MECHANICS OF MATERIALS. A Friendly Approach 9789811248450, 9789811248467, 9789811248474

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MECHANICS OF MATERIALS A Friendly Approach

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MECHANICS OF MATERIALS A Friendly Approach

Prashant Kumar

Indian Institute of Technology Kanpur, India

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Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Names: Kumar, Prashant, author. Title: Mechanics of materials : a friendly approach / Prashant Kumar, Indian Institute of Technology Kanpur, India. Description: New Jersey : World Scientific, [2023] | Includes index. Identifiers: LCCN 2022017127 | ISBN 9789811248450 (hardcover) | ISBN 9789811248467 (ebook for institutions) | ISBN 9789811248474 (ebook for individuals) Subjects: LCSH: Materials--Mechanical properties. | Strength of materials. Classification: LCC TA404.8 .K86 2023 | DDC 620.1/12--dc23/eng/20220624 LC record available at https://lccn.loc.gov/2022017127

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright © 2023 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

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Preface

I recall the days when I was a first year undergraduate student of engineering. During various discussions, I realized that my seniors found the field of mechanics of materials difficult. However, my classmates and I were lucky as the course was taught by a professor of MIT, Cambridge, USA who was visiting our Institute. He taught it so well that we found the subject very logical, interesting, and elegant. Probably, my seniors were not taught this subject properly. This introductory book on the mechanics of materials is the very foundation for advanced topics in mechanical, civil, aerospace, chemical, ceramic engineering, and materials science. The book provides a clear and in-depth understanding of the concepts involved. It explains the fundamentals with many illustrations and gradually increasing difficult levels. The currently available books in this field have not yet updated the contents to meet the modern requirements of computer-aided engineering. The available software packages have the capability to analyse sophisticated structural members with complex loading. A designer should be able to leverage the concepts of this field and the capabilities of the software packages rather than feel constrained only to simplistic analysis. Most books still deal with solving simple problems of two dimensions because the analysis can be managed with only three components of basic variables, stress, and strain. However, stress and strain are second-order symmetric tensors each with six components and cannot be treated like vectors with three components. In this book, stress and strain have been gradually developed in three dimensions; they are expressed with 3 × 3 matrices. This v

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enables designers to think in three dimensions and make good use of the available software packages. The language of mathematics is very concise and compact with its many symbols. To tackle it effectively, especial care has been taken in this book to carry out the algebra in small steps. This enables a reader to focus more on understanding the concepts rather than figuring out how to move from one equation to another. Many anecdotes and solved examples further facilitate the grasping of the concepts. The book is different from other books mainly on two counts. Firstly, it raises the bar in presenting the concepts suitable for modern requirements. Secondly, the material is presented in a friendly manner with difficulty levels increasing at a pace that a reader can teach himself through self-study. I have tried to fuse these two opposite looking aspects.

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About the Author Prashant Kumar worked with IIT Kanpur, India, for 30 years and taught courses related to solid mechanics and product design. He obtained his PhD in Solid Mechanics from Brown University, USA, MS in Design from the University of California, Berkeley, and BTech in Mechanical Engineering from IIT Kanpur, India. Earlier, he worked as a Design Engineer for two years with Cambar Manufacturing Company, Los Angeles, where he designed airbridges for Boeing 747 airplanes. He has published many research papers in the field of fracture mechanics and polymer composites, and helped many industrial companies as consultants. He has published two books: Elements of Fracture Mechanics and Product Design: Creativity, Concepts and Usability. He has six patents in his name. He has been working as Professor Emeritus, Department of Mechanical Engineering, College of Engineering Pune, India.

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Acknowledgements

It is a great pleasure and moments of immense satisfaction to express my gratitude towards my sons, Saurabh and Manu, and my daughters-in-law, Shailee and Shaphali, for their persistent encouragement to write this book. I owe my special thanks to: • Indian Institute of Technology Kanpur to provide me opportunities to teach mechanics of solids and other related courses many times • College of Engineering Pune for providing me with various infrastructure facilities • Abhishek Upadhye for helping me in preparing illustrations of this book • Many MTech students to provide beneficial feedbacks • Nikhil Pawar for critical reading of the initial chapters and providing valuable suggestions Finally, I profusely thank World Scientific for the friendly interactions and meticulous handling during the publication of the book.

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page xi

Contents

Preface

v

About the Author

vii

Acknowledgements

ix

1.

Getting into a New Ball Game

1

2.

Forces, Moments, and Equilibrium

7

3.

Reactions and Free Body Diagrams

23

4.

Internal Forces in Slender Members

55

5.

Stress Tensor

95

6.

Transformation of Stress Components

121

7.

Deformation and Strain Tensor

147

8.

Material Behavior: Stress–Strain Relations

175

9.

Axial Members: Stresses, Thermal Strains, and Compatibility

219

10.

Bending Moments and Shear Forces

257

11.

Deflection due to Bending

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12.

Torsion

377

13.

Combined Stresses and Yield Criteria

413

14.

Energy Methods

461

15.

Historical Perspective

503

Index

527

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Chapter 1

Getting into a New Ball Game

The game of football is very popular in almost all countries of the world. When I went to the USA to study, I found that the American football was quite different from the European football which is known as soccer in North America. Although the Americans call it football, it is played with hands and even the oblong shape of the ball is quite different from the spherical ball of the soccer game. Thus, the American football is a new ball game, but it is very interesting and exciting. Similarly, the field of mechanics of materials is quite different from other subjects. It is a new ball game, interesting and exciting. I always found the field of mechanics of materials logical and thought-provoking. I am quite confident that the readers will find this field exciting and enjoyable too. However, many students find it difficult to grasp the concepts. Probably, they were not introduced properly to this field which is of a very different nature. The concepts are presented very gradually in the book like the blooming of a flower, petal by petal from its bud. A solid member is made of atoms which are bound to each other through interatomic forces, resulting into some kind of equilibrium between the atoms. The changes in the interatomic forces due to externally applied forces on a structural member are known as internal forces. We are interested in determining the internal forces. If the internal forces are high, there is a possibility that bonds between the atoms are broken, resulting in the failure of the material. It is thus important to know the extent of the internal forces developed in the body due to the externally applied forces.

1

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(a)

(b)

Fig. 1.1 (a) A chair made of a steel tube and (b) an internal atom with forces from all directions.

1.1

Accounting for Internal Forces

Suppose we want to design the frame of a chair which is made of steel tubes as shown in Fig. 1.1(a). It should be designed to be safe even if a man of 120 kg sits on it. The internally developed forces may be too high to be safe. Therefore, we should determine the internal forces so as to choose the frame tube which would be strong enough to resist them. When the man sits on it, almost all molecules of the frame are subjected to internal forces. We, as beginners, may thus think that the task of evaluating the internal forces is difficult. In fact, it is not. We will be able to solve this problem quite easily once we learn the concepts covered in this book. An atom of a structural member feels internal forces from all directions, from left or right, from front or behind, or from top or bottom. Imagine that you are in a very crowded fair and people push or pull you from the behind, front, or side. Someone may even try to go forward by rubbing you from the side. An internal atom is in a similar situation as shown in Fig. 1.1(b). The various forces, in general, have different magnitudes and different directions. To account for the net effect of forces is quite complex and a vector is unable to achieve it. A question arises: how to account for various internal forces on the atom acting from all directions? Our forefathers have done an excellent job by managing such a complex situation with a second-order tensor.

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3

What is a tensor? A variable is expressed, in general, in terms of a tensor. In fact, tensors are of many kinds: zero order, first order, second order, third order, fourth order, etc. Some variables, known as scalars, are tensors of zero order. They deal with the magnitude only and are not associated to any space direction. Some of the wellknown scalars are as follows: mass, temperature, volume, time, and various kinds of energies. Vectors are first-order tensors and are associated with magnitude as well as one direction. Unlike a scalar, a vector cannot be described fully with one component. It is expressed in terms of three components. Some of the vectors relevant to this field are as follows: distance, velocity, acceleration, force, angular velocity, moment, momentum, angular momentum, etc. A second-order tensor is associated with two directions and nine components are required to describe it fully. The stress, which monitors the extent of the internal forces, is a second-order tensor and is expressed fully with nine components; it will be discussed in Chapter 5. There are higher-order tensors such as third order and fourth order. We will look into these higher-order tensors when need arises. 1.2

Accounting for Deformation

The distance between the atoms changes when internal loads are developed in a solid member. The cumulative changes alter its outside shape, which results in the deformation of the structural member. In fact, all real-life materials deform when loads are applied on a member. Depending upon the property of a material, some deform more and the resulting deformation can be sensed with our naked senses like sight or touch. Contrarily, the deformation is so small in some materials that it is hardly felt. For example, when we sit on a chair with a foam seat, the foam is compressed quite a lot and we can easily feel that the body goes down along with the compressing foam. In fact, the displacement is of the order of 5–20 mm. On the other hand, the steel frame of the chair is hardly compressed even when a heavy person of 120 kg weight sits on it. In fact, we may get the feeling that the chair does not deform at all. In fact, it also

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deforms, but the downward motion due to the deformation of the frame may be only a few micrometers, which is difficult to feel. The deformation of a structural member is also complex and needs special attention. Let us focus our attention on an internal point of the member and we imagine an infinitesimal cube around the point. Owing to various forces acting on the cube, the shape of the cube is altered. The distance between the opposite faces may decrease or increase and the cube may not remain a cube but changes into a cuboid. Furthermore, due to rubbing forces on its side faces, the right angles at the corners of the cube may not remain right angles after the deformation. How should we account for such a deformation? Our forefathers have formulated the theory of the deformation quite smartly with another second-order tensor, known as strain tensor. The strain tensor is also expressed with its nine components and will be explored in Chapter 7. 1.3

Formulating a Real-Life Situation

Many engineers face difficulties in formulating a new problem. We as engineers solve problems after problems of real-life cases. Formulation means that we identify the relevant variables of a problem and write appropriate mathematical equations describing their relationship. Then, we carry out the “solving” which means that we solve the equations developed during the formulation stage to obtain the results. In some fields like mathematics, the emphasis is more on “solving” and less on formulation. “Solving” has its own difficulties and some of the problems are quite difficult. There are several difficulties in the process of formulating a reallife situation. We develop an idealized model in the formulation stage. But, how do we make the idealized model? A real-life situation depends on many variables and it would be very difficult to account for all the variables. Usually, there are only a few prominent variables on which the results of the given problem depend heavily. They are addressed as primary variables. To make a good model, all primary variables should be identified and accounted for. Sometimes, we may miss an important primary variable and then the results will be unsuitable.

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5

There are some secondary variables whose influence on the reallife situation is marginally significant. For example, a secondary variable may influence the results by 4%. Now, it is up to our judgment whether to include this variable in our mathematical equations or not. If we can tolerate 4% error in our analysis, we may as well not include the variable in the analysis. The price has to be paid either way. If we include the variable, our formulation will become more difficult, making the mathematical equations more complex. Thus, it may take much more efforts and resources for “solving”. On the other hand, if we do not include it, the analysis is likely to be simpler, but the accuracy of the results is compromised. There are tertiary variables whose influence may be very small and they should be ignored right away without giving it much thought. We again consider the example of the chair to understand the secondary and tertiary variables. The entire weight of the chair is about 4 kg and the heaviest user weighs 120 kg. Thus, the own weight of the chair (known as dead weight) is a secondary variable. If we account for the loading due to its dead weight, the formulation and the subsequent solving will be quite complex. We would be better off by assuming (idealizing) that the chair is made of weightless material. Of course, during the design stage, we apply a suitable factor of safety which accounts for errors introduced during idealization. There are several tertiary parameters in this example. Small holes may be drilled in the steel tubes to weave the seat for the chair. The presence of the small holes may be considered as tertiary parameters and their effect may be neglected. Further, the steel tubes used in making the chair are commercially available materials and there may be variation in the outside diameter of the tubes or in the wall thickness. We do not account for such variations in our analysis and, therefore, tertiary parameters are ignored. A judgment is required to identify which variables are to be retained and which ones do not need to be considered. The decision may not be straightforward in certain cases. While solving the problems of dynamics, such as a rocket carrying a satellite to space, the deformation of the rocket body is neglected. In fact, the body is assumed to be rigid while formulating equations for the motion of a rocket. The primary interests in this scenario are to determine the motion of the body, its displacement, velocity, and acceleration.

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The focus is not on the deformation of the rocket body, but on how it moves from one place to another. In the field of mechanics of materials, we analyze the internal forces developed within the body of the rocket and how it deforms. The movement of the rocket from one place to another does not come under our interest and, therefore, it is thrown away during the formulation; only the deformation portions are retained. Once we know the internal forces, we can design the structure of the rocket body in such a way that it does not fail when it is launched into space. Thus, we make use of idealized models which depend on our requirements. 1.4

Concluding Remarks

There are two primary objectives: (i) to determine the extent of internal forces developed within a structural member and (ii) to determine the resulting deformation when external forces are applied. The loading at an internal point of a structural member is complex as forces act from all directions. It is accounted for by a second-order tensor, stress tensor. The deformation is accounted for by another secondorder tensor, the strain tensor. The stress tensor will be explored in Chapter 5 and the strain tensor will be covered in Chapter 7.

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Chapter 2

Forces, Moments, and Equilibrium

Broadly speaking, there are two types of diseases: physical and mental. Physical diseases such as typhoid, arthritis, and cancer are well-known and they are treated with appropriate medicines. However, many of us are not very much familiar with mental diseases. Mental diseases are as important as physical diseases and many doctors feel that several physical illnesses are caused because of mental illnesses. In the mechanics of materials, the status of forces and moments are similar. Most of us understand the forces reasonably well, but are not comfortable with the moments. Both are important; quite often, in comparison to the forces, the moments develop higher stresses. Various forces and moments act on a structural member and there has to be some kind of balance between the applied forces and the moments to keep the member in equilibrium. The balancing conditions are governed by equilibrium equations which will be developed in this chapter. 2.1

Newton’s Law of Action and Reaction

Newton’s law of “action and its reaction” is very powerful and we use it all the time. According to this law, for an action there exists a reaction which is equal in magnitude but opposite in direction. The magnitude of the forces is equal and both of them are aligned on the same line. However, the direction of forces is opposite to each other. If a person pushes a wall with his hand, the wall pushes back his hand with the same magnitude of force but in the opposite direction. 7

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Similarly, while sitting on a chair if he pushes the ground downwards with his foot with 50 N force, the ground pushes his foot upwards with the same 50 N force. The reactive force thus depends on what is applied. We will be using this law of action and reaction innumerous times in this book. 2.2

Laws of Motion

Newton’s law of motion is applicable when forces act on a point mass whose dimensions are theoretically zero. If this particle is moving with velocity V and its mass is m, the linear momentum is defined as mV . Newton realized that if a point mass is moving with linear momentum mV , it will keep on moving with the same momentum without any change. In other words, the linear momentum is conserved. If we apply a force on this point mass, its momentum will change. According to the law of motion, the rate of change of momentum is equal to the force applied. Mathematically, F =

d(mV ) dt

(2.1)

If there are several forces acting on the point mass, the equation is  ) modified to F = d(mV dt . If the mass is constant, Eq. (2.1) takes a more familiar form: F = m dV dt = ma where a is the acceleration. It is worth noting that it is a vector equation. However, it is a simple vector equation because the acceleration is in the direction of the force. 2.3

Units of a Force

We will now discuss the units of force. The units are an integral part of a variable or quantity. In our society, a person is not acceptable to interacting socially if he is not wearing any cloth on his body. Similarly, a parameter is not complete unless it is presented with its units; we should make it wear clothes. These days, most people in the world follow SI units, developed in 1960 after the French title Syst´eme international d’unit´es. Some of the units have been chosen to honor the great scientists. For example, Joule is honored for energy, Watt for power, Celsius for temperature,

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and Pascal for pressure. Newton played a very important role in understanding force and, therefore, we have honored him by assigning Newton (N) as the unit of a force. Let us go a bit deeper to understand what 1 N force is? A force that develops acceleration of 1 m/s2 on a body of 1 kg mass is called 1 N. This is the rigorous definition based on the equation F = ma . But it is difficult to get the feel of the force from this formal definition. We can have some feel by placing a small bag of 0.1 kg rice on our palm. The gravitational force attracts the mass with acceleration g (9.8 m/s2 ). Thus, the rice bag applies 0.98 N force in downward direction on your palm which is about 1 N. This is not a large force for many applications of engineering. My compact car weighs about 1,000 kg and therefore it applies 9,800 N force on the ground surface. Thus, we will be using force in N or kN. There is one more important aspect of the force unit N, that is, it is a composite unit. Since F = ma , 1 N = kg·m/s2 . Whenever we cancel units, we may have to replace N by kg·m/s2 . 2.4

Solid Members of Finite Size

As already mentioned, Newton’s law of motion was formulated for a point mass only. By definition, a point has no dimensions, that is, no length, no width, and no height. On the other hand, a real-life solid member is made of many molecules, attached to each other with interatomic forces. Therefore, solid members have a finite volume and finite dimensions. Now, a question arises: how do we apply Newton’s law of motion to real-life solid members having finite dimensions? Imagine a ballpoint pen is placed on a frictionless horizontal glass plate. This pen has its own weight and associated reaction in the vertical direction. In this example, we will not be considering the motion and forces in the vertical direction. We will instead focus our attention on horizontal forces and resulting motions in the horizontal plane (x−y plane). Figure 2.1(a) shows force F acting on a point mass. It is a simple case because F generates acceleration in its own direction. Consider a pen on which the same force is applied. Does it make a difference if the point at which the force is applied is changed? It does because if the force is applied at the lower portion of the pen, it is rotated in the counterclockwise direction and if the force applied is on the upper portion, the pen is rotated in the clockwise

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10

B F

CM

d

A

F F

Translation

θ

Rotation

(a)

(b)

Fig. 2.1 (a) A force applied on a point and on a pen and (b) the resulting translation and rotation.

direction. In addition, the pen also moves in the direction of the force. Therefore, the pen will have both kinds of motion, forward and rotational. The forward motion is called translation. Figure 2.1(b) shows both kinds of motion: (i) translation d from its original position, shown with the dashed lines with no rotation, and (ii) rotation from its original position with the angle θ without having any translation. The analysis of a finite body that has volume and finite dimensions is a lot more complex than that of a point mass. Things do become complex but we know how to tackle them. We take up the translation first. The center of mass (CM) of a body is a very important point. If force F is applied on the pen in such a way that its force line passes through the CM of the body, it translates the body and there is no rotation (Fig. 2.2(a)). It is important to realize here that the force line is considered and not the point at which the force is applied. In fact, in the case of the pen, the CM is located in the interior of the body and we cannot directly apply force on it. Also, theoretically, the force line extends from −∞ to +∞. Even for finding a moment caused by a force, the force line is important and not the point at which the force is applied. Figure 2.2(b) shows that the pen is rotated counter clockwise if its force line is below the CM. Of course, the pen is also translated. The pen is rotated clockwise if the force line is above the CM. Because of the finite dimensions of a solid member, moments are developed. In Fig. 2.2(b), the magnitude of the moment is Fd where d is the normal distance of CM from the force line.

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Forces, Moments, and Equilibrium

F

F

CM

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11

CM F

(a)

d

(b)

Fig. 2.2 (a) A force, passing through the CM, translates the member and (b) a force, not passing through CM, translates and rotates the member.

We will be dealing with moments all the time in this book and, therefore, it is important to look at its units. Its units are simple, Nm. Let us try to get some feel of its magnitude. Suppose a person is holding a very lightweight aluminum tube of length slightly longer than 1 m. He holds one of the ends of the tube with both hands and requests a friend to hang a bag containing about 0.1 kg of rice on the other hand (Fig 2.3). The moment acting on the palms of his hands is about 1 Nm. He will be able to resist the moment of 1 Nm but he will have to make some effort. If his friend hangs a bag of 1 kg of rice, the moment acting on his palms will be quite a lot. Thus, 1 Nm is a reasonably big unit; 1 N force is small but when it is multiplied by a large length of 1 m, the magnitude becomes large. A force acting on a point of the pen develops two kinds of generalized forces: (i) force F acting on the CM, and (ii) a moment F d about the normal direction (vertical in this case) where d is normal distance of force line from the CM, as shown in Fig. 2.4. The force translates the body while the moment rotates it. It is important to note here that if the distance d is large, the moment will be large too while the magnitude of force which 1m

100 g

Fig. 2.3

A small load applied at the end of an aluminium tube.

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F.d CM

F

CM

d

F

Fig. 2.4

Equivalent force and moment.

translates the body remains the same. The pen will rotate quite a lot while the translation will remain unaffected. Therefore, in many applications moments are more dangerous than forces. To squeeze the water out of a washed wet towel, we twist it between our two hands, thus applying a twisting moment to the towel. If we press the towel between our two hands, not much of the water will come out. The compressive force is not very effective here and, therefore, the force is applied in such a way that a high twisting moment is developed. Another example is of a woodcutter who makes a large tree fall. First of all, he ties one end of a long strong rope to a branch on the upper part of the tree. He then makes a groove on the trunk of the tree close to the ground by chipping off wood with his axe. When the groove is deep enough, he pulls the rope, as shown in Fig. 2.5, thus applying a moment on the tree. He succeeds in making even big trees fall on the ground. If the normal distance (d) between the groove and the rope is 8 m and the woodcutter applies 300 N force, the moment on the base of the tree is fairly large, 2,400 Nm.

90 d

o

Rope F

Fig. 2.5

Making a tree fall.

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Forces, Moments, and Equilibrium

Fig. 2.6

page 13

13

Analogy: Force like a female and moment like a muscular male.

We thus find that both, forces and moments, make a member move. They are quite different from each other in nature, even the units are different. However, both kinds are addressed as generalized forces. In the later chapters, we will find that a component may fail due to one of the generalized forces or through a combination of several generalized forces. Both can cause the failure of a component, but in many cases, moments are more dangerous. Let us consider an analogy. A force is analogous to a female while a moment to a male (Fig. 2.6). A male is born out of a female womb but it may grow to become a very muscular person. Similarly, a moment is developed through a force and may become a very powerful generalized force. The high potential of moments can be used both in positive and negative applications. A boy when grown up can become a powerful negative character like Hitler or he may become a powerful good man like Mahatma Gandhi. It depends on how a moment is used. Squeezing water out of a wet towel is a good use of moments and making a tree fall is a bad use of moments. Since moments are powerful and can become quite devastating, it is important for us to understand them well and identify them in real-life situations. 2.5

Euler’s Laws

Leonhard Euler (1707–1783), a Swiss mathematician and physicist, modified the form of Newton’s law of motion, about 50 years later,

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to have them in the forms which are more appropriate for studying the deformation of finite size members. The two laws are as follows: 1. Euler’s first law of the conservation of linear momentum deals with forces. 2. Euler’s second law of the conservation of angular moments deals with moments. 2.5.1

Euler’s first law of linear momentum

The law states that the sum of all external forces acting on a body is equal to the rate of change of linear momentum. It is expressed as 

F =

dG dt

(2.2)

In this relation, G is linear momentum. It is expressed in terms as mass m and velocity v o of the center of mass of the structural member as G = mv o . The form is similar to that of Newton’s law of motion, but it is more general because it is applicable also to members of finite dimensions. It is important to remember that velocity vo is the velocity of the center of mass. 2.5.2

Euler’s second law of angular momentum

A point is chosen in space about which we take the moments. The point can be ANY point, inside the structural member or outside the member. The point is usually chosen to suit our convenience, that is, it is chosen in such a way that it makes the formulation simple. The moment of a force F is defined in terms of cross-product as M =r ×F

(2.3)

where r is the distance vector from the chosen point O to any point lying on the force-line. Note that the moment is a vector and its direction is normal to the plane formed by the vector r and force F . The moment rotates the body with some angular velocity ω. Then, the angular momentum H is expressed as H = Iω where I is the mass moment of inertia. Euler’s second law states that the

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sum of all moments acting on a body is equal to the rate of change of angular momentum H , expressed as 

2.6

M=

dH dt

(2.4)

Equilibrium Equations

Both Euler’s laws are applicable to stationary as well as moving bodies. But in this introductory book on mechanics of materials, we will only be considering stationary bodies and therefore we will not be considering time-varying terms in Eqs. (2.2) and (2.4). However, we do consider dynamic cases of certain fields like impact loading and stress waves in solids. They are advanced topics and not included in this introductory book. The resulting equations, known as equilibrium equations, are as follows:  F =0 (2.5)  M =0 (2.6) These equations will be referred to as force balance and moment balance equations. These two equations are the backbone of this field because we will be solving many real-life problems by using these two equations. Since these two equations are of utmost importance to us, we will discuss them in detail, one by one, in the following two sections. 2.7

Discussion on Force Balance Equations

Consider three forces, F 1 , F 2 and F 3 acting on a solid member, as shown in Fig. 2.7(a). For equilibrium, F1 + F2 + F3 = 0

(2.7)

This is a vector summation and, in general, it is in three-dimensional (3D) space. It is important to note that only the force directions are required in the equation and not the location at which these forces

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z

F2

F1

F3

F1

F1z F1x

x

y

F1y

(a)

Fig. 2.7

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(b)

(a) Forces acting on a solid member and (b) components of a force.

are applied. This may look strange but the location of a force line is taken care of by the moment balance equation. We need to satisfy both force balance and moment balance equations for the equilibrium of a member. We often find it easier to deal with forces in components form. Force F 1 can be expressed in terms of its three components F1x , F1y , and F1z as (Fig. 2.7(b)): F1 = F1x i + F1y j + F1z k Similarly, F2 = F2x i + F2y j + F2z k F3 = F3x i + F3y j + F3z k Substituting them in Eq. (2.7), we obtain (F1x + F2x + F3x ) i + (F1y + F2y + F3y ) j + (F1z + F2z + F3z ) k = 0 Since i , j , and k are normal to each other, we obtain

For a general case, 

F1x + F2x + F3x = 0

(2.8a)

F1y + F2y + F3y = 0

(2.8b)

F1z + F2z + F3z = 0

(2.8c)

Fx = 0



Fy = 0



Fz = 0

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Thus, the sum of all force components in x-direction is zero, the sum of all force components in y-direction is zero, and the sum of all moments about z-axis is zero. 2.8

Discussion on Moment Balance Equations

In comparison to the force balance relations, the moment balance equations are a bit more involved. A moment is also a vector made of three components Mx , My , and Mz as shown in Fig 2.8(a). Thus, a moment in three dimensions can be written as M = Mx i + My j + Mz k On a structural member, forces as well as moments act. Forces acting on a structural member, of course, develop moments and we are quite familiar with it as M = r × F . But, in some cases, external moments may also be present on a member. For example, when we twist a wetted towel to squeeze the water out, we apply the twisting moment. Thus, the overall sum of the moments is the sum of moments developed by the external forces and the externally applied moments. We equate the overall sum to zero for stationary systems. Figure 2.8(b) shows a two-dimensional case with two forces F 1 and F 2 and two moments, M1z and M2z . When we use Eq. (2.3) to find the moment developed by a force, we have to choose a point about which the moment is determined. As already stated earlier, we can choose any point in the space. The point can be a point close to z

F1 M1z y

M

+

Mz Mx

x

d1

y

+

x F2F2

d2

M2z

My (a)

(b)

Fig. 2.8 (a) Three components of a moment vector and (b) forces and moments acting in a 2D case.

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the solid member, a point on the railway station, or a point on the airport of your town. From an analysis point of view, any one of these points will yield the right solution. But a point at the airport will involve large numbers and the algebra may become messy. Thus, we chose the point judicially to obtain simple moment balance equations. For taking the moment developed by a force, the force line (a line on which the force lies and it extends from −∞ to +∞) is important. Therefore, the direction of the force line and its location in the space is important. It is worth mentioning again that for the force balance, the location of the force line is not considered. In the two-dimensional case of Fig. 2.8(b), F 1 and F 2 are in the x − y plane only and moments M1z and M2z are about the z-axis. We can choose point O, about which the moment is taken, anywhere in the plane. The sum of the moments is then equated to zero, that is, Σ Mz = F1 d1 − F2 d2 + M1z − M2z = 0 It is worth noting that we keep track of the sign of moment while carrying out the sum. For a general case of three dimensions, all the three moment balance equations are as follows:    My = 0 Mz = 0 Mx = 0 Thus, the sum of all moment components in x-direction is zero, the sum of all moment components in y-direction is zero, and the sum of all moment components about z-axis is zero. Usually, forces and moments together are addressed as generalized forces. However, moments are quite different from forces. Even their units are different. Table 2.1 shows the equivalence between the two kinds of generalized forces. Table 2.1

Units Inertia Motion

Equivalence between force and moment systems. Force system

Moment system

N Mass Velocity Acceleration

Nm Mass moment of inertia Angular velocity Angular acceleration

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19

Some Comments on Solving Problems

We are learning, so far, by reading the material. But “reading” is not good enough; “doing” is better. However, “reading” is essential as a forerunner because “doing” is carried out based on what we read. We can compare it with learning how to swim. We can take many oral instructions from an expert or read many books on swimming, but we will not be able to learn swimming unless we get into the water of a pool and try to follow what we have learnt so far. It will require many attempts of trial and error before we master the skill of swimming. The same applies to learning an engineering subject. We should solve a large number of problems on our own. While solving a new problem, we usually struggle. Sometimes, we are on a wrong route and it takes a while to realize it. Sometimes, we do not know how to proceed. Then, we read and re-read our reference books to understand the theory in greater depth with the hope that it will enable us to find a lead. There will be times when we are not able to solve a problem at all and seek external help either from a teacher or one of our colleagues. However, the time we spend on solving a problem is good learning. We understand the concepts better and we learn why and where we get stuck. The problem-solving in this book is done through five steps: Given: We should exactly know what is given to us. A missing data, even a small one, may not allow us to proceed further or incorrectly understood data will put us on a wrong path. To find: It helps in formulating and solving if we know exactly what is to be determined. In fact, it guides us to move in the right direction. Strategy: Once we realize what to find, we look for a line of attack to crack the problem and develop appropriate equations. Often, we are required to create a model. As explained in the first chapter, we identify the primary parameters and take a judgment on whether we want to include secondary parameters in the formulation of the equations. Solution: Once the equations are formed, they are solved to get the desired results. In some problems the solving is simple, but there are problems when solving is a tedious task. In some cases, it is done with the help of a computer.

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Interpretation/Comments: Once we arrive at the results, it is recommended to spend some time to consider the impact of the results. For example, a variable z depends on two variables x and y and the √ result is z = y 2 x. Then, the effect of y is a lot more pronounced than that of x. Further, we should also explore whether the algebraic expressions of the results make sense. We often make mistakes in units and come up with a ridiculous looking number. For example, if the resulting force is 2.1 × 10−6 N, it is too small a force to make any sense for real-life problems. Note that 1 N is a small force and, therefore, the reported result is certainly too small to be correct.

2.10

Solved Examples

Two examples are solved including one of a 2D case and another of a 3D case. Example 2.1: On a bent bar, in-plane forces are applied at points A, B, C, D, and E, as shown in Fig. 2.9(a). Determine whether the bar remains in equilibrium. Solution Given: A 2-D problem of a bent bar; various in-plane forces acting on it, as shown in Fig. 2.9(a). To find: Whether the member is in equilibrium. Strategy: Check whether both force balance and moment balance equations are satisfied. Solution: Sum of all forces in x and y directions:  Fx : −800 − 100 + 900 → 0 Satisfied  500 − 500 → 0 Satisfied Fy : For the moment balance equation, point C is chosen because three forces pass through this point and, therefore, they do not contribute:  Satisfied MzC : 800 × 150 − 500 × 240 → 0 The beam is in equilibrium as all three equilibrium equations are satisfied.

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Forces, Moments, and Equilibrium z

800 N A

y

Mz=75 Nm x

500 N

500 N

150 N

150 100 N B

120

21

A 100 N

900 N C

240 mm

D

80

E

150 N

Fig. 2.9

y B 200 mm

100 N

Mx=50 Nm

(a)

x

500 mm

C

(b)

(a) The bent bar of Example 2.1 and (b) the bent bar of Example 2.2.

Interpretation/Comments: We will show that it makes no difference if we take the moment about some other point, say point A:  MzA : 500 × 120 − 100 × 150 − 500 × 360 + 900 × 150 → 0 Satisfied This shows that change in the moment point does not make any difference although the moment equation has more terms and solving becomes a little more involved. Example 2.2: Forces and moments applied on a bent bar are shown in Fig. 2.9(b). Determine whether the bar is in equilibrium. Solution Given: 3-D problem with applied forces and moments, as shown in Fig. 2.9(b). To find: Whether the member is in equilibrium. Strategy: Check whether all the force balance equations and all the three moment balance equations are satisfied. Solution: Checking force balance equations:  Satisfied Fx : −150 + 150→ 0  Fy : All terms are non-existent Trivially satisfied  Satisfied Fz : 100 − 100 → 0 Checking moment balance equations about point A:  Satisfied MxA : 50,000 Nmm − 500 × 100 Nmm → 0  Not Satisfied MyA : 100 × 200 Nmm = 0  Satisfied MzA : 75,000 Nmm − 150 × 500 Nmm → 0

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All six equations of equilibrium are not satisfied and, therefore, the bar is not in equilibrium. Interpretation: The bar will rotate about y-axis. 2.11

Summary

Newton’s law of action and reaction is very commonly used and it is as common as our breathing; many times, we do not even realize that we are using it. According to this law, if there is an action, there exists a reaction whose magnitude is the same, but the direction is opposite. Since a solid has a volume and dimensions, it is important to keep track of the direction and the location of the force line. In general, a force tends to translate the member and a moment tries to rotate it.  The force equilibrium equation F = 0, which is equivalent   is Fy = 0, and Fz = 0. For these equations, only to Fx = 0, magnitudes and directions of all forces acting on a member are used; it is not required to know the location of the force lines. The moment of all forces on a body can be taken atabout any point in the space. Themoment equilibrium equation M = 0, which is equivalent   is My = 0, and Mz = 0. There are six equilibrium to Mx = 0, equations for general cases of 3D problems and three equations for 2D cases.

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Chapter 3

Reactions and Free Body Diagrams

3.1

Introduction

A person presents his report on a new project to an audience and there is mild criticism. Some people like the criticism and thank the audience for taking pains to understand the project and come up with some points which can help them improve the work further. Some get hurt and take it as an attack on their personality. Some even get so aggrieved that they start defending their project with loud sentences and often illogical responses. In a similar fashion, when external forces are applied on a member, it is resisted by the supports and the supports are of several different kinds. The forces applied by the supports are known as reactions. One of the primary jobs of solving a problem is to find the magnitude and the direction of the reactions. Reactions are also generalized forces; they can be a mix of forces and moments. We will be looking into details of the various types of supports in this chapter. After a comprehension discussion of the various kinds of supports, we will learn how to separate a structural member out of the supports. On the separated member, we show all the external forces and the moments as well as reactions at the supports. The resulting diagram is called a free body diagram (FBD). The process of making an FBD is a very important activity of this field. Almost in all cases, we solve a problem by first drawing the FBD of a member and then by developing balance equations of both kinds, force balance and moment balance.

23

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3.2

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Reactions

We will only analyze static structures, as stated in Chapter 2. To make a structural member static, we must support it at some places. The member may be supported on ground or on some other members of a structure. When external forces are applied on the member, the supports must apply counter forces on the member to keep it in equilibrium. As already mentioned, the forces applied by the supports are called reactions which are generalized forces of both kinds, forces and moments. A question arises: Why do we call forces applied by the supports by a different name? A reaction is also a force but it is a bit different from external forces. For example, if we push a vertical wall with our thumb with a force of 10 N, the wall applies 10 N reaction, equal in magnitude but opposite in direction. If you increase the pushing force to 20 N, the reaction of the wall increases to 20 N. Thus, the reaction adjusts according to external forces in such a way that the structural member remains in equilibrium. In other words, reactions are unknown forces applied by the supports and we usually determine them before we start finding internal loads in the members of a structure. We consider another example. One end of a slender rod is supported through a hinge, attached to a wall, as shown in Fig. 3.1(a). A pin passes through the bracket and a hole in the rod’s end. This kind of a joint is usually called a hinged or pinned joint. The rod is free to rotate in the x − y plane due to the existence of the pin. Since the hinge support is not capable of restricting the rotational motion, moment reaction does not exist at location A. An axial force F is applied at the other end of the rod. The mass of the rod is quite small in comparison to the applied load and is neglected. (a)

(b)

A

B

F

x

R R A A

y

B

F

Fig. 3.1 (a) A slender member AB supported through a hinge and (b) FBD of member AB with reaction R at A when separated out from the support.

