B

~-P

A -

I-`

`\

D

60

\ ' I k1.A sin 8) cos 6*

[2F,=0]

+ (1rx,..4 cos H) cos al - (r,.,.4 sin 0) sin 0 Since the common term A can be canceled' and since r equal to r . , we use the relations

cos! 6 = I

+

2os 26

9

smog 6 = •

l

cos 26

2

.I 4

A

1.

(b)

is numerically

sin 6 cos 6

1

s

5811120

9/COMBINED STRESSES

324

'! r t

O

n

.

4

J

0

/

r

o

I

T mtg

Xl°.___.

kw

x

a ox

Ox

i

Tas' I

XI

'y

0y

(b) Stresses acting on wedge

{a) Original state of stress

Area of wedge face taken as A A / - Acos 8 to),A CO8

T_

Ao

A

O08

n

6

A:

e

6

\

UxA

.

I

fi*

(9-5a)

(9-6a)

Adding Eqs. (9-5) and (9-Sa) shows that the sum of the normal stresses on any + a . Also, comparison of

two perpendicular planes is a constant equal to fm

s

9/COMBINED STRESSES

326

Eqs. (9-6) and (9-6a) confirms the equivalence of shearing stress on perpendicular planes. •

9-7 MOHR'S CIRCLE

I 4

The formulas developed in the preceding article may be used for any case of plane stress. A visual interpretation of them, devised by the German engineer Otto Mohr in 1882, eliminates the necessity for remembering them.* In this interpretation a circle is used, accordingly, the construction is called Mohr's circle. If this construction is plotted to scale, the results can be obtained graphically, usually, however, only a rough sketch is drawn, analytical results being obtained from it by following the rules given later. We can easily show that Eqs. (9-5) and (9-6) define a circle by first rewriting them a.s follows: g -

ax

+ o", =

Ux

2

=

t

07x

2 08' cos 26 -

Try

sin 26

(of)

2 0'sin 26 +

Tx)-

cos 26

(b)

I I

I

I

1.

By squaring both these equations, adding the results, and simplifying, we obtain I

2

ax + o b i 2 + 72 =

l

+ (1'x,v)2

(c)

I

r !

Recall that ax, of, and Rx, are known constants defining the specified state of stress, whereas o and 1 are variables. Consequently, (u, + cr.v)/2 is a constant, say, C, and the right-hand member of Eq. (c) is another constant, say, R. Using these substitutions, we transform Eq. (c) into (U-€l)2+-r2=R2

I

N

(d)

This, being of the form (x - c)2 + y2 = R2, is readily recognized as a circle of radius 2

(

II

Bo

/

2

)

+ (1fxy)2

* Equations (9-5) and (9-6), as well as the succeeding variations of them, are identical to the equations that empress the variations in moments of inertia with respect to u and v axes inclined at an angle 6 to the reference axes x and y. Replacing normal stress by the moment of inertia I and the shearing stress by the product of inertia P, we obtain 1. and

p.,

v

1,+I, 2 +

2

/7 $ ,

1,

1N

-2 1, cos 26

20 +Pa y

.co

I

I l

a I

! !

5

I.

P sin 26 I. 1

s 28

A Mohr's circle treatment of these equations is described in detail in Appendix A, page 546.

i u

I

I

I

327

9-7 MOHH'S CIRCLE

whose center is offset rightward a distance

C=

0x

+ 'by 2

' . from the origin. Figure 9-14 represents Mohr's circle for the state of plane mess that was analyzed in the preceding article. The,center C is the average of the normal stresses, and the radius

2

Y

II

Q:

+ up'

2

is the hypotenuse of the right triangle CDA. How do the coordinates of points E, F, and G compare with the expressions derived in Eqs. (9-9), (9-10), and (9-lOa)'? We shall see that Mohr's circle is a graphic visualization of the stress variation given by Eqs. (9-5) and (9-6). The following rules summarize the con-struction of Mohr's circle.

T

ya

I

Ox

I

I

0x+07

Gy

751

I

'u

2

I

G \

I

I

Try 1

,r

I

x xy

OI

E

-l O

'S Figure 9-14 Mohr's circle for general state of plane stress.

Rules for Applying Mohr's Circle to Combined Stresses l

On rectangular 0r-r axes, plot points having the coordinates (ax, r,,,) and (try, ry). These points represent the normal and shearing stresses acting .on the x and y faces of an element for Which the stresses are known. In plotting these points, assume tension as plus, compression

E

328

9/COMBINED STRESSES

as minus. and shearing stress as plus when its moment about the center . of the element is clockwise.* 2. .loin the points just plotted by a straight line. This line is the diameter of a circle whose center is on the o axis. 3. As different planes are passed through the selected point in a stressed body. the normal and shearing stress components on these planes are represented by the coordinates of points whose position shi fts around

the circumference of Mohr's circle. 4. The radius of the circle to 'any point on its circumference represents the axis directed normal to the plane whose stress components are given by the coordinates of that point. 5. The angle between the radii to selected points on Mohr's circle is twice the angle between the normals to the actual planes represented by these points. or to twice the space angularity between the planes so represented. The rotational sense of this angle corresponds to the rotational sense of the actual angle between the normals to the planes, that is, if the n axis is actually at a counterclockwise angle 0 from the .x axis, then on Mohr's circle the n radius is laid ofilat a counterclockwise angle 26 from the x radius.

lLLUSTRATliVE PROBLEMS 923. At a certain point in a stressed body. the principal stresses are 0, = 80 MPa and Oj' = -40 MPa. Determine o and -r on the planes whose normals are at +30° and +120° with the .X axis. Show your results on a sketch of a differential element. Solution: The given state of stress is shown in Fig. 9-15a. Following the rules given previously, draw a set of rectangular axes and label them o and T as shown in Fig. 9-l5b. (Note that, for convenience. the stresses are plotted in units of MPa.) Since the normal stress component on the x face is 80 MPa and the shear stress on that face is zero, these components are represented by point A which has the coordinates (80, 0). Similarly, the stress components on the y face are represented by point B

e

(-40, 0)-

According to rule 2, the diameter of Mohr's circle is AB. Its center C, lying midway between A and B, is 20 MPa from the origin O. The radius of the circle is the distance CA = 80 20 = 60 MPa. From rule 4, the radius CA represents the x axis. In accordance with rules 4 and 5, point D represents the stress components on the face whose normal is inclined at +30° to the x axis, and point E represents the stress components on the perpendicular face. Observe that positive angles on the circle are plotted in a counterclockwise direction from the x axis and are double the angles between actual planes.

-

'

This special rule of sign for shearing stress makes Rx, = -r,, in Mohr's circle. From here on, we use this rule to designate positive shearing stress. However, the mathematical theory of elasticity uses the convention that shearing stress is positive when directed in the positive coordinate direction on a positive face of an element, that is, when acting upward on the right face or rightward on the upper face. This other rule makes °rxy = f , which is convenient for mathematical work but confusing when applied to Mohr's circle.

I

/

10

-3`

#1 -

329

ILLUSTFIATIVE pRO8LEm5

1(MPa)

1

O=50 y 1

I

"oV:-40 MPa

! ..4_....

x

L_

Ox

I

I l

1

20

B

I

4

EI

-=52

26`= 609

1

0 x = 80 MPa

A--ofMPa)

F

1

I

TI

.»'

oy

I

(a)

»


80

Tb)

Figure 9-15 From rule 3, the coordinates of point D represent the required stress components on the 30° face. From the geometry of Mohr's circle, these values are U

O F = O C + C`F= 20 + 60 cos 60° = 50 MPa

T

DF

=.

60 sin 60° = 52.0 MPa

On the perpendicular 120° face we have 0'=OG=OC-CG=20-60cos60° TI

= GE = -60 sin 60°

= -10 MPa

-52.0 MPa

=.

Both sets of these stress components are shown on the differential element in Fig. 9'-16. Observe the clockwise 8nd counterclockwise moments of a* and f', respect rely, relative to the center of the element (see rule I). Finally, note that a complete sketch of a differential element shows the stress components acting on all four faces of the element and the angle at which the element is inclined.

A

,4-

Qs

5 300

.

PROBLEMS 942. Using Fig. 9-l9a, indicate the planes on which each of the three shearing. stresses . . given in Eq. (9-I I) act; 943. For a state of plane stress with a, = 5000 psi and 02 = 2000 psi, determine pa) the maximum in-plane shearing stress and (b) theabsolute maximum shearing stress. Ans. (b)2500 psi .

944. Repeat Prob. 943 if o, = -40 MPa and 02 = -80 MPa 945. Find the absolute maximurri shearing stress for the state of stress shown in Fig. . . P-982 (page 333). 946-949. For each state of stress shown in Figs. P-945 through P-949, compute (a) the maximum in-plane shearing stress, and (b) the absolute maximum sheanlng stress.

al

i

y

I

t2

I

30 MPa

15 MPa

|-

L...

ii

1200 psi

1200 psi

- ---» x 3600 psi

L-._ l

i

an

x

80 MPa .

I

Figure P-946

Figure P»947

9/COMBINED STRESSES

340

"I

i"

30 MPa

inI

i

41°

MPa

q

ksi

I

I

L_

L..

-->-

I --x 75 MPa

-i!- - - x 10 ksi

- _

i

1

1

Figure P-949

Figure P-948

9-9 APPLICATIONS OF MOHR'S CIRCLE TO COMBINED LOADINGS

The most irnponant use of combined stresses ,is in the design of members sub jected to combined loadings, or the determination of safe loads. Here Mohr's circle makes possible a visualization of conditions that is superior to mere analytical manipulation. The usual procedure is to consider an element on which the effect of the three fundamental loadings-axial, torsional, and flexural--can be computed. A study of Mohr's circle for this element indicates the design criteria. The illustrative problems at the end of this article are typical of the procedures involved.

l

I

I

I.

I

Stress Trajectories

An element on the surface of the cylinder in Fig. 9-23a is subjected to the indicated torsional shearing stress. Fig. 9-23b shows Mohr's circle for this state of stress. The radius 0.4 specifies the X axis. The maximum tensile stress is denoted by point D, whose radius OD is 90° clockwise from OA. Hence the normal to the

J

4

x

x

0

* |

45°

P'

T '

/

Torsional shearing stress

(a)

'

Equivalent principal stresses

Figure 9~23 Cylinder subjected to torsion.

I

B .

To

(=)

0

(b)

Figure *9-28 The following variations of the flexure and torsion formulas as applied to a circular shaft are used: .

_ 4M

U/'

and

7I'f'3

2T 1

Substituting these values in Eq. (a) yield; 2

(b)

*

2

1

(5) +(2j;)

Max. 1 which reduces to

I

I

Max.

T

=

2 3

-of

Letting

or'

VM2 + T2

T, = VA/2 + T2, we obtain Finally I

Max.

f

35

I

(9-13)

or'

The similarity between Eq. (9~l3) andthe torsion formula in Eq. (b) suggests equivalan! torque as a suitable name for T,. An equation for maximum normal stress that is sim'ilar to the Flexure formula but involves an equivalent 'moment M,, is obtained as follows: In Fig. 9-28b, the maximum normal stress is max. o = a, = 301+ R

•

260 mm

1

5/ 1

I

v7777w7777.3 1-200 mm

»| 201 mm

Figure P-975 976. For the cantilever beam in Prob. 975, find the maximum shearing stress at point A. Show your results on a complete sketch of a differential element.

9-10 TRANSFORMATION OF STRAIN COMPONENTS Most of the problems encountered in engineering design involve a combination of axial, torsional, and flexural loads applied to homogeneous materials of` a prismatic shape. In such cases, the stresses may be computed as described in the preceding articles and maximum resultant stresses used as a criterion of design. Occasionally, however, irregularities in a structure, or conditions that violate the basic assumptions of the torsion or flexure theories, require us to resort to ex-

I

9-10 TRANSFOFIMATION OF STRAIN COMPONENTS

353

perimental methods of determining stresses. Since stress is a mathematical con-

cept that represents the intensity of force on a unit area, it cannot be measured directly. Nevertheless, the stress-strain relations defined by Hooke's law permit us to estimate stresses from strains that can be measured. In this article we stud y the transformati on of a given set of strains into principal strains. In the next

article we consider the application of strain measurements and their conversion into stresses. Consider now an element subjected to the general state of plane stress shown in Fig. 9-31a. The normal stresses (assumed to be tensile) elongate the element in the x and y'directions, and the shearing stresses distort the element through the shearing strain 'Y,xy as shown in Fig. 9-3lb. The effect of these strains on any line element OA in Fig. 9-3 la is shown in Fig, 9-3 lc where OA is elongated y TJ

4_ I

A

a

my

O

(H)

(b)

Al

/

A

B

4

ds

'Yr7

e, dx

/

J

do As "~f,, by

by

1

dy

8

A

8

6)

by

er do

0

A"

I

O

x

O --->-

dx

o

Figure 9-31

(c)

l . 1'z=.