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The rod at end A applies force of magnitude R in positive x-direction on the bracket of the hinge and, as a reaction, the bracket applies force of same magnitude R on the rod end in negative x-direction, as shown in Fig. 3.1(b). Considering the equilibrium of the rod, R = F . If F increases, R also increases accordingly. The reaction adjusts itself to the externally applied force F , but if we go on increasing the magnitude of force F , a stage may reach when the bracket will not be able to take the load and may get uprooted from the wall. This is a simple example of evaluating the reaction developed at the support, but in complex cases, it may be difficult to evaluate the reactions. There are several kinds of supports. Some supports are of very simple kind and are capable of applying only one reaction component. In some cases, there exist two reaction components at a support. A support can be more rigid, applying three, four, five, or six reaction components. We will start by considering only 2D structural cases and later on we will take up 3D cases. Let us take up the case of a support applying only one reaction component to the structural member under consideration. A roller support is shown in Fig. 3.2(a) in which a roller or a piece of pipe is placed under the structural member 1. The member can easily rotate in x − y plane and, therefore, moment reaction is not developed at point A. In addition, the roller can be easily translated as it does not resist any motion in the x-direction and, therefore, the reaction in x-direction is zero. Thus, the support applies only one reaction RAy in y-direction at point A. Figure 3.2(b) shows a different form of the same case. To the structural member, a small bracket is attached in which a wheel is installed. Because of this wheel, there is no resistance to rotation in x − y plane and motion in x-direction at point A, and, therefore, only one reaction RAy acts. There are many examples of this kind of support. This support is nothing but a wheel rotating freely on a pin such as the front wheel of a bicycle which is a freerolling wheel as long as the brake is not applied. Figure 3.2(c) is also a roller support and the inclination of the member 1 does not make a difference. In Fig. 3.2(d), the end of the member is resting on a smooth surface that applies very small resistance in x-direction. Therefore, it is modeled as a support that

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1

1

R Ay

R Ay

A

y

A

(a)

(b)

x

1

1

A

R Ay

A

(c)

Frictionless R Ay surface (d)

Fig. 3.2 (a) Roller support, (b) wheel or caster support, (c) a wheel attached to an inclined member and (d) support on a frictionless surface.

does not apply any resistance in x-direction and only one nonzero reaction RAy exists. Another kind of support is developed which is known as compliant member, as shown in Fig. 3.3(a). This compliant member is designed in such a way that it can shear easily, as shown in Fig. 3.3(b), which means that the motion of end A of the member is hardly resisted in x-direction and, therefore, the reaction RAx is negligible. The compliant member is designed to be fairly rigid in y-direction and thus only one reaction RAy is developed, as shown in Fig. 3.3(c). A good example is the supports of a modern bridge. If we peep under a modern bridge, say over a river, we will see a compliant block/plate placed between a pillar and the beam of the bridge. When the bridge expands in summer months or contracts in winter seasons, the compliant block does not provide resistance in the horizontal direction and then the beam is not loaded horizontally. The use of compliant members is a comparatively recent development and they are very handy to use at many places where we need to have only small adjustments. Consider the support surface as an inclined surface (Fig. 3.4(a)) with known inclination angle θ. The reaction RA is normal to the support surface, as shown in Fig. 3.4(b). The reaction RA can be split 1

A (a)

Fig. 3.3

y A (b)

A RAy

x

(c)

Support on a compliant member.

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27 y

A

θ

1

θ RA

(a)

(b)

RAx R Ay

x

(c)

Fig. 3.4 (a) Member supported on an inclined surface, (b) the reaction is normal to the surface, and (c) the reaction is split into its horizontal and vertical components.

into RAx (= RA sinθ) and RAy (= RA cosθ) as shown in Fig. 3.4(c). In this case also, the reaction is associated with only one unknown because the direction of the reaction θ is known. While making a diagram of a model, it is important to realize that we throw away the unnecessary hardware details and retain only the necessary aspects so that we can focus well on developing equilibrium equations. The model of all reactions shown in Figs. 3.2, 3.3 and 3.4 is just a vector RA depicted through an arrow and then we no longer keep track of whether it was a roller support or a frictionless support. We have discussed, so far, only cases in which reaction had only one unknown component. Now we will take up slightly more complex cases of 2D analysis having two unknown reactions at a support point. The first and the most common type of a support is a hinge, as shown in Fig. 3.5(a) for a 2D case. In this configuration, the member can easily be rotated in x − y plane and, therefore, the moment reaction is zero. The end A of the structural member cannot move in either x-direction or y-direction and, therefore, two reactions, RAx and RAy , are developed at point A, as shown in the figure. It is worth noting that we need not worry about the directions of reaction components. For example, RAx can be shown either along positive x-direction or negative x-direction. The reaction is still unknown at this stage. If the value of RAx turns out to be positive, our chosen direction of RAx was right. If the value turns out to be negative, we realize that the reaction component should be in the opposite direction. There is another equivalent way of showing the reaction through a single force RA and its inclination angle, θ. Both RA and θ are two unknown quantities and are determined using equilibrium equations. Most people prefer to use the first approach but the second approach is handy in some special cases.

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A

RAx

RA

RAy

(a)

A

1 Rough surface

RAx

θ

y x

A RAy

(b)

Fig. 3.5 (a) Member hinged at A with two unknown reactions and (b) resting on a rough surface.

Figure 3.5(b) shows another case in which the end A of the structural member rests on a surface that provides reasonably high friction. Member A can be easily rotated and thus there is no moment reaction. The friction does not allow the motion in x-direction and, therefore, reaction RAx is developed along with reaction RAy , thus having two unknown reaction components. However, this kind of support would work only in the case when there is compression between the member and the support. An ordinary ladder placed between a wall and a floor is a good example of this kind of support. Both the contacts points, at the wall and at the horizontal ground, belong to the category of two unknown reactions at a support point. We will now proceed further to the cases having three unknown reactions in 2D analysis. Figure 3.6(a) shows that the end A of the structural member is rigidly embedded within the surface of the support. This kind of support does not allow rotation of the member. Of course, the end A cannot move in x- or y-directions either. If only 2D analysis is considered, three unknown reaction components, RAx , RAy , and MAz are developed. It is worth noting here that the moment in 2D analysis is about z-axis. Figure 3.6(b) shows another configuration of fixed support for 2D analysis. In 3D cases, as shown in Fig 3.7, there can be six reaction components, three force types and three moment types. However, the concepts are same as of 2D; we only generalize them further. It is worth noting here that a 2D analysis deals with three degrees of freedom, two of translation in x- and y-directions, and one of rotation about z-axis. On the other hand, 3D analysis deals with six degrees

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29

y x

A

A RAx

A

MAz

RAx

RAy

A MAz RAy

(a)

(b)

Fig. 3.6 (a) A rigid support in 2D analysis with three reaction components and (b) another configuration of rigid support.

y

A

RAx

MAz

A

MAx

RAz

z

x

RAy MAy

Fig. 3.7

A rigid support in 3D analysis with six reaction components.

of freedom, three of translation in x-, y-, and z-directions, and three of rotation about x-, y-, and z-axes. If all the motions are resisted by a support, there are six unknown reactions, RAx , RAy , RAz , MAx , MAy , and MAz , as shown in Fig. 3.7. To determine these reactions, we use all six equilibrium equations, three of force balance and three of moment balance. There are many special cases used in the structural components of daily life. One of the most interesting and commonly used supports of 3D problems is a ball and socket joint, as shown in Fig. 3.8. In this kind of joint, the ball is free to rotate within its enclosure about all the three axes and, therefore, MAx = 0, MAy = 0, and MAz = 0. The end A of the structural member resists all the three translation motions and, therefore, reactions RAx , RAy , and RAz are developed. The ball and socket supports are used in some important applications. For example, our shoulder joint is a ball and socket joint, which allows rotation of the hand about all the three axes. In most automobiles, the steering of the front wheels is controlled by linkages using the ball and socket joints.

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y x z A R Ax

A

R Az

R Ay

Fig. 3.8

Ball and socket joint with three reaction components. 1

y RAy Pin

A

MAy A

RAz z

MAz

Bracket (a)

RAx x

(b)

Fig. 3.9 (a) A hinge joint in 3D and (b) five reaction components acting at point A of the member 1.

Figure 3.9(a) shows a hinge support in which member 1 is free to rotate about x-axis at point A and, therefore, MAx = 0. However, it resists rotation about the other two axes and, therefore, MAy and MAz are not zero. Of course, a hinge joint resists all three translation motions and thus RAx , RAy , and RAz are developed, as shown in Fig. 3.9(b). There are many special cases of 3D analysis. But it is not possible to consider all possible cases here. Whenever we come across a new kind of support, we just use our common sense and identify nonzero reaction components. 3.3

Kinds of Forces

On an FBD of a structural member, we show all forces acting on it which include all the external forces and the reaction forces developed at all support points. In this section, we will discuss the kinds of external forces. We broadly classify them into two categories: (i) body forces and (ii) surface forces.

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Body forces

Body forces act on each and every atom of a structural member. The gravitational force is the most common kind of body force we come across. Magnetic forces are also body forces and it is more difficult to account for them because a magnetic field usually changes from one point to another point of the body. On the other hand, there is hardly any change in the gravitational field within a structural component. To analyze the effect of the gravitation field, the concept of the center of mass (CM) is available and thus we do not account for the gravitational force atom by atom. We just show the entire gravitational force (mg) acting downwards at the center of mass. Of course, the unit of the gravitational force is Newton (N). In civil engineering structures, the gravitational force is quite important. Heavy steel girders or concrete beams, columns, and roofs are quite commonly used in making civil engineering structures like buildings, bridges, and dams. In fact, in some civil engineering structures, the dead weight (self-weight) is the main load. For some products, the consideration of the dead weight of a member may be important. In earth-moving equipment of mechanical engineering or process machines of chemical engineering, the dead weight of its members may be significant. However, many components of modern machines are quite lightweight (small deadweight) in comparison to the load they carry. In such cases, the dead load is neglected in the analysis. For example, a short length steel rod of 10 mm diameter is capable of carrying more than 1,000 N tensile force, while its own weight is less than 1 N and, therefore, we conveniently ignore its dead weight in the analysis. While making a model, we neglect less significant features to make the analysis simple. It is important to remember that we neglect in comparison only. For example, if A = a + b and a is quite small in comparison to b, then only we can neglect a. 3.3.2

Surface forces

The surface forces are of two kinds: concentrated force and distributed force. In addition, there may be a concentrated moment acting on a structural member. Let us discuss them one by one.

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Mechanics of Materials: A Friendly Approach y

w

w(x)

x (a)

(b)

Fig. 3.10 (a) Uniformly distributed load, w, per unit length and (b) distributed load with variation in x-direction.

Concentrated forces: If the surface area on which the force is applied is small, we can treat the force as a concentrated force. But how small is a small area? It should be compared with the area of the component. If the area on which the force is applied is a small percentage (say 10 as a good enough criterion. We now go back to answer the question raised: Why do we prefer cylindrical members? In cylindrical tanks, all stresses on the cross-section are in-plane tensile stresses. In-plane tensile stresses are resisted well by a material. Suppose the tank is made of flat surfaces in the shape of a rectangular box. The stresses developed in the wall of the container will no longer be in-plane tensile stresses but bending moments are developed. In Chapter 4, it was discussed that a bending moment is more dangerous than an axial force. We will show it very explicitly in Chapter 10 on bending moments. The walls of the rectangular tank will have to be made of a much thicker and the cost will shoot up. By choosing the cylindrical shape, we have smartly avoided the development of the bending moment. Examine carefully an empty aluminum can of a soft drink like a cola which is supplied under pressure. The cylindrical wall is much thinner than the flat end plates. The end plates are subjected to internal bending moments and therefore they are made considerably thicker. About 30–40 years back, the can of a cola drink was being made much larger in diameter and shorter in length for the same volume of the liquid. Now, the diameter has been reduced considerably so that the diameter of the end plate is much smaller. Then, the area of the plate is substantially reduced as it decreases with the square of the radius, requiring significantly less material. Further, the bending moment on the smaller plate is reduced and the plate need not be made very thick. In addition, the in-plane stresses in the smaller

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diameter can of a cola are also reduced, enabling us to make the cylindrical surface of very thin material. Example 9.7: Determine hoop and axial stress in z-direction for a cylindrical storage tank of diameter 500 mm and length 800 mm. It is made of a 5-mm-thick mild steel sheet. The cylinder is pressurized with 3 MPa pressure. Express the state of stress at an interior point and at an exterior point of the surface of the cylinder. Given: d = 500 mm; l = 800 mm; t = 5 mm; p = 3 MPa. To find: σθθ , σzz . Strategy: Invoke σθθ =

pR t

and σzz =

pR 2t .

Solution: The hoop stress is 3 × 250 MPa mm pR = = 150 MPa t 5 mm 3 × 250 MPa mm pR = = 75 MPa = 2t 2×5 mm

σθθ = σzz

On an internal point of the surface, the state of stress is ⎡ ⎤ −3 0 0 ⎢ ⎥ σ = ⎣ 0 150 0 ⎦MPa 0 0 75 On an external point of the surface, ⎡ 0 0 ⎢ σ = ⎣ 0 150 0 0

the state of stress is ⎤ 0 ⎥ 0 ⎦MPa 75

Interpretation/Comments: The magnitude of the radial stress σrr is negligible in comparison to the magnitude of σθθ and σzz . Thus, the state of stress of any point of the wall of the cylinder is taken as ⎡ ⎤ 0 0 0 ⎢ ⎥ σ = ⎣ 0 150 0 ⎦MPa 0 0 75

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Example 9.8: An assembly consists of two cylindrical rings (of small lengths). Ring 1 is snugly fitted within Ring 2 with no gap in between (Fig. 9.17). The coefficient of expansion of Ring 1 is higher than that of Ring 2. The thickness of rings is much smaller than their radii. The temperature of the assembly was increased by ΔT . The thickness t, the modulus E, and coefficient of expansion α are shown in the figure. Determine the following: (a) The pressure p developed between the rings. (b) Hoop stress developed in each ring. (c) Determine the pressure at the interface between the two rings and the hoop stresses developed if E1 = 70 GPa, E2 = 207 GPa, t1 = 2 mm, t2 = 1 mm, R = 150 mm and ΔT = 100◦ C, α1 = 22.2 × 10−6 (m/m)/◦ C, α2 = 12.0 × 10−6 (m/m)/◦ C. Given: Inner ring: E1 , t1 , α1 ; Outer ring: E2 , t2 , α2 ; R  t1 ; R  t2 , Temperature increase: ΔT . (1)

(2)

To find: Pressure p, hoop stress σθθ and σθθ . Line of attack : With the increase in temperature, both rings expand but inner ring tends to expand at a higher rate due to its higher α; consequently, a pressure develops between the rings which develop compressive hoop stress in the inner ring 1 and tensile hoop stress in y 2

r

1

E2, t2, α2

E1, t1, α1 x

Fig. 9.17

Cross-section of the assembly.

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248

the outer ring 2. Obtain strain in each case and then use compatibility relation. Solution: (a) We will first develop the relation between radial displacement and the hoop strain of a ring. The final circumference is the sum of original circumference and the increase of circumferential length due to θθ . Thus, 2π(R + ur ) = 2πR + 2πRθθ or,

ur = Rθθ

(9.19)

Thus, the radial displacement is obtained by multiplying radius with the hoop strain. The geometrical compatibility relation is (2) u(1) r = ur

Using Eq. (9.19), we have (1)

(2)

Rθθ = Rθθ

In the above relation, the radii of inner and outer rings have been taken to be the same as they are only marginally different. The radial stress σrr in a ring is negligible in comparison to corresponding hoop stress and σzz = 0 as the ring is free to expand in z-direction. Then, the substitution of stress–strain relations (Eq. (9.10a)) in the above equation yields 1 (2) 1 (1) σθθ + α1 ΔT = σ + α2 ΔT E1 E2 θθ (1)

(9.20)

(2)

pR But, σθθ = − pR t1 ; σθθ = t2 Substituting in Eq. (9.20), we obtain

−

pR pR + α1 ΔT = + α2 ΔT E1 t1 E2 t2

or,

p=

(α1 − α2) ΔT

R( E11t1 +

(9.21)

1 E2 t2 )

(b) Then, the hoop stresses become (1)

σθθ = − (2)

σθθ =

(α1 − α2) ΔT

t1 ( E11t1 +

1 E2 t2 )

(α1 − α2) ΔT

t2 ( E11t1 +

1 E2 t2 )

(9.22a) (9.22b)

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(c) Substituting the values in Eqs. (9.21) and (9.22), we obtain p=

(1)

σθθ

(2)

σθθ

N (22.2 − 12.0) × 10−6 × 100 [(m/m◦ C × ◦C)] = 0.568 2 1 1 2 mm mm 150( 70×103 ×2 + 207×103 ×1 ) mm[ N×mm ]

= 0.568 MPa

 N pR 150 mm = −42.6 MPa =− = −0.568 × t1 2 mm2 mm

 N pR 150 mm = 85.2 MPa = = 0.568 × 2 t2 1 mm mm

Comments/Interpretation: The thermal stresses are developed in this example as the rings were not free to expand during heating. If there were an adequate gap between the two rings, each would expand freely and no thermal stresses would be developed. These days many structures are designed in such a way that the members are allowed to expand freely to avoid the development of thermal stresses. Sometimes, we make positive use of thermal stresses. I have seen people fixing a mild steel tire (ring) over a wooden wheel of a horse cart by expanding the steel ring in a makeshift oven and slipping over the circumference of the wheel. On cooling, the steel tire grips the wooden wheel very tightly. 9.7.2

Internally pressurized thin-walled spherical shell

A thin-walled spherical shell, pressurized by an internal pressure p, is another interesting member (Fig. 9.18(a)). Some of the applications are soccer balls and tennis balls which are supplied with internal pressure. We note that the geometry of a spherical shell is very symmetric and when it is pressurized from inside with a fluid, its symmetry is not disturbed. Thus, all the three shear stress components are zero. Let us choose any point, say point H and cut a spherical shell of radius R with an imaginary plane passing through the center of the sphere and point H (Fig. 9.18(b)). Let us consider the equilibrium of the cut hemispherical shell. Stress σ on the cross-section acts normal to the cut surface of the area 2πrR; the projected area of the pressurized surface is the diametric

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r H

t p

H

σ

σ

H

σrr=0 pR σ = 2t σ

σ (c)

(b)

(a)

Fig. 9.18 (a) A spherical thin walled member with an internal pressure p, (b) a half-spherical sectioned shell, and (c) stress element at a point of the shell.

area of the cross-section (πR2 ). Thus, the force balance provides (2πRt)σ = (πR2 )p or,

σ = pR/2t

(9.23)

It is an in-plane tensile stress which tends to stretch the surface. Also, note that the spherical shell is a very symmetric structure and the in-plane stress will be the same, no matter how we cut the spherical shell through a diametrical plane. A stress element is shown in Fig. 9.18(c) with biaxial stress of the same magnitude. Stress in the radial direction is negligible for (R/t)  1 and all shear components are zero due to symmetry. Designers make good use of the spherical shell in the design of a cylindrical storage tank. The ends of high-pressure cylindrical tanks are usually made spherical. If the ends are made flat with a plate, the bending moment generated in the plate will be high; the plate has to be made thick and will cost more. On the other hand, only inplane tensile stresses are developed in the hemispherical shell, and, therefore, it can be designed with a thin sheet. 9.8

Summary

The stress distribution of an axial force on a slender member is uniform on its cross-section, that is, the stress tensor for all points is given by ⎤ ⎡P 0 0 ⎥ ⎢A ⎥ (9.24) σ=⎢ ⎣ 0 0 0⎦ 0

0 0

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where P is the axial force and A is the area of the cross-section and x-axis coincides with the axis of the member. For this simple case, all other components are zero. However, to have such simple uniform stress distribution, the axial force should act at the centroid of the cross-sectional area. If the force acts at some other point, one can transfer it to the centroid with a bending moment. Structures can be classified into determinate and indeterminate structures. In determinate structures, all the reaction forces of the support can be determined from the equilibrium equations alone. In indeterminate structures, equilibrium equations are not sufficient to determine all the reaction forces. Additional equations are obtained by considering the displacement of the members of the structure in such a way that geometrical compatibility is maintained. Whenever geometrical compatibility equations are needed, the material properties and the cross-sectional areas of the members are required to convert displacement components to strain components and then to stress components. On the other hand, the material properties and cross-sectional areas of the members of a determinate structure are not required to determine reaction forces at the supports and the internal loading at a section of a member. If the temperature of a member is raised or lowered, thermal strains are developed which are accounted for by the coefficient of thermal expansion. If the member is free to expand in all directions due to the temperature change, normal strains are developed in all the three directions but no shear strain is developed. The thermal strains caused even by moderate temperature changes are comparable to strains developed by external forces. Therefore, thermal effects should be accounted for whenever there is a temperature change. If thermal expansion or contraction is constrained, thermal stresses are developed. In a thin wall of a cylindrical member pressurized with an internal fluid or gas, the hoop stress σθθ is pR/t and the axial stress σzz is pR/2t. Both of them are in-plane tensile stresses and tend to stretch the cylindrical surface. The stress component in r-direction σrr is negligible. Also, all the three shear stress components are zero because of the symmetry of the member.

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9.9

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Problems

1. On a slender member with a triangular cross-section, the axial force acts at point Q, as shown in Fig. 9.19(a). The case is equivalent to the sum of an axial force acting at the centroid and a bending moment acting on the section. Determine the bending moment applied by the force on the slender member. 2. On a slender member made of a square cross-section, a force P acts at point Q which is at one of the edges, as shown in Fig. 9.19(b). Determine the bending moments, one about y-axis and another about z-axis, developed by force P . 3. A structure is made of square tube whose cross-section is 25 mm× 25 mm × 1.6 mm (Fig. 9.20(a)). Determine the strains developed at point Q if the structure is pulled by a force, as shown in the figure. The structure is then heated to raise the temperature of the entire structure by 80◦ C; determine the resulting strain y

a/2 a/3 a

Q

P

C

z

y

a

Q z C

x a

P

a/4

(a)

x

a (b)

Fig. 9.19

(a) Problem 1 and (b) Problem 2. y

E

z

x

8,000 N (a)

Fig. 9.20

150

Q

B 200 mm

0.18

D

J P

200

(b)

(a) Problem 3 and (b) Problem 4.

C 170

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4.

5. 6.

7.

8.

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253

at point Q (coefficient of expansion = 12.0 × 10−6 (m/m)/◦ C, modulus E = 207 GPa, Poisson’s ratio = 0.3). A thick and rigid member BC is supported by a hinge on its left end and a slender member ED at point D, as shown in Fig. 9.20(b). The right end C of member BC is 0.18 mm above a rigid platform. Determine force P applied at point J which just makes point C touch the rigid platform (modulus of member ED = 207 GPa). If the force P in Problem 4 is 15,000 N, determine the reaction developed at point C. A slender member is made of two segments, 1 and 2, as shown in Fig. 9.21. Its left end is supported by a hinge and the right end is rigidly mounted. Axial external forces are applied at location D symmetrically such that the resultant force of 15,000 N passes through the centroid of both members. The cross-sectional areas of portion 1 and portion 2 are 300 mm2 and 150 mm2 , respectively. Determine axial stress in both portions. A thin-walled brass tube is pressed between two rigid plates with the help of a central steel rod with threaded ends, as shown in Fig. 9.22(a). To apply the force, one of the two nuts was rotated by three quarters of a turn (pitch = 1 mm). Find the stresses in the tube and the rod (Ebrass = 110 GPa, Esteel = 207 GPa). A rigid member BG is supported on its left end with a hinge and two-force members 1 and 2 as shown in Fig. 9.22(b). The area of cross-section of member 1 is A1 = 50 mm2 and of member 2 is A2 = 100 mm2 . Members 1 and 2 are made of an aluminum alloy with a modulus of 70 GPa. If a force of 9,000 N acts at point G, determine axial stress in members 1 and 2 and deflection of point G. 1 A1= 300 mm

2 A2= 150 mm

2

15,000 N C

D

B 400 mm

Fig. 9.21

300

Problem 6.

2

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2

1 A1= 100 mm

H 2

500

6 mm dia.

2

1

15

1,000 mm

(a)

Fig. 9.22

15

B C 400 mm

D 500

300

G 9,000 N

(b)

(a) Problem 7 and (b) Problem 8.

9. A cylinder of wall thickness 4 mm and diameter 400 mm with hemispherical ends is pressurized with air of 1 MPa pressure. The cylindrical portion is 800 mm long. Determine the hoop stress and the axial stress in the wall of a cylindrical shell. Are they much larger than the radial stress = −1 MPa at the internal surface? What is σrr at any point of the external surface of the cylinder. Also, determine stresses in the wall of the hemispherical ends. 10. A designer proposes to replace the cylindrical pressure tank of Problem 9 with a thin-walled spherical tank whose total volume is the same as that of the cylinder and it is made from the same material and of the same thickness. Determine the following if the applied pressure is the same: (i) The radius of the equivalent spherical tank. (ii) The in-plane stresses in the wall of spherical shell. (iii) Percentage reduction of in-plane tensile stress in the spherical shell when it is compared with the hoop stress of the cylinder. (iv) Percentage reduction in the mass. 11. A designer proposes to replace the aluminum cylindrical can of a pressurized cola drink with a spherical tank such that the volume remains the same. Also, the in-plane stress of the spherical shell is the same as the hoop stress of the cylindrical portion of the can. Determine the thickness of the wall of the spherical shell and percentage saving in the mass if the pressure is the same. The cola can is made of 0.11 mm wall thickness with its end flat

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255

p

dia. 300 mm

1.5

p

800 mm

(a)

Fig. 9.23

(b)

(a) Problem 12 and (b) Problem 13.

plate of 0.28 mm thickness. The diameter and the length of the cola can are 65 mm and 160 mm, respectively. 12. A thin-walled cylinder with closed ends, shown in Fig. 9.23(a), is pressurized with 0.8 MPa internal pressure. The elastic constants of the material of the cylinder are E = 207 GPa and ν = 0.3. Determine the change in the diameter of the cylinder and the change in the length of the straight portion of the cylinder. 13. A cylinder with closed ends is fitted tightly within a rigid cylindrical cavity (no gap between the cylinder and the cavity) as shown in Fig. 9.23(b). The cylinder is made of a 1.0 mm thick wall and is pressurized with 1.2 MPa pressure. Determine the stresses developed in the wall of the cylinder (material properties: E = 70 GPa, ν = 0.3) and express it through a 3 × 3 matrix.

B1948

Governing Asia

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Chapter 10

Bending Moments and Shear Forces

10.1

Introduction

In a football match, if there is a superstar like Messi in the opposition team, we make a strategy to focus more on the superstar and restrict his performance. There are other good players and we make a strategy against them also. In our field, the dangerous superstar of the various loads is the bending moment and we focus more on it and, therefore, this chapter is one of the most important chapters of this book. We dealt with axial forces in Chapter 9 which are not difficult to analyze as a uniformly distributed normal stress is developed in the axial direction. In this chapter, we will be finding that a bending moment also develops normal stress in the axial direction, but the nature of stress distribution is not that simple. The stress varies from one point of the cross-section to another point. Also, we will find that a shear stress distribution is developed on the cross-section due to a shear force which also varies from one point to another. Thus, we will be developing an expression for finding non-uniform normal stress distribution due to a bending moment Mb and non-uniform shear stress distribution due to a shear force V. Also, a bending moment deforms an initial straight slender member to a curved member. We will find out how curved the member becomes. Why do we consider two generalized forces, Mb and V, in one chapter? Why do we not deal with one at a time, say the bending moment alone? To answer these questions, we realize that the bending moment and the shear force are inseparable. Recall the

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b expression we derived in Section 4.7 (Eqs. (4.6)): V = − dM dx . The equation tells us that if the shear force is zero in a segment, the bending moment remains constant. This particular case is known as pure bending moment, but such a case is rarely present in problems of practical importance. We can also write the equation in a different form as  Mb = − V dx

This form of the equation states that the bending moment changes with distance x if there exists a shear force. Thus, the interplay between Mb and V is always present, similar to the relation between the lightning and the thunder of a cloud burst. A slender member, subjected to bending moments and shear forces, is usually known as a beam. The nomenclature probably originated from the design of a building which is made of columns as vertical members and load-bearing horizontal members addressed as beams. Transverse forces, also known as lateral forces, are common external forces on a beam. Even if an external force acts at an angle, it can be split into an axial component and a transverse component. The internal stress developed by the axial component has been discussed already in Chapter 9. The transverse component gives rise to bending moments in the beam. Transverse forces are quite commonly applied on a slender member. Even when a torque is transmitted from one shaft to another, the bending moment is developed. The transmission of the torque is carried out with the help of hardware, such as gears, pulleys, chains and sprockets. In such hardware, a transverse force is involved. For example, in a bicycle, the chain applies tangential force on the sprocket which, in turn, develops torque. However, at the same time, this force acts as the transverse force on the shaft on which the sprocket is mounted, developing the bending moment. Thus, transverse forces are very common forces on slender structural members, resulting in the development of bending moments. We will discuss that the stresses developed by bending moments are quite high, so much so that they tend to cause the failure of the member. Bending moments are considered to be dangerous loads

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and many failures in real life are caused by them. This chapter is of utmost importance to us. The shear stresses developed on a cross-section by a shear force are small in comparison to bending stresses in many cases of slender members and we usually neglect them. But in certain cases when the member’s length is small, shear stresses are either comparable to bending moment stresses or higher. We should, therefore, understand how to account for them also. To neglect shear stresses, how long should a slender member be? There is no hard and fast criterion, but the effect of shear force is usually neglected if the length is eight times larger than the height of the cross-section. This criterion is usually met in most cases involving slender members. 10.2

Beams Loaded with a Pure Bending Moment

The case of pure bending moment (Mb ) is rare but its analysis is relatively simpler. The results are routinely applied to the more general case of combined loading of a beam with a bending moment and a force shear. The shear force does introduce its own shear stress field but even its largest magnitude is usually much smaller than the stress developed by the bending moment. In this chapter, we will restrict our discussion to beams which are straight before the bending moment is applied. Also, we will be considering only those slender members whose cross-section does not vary with the length. Such structures are usually known as prismatic bars. In addition, we will restrict our analysis to beams which are symmetrical either with respect to x−y plane (Fig. 10.1(a)) or x−z plane (Fig. 10.1(b)). They are often referred to as symmetrical beams. The analysis of other sections (non-symmetrical) is more complex and is beyond the scope of this book. We will find the stress distribution on a cross-section of a beam when bending moment Mb is acting on the cross-section, as shown in Fig. 10.1(a). We will be looking for two major aspects: 1. An initially straight beam becomes curved after deformation. What is the curvedness of the deformed beam? In other words, what is the curvature?

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Mb

Plane of symmetry (z=0) y y

Mb

C

z Plane of symmetry

z x

x

(a)

(b)

Fig. 10.1 (a) The member is symmetric across x−y plane and (b) symmetric across x−z plane.

2. Internal stresses are developed within the beam, usually known as bending stresses. Our interest is to determine the stress distribution on a section of the beam. It will enable us to determine the point at which the bending stress is maximum.

10.2.1

Neutral surface and neutral axis

An undeformed beam is shown in Fig. 10.2(a) and a deformed beam is shown in Fig. 10.2(b). Now, we focus our attention on a small segment of the beam, ABCD, shown in Fig. 10.2(a). When the bending moment is applied, the line AB at the top surface of the beam contracts to A B and the line DC at the bottom surface elongates to D C . If one moves from the top surface toward the bottom surface, the contraction gradually decreases. Similarly, moving from the bottom surface toward the top surface, the elongation decreases. A

y

N

N D

C (a)

Mb

B x

A

Mb

B

N

N

D (b)

C Neutral surface

Fig. 10.2 (a) An undeformed straight beam and (b) deformed beam with the neutral surface NN .

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261

y

Neutral axis

z

Fig. 10.3

C

Neutral axis on a cross-section.

We expect that there exists a point on the line AD at which a line in x-direction does not change its length. The surface made of such points is called neutral surface, shown as NN in the figure. When we focus our attention on a cross-section of the beam, the neutral surface looks like a line usually called a neutral axis (Fig. 10.3). A natural question is: Where is the neutral axis located? Its location will come out of the analysis invoking the boundary conditions. In fact, the analysis will show that neutral axis passes through the centroid of the cross-section. The beam becomes curved after the deformation. But how do we measure its curvedness? We will discuss it in the following section. 10.2.2

Curvature and radius of curvature

The curvedness is measured with a parameter curvature, κ. The slope of a continuous curve φ changes with distance s (Fig. 10.4(a)). Curvature κ is defined as the rate of change of slope with distance, that is, κ=

dφ ds

(10.1)

If dφ ds is constant, it is a special case, a circle. If φ changes quickly with s, the curvature is larger. The dimension of curvature is 1/length. To determine curvature at a point, we draw normal at two neighboring points which meet at point O, as shown in Fig. 10.4(b). The included angle between the two normal is Δφ. In fact, ρ is called

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O

s

90 90

o

o

s

(a)

(b)

Fig. 10.4 (a) The slope φ at a point of a continuous curve and (b) radius of curvature ρ.

radius of curvature and then distance Δs = ρΔφ. The definition of κ (Eq. (10.2)) yields κ=

Δφ 1 dφ = = ds ρΔφ ρ

(10.2)

Thus, the radius of curvature is the inverse of curvature. We will be using κ and ρ quite extensively in this chapter. Also, we will show later in this chapter that the curvature of a deformed beam is proportional to the bending moment at that cross-section. In most of the beams, the bending moment varies with x and, therefore, the radius of curvature also varies with the distance x along the beam. In most structures of real life, we are not able to detect, through our naked eyes, that a beam becomes curved. In fact, most structures are designed in such a way that they do not look to be curved when loaded. Once I designed a bookshelf that was made of a 20-mmthick wooden plank. When I placed my books on one of its shelfs, the midpoint of the shelf was sagged by about 20 mm. The sagging made me feel uncomfortable and, therefore, I bonded a 12-mm-thick plywood board underneath each plank. The resulting planks were satisfactory and did not look to be bent. However, the planks were still bent and the curvature could be measured with instruments. For example, if we place a sensitive spirit leveler, we can feel the variation in the angle.

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Two important assumptions

We make two important assumptions on the geometry of the deformation: (1) A sectional plane which is normal to the neutral surface remains plane after deformation. (2) The normal sectional plane remains normal to the neutral surface after the beam deforms. Figure 10.2 shows the outcome of these two assumptions. Consider a plane AD in the undeformed state which is normal to the neutral surface NN . After it is loaded with the bending moment, the plane AD moves to location A D , as shown in the figure. The plane A D is still a flat plane and normal to the neutral surface. The two assumptions are critical in the mathematical formulation to be presented subsequently. 10.2.4

Mathematical formulation

We start by finding stain at a point H of the beam which is distance y away from the neutral plane NN (Fig. 10.5). Consider a small line AB which passes through point H and is parallel to the neutral surface. The line AB deforms to A B with the radius of curvature ρ. We assume that the curvature of the beam is small and we can still use x-coordinate along curved A B . The normal strain in x-direction is given by xx =

A B  − AB AB

(10.3)

It is worth noting here that this relation could not be written without the assumption that a plane section which is normal to the neutral surface remains plane and normal after deformation. Since MQ is on the neutral surface, its length remains constant, that is, AB = M Q = M  Q = ρΔφ Then, the length of deformed A B  becomes A B  = (ρ − y)Δφ

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O

y A N

H

y M

B N Q x

Mb N

A y

H

B

N

Q

M

Deformed

Undeformed (a)

Fig. 10.5

Mb

(b)

Focusing on the change in length AB to determine xx .