L

7 '

I

9/COMBINED STRESSES

358

For the element rotated 60° from the x axis, the stress components are given . by points F and G, and the values are 'at F'

at G:

143

65.2 cos l5° = 80.0 MPa

0'

..':

'r

= 65.2 sin l5°

0'

= 143 + 65.2 COS 15° :z

4-0

:'-

16.9 MPa

206 MPa

Instead of using the transformed circle of stress, we can find the stresses directly from the strains, using Hooke's law for biaxial stress (see page 43) as expressed by E(ex+ve).). 'to

I..,,2'

5(¢,+n,).

a,=

1-»»*

'

1V.r>'=G7A'y

__

E 2(1

+ ») "7A'>'

On substituting the principal strains of 924 X l 0-6 and 76 X l 0-6 found from the strain circle in Fig. 9~33, we compute the principal stresses to be Ur

= (200 X 109X924 + 0.30 X 76XI0"*) = 208 MPa (0.30)2 l ¢

U2

= (200 x I0°X(6 + 0.30 x 924Xl0") = 77.6 MPa 1 - (0.30)2

These results agree with those found previously; consequently, if a Mohr's circle of stress is plotted using these principal stresses, the radius and center will have the values shown in Fig. 9-35. In a similar manner, the normal and shearing stresses on the element at 60° from the x axis can be determined from the corresponding strain components of Applying Hooke's law, = 90 X to Tb = 910 X 10'°, and 'Yah = 220 x 10

I

,

we obtain aa

(200 X 10'X90 + 0.30 x 910XI0'°) :: '

-

'Mb =

I

- (0.30)* x 10") ._

(200 x 109X220

zu + 0.30)

79.8 MPa

I

I I

16.9 MPa

A comparison of these two methods of computing stress components should convince you of the advantages of transforming the strain circle to the stress circle.

II

PROBLEMS 978. As shown in Fig. 9-32, prove that Eqs. (9-19) and (9~20) will ttanstbrm a strain circle into a stress circle.

'

979. Starting with an element subjected only to principal stresses, show that the angular deviation 8 ofa line element such as OA in Fig. 9-31 is equal to one-half the shearing strain 'Y¢/»»

?

_

a

980. A state of strain is defined by e, == -400 X 10", e, == 200 X 10*', and 'ny = 800 X IO". If E = 200 GPa and v == 0.30, determine the principal stresses and stress . components on the face whose normal is at +30° from the x axis. °..=

--400 981. A state of strain is defined by f, 600 X l 0-6, fy = -300 X 10'°, and 7 . . x I0°°. l f E = 30 x 10° psi and v 0.30, determine the principal stresses and the . maximum shearing stress. 17.8 ksi, 02 = -4.94 Lsi, 1 = I 1.4 ksi Are. 61

I

1

9-11 THE STRAIN FIOSETTE

359

982. The strain components at a given point are f. = -533 X 10*°, *y = 67 X l 0-6, and 'Yxy = -626 X l 0--6. IFE = 30 x 10° psi and v = 0.30, find the stress components on the face whose normal is at 45° from the x axis. Ans. a = -2800 psi, 1 = -6930 psi \

983. The strain components at a given point are Qs, = -800 X 10", e, = 200 x 10-6, and ,,,, = -800 X 10"'. Irs = 200 GPa and v = 0.30, fund the stress components on the face whose normal is at +20° from the x axis. Ans. a = -105 MPa, T = -96.6 MPa

9-11 THE STRAIN ROSETTE The stress in a bar subjected to biaxial stress can be determined experimentally by attaching a strain gage oriented in the direction of the stress. The stress is then computed, in terms of the strain, from a = EE. The strain i_s generally small (under I part in 1000), hence sensitive instruments are required for measuring it. Originally, strain gages were mechanical or optical, but these have now been almost completely replaced by electrical gages. This type of gage contains a wire or foil element whose electrical resistance varies with its deformation. The gage is cemented to the test specimen, the strain in the specimen being measured as a function of the change in the electrical resistance of the gage element. The. wire-type of strain gage has been brought to a high state ot' perfection by the Baldwin Southwark Division of the Baldwin Locomotive Works. These gages are marketed Under the well-known SR-4 trademark. As mentioned previously, a single strain gage oriented in the direction of a uniaxialstress is sufficient for computing the stress. For biaxial stress, we might suppose that two strain gages would be sufficient, this would be true if the directions of the principal stresses were known, but this is not usually the case. To determine the direction of the principal stresses in addition to their magnitudes, we need three values of strain. In the preceding article we showed how the values ex, by, and y , can be used for this purpose. Unfortunately, there is no equipment that conveniently gives a direct measurement of the shearing strain *fxIv» so other methods are required. We now show that a state of strain is uniquely determined by the measurement of three linear strains ea, et, and t, acting in three arbitrary directions Ha, 6,,, and 6, at the same point as in Fig. 9~36. By substituting these strains in Eq. (9-l7), we obtain the following set of simultaneous equations:

£+(V+ fa ~x

I

ex of)

(b) Equivalent wood section

(a) Original section

Figure 10-3

Solution: Although only rarely is steel used to reinforce one side of a timber beam, this problem illustrates many of the concepts encountered later in reinforced concrete beams. The first of these involves the location of the neutral axis. Since the neutral axis coincides with the centroidal axis of the equivalent section shown in Fig. 10-3b, the moments of area about an axis through the base gives

H45

x

10*) + (15 X I0')Iy'== (45 X II

2 ay]

"in

[As-=

Romeo; + ( I 5

X I0'x5)

121 mm

First finding the moment of inertia about an axis through the top of the flange and then using the transfer formula, we compute the moment of inertia about the neutral axis:

PROBLEMS

369'

1

I

Z

1

- Ad2]

3

1= 1NA

I 50(300)3

3

+ I500(l0)3

1350 X 10° mm'

3

= (1350 X 10')

(60 X l0'l(1II)' = 611 X 10° mm'

The resisting moment in terms of the maximum wood stress is

M.,

_

(8 x 10°x6I]

x

10'°)

189 X 104.

- 25.9 k n - m

In the wood equivalent of the steel, the maximum stress is u..

Us

n

120 20

6 MPa

Therefore the resisting moment that will not exceed the permissible steel stress is =

13.21 in.

To stress the concrete to its maximum limit will require a bending moment

[Mt =

Mb k d x j d ) ]

M , = §(l800)(lOX5.37)(l3.2l)

= 639 kip- in.

To stress the steel to its limit, the required bending moment is [AL =.g,4,id]

E

M , = 20(l.5X13.2l) = 396 kip in.

.

The safe bending moment is therefore 396 kip in. Since the steel governs, we conclude there is not enough steel, hence the beam is under-reinforced.

PROBLEMS 1024. In a reinforced concrete beam, b = 250 mm, d = 450 mm, A, = 1500 mm2. Find the values of kg' and did if (a) n = 6 and (b) n = 10. Ans. (b) kd = 148 mm;jd 401 mm 1025. In a reinforced concrete beam, b == 10 in., d = 16 in., and n = 8. The actual maximum stresses developed are f. = 1600 psi and L = 24 ksi. Determine the applied bending moment and the .steel area required. . Ans. M = 52.5 kip~flt, A, = 1.86 in.2 1026. Repeat Prob. 1025 i d = 18 in. 1027. Determine the maximum stresses produced in the concrete and steel of a reinforced beam by a bending moment of 70 kn~ m ifb = 300 mm, d = 500 mm, A, = 1200 mm2, and n == 8. Ans. L = 6.9! MPa,/`, :: 130 MPa

=

'E4

II

1

g

c!

377

10-5 DESIGN OF REINFORCED CONCRETE BEAMS

1028. In a reinforced concrete beam, b = 500 mm, d = 750 mm, A, :: 6000 mm2, and n = 10. What are the maximum stresses developed in the concrete and the steel by a bending moment of 270 kN - m? 1029. The dimensions of a reinforced concrete beam are b = 300 mm, d = 450 mm, A, = 1400 mm2, and n = 8. If the allowable stresses are); s 12 MPa and); S 140

_

MPa, determine the maximum bending moment that may be applied. In what state of reinforcement is the beam? Ans. M = 78.4 kN • m, under-reinforced

1030. In a reinforced beam, b = 10 in., d = 18 in., A, = 2 in.2, and n = 10. Determine the safe uniformly distributed load that can be carried on a simply supported span 12 ft long ilL s 1800 psi and f, s 20 ksi. Assume 2 in. of concrete below the reinforcing steel and include the weight of the beam. Concrete weighs 150 ib/ft'.

1031. In a reinforced concrete beam, b = 12 in., d = 18 in., and n = 8. If a maximum stress of 1400 psi is developed in the concrete when resisting a bending moment of 80 kip ft, what stress is developed in the steel? What area of reinforcing steel is required"

Ans. j; = 16.3 ksi, A, = 3.79 in.2

.

1032. Solve Prob. 1031 if'the bending moment is 60 kip-ft, all other data remaining unchanged. 1033. Solve Prob. 1027 by computing the moment of inertia of the transformed section and then applying the flexure formula according to the procedure in Art. 10~2. The distance of the equivalent concrete area from the neutral axis of the transformed section can be taken as its radius of gyration with respect to this axis.

1034. Solve Prob. 1029, using the procedure outlined in Prob. 1033.

10-5 DESIGN OF REINFORCED CONCRETE BEAMS

In the preceding article the dimensio.ns of the reinforced beam were specified. This fixed the location of the .neutral axis. Inasmuch as the stresses vary directly with their distance from the neutral axis, the applied bending moment may stress the concrete to its permissible limit while leaving the steel understressed-a condition known as over-reinforcement. The opposite condition, under-reinforcement, may occur when°the steel reaches its permissible limit first, the concrete remaining understressed. For maximum economy, both materials should reach their limiting stresses simultaneously-a condition known as balancedstress reinforcement. In designing a concrete beam with balanced-stress reinforcement, therefore> we start with the assumption that the position of the neutral axis is such that the maximum L and the maximum s t r e s s / n in the equivalent concrete occur simultaneously, this is shown in the stress diagram in Fig. 10-7. From this, by the proportional relations between the triangles ABC and .4DE, we obtain

LV = d

f

JC II

+L

378

10/REINFORCED BEAMS

I"f°*A

iI *

A

B

iv

QU..

I

I

d

I

I I I

D(

Ln

I > --__J;»_

,|_ __1

Figure 10-7 Stress distribution.

k == f

or

.i

L +

(10-7)

n

n

u

Having computed k in terms of the allowable stresses, we obtain the value of by canceling out the term d in Eq. (10-3); .

II I

1 I

II

f

1

-go

(10-8)

Once the values o f k and j are determined, Eq. (10-5) is used to compute the quantity bd2. The deeper the beam is, the greater will be the moment arm of the resisting couple and the smaller the force. A deep beam, therefore, requires less concrete and steel than a shallow beam. Because of practical limits to the depth, however, d is usually made about l.5b. From this and the now computed value of bd2, the dimensions b and d are found. As the final step, the area of the reinforcing steel is computed from Eq. ( I 0-6) or, preferably, from the condition that C = T = ASk, in which A., is now the only unknown. However, the reinforcing rods generally available are not of such size as to equal precisely the steel area required. As a consequence, balancedstress reinforcement can be only closely approximated. For most well-designed rectangular beams, the values of k and j are very closetok = § a n d j = g. If these values are used, dimensions may be rapidly estimated by substituting them in Eq. (10-5), thereby giving* bd2

= 6M f~

(10-9)

Having thus found bd2 and assigning values Io b and d, we compute the tensile force in the steel and the steel area from •

1

1'=

T and , 1 = -

* The similarity of this result to the flexure formula for rectangular beams, a ix simple to remember. J'

(l0~l0)

L

::

6Ai/h/12, makes

ILLUSTRATIVE PROBLEM

379

ILLUSTRATIVE PROBLEM 1035. Design a concrete beam with balanced-stless reinforcement to resist a bending moment of 90 kN • m. The allowable stresses a r e } = 12 Mpa,f, = 140 MPa, and n = 8. -

With balanced-stress reinforcement, the stresses in the concrete and the concrete equivalent of the steel have the values shown in Fig. 10-8. From the proportional relations between the similar triangles ABD and AEF, the value ofk is found to be " Solution:

L'

12 (140/8) + 12

d

4

k = 0.407

9

1

»

I

Lu

V

Z bed 2 f

I

1

A

td

1

i

»

/

A

I

iD

B

I

e e

d

ad

I

I

iH

I

l~=~~~ /

,.|

*vLr¢/r'/

f

I

I

f

I

I

/

f

E

n.-1s

l

-4-

_n

140

8

.L

.1

)

T A;

2-F

Figure 108 from which the value of is

j = 1

-

§/< = 1 - %(0.407) = 0.864 \

In terms of the concrete, the resisting moment is C(jd). SO

[m

:=

cud) = ( M b kdxidn

90 X 103 = %(I2 X 10°)(bd2>(0.407)(0.864l

bd' = 0.0427 m3 = 42.7 X 10° mm'

la)

Assuming that d = l.5b. we now obtain from Eq. (or). b = 26? mm and

d = 400 mm. The area of reinforcing steel is now the only unknown. Siam the compressive force C in the concrete equals the tensile force Tin the steel, we obtain

HM) kd = A,f.1 which gives

é(12 X 10">(0.267X0.407X0~.400)

A,

.-1,(140

x

10')

1.86 X 10" m2 = 1860 mm

Usually the available stock sizes of reinforcing steel do not produce exactly this area of steel, so the final design only closely approximates balanced-stress rein forccmcnt. *

J

.y

10/REINFORCED BEAMS

380

PROBLEMS 1036. A reinforced concrete bam is designed to reach f. = 12 MPa and f, = [40 MPa simultaneously. If n ==. 10 and d = 400 mm, compute the moment arm of the resisting couple. Ans. did = 338 mm I 03'7. In a reinforced concrete beam, d = 18 in. and n = 10. Find the dimensions b and A, that will resist a bending moment of60 ldp ft with balanced-stress reinforcement, iff; = 1600 psi and_)§ = 20 ksi.

1038. In a reinforced beam, b = 200 mm, d = 400 mm, and n = 10, the allowable stresses are _IQ = 10 MPa and f. = 140 MPa. Determine A, for balanced-stress design and the safe resisting moment. 1039. Design a reinforced concrete beam with balanced-stress reinforcement that will resist a bending moment of 140 kN ' m, assuming d = 1.5b,j2 = 12 MPa,L = 160 MPa, and n :'. 8. Ans. b = 316 m m , A , = 21 10 mrn2 1040. Solve Prob. 1039 if = Id. 1041. A simply supported beam 16 it long is designed to can'y a concentrated load of 20 kips at midspan. Compute b and A, for a depth d = 24 in., using balanced-stress reinforcement with f; = 1800 psi,f. = 20 ksi, and n = 10. Allow 2 in. of concrete below the steel and include the weight of the beam, assuming that concrete weighs 150 lb/ft'. (Hint: Assume an initial weight per foot for the beam and check this assumption after you have found dimensions.)

1042. A reinforced concrete beam 24 R long and perfectly restrained at the ends is to carry a live load of 1500 lb/ft in addition to its weight. Assuming b = 16 in., design 20 ksi and a beam with balanced-stress reinforcement, using j; :: i 600 psi, f, n = 12. Allow 2 in. of concrete below the reinforcing steel. Concrete weighs 150 lb/ft'. (Hint: Use the Hint in Prob.ll04l.) Ans. d : 13.9 in., A, 4.37 in.2 =.'

1043. Design a balanced-stress reinforced concrete beam to calTy a uniformly distributed live load of 1000 lb/R over a simple span 18 ft long. UsejQ = 1800 psi,f, = 20 ksi, and n = 10. Allow 3 in. of concrete below the steel and include the weight of the beam, assuming that concrete weighs 150 lb./ft'. (See the Hin! in Prob. l04l.)

10-6 TEE BEAMS OF REINFORCED CONCRETE The method used for rectangular reinforced concrete beams becomes quite involved when applied to T beams. Because of the flange of the T. the centroid of the compressive area is no longer §kd from the neutral axis, nor is the line of action of the resultant compressive force ikd from the top of the beam. As a consequence, it is cumbersome to use the basic procedure described in Art. 10-4, although textbooks on reinforced concrete do develop equations in terms of k and j. It is preferable for the beginner to apply the flexure formula directly to an equivalent section, as indicated in Art. 10-2 and as illustrated in the folu lowing problem.

I

I

*

I

~

I4 I

I

ILLUSTRATIVE PROBLEM

381

ILLUSTRATIVE PROBLEM 1044. The T beam in Fig. 10-9 is reinforced with 2400 mm' of steel. Assuming n = 8, _L s 12 MPa, /, 5 140 MPa, determine the maximum safe resisting moment.

I•

.1

750 mm

I ;

1(»(»*w§(1~l'0

Below this value, as shown in Fig. l 1-7 by the dashed portion of Euler's curve, the Euler unit load exceeds the proportional limit. Hence for L/r < 100, ELder's formula is not valid, and the proportional limit is taken as the critical stress. The curve also shows that the critical or allowable stress on a column decreases rapidly 3.S the slenderness ratio increases, hence it is good design to keep the slenderness ratio as small as possible. \

II

\

\

\ \

200 MPa (P.L.)

\

r

g

I I I I 100

1

Euler's curve

I

p _ En*

*Z-l(L/r)2

L/f

Figure 11-7 Critical or allowable slrcss is given by the solid line. Dashed portion of Euler's curve is not valid. I

Finally, remember that Euler's formulas determine critical loads, no t working loads. It is therefore necessary to divide the right side of each formula by a suitable factor of safety--usually 1,7 to 2.5, depending on the materialin order to obtain practical allowable values.

I

I I { l

l

L

395

ILLUSTRATIVE PROBLEM

I

ILLUSTRATIVE PROBLEM

w shape that can be used as a column 7 m long to support an axial load of 450 kN with a factor of' safety bf 3. Assume (a) both ends hinged and (b) one end fxxed and the other hinged. Use an = 200 MPa and E == 200 GPa.

1101. Select the lightest

Solution: Part a. For steel with a proportional limit of 200 MPa, the specifications for Euler's formula with hinged ends require that L/r 2 100. Il'L/r < 100, the limiting stress is the proportional limit. The specified working load, when multiplied by the factor of safety, gives a critical Euler load of 1350 kn. Applying Euler's formula and solving for I, we obtain n

P

-. EIT2 L2

PL*

I

(1350 X 103)(7)1 4

E#

(200 X 10°)ur2)

= 33.5 X 10-° m4 = 33.5 X 106 mm' Also, the slenderness ratio L/r z 100, from which the least r is L 7000 r s 100: 100 =70•Q0 rnm J

These criteria establish that the section must have a least I 2: 33.5 X 105 mm4 and a least r 5 70.0 mm. This is satisfied by choosing a W250 X 73 section with a least I = 38.8 X 106 mm' and a least r = 64.7 mm. If the selection were based on the proportional limit, the section must have a minimum area of 6750 mm2 (obtained by dividing the load of 1350 iN by the proportional limit of 200 MPa) and a least r greater than 70.0 mm- These conditions are satisfied by a W310 X 97 section with A = 12 300 mm2 and least I' = 77.0 mm. The lightest section therefore is the W250 X 73 section. x Parr b. The critical Euler load is 1350 kn, as before. With one end fixed and the other hinged, the effective length of an equivalent hinged column is 0.7L :: 0.7(7) = 4.9 m. Using this effective length in place of the actual length, we find that the criteria for Euler's formula are

I

2.

PL* E1r2

(1350 (200

x 103X4.9)2 x 1052 =

16.4 x 10-°

in*

2 16.4 X 10° rum"

and

r s

L 4900 = = 49.0 mm 100 100

The lightest section that satisfies these conditions is the W360 X 64 with least I = 18.8 X l0° mm4 and least r == 48.1 mm. The other set of criteria based on the proponionai limit are

1350 X l03 = 6.75 X l0" m* = 6750 mm* 200 X 10

I

396

11/COLUMNS

and

r 2 49.0 mm

for which the lightest section available is the W250 X 58 with A = 7420 mm' and f = 50.3 Mm. Comparing the two sets of criteria, we see that the proper section is the W250 X 58. . Unwary students might be tempted to select a section based Only on I without checking r and thereby choose a W200 x 52 section with least I = 17.8 X 106 mm4. However, this section has a least r = 51.7 mm and an area of 6660 mm2, which results in a stress exceeding the proportional limit of 200 MPa. It is, therefore, not acceptable because it violates the stress-strain proportionality on which Euler's formula is based. This problem demonstrates the importance of the slenderness ratio i n column analysis. In Part a, the selection is governed by elastic stability (i.e., the use of Eulerls formula), whereas in Part b the selection is determined by the proportional limit.

I

I

I t

I

: 1

4

1

I

PROBLEMS 1102. A 50-mm by l 00-mm timber is used as a column with fixed ends. Determine the minimum length at which Euler's formula can be used if E = IO GPa and the proportional limit is 30 MPa. What central load can be carried with a factor of safety of2 if the length is 2.5 m? Arzs.

I103. An aluminum strut 6 ft long has a rectangular section

L

:=

1.66 rn, P = 32.9 kN

i in. by 2 in.

A bolt through

each end secures the strut so that it acts as a hinged column about an axis perpendicular to the 2-in. dimension, and as a fixed-ended column about an axis perpendicular to the 4-in. dimension. Determine the safe central load, using a factor

of safety ofl2 and E = 10.3 X 10° psi. Ans.

P = 2770 lb

1104. A square steel bar is to support a load of 20 kips on a length of I O ft. Assuming rounded ends, determine the length of each side. Use E = 29 X 106 psi. Ans. 1.86 in. 1105. Repeat Prob. 1104 if the column is made of wood for which E = 1.6 X 106 psi. 1106. Two C310 X 45 channels are latticed together so they have equal moments of inertia about the principal axes. Determine the minimum length of a column having this section, assuming pinned ends, E = 200 GPa, and a proportional limit of 240 MPa, What safe load will the column carry for a length of 12 m with a factor of safety of 2.5? Ans. L = 9.89 m, P = 742 kN 1107. Repeat Prob. I 106 assuming that one end is fixed and the other hinged. 1108. Select the lightest w shape that will act as a column 8 m long with hinged ends and support an axial load of 270 kN with a factor o f safety of 2.5. Assume that ' the proportional limit is 200 MPa and E = 200 GPa. Ans. W250 X 67

1109. Select the lightest W shape that will act as a column 40 ft long with fixed ends and support an axial load of ISO kips with a factor of safety of' 2. Assume that the proportional limit is 30 ksi. Use E = 29 X l0° psi. What set of criteria determines 'he selection" I I

J

397

1 1»5 INTERMEDIATE COLUMNS: EMPlFllCAL FORMULAS

11-5 INTERMEDIATE COLUMNS; EMPIFHCAL FORMULAS 1.

The preceding discussion showed that long columns can be treated by Euler's formula provided that the slenderness ratio is larger than the value at which the average stress reaches the proportional limit. For hinged steel columns, this limit 100 at 200 MPa. Euler's formula is not valid for smaller slenderness is L/r ratios. The definition of a short column as one whose length does not exceed 10 timeslthe least lateral dimension sets the upper limit of the slenderness ratio at about 30 for a rectangular section. For practical purposes, the limiting stress on a short column has been found to be the stress at the yield point. Figure I 1-8 shows these conditions for steel having a yield point of 280 MPa. and a proper» tonal limit of 200 MPa.

-

350

V

300 Uyp = 280

250

iv

Tangent-modulus curve n

0

o

..1

II

c\1

,g-1

(edo)V/d

!

1

Euler's curve

150 -

P' _ "

'

1

A

100

50

Short columns

4

Intermediate columns

E322

L

(7)

Long columns

>

|

30

i

100

200

L/'r

Figure 11-8

Various methods have been proposed for bridging the gap between ,*he short column range and the long column range. However, none of them has been accepted universally for intermediate columns partly because of their departure from the stress-strain relationship when the stresses exceed the proportional limit, and partly because of their indeterminate mixture of direct and flexural stresses. Most empirical formulas for intermediate columns have been developed for steel because it is such a common structural material. We discuss these first and 'hen indicate their extension to other structural materials. ,

r 1

11/COLUMNS

398

In one proposed method--that of the tangent-modulus theory-the Euler formula is extended to intermediate columns stressed above the proportional limit by replacing the constant modulus E by a tangent modulus E,, namely, II

Q..

Apr2

we

(ll-6)

(L/f)2

The tangent modulus E,, also called the effective modulus, is obtained by using for E, the slope of the tangent to the stress-strain diagram at the point conesponding to the average stress in the column. This yields a curve that connects the curves in Fig. l 1-8 representing the short and long column formulas. Although this method is empirical because it violates the stress-strain proportionality assumed in the derivation of Euler's formula, actual tests show close agreement with the theoretical curve.* Other methods are, frankly, empirical. Design equations for a variety of engineering materials can be found in most engineering handbooks. In general, empirical equations are either linear or nonlinear depending on the best agreement that can be found with data from tests of real columns. Modern column studies which include the effects of residual stresses, will undoubtedly lead to a more complete understanding of the behavior of intermediate columns. The American Institute of Steel Construction (AISC) defines the limit between the intermediate and long columns to be the value of the slenderness ratio C, given by .

I

I

i

I

I l

C

in which E is the modulus of elasticity (200 GPa or 29 X 106 psi, for most grades of steel) and "yr is the yield stress for the particular grade of steel being used. For columns of effective length L, and minimum radius of gyration r, AISC specifies that for L,/r > Cf. the working stress, aw, is given by I

12125 07»'

(ll-'7)

(Note that this is Euler's formula with a factor of safety of 23/12

II

23(L,If)2

L92.) For

L,/r < Cc, AISC specifies the parabolic formula 1

Up

'r'

(L¢/f)2` Ora

2c2

FS

(ll-8)

where the factor of safety, FS, is given by

FS

§

3+

3 ;

0

0

25

50

100

7o

125

150

z

I ITS

I s

i

200

Le `

r

for several gxades of steel. Figure 11-9 Working stress for columns (AISC specifications)

We now consider column formulas for some materials other than steel. of the The Alumin um Association, Inc., lists column specifications for each various types of aluminum alloys. In these specifications, the allowable stress is a constant for short columns, a straight-line relation approximating the tangentmodulus formula is used for intermediate columns, and Euler's formula is used for long columns. For example, the specifications for 2014-T6 aluminu m alloy

are*

* The values quoted here in U.S. Customary Units are taken from Specihcazionsfor Aluminum

Association, Inc., Washington, .S`truc°turev, 4th ed., Aluminum Construction Manual, Set;. l. Aluminum ns. U.('., April l982. p. 21. The Sl numbers are approximate conversio

r I

11/COLUMNS

400

SI Units

U.S. Customary Units 0,

£S

12

0,

L 12 < -
' 862 so 3

_

_

_

--»

' from

which

aw =

u,

{l

[I -(L,/r)2/2C2]¢,, FS [(79.4)2/2(I02)2]} (380 X 10'~) 1.90

= 139 MPa

Finally, the safe axial load is [P = UA]

P = (I39 X 10°X15 500 X 10-6~} = 2150 kN

Ans.