Substituting in Eq. (10.3), we have xx =

y (ρ − y)Δ∅ − ρΔ∅ =− ρΔφ ρ

(10.4)

The form of strain turns out to be very simple; strain is proportional to distance from the neutral surface, y. However, the result is expressed in terms of the radius of curvature ρ, which is still unknown to us. The expression has been obtained just from geometrical considerations. We still have not used the bending moment, the area properties of the cross-section, and the material properties. We still do not know the location of the neutral axis and the radius of curvature. They will be shortly evaluated using the boundary conditions on a cross-section of the beam. In the expression of strain xx = − yρ , the negative sign appears because the stain is compressive for positive y. Since the relation is linear with y, the outermost points of the beam are subjected to the largest strains. Now, we will show that normal stresses in y- and z-directions are negligible too. A beam is a slender member whose lateral dimensions are much smaller than the length. The lateral surfaces are free to expand or contract in the lateral directions and, therefore, we can take σyy and σzz to be negligible at all points of the beam. It can be shown that for sections having one of the two symmetries shown in Fig. 10.1, all the three shear stress components are zero for a member subjected to a pure bending moment. Consequently, the stress tensor

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for all points is written as ⎤ 0 0 ⎥ 0 0⎦ 0 0

⎡

σxx ⎢ σ=⎣ 0 0

Invoking the stress–strain relation, we have xx =

1 [σxx − ν(σyy − σzz )] E

Since σyy and σzz are negligible, we have σxx = Exx Substituting xx = −y/ρ, we obtain σxx = −

Ey ρ

(10.5)

The important boundary conditions for this pure bending case are as follows: BC1 : Net axial force = 0 BC2 : Moment about z-axis = Mb In fact, we will show that BC1 provides the location of the neutral of curvature ρ. axis and BC2 determines the radius The summation of σxx is A σxx dA, where A is the area of the cross-section (Fig. 10.6). Thus, BC1 makes  A

σxx dA = 0

Substitution of σxx = −Ey/ρ yields 

A

Ey − ρ

 dA = 0

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z y

xx

C

y z

x

Fig. 10.6

On a cross-section.

Since E and ρ are constant on a cross-section, we take them out of the integral sign to have  y dA = 0 (10.6) A

The centroid of a section is evaluated as ydA =0 y¯ = A A dA

(10.7)

where y¯ is the distance of the centroid from the neutral axis. Thus, the neutral axis must pass through the centroid of the section to meet the first boundary condition. The net moment of the normal stress σxx about z-axis (Fig. 10.6) is equal to the bending moment at the section, that is,  y(σxx dA) = Mb − A

Substituting σxx = −Ey/ρ, we obtain  

yE dA = Mb y − − ρ A  E or, Mb = y 2 dA ρ A

(10.8)

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In fact, A y 2 dA is an important property of the area and is called moment of inertia I. Thus,  y 2 dA (10.9) I= A

A query may be raised here: Why should we call it a moment of inertia while it is an area property? One of my smart postgraduate students used to call it the second moment of area which I felt was more appropriate. Since everybody calls it a moment of inertia, we will also be addressing it by the same name. The dimensions of the moment of inertia (MOI) are (length)4 . It is a very widely used property of an area and we will be using it extensively in this book. Equation (10.8) then becomes Mb = or,

EI ρ

Mb 1 = ρ EI

(10.10)

Since κ = 1/ρ κ=

Mb EI

(10.11)

This equation gives one of the two important results we have been seeking. To obtain curvedness of the beam, κ is obtained just by dividing the bending moment by EI. It is worth noting that EI appears as a group in the analysis for many practical cases. The group, known as flexural rigidity, is made of the material property E and the area property I. Let us spend a few minutes to explore whether the result of Eq. (10.11) makes sense. We expect higher deformation (κ) with the higher bending moment in the beam. At the same time, κ is inversely proportional to modulus E; it looks reasonable as we expect the curvature to be small with high modulus materials. If the moment of inertia is large, we expect small deformation which sounds reasonable. We can have small bending deformation either by choosing a material of high modulus or increasing the MOI. The modulus of

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steel is one of the highest among commonly used materials. On the other hand, the modulus of a polymer is only 1–2% of that of steel. If we are designing some component of a polymer, we cannot play much with modulus. We then choose the cross-section with a high MOI. These days, polymers are reinforced with high stiffness and high strength manmade glass or carbon fibers. These materials are commonly known as polymer composites. Ey b We will now find σxx by substituting 1ρ = M EI in σxx = − ρ to have 

Mb σxx = −Ey EI σxx = −

or,

Mb y I

(10.12)

It is to be noted that material property does not appear in the expression. The negative sign appears because σxx is compressive for positive y and tensile for negative y. The stress distribution is linear with respect to y for linear elastic materials, as shown in Fig. 10.7. Further, for positive bending moment, the maximum compressive stress is at the top free surface and maximum tensile stress at the bottom surface. The magnitude of the maximum compressive stress at the top surface need not be the same as that of maximum tensile stress at the bottom surface. Also, it is worth noting that the normal components yy and zz are finite due to Poisson’s effect. In fact, σxx E σxx = −ν E

yy = −νxx = −ν zz = −νxx

y Mb N

N

N

N

x

Fig. 10.7 Linear stress field within the material, compressive for positive y and tensile for negative y.

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Thus, the material above the neutral axis expands laterally and the material below contracts for the positive bending moment. The deformation in y and z directions is not important for most cases of real life. The bending moment makes the material move in y-direction, caused by the curvature developed in the beam. When a lateral force is applied on a beam, it develops a bending moment which is an internal generalized force. The internally developed bending moment develops only σxx within the beam and makes the beam curved with curvature κ. Many years back, when the use of computers was rare in carrying out the stress analysis of structural, I worked as a design engineer in a small firm in Los Angeles. My boss, who was the owner of the company also, remembered only the bending formula σxx = −Mb y/I and forgot everything else although he had his bachelor’s degree from a reputed engineering college. He made a lot of money designing his products just using this bending formula. Since then, I call this relation “a six-million-dollar formula”. In our field, probably this is the most widely used relation because many critical components are subjected to bending moments. Either the stress developed within a beam becomes too high to cause its failure or the curvature of the deformed beam becomes so high that it is not acceptable. In the bending formula, σxx = −Mb y/I, there are two variables, y and I, which basically deal with the geometry of the cross-section. Can we combine them to make one? Yes, the resulting variable is section modulus and is the topic of the following section. 10.2.5

Section modulus

For designers and other field persons, the failure of the critical point (the worst loaded point) is important and the critical point corresponds to the largest |y|. It is either at the top point or the bottom point of a beam. Let us call this dimension ym . Then, the section modulus S is defined as S = I/(|ym |)

(10.13)

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(d) (b)

(c)

(a) (h) (f)

(g)

(e)

Fig. 10.8

Various sections of slender members of steel available in markets.

The dimension of section modulus is (length)3 . The bending formula simplifies to σxx = Mb /S

(10.14)

This equation is quite handy to use provided S is easily available. Beams are available in various cross-sections such as round tube, rectangular tube, I-section, angle iron, T-section, and round flat, as shown in Fig. 10.8. These slender members are available in various cross-sectional sizes. Design handbooks or product catalogs list the area properties, I and S, through tables. To find bending stress σxx , the designer takes the values for the selected cross-section from the relevant table. The range of sections available for mild steel is limited mostly to sections shown in Fig. 10.8 due to problems involved in shaping mild steel into more complex shapes. However, many varieties of sections of aluminum metal or polymers are available as they can be extruded easily. 10.2.6

Magic of moment of inertia

In order to have a feel of moment of inertia (MOI), we will determine it for a rectangular section, as shown in Fig. 10.9(a). This is a commonly used section if wood or stone is used as a structural member.

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The MOI of this section is determined easily by considering a strip b × Δy at a distance y from the neutral axis NN as  h/2  2 y dA = y 2 (b × dy) = bh3 /12 I= −h/2

A

Note that the height h of the section plays a dominant role as I ∝ h3 . Let us consider the bending of a steel measuring scale of 300 mm long commonly used in workshops whose cross-section is 25 mm × 1 mm. We will determine its deformation in two configurations, vertical (Fig. 10.9(b)) and horizontal (Fig. 10.9(c)). In the vertical configuration, the moment of inertia Iv is Iv =

1 mm × 253 mm3 = 1,302 mm4 12

In the horizontal configuration, Ih is Ih =

25 mm × 13 mm3 = 2.08 mm4 12

Then, the ratio Iv /Ih is given by 1,302 Iv = 626 = Ih 2.06

y y

P y

h

N

N

P

z

b (a)

Fig. 10.9

x (b)

A beam with rectangular cross-sections.

(c)

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It is a very large number. Since the curvature is inversely proportional to MOI, the measuring scale in horizontal configuration bends to a high curvature, while the vertical scale bends so small that we cannot detect the bending with our naked eyes. You can do this simple experiment and convince yourself by taking a measuring scale of any material, holding one of its ends tightly in one hand and pressing the other end with the other hand to make the scale work like a cantilever. The difference between the two configurations is amazing. The sectional modulus of the two sections is Sv =

Iv 1,302 = 25 = 104.2 mm3 ym (2)

Sh =

Ih 2.08 = 1 = 4.16 mm3 ym (2)

and

Since maximum bending stress developed is inversely proportional to section modulus, the ratio of two stresses is h 104.2 σxx = 25 = v σxx 4.16

Thus, stress developed in the horizontal configuration is 25 times higher than that in vertical member. The difference in the performance of the two configurations is amazing. When we increase MOI of a beam, stresses as well as curvature are reduced, but the reduction in curvature is higher. When we look carefully into the relation I = A y 2 dA, we realize that we should design our beam in such a way that the major portion of the area of a cross-section is as far away as possible. I-section, shown in Fig. 10.8(c), is a good example of the optimum use of material. The horizontal portions, known as flanges, are made comparatively thicker than the central portion known as web. The flanges are quite far away from the neutral axis and, therefore, constitute the major part of the overall MOI. On the other hand, the web is made comparatively thinner and does not contribute much to MOI. I-sections of mild steel are routinely used in civil engineering applications.

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The MOI of a solid round bar is I=

πd4 πr 4 = 4 64

(10.15)

where r and d are the radius and diameter of the round rod. The proof of the expression is presented in Section 10.5.3, Appendix A, which also covers the MOI of some other sections. Example 10.1: A slender member is subjected to a four-point loading, as shown in Fig. 10.10(a). Draw BMD and SFD and show that the middle portion BC of the beam is loaded with pure bending moment (no shear force). Initially, a solid round rod of 12 mm diameter was chosen by the designer. He was suggested to replace the solid rod with a circular tube of 25 mm outside diameter whose mass was the same as that of the rod. Compare the curvature and the maximum stress developed in both members (E = 207 GPa). Given: A four-point loading shown in Fig. 10.10(a); diameter of rod Ds = 12 mm; outside diameter of tube Do = 25 mm; mass of tube = mt ; mass of rod = mr ; E = 207 GPa. To find: SFD and BMD; curvature κs of solid rod and curvature κt of tube and compare them; maximum stresses in each member and compare them. Strategy: Find internal diameter of the tube by equating the mass; max for both determine MOIs of both members to evaluate κs , κt , σxx cases. 300 N

(a)

A

300

B 200

C 800 mm

(b)

(c)

x

SFD –300

Fig. 10.10

(f)

G

300

V Mb 300 s

E 300 V M b

(d) 300 N

V

E 300

200

y

(e)

300

300 D

s

G 60,000 Nmm

Mb A

B

C

D

A slender member with a four-point loading.

BMD

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Solution: Figure 10.10(b) shows the FBD of the beam. Taking a section at point E, as shown in Fig. 10.10(c) and evaluating shear force and bending moment, we obtain V = −300 N Mb = 300 s The bending moment increases with s to 60,000 Nmm at location B of the beam. We will determine Mb and V at location G, as shown in Fig. 10.10(d). V = −300 + 300 = 0 Mb = 300s − 300(s − 200) = 60,000 Nm Thus, the shear force in the middle span BC is zero (Fig. 10.10(e)) and the bending moment is 60,000 Nmm (Fig. 10.10(f)) which is constant in the middle segment. Whenever we want to test a beam under a pure bending moment, we find the four-point loading system handy. If Do and Di are external and internal diameters of the tube and Ds is the diameter of the solid rod, we equate their area of crosssection to have π π 2 (Do − Di2 ) = Ds2 4 4 √ or, Di = Do2 − Ds2 = 252 − 122 = 21.93 mm The MOI of the solid bar is Is =

π(12)4 πDs4 = mm4 = 1,018 mm4 64 64

While the MOI of the tube becomes It =

π(254 − 21.934 ) π(D04 − Di4 ) = = 7,821 mm4 64 64

The ratio R of the two MOIs is R=

7,821 = 7.7 1,018

The increase in MOI is 7.7 times when a solid rod is converted into a tube which is a tremendous advantage. The curvature and the radius

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of curvature of the solid bar in the central portion are evaluated as

and,

κs =

Mb 60,000 Nmm 1 = = 284.7 × 10−6 N 3 4 EIs mm 207 × 10 ( mm2 ) × 1,018 mm

ρs =

1 1 = mm = 3.51 m κs 284.7 × 10−6

For the tube, κt and ρt are determined as

and,

κt =

Mb 60,000 Nmm −6 1 = = 37.06 × 10 N EIt mm 207 × 103 ( mm2 ) × 7,821 mm4

ρt =

1 1 = mm = 26.98 m κt 37.06 × 10−6

Comparing them, we obtain κs = 7.7 κt

and

ρs = 0.13 ρt

Since the central span of the beam is subjected to maximum bending moment, the critical stress is also in the central span. The magnitude of stress at the top point and the bottom point is the same because the member is symmetrical about y = 0 plane. We use the bending s in the solid formula σxx = −Mb y/I to evaluate maximum stress σxx t bar and maximum stress σxx in the tube as s =− σxx

60,000 × ( 12 Mb ym 2 ) Nmm·mm =− = −353.6 MPa Is 1, 018 mm4

t =− σxx

60,000 × ( 25 Mb ym 2 ) Nmm·mm =− = −95.9 MPa It 7, 821 mm4

The ratio of the two stresses is s −353.6 σxx = 3.7 = t σxx −95.8

Comments/Interpretation: In the solid bar, most of the material is close to the neutral axis. When the same material is moved away by shaping into a tube, the MOI increases by a factor of 7.7 which in turn decreases the curvature by 7.7 times. The decrease in stress is 3.7 times.

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Can we make the diameter of the tube still larger to increase MOI further? The MOI, of course, will be higher. But there are other considerations. A larger diameter tube means a thinner wall of the tube for the same mass. When the wall is too thin, a stone falling on it will make a dent easily or just a push by a person will deform the material. There must be a minimum thickness of a structural member to avoid its undesirable deformation under adverse conditions. Parallel axis theorem to determine moment of inertia

10.2.7

The parallel axis theorem is very handy to determine MOI of beams with complex sections. It helps in evaluating MOI of a subarea whose centroid does not lie on the neutral axis. Figure 10.11(a) shows an area (A) whose own centroid C is d distance away from the neutral axis. Then, the MOI of this area (IA ) about the neutral axis NN is given by IA = Ic + Ad2

(10.16)

where Ic is MOI of the area about its own axis which is parallel to neutral axis NN and passes through its own centroid C . The proof of the theorem is presented in Section 10.5, Appendix A.

A

tf

C y

tw

d h N

N

2 Web

z

x

b (a)

Fig. 10.11 I-section.

1 Flange

tf

3 Flange

(b)

(a) Area (A) located at distance d from the neutral axis and (b)

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Example 10.2: Determine (i) MOI of an I-section shown in Fig. 10.11(b) and (ii) compute MOI for h = 100 mm, b = 50 mm, tf = 6.4 mm, and tw = 4 mm. Given: (i) Geometry of I-section shown in Fig. 10.11(b) and (ii) h = 100 mm, b = 50 mm, tf = 6.4 mm, and tw = 4 mm. To find: MOI of the section. Strategy: Consider flange and web separately. Invoke the parallel axis theorem. Solution: (i) If I1 , I2 , I3 are the MOI of the three portions shown in the figure, the overall moment of inertia becomes I = I1 + I2 + I3 Portions 1 and 3 are symmetrically placed about y = 0 plane. Also, MOI is an even function of y, that is, its value remains unchanged if y is replaced by −y. In addition, Ad 2 term in Eq. (10.16) is also an even term with respect to distance d. Thus, I1 = I3 and the overall MOI becomes I = 2I1 + I2 I1 is determined using Eq. (10.16) as I1 =

bt3f 12

+ btf

h tf − 2 2

2

The MOI of portion 2 is I2 =

tw (h − 2tf )3 12

Then,

I = 2I1 + I2 = 2 ×

bt3f 12

+ btf

h tf − 2 2

2  +

tw (h − 2tf )3 (10.17) 12

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(ii) Substituting the given dimensions, we have

50 × 6.43 + 50 × 6.4 × I =2× 12



100 6.4 − 2 2

2 

4 (100 − 2 × 6.4)3 + 12 = 1,404 × 103 + 221 × 103 = 1,625 × 103 mm4 Comments/Interpretation: The parallel axis theorem is simple to use and very handy. In fact, in many sections, the part of MOI determined through the parallel axis theorem is a major contribution to MOI. In this example, the contribution of flanges is 86.4%. Sometimes, we neglect the MOI of the flanges about its own centroid because its contribution is usually small. In this example, it contributes only 1.3% of the total value. Example 10.3: A T-section beam, made of an aluminum alloy, is loaded, as shown in Fig. 10.12(a). The details of its cross-section are shown in Fig. 10.12(b). Determine bending stress σxx at the top and the bottom points on the section where the stress is maximum. Also, determine the radius of curvature at this section. Ignore the effect of shear force while evaluating the stresses developed by the bending moment (E = 70 GPa). y 4 2,400 N B

1

D

60 mm

y

C1

x 300

z

4 (a)

C

32 17.85

1,200 mm

2 50

C2

(b)

Fig. 10.12 (a) The beam with supports and an applied load and (b) the details of its section.

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Given: The beam with loads and supports shown in Fig. 10.12(a); dimensions of its cross-section in Fig. 10.12(b); E = 70 GPa. To find: Location of maximum bending moment; bending stress σxx at the top and bottom points; radius of curvature. Strategy: Plot BMD to determine the location of maximum moment; invoke σxx = −Mb y/I to determine stresses; invoke κ = Mb /EI to determine the curvature and the radius of curvature. Solution: Since the cross-section is symmetric about the vertical plane, we can use the bending formulas developed in this chapter. The FBD of the beam is shown in Fig. 10.13(a). Taking moment about point B, we have ΣMzB : −1,200 RA − 300 × 2,400 = 0 → RA = −

300 × 2,400 = −600 N 1,200

Taking a section at distance s from point A (Fig. 10.13(b)), we determine bending moment as Mb = −600s It is a linear variation in segment AB with its highest value of −720 × 103 Nmm at point B. Taking a section in segment BC, 2,400N y B

A

(a)

1,200 mm

300 RB

RA (b)

C

x

600

V Mb A s

(c)

2,400N Mb

C V

Mb

r BMD

(d) –720 Nm

Fig. 10.13 (a) FBD of the beam, (b) section in segment AB, (c) section in segment BC, and (d) BMD.

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at distance r from point C (Fig. 10.13(c)), we obtain bending moment as Mb = −2400r This is also linear in segment BC with its highest value at point B. The full BMD is shown in Fig. 10.13(d) with maximum bending moment of −720 × 103 N·mm at point B. To determine MOI, we will now determine the centroid C of the entire section. C1 and C2 are the centroid of portions 1 and 2 (Fig. 10.12(b)) and they are located at distances y1 and y2 from the bottom surface. The centroid of the entire section is given by 32 × (4 × 56) + 2 × (50 × 4) mm·(mm·mm) y1 A1 + y2 A2 = A1 + A2 4 × 56 + 50 × 4 mm·mm = 17.85 mm

y¯ =

The moment of inertia is evaluated as I = Contribution of area 1 + contribution of area 2   4 × (60 − 4)3 2 + 4 × (60 − 4) × (32 − 17.85) = 12   50 × 43 2 + 50 × 4 × (17.85 − 2) + 12 = [58.5 × 103 + 44.8 × 103 ] + [0.3 × 103 + 50.2 × 103 ] = 103.3 × 103 + 50.5 × 103 = 153.8 × 103 mm4 t at the top point at location B of the beam is Stress σxx

(−720 × 103 ) × (60 − 17.85) (Nmm) × mm Mb ym =− I mm4 153.8 × 103 = 197.3 MPa

t =− σxx

At the bottom point: (−720 × 103 ) × (−17.85) (Nmm) × mm Mb ym =− I mm4 153.8 × 103 = −83.6 MPa

b =− σxx

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The curvature at the location of the maximum bending moment is κ=

−720 × 103 Mb = EI (70 × 103) × (153.8 × 103)

Nmm N 4 mm2 mm

= −66.8×10−6 1/mm

The radius of curvature is ρ=

1 1 = = 14.95 × 103 mm = 14.95 m κ 66.8 × 10−6

Interpretation/Comments: The stress at the top points of the beam is different from that at the bottom points. The curvature is maximum at the support point B because the bending moment is largest; the negative value means that the beam shape is convex at B. Further, it may look strange that the curvature is maximum under the right support while the deflection is zero. In this problem, it so happened that the beam is not allowed to have vertical displacement at the support point B where the curvature is maximum. But we should note that there is no constraint imposed by the support at location B on the bending rotation of the beam, that is, the beam is allowed to rotate about the z-axis at point B. In the following chapter, we will discuss the deflection of a point due to the beam getting curved by a bending moment. In the bending formula, σxx = −Mb y/I, material property E no longer appears. This means the stress is the same no matter whether we have a beam of very stiff material like steel or soft material like a polymer. But the modulus E appears in the evaluation of curvature using κ = Mb /EI. The beam of nylon will flex much more than a steel beam. However, if we have a beam made of a composite of two or more materials, the modulus of the materials does appear in the expression of the developed stress field. This point will be understood better after we are through solving Example 10.4 which deals with a beam made of two materials. Example 10.4: A beam is made by bonding strips of two materials (1 and 2) as shown in Fig. 10.14. The centroid of both cross-sections is at distance s¯1 and s¯2 and the neutral axis is at distance sn as shown. The beam is subjected to a load of pure bending moment Mb . (a) Show that the neutral axis is given by sn =

E1 A1 s¯1 +E2 A2 s¯2 E1 A1 +E2 A2 .

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y

C1 C

x

N 2

C2

s2

s

sn

N

s1

1

Fig. 10.14

A beam made of two different materials.

b (b) Show that the radius of curvature is 1ρ = E1 I1M +E2 I2 where the moments of inertias are I1 and I2 with respect to the neutral axis. (c) Bending stress in materials 1 and 2 are given by

E1 yMb E1 I1 + E2 I2 E2 yMb =− E1 I1 + E2 I2

(1) =− σxx (2) σxx

where y is measured from the neutral axis. Given: Beam made of two materials 1 and 2 as shown in Fig. 10.14 with centroid C1 at distance s¯1 and centroid C2 at distance s¯2 , subjected to pure bending moment Mb . To find: Location of neutral axis, curvature, and bending stress in each material for the general case. Strategy: Since both materials are deformed with the same curvature, the expression of xx = − yρ is still the same. Find bending stress in each material and then the net axial force is equated to zero to obtain the location of neutral axis. The net bending moment on the section yields ρ. Solution: (a) The strain at a point of the beam is xx = − yρ where y is measured from the neutral axis which is to be determined. Also, ρ is unknown and will be determined. The stresses in both materials

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283

(1) =− σxx

E1 (s − sn) E1 y =− ρ ρ

(10.18a)

(2) =− σxx

E2 (s − sn) E2 y =− ρ ρ

(10.18b)

become

Since there is no net axial force on the section, the force balance in x-direction yields   (1) (2) σxx dA+ σxx dA = 0 A1

A2

We still do not know the location of the neutral axis and, therefore, we will be initially making calculations with the bottom edge of the cross-section as the reference and expressing distances in terms of s, as shown in Fig. 10.14. Substituting the values of stresses, we obtain   E1 (s − sn) E2 (s − sn) dA+ dA = 0 − − ρ ρ A1 A2     E1 sn E1 s E2 sn E2 s dA− dA+ dA − dA = 0 or, ρ ρ A1 A1 ρ A2 A2 ρ   sdA + E2 sdA or, E1 sn A1 + E2 sn A2 = E1 A1

A2

(10.18c) The integrals in the above equations are the same as those used in finding the centroid of a section. Thus,  sdA = A1 s¯1 A1



A2

sdA = A2 s¯2

Substituting these expressions in Eq. (10.18c), we obtain the location of the neutral axis: sn =

E1 s¯1 A1 + E2 s¯2 A2 E1 A1 + E2 A2

(10.18d)

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(b) To determine the radius of curvature, we use the moment equation. We determine the net moment about the neutral axis caused by the stresses and equate it to the bending moment Mb . Also, note that further work on this problem will be carried out with origin at the neutral axis. The moment balance on the entire cross-sectional area gives   (1) (2) σxx ydA − σxx ydA = Mb − A1

A2

Substitution of the stress from Eqs. (10.18a) and (10.18b) yields   



E1 y E2 y ydA − ydA = Mb − − − ρ ρ A1 A2     1 2 2 E1 or, y dA + E2 y dA = Mb ρ A1 A2 But the moments of inertia of both areas are defined with respect to the neutral axis as  y 2 dA I1 =  I2 =

A1

y 2 dA A2

Then, the expression is simplified to Mb 1 =κ= ρ E1 I 1 + E2 I2

(10.18e)

Thus, we obtain the radius of curvature ρ of the deformed beam. (c) We obtain the stress field by substituting ρ from the above expression in Eqs. (10.18a) and (10.18b) to have (1) =− σxx

E1 Mb y E1 I 1 + E2 I2

(10.18f)

(2) =− σxx

E2 Mb y E1 I 1 + E2 I2

(10.18g)

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285

y

C2

20

1

C C

s

C1

20 mm

x

2

(a)

Fig. 10.15

33.5 20.1

10

z

16.54

69.0

50.9 MPa

xx

(b)

A beam made by bonding two rectangular sections.

Comments/Interpretation: The expressions of bending stress involve the modulus of both materials. This is expected as the material with higher stiffness tends to resists the deformation, while the material with lower modulus tends to bend easily. Example 10.5: A special case of the beam of Example 10.4 is shown in Fig. 10.15 with two rectangular sections joined together. A pure bending moment of 180 Nm is acting on the section. Determine the location of the neutral axis, radius of curvature, as well as stress field in each portion and plot the stress distribution on the entire section. Given: Dimensions of each rectangle shown in the figure; modulus E1 = 200 GPa; modulus E2 = 120 GPa; Mb = 180 Nm. (1)

(2)

To find: sn , ρ, σxx , σxx , and plot σxx (y). Strategy: Use equations developed in Example 10.4. Solution: The neutral axis is determined using Eq. (10.18d) as sn = =

E1 s¯1 A1 + E2 s¯2 A2 E1 A1 + E2 A2 (200 × 103 ) × 20 × (20 × 20) + (120 × 103 ) × 5 × (20 × 10) (200 × 103 ) × (20 × 20) + (120 × 103 ) × (20 × 10) ×

(N/mm2 )mm(mm2 ) (N/mm2 )(mm2 )

= 16.54 mm

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286

Also, the moment of inertia for each rectangular section with respect to neutral axis is determined as I1 =

bh3 20 × 203 + Ad2 = + (20 × 20) × (20 − 16.54)2 12 12

= 18.12 × 103 mm4 I2 =

bh3 20 × 103 + Ad2 = + (20 × 10) × (16.54 − 5)2 12 12

= 28.30 × 103 mm4 They are substituted in Eq. (10.18e) to determine radius of curvature as Mb 1 = ρ E1 I 1 + E2 I2 =

180 × 103 (200 × 103 ) × (18.12 × 103 ) + (120 × 103 ) × (28.30 × 103 ) ×

Nmm (N/m2 ) × mm4

= 25.64 × 10−6 or,

1 mm

ρ = 39.0 × 103 mm = 39.0 m

The stress field is obtained by substituting values in Eqs. (10.18f) and (10.18g) as (1) =− σxx

=− ×

E1 Mb y E1 I 1 + E2 I2 (200 × 103 ) × (180 × 103) × y (200 × 103 ) × (18.12 × 103 ) + (120 × 103 ) × (28.30 × 103 ) (N/mm2 ) × Nmm × mm (N/mm2 ) × mm4

= −5.128y MPa (2) =− σxx

E2 Mb y E1 I 1 + E2 I2

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287

(120 × 103 ) × (180 × 103) × y (200 × 103 ) × (18.12 × 103 ) + (120 × 103 ) × (28.30 × 103 ) (N/m2 ) × Nmm × mm (N/m2 ) × mm4

= −3.077y MPa For sketching the distribution of bending stress in y-direction, we note that the variation is linear and thus we need to determine only stress at the top and bottom point of each material. The above equations yield the bending stresses as (1) Stress at the top point of material 1: (σxx )A = −5.128×(30−16.54) = −69.0 MPa (1) Stress at the bottom point of material 1: (σxx )B = −5.128 × (−16.54 + 10) = 33.5 MPa (2) Stress at the top point of material 2: (σxx )B = −3.077 × (−16.54 + 10) = 20.1 MPa (2) Stress at the bottom point of material 2: (σxx )C = −3.077 × (−16.54) = 50.9 MPa Figure 10.15(b) shows the bending stress distribution. Comments/Interpretation: Across the interface, there is a jump in the bending stress. The strain xx is continuous across the interface and when we multiply it with different values of modulus, the resulting stresses show a jump. Thus, the jump exists because of the jump in the modulus.

10.3

Shear Stress Developed due to a Shear Force

We analyzed the case of a beam subjected to a pure bending moment in the previous section. However, almost all beams are subjected to the combination of a bending moment and a shear force. In the 2D analysis that we will be carrying out in this section, the shear force Vy develops predominantly shear stress τxy which is not uniformly distributed on a cross-section of the beam. In this section, we will determine the shear stress distribution on the entire section. Within a slender member, the magnitude of normal stress developed through the bending moment is substantially higher than that

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of shear stress developed through the shear force. Therefore, we assume that the shear force does not appreciably change the results of the case of pure bending. The assumption is quite reasonable for a beam whose length is eight times or more than the largest dimension of its cross-section. We will be showing that the assumption works well. There are some useful cases where the length of a member is of the order of the lateral dimensions or smaller. Some people like to call them short beams. An excellent example is the pin of a hinge shown in Fig. 10.16. The distance between points A and B of the pin is very small, much smaller than the diameter. The force of the upper member is first passed to the pin and then the pin transfers it to the lower member. The portion AB is primarily loaded with a shear force. Some designers manage by the average shear stress obtained by dividing the shear force with the area of the pin’s cross-section. We will be showing that this average value is smaller than the actual shear stress at the critical point. 10.3.1

The nature of shear stress

In comparison to a normal stress, a shear stress is more complex to visualize. Thus, a shear stress component is more abstract and it takes a while to be familiar with it. In this section, we will look into Upper link

A B

Pin Lower link

Fig. 10.16

The pin of a hinge.

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some important aspects of shear stresses which will facilitate us to derive the expression for the shear stress distribution on a section of a beam. Let us consider a 2D stress element shown in Fig. 10.17(a). The shear stress components work simultaneously on all the four faces of the element. If its magnitude on one of the four faces changes, the magnitude of the other three faces changes along with it. It is like water level within a partly filled tank. If the level of water is increased by 20 mm at one corner of the tank, the water level in the other three corners also increases by the same height. Thus, we can measure the height of the water level at any one of the four corners. Similarly, we can determine the shear stress on any one of the four faces, whether it is an x-face or a y-face. In Section 5.9, we have shown that shear stress at a point on the plane of symmetry is zero. We will be using this realization in this section. Figure 10.17(b) shows how a stress element deforms due to a shear stress τxy . The angle ABC, originally at 90◦ , deforms to angle A B C which is less than 90◦ . Similarly, the angle DAB increases from 90◦ to angle D A B . If there is no shear stress, the 90◦ angles remain 90◦ even after deformation which is an important observation. We now consider a beam with shear force V and bending moment Mb acting on its section, as shown in Fig. 10.18(a). The top surface AB and the bottom surface DC are free surfaces on which all the three stress components, σyy , τxy , and τyz are zero. Let us focus our attention at corner point B and draw an infinitesimal stress element

y-face

B

A

B A y

y

x x

C D

x-face (a)

C

D (b)

Fig. 10.17 (a) Shear stress on a 2D stress element and (b) deformation of the stress element.

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290 A

B y x

D

(a)

C

xy=0

V Mb E

B

K

J

H

G

L

C

(b)

xy=0

(c)

Fig. 10.18 (a) A beam with shear force, (b) stress element at corner point B, and (c) stress element at corner point C.

BGHE, shown in Fig. 10.18(b). The shear stress τxy on face EB is zero as it lies on the top free surface. Thus, shear stress on all the four faces is zero. Then, we argue that the angle EBG remains 90◦ even after deformation. Similarly, at the bottom corner point C, the shear stress is zero, as shown in Fig. 10.18(c) and the angle LCJ remains 90◦ after deformation. Figure 10.19 shows how BC plane, which is normal to the neutral surface before deformation, deforms to surface B C due to shear force V. The angle at points B and C remain 90◦ as the shear stress is zero. However, as one moves from point B toward the neutral surface, the magnitude of the shear stress starts increasing. Thus, the deformed surface is no longer a plane. Recall that to solve the problem of pure bending moment, assumptions were made that a plane section, which is normal to the neutral surface, remains plane and normal after deformation. The assumptions are no longer rigorously correct for analyzing problems having both bending moment and shear force acting on the section. However, the presence of shear force does not alter the results of pure bending analysis if the shear force does not vary along the length of the member. In Fig. 10.19, the deformed shape B C is the same as deformed shape D F at some other location. Consequently, the fiber length HK at point H is shifted to H K having the same length. Thus, shear deformation does not affect the strain xx developed through the bending moment. However, if shear force V varies along the length, D F will no longer be identical to B C and the bending strain will be affected. For a slender member, the error involved is small. Thus, we continue using the bending formulas for the mixed loading of bending moment and shear force.

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y D H

B B 90o K K V(constant)

H

x 90o C C

F

Fig. 10.19 Non-planer deformation due to shear force V which does not vary along the length of the member.

10.3.2

Mathematical formulation

Our aim is to determine shear stress distribution developed by a shear force V on a cross-section (Fig. 10.20). Let us determine τxy at a point H at distance y = y1 shown in the figure. It is noted that τxy varies with y with its value zero at the top and the bottom points of the section. In fact, we expect that τxy increases as one moves from top point toward the neutral axis NN . Now, the question is: How can one find the stress distribution at any point of the cross-section?

y xy

T R Q H N z

y1

C V Mb B

Fig. 10.20

N

x

A cross-sectional plane with shear force V and bending moment Mb .