Part c. Braced at the midpoint, the column is equivalent to one having a length = of 5 m, fixed at one end and hinged at the other. The effective length is L, than less is which 55.6, :: 3500/63.0 = L,/r Therefore m. 3.5 0.7L = 0.7(5) = Cc :: 102. Proceeding as in Part b, we find FS = 1.85 and v. = H5 MP3, Hence the safe axial load is [P

::

UA]

P

(is X

toms 500 X

2710 kN

10'°)

And.

can This problem illustrates the increased strength of a column whose ends better is it practice, in realized never is condition this Since rigidified. be perfectly be more when determining allowable loads always to assume hinged ends, or Lu instead realistic in selecting the effective length with fixed ends as about 0.75i'.

of 0.5L. Ill

Select the lightest W shape that will support an axial load of 90 kips on an effecand 1-' = tive length of 15 ft. Use AlSC column spec fications with "in = 36 si psi. 10° X 29

1

V I

402

11/COLUMNS I

Solution: Since both the area A and the least radius of gyration r are unknown and no convenient relation between them can be set up, the selection of the lightest W shape involves a trial-and-error procedure. The steps are (I) assume a working stress; (2) calculate the area required; (3) select the lightest appropriate section based on the area required, and (4) for the section selected, 'calculate the allowable load based on the column specifications. If the allowable load equals (or is slightly larger than) the applied load, the section selecte is the appropriate one. If the d allowable load is less than the applied load, a heavier section must be selected and the procedure repeated. Clearly, the number of trials that must be attempted before the correct section is determined depends on how close the initial assumed stress is to the actual stress. One suggestion is to assume an initial working stress of 80% of the stress at L/r = 0 determined from the column specifications. For steel with Uyp = 36 ksi, the limiting slenderness ratio is found to be

C¢ =

'2~»t=(29 x

21r2E =

10°)

36 X 103

'iv

I

126

1

I

g,

First Try. At L,/r = 0, FS = and 0, = ay,/FS = 36/§ = 21.6 ksi. Assuming an initial stress of`0.80(2I.6) = 17.3 ksi, the required area is

A

_.p 0

90 X 103 17.3 X 103

_

aw =

]

I212E 23(Le/r)2

.

Up

23(143)2

7.30 ksi

Second Try. Next, we select a $8 X 28, which has a larger area and a larger least A For this section, A = 8.25 in? and least r = 1.62 in. The slenderness ratio is L,/r = 15(l2)/1.62 = I l l , which is less than C, = 126. The working stress for this section is determined as follows: 5 3(L,/r) 3 + 8Cc

(L,/r)3

8c'

which gives

I

of.,

=

[1

(L./f)'/202] "w

FS

[I Up

(1 1 u2/2(126)21 36 1.9 I

=:

11.53 ksi

Then, the allowable load for this section is P

I

I

l

121#(29 X 106)

Then, the allowable load is P = JA = (7.30X6.l6) = 45.0 kips. Since this is less than the applied load of 90 kips, the section is inadequate.

FS

I

5.20 in_2

Therefore, from Table B-2, Appendix B, we select a $8 X 21 with A = 6.16 in.2 and least r = 1.26 in. For this section, the slenderness ratio is L,/r = 15( l2)/l .26 = 143 which is greater than Cc :: 126. Therefore the working stress for this section

is

I

= HA = (l 1.53X8.25) = 95.1

kips

Since this load is only slightly larger than the applied load of 90 kips, the W8 X 28 . is the appropriate section. The procedure of selecting a section is greatly simplified by using tables /

c

PROBLEMS

403

giving the allowable axial loads for different sections of various lengths. Such tables are found in a steel handbook such as that published by the AISC. However, this problem illustrates the trial-and-error method that arises frequently in structural ` design.

PROBLEMS 1112. Determine the slenderness ratio of a 5~m column with built~in ends ii its crosssection is (a) circular with a radius of 40 mm and (b) 50 mm square. Use the concept of effective length.

(a) l25,€b} £73

Ans.

end fixed and the other hinged fits cross section is (a) circular with a radius o f 2 in. and (b) 2.5 in. square.

1113. Find the slenderness ratio of a 12-ft column with

OI1€

1114. Determine the maximum length of a W250 X 167 section useclas a hirigedend column to support a load of 1600 kn. Use AISC specifications with "iv = 350 MPa and E = 200 GPa. 1115. A WI4 X 82 section is used as a column with an efi'ective length of 30 ft. Using AISC specifications, compute the maximum load that can be saIlel§' applied. Use Uyp = 50 ksi and E = 29 X 106 psi. I

l?I leaps

.4n5.

-

1116. A W3l0 X 52 section is used as a column with hinged ends. Using AISC specifi*-1 o=. cations, determine the maximum load that can be applied if pa) I. = IG L = 14 m. Use 07yp = 250 MPa and E = "OO GPa. 1117. Four 4 X 4 X I in. angles are latticed together to form the column section shown in Fig. P-1 l 17. Use AISC specifications to determine the maximum length at which to a 200-kip load can be safely supported. What should be the spacing.?~etwee:~ Il'.:...lL~.. three-quarters exceed to not is angle separate each of ratio slenderness the bars if of that of the fabricated section" Use 0>, = 60 ksi and E = 29 X IO6 `l

»g

F

*Go

10 in

ll` n

-Ni.

n'

- -

:EII

lb »

n

,

s s I

F L

IU

:.

;

E

F10 in 11

I

E l

\

|.

I

r

rl

F

.

1

Figure P-l 117

-

._

__

__

w-»

I I ] 8. A steel column with an effective length of 10 m is lhtvricated from two C?S0 x 45 channels latticed together so that the section has equal moments ollinertia about the principal axes. Determine the safe load using AlSo specifications. Use t o , 1

380 MPa and E = 200 GPa. .4 PS.

404

t 1/COLUMNS

1119. In the bridge truss shown in Fig. P-I l 19. the member A C is composed of Iwo C9 X 20 channels latticed together so that the fabricated section has equal moments of inertia about the axes of symmetry. If the safe load P on the truss IS governed by the strength of the member AC. determine P using AISC sD€c1f1cat1ons with Uvp :: 36 ksi and E = 29 X 10° psi. Ans. P = 167.0 kips

I I

¥ 1

5

I

B

a 22.5 ft C`

30 ft

30 f t *

P

P

r

1

30 fn

30 fn

1r

1

1

P

E

Figure p-1119

|

s

1120. Rework Prob. I 1 19 using two C10 X 30 channels and do == 50 ksi.

1

1121. For the truss in Fig. P-1 I 19, find the lightest W shape for chord A B imp = 60 kips. Use AISC specifications with oyt; = 50 ksi and E = 29 X 10° psi. 1122. Select the lightest

w shape that can be used as a column to support an

axial load

of 4'70 kN on an effective length of-4 in. Use AISC specifications with ay, = 250 MPa and E = 200 GPa. Ans. W200 X 36

I

1123. Select the lightest W shape, according to AISC specifications, that can be used as a column to support an axial load of 700 kN on an effective length of 5.5 in. Assume UYP = 250 MPa and E = 200 GPa.

1124. Repeat Prob. l 123 assuming that the axial load is 690 kN and aw = 345 MPa. 1125. A hinged-end steel column 30 ft long is fabricated from a W8 X 31 beam and two C12 X 30 channels arranged as shown in Fig. P-1 125. Determine the safe axial load using AISC specifications with Up = 36 ksi and E = 29 X 106 psi. Ans. P = 354 kips s

l I

H

r

I I

W8x 31

p

'C12X3g

I

II

Figure P-I 125

1126. Derive a parabolic formula of the general type P/A = U - c(L/r)'>a that will be applicable to aluminum alloy columns with hinged ends. Assume that theparabolic formula will be tangent to an Euler formula with a factor of safety of` 2. Use a = I 10 MPa and E = 70 GPa. (Hint: For the two formulas, equate their unit loads and also equate their derivatives with respect to the slenderness ratio.) Ans. p/.4 = (1 10 x 106) 8760(L/t)-' for 1./t < 79.3

-

I

405

1 1-6 ECCENTFIICALLY LOADED COLUMNS

! i I

I

I

1127. Four 4 X 4 X I in. angles are bolted back to back as shown in Fig. P-l 127. Determine the safe load when they are used as a hinged-end column 12 ft long. Use AJSC specifications with aw = 36 ksi and E = 29*.X 10° psi.

Figure P-l127

1128. Determine the safe axial load that can be applied to a 2014-T6 aluminum alloy column if its length is (a) I m and (b) 3 m. Assume the geometric prop-enies of the cross section are identical to those of an S310 X 52 steel section. A/15.

(a) 984 kn; (b) 172 kN

1129. Determine the safe axial load that can be applied t o a 2014-T6 aluminum alloy column if its length is (a) 4 f`t and (b) 10 ft. Assume the geometric properties of the cross section are identical to those of an S12 X 35 steel section. Ans. (a) 200 kips, (b) 32,4 kips »

1130. Repeat Prob. l 129 using the properties of an S l 2 X 50 steel section. 1131. Compute the safe axial loads for an oak column 150 mm by 200 mm ill the length is (a) 2 m and (b) 4 m. Use E = 11.5 GPa. Ans. (a) 582 kn; (b) 146 kN

1132. Determine the safe axial loads for a pine column 2 in. by 8 in. if the length is (a) 6 ft and (b) 12 n. Use E = 1.6 >( 10° psi.

11

ECCENTRICALLY LOADED COLUMNS

Columns are usually designed to support axial loads, and the preceding I`orrnulas have been presented with this in mind. Under certain conditions, however. columns are subject to loads having a definite eccentricity. This occurs, For example, in the case of a beam connected to the column flange in a building. In the maximum stress approach, the eccentrically loaded column is treated as if it were an eccentrically loaded short strut (see Art. 9-3). However, to eliminate the possibility of buckling so that the effect of detlection on the moment arm of eccentric loads may be neglected, the maximum compressive stress is limited to the value computed from the specified column formula. This approach is valid only for moderate slenderness ratios. Applying this procedure to the column in Fig. 1 1-10 that supports an axial load P0 and a load P at an eccentricity e, we Lind the design criterion to be O'

2

2P A

Mc I

p 0 + p + Pe A S

(II-Is)

|\

1 1 /COLUMNS

406

P0

~. / Y \

_

!2

x

J

.r

ii*

\ / L

1

\

I Figure 11-10 Axial load Po and eccentric load P on column.

II

e I I

Here O' is the stress computed from the specified column formula (always use the least radius of gyration to determine the slenderness ratio), 1 is the moment of inertia with respect to the axis about Which the eccentric load causes bending (axis X-X in Fig. ll-l0), and S is the section modulus with respect to that axis. Modern design criteria have refined the maximum stress approach to include themoments, called secondary moments, that are also introduced because the neutral axis is deflected (the so-called P-5 effect). These criteria most often take the form of interaction equations that attempt to "weigh in" the relative importance of the axial stress and the bending stress. For example, AISC* recommends that, when the computed axial stressf, is less than 15% of the actual stress .Fa that would be permitted if only axial stress were acting, the secondary moments may be neglected and the member must satisfy the following criterion:

_t;*,n, + f be. Fc

9

F,,,.

.Fbx

S

1.0

(it)

Whenf, > 0. I5F4, secondary moment effects cannot be neglected. In these cases, AISC requires the following formulas to be satisfied;

a.Fa +

+

C"f)1¥_/bf

(I

"fllFex)Fb.~(

f,

}

0.60F,,

Cm,f,,

5

1.0

(to

5

1.0

(c)

-for/Fey)-Fby

+ fn + Tb" Fex

Fly

In Eqs. (a), (b), and (c), the various terms are as follows: /Q = computed axial stress ` c'!ion.American * The* notation used here is that used in 'Manual of 91001 C`ons1ru -

Steel Construction, New York, 8th ed., P980.

I

Inst ' rate -'1"

I

I

I

i I

T

I

I

1 1-6 ECCENTRICALLY LOADED COLUMNS

407

Fa == allowable axial stress if axial force alone were acting = computed bending stress about the major axis disregarding the sec-

f,.

ondary moment

Cm.:

|

.

= computed bending stress about the minor axis disregarding the secondary moment " Fb.r = allowable compressive bending stress about the major axis if moment alone were acting F,,,. = allowable compressive bending stress about the minor axis if moment alone were acting . Few = Euler buckling stress for buckling about major axis Fw = Euler buckling stress for buckling about minor axis cm, = reduction factors to correct for overconservatism in some cases of the amplification factor [ l - (L,/F,)] F, yield stress ..

For compression members in frames subject to joint translation, or sidesway, Cm may be taken as 0.85. For compression members in frames braced against sidesway and subject to end moments (but not transverse loads between supports). use C,,, = 0.6 - 0.4(M'1 Ill/12) 2. 0.4, where M, /M2 is the ratio of the smaller to the larger end moment. This ratio is positive when the member is bent in reverse curvature and negative when it is bent in single curvature. For compression members in framesbraced against sideway in the plane of the loading and Subjected to transverse loading between the supports. C°.,,, may be taken as 0.85 for members with restrained ends and as unity for members with no end restraints; C,,, may also be determined by rational analysis in this case. AISC specifications also 'include formulas for determining the allowable bending stresses pa as a fraction of the yield stress. The value of F, depends on the properties of the cross section and the bracing intervalsInteraction formulas similar to the AISC formula have been adopted for other structural materials such as wood and aluminum. The design of members to can'y both axial and bending loads is essentially an iteration procedure. An assumed section is checked t`or adequacy using the appropriate criteria. This procedure is greatly simplified by the many tables and graphs that are available to assist the designer. Computer programs are also i

available that will assist the designer in selecting the optimum section that satisfies the interaction formulas.

l

The problems that follow will illustrate the application of the maximum stress approach. Forapplications of the interaction equations, refer to any modern structural design text.*

I

I1 1 1

i

* Sec, for example, L. A. Hill. Jr.. Fzzndanlezzfuls QtlS!ru

i

I

The effect of the central load P is resisted equally by the direct load Pa = P/n acting on each of the /I connectors, as shown by the free-body diagram of the plate in Fig. 12-l4a. The torsional couple T is resisted by torsional loads P, (Fig. 12-l4b), which act perpendicular to the radius p from the centroid of the group and vary directly with .the distance of the connectors from the centroid. To determine the torsional load on any rivet or bolt, we may consider that the connection is` equivalent to a flanged coupling consisting of three concentric circles of connectors and use the method outlined in Art. 3-3 (see page 76). Then the resultant load on any rivet or bolt is the vector sum of the direct and torsional loads on that rivet, and appears as shown in Fig. I2-l4c. A better method of determining the torsional load is to apply the torsion formula T = Tp/J. Here r represents the average shearing stress on any connector, p is the radial distance from its center to the centroid of the connector group, and J may be expressed as E A0'an (0) Since all the connectors have the same area A, and since p2 may»-be expressed in terms of the x and y coordinates of any connector so that p1b == Hz 51- y2 (see Fig. I2-l4b), we may rewrite Eq. (a) as

J

A(E x'

+ Z v2)

(b)

12/RNETED, BOLTEO, AND WELDED connecTions

426

y

H P é

5-

+

T= Pei

'a'

I

\

x

I lg

T

>

I

I '

./

0-14

°'

x

x lill-1in111l.

» /

P

(a) Equal direct loads

5

1P//B

Q

C*\

1

1:

I

(b) Distribution o f torsional loads

\

2*

I

x

.D 10

~\i

p

C+ g,

.1

\

\

f I

I

\

(c) Resultant connector loads I

Figure 12-14 Analysis of eccentrically loaded connection.

l

n

from which the torsion formula becomes T'

=

E

To

.

I4(z X2 + 2 y2)

(c)

Transposing A to the left side of this equation determines the torsional load P, . on any rivet from P, = As, so we finally obtain

p,

To 2

(12-S)

+ 2 y2

/2

The resultant load on a typical rivet or bolt is obtained as the vector sum of Pd and P,. See Fig. ,l2-140. This vector addition is easily performed analytically by resolving Pa and P, into x and y components. The components Pd, and Pd of the direct load are constant for all connectors. The components of the torsional load P, are obtained by observing from Fig. I2-l4b that the angle a between the radius p and the x axis equals the angle between P, and the y axis: hence

and

11,

P, sin a

p,,

P, cos a

p

»

since sin a = y/p and cos a 'r

from Eq. (12-5), we obtain

p, -y :

P,-x p

x/p. Replacing P, in these relations by its value

u

i

427

ILLUSTRATIVE PROBLEM

p,, 1

I

E

+ 2 y2

X2

(12-6)

To ~. P,, = E x 2 + Z y I

I

where x and y are the coordinates of a connector measured from the centroid of the connector group. The maximum load on any connector occurs when Pd, and maximum PI as well as Pd, and maximum Pfy are additive, as at the upper right comer.

I

The resultant rivet load is found from p, = V(p,,, + p i '

I

I

+ ( p , + Po2

£12-7)

The use of these equations is shown in the following illustrative problem.

ILLUSTRATIVE PROBLEM E

1229. On the connection of 12 rivets shown in Fig. 12-15, the load P = 48 kips- Passes through the center of rivet B and has a slope of 4 to 3. Determine the resultant load on the most heavily loaded rivet. y

s

_

Q

6

21+ 8,

F

I;,+

A

8,

I

I Il

I I

4 in.

F; :: 8 -_-._,____

c+-~@

:*

25

kips

x

i

! I

Pa/

I

4 in. 38.4 kips

l

+

3

1 P = -LS kips

l

1,

€>~

°£ |. Iu

4

i

3 in.->-H ~~3in

>

.Q

ex

•

u

I 1I

Q

I

Figure l 2-15

r

I 1

,a I

Solution' The clTect of the applied load is equivalent to an equal central load acting through the centroid C of the rivet group plus a torsional couple equal to the rivet group. Replacing P by its CO!T1~ the moment of P about the centroid of that the moment of' P is pon ens P = 28.8 kips and P, = 38.4 kips, and noting s, we find that the torsional couple is onent comp its of sum ent mom the to equal 1' :.° 38.4(4.5) = 172.8 kip-in.

428

12/RIVETED. BOLTED, AND WELDED CONNECTIONS

. Before applying Eq. (l2-6) we compute the value of Z x2 and 2 y2_ There are six rivets whose x coordinate is L5 in. and six rivets whose x coordinate is 4.5 in. Also, there 8IIC eight rivets whose y coordinate is 4 in. Therefore 2 x' + Z y2 = 16(I.5)2 + 6(4.5)21 + 8(4)2

::

263 in?

Applying EQ- (l2-6) gives the maximum components of the torsional load as

Ty 2.x2+ Zy2

] Z x2 + Z V2] To

(172.8

Ply

(172.8

p,,

x 103X4) = 2620 lb 263 x

10'X4.5) =

263

2950 lb

The hand ycomponents of the direct load On any rivet are found by dividing the x and _v components of the applied load by the number of rivets. Thus

p, n

Pd,

Pd, _ & _ n

and

I

28.8 X 103 = 2400 lb 12

38.4 10J x 12

I

3200 lb~

The most heavily loaded rivet is at A, where the fhaximum components of the direct and torsional loads are additive as shown. Applying Eq. (l2~7), we have

p, = =

l i

I I

v'(p.,, + pa + (pd, + p,.)= V(2400 + 262012 + (3200 .+ 2950)2 = 7950 lb

v

i

I

A ns.

PROBLEMS

1

1230. Compute the resultant load on the least loaded rivet in Illustrative Problem 1229. 1231. A gusset plate is riveted to a larger plate by four 2.2-mm rivets arranged and loaded as shown in Fig. P-1231. Determine the maximum and minimum shear stress developed in the rivets. 22.4 MPa Ans. M3X.'1 = 37.2 MPa, min. 'r

| I

n I

I

60 'mm r~£°**'°*1 I _.*.___~#'.__1*._ __*_ 80 mm L..., 'I '\

I

In

f

in

~

.

P

30 kN

o

.

Figure P-l231

i

-1 \

1232. In the gusset plate connection shown in Fig. P- l232, each rivet has a cross-sectional area of 0.50 in.2. The allowable load P was designed for a shear fig stress of' IO ksi. What will be the maximum shearing stress in the rivets if the rivet at A is improperly driven so that it cannot carry any load? s I

I a

i l

I

PROBLEMS

429

in.

a

I

.

in,

Q

Q

••

rn.

+-

.

.\

r w

x

~»jI~~» "'T .

-»--1-. -1-

-=>+.

-4-.

I

Ag

_V___Y__

I

J

Ip

Figure P-I232

1233. If the maximum load permitted on any single rivet in the connection shown in Fig. P-l233 is 3500 lb, compute the safe value ofP.

»

P = 8860k lb

A115.

1

I I I I

I I |

I

I s

r

-

p ,

" ~C J P :g J .

I

'

(=)' U

` ¢1

P w\.

I I

I

(c) Tordond force q

Direct force 7;

I

Part (a) is the vector Figure 12-21 Analysis of eccentrically loaded welded connections. . sum of parts (b) and (c).

r

y I

_

/"L

9 / 1 ?

c,/_az

.

`7

/

I-

x

I.i i

I

I

Figure 12-22 Evaluation ofJ.

directed along and perpendicular to its length. These values are respectively zero y Applying the transfer formula, we obtain, with respect to the cenu'o'd and ' . . C of the weld group J = 1-13 + LE2 =

[J = l + Ld21

-5/9 + L(i2 + J?)

Repeating this computation for .every weld in the connection and adding the results, we find that the modified J of the torsion formula becomes. II

*s

Z L ( ~ L 2 + P + ,We

Applying the torsion formula gives the torsional force q, acting perpendicular to the radial location p of any point on a weld as

l`

Q: i

~ 1L1+*"* _. ` 2+*1)

};L(i'i

(12-9)

12/RIVETED, BOLTED. AND WELDED CONNECTIONs

438

More useful are the following expressions for the components of q,, obtained 35 in the analysis of riveted connections: 91,

of,

2 M513 + P _

+ 52)

(12-10)

To

2 L(i%L2 + P

+ W)

in which x and y are the coordinates of any selected point of any weld. The maximum intensity of the weld force occurs at the point where Qu, and maximum q,,, as well as Qu, and maximum qly# are additive. Combining

I

these values vectorially gives q=

VUI4, + q.,)' + (nu, + a,,>'

(12-11)

Frequently this value of q is used to determine the size of all welds, but occasionally the size of each individual weld is based on the value of the highest stressed point in that weld.

ILLUSTRATIVE PROBLEM 1248. A plate is attached to the frame of a machine by two side fillet welds as shown in Fig. 12-23a. Determine the size of the welds tO resist a vertical load of 10 kips. Assume that the allowable shearing stress through the throat of the weld is 21 si.

p = 10 kips

Q.

P

qd

E

D

B

A

B

6 in

Q.

¢14

A

-4 1.6 in

3.4 in

\7

on

(b)

(=) Figure 12-23

|

The centroid of the weld lines, with respect to an origin at A, is found . ' . ' if = 2.60 in. (4 +6)x" = 4(2) + 6(3) [Lf-= Z/XI 1.60 in. l 07 =.4(4) [W=ZM v

II

SoluH6n~ to be

f

l

439

ILLUSTRATIVE PROBLEM

Using these values, we locate the centroid C as shown in Fig, I 2-23a. The moment of P about this centroid determines the torsional couple to be I

N

T = Pe = l0(2.6 + ) = 66.0 kip-in.

'

The modified .I for the weld group is the sum of the J's for each weld. Remembering that if and }7 are the coordinates of the center of each weld relative to the common centroid C, we obtain

J

L

(Q__ L2 + £2 + l

H

-12 + (0.41' + um'] = 34.3 in? (6)'

1,8 JDE

=

i

4{ 12 + (4)'

(0.6)'

+ (2.4)'] = 29.8 in."

Their sum determines the modified J to be

J=

Z J-= 34.3 + 29.8

:

64_1. in_3 s.

The components of the direct load are Go,

P .2 L

-

10 kips 10 in.

1.00 kips/in. T and

-0

Go, -

These values are to be combined with the components of the torsional forces at A

land E. These are the highest stressed points in_the welds AB and DE as revealed by an inspection of.Fig. I 2-23b. Applying Eq. (12-10), we obtain

2

l Q

Combining the direct and the torsional force components shows that the -highest stressed values in the welds are

[Q = i/(z QUO2 + ( Z '*)

in which x and y are the coordinates of any selected point of any weld. The maximum intensity of the weld force occurs at the point where Go, and maximum q , as well as Go, and maximum 1N

A

A `800

RD)

n

1 \ 2

RA=8OON

1 N 2 (c) Moment diagram

( b ) Dummy loading

( a ) Actual loading

of actual loading

Figure I3-10° II

50

f

B

A

L

Aim do + EI

C

Mm alt EI

D1U!72 C f x

+

(fol

EI

C

The various values of m are found from the free-body diagram of the frame (Fig. 13-IOb) in which a unit dummy load has been applied at D in the direction of the desired deflection. The conditions of static equilibrium determine the re~ actions to this unit load to be as shown. The values ofm shown are now determined from the definition of bending moment, which specifies that we may take moments about' an exploratory section of the forces lying to one side of the section. Substituting these values of m in Eq. (a) and noting that £1 is constant, we obtain C`

8

II

E150

f

XA M Dr

+

A

L

(3

+ ex¢)M Dr +

D

f

XDM

Dr

(be

C'

These integrals have the same physical meaning that we encountered in our earlier study of the area-moment method (page 196), and hence we may express them in the equivalent form END = (area),48-i4 + 3(area)B¢ + l(area)¢B~i'¢ + (a;ea)D¢-i>

in which, for example, (area),48~ x] represents the moment of area about A of the actual moment diagram for the segment AB. Observe that moments of area are taken about the point from which .X was measured in finding the various values

of m in Fig. I3-lo b. The moment diagram by parts of the actual loading is now drawn as in Fig. I3-lOc. Observe that no moment exists in the legs A 8 and CD; hence in Eq, (c) the first and last terms on the right side are zero. Evaluating Eq. (c), we now obtain i I

/60

3

•

-

2

•

0-0

from which

~u

+I

1600 X I

2

3

2

e

5/6n

1867 N-nl3

Ans.