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The analysis is not straightforward. We will use the relation dMb dMb b V = − dM dx . We first relate τxy with dx and then replace ( dx ) with −V . This means that we should consider the specimen in the length direction, as shown in Fig. 10.21(a). We consider a thin slice JKLM of length Δx. On its left face, bending moment Mb and shear force V are present, while on its right face the bending moment is slightly increased to Mb + ΔMb . We further focus on a portion JKTS of the slice, as shown in Fig. 10.21(b). The magnitude of σxx on the right face is slightly higher than that on the left face although we have exaggerated the difference in the figure to make our point. We also show a 3D sketch of the portion JKTS in Fig. 10.22. At point H, we can either deal with τxy on x-face or τxy on negative y-face. Since we will do the force balance in x-direction, we will consider τxy on the negative y-face. Note that in Fig. 10.22, we have omitted showing σxx on both the x-faces just to avoid overcrowding in the figure. On the negative y-face, the total force is ΔF = τxy (Δx)b where b is the width of the section at y = y1 . All the forces acting in x-direction are shown in Fig. 10.21(b). The total force acting on an x-face is obtained by accounting for σxx . The integration is carried out on area Ab which is the area beyond y = y1 , as shown in Fig. 10.22. Thus, the force balance in x-direction yields 

 Ab

σxx dA − τxy (Δx)b +

Ab

y Mb

σxx

K

J

(σxx + σxx) dA = 0

K (σxx+ Δ σxx)

J

V S

T

N

Mb+ΔMb N

x

S ΔF T N

y2

−

N

M Δx L (a)

(b)

Fig. 10.21 (a) A thin slice in the length direction and (b) choosing a portion JKTS of the slice toward its top.

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J

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293

Ab xy on negative y-face

K

xy on x-face

S T

H

b y1

y

N N Fig. 10.22

z

x

3D sketch of portion JKTS showing only the shear stress.

Substituting σxx = −Mb y/I, we have   

(Mb + ΔMb )y Mb y dA − τxy (Δx)b + dA = 0 − − − I I Ab Ab Note that Mb and I do not depend on y of a cross-section and thus they can be taken out of the integral. Thus, the equation then simplifies to

 dMb 1 − y dA τxy = bI dx Ab b Substituting − dM dx = V, we obtain  V y dA τxy = bI Ab

We define

 y dA

Q= Ab

where Q is the first moment of area carried out on the area beyond y = y1 . Then, τxy =

VQ bI

(10.19)

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We now have a relation which enables us to find shear stress at any point of the cross-section. We will now discuss the resulting expression τxy = VQ/bI. It looks reasonable that the shear stress at a point is proportional to the shear force V at that section. Further, note that Q is zero at the top and the bottom point of a section, giving τxy = 0. They, in fact, are our boundary conditions as discussed in the previous section. As one moves from the top point toward the neutral axis, the value of Q increases up to the neutral axis. However, beyond the neutral axis, the value of Q starts decreasing as y becomes negative; eventually, its value becomes zero at the bottom point. Thus, the shear stress attains its maximum value at the neutral axis. To continue our discussion, τxy is inversely proportional to the moment of inertia I. On a section with a high I, the shear stress is low. Further, τxy is inversely proportional to local width b of the cross-section. It is because if a section is narrow at point H, it carries the entire unbalanced force developed through the difference in σxx across the Δx slice of Fig. 10.21(b). Some people like to define shear flow as q = bτxy . In fact, we have assumed that shear stress is uniform along the width of the cross-section at y = y1 . This assumption works well for many sections used in the structural applications. Then, Eq. (10.19) leads to q = VQ/I which is more rigorous. It is worth noting that the proof of the shear stress expression of Eq. (10.19) is not straightforward. To determine shear stress τxy on the x-face, we did the force balance using τxy on the negative y-face of the portion separated out beyond y = y1 . Historically also, this formulation was developed much later than the major development of this field. In fact, Russian engineer D. J. Jourawski observed in the field applications that a built-up wooden section (a beam made by bonding wooden boards together) was failing in shear close to the neutral surface. He then developed his analysis in 1844 by separating the top portion above the neutral surface and showed that high shear stress existed at the neutral axis. It then generated interest among the experts and the rigorous analysis was developed by Saint-Venant in 1856.

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Beam with a rectangular cross-section

10.3.3

A beam with a rectangular cross-section is an interesting case as it gives a good insight into the relation τxy = VQ/bI. In fact, we will find that the magnitude of the shear stress distribution is parabolic in nature with its value zero at the top and the bottom points. Consider the rectangular section beam shown in Fig. 10.23. The MOI of this beam is bh 3 /12. We want to determine τxy at any point, say point H, located at distance y from the neutral axis. The distance of the centroid of area beyond point H is yc = 12 ( h2 + y). Then, Q becomes 1 Q = yc Ab = 2







 h h b h2 2 +y × −y b= −y 2 2 2 4

Thus,   τxy =

VQ = bI

V

b 2

h2 4

− y2  

b

bh3 12

 3 V N × mm3 = (h2 − 4y 2 ) 4 mm×mm 2 bh3 (10.20)

The variation in the magnitude of τxy is parabolic with distance y. At y = ± h2 , shear stress τxy is clearly zero. The distribution on τxy is shown qualitatively in Fig. 10.24(a). The variation of the magnitude

y Ab H

h

y N

N z b

Fig. 10.23

x

A beam of rectangular section.

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y 3V 2A

y x z x

Magnitude

(a)

(b)

of

xy

Fig. 10.24 (a) Shear stress distribution on a cross-section and (b) the variation in the magnitude of shear stress (parabolic).

in y-direction is shown in Fig. 10.24(b). The maximum value of τxy is at y = 0 and is given by 3V 3V = (10.21) 2bh 2A where A is the area of the entire section. The shear stress is maximum at the neutral axis and its magnitude is 50% higher over the average shear stress. Evaluating shear stress by dividing the shear force with the area is too simplistic, incorrect, and optimistic. We also note another important point. The bending stress σxx is maximum where the shear stress is zero. It can be restated that shear stress τxy is maximum where bending stress σxx is zero. We will show that the magnitude of the maximum of τxy in a slender member is substantially smaller than that of σxx . Take an example of a beam supported at its end as shown in Fig. 10.25 whose cross-section is rectangular with height h and width b. The FBD of the beam is shown in Fig. 10.25(b) and the resulting SFD and BMD are shown in Figs. 10.25(c) and (d). The bending moment is maximum at point B and the resulting maximum bending stress in the beam is max = τxy

max =− σxx

PL h ( ) Mb ym 3P L = − 4bh32 = − I 2bh2 12

The maximum shear stress is max τxy

3 × P2 3P 3V = = = 2A 2bh 4bh

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P A

B

C

A

L/2

297 P B L

P/2

page 297

C P/2

L

(a)

(b) PL/4

P/2 0

SFD

BMD

0

P/2

(c)

(d)

Fig. 10.25 (a) A beam of rectangular cross-section is loaded at its end points, (b) FBD of the beam, (c) SFD, and (d) BMD.

The ratio of the magnitude of the two stresses is max τxy = max σxx

3P 4bh 3P L 2bh2

=

h 2L

Now we recall that a slender member is called beam for L ≥ 8h. Then, max τxy 1 ≤ max σxx 16

Thus, the shear stress is substantially smaller and we neglect the effect of shear stress in many practical applications. 10.3.4

Slender member with circular cross-section

The case of circular sections is quite important to us for certain applications. The pin of a hinge shown in Fig. 10.16 is subjected to high shear stress developed due to the shear force acting on it. The analysis of determining shear stress in a circular section is not as simple as that of a rectangular cross-section beam. Consider a 3D stress element at point H on the circumference of a circular cross-section of Fig. 10.26(a) whose blown-up view is shown in Fig. 10.26(b). Due to lateral free surface, the shear stress in radial direction is zero. The non-zero shear stress component is

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Free surface

y

y

=0

H

4r 3

Ab C

N

=0

N

N z N

z

x

x

(a)

(b)

(c)

Fig. 10.26 (a) Point H on the circumference, (b) stress element at H, and (c) shear stress at the neutral axis.

aligned in θ-direction, that is, it is in the tangential direction to the circumference at point H. This shear stress has components in y- and z-directions. Thus, it is made of the combined action of τxy and τxz . In this book, we will not determine shear stress developed due to a shear force at all points of a section. We will just find shear stress only at the neutral axis NN on which τxz is zero and τxy is maximum. Thus, we consider only the top half of the circular section to determine τxy . The centroid of a semicircular area is at 4r/3π distance from the center (Fig. 10.26(c)). Thus,

2  4r 2 πr × = r3 Q = Ab yc = 2 3π 3 Then, the maximum shear stress is given by   V 23 r 3 VQ 4V max  = = τxy = 4 bI 3πr 2 2r πr4 Since the area of the circular cross-section is A = πr 2 , the expression simplifies to 4V (10.22) 3A Thus, the maximum shear stress is 33% higher than the average shear stress. In this case also, the average shear stress is on the optimistic side. max = τxy

Example 10.6: Consider a beam supported at its end points as shown in Fig. 10.27(a) with a center load P = 24,000 N. The crosssection of the beam is I-section and it is the same as used in

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P=24,000 N B

C

y

100 mm

A

z

L/2

2

y

c1

6.4

c2

50

6.4

299

N z

4

4 L=1,000 mm 50 (a)

N

6.4

(b)

(c)

Fig. 10.27 (a) I-beam supported at its end points, (b) the cross-section, and (c) considering the upper half to determine τxy at the neutral axis.

Fig. 10.11(b), which is reproduced here in Fig. 10.27(b) for our convenience. (a) Determine the maximum tensile stress. (b) Determine the maximum shear stress. (c) Show a stress element at the neutral axis in lengthwise view of the beam and rotate the stress element to obtain the maximum compressive stress in the web. Given: Loading on the beam shown in Fig. 10.27(a); dimensional details of the section shown in Fig. 10.27(b), load P = 24,000 N. max , (b) maximum shear stress To find: (a) Maximum tensile stress σxx max τxy , and (c) maximum compressive stress in web at the neutral axis.

Strategy: Determine maximum bending moment and shear force in VQ max = − Mb ym to determine σ max ; use τ the beam; use σxx xy = bI xx I max to determine τxy which is at the neutral axis; rotate axes at the neutral axis to find maximum compressive stress and its direction. Solution: We have determined SFD and BMD for these beam problems several times. The maximum bending moment is under the load and is equal to PL/4. MOI of this case was determined as 1625 × 103 mm4 earlier in Example 10.2. The maximum tensile stress is at y = −50 mm given by max =− σxx

PL ( (− h2 ) Mb ym =− 4 =− I I

= 184.6 MPa

24,000×1,000 ) 4

× (−50) Nmm·mm mm4 1,625 × 103

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(a) The maximum shear force is V = P/2 in the right half of the beam and V = −P/2 in the left half; let us take V = P/2. We consider the top half of the I-section to evaluate the maximum shear stress on the cross-section at the neutral axis. To find Q, the top half portion of the I-section is shown in Fig. 10.27(c). C1 and C2 are the centroids of portions 1 and 2. Then, Q is determined as Q = Σyci Ai = y1 A1 + y2 A2



= (50 − 3.2) × (50 × 6.4) +

50 − 6.4 2



× [(50 − 6.4) × 4] mm·mm2 = 18.78 × 103 mm3 (b) τxy is calculated as max = τxy

3 ( 24,000 VQ 2 ) × (18.78 × 10 ) = 34.7 MPa = bI 4 × (1,625 × 103 )

(c) A stress element at the neutral surface is shown in a lengthwise view of the beam in Fig. 10.28(a) with a blown-up view in Fig. 10.28(b). Note that the bending stress σxx is zero at the neutral surface. The corresponding Mohr circle is shown in Fig. 10.28(c). We rotate the axes by −45◦ so that the axis x corresponds to maximum compressive stress. The state of stress with respect to x −y  axes is shown in Fig. 10.28(d). The maximum compressive stress is 34.7 MPa. If we choose the web to be too thin, the material at the neutral surface may buckle in x -direction and the beam will fail. y

34.6 MPa

34.6

y 34.6

H

y

x

34.6

y H x

H x

y

x 45

o

x

(a)

(b)

(c)

(d)

Fig. 10.28 (a) Point H at the neutral axis, (b) stress element at point H, (c) Mohr circle, and (d) stress element at −45◦ inclination.

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Interpretation/Comments: The maximum shear stress of 34.7 MPa is not as large as σxx = 184.6 MPa but may not be negligible. This is because I-beam is optimized to have more material far away from the neutral surface and the web is chosen to be thin. Example 10.7: A T-beam is made of wood by nailing a 25 mm × 100 mm rectangular section over another 25 mm × 100 mm section, as shown in Fig. 10.29(a). They are nailed together with steel nails at the regular distance d as shown. If a nail is capable of resisting a shear force of 700 N, determine d. A shear force V = 2,000 N acts on this beam at this section. Given: The dimensions of T-section in Fig. 10.29(a); maximum shear force capability of the nail = 700 N; V = 2,000 N. To find: Distance d between nails. Strategy: Find the shear stress τxy at the joint plane; shear force on one nail is equal to τxy × A where A is the area at the joint plane corresponding to one nail (25 × d). Solution: We will first find the centroid C of the section with respect to the bottom surface of the beam (Fig. 10.29(b)). C1 and C2 are the local centroid of areas 1 and 2 as shown in Fig. 10.29(b). The y 1 1

J z

c

c2 50

x

2

25

(a)

12.5

c1

25

c1 y J

100 mm

z

100

100

112.5 81.25

V

100 mm

25

d

z

J

y

C

J

c2

2

25

(b)

(c)

Fig. 10.29 (a) A beam made by nailing two rectangular wooden sections, (b) determining the centroid, and (c) finding shear stress at the joint plane.

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distance yc of the centroid of section C is determined as yc =

yc1 A1 + yc2 A2 112.5 × (100 × 25) + 50 × (25 × 100) = A1 + A2 (100 × 25) + (25 × 100)

mm · mm2 = 81.25 mm mm2 The moment of inertia I of the beam with origin at point C is ×

I = Contribution of area 1 + contribution of area 2   100 × 253 + 100 × 25 × {112.5 − 81.25}2 = 12   25 × 1003 2 + 100 × 25 × (81.25 − 50) + 12 = 7.097 × 106 mm4 Now we find shear stress at joint plane JJ (Fig. 10.29(b)). To evaluate it, we separate area 1 to determine Q as Q = (Area 1) × y¯ = 100 × 25 × (12.5 + 18.75)mm2 ·mm = 78.13 × 103 mm3 Now it is a simple matter of finding τxy at the joint plane JJ as τxy =

2,000 × (78.13 × 103 ) N·mm3 VQ = = 0.881 MPa bI 25 × (7.097 × 106 ) mm·mm4

The shear force S taken by one nail is S = τxy × 25 × d Equating S to the capability of a nail, we have

or,

τxy × 25 × d = 700 N 700 = 31.78 mm = d= N 25 × 0.881 mm × ( mm 2)

Interpretation/Comments: This is a good example of the importance of determining shear stress developed by a shear force in a beam.

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303

Summary

In this chapter, we have developed the relations to determine stress distribution caused by a bending moment Mb and a shear force V. These two generalized forces are addressed in this chapter because b they are coupled with the important relation V = − dM dx . We have carried out the analyses in 2D in this chapter; the beam is aligned along x-axis, the shear force is in y-direction, and the bending moment is about z-axis. The bending moment develops normal stress σxx . There exists a neutral surface on which σxx is zero. The neutral axis passes through the centroid of the section and thus the centroid is used as the origin of the coordinate system employed. The resulting expression for the stress developed by the bending moment is σxx = as moment of inertia which is an − MIb y . In this relation, I is known area property given by I = A y 2 dA. The variation of σxx is linear with y. The bending stress σxx is tensile on one side of the neutral surface and compressive on the other. The maximum magnitude of σxx is developed at the farthest point from the neutral axis of the section. The formulas developed in this chapter are valid only for the sections which are symmetrical either across the vertical plane or the horizontal plane passing through the centroid. The bending moment deforms a straight beam to a curved beam. The curvedness is measured through the parameter curvature κ which is evaluated using the expression κ = Mb /EI. The product EI is known as modulus of rigidity. The radius of curvature ρ is related to curvature κ as ρ = 1/κ. The shear force on a section develops shear stress on the crosssection. Its value is zero at the top and the bottom points of the section. Its magnitude is highest at the neutral axis. The resulting expression to determine shear stress is τxy = VQ/bI where Q is the first moment of area (Ab ) beyond the point at which τxy is being determined. It is defined as Q = Ab ydA. In most cases of beams, the magnitude of maximum shear stress developed by a shear force is substantially smaller than the magnitude of the maximum stress developed by a bending moment on a section.

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10.5

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Appendix A: Moment of Inertia — An Important Area Property Definition

Moment of inertia (MOI) of an area is an important property. In fact, it is a second moment of the area and is needed to determine stress developed by a bending moment, and shear stresses developed by a shear force. MOI is a second-order symmetric tensor with six independent components. However, in this chapter, we need to know only three of the six components and, therefore, we will not be looking into a comprehensive coverage on MOI. The components of this secondorder tensor, needed for our analysis, are as follows (Fig. 10.30):   r 2 dA = (y 2 + z 2 )dA (about x-axis) (10.23) Ixx =  Iyy =  Izz =

A

A

z 2 dA (about y-axis)

(10.24)

y 2 dA = I

(10.25)

A

(about z-axis)

A

Ixx is about x-axis and is known as polar moment of inertia and it will be addressed by symbol J; we will need it to solve the torsion problems to be discussed in Chapter 12. Iyy is the second moment of area about y-axis and we need it for bending problem in which the bending moment (My ) about y-axis exists. In this chapter, we have not solved problems involving My as the generalized force because its solution is similar to that of z

y

dA y

z

Fig. 10.30

x

Determination of moment of inertia.

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305

bending moment Mz which has been addressed as Mb in this chapter. However, Problem 10.16 deals with both My and Mz . Izz is the second moment of area about z-axis. It is an important element of this chapter to evaluate σxx from Mz and τxy from V and we just call it I. Why do we use double subscript? Whenever we use MOI, we should keep in mind that it is a second-order tensor. There are three other components of this second-order symmetric tensor, Ixy , Ixz , and Iyz . In fact, if a cross-section is not symmetric either about the vertical plane or horizontal plane, Iyz plays an important role. However, we will not be using these three components of moment of inertia in this book and, therefore, will not discuss them. 10.5.2

Parallel axis theorem

If MOI is known about an axis, it can be determined about any other parallel axis. The theorem is quite handy in determining the MOI of a beam with a complex cross-section. For example, consider a section shown in Fig. 10.31 whose MOI is to be determined. This section can be divided into three portions, as shown in the figure. Portion 1 is a rectangle whose MOI is known about its own centroid C1 . But the centroid C of the entire section does not coincide with the local centroid C1 . The parallel axis theorem enables us to transfer the local MOI to the axis passing through the centroid C of the overall crosssection and thus determine the contribution of portion 1 to the overall MOI of the section. Similarly, we find the MOI of other portions and transform them to the axis passing through overall centroid C. Consider an area A shown in Fig. 10.31(b) whose MOI is known about axis BB passing through its own centroid C as IBB  . Our interest is to find MOI of this section about another parallel axis DD which is distance d away as shown. Then, the theorem states that IDD = Ad2 + IBB 

(10.26)

To prove this relation, let axis DD be z-axis and y-direction is normal to it, as shown in the figure. Consider an infinitesimal area dA at a distance y from DD . Then, the MOI of area A about axis DD is  y 2 dA IDD = A

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306 b1 h1

B

C1 y

z

y

dA

1

2

y1

B

C y

C

d

y 3

D

D

z

(a)

(b)

Fig. 10.31 (a) Dividing a complex section into simple portions and (b) parallel axis theorem.

Substituting y = d + y  , where y  is respect to axis BB , we obtain  IDD = (d + y  )2 dA A



2



(d) dA +

= A



 2

(y ) dA + A

(2y  d)dA

A

Since d is constant, it is taken out of the integrals. Also, IBB  . The equation is simplified to  2 y  dA IDD = Ad + IBB  + 2d

A

(y  )2 dA =

A

We note that y is measured from the centroid of area A and invoking the definition of centroid we have A y  dA = 0. Thus, the above relation is simplified to our desired result as IDD = Ad2 + IBB  This relation is extremely useful to determine MOI of a complex cross-section. Example 10.8: Determine MOI of a rectangular tube shown in Fig. 10.32 through two approaches: (a) by deducting MOI of the hollow portion from MOI of the outside rectangle and (b) by dividing the tube into four rectangles. Given: A rectangular cross-section is shown in Fig. 10.32.

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307

20 B

A

A A

B B

60 mm

y z

D D D

C C C 2 mm all over

Fig. 10.32

A rectangular tube.

To find: MOI. 3

Strategy: (a) I = Iout − Iin ; invoke I = bh 12 . (b) Divide into four portions, find MOI of each portion using parallel axis theorem, and add MOI of all the four portions. Solution: (a) The MOI of the outside area ABCD is Iout =

20 × 603 bh3 = = 360 × 103 mm4 12 12

The MOI of the hollow portion EFGH is Iin =

16 × 563 (b − 2t) × (h − 2t)3 = = 234.2 × 103 mm4 12 12

Then, the MOI of the rectangular tube becomes I = Iout − Iin = 360 × 103 mm4 − 234.2 × 103 mm4 = 125.8 × 103 mm4 (b) The wall of the tube is divided into two horizontal strips, from left edge to the right edge, and two vertical strips, each of 56 mm height. The horizontal strips are identical in shape and are located

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at the same distance from the z-axis. Similarly, vertical strips are identically placed. Thus, the overall I of the section is expressed as I = 2I h + 2Iv where subscripts h and v represent horizontal and vertical strips. Invoking the parallel axis theorem to horizontal strips, we have

 

2  h t 2 60 bt3 20 × 23 + bt − + 20 × 2 −1 = Ih = 12 2 2 12 2 = 33.7 × 103 mm4

 2 × (60 − 4)3 = 29.27 × 103 mm4 Iv = [(t(h − 2t) )/12] = 12 

3

I = 2I h + 2I v = 2 × 33.7 × 103 + 2 × 29.27 × 103 = 125.9 × 103 mm4 Interpretation/Comments: We thus obtain the same value, within the rounding errors, from both methods. 10.5.3

Polar moment of inertia

The polar moment of inertia J is defined for x-axis as (Fig. 10.33(a))  r 2 dA (10.27) J= A

The polar moment of inertia is needed to determine shear stress developed due to a torsional moment whose analysis will be taken up in Chapter 12. However, we find the evaluation of J convenient to determine Izz or Iyy for some sections having symmetry with both y- and z-axes. A circular section is one such section. We will show that J = Iyy + Izz . In the above equation, we substitute r 2 = y 2 + z 2 to have    2 2 2 J = (y + z )dA = (z )dA + (y 2 )dA A

or,

J = Iyy + Izz

A

A

(10.28)

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y

y

dA r

page 309

dr

R

r

z

C x

z (a)

(b)

Fig. 10.33 (a) Determination of polar moment of inertia and (b) polar moment of inertia of a circular area.

10.5.4

Moment of inertia of a circular cross-section

For a circular solid section of radius R (diameter d), we will show πd4 πR4 = J = Polar moment of inertia = 2 32  4 4 πd πR = y 2 dA = I = Izz = 4 64 A  πd4 πR4 = z 2 dA = Iyy = 4 64 A

(10.29) (10.30) (10.31)

To obtain these relations, we first evaluate polar moment of inertia J. We will then invoke Eq. (10.28) to determine MOI about z-axis. Consider a ring of Δr thickness at distance r, as shown in Fig. 10.33(b). The J of this ring is r 2 (2πrΔr) as all points of this ring are at distance r. The overall J then becomes 

R

J=

r 2 (2πrΔr) =

0

πR4 2

In the relation J = Iyy + Izz , note that Iyy is the same as Izz because of the circular symmetry. Thus, Izz

1 J =I= = 2 2



πR4 2

 =

πR4 πd4 = 4 64

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If a structural member is a circular tube of radius R and wall thickness t where t R, we have J = R2 [(2πr)t] = 2πR3 t =

πd3 t 4

Then, Izz = 10.6

J πd3 t = πR3 t = 2 8

Problems

1. A cantilever, made of aluminum (E = 70 Gpa), is loaded in such a way that a pure bending moment acts at its free end, as shown in Fig. 10.34. Determine the following: (a) Radius of curvature of beam AB. (b) Maximum bending stress. 2. A wire of 4 mm diameter is made of a hardened alloy steel (E = 207 GPa) with a high yield point. It is bent to form a complete circle of radius 500 mm, as shown in Fig. 10.35(a). Determine the maximum stress generated within the material of the wire. The wire is deformed through elastic deformation only. 3. Consider a slender member with its cross-section, as shown in Fig. 10.35(b). If the bending moment at a section of the member is 300 Nm, determine the bending stress distribution on the

x

S

B

200 mm

SS

9

20 mm

S

A

100

y

100

200 N

200 N (a)

Fig. 10.34

(b)

Problem 1.

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311

y

10

0

50

30 mm

4 m

m

10

z

30

(a)

(b)

Fig. 10.35

(a) Problem 2 and (b) Problem 3.

50

6

15

z

8

80 mm

6

15

y

6

50

30

6

b

3

30

h

R10

20

Fig. 10.36

10 mm

Problem 4.

section. If the maximum allowable stress is 90 MPa and the bending moment is increased further gradually, which side of the section will yield first? 4. Determine the centroid and the moment of inertia for each section of Fig. 10.36. 5. The I-beam of Example 10.2 is reinforced by welding flat bars of 40 mm × 6 mm section on the top and bottom surfaces symmetrically, as shown in Fig. 10.37(a). Determine the percentage decrease in the maximum bending stress.

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6

6.4

H

4

R R z

6.4

40 50

6

Mb

x

Welded

(a)

Fig. 10.37

y

(b)

(a) Problem 5 and (b) Problem 6.

6. A thin wall tube of radius R and wall thickness t (Fig. 10.37(b)) is subjected to a bending moment Mb . Determine bending stress σxx as a function of θ if Mb = −50 Nm, R = 15 mm and t = 1.5 mm. Plot σxx for 0 ≤ θ ≤ 180◦ . 7. Photo-elasticity is a technique to determine stress field within a specimen made of a transparent material like epoxy. Consider a specimen of a rectangular cross-section of 20 mm × 6 mm which is supported on its end points loaded with a four-point loading fixture to have a pure bending moment in portion CD of the specimen (Fig. 10.38). A loading bridge is placed between the loading arm and the specimen as shown to facilitate the development of a pure bending moment of constant magnitude (no shear force) in the central portion CD of the specimen. If the weight of the loading arm is 5 kg and a dead weight of 20 kg is applied at the free end of the loading arm, determine the maximum stress developed in the central portion of the specimen. 8. A wooden plank is to be designed to place books whose weight is 32 kg per meter length, as shown in Fig. 10.39. The supports are at a distance s from the ends of the plank. A thin steel plate is screwed at each end to stop the books from falling; the weight of the plank and the thin plates can be neglected.

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313

800 mm 300

Loading arm

Bridge

D

C

20

Specimen

60

20 Kg

100

Fig. 10.38

Photo-elastic setup of Problem 7. Books

s L

Wooden plank

Fig. 10.39

s

b

Wooden plank for books of Problem 8.

(a) Show that the maximum bending moment developed is minimum corresponding to s = 0.207L. (b) Determine the maximum bending moment if L = 1,400 mm and b = 250 mm. (c) Determine the safe thickness of the plank from the point of view of strength if the allowable stress of the critical point is 10 MPa. 9. A circular rod of mild steel is bent, as shown in Fig. 10.40(a). Its end at A is rigidly clamped. (a) Determine bending moment diagram for all the three portions of the structure. (b) Determine maximum bending moment.

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y 100 N

D

P

100

2R

150 mm

A

6R

8 Dia. (all over)

x

C

B

L

60 60

(a)

Fig. 10.40

(b)

(a) Problem 9 and (b) Problem 10.

(c) Determine the maximum stress (critical point) in the structural member. Note that member BC carries two kinds of generalized forces, bending moment and tensile axial force. Invoke the principal of superposition to determine the critical point/points. 10. A cantilever is made of a truncated cone, as shown in Fig. 10.40(b). (a) Show that the bending stress is maximum at x = (3/4)L. (b) If L = 120 mm, R = 20 mm, and P = 5,000 N, determine the maximum bending stress. (c) Plot σxx vs. x to have some feel of the bending stress variation. 11. A beam is made of two wooden bars of cross-section 100 mm × 50 mm each. They are bonded together with an adhesive (Fig. 10.41(b)). The resulting beam is loaded in a three-point loading system, as shown in Fig. 10.41(a). Determine shear stress at the bonded interface. 12. The wooden bars of Problem 11 are not bonded but bolted together with 6 mm diameter bolt with configuration shown in Fig. 10.41(c). Each bolt is capable of resisting shear force of 1,000 N. Determine the spacing d between the bolts. 13. A flat section of 80 mm × 5 mm is joined to a thin wall circular tube through brazing (Fig. 10.42). The resulting section is subjected to a shear force of 5,000 N.

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315

10,000 N (a) 1,000 2,000 mm

10

0

50

50

d

(b)

(c)

Fig. 10.41 (a) A beam supported at its end point, (b) bonded wooden bars of Problem 11, and (c) bolted wooden bars together of Problem 12.

(a) (b) (c) (d)

Determine the location (¯ s) the neutral axis. Determine moment of inertia I. Shear stress stress at the brazed interface. Determine maximum shear stress on the sectional plane.

14. A boxed section is made from 10-mm-thick wooden planks through two configurations, (a) and (b) shown in Fig. 10.43. In the configuration (a), the planks are bonded at the interface AA and BB , while in configuration (b) planks are bonded at the interfaces DD and EE . The shear force on each cross-section is V = 1,000 N. Determine the following: (i) shear stress at interface AA . (ii) shear stress at interface DD . Which interface is subjected to smaller shear stress? 15. (a) A beam is made of a sandwich configuration with its outer layers (known as skins) made of a high strength and high stiffness polymer composite material (E = 100 MPa) and center core from a foam of very low density and a low stiffness (E = 0.4 GPa) as shown in Fig. 10.44(a). It can be assumed that the foam takes up negligible bending stress. If the shear force on the cross-section is

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3

50 dia.

80 mm

s

Brazed

5

10

Problem 13.

10

Fig. 10.42

V

A

10 100 V

B

E

B

10

E

D

10 100

A

10

10 m 0m

(a)

Fig. 10.43

D 10

10 mm 00

1

(b)

Beams of boxed cross-section of Problem 14.

V = 2,000 N, determine the shear stress at the interfaces between the skins and the foam. Does the shear stress remain constant through the foam material? 16. A beam is made of a rectangular cross-section, as shown in Fig. 10.44(b). It is subjected to two bending moments,

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3

V

x

40 (a)

Fig. 10.44

My Mz

z

3

40 mm

y

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317

50 mm

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x

30

(b)

(a) Problem 15 and (b) Problem 16.

Mz = 120 Nm and My = 80 Nm. Determine the stress field. Also, determine the location and the magnitude of the maximum tensile and compressive bending stresses.

B1948

Governing Asia

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Chapter 11

Deflection due to Bending

11.1

Introduction

In many situations, the bending moment is the most dangerous force out of the four internal forces as it develops high stresses. I, therefore, sometimes call it a monster. This monster not only develops high stresses but also causes large transverse deflections too. If we pull a bicycle spoke by holding its ends between the two hands, we will hardly notice the extension. In fact, the extension is so small that we need a sensitive measuring device to detect it. When one end of the spoke is held rigidly with one hand and a transverse force is applied at the other end, the spoke is bent easily. In fact, the spoke takes a curved shape and the resulting maximum transverse deflection is several millimetres or several centimetres and easily seen through our naked eyes. One of my friends was involved with the design of a railroad coach which is supported on beams resting on the front and rear sets of wheels. The deflection of the coach is maximum at the mid-length, caused by the bending moment. The specifications did not allow the deflection to exceed 18 mm, that is, if the deflection exceeds 18 mm, it is considered to be a failed design. The beams used in designing the coach were quite heavy and strong, but still, the deflection was considerable. The design was stiffness-based and not strength-based. The presently used materials have been developed to have high strength by adding suitable alloying elements and carrying out appropriate heat treatments. For example, when the carbon concentration is increased in steel from 0.2% to 0.5%, the yield stress increases

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Mechanics of Materials: A Friendly Approach q B

C

A

Fig. 11.1

Large deflections due to rotation of the member.

substantially but the modulus is hardly affected. It is difficult to enhance the modulus of a material which controls the deflection of a structure. Thus, often the deflection of a structural member is a major concern in many designs and then the design is stiffness based. Why does bending moment cause high deflection? To answer it, let us consider an example of a cantilever ABC shown in Fig. 11.1. The distributed force is applied on a small segment AB of the member. In this problem, the bending moment is developed only within segment AB and there is no bending moment in segment BC. Segment AB deforms to a curve whose slope at point A is zero as the beam is fixed at this point. At point B, there is a finite slope θ. But segment BC remains straight as it is not subjected to any bending moment. However, it rotates by angle θ and point C deflects by a large distance. Thus, we find that the slope rotates a member causing large displacements, addressed as distance leverage in this book. We sometimes do not like a phenomenon of nature and start cursing it. But the same phenomenon may be used to our advantage. For example, we make various kinds of springs based on the large deformation caused by bending. A leaf spring used under a car or truck is made of thin strips of a high strength steel. A force, applied at the centre of the leaf spring, moves by a considerable distance and thus a large quantity of energy is stored which is known as strain energy. The bow of archery also works on the same principle. When the archer pulls the string, the bow bends to a higher curvature and thus stores strain energy. When the archer withdraws the pulling force, the arrow acquires a high velocity. Another good example is of a pole vault used by an athlete who runs and picks up speed while holding the pole in his hand. Just before the cross bar, he butts (stops) one end of the pole on the ground for which an appropriate provision is made. He stores the entire kinetic energy of his body by letting the pole curve quite a lot while he supports his body by hanging on the other end of the pole. The stored strain energy is

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then released and facilitates the pole vaulter to go over the high bar which is placed at heights usually greater than 5 m. In this chapter, we will determine the deflection caused by the internal bending moment. Most books call a slender member a beam and external forces as loads when it is subjected to bending moments. This terminology is highly influenced by civil engineering applications. But the concepts being developed here are applicable to other fields like mechanical, aerospace and chemical engineering, material science, and bio-engineering. We will address the structural member by a more general term, a member, in place of a beam and forces in place of loads.

11.2

Governing Differential Equation

Both kinds of internal forces, shear forces and bending moments, cause transverse deflection of a member. However, deflection developed by a shear force is much smaller than the deflection caused by the bending moment in a slender member and, therefore, we neglect the deflection by the shear force most of the time. Why do we neglect small quantities regularly? Some like to call it the art of approximation or neglecting. A natural phenomenon, if formulated very factually, is quite complex and our calculations may be so involved that we may not be able to solve the equations. However, we neglect judicially and always in comparison. An initially straight member becomes curved due to the internal bending moment. We keep track of the deformation of the neutral surface which curves along the length of the member. The deflected curve is known as the elastic curve of the member. In this book, we align the axis of the member with x-axis and in most problems the external forces are in x–y plane. The displacement of a point is addressed by the vector (u, v, w). We usually do not determine displacement u is in the axial direction because it is quite small in comparison to the transverse deflection. We focus on finding the lateral deflection v in y-direction. In z-direction, the deflection w, developed mostly due to Poisson’s effect, is quite small in comparison to deflection v. We are now ready to formulate the governing differential equation. In Chapter 10, we did develop a relation to determine the

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curvature from the bending moment. The curvature represents a kind of deformation. The local curvature κ was obtained from the bending moment Mb as κ=

Mb EI

(11.1)

It is worth noting that a bending moment, in general, varies with x and, therefore, curvature κ also varies with x. EI is called flexural rigidity and gives an idea of how curved a deformed member is. For example, the resulting deflection is small if its flexure rigidity is high. The relation between curvature κ of a curve and deflection v is geometric and is well known as d2 v dx2

κ=   dv 2 3/2 1 + dx

(11.2)

This is a complex relation, but we can simplify it for our applidv of the member is much smaller than unity for cations. The slope dx most cases. Its square makes it further smaller than unity. For example, a slope of 5◦ is 0.087 in radians which is quite small. Its square 0.0076 is still smaller in comparison to unity. Thus, in the denomina dv 2 is neglected in comparison to unity tor of the above equation, dx and then the relation is simplified to κ=

d2 v dx2

(11.3)

We eliminate κ from Eqs. (11.1) and (11.3) to have the governing differential equation (GDE) as EI

d2 v = Mb dx2

(11.4)

To determine the transverse deflection v, we integrate GDE twice with two integration constants. We will discuss the various boundary conditions in the following section to evaluate the value of the integration constants.