_

I60€>x° 3

I

'Kill 7

13/SPECIAL TOPICS

456

n at D was If this result had been negative, it would merely mean that the deflectio opposite to the direction of the unit load at D.

PROBLEMS length L care/ing 1303. Determine the midspan value ofE[6 for a simply supported beam half. right its over to a uniformly' distributed load of` intensity rigid supports, 1304. As shown in Fig. P-1304, two aluminum rods AB and BC, hinged to has a crossrod each If lb. 6000 of load vertical a carry to 8 at are pinned together vertical sectional area of0.6 in and E = 10 X l0° psi, compute the horizontal and 30°. = 6 and 30° = a Assume deflections of point B.

Ans.

$h

0.016 fl

0.0023 ft, 6,,

H

I

L=loft

Bf \

P

E: 6 ft

I

C

Figure P-1304 and P-1305

a = 45° 1305. Solve Prob. 1304 if rod AB is of steel, with E = 29 X 106 psi. Assume . unchanged remain and 8 = 30°, all other data 1306. Use =Castig,lianols theorem to solve Illustrative Problem 202 (page 37). plane 1307. A circular bar is bent into the shape of a half-ring and supported in a vertical the and C point of nt moveme horizontal the Determine as shown in Fig. P-l307. vertical movement of point 8.

_

Ans.

66,,

PR3'1r 2/£1

7

580

pRo 25/

rd at C. i308. Repeat Prob. 1307 assuming that the load P is applied vertically downwa of plane the to icular perpend and C at 1309. In Prob. 1307, let the load P be applied load? the of direction the in C of ent displacem the be will What AEC. + PR' PR' A is.

JG

(g )

El

.)

'7

I !,

E

I I

1

I a

3

1

Q

F I

n

+

I

Fa.)

l

Figure P-1307, P~l308, and P-1309 v

PROBLEMS

457

1310. A vertical load P is applied to the rigid cantilever frame shown in Fig. P~l3l0. Assuming El to be constant throughout the frame, determine the horizontal and vertical displacements of points B and C. Neglect axial deformations.

68

Ans.

|

0, 55/l

_ 6 _ Pba2 . _ pi:2 c.

25/

1

$c.

E/

.

~-1

'C

B

P

a

Figures P-1310 and P21311,

131 l . In Prob. 1310, let the load P be applied at C and pemerzdicular to the plane of ABC Find the displacements of points 8 and C in the direction of the load.

A

'no.

Pub* + P la' 3EI

.L

ac = JG

u

53)

1312. The rigid frame in Fig. P-I3 12 is supported by a hinge at A and a roller at D and carries the loading shown. Assuming constant EI, compute E15 at the roller D; Neglect axial deformation. An5.

EIRE = 31.4 kn»m3

600 N.~'m

800

n---> B

1r

F

1

|

l

450 vi 1

|

.C

-I

:

n Q

I Ill

' l `

65:

n J.. I l l

1

600 !b

3

P

L

I

B E:

3m 5m 12 fc

A\ ' \ I

o

D

€€~79¥.;

_-1 o u'

_T

Figure P-1313

Figure P-1312

1313. The rigid frame in Fig. P-l 3 1 3 is subjected to a horizontal load and a vertical load as shown. Assuming constant EI and neglecting axial deformations, determine the value of E16 at the roller support D.

v

13/SPECIAL TOPICS

458

13-6 IMPACT OR DYNAMIC LOADING |

I I

The deformations produced in elastic bodies by impact loads cause them t o act as springs, although that is not their designed function. If the equivalent spring constant for such members is defined as the load required to cause a unit de~ formation, the spring-constant in each case can be determined from our earlier study of deformations. Actually. however, as we shall see, it is unnecessary to determine the equivalent spring constant. For the present, we may consider the problem of impact as analogous to that of a falling body stopped by a spring with constant if (Fig. 13-1 1). The mass m has zero velocity when first dropped.

m

i

-f *T*

1

6

+

I//

I7/7////' /

Figure 13-1 I

and also when the spring is deflected through the maximum dynamic deflection 5. Equating the resultant work done on mass m to the zero change in kinetic energy, we therefore obtain*

mg(l1 + al - go# = 0

(or)

where mg(h + 6) is the work done by gravity on the body and §k52 is the resisting work done by the equivalent spring. If Eq. (a) is rearranged in the form mg mg h= 0 5 2 - 2 -- ' - 2

If o

k

and milk is replaced by 5 , which is the static deformation produced by a gradual application of the weight mg, the following general value of 5 is obtained:

6 = BSl +

IliuM +

25,,h

(b)

Two extreme cases are of interest. If h is large compared with_5, we ma y neglect the work mgt in Eq. (a), which then reduces to

5

2 me n

l/265,/1

(6)

1

l In the other extreme case, h Eq. (a) reduces to

'

::

0 (i.e., the load is suddenly applied) and

The mass is assumed to remain in contact with the spring. Also, some energy is dissipated by the impact, so the actual deflection is always less than that given by Fq. (a).

459

6

II

13-6 IMPACT OR DYNAMIC LOADING

2 Q35

25,,

k

(d)

Because of a suddenly applied load, the deflection and consequently the stress which is directly proportional to it are therefore twice as great as that caused by the same load gradually applied. The ratio of the maximum dynamic deformation 5 to the static deformation 6st gives a value that may be called the impaclfactor. This is easily determined by rearranging Eq. (b) in the form

6=

6SI

+ ex.

5,,II+

Hence the impact factor is 6 =1+

1

+

(1 7)

5st

Multiplyi ng mg by this factor gives an equivalent impact load P, which may be used in the formulas for static loading to compute the maximum stress and deflection . Or, if preferred, the static stress due to a gradual application of mg may be multiplied by the impact factor to give the maximum stress; Ural.

Us!

II +

1+

2h

(13-8)

W e now apply these results to various types of impact loading. Tension

The most usual type of impact loading is shown in Fig. l3~l2. A mass m drops freely through a height h before striking a stop on the end of the rod, thereby producing the dynamic deflection 6. Assuming that the stresses remain within f°».»/¢v~/{,'{/»'/

'I

I

I ! I l

L

w

Figure 13-12 Impact loading of a. rod.

no

I

f:::1

.J

I

T, i

the elastic range and that é is negligible compared with iz. we replace 6Sl in Eq. 5st = moL/AE, from which we

(c) by its value from the deformation equation obtain

.p

13/SPECIAL TOPICS

460

I i

,lAw

-

ZL

-100 mm*-I f

I

/

,,._q8=5646

n/m

B _IDJ.>.A

4e

ii;

-4

"I

250 mm

NA

. `|

NA

_go

NA C

I

I

1

-4-l=2.5 mm

1.lr

V,=2000 N

V*=2000 N II

=5

I

I

*Q*-

I

n

ac ='9l'15 N/[n \

H=282 N

l

Hz 282 N (c)

lb)

(8)

Figure 13-26

Solution: The moment of inertia about the neutral axis is computed from

P

2 eh' +Ad-,

)]

(

1

(2.5)(250)3

+ 2(2.5)(100)(I 25)2

12

x 106 mm" 11.07 x 10" m4

11.07

Because of the thin wall, the shear flow may be assumed to act along the centerline ABCDE of the section, as shown in Fig. I 3-26b. At A, the shear flow is zero, at 8, Eq. (5-4a) gives QUO

V 2000 = -IQAB = 1I.07X 10-6(2.5x 10-3)( 100 X 10-3 ) x ( I 2 5 X 10 -3 ) = 5646 N/m

The shear flow at any other point C may be found directly from Qc = (V/I)Q,4€. However, since $46 is the moment of area from A to C, which is equivdent to the sum of the moment of area from A to B plus that from 8 to C, we may write Qc = (V/IXQB + Q8€-), which then reduces to the more convenient form Qc = is + (V/I)Q8¢. Thus the shear flow at C is' V

2000

§ c = ¢ 1 s * ' ' Q a c 6 5646 + 11.07 X 10-6 (2.5 X 10"3) X (125

x

(

!0-3)

125 X I0-3

2

)]

=,5646 + 3529 = 9175 N/IN As shown in Fig, I3-26b, the shear flow from A to B varies directly with the distance from A; but from B to C to D it varies along a parabolic arc. The average shear flow in the web is therefore 5646 + §(3529) = 8000 N/m. The shear force with the "or"" *u* applied in the web is IC. == gaveL = 8000(0.250) = 2000 N, which agrees vertical shear V == 2000 N. The shear force in each flange is H = q,vi L ==

(l X 5646X0.l00) = 282 N.

-.

471

PROBLEMS

To avoid twisting of the section, the external shear V must lie a distance e exerted by the internal to the left of' C' (Fig. I3-26c) so that the twisting moments Hence balanced. be shear forces will l

[ 2 mC =:: 0]

2000e

28212501

e = 35.3 mm

The value of e can be computed more easily from Eq. (13-9):

e=

h2b2r = l250)2(100)2(2.5l = 35.3 mm 4(1 1.07 x 10°) 41

However, the above numerical computations are presented in order to emphasize the principle of shear flow and to indicate its extension to the more complex problems that follow.

PROBLEMS 12 = 1322. Locate the shear center for the section shown in Fig. 13-22 (page 468) if/1 = mm. 200 = /13 mm, 140 = /12 mm, /3 = IO mm, h, = 180 1323. Determine the position of the shear center for a section composed ofa thin-wailed cylinder of thickness I and mean radius r which is split along one longitudinal element, as in Fig. I3-17c. ,4 ns. e = Zr measured along the axis of symmetry from the center ' of the cylinder in a direction opposite to the split element. 1324. Show that the position of the shear center for the semicircular thin ring in Fig. P-1324 is e = 4r/1r to the left of O.

O

1

Figure p-1324

1325. The thin-walled section shown in Fig. P-1325 consists of a semicircular ring of mean radius r and t.we straight pieces of length r. Show that the shear center is e = (/r"/I)(7r + 3) to the left of O, and hence for r = 50 mm and I = 2.5 mm. that e = 86.0 mm. Need the value of! be specified"

r

I

! to I

Figure p-1325 s

13/SPECIAL TOPICS

472

1326. If the vertical shear on the section shown in Fig. P-1326 is 3600 lb, construct a shear flow diagram and locate the shear center.

-4 in. ._ .I

up

= 0.10 in.

u

8 in.

I T+ I

2 in.

I

Figure P-1326

1327. If the vertical shear on the section shown in Fig. P-1327 is 3000 N, construct a shear How diagram and locate the shear center. r Ans. QB = QE = 3.44 kN/rn, Qc = Go = 12.1 kN/m, e = 44.7 mm to the left of web center on MM-

+

4

_

CL

x

40 mm

RI

1

A 250

*-10 mm

0-

mm

3

i

F

•

40 mm D

I

Figure P-1327

1328. Locate Lhe shear center for the thin-walled section shown in Fig. P-1328. Ans. 1.05 in. to the left of the web

P*-6 in.

.

l

In

6 in.

.

f

~"'

l

¢

%+

3 in.

NA (II)

I

llllllulllh

i

1+

3 in,

I

Figure P~l328 6

1329. Show that the shear center for the thin-walled split square section shown in Fig. P-I 329 is b/(26) to the left of the corner opposite the split.

473

13-9 UNSYMMETFNCAL BENDING

NA

mm

NA

mm

Qo mm . Figure p-1330

Figure P-1329

1330. A thin-walled section has the shape shown in Fig. P-1330. The outstanding legs have a slope of 3 vertical to 4 horizontal. Locate the shear center. Ans. 21.8 mm to the left of web

13-9 UNSYMMETRICAL BENDING

The theory of flexure developed in Chapter 5 was restricted to loads lying in a plane that contained an axis of symmetry of the cross sectionfWith this restriction, the neutral axis passes through the centroid of the section and is perpendicular to the plane of loading. The preceding article extended the application of the flexure formula to sections with only one axis of symmetry that were loaded so that this axis became the neutral axis. In either case, bending without twist is possible only if the plane of loading contains the shear center, a requirement that is automatically satisfied when the axis of symmetry coincides with the . plane of loading. There is a further restriction that SO far has been adhered to, although not emphasized. The plane of loading must be parallel to or contain a principal axis of inertia of the beam cross section. We first consider the case in which the plane of loading contains an axis of symmetry, such as the y axis in Fig. 13-°?. in deriving the flexure formula (Art. 5-2), we applied the condition 0-feqtiilibrium that the applied bending moment about the x axis is balanced by the resisting moment exerted by the flexure stresses, that is, M , = Ytft ci4). If the bending loads are restricted so that they lie in the longitudinal plane containing the y axis, the external moment My must be zero.* However, the flexure force O' dA on a typical element of the section has a moment x(cr act) about the y axis. If the y axis is an axis of symmetry, this moment about y is neutralized by an identical force (not shown) acting through the point of symmetry. For sections that do not have an axis of symmetry, the resultant moment of the flexure forces about the y axis is

/

'

When the section has only one axis of symmetry that becomes the neutral ates the plane

of loading must be offset from, but parallel to, the longitudinal centroidal plane seas to pass through the shear center; but even then M, is zero. LE

13/SPECIAL TOPICS

474 Longitudinal plane of bending loads

_;p ,-B' :_

. xx." . ~lI . . \\ _

' .

-

g"

\...

\.

'

*\

°§"

ii(

=i,. e

**u\s§*.

a S' -.~~t.;.g .

.3

./" i I

I

0

_,/

0 d A =E - _}'dA p

Figure 13-27 Flexure stress causes resisting moment about y axis as well as about .r axis. axis if there is no axis of symmetry).

f

A°(a d.~1)

II

m,

E

v dA (-.) p

x

II

m., = 0 if y is an axis of symmetry (or a principal

E p

by

do

I

Therefore 1*//,_ will be zero and equilibrium satisfied only if the integral by dA is zero. This integral is the product of inertia Pxy, which is zero only if x and y are the principal axes of inertia of the section. We conclude that the flexure formula may be applied only if the bending loads act in a longitudinal plane parallel to or containing one of the principal axes of the section. These planes are called the principal planes of bending. We are now ready to discuss zuzsymmetrical bending, which is defi ned as bending caused by loads that are inclined to the principal planes of bending. Examples of unsymmetrical bending are roof purlins that, because of the inclination of the roof, are subjected to loads whose planes make large angles with the principal axes of inertia of the section, and beams in structures and machines that are subjected to loads that, because of deformation or design, are inclined . to the principal planes. We consider first the case shown in Fig. 13-28, in which a symmetrical section is subjected to loads inclined to the axes of symmetry. Resolving the loading into horizontal and vertical components, we obtain the two loading conditions shown in parts (b) and (c), which can each be solved directly by the flexure formula. In part (b), the x axis is the neutral axis, whereas in part (c) the y axis becomes the neutral axis. Each of these conditions produces flexure stresses that are normal to the cross section, hence the resultant stress at any point is the algebraic sum of the stresses at that point caused by each case considered separately, that is M,y Myx + (13-ll) 0 I 1;

_

v

where M, is the bending moment about the x axis caused by P cos 0, and My is the bending moment about they axis due to Psin 6. In terms of the total bending

13-9 UNSYMMETRICAL

BENDING

4/o

y

VP

)'|

/I - A.

\\

I

!

!

` l

Yi

x

In

1I

\

I

cos 6, I

II

I

v

I

I

>
ydA =0 odA (c) d8 R-e+y

f

13-10

cunveo

483

BEAMS

Since E d¢/d6 cannot be zero, we obtain V rL4

f R - e + y*- 0

(d)

'

in which e is the only unknown. Its value may be found from Eq. (of) by letting v denote the distance from the axis of curvature to the element (Ll. Then y = v = (R - e), and Eq. (cl) is rewritten as

e + _v

f

II

fR

ydA

v-(R-e)d4=

f

v

d.4-(R-€)

dA U

=0

from which we obtain

A

(13-13)

n 1'

Equating the applied bending moment to the resisting moment gives M

f

' DA

yo'

E do)

== --dl)

}.2

R

-

fig e+y

Fe)

This integral is simplified by adding and subtracting (R - e) to one of the two y`s in' the numerator so that y = ( R - € + y ) ( R - e). The integral is than rewritten as

fR

.v2 DA

WIN - ( R

e+y

f 5:

.1' ci-I

1)

e+ y

of)

The first integral on the right side of Eq. (f) is the moment of the entire cross-sectional area about the neutral axis and equals As. The second integral, from Eq. (d), equals zero. Equation (e) can now be rewritten as

E do d6

M .48

This value onE d¢>/(16 is then substituted in Eq. (Iv) to yield fmallf. O'

M AeR

y e+y

My As v

U3-14)

Equations ( l 3- l3) and ( l3-14) are theoretically adequate to determine the stresses in curved beams but are limited in usefulness by the ditlieultyof computing the value of' e. This difficulty may be avoided by means of a study made by B. J. Wilson and J. F. Quereau.* These investigators computed the extreme fiber stresses in curved beams of various cross sections with the curved beam theory and with the ordinary flexure formula. From a comparison, of these results Wilson and Quercau determined values ofa correction fetor K by which stresses

• See "A Sxmplc Method of Determining Stresses in Curred Beams," Curculur 16, Engineering Experiment Station. Univcrsuty of' Illinois, Urbana. Ill., UPS.

13/SPECIAL TOPICS

484

computed with the flexure formula can be multiplied to give the actual stress in a curved beam. A modified equation for computing the extreme fiber stresses in curved beams is therefore

K McI

0

(13-15)

Values of K in Eq. (13-15) vary with the ratio R/c, where R is the radius of curvature of the centroidal axis and c is the distance from the ccntroidal axis to the inner fiber. As Fig. 13-33 shows. these stress correction factors are greater than unity for the inner fibers and less than unity f`or the outer Fibers. A t values ofR,i'r greater than ')0). these factors approach unity and the flexure formula may be applied directly to such slightly curved beams. Table l 3-2 lists correction factors for various cross sections.

TABLE 13~2

CORRECTION FACTORS K FOR CURVED BEAMS

For USE IN EQ (1345) .......-...

r..-

I-1111l1l1n1

q

I

+

1.,€/

||

l

I

I

/

I',/ ./ y;;>`

I T Cross sectloru

a

g

E

I

>.-\`

N

s

1.1 l -2 1.f~ 1.8

_

".0 3.0 4.0

6.0

inside

5 l

a

140

1.96

1 I |

5

Circle or ellipse

1

1

I _r

1 _|

i Rec

~i

|

l

5

I 5

TO e I

c

1

1

1

I I

i

|

I

1'%/ ,


C

e

(be A

Figure 14-5 Unloading and reloading of(a) actual ductile material and (b) elastic-perfectly plastic material. In part (b) the unloading and reloading lines actually coincide but are shown slightly separated for better comparison with part (a). *

The principal effect of unloading a material strained into the plastic range create a permanent set. If this permanent set is restricted in any manner, there is created a System of self-balancing internal stresses known as residual stresses; The magnitude and distribution of these residual stresses may be determined by combining the stress pattern (partly or fully plastic) caused by the given loading with the stress pattern created by a load equal to but of opposite sense to the original load. The effect of applying a load equal but opposite to the given load is equivalent to unloading the structure. This unloading creates a stress pattern that is assumed to be fully elastic, as indicated in Fig. 14-5. However, this method of superposition cannot be used if the residual stresses thereby obtained exceed the yield stress. As a first example of residual stress, we consider a circular bar strained into the fully plastic state by the limit torque. As we saw in Art. 14-2, the limit torque is § the yield torque, and the stress distribution is as shown in Fig. I 4-6a. To unload the Ba we now apply an opposite torque as shown in Fig. 14-6b. Recalling that the unloading is assumed to be elastic, we obtain the stress disis to

14/INELASTIC ACTION

502 4 yp T-5T

TL " éfyp

T-0

-

>-

;7

II

h

>-

r>

7

a up

(a) Plastic loading

3 'in

1

'§ Up (c) Residual stresses

(b) Elastic unloading

Figure 14-6 Residual stresses in torsion.

tdbution shown. Superimposing the loadings and stress patterns of parts (a) and (b), we obtain the unloaded bar with the residual stress distribution shown in ` part (c). An interesting phenomenon of residual stresses is that the bar now behaves elastically. if the original limit torque is now reapplied, as shown in Fig. 14-7. Combining parts (a) and (h) of Fig. 14-.7, we obtain the Original plastic state shown in part (c). On the other hand, with reversed reloading, the residual stresses

.

4

4

T-0 in

1 'l

.

.

4

II

1

4

4

u

Top

._

'n

r

I / '

i3f

1 UP

(a) Elastic stresses

Top

a 71p

°"'

(b) Residual stresses

(c) Fully plastic

Figure 14-7 Original torque ,reapplied to part (c) of Figure 14-6.

are unfavorable and no more than two~thirds of the yield torque can be reapplied in the opposite sense before the bar reaches its plastic limit, this is shown iN Fig. 14-8. Evidently, further plastic yielding occurs if the sum 'of the original and reversed loading exceeds twice the value of the maximum yield torque. Under these conditions, the additional yield that occurs after each cycle soon results in J

14-4 RESIDUAL sTRessEs

T

Q

503

2 --T a yp

Z

T-0

.L

~-5 n.

l v

s

-ETyp

1yp

(b) Residual atreases

(a) Elastic stresses

(c) Start of yielding

Figure 14-8 Effect of reversed loading on pan (c) in- Figure 14-6.

rupture, as we know from bending a strip of metal back and forth plastically for a few times.* As a second example of residual stress, we consider a rectangular section of a beam that is strained into the fully plastic state by the limit moment. For the rectangular section, the limit moment is 22 the yield moment and produces the stress distribution shown in Fig. 14-9a. Releasing the load is equivalent to adding the equal but reversed loading of part (b) to pan (a), this results in an unloaded bar having the residual stress distribution shown in part (c). Remember that the unloading is assumed to be elastic, which produces the linear stress distribution shown in Part (b). Observe that although residua]_stresses are selfbalancing, if some of the material is removed, an unbalance's created. This explains why members that are cold-formed distort amer machining. '

M-

3

5 Map

> > >

»

¢

-t

-as

M-0

11

h/2 or

n

II

S

h/2

i

-Q

"in (a) Fully plastic

L

3

"

E "yr

(b) Elastic unloading

"

2 "in

(c) Residual stresses

Figure 14-9 Residual stresses in flexure.

As we saw in the case of torsion, a beam that has been unloaded from the fully plastic state may be reloaded in the same sense, the beam now remaining elastic until the limit moment is reached. For reversed reloading, the beam also * For further detail, see J. A. Van den Brock, Theory of Limit Design. Wiley, New York, 1948, especially pp. 23-25.

14/INELASTIC ACTION

504

remains elastic, the only restriction being that the sum of the original and reversed loads must not exceed twice the maximum elastic moment My if further yielding is not to occur. Since the limit moment is aM,,,, this restriction makes the max4 imum reversed moment by, for rectangular sections. of bending a effect the As a final example of residual stress, we consider When the I4-lOa. Fig. in shown as die circular a about bar straight rectangular bar is released, it springs back through the angle 65 as shown in Fig. 14- lob. This springbok angle is of great importance in metal-forming operations. Its value and the relation between the forming radius of curvature RO and the final radius of curvature Rf c3l1 be found by combining the plastic strain caused by loading with the elastic strain caused by unloading. This procedure duplicates that used ` to determine residual stresses.

»~

IN

.la

h

I

s

+

+

___

\

\

'\

l

Die

R/ Y

\

A

I

6, in)

all

'

0/00

(b)

Figure 14-10 Springback.

As mentioned in Art. 5-2, the strain in bending is e = y/p, hence at an outside element of the bar, the initial pMtic strain in part (a) is

n/2 £0

RE

and the final residual strain in part (b) is h/2 of:

R/ Unloading the bar is equivalent to applying a moment opposite to the limit moment to the -deformed bar in part (a). As seen in Fig. 14-9b, the maximum stress in this unloading is §'a,, and hence the corresponding elastic strain is f.

'

0'

::

-- =

E

3/Z0 yp E

Superimposing these strains gives the residual strain as s

6/

or

Fu

-

e

n/2

h/2

_ 3/Zan,

R/

IL,

ii

505

ILLUSTrATIVE PROBLEM

which reduces to

1.

I

R,

30" Eh

R.,

(14-6)

The angular change 6/ caused by the final radius of curvature R/is found . ' by applying ds = Rf do, rewritten as

I --d .s Rf

d0 and then integrated to give

I'

of

f

-

.5

0

Since 1/R/ is constant and the length ROflo, we obtain

1

- ds R/ of the bend as shown in pan (a) is 5

S

_ 6f = RE _ 90

Rf

The springbok angle may now be evaluated as H.,

0O

"

9/ =

a

I

of/

from which, by using Eq. (14-6), we finally obtain 6,

6,

RE

f x4-7)

This result indicates that the relative amount ofspringback may be reduced by using a smaller forming radius, or thicker bars, OI' material having a low yield strain of,/E. It also indicates the amount by which the forming angle 3cz must be modified to produce a final bend of a specified amount. In circular bars twisted into the plastic state, springback also- occurs alter the torque is removed. In this case, the elastic springbok is equal to the angle of twist caused by elastic unloading. ILLUSTRATIVE PROBLEM 1414. The outer bars in Fig. 14-1 1 are of 20»2¢FT4 aluminum alloy for which 5rn=: 330 MPa, the center bar is of steel for which "iv = *9O MPa. The cross-sectional area

of each aluminum bar is 600 mm2. for the steel bar; it is 900 mm1. l f p is the limit load, determine the residual stress after P is remove. Assume that the bars are securely attached to the rigid end plates and that lag, = 70 GPA and ESI =.200 GPa. Solution: The limit load is the load~al which yielding begins in both materials. Its value is

[P

X A "vol

p = 2(600

l0l°X3-30 x I 0°) + (900 x 10-°x290 x 10°)

_ [°,. = 54.7 k N

PROBLEMS

S07

1416. The bar shown in Fig. p~1416 is firmly attached to rigid supports. The yield strengths for steel and aluminum alloy are, respectively, 42 ksi and 48 ksi. Determine the residual stresses if the limit load is applied at P and then removed. -10 in.

-1-

\

Aluminum

is

in.-

P Steel

E= 10X 106 psi e=»29 X 106 psi Figure~P-1416 and P-1417

A = 1.5 in?

A = 2 in.2

1417. Solve Prob. 1416 if a load P = 132 kips is applied and then removed.

.