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Segments and Boundary Conditions 2

d v Before we solve EI dx 2 = Mb , we should explore its domain within which it can be integrated. If the term Mb /EI is not continuously differentiable at a point, we cannot integrate it through that point. We call this point the transition point. We will now discuss various kinds of transition points. On the member AB, shown in Fig. 11.2(a), an external moment M acts at location H. In its bending moment diagram, the bending moment would show a jump at point H. Clearly, the GDE is not defined at location H. We must solve the GDE separately in segments AH and HB. Thus, location H is the transition point that divides the member into two segments. Similarly, force P at location Q in Fig. 11.2(b) also separates the member into different segments. The bending moment at Q is piecewise continuous but it is not differentiable. Thus, Q is the transition point, dividing the member into different segments. Figure 11.2(c) shows a roller support which is similar to Case (b) and location R is the transition point. A hinge support, shown in Fig. 11.2(d), is a transition location where forces are applied in x- and y-directions. The distributed force shown in Fig. 11.2(e) has transition points at locations A and B. In the case of the distribution force, shown in Fig. 11.2(f), q(x) is only piecewise continuous at location J. This, in turn, does not make the bending moment differentiable at location J. Thus, J becomes the transition point.

P

R

M A

C

B

H (a)

D

Q

y

(c)

(b)

J

q(x)

x H E

(d)

Fig. 11.2

A

B

F

G

H

(e)

Transition points of various cases.

K (f)

L

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324

y P1 B A

C

B

D

A

(a)

G

x C

D

E

H

J P1 K

F

(b)

Fig. 11.3 (a) Transition points when the moment of inertia is not continuous and (b) various types of transition points on a slender member.

Figure 11.3(a) shows a member whose cross-section changed gradually from location B to location C and there is step change in the cross-section at location D. Since the moment of inertia depends on the geometry of the cross-section, the term Mb /EI of the GDE is not differentiable at locations B, C, and D and, therefore, they are the transition points. There are many transition points on the slender member of Fig. 11.3(b) as follows: (i) support points at locations A, C, F; (ii) external forces at locations B, D and K; (iii) change in the crosssection at location E; (iv) discontinuity of the distributed force at G, H and J. Thus, there are nine segments and we should solve the GDE separately in each segment. The member does not bend sharply across a transition point if the material remains linear elastic at all points. In fact, deflection v and dv are the same across a transition point. They are the useful slope dx boundary conditions at the junction point. In Fig. 11.4, the elastic curve of the deformation is shown with a dashed line. If we imagine a point H − just left of H and a point H + just right of point H, the two boundary conditions at the transition point are as follows: BC1 : vH − = vH +     dv dv BC2 : = dx H − dx H + The GDE can be solved in various ways. We will cover only two methods in this chapter: 1. Direct integration method; 2. Method of superposition.

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Elastic curve H

Transition point

Fig. 11.4 The elastic curve of the deformed member with the same value of deflection and slope at the transition point H.

11.4

Direct Integration Method 2

d v The integration of the GDE, EI dx 2 = Mb , determines the deflection of a bent member. We will now solve a few problems.

Example 11.1: Consider a slender member whose one end is rigidly fixed, as shown in Fig. 11.5(a). A moment M acts at a distance x = a. Determine the deflection at the free end B of the member. Given: Member length= L; fixed at end A; Moment M applied at x = a. B

y (a)

Ax

a

C

B

M L

Mb (b)

x

C

M

M (c) A

C

BMD B

(d)

Fig. 11.5 (a) The slender member AB, (b) finding bending moment in segment AC, (c) bending moment diagram, and (d) the deformed shape of the member.

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To find: vB . Strategy: Integrate the GDE and apply boundary conditions at point A. Solution: There are two segments, AC and CB, and two following boundary conditions at point A: BC1 :

v = 0 at x = 0 dv = 0 at x = 0 BC2 : dx We now take a section in segment AC at distance x, as shown in Fig. 11.5(b). To find the bending moment, save the right portion, as shown in the figure. The bending moment in segment AC is Mb = M

0≤x≤a

The bending moment diagram is shown in Fig. 11.5(c). The GDE then becomes d2 v =M 0≤x≤a dx2 Integrating the GDE in segment AC with c1 as integration constant, we obtain EI

dv = M x + c1 dx We now invoke BC2 at x = 0 to have EI

c1 = 0 Thus, dv = Mx (11.5a) dx Integrating again with c2 as integration constant, we obtain EI

M x2 + c2 2 Invoking BC1 at x =0, we have EIv =

c2 = 0

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Thus, the equation of elastic curve in the segment AC is given by v=

M x2 2EI

(11.5b)

The dashed line in Fig. 11.5(a) shows the elastic curve. The segment AC deforms to take the shape of a parabola, but the segment CB does not deform as there is no bending moment within it and remains straight. The slope of this segment is the same as the local a slope at point C of the parabolic curve ( M EI ). Thus, the deflection of a2 point B is the sum of deflection ( M 2EI ) at point C and the deflection due to the rotation of the segment (vB/C ). Thus, vB = vC + vB/C

(11.5c)

M a2 M a + (L − a) 2EI EI M aL M a2 − = EI 2EI

=

Figure 11.5(d) shows the view of the deformed member; it is exaggerated in y-direction. In case a = L, we obtain vb =

M a2 2EI

Interpretation/Comments: The deflection is proportional to the applied moment as per our expectations. However, it is proportional to the square of the length of the member and it is inversely proportional to flexural rigidity EI. Example 11.2: Consider a slender member AB with its end A fixed, as shown in Fig. 11.6(a). A force P acts at distance x = a. Determine deflection of the end point B. Given: Member fixed at A; member length= L; force P applied at x = a. To find: vB . Strategy: Integrate the GDE in segment AC and apply boundary conditions at A.

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Solution: There are two segments AC and CB. There is no loading in segment CB and, therefore, it remains straight although it is rotated due to local slope at point C. To find the bending moment in segment AC, we take a section, as shown in Fig. 11.6(b). The bending moment at distance x in Segment AC is given by   Mb = P (a − x) 0 ≤ x ≤ a Then, the GDE becomes EI

  d2 v = P (a − x) 0 ≤ x ≤ a 2 dx

The boundary conditions at point A are as follows: BC1 : v = o at x = 0 dv = 0 at x = 0 BC2 : dx The integration of GDE with c1 as constant of integration yields EI



P x2 dv = P ax − + c1 dx 2

0≤x≤a



Applying BC2 yields c1 = 0 y (a)

A

P

x

a L Mb

(b)

x

C

B

C

B

P

a

(c)

Fig. 11.6 (a) The member, (b) sectioning to find bending moment in AC, and (c) the deformed shape.

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Then, EI



P x2 dv = P ax − dx 2



0≤x≤a

(11.6a)

Integrating again with c2 as integration constant, we obtain EIv =

P ax2 P x3 − + c2 2 6





0≤x≤a

Invoking BC1 , we have c2 = 0. Then,     P ax2 P x3 1 − 0≤x≤a v= EI 2 6

(11.6b)

The dashed curve in Fig. 11.6(a) corresponds to the elastic curve. The deflection of point C becomes  3 Pa 1 vC = EI 3 The deflection of point B is given by   a≤x≤L vB = vC + vB/C vB/C is given by vB/C = (Slope at point C from Eq. (11.6a)) × (L − a)   P a2 1 2 Pa − × (L − a) = EI 2 =

P a2 (L − a) 2EI

Thus, vB =

1 EI



P a3 3

 +

P a2 (L − a) 2

(11.6c)

The deformed shape of the member is shown in Fig. 11.10(c) which is exaggerated in y-direction.

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When the load is applied at the free end (a = L), we obtain 1 vB = EI



P L3 3



The deflection of the cantilever is widely used and, therefore, this result is worth remembering. Interpretations/Comments: We will now discuss the nature of deflection due to bending. In all problems dealing with members made of linear elastic materials, the deflection at a point of the member is proportional to the applied force and it sounds reasonable due to the linearity of all equations involved. However, the deflection is proportional to the square of the length for the applied moment, as we found in Example 11.1. It is proportional to the cube of the length, as shown in this example. We will show in Example 11.3 that it is proportional to the fourth power of length if the applied force is distributed. Thus, we should be careful while designing members of a long length. In fact, if a member is quite long, it is worth exploring the possibility to have more supports at the intermediate points. The deflection is inversely proportional to the modulus of the material chosen. Mild or medium carbon steels and alloy steels are widely used materials in the industry. There are several reasons for it but its high modulus is an important factor. In fact, the modulus of steel is one of the highest moduli of materials used in most applications. However, as discussed in Chapter 8, we cannot modify the modulus of a metal significantly. On the other hand, the strength of a metal is quite often enhanced considerably by alloying it with small percentages of other metals or through a suitable heat treatment. Thus, we do not have much of a choice in choosing a material whenever we look for a high modulus material. The modulus of polymeric materials is about 1% of steel’s modulus. The strength of a polymer is not too much less than that of mild steel but the modulus is too small in comparison. The low value of modulus of most polymers is the major drawback. These days, we like to reinforce a polymer with high stiffness and high strength fibres made of glass, carbon, or Kevlar fibres. These materials are known as polymer composites. We can make the deflection smaller by increasing the moment of inertia (MOI) of the member and designers do it all the time.

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The MOI is increased by controlling the section of the member in such a way that most of the material is placed far away from the neutral axis. Sections such as I-section, channels, Z-sections, and tubes are good examples of high MOI. A plate is not a suitable member from both stiffness and strength point of view and, therefore, we try to avoid using it as a structural member. We thus play with the modulus and the MOI of a structural member to restrict its deflection. Example 11.3: Find deflection of a member with a distributed force q, as shown in Fig. 11.7(a). Determine the elastic curve of the deformed member, the maximum deflection, and the maximum bending moment. Given: Length= L; pin supported at the left end; roller supported at the right end; constant distribution force q on the entire length. To find: (i) The elastic curve, (ii) maximum deflection, and (iii) maximum bending moment. Strategy: Integrate the GDE, apply boundary conditions at points A and B, and make use of the symmetry of this case. y (a)

q x

A

L

B q

(b) RA= qL/2

RB= qL/2

q Mb

(c)

x

qL/2

(d) A

B

Fig. 11.7 (a) The member, (b) reactions, (c) bending moment at distance x, and (d) deflected member.

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Solution: This is a single segment problem with the boundary conditions: BC1 : v = o at x = 0 BC2 : v = o at x = L Also, note that it is a statically determinate problem as we can determine all the reactions just by using the equilibrium equations. Since there is no force applied in x-direction, the reaction in x-directions at point A is zero. This problem becomes symmetric with respect to forces in y-direction and, therefore, the reactions in y-direction (Fig. 11.7(b)) are as follows: qL 2 qL RB = − 2 The bending moment at x-distance is (Fig. 11.7(c)): RA = −

Mb = −

x qLx + (qx) 2 2

(11.7a)

The GDE then becomes x qLx d2 v + (qx) EI 2 = − dx 2 2 The integration of the GDE with c1 as constant of integration yields qLx2 qx3 dv =− + + c1 dx 4 6 Integrating again with c2 as integration constant, we have EI

qLx3 qx4 + + c1 x + c2 12 24 Invoking BC1 at x = 0 yields EIv = −

c2 = 0 Invoking BC2 at x = L gives c1 =

qL3 qL3 qL3 − = 12 24 24

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Substituting the value of c1 and c2 , we obtain the elastic curve as v=

q (x4 − 2Lx3 + L3 x) 24EI

(11.7b)

Since the deflection is symmetric, the maximum deflection is at x = L/2. It turns out to be vmax =

5 qL4 384 EI

(11.7c)

The deformed shape is shown in Fig. 11.7(d) (shown exaggeratedly). The maximum bending moment Mbmax is at x = L/2 and is obtained from Eq. (11.7a) as      L qL L2 L qL2 max 2 + q =− =− Mb 2 2 2 8 Interpretation/Comments: In this case of distributed force, the deflection is proportional to the fourth power of length. Should we support the members at very short intervals? If we look into the design of modern bridges made by supporting its beams on pillars, we find that pillars are erected at short intervals. In some cases, it costs a lot of money to erect pillars, especially in the middle of deep water of a river or in shallow sea water. If making pillars is prohibitively expensive, the civil engineers like to avoid using beams and resort to a suspension bridge in which the major forces are tensile and are born by thick steel cables. Example 11.4: A slender member is supported at its ends, as shown in Fig. 11.8(a). A force P acts at a distance x = a. Determine the deflection of the member. Given: Length= L; pin joint at A; roller support at B; force P at x = a. To find: Deflection of the member, v. Strategy: Integrate the GDE separately in segment AC and CB with appropriate boundary conditions at A, C, and B. Solution: This is a statically determinate case as all reaction components can be determined using equilibrium equations only. There is

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no force in x-direction and, therefore, the reaction in x-direction at point A is zero. Force balance in y-direction yields (Fig. 11.8(b)). ΣFy :

RA + RB = P

Moment about point B gives ΣMB : −RA L + bP = 0 Pb RA = L

or,

Substituting in ΣFy equation, we obtain RB = P −

P aP bP = (L − b) = L L L

The nature of the elastic curve is shown with the dashed line in Fig. 11.8(a). The member does not bend with a kink at point C and, therefore, all the four boundary conditions are as follows: BC1 : v = 0

at x = 0

BC2 : v = 0 at x = L     dv dv = BC3 : da C − da C + y (a)

P A

x

P B

C a

L

b

a R A= Pb/L

Pb/L

R B= Pa/L Mb

A

(c)

(e)

B

C

A

(b)

x

P C

(d) Pb/L

Mb

x

Fig. 11.8 (a) The member AB with external force P, (b) free body diagram of AB, (c) bending moment in segment AC, (d) bending moment in segment CB, and (e) the deformed shape.

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where C − is just left of point C and C + is just right of point C as the slope of the two segments is the same at point C. BC4 : vC − = vC + because the deflection is the same at point C. We will first solve the GDE in segment AC. The bending moment in the segment is given by (Fig. 11.8(c)). Mb =

Pb x L

(0 ≤ x ≤ a)

Then, the GDE becomes EI

Pb d2 v x = 2 dx L

(0 ≤ x ≤ a)

Integrating with c1 as integration constant, we have EI

P bx2 dv = + c1 dx 2L

(0 ≤ x ≤ a)

(11.8a)

Integrating again with c2 as integration constant, we obtain EIv =

P bx3 + c1 x + c2 6L

(0 ≤ x ≤ a)

Invoking BC1 at x = 0, we have c2 = 0 Then, we are left with EIv =

P bx3 + c1 x 6L

(0 ≤ x ≤ a)

(11.8b)

Now we develop relations in segment CB. The bending moment is given by (Fig. 11.8(d)). Mb =

Pb x − P (x − a) L

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The GDE becomes EI

Pb d2 v x − P (x − a) (a ≤ x ≤ L) = 2 dx L

Integrating with c3 as integration constant, we have EI

P bx2 P (x − a)2 dv = − + c3 dx 2L 2

(a ≤ x ≤ L)

(11.8c)

Its integration one more time yields with c4 as integration constant EIv =

P bx3 P (x − a)3 − + c3 x + c4 6L 6

(a ≤ x ≤ L)

(11.8d)

Applying BC2 at x = L yields P bL3 P (L − a)3 − + c3 L + c4 = 0 6L 6

(11.8e)

BC3 equates the slope at x = a. Thus, Eqs. (11.8a) and (11.8c) yield P ba2 P ba2 + c1 = + c3 2L 2L yielding c1 = c3 Since BC4 equates the deflection at x = a, Eqs. (11.8b) and (11.8d) yield P ba3 P ba3 + c1 a = + c3 a + c4 6L 6L Since c1 = c3 , the equation makes c4 = 0

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Then, Eq. (11.8e) is simplified by putting L − a = b and c4 = 0 to P bL3 P b3 − + c3 L = 0 6L 6 yielding c3 =

Pb 2 (b − L2 ) = c1 6L

Substituting c1 and c3 in Eqs. (11.8b) and (11.8d), we obtain the required elastic curves as v= v=

Pb [x3 − (L2 − b2 )x] (0 ≤ x ≤ a) 6EIL  L Pb x3 − (x − a)3 − (L2 − b2 )x 6EIL b (a ≤ x ≤ L)

(11.8f)

(11.8g)

dv = 0 and it is interestThe maximum deflection corresponds to dx ing to determine whether the deflection is minimum under the load point. Let us determine the slope at point C from Eq. (11.8a) as    1 P ba2 P b 2 P bL 3a2 b2 dv 2 = + (b − L ) = + 2 −1 dx EI 2L 6L 6EI L2 L

Note that the slope is not zero at point C unless a = b = L/2. In fact, the maximum deflection is at some other point as can be seen from the elastic curve shown in Fig. 11.8(a). Another interesting case is to find the maximum deflection under the applied force for the case a = b = L/2. Then, Eq. (11.8f) yields v=−

P L3 48EI

(11.8h)

Interpretation/Comments: The algebra becomes quite involved when we have several segments because we then have to solve our GDE separately in each segment with a large number of integral constants. Further, in comparison to the deflection of a cantilever (Example 11.2), the deflection of a simply supported member is much

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reduced and, therefore, we prefer to support a force through a member with a support on both sides of the applied force. However, in certain problems, we are constrained to use the cantilever mode in supporting the external force. If we want to use a member as a spring, where we need to have large displacement, we prefer to go for a cantilever kind of loading. Leaf springs and pole vaults are good examples of the cantilever type of loading.

11.5

Solution of Indeterminate Cases through the Integration Method

We should determine all reaction components at the support points to develop expressions for internal bending moment and deflection of a member. If it is a statically determinate case, we can determine all the reaction components using only the equilibrium equations and we need not use the material properties and the area properties of the cross-section. It does not make a difference whether a member is made of wood or a metal. In a statically indeterminate structure, we need to develop additional equations. They are obtained by solving the deflection relations and applying geometrical constraints. However, in doing so, the material properties and the area properties of the member’s cross-section (moment of inertia for bending problems) may be required. If the member is made of a homogeneous material and its crosssection is uniform throughout the length, the expression EI may cancel out in the mathematical analysis and the material or area properties may not appear in the expressions of reactions. However, they will appear in the expressions if the cross-section or modulus varies with length. For example, when a slender member is made by joining two materials through a butt joint, the material properties will appear in the expressions of the reaction components. If the left half is made of a steel bar and the right half of a brass bar, the expressions of the reactions of an indeterminate structural member will have the modulus of both materials. The value of reactions will be different from those of a member made entirely from steel. On the other hand, if the supports make the member statically determinate, the reactions will be independent of the materials used.

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It is important here to discuss the degree of a statically indeterminate structure. In 2D structures, we have three equilibrium equations. If the number of unknown reaction components is four, we call the structure to be statically indeterminate of degree one. If the unknown reactions are five, it is of order two. Thus, if there are nr unknown reaction components and ne equilibrium equations, the statically indeterminate structure is of (nr − ne ) degree. We will be solving simple problems of statically indeterminate cases in this section using the direct integration method. More complex problems are solved effectively through the technique of superposition or other methods. Example 11.5: A slender member is rigidly supported at one end and roller supported on the other end, as shown in Fig. 11.9(a). The member is loaded with a distributed force of constant magnitude q N/m as shown. Determine the reactions at the support points, the elastic curve, and the maximum deflection. Given: Member length L; fixed support at A; roller support at B, distributed load q N/m. To find: (i) Reaction components MA , RA and RB (Fig. 11.9(b)), (ii) elastic curve v(x), and (iii) maximum deflection vmax . Strategy: Invoke equilibrium equations; solve the GDE and apply geometric compatibility relations at the supports to obtain an additional equation. Solution: There are no forces in x-direction and, therefore, the x-component of reaction at point A is zero. Thus we are left with y (a)

q

q B

A

x

(b)

B

MA A RA

L

RB q

(d)

(c)

Mb

MA A

RA

x

Fig. 11.9 (a) The member, (b) reaction components, (c) bending moment at distance x, and (d) the deformed shape.

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two equilibrium equations and three unknown reactions, MA , RA and RB , as shown in Fig. 11.9(b). This problem is statically indeterminate of degree one. There are three constraint boundary conditions at the supports. The boundary conditions are as follows: BC1 : v = 0 at x = 0 dv = 0 at x = 0 BC2 : dx BC3 : v = 0 at x = L The two equilibrium equations provide (Fig. 11.9(b)) ΣFy : RA + RB = qL

(11.9a)

ΣMA :

(11.9b)

  L =0 − MA + RB L − qL 2

The bending moment at x-distance is (Fig. 11.9(c))

x Mb = MA + RA x − qx 2 Then, the GDE becomes EI

x d2 v = M + R x − qx A A dx2 2

Integrating it with c1 as integration constant, we have EI

RA x2 qx3 dv = MA x + − + c1 dx 2 6

(11.9c)

BC2 at x = 0 makes c1 = 0. Integrating again with c2 as integration constant, we have EIv =

MA x2 RA x3 qx4 + − + c2 2 6 24

(11.9d)

BC1 at x = 0, makes c2 = 0. BC3 at x = L yields MA L2 RA L3 qL4 + − =0 2 6 24

(11.9e)

This is the additional equation we have been looking for. We now have three equations (a, b, e) for three unknowns, RA , RB , and MA.

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The solution of these three simultaneous equations yields RA =

5 3 qL2 qL RB = qL MA = − 8 8 8

Substituting the value of RA and MA in Eq. (11.9d), we obtain the elastic curve as  

x 3

x 2 x 4 qL4 −2 v(x) = +5 −3 +0 (11.9f) 48EI L L L Now we will determine the maximum deflection whose location is dv = 0. We differentiate the above equation to have given by dx  

x 2 qL3 x 3 x dv = −8 + 15 −6 +0 =0 dx 48EI L L L It has three roots of x/L as 0, 0.578, and 1.297. The first one, of course, is the slope at x = 0, our BC2 . The third one is beyond point B and is not relevant to us. Thus, the maximum deflection is at x/L = 0.578. Substituting this value in Eq. (11.9f), we obtain the maximum deflection as vmax = −0.00542

qL4 EI

Interpretation/Comments: The direct integration method enabled us to generate one additional equation using the geometric constraint at x = L to determine all the three reaction components. Example 11.6: A moment Mo is applied to a slender member, fixed at one end and roller supported at the other end, as shown in Fig. 11.10(a). Determine the reactions and the elastic curves. Given: Length= L, fixed support at A; roller support at B; external moment Mo at mid-length. To find: Reaction components MA , RA and RB (Fig. 11.10(b)); elastic curves v(x) in segments AC and CB. Strategy: Solve the GDE of deflection and invoke all the three geometrical constraints conditions at the supports and continuity conditions at point C to develop an additional equation.

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y (a)

A

B

C

x

L/2

(b)

MA

M0 L

(c)

(d)

C

RA

M0

MA RA

(e)

A

RB

Mb

x C

MA RA

B

x

Mb

M0

Fig. 11.10 (a) the member, (b) unknown reactions, (c) bending moment in segment AC, (d) bending moment in segment CB, and (e) the deformed shape.

Solution: This problem has two segments, AC and CB. Also, it is statically indeterminate of degree one as only two equilibrium equations are available and there are three unknown reaction components, MA , RA , and RB . There are five boundary conditions and four integration constants, two in each segment. Thus, we will obtain one additional equation needed to determine the reaction components. The boundary conditions are as follows: BC1 : v = 0 at x = 0 dv = 0 at x = 0 BC2 : dx BC3 : v = 0 at x = L     dv dv = BC4 : da C − da C + where C − is just left of point C and C+ is just right of C; the slope of the two segments is the same at point C BC5 : vC − = vC + The deflection of the two segments is the same at point C. The two equilibrium equations provide (Fig. 11.10(b)) ΣFy : RA + RB = 0

(11.10a)

ΣMB : −M A − RA L + Mo = 0

(11.10b)

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We will first solve the GDE in segment AC. The bending moment at distance x is (Fig. 11.10(c)). Mb = MA + RA x

(0 ≤ x ≤

L ) 2

The GDE becomes EI

d2 v = MA + RA x dx2

(0 ≤ x ≤ L/2)

Integrating with c1 as integration constant, we have EI

RA x2 dv = MA x + + c1 dx 2

0 ≤ x ≤ L/2

(11.10c)

BC2 at x =0 makes c1 = 0. Integrating again with c2 as integration constant, we have EIv =

MA x2 RA x3 + + c2 2 6

(0 ≤ x ≤ L/2)

(11.10d)

BC1 at x = 0 makes c2 = 0. In segment CB, the bending moment is (Fig. 11.10(d)) Mb = MA + RA x − Mo

(L/2 ≤ x ≤ L))

The GDE becomes EI

d2 v = MA + RA x − Mo dx2

(L/2 ≤ x ≤ L)

Integrating with c3 as the integration constant, we have EI

x2 dv = MA x + RA − Mo x + c3 dx 2

(L/2 ≤ x ≤ L)

(11.10e)

Its integration one more time yields with c4 as integration constant EIv =

x3 Mo x2 MA x2 + RA − + c3 x + c4 2 6 2

(L/2 ≤ x ≤ L) (11.10f)

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Since the slope of left side is the same as that of right side at point C, we invoke BC4 by using Eqs. (11.10c) and (11.10e) to have L RA L2 L RA L2 Mo L + + 0 = MA + − + c3 2 8 2 8 2 It simplifies to MA

L 2 The deflection of left side is the same as that of the right side at point C and, therefore, we invoke BC5 at x = L/2 using Eqs. (11.10d) and (11.10f) (with c3 = Mo L2 ) to have c3 = Mo

MA L2 RA L3 Mo L2 Mo L2 MA L2 RA L3 + +0= + − + + c4 8 48 8 48 8 4 yielding Mo L2 8 Substituting c3 and c4 in Eq. (11.10f), we have c4 = −

x3 Mo x2 Mo Lx Mo L2 MA x2 + RA − + − (L/2 ≤ x ≤ L) 2 6 2 2 8 (11.10g) Invoking BC3 at x = L, the expression yields

EIv =

L3 Mo L2 Mo L2 Mo L2 MA L2 + RA − + − =0 2 6 2 2 8 Substituting RA =

Mo −MA L

from Eq. (11.10b), we have   Mo − MA L3 Mo L2 MA L2 + − =0 2 L 6 8

It yields MA = −

Mo 8

From Eq. (11.10b), we have RA =

9Mo 8L

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Then, Eq. (11.10a) gives RB = −

9Mo 8L

The elastic curve in segment AC is obtained by substituting RA and MA in Eq. (11.10d) as     x 3 x 2 L Mo L2 3 − 0≤x≤ v= 16EI L L 2 The elastic curve in segment CB is obtained from Eq. (11.10f) as  

x 2

x x 3 Mo L2 3 −2 (L/2 ≤ x ≤ L) −9 +8 v= 16EI L L L The deformed shape of the member is shown in Fig. 11.10(e) in which the lateral deflection is exaggerated in y-direction. Interpretation/Comments: We can solve the statically indeterminate problem of several segments by invoking appropriate boundary conditions at the meeting points of the segments. The direct integration method is powerful enough to solve statically determinate and indeterminate problems. However, if the number of segments is more than two, the calculations become cumbersome as each segment provides two integration constants and two continuity conditions at each transition point. Even with two segments, there are a large number of calculation steps. The method of superposition is much faster. 11.6

Method of Superposition

We make extensive use of the principle of superposition in the field of mechanics of materials. A complex problem is split into several subproblems whose solutions are already known to us. We then add up the solutions of the subproblems to have the solution of the complex problem. We use the principle of superposition frequently knowingly or unknowingly to solve problems. However, it pays to know this principle articulately so that we can exploit this powerful tool more often

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to our advantage. I like to attach a well-known fable to it. A rich man had seven sons and he wanted to teach them an important lesson when he was on his death bed. He took seven thin wooden sticks and tied them together tightly with a strong string. The eldest son was asked to break the bundle. He tried but could not do it. Other sons also tried it but failed. The old man then untied the bundle and distributed the individual sticks, one to every son. Everybody broke their stick easily. I use this fable whenever I face a complex problem and I try to break the problems into subproblems. Does the principle of superposition work for all kinds of materials? The principle of superposition works well in solving problems when the material behavior of the member is linear elastic. In other words, external forces should not be high enough to cause plastic deformation. In most designs of real life, we do not let the material yield and then the stress–strain relations remain linear. We will now explore mathematically how linear equations enable us to use the principle of superposition. Consider two subproblems, 1 and 2, of a problem. We have the stress–strain relations for both parts as 1 E 1 = E

(1) xx = (2) xx



 (1) (1) (1) σxx − ν σyy + σzz



 (2) (2) (2) σxx − ν σyy + σzz

Summing them, we have (2) ((1) xx + xx ) =





  1  (1) (2) (1) (2) (1) (2) (σxx +σ xx ) − ν σyy + σyy + σzz + σzz E

Thus, the solution of the sum of the individual problems is the solution of the more complex problem. Now, we will explore how the superposition works for our bending equation for deflection (the GDE). The equations for the two subproblems are as follows: (1)

M d2 v (1) = b dx2 EI

(2)

M d2 v (2) = b 2 dx EI

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The summation of these equations yields (1)

(2)

(Mb + Mb ) d2 (v (1) + v (2) ) = dx2 EI Thus, the principle of superposition works for our GDE. It is worth noting that all equations we use for linear elastic materials are linear and we can make good use of the principle. We made several assumptions, like neglecting the effect of the shear force on the deflection of the member and neglecting the slope of the member in the curvature-displacement relation (Eq. 11.2). We should always remember our assumptions made while formulating a mathematical model and check whether the assumptions are valid for the problem at hand. When the deflection of a member is much smaller than the length, our assumptions are quite reasonable. One of my civil engineering friends told me that his team designs structures in such a way that the maximum deflection of a slender member is less than L/300 where L is the length of the member. People use different “verbs” for obtaining subproblems from the main problem such as to divide, to break, and to split. In the context of this book, I like to use “to split” because I feel it shows more action. The concept of splitting a complex problem into smaller problems is extensively used in all branches of engineering. In fact, I will go a step further to state that it is used to solve many problems of dayto-day living. We have already used the principle of superposition several times in this book. In Section 8.6.1, we used the principle of superposition to develop the relations between normal strains and normal stresses. We divide the general loading on a slender member into four types of internal generalized forces, axial, bending (two kinds in 3D), shear (two kinds in 3D), and torsion. We then find stresses and deflection developed by each internal force separately. In Chapter 9, we considered the case of axial force. In Chapter 10, we discussed how to determine stresses developed by a bending moment and also through a shear force. We will discuss the stresses developed by a torsional moment in Chapter 12. In Chapter 13 on combined stresses, we will discuss in detail how to sum up the stresses developed by each generalized internal force. We will now solve a simple problem to appreciate how powerful the method of superposition is.

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Example 11.7: Two external forces P and 2P are applied on a cantilever, as shown in Fig. 11.11. Determine the deflection of point B using the superposition method. Given: Cantilever length L, force P acts at x = L in negative y-direction, force 2P acts at x = L/2 in positive y-direction. To find: vB . Strategy: Split into two subproblems, each one having only one external force and use the results of Example 11.2. Solution: We split the problem into two subproblems, 1 and 2, as shown in Fig. 11.11. In Example 11.2, we obtained the general expression to determine the end deflection for the case when force P is applied at x = a as (Eq. (11.6c)):  3 Pa P a2 1 + (L − a) vB = EI 3 2EI The subproblem 1, in which a = L, yields (1)

vB = −

P L3 3EI

The subproblem 2 yields  2   3  (2P ) × L2 2P L2 L (2) + L− vB = 3EI 2EI 2 =

5 P L3 24 EI 1 A

y A

L/2

2P

2 L

A

C L/2 2P

Fig. 11.11

P B

L

P B

C

x

C

A cantilever loaded with two forces.

B

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The net deflection at point B is obtained by superposing the two results as   5 P L3 1 P L3 1 (1) (2) =− vB = vB + vB = − + 3 24 EI 8 EI Interpretation/Comments: It is a two segments problem and calculations are quite involved if we solve it through the direct integration method. Also, note that the geometrical compatibility conditions (displacement constraints) at the support are met and the sum effect of the externally applied forces is the same as that of the original problem. We need a library that lists the deflections and the slope of the deformed member of simple problems of a slender member for various kinds of loadings. In Section 11.10, Appendix A, the results of some basic configurations are presented. 11.7

Method of Superposition to Solve Determinate Cases

We recall that statically determinate cases are the ones in which all reaction components of the supports can be determined using equilibrium equations alone. The method of superposition is very effectively applied to solve the problems of these kinds. All we do is that we split a given problem into subproblems in such a way that there is only one external force acting on each subproblem. We lift the results of each subproblem from our library of Section 11.10, Appendix A, and they are summed together to obtain the solution of the problem. We will solve a few problems to demonstrate the effectiveness of the method. Example 11.8: Figure 11.12 shows a cantilever on which three different kinds of forces are applied. Determine the deflection of points D and B (flexural rigidity = EI ). Given: Cantilever length= L; moment M acting at x = L/2; a triangular distributed force with q at x = 0 and zero at x = L; force P acting at x = 3L/4; flexural rigidity=EI.

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To find: vD ; vB . Strategy: Split the problem into three subproblems; find vD and vB for each subproblem and then add them. Solution: The three subproblems, 1, 2, and 3, are shown in Fig. 11.12. For subproblem 1, the elastic curve from Fig. 11.20(c) is v (1) =

(−qo )x2 [10L3 − 10L2 (x) + 5L(x)2 − (x)3 ] 120LEI

We obtain displacements of point D (x = 3L/4) as  2  3   2    qo 34 L 10L3 − 10L2 34 L + 5L 34 L − 34 L (1) vD = − 120LEI 4 qo L = −0.0229 EI The displacement of point B is directly obtained using the expression of the maximum deflection from Fig. 11.20(c) as (1)

vB =

q o L4 (−qo )L4 = −0.0333 30EI EI q0 B

A C

y q0 A

x

C L/2 L

B

1

M

B

A C

L/2

M D L/4 P

2

L

B

A 3L/4

C

D P

3

Fig. 11.12 The loaded member of Example 11.8 with three forces and the three subproblems.

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For subproblem 2, we obtain the elastic curve from Fig. 11.20(f) as

v

(2)

   M L2 L 2x − = 2EI 2

L/2 ≤ x ≤ L

This expression yields displacement at points D and B as (2) vD

and,

(2)

vB

     M L2 3 L 1 M L2 2 L − = = 2EI 4 2 4 EI L   M 2 L 3 M L2 2L − = = 2EI 2 8 EI

For subproblem 3, the elastic curve from Fig. 11.20(e) is

v

(3)

=

P

 3 2   3x − 34 L 4L 6EI

3 L≤x≤L 4

The expression yields

(3) vD

and,

(3)

vB

 3 2   3  3  3 4L − 4L P L3 4L = 0.141 = 6EI EI  3 2   3 3 P 4L 3L − 4 L PL = 0.211 = 6EI EI P

Summing the solutions of all the three subproblems, we obtain M L2 P L3 q o L4 + 0.250 + 0.141 EI EI EI 4 2 ML P L3 qo L + 0.375 + 0.211 vB = −0.0333 EI EI EI

vD = −0.0229

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Interpretation/Comments: We solved this problem of three segments easily with the help of Section 11.10, Appendix A, by invoking the method of superposition just by splitting the complex problem into three subproblems with each one having only one external force. There is another advantage of this method. The chances of making a mistake are reduced drastically as we do not have to keep track of many integration constants and a large number of equations. We just add up the known solutions. Example 11.9: A slender member is supported through a hinge at point A and a roller support at point B, as shown in Fig. 11.13(a) with forces 3P and P acting at points C and D. Determine the deflection of points C and D (flexural rigidity=EI ). Given: Member AD of length 3L/2; supported at A with a hinge; supported at point B with a roller support; force P acts at x = (3/2)L; force 3P acts at x = L/2; flexural rigidity=EI. To find: vC , vD . Strategy: Split into two subproblems and account for slopes of the members. 3P

y (a)

3P A

P

C x L/2

B

L

A

D

1 C

A

L/2

B

C

B

D

P

2 A

(b)

C

A 2

B

P

2a

B PL/2

C

P B

D

2b

Fig. 11.13 (a) The loaded member AD with its two subproblems 1 and 2 and (b) the subproblem 2 is further divided into two sub-sub-problems, 2a and 2b.

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Solution: In subproblem 1, the displacement of point C is obtained directly from Fig. 11.20(h) as P L3 (−3P ) L3 =− 48EI 16EI To find the displacement of point D of subproblem 1, it is realized that segment BD rotates anticlockwise as can be seen through the elastic curve shown with the dashed line. The angle of rotation is obtained from Fig. 11.20(h) as (1)

vC =

(1)

θB = −

(3P )L2 (−3P )L2 = 16EI 16EI (1)

The point D moves up by the distance θB × BD and thus 3 P L3 (3P )L2 L × = 16EI 2 32 EI Subproblem 2 is further split into two sub-sub-problems, 2a and 2b, as shown in Fig. 11.13(b). The portion AB bends convex up as shown by the elastic curve shown through the dashed line in subproblem 2. To determine the deflection of point C, note that segment AB is loaded with a moment PL/2 at point B as shown for sub-sub-problem 2a. Then, the deflection of point C is determined from Fig. 11.20(j) as PL L  2  L   L 2  − 3L × × 2L 2 2 2 + 2 P L3 (2) = vC = 6LEI 32EI The deflection of point D is the sum of two deflections: (i) due to rotation of segment BD in clockwise direction caused by moment  PL − 2 in sub-sub-problem 2a and (ii) due to deflection of segment BD working as cantilever in sub-subproblem 2b. The rotation of point B is determined using Fig. 11.20(j) as PL   L P L2 (2) 2 = clockwise θB = 3EI 6EI (1)

vD =

Then, the deflection of point D due to the rotation of segment BD becomes   L 1 P L3 P L2 (2),rot × =− =− vD 6EI 2 12 EI

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The deflection due to cantilever action is obtained from Fig. 11.20(a) as  3 P L2 1 P L3 (2),can =− =− vD 3EI 24 EI Thus, deflection of point D of subproblem 2 becomes 1 P L3 1 P L3 1 P L3 − =− 12 EI 24 EI 8 EI Summing up the results of both subproblems, we have (2)

(2),rot

vD = vD

(2),can

+ vD

=−

P L3 P L3 P L3 + =− 16EI 32EI 32EI 3 3 1 PL 1 P L3 3 PL − =− and vD = 32 EI 8 EI 32 EI vC = −

Interpretation/Comments: We should keep track of the nature of elastic curve so that the effect of the member’s rotation and its rotational direction can be easily accounted for. Example 11.10: Consider a slender member of length L which is fixed at point A, as shown in Fig. 11.14(a). A distributed force of constant magnitude q acts from x = L/3 to x = 2L/3. Determine the deflection of point B (flexural rigidity = EI ). Given: Length = L; distributed force q from x = L/3 to x = 2L/3; flexural rigidity = EI. To find: vB . Strategy: We will use the Green function approach. We will first find vB for a point load qdx at distance x and then let this load point move from point C to point D by integrating the result of the point load. Solution: Consider a very thin slice of the distributed force at distance x shown in Fig. 11.14(b). The force contribution of the thin slice, qdx, can now be considered as a concentrated load as dx can be made as small as we want. The deflection of point B is determined from Fig. 11.20(e) as vB =

(−qdx)x2 (3L − x) 6EI

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q A

C

xxx L/3

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355

qdx D

L/3 L (a)

B x

dx (b)

Fig. 11.14 (a) Member AB with distributed force q and (b) the member with a concentrated force q dx at distance x.

The deflection due to the entire distributed force becomes q vB = −

 2L 3 L 3

x2 (3L − x)dx 6EI

 2L/3 q  3Lx3 x4  0.0355qL4 =− − = − 6EI  3 4 L/3 EI

Interpretation/Comments: The Green function approach is a very powerful technique to solve many problems in all technical fields. If we know the response of an action at a point, we can let the point move on a finite length and integrate the response. The force distribution need not be constant. For example, if the force distribution is trapezoidal, we could still solve the problem through the appropriate integration (Problem 8). We have solved several different kinds of problems in this section. In the first problem of this section, we showed how to split a problem with multiple loads. In the second problem, we showed how to account for the bending rotation of the member. In the third problem, we showed how to use the Green function approach to solve problems of a varying distributed load. 11.8

Method of Superposition to Solve Indeterminate Cases

A problem of indeterminate structures is a bit more involved to solve because the number of unknown reactions exceeds the number of available equilibrium equations. Besides equilibrium equations, we need to generate additional equations to solve such problems. It should be noted here that a statically indeterminate structure applies

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a higher number of geometrical constraints which, in turn, supply us with the additional equations. However, in order to use the geometrical compatibility, we must determine the deflection at the constraint points. We will now solve a few examples to show how to use geometrical compatibility to find all reaction components of statically indeterminate cases. Example 11.11: A slender member AB of length L is fixed at point A and is roller supported at point B, as shown in Fig. 11.15(a). The member is loaded with a distributed load w N/m. Determine the reaction components (flexural rigidity: EI ). Given: Length = L; fixed at A; roller supported at B; distributed load w on the entire length; flexural rigidity = EI. To find: RA , MA , RB . Strategy: Consider RB as the redundant force (Fig. 11.15(b)) and split the problem into two subproblems, one with w load and another with RB force. Determine deflection at point B for each subproblem and equate their sum to zero to obtain the geometrical compatibility equation. Solution: There are three unknown reactions and two available equilibrium equations. The degree of indeterminate structure is one. Thus, we need to develop one compatibility equation to solve for y (a)

w

w A x

B

A

B 1

2

RB

w

(b) Mb

B

A RA (c)

B

A

A

RB

B

Fig. 11.15 (a) Member AB with supports and external load, (b) free body diagram, and (c) the deformed shape.

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all the three reaction components. We can choose any one of the three reaction components as a redundant force. Let us treat RB as an external force and then the problem looks like a statically determinate problem and we can evaluate the deflections using Section 11.10. Both subproblems are shown in Fig. 11.15(a). In subproblem 1, we find deflection from Fig. 11.20(b) at point B as (−w)L4 8EI In subproblem 2, we find deflection due to RB Fig. 11.20(a) as (1)

vB =

using

RB L3 3EI In the given problem, the net deflection at point B is zero. Thus, the sum of the two displacements must be zero, yielding the compatibility equation as (2)

vB =

(1)

(2)

vB = 0 = vB + vB = −

wL4 RB L3 + 8EI 3EI

yielding 3 wL 8 Now we apply the equilibrium equations to obtain the other two reaction components. The equation in y-direction yields (Fig. 11.15(b)). RB =

ΣFy : RA + RB = wL Substituting the value of RB , we obtain 5 3 RA = wL − wL = wL 8 8 The moment equilibrium equation about point A yields L + RB L = 0 2 wL2 3 1 − wL2 = wL2 MA = 2 8 8

ΣMA : MA − (wL) or,

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Interpretation/Comments: Let us review the procedure adopted. We chose one of the reaction components RB and called it the redundant force which is now treated as an external force. Then, we invoked the method of superposition to find displacement at B. The sum of the displacement was equated to zero and thus we obtained an additional relation known as compatibility relation. Note that the sum effect of the two subproblems applies the same external forces and meet all displacement constraints. Do we get the same results if we choose some other reaction component as a redundant member? Yes, we can choose any one of the three unknown reactions as the redundant force. For example, we now solve the same problem by choosing MA as the redundant force. The problem is split into two subproblems, as shown in Fig. 11.16. Fig. 11.20(j) provides the rotational angle at point A for subproblem 1 as (1)

θA =

MA L 3EI

counterclockwise

Fig. 11.20(g) provides the rotational angle at point A for subproblem 2 as 3

wL clockwise θA = − 24EI (2)

The geometrical compatibility states that net rotation at point A is zero. Thus, (1)

(2)

θA + θA =

wL3 MA L − =0 3EI 24EI

yielding, wL2 8 Thus, we obtain the same value of the moment reaction. We can now easily show that the other two reaction components come out to be the same. MA =

y

w

MA

w

MA

x A

L

B

A

1

B

A

2

B

Fig. 11.16 Treating MA as the redundant force, we split the problem into two subproblems.

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Example 11.12: A slender member (EI ) is supported at three points, as shown in Fig. 11.17(a). Determine the reaction components at the supports and the deflection at point C. Given: Total length = 2L; hinge support at A; roller support at B; roller support at C; external force P at C; flexural rigidity = EI. To find: (i) RA , RB , RD , (ii) vC . Strategy: Consider RB as redundant force (Fig. 11.17(b)), invoke superposition, determine displacement at B from each subproblem, and equate the sum to zero to obtain the compatibility relation. Solution: Since there are no forces in x-direction, the x-component of reaction at A is zero. We are now left with three unknown reactions, RA , RB , and RC (Fig. 11.17(b)) and we have two equilibrium equations, ΣFy = 0 and ΣMA = 0. Thus, the degree of the statically indeterminate structure is one. To obtain the compatibility relation, we can choose any one of the three reactions as the redundant force. We choose RB as the redundant force that is treated as the external force for further calculations. The problem is split into two subproblems, as shown in Fig. 11.17(a). The deflection at point B of subproblem 1 is determined from Fig. 11.20(h) as (1)

vB =

RB L3 RB (2L)3 = 48EI 6EI

(a)

(b)

(c)

Fig. 11.17 (a) The member AB with supports and external load, (b) free body diagram, and (c) deformed shape.

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To determine deflection at point B in subproblem 2, the deflection in portion AC is obtained from Fig. 11.20(h) as   2    (−P ) L2 x (2L)2 − L2 − x2   0 ≤ x ≤ 3L/2 v (2) = 6 × (2L)EI The deflection at point B (x = L) becomes      2 P L2 L (2L)2 − L2 − L2 11 (2) = − P L3 vB = − 6 × (2L)EI 96 Since the net deflection at point B is zero, we have the compatibility equation as (1)

(2)

vB + vB = 0 RB L2 11 − P L3 = 0 6EI 96 11 RB = P 16

or, or,

To determine other reaction components, we use the moment equilibrium equation in Fig. 11.17(b) as 3 ΣMA : LRB − LP + 2LRC = 0 2 Substituting the value of RB , we have RC =

13 P 32

Invoking the force equilibrium equation, we obtain ΣFy : RA + RB + RC = P yielding RA = P −

13 3 11 P− P =− P 16 32 32

We have evaluated all the three reactions. The member tends to move up at A and, therefore, RA is negative (downwards).

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Now we will solve the second part of this problem. We obtain the deflection of point C from subproblem 1 using Fig. 11.20(h). For ease of calculations, x is measured from end B to obtain deflection at point C as (1)

vC =

11 16 P L

L  2

(2L)2 − L2 −

 L 2  2

= 0.0788

6 × (2L)EI

P L3 EI

In subproblem 2, the deflection at point C is again determined from Fig. 11.20(h) as

(2)

vC =

(−P )

 L   3L   2

2

(2L)2 −

 L 2

6 × (2L)EI

2

−

 3L 2  2

= −0.0938

P L3 EI

The deflection at point D is the sum of the two displacements. Thus, vc = vc(1) + vc(2) = 0.0788

P L3 P L3 P L3 − 0.0938 = −0.015 EI EI EI

Interpretation/Comments: We treated one of the three reaction components as redundant force and then by invoking the geometrical compatibility we obtained the required additional equation. The deformed shape of the member is shown in Fig. 11.17(c). We have so far solved problems of statically indeterminate cases of one degree and we developed one geometrical compatibility equation to determine all reaction components. Now, we will solve a problem of two degrees of a statically indeterminate structure in which we will formulate two compatibility relations to have enough equations to determine all the reaction components. Example 11.13: A slender member is rigidly fixed at both ends, as shown in Fig. 11.18. It is loaded with a distributed force −w . Determine all the reaction components, maximum bending moment, and maximum deflection (flexural rigidity = EI ). Given: Length L; fixed support at A; fixed support at B; distributed force = −w; flexural rigidity = EI.

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w y

w

(a) A

A

x L/2

B

C

A

2

L A

w MA

(b)

B

1

3

B RB

B

MB

MB

C

RA

RB

w MC

2

(c) wL /12

C wL/2

(d) L/2

C

Fig. 11.18 (a) The slender member fixed at both ends, (b) free body diagram, (c) bending moment at point C, and (d) deformed shape.

To find: (i) RA , MA , RB , and MB, (ii) Maximum bending moment, (iii) Maximum deflection. Strategy: Treat RB and MB as redundant forces (Fig. 11.18(b)) and formulate both compatibility relations at B. Solution: The degree of indeterminacy is two because there are four unknown reaction components and two equilibrium equations. Due to deformation of the member and the rigid supports at the ends, the member AB is likely to get elongated which should develop axial forces in the member. However, we will solve the problem with small bending deflections. If we denote s along the length direction of the deformed member, dx = cos θds where θ is the angle between

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ds and dx which varies along the length. In this example, θ is zero at points A, B, and C. Writing cos θ in series form, we have cos θ = 1 −

θ4 θ2 + − ··· 2 24

where θ is in radians. If maximum angle of the deformed member is small say θ = 5◦ (or θ = 0.0873 radian), the second term in the above series is only 0.0038 which is much smaller than unity. The third and other terms are still smaller. Thus cos θ is very close to unity and there is hardly any change in length of the member. However, if the flexural rigidity is small and applied force is high, our assumptions will no longer be valid and the problem is solved by accounting for axial force developed in the member. Since RB and MB have been chosen as redundant forces, they are now treated as external forces. We then split the problem into three subproblems, each one having one force only −w, RB , or MB , as shown in Fig. 11.18(a). Realizing that the displacement in y-direction and rotation of the member are zero at point B, we formulate the following compatibility relations: (1)

(2)

(3)

(1)

(2)

(3)

vB + vB + vB = 0 θB + θB + θB = 0 For subproblems 1, Fig. 11.20(b) provides (1)

(−w)L4 8EI wL3 clockwise =− 6EI

vB = (1)

θB

For subproblem 2, Fig. 11.20(a) gives (2)

RB L3 3EI RB L2 = 2EI

vB = (2)

θB

counterclockwise

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For subproblem 3, Fig. 11.20(d) yields (3)

−MB L2 2EI (−MB )L = EI

vB = (3)

θB

clockwise

Substituting them in the compatibility equations, we have RB L3 MB L2 (w)L4 + − =0 8EI 3EI 2EI RB L2 (MB )L (w)L3 + − =0 − 6EI 2EI EI −

We have two simultaneous equations for two unknowns. Solving the equations, we have RB =

wL 2

MB =

wL2 12

In this particular problem, RB should come out to be wL/2 because it is a symmetric case. Also, MA should be equal to MB . Thus, RA =

wL 2

MA =

wL2 12

To solve the second part of finding the maximum bending moment, it is likely to be maximum at one of the several points, points A, B, or C. We already know the bending moments at points A and B. To determine bending moment at midpoint C, consider a section at C and the free body diagram of portion AC, as shown in Fig. 11.18(c). The bending moment at point C consists of three terms: (i) due to distributed force w, (ii) due to bending moment MA at support A, and (iii) due to reaction at RA . Thus,     wL L wL2 wL L wL2 − + = Mc = − 2 4 12 2 2 24 2

2

wL At the ends, the moment is − wL 12 and at the centre it is 24 . Thus, the magnitude of the bending moment is maximum at the 2 ends with its magnitude being equal to wL 12 .

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To find the maximum deflection, we note that the problem is symmetric and thus the maximum deflection is at the centre point C. We determine the deflection for each subproblem and then sum them. For subproblem 1, we obtain the deflection using Fig. 11.20(b) as    3  2  17 wL4 L 4 −w L (1) 2 L =− − 4L + 6L vc = 24EI 2 2 2 384 EI For subproblem 2, we obtain the deflection using Fig. 11.20(a) as vc(2)

 wL   L 2  =

2

2

3L − 6EI

L 2

 =

5 wL4 96 EI

For subproblem 3, we obtain the deflection using Fig. 11.20(d) as

  −wL2 L 2 12 2 1 wL4 (3) =− vc = 2EI 96 EI The deflection at point is the sum as vc = vc(1) +vc(2) +vc(3) = −

17 wL4 5 wL4 1 wL4 + − 384 EI 96 EI 96 EI

=−

1 wL4 384 EI

The deformed shape is shown in Fig. 11.18(d). Interpretation/Comments: We generate as many compatibility equations as the degree of indeterminacy. We now compare the maximum bending moment and maximum deflection of the two cases shown in Table 11.1. The maximum deflection of the member with simple supports is found to be five times higher than the deflection with fixed supports at both ends; a fixed support decreases the deflection of a slender member considerably. The maximum bending moment in simply sup4 wL2 ported member is wL 8EI at the center point C, while it is 12EI at the ends in fixed support case. In the old days when the theory was not well developed, the engineers used to model a slender member as simply supported beam at the ends even for the case when the member was rigidly supported. It was a conservative design. However, we now like to know the factual internal forces on a member and then apply an appropriate factor of safety.

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Table 11.1

Comparison of maximum bending moment and deflection. Maximum bending moment

S. No.

3

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1

9in x 6in

Beam simply supported at both ends with uniformly distributed load w (Example 11.3) Beam with fixed support at both ends with uniformly distributed load w (Example 11.13)

Maximum deflection

wL2 at the 8 center point

5 wL4 384 EI

wL2 at the 12 ends

1 wL4 384 EI

In cases like bridges, we like to have a beam simply supported at its end because the length of a bridge increases or decreases with temperature. A simply supported beam can expand or contract freely. If it is fixed at its ends, thermal axial strain will also be developed. We had discussed in Chapter 9 that thermal strains are comparable to the strains of external loads. We thus avoid the development of thermal strains within the beams of a bridge. One of the design principles is to yield against undesirable developments. Example 11.14: Consider a slender member AB whose end A is fixed. Under the unloaded conditions, there is a roller support, 1 mm below the free end B, as shown exaggeratedly in Fig. 11.19(a). When a load of 10,000 N is applied at the midpoint C, the member touches the roller and then a reaction starts acting at point B in y-direction (Fig. 11.19(b)). Determine the reaction at B and deflection of the midpoint C (E = 70 GPa, I = 100 × 103 mm4 ). Given: Member length = 240 mm, fixed support at A; roller support at 1 mm below the bottom surface of the member at point B; external force 10,000 N at x = 120 mm; E = 70 GPa, I = 100 × 103 mm4 . To find: (i) RB ; (ii) vc . Strategy: First check whether end B touches the roller support for the given load. If yes, invoke the method of superposition with an appropriate geometrical compatibility at point B incorporating the 1 mm gap.

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(b) (a)

(c)

Fig. 11.19 (a) The undeformed member with a gap of 1 mm at B, (b) deformed member touching the roller support, and (c) two subproblems.

Solution: To check whether the end B touches the roller, we consider the member as a cantilever in which the end B is free to move up and down. Assuming that the roller support does not exist, we determine the deflection at point B using the expression of deflection in segment AC from Fig. 11.20(e) as free = vB

(−10,000) × (120)2 × (3 × 240 − 120) 6 × 70 × 103 × (100 × 103 )   N N mm4 × (mm2 )mm mm2

= −2.057 mm The displacement exceeds 1 mm and, therefore, the end B touches the roller and reaction RB develops on the member. We now split the problem into two subproblems, 1 and 2, as shown in Fig. 11.19(c). The compatibility relation becomes (1)

(2)

vB − vB = −1

(11.11)

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For subproblem 1, we have already calculated the deflection as (1)

vB = −2.057 mm For subproblem 2, we obtain deflection using from Fig. 11.20(a) as (2)

vB =

Nmm3 RB (240)3  N  3 3 3 × (70 × 10 ) × (100 × 10 ) mm2 mm4

= 0.658 × 10−3 RB mm The compatibility equation (11.11) yields −2.057 + 0.658 × 10−3 RB = −1 This yields RB = 1,606 N For the second part, the deflection at point C is given by (1)

(2)

vC = vC + vC

We obtain deflection of subproblem 1 from Fig. 11.20(e) as (1)

vC =

N(mm3 ) (−10,000) × (120)3   N 4 3 × (70 × 103 ) × (100 × 103 ) mm 2 mm

= −0.823 mm For subproblem 2, Fig. 11.20(a) yields (2)

vC =

(1,606) × (120)2 × (3 × 240 − 120)) N(mm2 ) mm  N  6 × (70 × 103 ) × (100 × 103 ) mm4 mm2

= 0.330 mm Then, the displacement of point C is given by vC = −0.823 + 0.330 = −0.493 mm Interpretation/Comments: The geometrical compatibility need not be zero displacement. It should be formulated with care. There are several other methods to determine the deflection of a member like moment area method and through the use of singularity functions. I feel that the two methods discussed in this book are

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appropriate for this introductory course. There is another powerful method through the strain energy; we will discuss it in Chapter 14. 11.9

Summary

Let us have a bird’s-eye view of this important chapter. A bending moment is a very powerful internal force which not only develops high stresses but also causes large lateral deflection. In Chapter 10, we developed a relation, κ = Mb /EI, between curvature κ and bending moment Mb . Although κ gives some idea about the deformation, we would rather know the lateral deflection v. We use a geometrical relation between deflection and curvature from calculus. Although d2 v its form is not simple, it becomes simpler, κ = dx 2 , if the slope of the deformed member is small. Eliminating κ from both the relations, d2 v we obtain the required bending differential equation, dx 2 = Mb /EI. This relation is addressed as governing differential equation (GDE) in this chapter. Several techniques have been developed to solve the GDE. We have discussed two of them in this chapter. The first technique is the direct integration method. We integrate the GDE twice; two integration constants are developed which are evaluated from the boundary conditions. The continuity of GDE in the length direction is disrupted by the applied forces like concentrated forces and reactions and an abrupt change in distributed forces. Thus, a slender member is divided into several segments separated by these discontinuities. The GDE is solved only within a segment. It is solved separately for each segment with its own integration constants. However, at a trandv are sition point between adjacent segments, deflection v and slope dx continuous. This helps in evaluating the integration constants. The direct integration method is convenient only for a structural member having one or two segments. The GDE also helps us in solving statically indeterminate cases in which the number of unknown reaction components exceeds the number of equilibrium equations. The additional equations are developed through the geometrical constraints of the problem. They are known as geometrical compatibility relations. The method of superposition is powerful and can solve many problems. In statically determinate cases, we split the problems into subproblems, each one having an externally applied force. A library is

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available, which provides the solutions to some basic simple problems. For each subproblem, we lift the solution from the library and then add the results to obtain the solution of our problem. The method of superposition is also handy in solving statically indeterminate cases. If available equilibrium equations are ne and the number of unknown reaction components is nr , we need (nr − ne ) additional equations. They are obtained from geometrical compatibility relations. We choose any (nr − ne ) reaction components and call them redundant forces. These redundant forces are now treated as external forces. We then invoke the method of superposition and evaluate the displacements required in solving the geometric compatibility relations. The deflection v of a slender member is found to depend on a group of variables as follows: P L3 for concentrated external force P EI qL4 for distributed external force q(N/m) v∝ EI M L2 for concentrated external moment M v∝ EI

v∝

It is worth noting that the deflection depends heavily on the length between two adjacent supports. It is inversely proportional to modulus E and moment of inertia I. Thus, to restrict the deflection, we choose the material with high modulus like steel. In the case of polymers, we may reinforce it with high modulus glass or carbon fibres. The cross-section of the member is also optimized to have a high moment of inertia. 11.10

Appendix A: Deflections and Slopes of Various Cases

The deflections and the slopes of the deformed member of simple cases of a slender member for various kinds of loadings are presented (see Fig. 11.20). These simple cases are useful to find deflections of more complex members under bending loads through the method of superposition.

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No.

Member with forces and supports

(a) A

B

(b)

y

(c)

qx2 (6L2 − 4Lx + x2 ) 24EI qL4 vmax = v(L) = 8EI qx(3L2 − 3Lx + x2 ) dv = dx 6EI qL3 θB = 6EI v=

q A

B

L

x

y

qo x2 (10L3 − 10L2 x + 5Lx2 − x3 ) 120LEI q o L4 vmax = v(L) = 30EI qo x(4L3 − 6L2 x + 4Lx2 − x3 ) dv = dx 24LEI q o L3 θB = 24EI

v=

q0 A

(d)

P x2 (3L − x) P L3 ; vmax = v(L) = 6EI 3EI dv P x(2L − x) =− dx 2EI P L2 θ(L) = 2EI

v=

x

L

371

Elastic curve, maximum deflection and slopes

P

y

page 371

B

L

y

x

v=

M A

B

L

x

M x2 2EI

vmax = v(L) =

M L2 2EI

Mx dv = dx EI ML θB = EI (e)

y A

a

C

L

b

P x2 (3a − x) ; 6EI 2 P a (3x − a) ; v= 6EI P a3 vC = 3EI P a2 (3L − a) vB = ; 6EI v=

P B

x

dv P x(2a − x) = 0≤x≤a dx 2EI 2 dv Pa = a≤x≤L dx 2EI

θB =

P a2 2EI

Fig. 11.20 (a)−(k) The deflections and the slopes of the deformed member of simple cases of a slender member for various kinds of loadings.

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No.

Member with forces and supports

Elastic curve, maximum deflection and slopes

(f)

y

v=

M A

(g)

a

x

B

C

L

y

q0

A

(h)

y

x

a

C

L

b

P bx(L2 − b2 − x2 ) dv P b(L2 − b2 − 3x2 ) ; = 6LEI dx 6LEI ( 0 ≤ x ≤ a)     L P b (L2 − b2 )x − x3 + (x − a)3 b v= 6LEI (a ≤ x ≤ L)

v=

P A

q0 x(L3 − 2Lx2 + x3 ) 24EI 5q0 L4 vmax = v(L/2) = 384EI q0 (L3 − 6Lx2 + 4x3 ) dv = dx 24EI q 0 L3 q 0 L3 θA = ; θB = − 24EI 24EI

v=

B

L

M x2 dv Mx ; = 0≤x≤a 2EI dx EI M a(2x − a) dv Ma v= ; = a≤x≤L 2EI dx EI 2 Ma vc = 2EI M a(2L − a) Ma ; θB = vB = 2EI EI

B

x

P ab(L + b) P ab(L + a) ; θB = − 6LEI 6LEI When a = b = L/2

θA =

P x(3L2 − 4x2 ) ; 48EI   L 0≤x≤ 2

dv P (L2 − 4x2 ) = dx 16EI

v=

vmax =

Fig. 11.20

P L3 ; 48EI

θA =

P L2 ; 16EI

(Continued )

θB = −

P L2 16EI

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Member with forces and supports

No.

y

(i)

q L/2

x

C

vc =

y

M

5qL4 768EI

A

B

x

y

Mo A

C

L/2 L

3qL3 ; 128EI

0≤x≤ 

L 2

0≤x≤ θB = −



L 2



7qL3 384EI

M x(2L2 − 3Lx + x2 ) 6LEI dv M (2L2 − 6Lx + 3x2 ) =− dx 6LEI   M L2 1 vmax = √ at x = L 1 − √ 9 3EI 3 ML ML θA = − ; θB = 3EI 6EI v=−

B

x



M0 x(L2 − 4x2 ) 24LEI

vc = 0 M0 L θA = ; 24EI

Fig. 11.20

11.11

θA =



v=−

L

(k)

qx(9L3 − 24Lx2 + 16x3 ) 384EI

q(9L3 − 72Lx2 + 64x3 ) dv = dx 384EI

L

(j)

373

Elastic curve, maximum deflection and slopes

v=

B

A

page 373

θB = −

0≤x≤

L 2



M0 L 24EI

(Continued )

Problems

1. A slender member AB (flexural rigidity = EI ) is fixed at end A. On the other free end B, a force P and a moment M are applied, as shown in Fig. 11.21(a). Using the method of direct integration, determine the following: (i) elastic curve, (ii) deflection at midpoint C, and (iii) slope of the member at point B. 2. Consider a simply supported slender member (EI ) loaded with a triangular distributed force, as shown in Fig. 11.21(b). Determine the following: (i) deflection at x = 2L/3 and (ii) slope at point B. Use the method of direct integration.

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374 y

P A

B

C x

y

w

M

x

A

L/2

L

2L/3

C

B

L

(a)

(b)

Fig. 11.21

(a) Problem 1 and (b) Problem 2.

3. A simply supported member AB is loaded with a moment Mo at its mid-length point C, as shown in Fig. 11.22(a). Use the method of the direct integration to determine the following: (i) elastic curve of segment AC, (ii) rotation of member at point A, and (iii) rotation at point B. 4. A slender member is made of two cylindrical bars of the same diameter which are brazed together at mid-length C. It is simply supported at its ends, as shown in Fig. 11.22(b). The modulus of the right portion is double that of the left portion’s modulus. An external force P is applied at point C. Determine elastic curve of segment AC and deflection of point C (flexural rigidity of left portion =EI ). 5. A slender member (EI ) is supported on three supports, as shown in Fig. 11.23(a). A distributed load of constant magnitude w acts on the entire length. Determine the reactions at the supports and bending moment at support point C and point D. 6. A slender member AB of length L and rigidity modulus EI is fixed at point A and roller supported at point B, as shown in Fig. 11.23(b). A force P is applied at distance L/3. Determine all reaction components and deflection at point C.

y

A

y

Mo x L/2

C L (a)

Fig. 11.22

B

P

A

B

C

x L/2

2

1 L (b)

(a) Problem 3 and (b) Problem 4.

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Deflection due to Bending y

y

w

A

L/2 L

x D

375

P A

C

B

x L/3

B

C

page 375

L

L (a)

(b)

Fig. 11.23

(a) Problem 5 and (b) Problem 6.

7. A slender member (EI ) is simply supported, as shown in Fig. 11.24(a) with various external forces. Determine deflection at point C using the method of superposition. 8. A cantilever is loaded with a triangular distributed load, as shown in Fig. 11.24(b). Determine the deflection at point D using the Green function approach. 9. On a pole AB of flexural rigidity EI, a signboard of weight P is supported through a crossbar CD, as shown in Fig. 11.25(a). Determine the lateral deflection of point B. y

y

w

A

C

x

w

B

x C

D

L/4

L/4

2P L/4 L/2

L/4

P

D

B

L/4 L

L

(a)

Fig. 11.24

(b)

(a) Problem 7 and (b) Problem 8. L

B L/2 D

C L

L/2 P

A

2L 3

1

C

A (a)

Fig. 11.25

2

B

E

P

L

(b)

(a) Problem 9 and (b) Problem 10.

D

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E2=50 GPa 3 4 I2=70x10 mm

2 3

y

P A

C

B

4

1

B

P H

20 0

E1=70 GPa I1=60x10 mm C

x

m

D L

150

(a)

40

0

m

L/2 A 300 m m

(b)

Fig. 11.26

(a) Problem 11 and (b) Problem 12.

10. Consider two slender members, 1 and 2, as shown in Fig. 11.25(b). They are attached to each other through a ball. The flexural rigidities are E1 I1 and E2 I2 of members 1 and 2. An external force P is applied at point E. Determine all reaction components and deflection at point B. 11. Consider a slender member AB whose one end is fixed and an external force P is applied at the free end, as shown in Fig. 11.26(a). The member is supported at its mid-length point C with a compression spring whose spring constant is k. Determine the reactions at point A and deflection at point C. 12. A slender member CD is supported on rollers at its ends. It is also supported at the midpoint by a cross member AB which is supported at its ends on rollers, as shown in Fig. 11.26(b). The bottom surface of member CD is just touching the top surface of member AB when there is no external force on the unit. A force P = 12,000 N is applied at the mid-length point H of member CD. Determine the deflection of point H. The modulus and moment of inertia of both members are shown in the figure.

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Chapter 12

Torsion 12.1

Introduction

When I pluck a mango from a tree, I find it difficult to separate it by pulling. A clipper can pluck it, but I do not usually carry one. I hold the mango and start twisting it using the fingers of both hands and after a few turns, it neatly separates out. An applied torque separates the mango easily. There are many other examples of daily life where we apply torque. For example, we apply a torsional moment on a screwdriver. Most caps of bottles we use in our homes are based on twisting and they are designed to withstand the stresses developed during tightening. Another less obvious application is the use of tension or compression helical springs to store energy. The wire of the spring is subjected to an internal torsional moment. An externally applied torque develops an internal twisting moment and causes the member to rotate by an angle, known as the angle of twist. The internally developed twisting moment is one of the four internal generalized forces. The other three have already been discussed, that is, axial force in Chapter 9 as well as shear forces and bending moments in Chapters 10 and 11. On a cross-section, each generalized force develops its own stress distribution. The internal torsional moment may vary along the length of a member. We often are interested in knowing the cross-section on which the internal twisting moment is maximum. This is obtained by drawing a torsional moment diagram (TMD) as discussed in Chapter 4. In this book, we use symbol T for an external torque applied on a shaft. An internally developed torsional moment is addressed 377

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by symbol Mt . We will explore the nature of stress distribution on a cross-section which is caused by the internal torsional moment. Mathematical expressions will be developed to determine stress at any point of the cross-section and the angle of twist per unit length, α. The mechanical power is quite commonly transmitted from one member of circular cross-section to another. Such circular solid slender members are known as shafts. These shafts are widely used elements in machines like electric motors, gearboxes, drill machines, locomotives, autos, power generation plants, and fans. We transmit power from one shaft to another shaft using elements like gears, sprockets and chains, pulleys, and belts. Figure 12.1(a) shows the transmission of torque T from shaft A to shaft B through a pair of meshing gears. At the contact point, a force acts in the tangential direction of the pitch circles of the gears. The tangential force on gear B is P is acting downwards, as shown in Fig. 12.1(b) and at the distance r away from the axis of the shaft. It is convenient to consider an equivalent loading at location Q of shaft B. It consists of torsional moment equal to P r and a transverse force P , as shown in Fig. 12.1(c). For the shaft, the moment can be considered as an externally applied torque, T = P r. The torque T develops internal torsional moment in shaft B which, in turn, gives rise to stress field within the shaft. We found in Chapter 10 that the normal stress on a cross-section developed by an internal bending moment is not uniformly distributed. Similarly, we will find that the stress distribution on a cross-section by an internal torsional moment is not uniform. Some of the readers find it difficult to visualize the deformation caused by

(a)

(b)

(c)

Fig. 12.1 (a) Power transmission through meshing gears, (b) force acting on the gear of shaft B, and (c) equivalent forces on the shaft.

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379

an internal torsional moment and the stress developed. It is difficult to visualize the deformation with 2D views and, therefore, we need to visualize in 3D. The cylindrical coordinates (r, θ, z) are convenient to analyse torsional problems of members with circular cross-section because the value of stress does not depend on θ and the analysis becomes a lot easier. We choose z-axis aligned with the axis of the cylindrical member (Fig. 12.2(a)). Note that r-direction is radially out and the distance is measured from the z-axis; θ-direction is normal to r-direction and the value is expressed with angle θ measured from a reference direction, as shown in the figure. A stress element at an internal point H of the cylindrical member is shown in Fig. 12.2(b). In this chapter, we will focus our attention on such an element to explore how it deforms and how the stresses are developed. In this chapter, we limit our analysis only to members with circular cross-section, both solid and tubular members. The analysis of noncircular members is more complex because a cross-sectional plane no longer remains plane after it is deformed by an internal torsional moment. In fact, the cross-section warps, that is, some points of the cross-section move in positive z-direction and others in negative z-direction. On the other hand, a cross-section of a circular member does not warp at all and the analysis becomes simpler. z-direction Reference direction

r-direction

z H

zz

-direction rz

r (a) Fig. 12.2 point H.

rr

(b)

(a) Cylindrical coordinates r, θ, z and (b) a stress element at an interior

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The analysis in this chapter will be limited to homogeneous, isotropic, and linear elastic materials. Further, the analysis will be limited to small rotations because our linear elastic relations are applicable only to small displacements.