Ana. Us = 7.40 ksi, J, = 5,55 ksi

1418. A sandwich beam is made by bonding Q-in. strips of 2024-T3 aluminum alloy between layers of foam plastic to form the section shown in Fig P-14 18. The foam plastic acts only to separate the aluminum strips, its effect on bending resistance is negligible. A moment of 10 kip ft is applied and then removed. Determine the residual stresses if "iv = 44 ksi.

-

_

_

Ans. ` $4000 psi on outer strips, £8000 psi OD 1R1'1€1' strips

121»1_1_ r

.'r re" "A*

.g

- ,o.*l

• ,?"» . 1

.'

as - ' °.•-5.9®o~ o r * »*~ r . s*"' a -2

I

o» ._ o** : I -

lai:$l

o

T

~*.°~°° »4»-

.lv

3'r*1 1 - ' • .,

°"U°»

'-*;

-Q,

2 in.

T

Figure P-1418

1419. The outer diameter of a hollow circular shaft is twice the inner diameter. Determine the residual stress pattern after the limit torque ha been applied and removed. 1420. The torque applied to `a solid circular bar of radius r causes the elastic region to extend to one-halfthe radius. Determine the residual stress pattern alter the torque

is removed. What will be the residual angle of twist? .

1421

Ans.

a, =

gi-f,.,/Gif

. A 200-mm length at each end of a straight shalt l rn long and 10 mm in diameter is bent at right angles to the shaft. Determine the angle through which one of the bent ends must be twisted relative to the other so that they will be exactly 90° apart after the twist load is removed. Assume that TYD = 120 MPa and G = 80 • GPa. 1

1422. A rectangular bar 50 mm wide by 90 mm deep is subjected to a bending moment

u

that makes 80% of the bar plastic. lf` aw = 260 MPa, determine the residual stress . pattern after the moment is removed.

1423. ' A rectangular bar 30 mm wide by 60 mm deep is loaded by a bending Moment of 6 kN m, which is then removed. If "in = 280 MPa. determine the residual stress at 20 mm from the neutral axis. (Him: Refer to Prob. l4l3.)

-

' u

Ans.

a, = 15.1 MPa

1

s

508

14/INELASTIC ACTION

1424. If a beam of unsymmetrieal section is loaded into the fully plastic state, show that any release of the load will cause a residual stress in excess of the yield stress which, being impossible, means that the theowr of elastic unloading cannot be applied to

beams of unsymmetrical section.

1425. The section shown in Fig. P~l425 is loaded with a moment that causes the elastic region to extend for 1.5 in. from the neutral axis. If "re residual stress pattern after this moment is removed.

::

40 ksi, determine the

3 in. l \ l \

\

\ l

3 in.

4 .

!

2 in

Sin.

Figure P-1425 1426. A sheet of steel g in. thick is bent over a 90° arc of a circular die go in. in radius. If Uya = 40 ksi and E = 29 X 10° psi, determine the residual radius of curvature. or, An5. R = 4.28 in.

1427. Determine the angle of contact with the circular die so that the sheet described in Prob.. 1426 will have a permanent bend angle of 90°. . 1428. A circular die with a radius of250 mm is used to bend a 2024-T4 aluminum alloy plate 10 mm thick. Determine the angle of contact so that the plate will have a. permanent bend angle of l80°. Assume that "in == 330 MPa and E = 70 GPa. . Ans. 0o 282°

14-5 LIMIT ANALYSIS

u

We now consider the application of limit load, limit torque, and limit moment to the analysis ollstatically indeterminate structures. This procedure, known as l i m ! anal)/sis,* is the method of determining the loading that causes actual collapse of the structure to impend or results in excessively large deformations. It is applicable only to ductile materials, which in this simplified presentation are assumed to be elastic-perfectly plastic (see Fig. 14-1b). The method is surprisingly simple, since it consists of only two steps. The first step is a geometric study of the structure to determine what part or parts of it must become fully plastic to permit the structure as a whole to undergo large deformations. The second step is an equilibrium analysis to determine the external loading that creates such localized fully plastic Parts. - . One example of limit analysis applied to axially loaded members has already

'

. See J. A. Van den Brock, Theory of Limit Design, Wiley, New York, 1948 l`orla concise Justification of the principles and an extended application to redundant beams and other structures.

I

509

14-5 LIMIT ANALYSIS

been presented in Illustrative Problem 1414. As another example, consider the rigid beam in Fig. 14- l 2 which is supported by two steel rods of different lengths. In the elastic solution, the deformations of the rods are proportional to the distances of the rods from the hinge. This condition results in one relation between P.4 and PB, after which the equation of static equilibrium Z MR = 0 can be applied to determine the maximum load that will not overstress either rod.

B 2m

L=3m

*PA 2m

2m

5>

~.;°

Ju.u.

lm |

I

1-2 m 4-

2m

>ln1 m>

s

w

W

R

l

Figure 14-12

In limit analysis, however, the 'capacity of each rod is determined be the load'at which yielding begins, that is, P = Aaw.-Thus if we assume the areas of' the steel rods A and 8 to be 300 mm2 a.nd 400 mm2, respectively, and the yield stresses to be 330 MPa and 290 MPa, respectively, a moment summation about the hinge determines the Maximum value of Wto be

[Z

mR = 01

5w'= 2(300 X 10-')(330 X 106) + 41400 X 10")

c

/

/`

Aluminum

Steel -

I

I I

I1 h

_\\\\\\

r' .r

Ii.. ,_

_

s

-

-

;

-

../

f

. I

_z

'

/1

d

Q\

\\\

/

50 mm die.

70 mm die.

G = 28 GPa 'in

:

160 MPa

T

/ /° / /'

G = 80 GPa Top : 140 MPa

To:

Figure 14-13 1 I

510

14/INELASTIC ACTION

Since We are assuming the materials to be elastic-perfectly plastic, remember that yielding is assumed to progress unirnpeded after the yield stress is reached. While either segment remains in the elastic range, angular deformations will -remain small. Hence excessive yielding will not OCCU-I' until both segments have reached their limit torques. The limit torque is I

.

4

f

T O = T y p

3

11'/°3

)

2 To

Hence a moment summation about the axis of the shaft gives 4 .~ 4 'II' 3 s I T §(;)(0.035) (160 X 10 ) + )(0.025) 3(140 x 106) [*;T=0] 2 II

(

18.9 kN-m

Arzs.

For comparison, if both materials must remain elastic, the maximum torque will be reduced to 9.60 kN in. Finally, we discuss the limit analysis ofbearns. If we reconsider the cantilever beam in Fig. 14-3 (page 498), the most highly stressed section at the wall changes successively from fully elastic to partly plastic and finally to fully plastic as the load is increased. Other sections, such as b-b, become partly plastic, to the left of section a-a the beam remains fully elastic. Only the wall section approaches the fully plastic state because, although ductile yielding has begun between sections a-a and c-c, the end of the beam will not deflect uncontrollably as long as the wall section can absorb an increase in bending moment. Once a section becomes fully plastic, all its fibers yield without further increase of stress, thereby permitting the parts of the beam on either side of this section to rotate relative to each other.* For this reason, a fully plastic section is called a plastic hinge and the bending moment that creates it is assumed to be the limit moment ML. Collapse of a statically determinate beam is considered to be synonymous with the formation of a plastic hinge. The outline of a beam when uncontrolled deformation can occur is called a collapse mechanism. Several ecamples are shown in Fig. 14-14. In each case the dashed outlines represent the collapse . mechanism. =in general, a plastic hinge will form at a section of zero shear, that is, where the bending mornentis a maximum. The location of plastic hinges is therefore obvious for beams subjected to concentrated loads and reactions. For indeterminate beams carrying distributed loads, the location of plastic hinges is more complex. Sometimes more than one collapse mechanism is possible, in which case we must compute the limit load for each possibility and then use the smallest limit load. These concepts are discussed in the following illustrative

-

problems.

4

"

'

'

'

We exclude strain n-hardening by assuming the material to be elastic-fpcrfcclly'plastic.

511

ILLUSTRATIVE PROBLEMS

P ¢

I

N

s

°5 *v

`~

'~-x

.1

¢*'

a* ,»

*

1'

p*

132

R1 P

.r

I

ad

l

~:~. "~"t\

. r 1 "

Q

¢ /

J '

\:'¢"

\.\

H,

R1 A

P D

II

\\

\l

»PA

/I

\\ \\ \\

H

. n/ I

Figure 14-14 .Plastic hinges H form at sections of maximum moment.

is:

1'

ILLUSTRATIVE .PROBLEMS 1429. A beam perfectly restrained at the ends carries a uniformly distributed load 3" intensity.w0 over its length, as shown in Fig. 14-15; Determine the limit Egad and compare it with the maximum elastic load. "

l l l l l l l l l l l l l l I lH - I

MA

C

I - '

U

_*E

.\

D

MB

L

al:

1

2

B

2

VA

V8 Figure 14-15

Solution' From symmetry, it is evident that sections of zero shear and of maximum moment occur at midspan and at the ends. The collapse mechanism shown by dashed lines will occur when plastic hinges Nomi at .4 8, and Cas the moments at these positions each approach the limit moment Alf, The redundant supports V,4 and M can now be found by applying the equations of static equilibrium to this collapse mechanism. However, an equivalent and preferable procedure is to apply the definition of bending moment. I

Q

512

14/INELASTIC ACTION

Since the values ofM,,. Mg, and MC all become ML. but the sign convention of bending moment makes M and M, negative, we obtainQ (Q)

VOL - ML

(WQL)

'

(b)

from which

v,

weL

and

2

ML

LE

l4'0

(c)

16

The elastic solution for this case, previously solved in Chapter 7 and listed in Table If'-2, gives M, = -v.i0L2/12. Ignoring the Minus sign (which merely means that the end moment was originally assumed to be positive) and letting MA be the maximum elastic moment My and w, be the elastic load, we obtain

My, =

weL2

o

2»

(d)

l2

On dividing (c) by (d), we find that the ratio between the limit load and the elastic

load is

f

Wo

4 it.

We

3 Map

Ans.

The ratio M.t_/M>, (listed in Table I4-1) is appreciable for rectangular or circular sections, but For structural sections it is so close to unity that ML is usually taken as equal to My'

.

_

1430. A propped beam carries a distributed load of intensity Wo over its length, as shown in Fig. I4-l6a.Determine the relation between the limit load and the limit moment.

Solution: The collapse mechanism is shown in Fig. l4-l6b. The location of the plastic hinge a C is unknown, but it may be found from the condition that when the moment at C is a maximum, the vertical shear is zero. This condition was developed from statics (see Art; 4-4) and is valid whether or not the stresses or suainsare in the elastic range. Drawing the free-body diagrams of the beam segments as in Fig. I4-l6c, where the moments at C and B are the limit moments, we have from the segment AC J

[MA = ( Z MM]

0 = M1.

W0X2

(cl)

2

and from the segment CB

[Ms

::

(2 Mk]

"ML=ML

Wo

2

(L

x)2

(D)

* It is unlbrlunate that there are so many L's in Eqs. (a) and (b). Because of the plan of using the first letter of a word as a symbol, L here has three meanings: it stands for /of! in ( Z M),., for l i m ! in ML, and for length of beam. We hope this explanation will eliminate confusion.

' See Van den Brock, Theory of Limil Design. p. 39, I

513

ILLUSTPATIVE PPOBLEMS

Wo

Allllllllll

r

1y

4

Ms

\ \.

m r

,QL

pa.

V8 '(a)

I.

L-x

x

AS v

Wo l

A1

I

) ML

CO

I

x

-I

>

v-0

RA

wo

`

-4

I-~1 ' l

'v

Y0

L,

0.264r

1l)

low

low

m

,

run

Mug

L-8 31am

WWW

'uoqsna %;~0 Waals

P3\[O.l lOLl

'uoqaea 't)Z'0

'uoqnza we P31101 Flo:>

pallo.x

`LlOQlUJ "uv8l0

pa\loJ

UOJI ISO.)

uoJg rosco °lq1==112w

.LSU .Z

0Zt>

OZZ 09 OIZ

0zg

of I 058

0§i'

ozv

81

08

06 Oss

p

Off

q

069

oz

ova

so

OZ SE

IL 01. 061 01. I

of

q

0§S

OS I

UOlSU8_L

'dwog

oz

OOI OOZ

go;

q

of

Jeaqg

Jeaqg

(How) uxbuens 6m8wmn

Jvdw) Iii-UH luuoluodoad

p

ooz

ozv

q

uogsuaJ,

3 'uolsuaJ_ .(2d'o)

9 ueaqg

(we of up) UOll96UO|9

*O 968)U80J6d

Azwsxsvla ;o srqnpow

d BSVUEAV S.LINIll IS -°S'W.1.3V\l NOWWOO :JO S3LLH3dOUd "IVOlSAH

552

490

490

palloJ low{

Weight. (lb/f1°)

25 20 38

55 33 55

Shear, G

120 20 54 50 13 56

Modulus of elasticity (psi)l

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The ultimate compressive strength for ductile materials may be taken as the yield point, which is slightly greater than the proportional limit in tension. c Not well defined, approximately 6 si. 4 Castiron fails by diagonal tension.

A

limit and modulus of elasticity for compression may be assumed to equal these values for tension except for cast iron where proportional limit

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