12.2

Stress and Displacement Fields

The results will be presented prior to the analysis and they will be discussed with a solved example. Once some feel of the torsion problem is acquired, we will go ahead and carry out the analysis in detail in the following section. This will facilitate the visualization of how a stress element in cylindrical coordinates deforms. We now consider a solid circular member of radius R subjected to an internal torsional moment, as shown in Fig. 12.3(a). The stress tensor at a point of the member has the following form: ⎤ ⎤ ⎡ ⎡ 0 0 0 σrr τrθ τrz σ = ⎣ τrθ σθθ τθz ⎦ = ⎣0 0 τθz ⎦ τrz τθz σzz 0 τθz 0

z Mt

O Reference direction

C

B

A

O C

r

B

A

r (a)

(b)

Fig. 12.3 (a) A cross-section of a cylindrical member with Mt as internal torsional moment and (b) linear shear stress distribution.

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381

There is only one nonzero stress component, τθz . To visualize it, we focus our attention on a z-face with the stress element at point C. The normal of the cross-section is in z-direction and, therefore, one of the subscripts of the stress component is z. Since the shear stress is along θ-direction, the second subscript is θ. Further, the shear stress τθz increases with increasing r, as shown through stress elements at A to B and then to C in the figure. In fact, we will find that the variation is linear with r, as shown in Fig. 12.3(b) and is given by τθz =

Mt r J

where J is the area property of the cross-section and is known as the polar moment of inertia. It has been defined in Sections 10.5.3 and 10.5.4, Appendix A, of Chapter 10. If R is the radius of the circle, J = πR4 /2. The torsional moment makes the member rotate. To feel how the rotation deforms a circular member, we take a circular rod of a soft material like rubber and mark a square grid on its lateral surface with a thin tip permanent black pen, as shown in Fig. 12.4(a). Now, we hold the ends of the cylinder between our two hands and twist the rod. The top cross-sectional face rotates with respect to the bottom cross-sectional face. We note that the triangle OQS on the top crosssectional face rotates to OQ S  . We will now study how the strain is developed. The lengthwise grid lines get inclined, as shown in Fig. 12.4(b). However, the circumferential grid lines on the side cylindrical surface, which are in θ-direction, do not get inclined, that is, they remain normal to the axis. Also, they do not get compressed or stretched as a crosssectional plane rotates about its axis, that is, Q S  = QS. The square element CDEF in the undeformed condition becomes a parallelogram, C D E F , after the deformation. Note that the original 90◦ angles of element CDEF do not remain 90◦ after the deformation. Thus, the element CDEF is deformed in shear in θ−z plane. The angle CFE, which was 90◦ in the undeformed condition, is reduced by angle C F C . Thus, the inclination angle of the lengthwise grid lines becomes the shear strain γθz . The angle of twist of the torsional member (rotation of the top section) is monitored conveniently through an important parameter,

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382

Mt

O

O

S

Q

Q

S

C

D

C

F

E

F

D

C E

z

Fig. 12.4

Undeformed

Deformed

(a)

(b)

(a) Grid on the external surface and (b) deformed cylinder.

α, which is the angle of twist per unit length of the member. If G is the shear modulus of the material, we will prove in the following section that α=

Mt GJ

If the length of the member is L, the relative rotation of the two end faces is αL or Mt L/GJ. Example 12.1: A straight cylindrical member is subjected to an internal torsional moment Mt = 500 Nm. Determine the shear stress distribution on a cross-section and the angle of twist φ between two locations, 800 mm apart in the axial direction. The diameter of the member is 30 mm and shear modulus 27 GPa (Fig. 12.5). Given: Mt = 500 Nm;

R = 15 mm; L = 800 mm; G = 27 GPa.

To find: τθz , φ. Strategy: Invoke torsional formulas.

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Torsion

383

15mm

z

m

800m

(a)

(b)

Fig. 12.5 (a) The straight solid cylindrical member and (b) shear stress distribution on a radial line.

Solution: Polar moment of inertia J for the member is J=

π(15)4 πR4 = mm4 = 79.52 × 103 mm4 2 2

The maximum shear stress is at the periphery of the circular cross-section and is given by max = τθz

(500×103 )×15 (Nmm) × mm Mt R = = 94.3 MPa J mm4 79.52×103

The angle of twist per unit length, α, becomes Nmm 500×103 Mt  N  = 3 3 GJ (27 × 10 ) × (79.52 × 10 ) mm2 × mm4 rad = 0.233 rad/m = 0.233 × 10−3 mm  180◦ = 13.35◦ /m = 0.233 × π

α=

The relative rotation φ between two cross-sections, 800 mm apart, is  1 m = 10.68◦ φ = αL = 13.35 × 0.8 m Interpretation/Comments: Figure 12.5(b) shows the distribution of τθz which is zero at the centre and maximum at r = R. We prefer to find rotation in terms of α because it is independent of length.

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If we want to find the relative rotation between two cross-sections, we just multiply it with length L. Also, note that most engineering calculations are performed with angles in radians. However, we as humans are taught to think in terms of degree and, therefore, we convert it into degrees for having a feel whenever we desire.

12.3

Determination of Stress Field and Angle of Twist

The analysis of the torsion of a slender member with circular crosssections is relatively simpler because of the θ-symmetry, that is, stress, strain, and displacement do not depend on coordinate θ of the r, θ, z coordinate system. The analysis of a member with any other cross-section (non-circular) is comparatively more complex and beyond the scope of this book. The symmetry of a circular cross-section is useful and does provide a lot of information on the nature of the deformation caused by a torsional moment. However, the arguments dealing with such symmetry do become a bit involved and we will take a simpler approach by making several assumptions on the deformation. A cross-sectional plane which is normal to the axis of the member remains plane and normal after the torsional moment is applied. Figure 12.6(a) shows a normal plane N which remains normal after the deformation, as shown in Fig. 12.6(b) as plane N . However, this plane rotates about the axis. A sector QOS is rotated to Q OS as shown in the figure. The radial lines remain radial and straight after the deformation. A cross-sectional plane, which is a circle, remains circular with no change in its radius. Thus, the cross-section is not deformed on its own plane. It only gets rotated. This makes rr = 0 and θθ = 0. In addition, the length of the member does not change in the deformed configuration which makes zz = 0. These assumptions can be easily verified by simple experiments. In addition, the θ-symmetry is also handy to justify them. For example, if a sectional plane becomes inclined or warped, it will no longer have the θ-symmetry. Further, the advanced and rigorous analysis, which is beyond the scope of this book, does not make such assumptions on the deformation of a torsional member. The results do show that

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Torsion

385

N

N O Q

S

Undeformed (a)

Fig. 12.6

page 385

O Q

S

Deformed (b)

(a) Undeformed member and (b) deformed member.

the assumptions are valid for the torsion of a member with circular cross-section. We have already shown that all the three normal strains are zero. We will now focus our attention on a shear stress element in r, θ, z coordinates and will show that rθ = 0 and rz = 0. However, 3D visualization is required in the cylindrical coordinates. We consider a thin slice of thickness Δz of a torsional member shown in Fig. 12.7(a). In its magnified view, shown in Fig. 12.7(b), we consider a thin annular tube of thickness Δr in the interior of the member, say at radius r. On this thin tube, we consider a 3D stress element shown with solid lines in Fig. 12.7(b) whose top face is ABCD and the bottom face is EFGH. All angles of this element are 90◦ . With deformation, the top face ABCD (on r−θ plane) moves to A B C D as shown in Fig. 12.7(c). Since all the four angles of face A B C D are 90◦ even after deformation, strain component γrθ = 0. Similarly, all angles of face A B F E (on r−z plane) remain 90◦ after deformation and, therefore, γrz = 0. The shear deformation takes place only on face A D H E of the stress element which is on θ−z plane and thus only nonzero shear strain component is γθz .

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386

(a)

(b)

(c)

Fig. 12.7 (a) A round torsional member, (b) an undeformed stress element within the material at radius r, and (c) the deformed stress element.

The deformation can be explained with an analogy that a short length circular disc material is bonded between two rigid and parallel plates, as shown in Fig. 12.8. The top plate is rotated with respect to the bottom plate. The stress element ABCD (on θ−z plane) deforms to AB C D. The top face with its normal in z-direction of the element does not get tilted. Only the side face gets tilted, causing shearing. Let us look into Fig. 12.7(c) again carefully. The angle AE A is the change from the original 90◦ and thus this angle is shear strain γθz . Therefore, the strain tensor has the following form: ⎤ ⎡ 0 0 0 = ⎣0 0 θz ⎦ 0 θz 0 where θz = γrθ /2. We can express the shear strain γrθ as (Fig. 12.7(c)) γrθ =

AA AA = AE  Δz

(12.1)

AA can be expressed in terms of the angle of rotation Δφ as AA = rΔφ

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Torsion

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387

B B C C Rigid blocks

z A

D

Torsional member

Fig. 12.8

An analogy to show the nature of deformation.

Substituting it in Eq. (12.1), we obtain γrθ = r

dφ dz

(12.2)

dφ dz

is the angle of twist per unit length. We are thus able to express shear strain in terms of rotation of the cross-section. The angle of rotation per unit length is an important parameter and we denote it by α. Thus, α=

dφ dz

(12.3)

Then, we have γθz = rα

(12.4)

On a cross-section γθz is proportional to r with its maximum value at the periphery of the circular cross-section. If we look carefully at Fig. 12.7(c), it makes sense because the farther the stress element is, the higher the shearing is. In the relation of Eq. (12.4), α is still unknown. In fact, the expression is obtained purely on geometrical considerations and we still

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have not considered the torsional moment Mt . The angle of rotation per unit length depends on Mt ; a higher value of Mt should give higher α. How should we relate α with Mt ? It is determined by carrying out the equilibrium. We will find the stress field so as to develop the required equilibrium relation. We, therefore, use stress–strain equations to evaluate stress tensor. All normal stress components are zero because normal strains are zero. The shear stress components are as follows: τrθ = Gγrθ = 0

(12.5)

τrz = Gγrz = 0 τθz = Gγθz = Grα Thus, the resulting stress field, still in becomes ⎤ ⎡ ⎡ 0 0 σrr τrθ τrz σ = ⎣ τrθ σθθ τθz ⎦ = ⎣0 0 τrz τθz σzz 0 τθz

terms of unknown α, ⎤ 0 τθz ⎦ 0

Now, we consider a cross-section with its normal in z-direction, as shown in Fig. 12.9. On this plane, the internal torsional moment Mt is acting. We will carry out the force balance on this cross-sectional plane. Consider a stress element A B C D on the top section on which the shear stress τθz is acting in the θ-direction. The force on this element is τθz dA, acting in θ-direction at a distance r from the axis and thus contributes toward torsional moment as r(τθz dA). The sum effect of this moment is carried out by integrating on the crosssection, yielding

r τθz dA Mt = A

Substituting τθz from Eq. (12.5), we obtain



r(Grα)dA = Gα r 2 dA Mt = A

A

(12.6)

 We are familiar with A r 2 dA which is known as polar moment of inertia and is usually addressed with symbol J. Thus, Eq. (12.6)

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389

z Mt

r

r

A

B D

C

z

Fig. 12.9

The shear stress on a cross-sectional plane.

yields the required relation between the torsional moment and the twisting angle per unit length as α=

Mt GJ

(12.7)

This is an important relationship between the torsional moment and the angle of twist per unit length. The result is rather a simple one; we obtain α just by dividing Mt with GJ. The product,GJ, is known as torsional rigidity; it consists of a material property, G, and area property, J. If we want to know relative rotation φ of two crosssections, L distance apart, we easily obtain it by multiplying α with L to have φ = αL =

Mt L GJ

(12.8)

Some people like to work with this form. I personally prefer Eq. (12.7) which relates α with Mt because α is independent of the length parameter.

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To obtain stress field within a cylindrical torsion member, we substitute α from Eq. (12.7) into Eq. (12.5) to have  τθz = Gr or,

τθz =

Mt GJ



Mt r J

(12.9)

This is the important relation we have been seeking for. We will be addressing it as the stress equation of torsion. The stress tensor is then written as ⎤ ⎡ 0 0 0 σ = ⎣0 0 MJt r ⎦ 0 MJt r 0 It is important to realize that the torsional moment is the internal force on the cross-section, while stress is at a point which varies with r. Further, the stress does not depend on the material properties as long as the stress is smaller than the yield stress, the deformation is linear elastic, and the material is homogeneous and isotropic. Let us have some more discussion on the stress equation τθz = Mt r/J. Substitution of the expression of J of a circular bar into this equation yields Mt r 32Mt r 2Mt r = τθz = 4 = 4 πR πR πD 4 2

where D is the diameter of the member. The maximum shear stress is at the periphery of the cross-section (r = D/2) and is given by max = τθz

16Mt πD 3

(12.10)

The direction of the shear stress at any point is normal to the radial line on the face of the cross-section, as shown in Fig. 12.10(a). As expected, it is not dependent on coordinate θ. Figure 12.10(b) shows the shear stress distribution on the entire face of a cross-section with the length of the arrows representing the magnitude of the shear

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391

r

r

(a)

(b)

Fig. 12.10 (a) Shear stress distribution on a cross-section and (b) shear stress is normal to radial direction at all points. max is proportional to 1/D3 . If the diameter is stress. Note that τθz doubled, the maximum stress is reduced by eight times. It should be remembered that the equations developed for cylindrical members cannot be applied to cases with non-circular crosssections. In some of the non-circular members, the farthest point is not subjected to maximum shear stress. We will now have some discussion on the relation of the angle of twist, α = Mt /GJ. Substitution of the expression of J into this equation yields

α=

32Mt 2Mt M t = = 4 4 πGR πGD 4 G πR 2

(12.11)

Thus, the angle of rotation per unit length is proportional to 1/D 4 . If the diameter is doubled, the angle of rotation is decreased by 16 times. In other words, the member becomes 16 times stiffer. The maximum stress is proportional to 1/D 3 , while the angle of rotation is proportional to 1/D 4 . Thus, with the increase of the diameter of a circular member, the rotation and the stresses both decrease substantially; however, the rotation decreases at a faster rate. We now look back and take a bird’s-eye view of the analysis we have carried out in this section. Based on geometric considerations, we make certain assumptions for a torsional member with circular cross-section. The only nonzero strain component was γθz . It was expressed in terms of the angle of twist per unit length α which was

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unknown at this stage. We invoked stress–strain relations to obtain τθz . The equilibrium condition was applied on the cross-sectional face by finding the sum effect of moment caused by the shear stress and equating it to the torsional moment Mt . This yields the expression of α and then the shear stress distribution in terms of Mt . The formulation of bending analysis in Chapter 10 was carried out in a similar way. Recall that the form of normal strain xx was determined in terms of still unknown curvature κ, based on geometric considerations only. Using the stress–strain relations, the normal stress σxx was determined. The sum effect of moment caused by the normal stress was equated to the bending moment Mb to express κ and σxx in terms of Mb . At the location where the external torque is applied, the simple shear stress field of linear variation with radius does not exist. Instead, the stress field is quite complex and its distribution depends upon how the external torque is applied. However, Saint-Venant’s principle works and, at a short distance away, the linear variation of stress distribution with r is attained within the material. On slender members, these local effects are ignored. Similarly, if the diameter of the member changes abruptly, there is a local disturbance in the stress field which is usually ignored in the analysis of slender members. Example 12.2: Design the diameter of a shaft to transmit a torque of 400 Nm in such a way that the maximum allowable angle of twist is 2.5◦ /m and the maximum allowable shear stress is 75 MPa. Shear modulus G = 80 GPa. Given: Mt = 400 75 MPa.

Nm; G = 80

GPa; αmax = 2.5◦ /m;

τa =

To find: Diameter of shaft: 2R. max = τ Strategy: Solve in two phases: (i) design for stress to have τθz a and (ii) check whether α is within the limit. If not, design for stiffness. max = τ , we have Solution: Designing for τθz a max = τθz

Mt R Mt R 2Mt = 4 = πR J πR3 2

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Equating it to τa , we obtain

or,

2Mt = τa πR3 1/3 1/3    Nmm 2Mt 1/3 2 × 400×103 R= = N πτa π×75 mm2 = 15.03 mm = 15 mm (rounded)

Now, we check for stiffness. The angle of twist per unit length α becomes α=

400×103 Nmm Mt



= 4 2 π×15 N GJ (80 × 103 ) × × mm4 2 mm

= 0.0629 × 10−3 rad/mm = 0.0629 rad/m = 3.60◦ /m This value of α exceeds the allowable 2.5◦ /m. We, therefore, design the shaft for stiffness to have αmax =

Mt Mt 2Mt = 4 = 4 πR GJ πGR G 2

or,

⎛

⎞1 4 3 2 × 400×10

⎠ R= =⎝ 2.5×π π×(80 × 103 ) × 1,000×180  1 4 Nmm ×  N   1  × mm mm2 

2Mt πGαmax

1 4

= 16.43 mm = 16.50 mm (rounded) The designer should choose a shaft whose diameter is not less than 33 mm. Interpretation/Comments: In many such designs, either we design for strength or stiffness. These days, the strength of most available

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structural materials has been enhanced considerably through alloying or heat treatment. On the other hand, there are limitations on how much enhancement in stiffness we can achieve. We can play with area properties like moment of inertia for bending problems and polar moment of inertia for torsion problems to a certain extent only. It is not possible to enhance the modulus of a material through heat treatments or through alloying with a small percentage of another material. Thus, some designs these days are stiffness based in which the maximum stress is less than the allowable stress. Example 12.3: Three gears are mounted on a shaft within a gearbox, as shown in Fig. 12.11. A torque is applied to the centre gear B through a tangential force of 6,000 N. The torque is resisted by gears A and C as shown. Determine the following: (i) force Fc that balances the torque on the shaft, (ii) maximum shear stress in the shaft whose diameter is 24 mm, and (iii) the angle of twist of gear C with respect to gear A. The shear modulus of shaft is G = 80 GPa. Given: Shaft radius R = 12 mm; gear A (rA = 40 mm, FA = 4,000 N); gear B (rB = 50 mm, FB = 6,000 N ); gear C (rC = 28 mm); G = 80 GPa.

Fig. 12.11

Shaft with three gears.

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max ; rotation of gear C with respect to gear A (φ To find: Fc ; τθz C/A ).

Strategy: Invoke equilibrium of torque to determine Fc ; invoke the stress and the angle of twist equations. Solution: The support bearings do not resist any torque and the torques acting on the shaft by Gears A and B are resisted by torque developed through Gear C. The equilibrium of external torques yields ΣMz :

MA + MB + MC = 0 −40 × 4,000 + 50 × 6,000 − 28 × Fc = 0 → Fc = 5,000 N

or,

Now, we draw a torsional moment diagram (TMD) to determine internal torsional moments in the various segments of the shaft. Taking a section in segment AB (Fig. 12.12(a)), we have Mt = 40 × 4,000 = 160,000 Nmm = 160 Nm A section in segment BC gives (Fig. 12.12(b)) Mt = 40 × 4,000 − 50 × 6,000 = −140,000 Nmm = −140 Nm The resulting TMD is shown in Fig. 12.12(c). The maximum magnitude of torsional stress is in segment AB. The polar moment of inertial of the shaft is J=

π × 124 πR4 = = 32.57 × 103 mm4 2 2

0N

0N

00

6,0

Mt A

40 mm

(a)

0

4,00

4,00

N

A

160Nm

A

B C

Mt B

–140Nm

TMD

50 mm

(b)

(c)

Fig. 12.12 (a) A section between gears A and B, (b) between gears B and C, and (c) torsional moment diagram.

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Invoking the stress formula of torsion, we obtain the maximum shear stress corresponding to Mt = 160,000 Nmm as max = τθz

160,000 × 12 (Nmm)mm Mt R = = 58.95 MPa J mm4 32.57×103

To determine the relative rotation of gear C with respect to gear A (φC/A ), we can write φC/A =φC/B +φB/A We determine φB/A using the angle of twist equation as φB/A =

(Nmm)mm 160,000 × 70 Mt L  N  = 3 3 GJ (80×10 )×(32.57×10 ) mm2 ×mm4

= 4.298 × 010−3 rad = 0.246o

(in positive z-direction)

Similarly, we have φC/B =

(Nmm)mm (−140,000) × 60 Mt L  N  = 3 3 GJ (80 × 10 ) × (32.57 × 10 ) mm2 × mm4

= −3.224×10−3 rad = −0.185o

(in negative z-direction)

Thus, ΦC/A = 0.246◦ − 0.185◦ = 0.061◦

(in positive z-direction)

Interpretation/Comments: It is important to draw a TMD to determine the maximum internal torsional moment. We then determine max for the maximum magnitude of the internal torsional moment. τθz Also, the angle of twist is determined for each segment, keeping track of the direction of the angle of twist. A comprehensive analysis of finding the stress field has not been done in this example. In fact, the tangential force acting on a gear develops three kinds of internal forces on a shaft, bending

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moment, shear force, and torsional moment. Each one, in turn, develops different kinds of internal stresses. In this problem, we have determined the stress field developed only by the internal torsional moment. Further, if helical gears are used, an axial force is also developed in the shaft. In such a scenario, all the four kinds of generalized forces are present in the shaft. To solve the problem, we invoke the principle of superposition. For each generalized force, the stress field is determined and the results of all the cases are added up to obtain the total stress at a point. This topic will be discussed further in detail in Chapter 13. 12.4

Circular Tubes

Circular tubes are commonly used in many structures and they are often subjected to torsional moments, and, therefore, we should understand how the shear stress is developed in the material of the tube and how much the angle of twist is. Since θ-symmetry is present in a tubular cross-section, all assumptions we made for circular solid members are valid for circular tubes. Thus, γθz is the only nonzero strain component. Also, τθz is also the only nonzero stress component and it varies linearly with radius r. The stress equation of torsion and the equation to determine the angle of twist remain the same; only the polar moment of inertia changes. The polar moment of a tube becomes J=

π(Do4 − Di4 ) π(Ro4 − Ri4 ) = 2 32

(12.12)

It is worth noting here that J of a solid circular member is proportional to D 4 , while the mass is proportional to D 2 . Thus, if a solid member is made hollow, the percentage decrease in J is less than the percentage decrease in mass and it is beneficial to use hollow circular members. In the following example, we will elaborate on this point. Example 12.4: The performance of a solid member is compared with that of a tube whose cross-sectional area is the same and the tube is made of thickness Do /10 where Do is the outer diameter of the tube. If subscripts (or superscripts) t and s refer to the tube and

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solid, determine the following: (a) the ratio of the carrying capacity of torsional moment (Mtt /Mts ) if the maximum allowable shear stress τa is the same and (b) the ratio of the angle of twist αt /αs . Given: Cross-section diameter of solid member = Ds ; tube thickness = Do /10; tube is having the same cross-sectional area. To find: Mttube /Mts ; αt /αs . Strategy: Determine Do and Di for the same area of cross-section; invoke the stress and the angle of twist equations. Solution: The outer diameter of the tube Do is obtained by equating the cross-sectional area. Thus, πDs2 4

=

  π Do2 − Do −

 2Do 2 10

4 Ds = 1.667Ds Do = √ 0.36 Di = 0.8Do = 1.334Ds

and,

 =

π(0.36Do2 ) → 4

(a) Maximum capability of torsional moment in solid member is evaluated by considering the stress equation as max τθz

  Mts D2s 16Mts = 4 = = τa πDs3 s π D 32

Mts =

or,

πDs3 τa 16

Maximum capability of torsional moment in tube is also determined using the stress equation as max τθz

or,

    s Mttube D2o 32Mttube × 1.667D 2 = τa = D4 −D4 = π[(1.667Ds )4 − (1.334Ds )4 ] o i π 32

Mttube =

π [(1.667Ds )4 − (1.334Ds )4 ]τa π = (2.733Ds3 )τa 16 1.667Ds 16

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Thus, the ratio of the two maximum torsional moments is   π(2.733Ds3 )τa tube 16 Mt  3  = = 2.733 s πDs τa Mt 16

(b) The ratio of angle of twist per unit length is determined for a given torsional moment Mt as (Mt /(GJt )) Ds4 αt = Js /Jt = = αs (Mt /(GJs )) (1.667Ds )4 − (1.334Ds )4 1 = 0.220 = 4.56 Interpretation/Comments: For the same mass per unit length, the tube of this example can take 2.732 times more torsional moment and its angle of twist is only 22% of that of the solid circular member. A thinner tube will improve the results further due to a higher polar moment of inertia. Why should we limit the thickness to some finite value, 10% of the outside diameter in this example? We can choose tubes thinner than Do /10, but we cannot make them very thin because of several other design considerations. One of them is that a thin wall tube is likely to buckle due to stress field developed in the wall of the tube. Why does a thin wall tube buckle? The internal torsional moment also develops compressive stress in the wall of a tube in a particular direction which may cause the buckling of the thin wall of the tube. To understand it, let us consider a thin wall tube subjected to an internal torsional moment, as shown in Fig. 12.13(a). The shear stress developed is shown on the stress element. For better visualization, we call the axes of this stress element as y and z as shown. The stress element is subjected to pure shear stress τyz . We will now explore the orientation of the axes that gives maximum compressive stress. The Mohr circle of the stress element is shown in Fig. 12.13(b). It is clear from the Mohr circle that if we rotate the axes by 45◦ , point y goes to point y  and point z goes to point z  . The maximum normal compressive stress is represented by point z  . The stress element with 45◦ rotation is shown in Fig. 12.13(c). If the tube is made of a thin wall thickness, this compressive stress is likely to cause buckling.

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400

z

z

y

z

z

y y y

y

yz

zz

y

(a)

(b)

(c)

Fig. 12.13 (a) A stress element on the wall of a thin tube, (b) the corresponding Mohr circle, and (c) stress element with respect to ±45◦ axes.

12.5

A Torsional Member with Segments

If a discontinuity exists at a location of a torsional member, the torsional stress equation and the equation to determine the angle of twist are not applicable through the discontinuity. Figure 12.14(a) shows the discontinuity between two segments 1 and 2 at a location where an external torsional moment T is applied. In Fig. 12.14(b), the change in radius creates a discontinuity at the junction of segments 1 and 2, and in Fig. 12.14(c), two materials of different modulus are joined to have a discontinuity between two segments 1 and 2. At the transition location, the angle of twist of the left segment is the same as that of the right segment. Material A

2

1 (a)

2

1 (b)

2

Material B

1 (c)

Fig. 12.14 Discontinuity due to (a) an external moment and (b) change in radius and (c) joining of two materials.

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Example 12.5: One end of a circular solid member is rigidly clamped and two external torsional moments are applied, as shown in Fig. 12.15(a). The diameter of the member is reduced from 30 mm to 24 mm at location D. Two different materials are joined at location E. The shear modulus of segments 1, 2, and 3 is 80 GPa, while it is 50 GPa for segment 4. Determine the following: (a) TMD, (b) angle of twist at location B, and (c) maximum shear stress. Given: Rigid clamping at A; TC = 200 Nm; TB = 300 Nm; diameter change at location D from 30 mm to 24 mm, G1 = G2 = G3 = 80 GPa, G4 = 50 GPa. max . To find: (1) TMD, (2) φB , and (3) τθz

Strategy: Plot the TMD and then determine shear stresses and the angle of twist segment-wise. Solution: (1) It is a statically determinate case and we can determine the internal torsional moment by taking a section at relevant locations. The internal torsional moment is the same in segments 2, 3, and 4 and thus we can take a section at any place. Figure 12.15(b) shows A

C

D

(a) 1 200 Nm 160 mm 120

2

140

B

3

180

Mt

(b) Mt

(c)

C

D

E

500 Nm 300 Nm

A

4 300 Nm

B B 300 Nm

200 Nm

(d)

300 Nm

E

C D

E B

TMD

Fig. 12.15 (a) A member with four segments, (b) section in CB portion, (c) section in AC segment, and (d) TMD.

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a section that gives internal torsional moment as Mt = 300 Nm Taking a section in segment 1 (Fig. 12.15(c)), we have Mt = 200 + 300 = 500 Nm Figure 12.15(d) shows the TMD. (2) To determine the angle of twist at point B, we can write φB = φA + φC/A + φD/C + φE/D + φB/E Various φ are evaluated as φA = 0 φC/A =

500×103 ×160 (Nmm) mm Mt L 4  N  = GJ mm4 (80 ×10 3)× π30 mm2 32

= 12.58 × 10−3 Radian = 0.721◦ φD/C =

300×103 ×120 4 = 5.66 × 10−3 Radian = 0.324◦ π30 3)× (80 ×10 32

φE/D =

300×103 ×140 4 = 16.12×10−3 Radian = 0.924◦ π24 3)× (80 ×10 32

φB/E =

300×103 ×180 4 = 33.16 × 10−3 Radian = 1.900◦ 3) (50 × 10 × π24 32

Thus, φB = 0.721◦ + 0.324◦ + 0.924◦ + 1.900◦ = 3.869◦ (3) To find the maximum shear stress, we try to judge which segments are likely to have the maximum shear stress. We should explore segment 1 because it carries the highest internal torsional moment. We should also explore the portion with the smaller radius. Segments 3

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and 4 have the same radius and carry the same torsional moment. Since shear stress does not depend on the material property, we can determine shear stress in any one of the twos. In segment 1, the maximum shear stress is (1) τθz

 D  16× 500×103 Nmm 16Mt 2 = = 94.3 MPa = πD 4 πD 3 mm3 π×303

Mt =

32

In segments 3 and 4, the maximum shear stress is (3,4)

τθz

=

16×(300×103 ) = 110.5 MPa π×243

The maximum shear stress is within segments 3 and 4. Interpretation/Comments: To find the maximum shear stress, sometimes we have to evaluate at several places and then figure out the point that is stressed the most. 12.6

Statically Indeterminate Torsional Member

We have only one equilibrium equation to account for torsional moments acting on a slender member. If there is more than one support which resists torsional moments, we need to develop additional geometric compatibility relations. To develop them, we need a force–displacement relation which has already been developed as tL φ= M GJ . An example is presented to demonstrate how to solve an indeterminate structural problem. Example 12.6: A solid circular slender member is rigidly supported at each end, as shown in Fig. 12.16(a). An external torsional moment T is applied at a distance a from end A, as shown in the figure. Determine the torsional moment resisted by each support. Given: Length = L; rigid support at A; rigid support at B; external torsional moment T at location C. To find: MtA ; MtB . Strategy: Consider one of the two supports as redundant.

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C

A

B (b)

(a) a

T L

MtA A

D

C

E

MtB

T

b (c)

Mt D

C

E

T (b/L)T (e) 0

(d) -(a/L)T

B

E

Mt

B MtB B MtB

Fig. 12.16 (a) A solid circular member supported rigidly at its ends, (b) free body diagram, (c) finding Mt in segment AC, (d) finding Mt in segment CB, and (e) torsional moment diagram.

Solution: It is a case of indeterminacy of degree one because there are two unknown supports and only one equilibrium equation (Fig. 12.16(b)). Treating MtB as a redundant force, the corresponding compatibility condition is φB =0. Note that a torsional moment diagram (TMD) should be drawn to know the internal torsional moment carried by each segment. The internal torsional moment in segment AC is determined by taking section at any location, say at D (Fig. 12.16(c)). Then, Mt becomes Mt = T + MtB This twisting moment is constant on segment AC. The angle of twist at location C with respect to location A becomes φC/A =

(T + MtB )a GJ

By taking a section in segment CB (Fig. 12.16(d)), the internal twisting moment is determined as Mt = MtB The angle of twist of B with respect to C is φB/C =

(MtB ) × (L − a) GJ

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The compatibility condition yields φB = φC/A + φB/C = 0 (T + MtB )a MtB ×(L − a) + =0 GJ GJ

or,

T a + MtB L = 0 a T MtB = − L

or, or,

(12.13)

The equilibrium equation gives − MtA + T + MtB = 0 (12.14) a (L − a) b T = T = T MtA = T + MtB = T − L L L

ΣMz : or,

The torsional moment diagram (TMD) is shown in Fig. 12.16(e). Interpretation/Comments: The approach is similar to the one we adopted to solve cases of statically indeterminate problems of bending moments. There also, we treated a reaction support as a redundant member and invoked the geometrical constraint to develop an additional equation. We will try to weigh whether our results make sense. One way to check is that we consider a symmetrical case whose solution we can predict without solving. For example, when a = b = L/2, we predict MtA = T /2 and MtB = −T /2. These values should match with our solution. We find that Eqs. (12.13) and (12.14) yield the same results. The matching of the two results is a necessary condition and by no means is the sufficient condition. However, if the results match, it boosts our confidence and there is a good chance that our results are correct. Another way to check our results is to take the problem to an extreme case whose solution we can predict. In this problem, a =0 is one of the extreme cases. The entire external moment is applied at the end A. The applied moment is then fully resisted by the support at A with no role of the support B. Thus, MtA should be equal to T . Equation (12.14) gives the same answer. Our confidence in the results is further enhanced. In real-life cases, we often solve problems whose results are not known. We try to invent ways to weigh the creditability of our results.

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Comments on Determination of Stresses at a Point

We have completed the discussion on stresses developed by each generalized force and it is the right time to look back to realize what has been done so far and what are the general aspects. All the four types of internal generalized forces on a section of a slender member are as follows: (i) axial force, (ii) bending moment, (iii) shear force, and (iv) torsional moment. In Chapter 9, we discussed the stress developed due to the internal axial force on all points of a cross-section. In Chapters 10 and 11, we discussed stress developed due to bending moment and shear force on a cross-section. Finally, the stress developed due to the torsional moment on a cross-section is presented in this chapter. If more than one generalized internal forces act on a section, we can invoke the principle of superposition to split the problem into several subproblems. We solve each subproblem separately and, in the end, we sum the results of the subproblems to have the solution for the complex problem. We will further discuss this procedure in Chapter 13 and solve several problems.

12.8

Summary

An analysis has been developed to determine the stress distribution on a cross-section of a slender member and the angle of twist caused by an internal torsional moment. The results of the analysis are confined only to circular cross-section, solid or tubular. The results are not applicable to noncircular cross-section whose results are quite different from those of circular members and the analysis is beyond the scope of this book. The analysis is convenient to solve with cylindrical coordinates, r, θ, z with z-axis aligned with the length of the slender member. The analysis of a circular member is simpler because θ-symmetry exists and there is no warping of the cross-section. There exists only one nonzero strain component, γθz , which is equal to rα where α is an important parameter, the angle of twist per unit length. This makes only one stress component nonzero, τθz = Grα. Finally, equilibrium analysis is carried out on the cross-section which yields the useful torsional stress equation as τθz = MJt r where J is the polar moment Mt and the angle of inertia. The angle of twist per unit length is α = GJ of twist for a member length L becomes φ = Lα = Mt L/GJ.

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12.9

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407

Problems

1. A circular bar AB of diameter 30 mm is rigidly supported at its end A and the end B is welded to tube BC at location B, as shown in Fig. 12.17(a). The outside diameter of tube BC is 36 mm and inside diameter 30 mm. End C of the tube BC is twisted with torque 200 Nm. Determine the maximum shear stress in both portions. 2. A circular shaft AB of diameter 25 mm is rigidly fixed at its end A and a torque of T = 80 Nm is applied at end B, as shown in Fig. 12.17(b). Its performance is compared with a tube CF of outside diameter do having its mass same as the shaft AB. The wall thickness of the tube is do /10. It is rigidly fixed at end C and the same torque of 80 Nm is applied at end F. (a) Determine maximum shear stress in AB and the angle of twist at location B. (b) Determine do. (c) Determine maximum shear stress in tube CF and the angle of twist at location F. (d) Compare the two shear stresses and the angle of twist. Comment on the results. 3. Shaft (1) is joined with another shaft (2) of the same diameter through a coupling, as shown in Fig. 12.18(a). The coupling is made of a disk plate attached to each end of the shaft. The

A

A B

30

500

0m

m 50

C

0m

B

mm

T=80 Nm

C

m

(a)

F

200 Nm

500

mm T=80 Nm

(b) Fig. 12.17

(a) Problem 1 and (b) Problem 2.

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disk plates are joined with six screws on the nominal diameter of 90 mm. A torque of 300 Nm is transferred from one shaft to another through bolts loaded in shear. The maximum allowable shear stress in the shafts is 50 MPa. Determine the following: (a) Shear load carried by a bolt. (b) Diameter of shafts. 4. A cross-spanner with its all arms of 250 mm length tightens a bolt by applying force 120 N each at points C and E, as shown in Fig. 12.18(b). The forces are applied normal to the plane of the cross-spanner. Determine the diameter of the arm AB if maximum allowable shear stress is 40 MPa and the rotation of the spanner does not exceed 0.5◦ (shear modulus = 79 GPa). 5. An aluminium rod of circular cross-section and diameter 32 mm is bent, as shown in Fig. 12.19(a). Its end A is rigidly fixed to a wall and a force of 500 N is applied at the free end as shown. The modulus of the aluminium bar is 70 GPa and Poisson’s ratio is 0.3. Determine the following: (a) Maximum shear stress in portion AB developed by the torsional moment. (b) Deflection of point E due to the torsional rotation of portion AB. 6. A load W (= 1,800 N) is lifted using a rope that is wound on a drum of diameter 150 mm, as shown in Fig. 12.19(b). The drum is attached to an axle of 28 mm diameter as shown and the axle is free to rotate at both support points. The load is lifted by applying a force on the handle of the lever arm. Determine the z y

2 300 Nm

x C

1

P

250

A mm

Fig. 12.18

250

mm

P E

Coupling (a)

B

(b)

(a) Problem 3 and (b) Problem 4.

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500

409

mm

A B

P

z

500 N 40 m 0m

C E

300

mm

(a)

Fig. 12.19

y

m

0m

45

W

x (b)

(a) Problem 5 and (b) Problem 6.

required force P on the handle to raise the load. Also, determine the maximum shear stress on the axle developed by the applied torque. 7. A shaft AB is attached to an electric motor at its end A through a coupling, as shown in Fig. 12.20. The shaft is supported at supports C and B which do not provide any resistance against the rotational motion. Two gears at locations E and F are attached to this shaft. The electric motor is stalled due to some malfunctioning and the machine stops. Gears at E and F develop torques of 300 Nm and 400 Nm in the direction shown in the figure. Determine the following: (i) TMD, (ii) rotation of shaft at end B with respect to location A, and (iii) maximum shear stress in the shaft due to the torsional moment. The diameter of the shaft is 22 mm, the modulus of its material is 207 GPa, and Poisson’s ratio is 0.29. 8. In an experimental setup, a pure torsional moment is developed in a circular bar BF of diameter d, as shown in Fig. 12.21(a). The bar is rigidly fixed at end B. It is supported by a ball bearing at location C within which the bar rotates freely. Through a thin steel cable, a pure torque is applied to the load pulley using a deadweight W and an appropriate configuration of cable loading. One end of the steel cable is gripped securely to point M on the circumference of the load pulley. It is wound over the load pulley to idler pulley G and then to a floating pulley H to which the dead load is suspended. Then, the cable passes over the idler pulleys, J and K. The cable wraps over the load pulley and the end of the cable is firmly secured at point N of the circumference

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410

A

C E

F

1m

1.2

B m 0.8

Fig. 12.20 B

z

y x

m

Problem 7.

K

Load Pulley

C

M

L

2

1

R F N G

J

H A W

(a)

Fig. 12.21

300 mm

B

(b)

(a) Problem 8 and (b) Problem 9.

of the pulley. All the three idler pulleys do not exert any rotational resistance and thus the tensile force in the entire cable is the same. Determine the tension in the cable, the pure moment applied on the bar BF, the maximum shear stress in the bar and the rotation of the load pulley.

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page 411

Torsion

411

9. Two circular concentric tubes are rigidly fixed at their ends at location B, as shown in Fig. 12.21(b). The other ends are fixed to a rigid plate at location A. The outer tube is made of aluminium (shear modulus = 27 GPa) with its outer diameter of 60 mm and wall thickness of 3 mm. The inner tube is made of mild steel (shear modulus = 80 GPa) with its outer diameter of 50 mm and wall thickness of 4 mm. A torque of 800 Nm is applied to the plate at location A. Determine the angle of twist of the plate and the torsional moment carried by each tube. 10. Shaft AB is rigidly fixed at end A and is supported at location C which allows the shaft to rotate freely (Fig. 12.22(a)). A gear of 48 teeth is mounted at the end B of the shaft. Another shaft EF is supported on two supports M and N, which allow free rotation of the shaft. A gear of teeth 18 is mounted at end E on shaft EF in such a way that it meshes with the gear of shaft AB. At end F, a torque of 225 Nm is applied. Determine the diameter of both shafts if the shear stress developed by the torsional moments does not exceed 50 MPa. Determine the angle of twist at location F (shear modulus of shafts = 80 GPa).

1,2

00

A

mm

C

F

B M 1

N 80 0m

m

E

A

2

100

400 mm B

(a)

Fig. 12.22

(b)

(a) Problem 10 and (b) Problem 11.

C

E

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11. The ends of two circular rods, AB and CE, are brazed to a central plate, as shown in Fig. 12.22(b). The diameter of the centre plate is large and can be considered as a rigid member. The other ends of the rods are rigidly fixed. Rod AB is made of 40 mm diameter and rod CE of 25 mm. A torque of 200 Nm is applied to the central plate. Both bars are made of brass with its shear modulus of 37 GPa. Determine the following: (a) The angle of rotation of the central plate. (b) Torsional moment carried by each bar. (c) Maximum shear stress in each bar.

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Chapter 13

Combined Stresses and Yield Criteria

A fable was narrated in Chapter 11 about the seven sticks bound together with a string and seven sons. I want to refer to it again. To break the bundle, the sons first figured out how to untie the binding string and separate the sticks out. Once they were separated, individual sticks were broken easily. Similarly, we have learnt to split the analysis of a slender member into four kinds of internal generalized forces and learnt to solve each subproblem in Chapters 9–12. In the fable, the broken sticks were thrown away. An analogy usually works only up to a certain extent. In our analysis, we take the results of individual subproblems and superpose (add) them to obtain the combined stress at a point by applying the principle of superposition. 13.1

The Principle of Superposition

We discussed this principle in Section 11.5 but it will be discussed again as it is the heart of this chapter. The principle of superposition works if the deformation of all the points of the structural member remains elastic and follows the linear elastic behavior, that is, none of its points is yielded. We split the problem into subproblems, each having a smaller number of applied forces, usually one in each subproblem. We solve all the subproblems separately and find the stress at a point of interest. The principle states that the combined (overall) stress at the point is the sum of the stresses of all subproblems at that point. Also, the combined displacement of the point is the sum of the displacement of the subproblems.

413

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Mechanics of Materials: A Friendly Approach

414

P1 A

P1 A

Fig. 13.1

H

P2

P2 H

B

B

P3

A

H

B

A

H

B

P3

Splitting a problem of three external forces into three subproblems.

The principle of superposition works for external forces as well for internal generalized forces. Figure 13.1 shows the problem with three external forces. The problem is split into three subproblems, each having only one external force. If our interest is to determine stress at point H, we find stress at this point for each of the three subproblems. The sum of the three stresses is the combined stress at point H. Figure 13.2(a) shows the free body diagram of the member with three external forces P 1 , P 2 , and P 3 and the objective is to determine stresses at the internal point H. There are three reaction forces, VA , HA , and VB , developed by the supports. In this 2D problem, there are three internal forces shown on the cross-sectional plane passing through point H in Fig. 13.2(b), the axial force F , the shear force V, and the bending moment Mb . In fact, the generalized internal forces can be treated as external forces on the saved part. We split the problem into subproblems (Fig. 13.2(c)) and determine stress at the point of interest, say at point H. Then, we invoke the principle of superposition by summing the stresses of the subproblems to have the combined stress at point H. Once we have the combined stress at a point, we can find whether the external forces develop high enough internal stresses at this point to cause material failure. These aspects will be addressed through yield criteria in Sections 13.7–13.9. Also, if the combined

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Combined Stresses and Yield Criteria

415

P2

P1 D H

HA

P3 VB

VA (a)

D H

F

P1 V Mb D H

HA

VA

V D H

F

Mb D H

(b) (c)

Fig. 13.2

Splitting three internal forces into three subproblems.

displacement of a point of the member is more than the displacement permitted, the member is regarded as failed. This will be discussed in Section 13.10. 13.2

Overall Procedure to Find Combined Stress at a Point

Before we proceed further, we first review the main aspects of the first 12 chapters and take a bird’s eye view. In this book, we have mainly focussed our attention on how to find stress at an internal point H of a slender member. The net stress is determined through the following procedure. We separate the member from other members and the supports and draw its free body diagram. In fact, we identify all external forces acting on the member. Also, we identify all nonzero reactions at the support points. We then invoke equilibrium equations to determine the value of all reaction components. This is easily done for a statically determined structural member. For the statically indeterminate member, the procedure is a bit more involved but the strategy of solving is the same.

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We then section the member through an imaginary plane which is normal to the axis of the member and passes through point H. We save one of the two separated parts. On the cross-sectional plane of the saved part, we determine all the four kinds of internal generalized forces, that is, axial force F , bending moment Mb , shear force V and torsional moment Mt using equilibrium equations in 2D members. In 3D members, there is one axial force F , two bending moments My and Mz , two shear forces Vy and Vz , and one torsional moment Mt . We then focus our attention only on the sectioned plane. We consider all nonzero generalized internal forces, one at a time. For each generalized force, we determine the stress component at any point, say point H. It is worth noting that stress, in general, varies from one point of the section to another. We write the stress tensor at the point in 3 × 3 matrix form and also draw a 3D stress element, clearly showing the stress component on all the relevant faces and in correct directions. It is important to use the full 3 × 3 matrix to facilitate the superposition. The superposition is carried out by summing all stress tensor matrices to obtain the net stress tensor at point H and also draw a 3D stress element. The procedure may seem long and requires the sketching of several 3D stress elements and the use of 3 × 3 matrices. However, I have realized through my long teaching experience that students make many mistakes by not showing a stress component on the correct face and in the correct direction of a stress element. If the long procedure is systemically followed, the chances of making mistakes are considerably reduced. A symmetrical stress tensor with six components is more complex than a vector with three components and a disciplined analytical methodology is strongly recommended especially for beginners. However, engineers in the field may not solve new problems in their routine work and in that case, the long and thorough procedure may be shortened. However, if they solve new problems, it is recommended that they go through the long process of working on one step at a time. Why do we emphasize on finding a stress value at a point? The stress tensor with all of its six components varies, in general, from one point of the member to another. Only in some special cases, its value is uniform over the cross-section of the member.

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page 417

417

Most books on the introductory level deal with only three stress components of 2D analysis. But the world is three dimensional and sometimes a 2D analysis leads to serious errors. In fact, I personally do not like to represent stress tensor with one subscript as σx and σy even for 2D analysis. We should call a spade a spade. If the parameter is a second-order tensor, we should not treat it as a component of a vector. 3D analysis has now become more important as we can solve very complex problems of stress analysis using a computer employing any one of the several good standard software packages. If a powerful solving tool is available to us, it makes sense to upgrade our analytical tools also. We will now look into several solved examples, starting from a simple problem of 2D and ending up with a relatively complex 3D problem. Example 13.1: Consider a solid round slender member of diameter 20 mm whose end B is rigidly fixed, as shown in Fig. 13.3(a). Three external forces act at location A; an axial force of magnitude 2,500 N at the centre of the round member and two forces, 120 N each, act on a rigidly fixed cross-member, as shown. Determine the combined stress tensor at point H which is 4 mm away from the centre and sketch the stress element. Given: A solid slender member AB; diameter d = 20 mm; rigidly fixed at end B; at location A, axial force = 2,500 N and two forces, 120 N each, acting on a cross member at distance 50 mm each. To find: Combined stress at point H; sketch stress element. Strategy: We first determine generalized internal forces at the crosssection passing through point H and then invoke superposition. Solution: In this simple problem, we need not evaluate the reactions at support A. Figure 13.3(b) shows the internal generalized forces at the cross-section passing through point H. The axial force F , the bending moment Mb , and shear force V are as follows: F = 2,500 N Mb = 120 × 50 + 120 × 50 = 12,000 Nmm V =0

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418

120 N 2,500 N

H x

H

Ø20

4

B

50

A

Q

60 50

y (a)

9in x 6in

120 N

200 mm

y

F

x 50

A

M H V b

x

120 N

Fig. 13.3

140 mm

Determination of net stress at point H.

H

+

4

Bending

Tension (a)

Fig. 13.4

H

(b)

2,500 N

50

120 N

(b)

Subproblems of Example 13.1.

We now focus our attention on the cross-section and split the problem into two subproblems to invoke the principle of superposition (Fig. 13.4). Figure 13.4(a) shows the first subproblem with only the internal axial force. Since the axial force acts at the centroid, the stress is uniform, that is, the stress is the same at all points of the crosssection. Thus, stress at point H is   2,500 P N a = π(20)2 = = 7.96 MPa σxx A mm2 4

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Combined Stresses and Yield Criteria

The stress tensor at point H in 3 × 3 ⎡ 7.96 0 ⎢ a 0 σxx = ⎣ 0 0 0

page 419

419

matrix form is ⎤ 0 ⎥ 0⎦MPa 0

The corresponding stress element is shown in Fig. 13.5(a). The second subproblem is shown in Fig. 13.4(b) with the bending moment acting on the cross-section. The resulting stress distribution b is triangular, as shown in the figure. At point H, the bending of σxx stress is obtained as b =− σxx

Mb (y) Mb y 64 × 12,000 × 4 Nmm·mm = − 4  = − I π(20)4 mm4 πd 64

= −6.11 MPa The stress tensor for this subproblem in 3 × 3 matrix form is ⎡ ⎤ −6.11 0 0 ⎢ ⎥ b 0 0⎦MPa =⎣ 0 σxx 0 0 0 The corresponding stress element is shown in Fig. 13.5(b). The net stress at point H is the sum of the two results. Thus, ⎡ ⎤ ⎡ ⎤ 7.96 0 0 −6.11 0 0 ⎢ ⎥ ⎢ ⎥ net a b 0 0⎦ + ⎣ 0 0 0⎦ = σxx + σxx =⎣ 0 σ11 0 0 0 0 0 0 ⎡ 1.85 ⎢ =⎣ 0 0

⎤ 0 0 ⎥ 0 0⎦MPa 0 0

The stress element of the combined stress is shown in Fig. 13.5(c). Interpretation/Comments: The axial force and the bending moment are different kinds of generalized internal forces, but both of them contribute toward normal stress σxx . Let us find the combined stress

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Mechanics of Materials: A Friendly Approach

y 4

H

(a)

2,500 N a xx = 7.96 MPa

z

12,000 Nm

x

y

y

(c)

H

net

xx = 1.85

z

(b)

x

b xx= 6.11

z

x

Fig. 13.5 (a) Stress due to axial force, (b) stress due to bending moment, and (c) combined stress at point H.

at the top point Q (Fig. 13.3(a)). The stress of the first subproblem will remain unchanged. However, the bending stress at point Q becomes Mb d2 64 × 12, 000 × 20 Mb y Nmm·mm b 2 = − 4 = − σxx = − 4 I mm4 π (20) πd 64

= −15.28 MPa The stress tensor at point Q is ⎡ 7.96 ⎢ net a b σxx = σxx +σxx = ⎣ 0 0 ⎡

−7.32 ⎢ =⎣ 0 0

⎤ ⎡ 0 0 −15.28 ⎥ ⎢ 0 0⎦ + ⎣ 0 0 0 0

⎤ 0 0 ⎥ 0 0⎦ 0 0

⎤ 0 0 ⎥ 0 0⎦MPa 0 0

Although point Q is on a free surface, only one face of the stress element lies on the free surface; all other five faces are located inside the material of the member. Example 13.2: A slender member AB is made of a rectangular cross-section. It is supported through a hinge at end A and is roller

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Combined Stresses and Yield Criteria 3

P=10,000 N

(a)

x

B D

E

100

12

150 300 mm

VA

(c)

E

A

HA

Py=10,000/ 2 N

Px=10,000/ 2

10,000/ 2

H 8

45o

A

(b)

421

32 mm

y

page 421

Cross-section at D B

VB

H V Mb F D

5,000/ 2

Fig. 13.6 (a) Slender member, (b) free body diagram, and (c) internal generalized forces at the cross-section passing through point H.

supported at end B, as shown in Fig. 13.6(a). A force P = 10,000 N is applied at location E at an angle of 45◦ as shown. Determine the combined stress at point H which is 100 mm away from point A and whose location on the cross-section is shown in the end view. Also, sketch the stress element of the combined stresses. Given: AB = 300 mm; hinged at A; roller supported at B; P = 10, 000 at 45◦ angle applied at x = 150 mm; cross-section 32 mm × 12 mm. To find: Combined stresses at point H. Strategy: Determine reactions at points A and B; section the member at a place in such a way that it passes through point H. Split into subproblems and superpose the results. Solution: It’s a statically determinate structure and we can find reaction components using only equilibrium equations. It is convenient to consider components of external force P in x and y directions, as

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shown in Fig. 13.6(b). The force balance in x-direction gives √ √ ΣFx : HA + 10,000/ 2 = 0 → HA = −10,000/ 2 The force balance in y-direction gives ΣFy :

√ VA + VB = 10,000/ 2 N

Taking moment about point A, we obtain   √ 10,000 √ = 0 → VB = 5,000/ 2 ΣMA : 300VB − 150 × 2 Then, VA =

√ 5,000 10,000 √ − √ = 5,000/ 2 2 2

To obtain internal generalized forces at the cross-section passing through point H, we make an imaginary cut and save the left portion, as shown in Fig. 13.6(c). The generalized internal forces on the section (F, Mb and V ) are determined using equilibrium equations as √ F = 10,000/ 2 N √ V = −5,000/ 2 N 5,000 Mb = 100 × √ = 353.6 × 103 Nmm 2 Now, the problem is split into three subproblems, as shown in Fig. 13.7 to invoke the principle of superposition. We consider the first subproblem shown in Fig. 13.7(a) with only axial force F . Since the force is applied at the centroid of the section, the developed stress (1) σxx is uniformly distributed on the cross-section and is given by 10,000

(1) = σxx

√ N F 2 = = 18.41 MPa A 32 × 12 mm2

Its stress element is shown in Fig. matrix form is ⎡ 18.4 ⎢ (1) σH = ⎣ 0 0

13.7(a). The stress tensor in 3 × 3 ⎤ 0 0 ⎥ 0 0⎦MPa 0 0

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Combined Stresses and Yield Criteria

423

y y

(a)

H

N

Nl (1)

F x

z

z y

y

(b)

N

H

N z

= 18.4 MPa

x y

l

(2)

x z

= 10.4

= 86.3

x

= 67.9

y C

y

(c) NH

z

8 Nl 8 N

V

H

12

(3)

z

x

= 10.4

l

N

(d)

z

x

Fig. 13.7 (a) Axial force, (b) bending moment, (c) shear force, and (d) stress element of combined stress tensor.

For the other two subproblems, we will need the value of the moment of inertia I which is determined as I=

12 × (32)3 bh3 = mm·mm3 = 32.77 × 103 mm4 12 12

Figure 13.7(b) shows the second subproblem with internal bending moment acting on the section. This generalized force develops σxx at point H as (2) =− σxx

(353.6 × 103 ) × 8 (Nmm) × mm Mb y =− = −86.3 MPa I mm4 32.77 × 103

The corresponding stress element is shown in Fig. 13.7(b) and the stress tensor is ⎡ ⎤ −86.3 0 0 ⎢ ⎥ (2) 0 0⎦MPa σH = ⎣ 0 0 0 0 The third subproblem is shown in Fig. 13.7(c) with shear force V . The shear stress at point H is determined by the formula τ = V Q/bI.

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Mechanics of Materials: A Friendly Approach

424

To determine Q, we consider the area of the section beyond the point H. Q for this case is found easily by taking moment of the area, that is, we multiply the area with the distance of its centroid C from the neutral axis NN . Thus, Q = A¯ y = (8 × 12) × (8 + 4) = 1.152 × 103 mm3 Then, (3) τxy

 5,000 √ − × (1.152 × 103 ) N·mm3 VQ 2 = = = −10.4 MPa bI 12 × (32.77 × 103 ) mm·mm4

The stress element is shown in Fig. 13.7(c) and the stress tensor is ⎡ ⎤ 0 −10.4 0 (3) 0 0⎦MPa σH = ⎣−10.4 0 0 0 We obtain the combined stress at point H by adding all the three matrices to have (1)

(2)

(3)

σH = σH + σH + σH ⎡ ⎤ ⎡ 18.4 0 0 −86.3 0 0 0⎦ + ⎣ 0 0 =⎣ 0 0 0 0 0 0 ⎡ ⎤ −67.9 −10.4 0 0 0⎦MPa σH = ⎣−10.4 0 0 0

⎤ ⎡ 0 0 0⎦ + ⎣−10.4 0 0

−10.4 0 0

⎤ 0 0⎦ 0

The stress element of the combined stress is shown in Fig. 13.7(d). Interpretation/Comments: The problem with three internal forces was solved by considering the solution of each one separately and then superposing all the three solutions. Example 13.3: A shaft AB of 25 mm diameter is supported on ball bearings at locations A and B (Fig. 13.8(a)). The bearing at location A is capable of resisting radial as well as axial load, while the bearing at location B does not resist axial motion. At location D, a helical spur gear applies three external forces: (i) torque T = 80 Nm, (ii) transverse force P = 2,700 N, and (iii) axial force F e = 2,200 N, as shown in the figure. On the left end of the shaft at location C, a diesel

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Combined Stresses and Yield Criteria

y

C

C

(c)

x

A

2,700 N 80 Nm 2,200 N D

RA (d)

2,200 N

C

B

RB

V Mb

F s

Mt

540 N (e)

(b) L

C

80 Nm

U

2,200 N B D

200 250 mm

HA

Mde

425

2,700 N 80 Nm

(a)

page 425

F

Mt

Mb V

r

B 2,160

Fig. 13.8 (a) Shaft with forces and supports, (b) the cross-section just left of point D, (c) free body diagram, (d) section in AD portion, and (e) section in DB portion.

engine is attached with a flexible coupling in between in such a way that it resists torque developed in the shaft by the external forces but does not resist axial and transverse forces. Determine the following: (i) TMD, AFD, SFD and BMD and (ii) combined stresses at the uppermost point U and at the lowest point L of the cross-section just left of location D (Fig. 13.8(b)). Given: AB = 250 mm; shaft diameter d = 25 mm; bearing at A resists radial and axial forces; bearing at B resists only radial force; external forces at location D are as follows: T = 80 Nm; P = 2,700 N; Fe = −2,200 N; only torque is resisted at location C (Fig. 13.8(a)). To find: (i) TMD, AFD, SFD and BMD, (ii) σ at the uppermost point U, σ at the lowest point L of cross-section just left of location D. Strategy: Determine reactions and draw all four diagrams; invoke superposition to find the net stresses at points U and L.

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426

Solution: It is a statically determinate case and the free body diagram is shown in Fig. 13.8(c). We now determine the reactions at the supports using equilibrium equations: ΣFx :

HA − 2,200 = 0

ΣFy :

RA + RB = 2,700

ΣMzA ΣMxC

:

250RB − 200 × 2,700 = 0

:

C −Mde + 80 = 0

These equations yield HA = 2,200 N;

RA = 540 N;

RB = 2,160 N;

C Mde = 80 Nm

To determine internal generalized forces in the AD segment of the shaft, we take a section at distance s and save the left portion, as shown in Fig. 13.8(d). The internal forces at the section are as follows: Mt = 80 Nm;

F = −2,200 N;

V = −540 N;

Mb = 540s

We section the DB portion of the shaft at a distance r from location B and save the right portion, as shown in Fig. 13.8(e). The internal forces at the section are as follows: Mt = 0;

F = 0;

V = 2,160 N;

Mb = 2,160r

The resulting TMD, AFD, SFD, and BMD are shown in Fig. 13.9. We will now determine the combined stress at points U and L. Figure 13.10 shows the section just left of point D with all the internal forces. We split the problem into four subproblems. Before we move further, we first find out the area properties needed to solve the subproblems as follows: π(25)2 = 490.9 mm2 4 π × (25)4 πd4 = = 19.17 × 103 mm4 Moment of inertia I = 64 64 π × (25)4 πd4 = = 38.34 × 103 mm4 Polar moment of inertia J = 32 32 The first subproblem with torsional moment Mt = 80 Nm is shown in Fig. 13.11. The shear stress τrθ is determined at all points of the Area A =

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Combined Stresses and Yield Criteria

page 427

427 2,160 N

80 Nm 0

A

D

B

TMB

0

B

A

D

SFD

–540 N 108 Nm 0A

D

B

AFD

0

D

A

B

BMD

–2,200 Nm

Fig. 13.9 Torsional moment diagram (TMD), axial force diagram (AFD), shear force diagram (SFD), and bending moment diagram (BMD).

y U Mb=108 Nm F= 2,200 N

z

L Mt=80 Nm

V=540 N

x

Fig. 13.10 All the four generalized internal forces on a cross-section just left of location D.

circumference of the section as τrθ =

(80 × 103 ) × ( 25 Mt r 2 ) Nmm·mm = = 26.08 MPa 3 J 38.34 × 10 mm4

The direction of shear stress τrθ is shown on the circumference of the section (x-plane) in Fig. 13.11(a). Its direction is normal to the radial direction everywhere. Since all other stress components of this problem are expressed in Cartesian coordinates, we will express the shear stress components in Cartesian coordinates too. At points U and L, the stress elements are shown in Figs. 13.11(b) and 13.11(c).

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428

y

y

z

y

Shear Stress

U

L

x

xy=26.08

z

x

xy=26.08

z

x

Point U

Point L

(b)

(c)

(a)

Fig. 13.11 (a) Shear stress on the cross-section, (b) shear stress at point U, and (c) shear stress at point L.

We now focus our attention on x-face of the stress element. The shear stress is in positive z-direction at point U and negative z-direction at point L. They are expressed through matrices as ⎡

(1)

σU

0 =⎣ 0 26.08

⎤ ⎡ 0 26.08 0 (1) 0 0 ⎦MPa; σL = ⎣ 0 0 0 −26.08

⎤ 0 −26.08 0 0 ⎦MPa 0 0

The second subproblem is shown in Fig. 13.12(a) with axial internal force F = −2,200 N. The stress field developed is uniform over the cross-section and is given by σxx =

2,200 N F =− = −4.48 MPa. A 490.9 mm2

The stress tensors are as follows: (2)

σU

⎡ −4.48 =⎣ 0 0

⎤ 0 0 0 0⎦MPa; 0 0

(2)

σL

⎡ −4.48 =⎣ 0 0

⎤ 0 0 0 0⎦MPa 0 0

The third subproblem, shown in Fig. 13.12(b), deals with shear force V = −540 N. In this case, the shear stress is zero because Q in shear stress formula τ = VbIQ is zero at points U and L. Alternatively, we need not invoke the shear stress formula and directly state that they are zero at points U and L because y-faces are lying on the lateral surfaces which are traction free (Fig. 13.12(b)).

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y

429

y

U

(a)

page 429

F=2,200 N (L)

z

L

x

z

y

U

L

x

xx= 4.48 MPa

y xy= 0

x z

xx= 4.48

y

U

(b)

z

x(U)

L V=540 N

xy= 0

z

y

U

z

x

L

x

y

y

U

(c)

Mb=108

z

L

x

(U)

z

xx= 74.2

U

x

(L)

xx= 70.42

z L

x

Fig. 13.12 (a) Subproblem with the axial force, (b) with the shear force, and (c) with the bending moment.

Thus, stress tensors ⎡ 0 ⎢ (3) σU = ⎣ 0 0

at points U and L are as ⎤ ⎡ 0 0 0 ⎥ ⎢ (3) 0 0⎦MPa; σL = ⎣0 0 0 0

follows: ⎤ 0 0 ⎥ 0 0⎦Mpa 0 0

The fourth subproblem with Mb = 108 Nm is shown in Fig. 13.12(c). The resulting stress component at points U and L are as follows:

108 × 103 × 25 Mb y (U ) 2 Nmm·mm =− = −70.42 MPa σxx = − 3 I mm4 19.17 × 10 (L) σxx





108 × 103 × − 25 Mb y Nmm·mm 2 =− =− = 70.42 MPa I mm4 19.17 × 103

The stress elements at points U and L are as follows: ⎡ ⎤ ⎡ ⎤ −70.42 0 0 70.42 0 0 ⎢ ⎥ ⎢ ⎥ (4) (4) 0 0⎦MPa; σL = ⎣ 0 0 0⎦MPa σU = ⎣ 0 0 0 0 0 0 0

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The combined stresses at points U and L are obtained by just summing the matrices of all the four cases as follows: ⎡ ⎤ ⎡ ⎤ (−4.48 − 70.42) 0 26.08 −74.90 0 26.08 ⎢ ⎥ ⎢ ⎥ 0 0 0 ⎦= ⎣ 0 0 0 ⎦MPa σU = ⎣ 26.08 0 0 26.08 0 0 ⎡ ⎤ (−4.48 + 70.42) 0 −26.08 ⎢ ⎥ 0 0 0 ⎦ σL = ⎣ −26.08 0 0 ⎡ ⎤ 65.94 0 −26.08 ⎢ ⎥ 0 0 ⎦ MPa =⎣ 0 −26.08 0 0 Interpretations/Comments: In this 2D problem, all four generalized internal forces act on the cross-section. We managed it by following the procedure of taking one subproblem at a time and drawing the required sketches. Also, note that the stress components are in 3D form although the problem belongs to 2D. Thus, it is recommended strongly that we should use 3D analysis even if the problem looks to be 2D analysis. It is important to write stress components in matrix form for all problems, no matter how simple a case looks. The zeros in a 3 × 3 matrix of a stress component make us feel good because we are assured that there is no other stress component that has been left out. Example 13.4: A solid round member AB of diameter 28 mm is fixed at end A. BD is a cross member welded to member AB at point B, as shown in Fig. 13.13(a). Two forces, P 1 = 400 N and P 2 = 300 N, are applied at location D as shown. Determine combined stress at points H and K of the cross-section at location S. Given: AB = 500 mm; AS = 300 mm; BD = 300 mm; diameter of AB = d = 2r = 28 mm; P 1 = 400 N; P 2 = 300 N. To find: Combined stress at point H and point K of the cross-section at location S (x = 300 mm). Strategy: Determine all generalized forces on the section at location S, split into subproblems and invoke superposition.

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431

y 5 300 mm 00 mm

x

A

B

S

y My=80 Nm

z K

Vy=300 N Vz=400 N Mt=90 Nm

m

D P1=400 N

H

0m

K

30

y

P2=300 N

z Section at S

(a)

H Mz=60 Nm

x

z (b)

Fig. 13.13 (a) Structural member and (b) internal generalized forces on the cross-section at location S.

Solution: In this statically determinate case, we need not determine the reaction components at location A. The internal generalized forces are directly determined on the cross-section at location S as (Fig. 13.13(b)): Vy = 300 N;

Vz = −400 N

Mt = −300 mm × 300 N = −90,000 Nmm = −90 Nm My = (500 − 300) mm × 400 N = 80,000 Nmm = 80 Nm Mz = (500 − 300) mm × 300 N = 60,000 Nmm = 60 Nm Before we start determining the stress components, we evaluate the area properties of the cross-section. π × (28)2 πd2 = = 615.8 mm2 4 4 π × (28)4 πd4 = = 30.17 × 103 mm4 Moment of inertia I = 64 64 π × (28)4 πd4 = = 60.34 × 103 mm4 Polar moment of inertia J = 32 32 Area A =

We take up subproblem 1 with Vy = 300 N (Fig. 13.14(a)). This shear force does not develop shear stress at the top point K.

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y (a) H

z H

Q x

z

y

Hl Vy=300 N

K

K

Hl

x

0.65

z

Point H

x

y

y

H

H

z

4r 3

C

y (b)

y

K

K

z

Q Vz= 400 N

H

C Q

z

4r 3

0.87 Point K

x

Fig. 13.14 (a) The first subproblem with Vy = 300 N and (b) the second subproblem with Vz = −400 N.

The shear stress at point H is determined using τxy = V Q/bI where b = 28 mm. Q is the first moment of the area above HH , a semicircle (Fig. 13.14(a)). It is determined by multiplying the area with the distance of its centroid C from the neutral axis HH . The distance of C is taken from a table of properties of an area and is equal to 4r/3π. Thus,  Q=

πr 2 2



 ×

4r 3π



2r 3 2 = = 3 3



28 2

3

= 1.829 × 103 mm3

The shear stress at point H is τxy =

(300) × (1.829 × 103 ) N·mm3 VQ = = 0.65 MPa bI 28 × (30.17 × 103 ) mm·mm4

The corresponding stress tensors at points H and K are as follows: ⎡

(1)

σH

0 ⎢ = ⎣0.65 0

0.65 0 0

⎤ 0 ⎥ 0⎦MPa; 0

(1)

σK

⎡ 0 ⎢ = ⎣0 0

⎤ 0 0 ⎥ 0 0⎦MPa 0 0

The second subproblem with Vz = −400 N is shown in Fig. 13.14(b). This shear force does not develop any shear stress at point H. The analysis to determine τxz at point K is similar to that of

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subproblem 2. The value of Q of the left portion of the section (Fig. 13.14(b)) is 1.829 × 103 mm3 . The shear stress then becomes τxz =

−400 × (1.829 × 103 ) N·mm3 VQ = = −0.87 MPa bI 28 × (30.17 × 103 ) mm·mm4

The stress tensors are as follows: ⎡ ⎤ ⎡ 0 0 0 0 ⎢ ⎥ ⎢ (2) (2) σH = ⎣0 0 0⎦MP; σK = ⎣ 0 0 0 0 −0.87

⎤ −0.87 ⎥ 0 ⎦MPa 0

0 0 0

The third subproblem is shown in Fig. 13.15(a) for torsional moment Mt = −90 Nm. The shear stress on all circumferential points is (−90 × 103 ) × 28 Mt r 2 = = −20.88 MPa τrθ = J 60.34 × 103 On x-face, it develops τxy = 20.88 MPa at point H (Fig. 13.15(b)) and τxz = −20.88 MPa at point K (Fig. 13.15(c)). The tensorial forms of stress are as follows: ⎡ ⎤ ⎡ ⎤ 0 20.88 0 0 0 −20.88 ⎢ ⎥ ⎢ ⎥ (3) (3) 0 0⎦MPa; σK = ⎣ 0 0 0 ⎦MPa σH = ⎣20.88 0 0 0 −20.88 0 0 y

y

K

z

H

z

Q

x Mt= 90 N (a)

K

H

Section at D

y

y

20.88

20.88

x

z Point H

(b)

z

x Point K (c)

Fig. 13.15 (a) The third subproblem with torsional moment, (b) stress element at point H, and (c) stress elements at point K.

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y

y K

H

z

Q

H My=80 N

37.12

z

x

x

(a)

Point H

y K

y H MZ=60 N

H

z

Q

x

27.84

z (b)

xx

x

Point K

Fig. 13.16 (a) Fourth subproblem with My = 80 Nm and (b) fifth subproblem with Mz = 60 Nm.

The fourth subproblem, shown in Fig. 13.16(a), deals with bending moment My = 80 Nm. The neutral axis coincides with y-axis and, therefore, the bending stress at point K is zero. At point H, σxx is determined as (80 × 103 ) × 28 My z (Nmm) mm 2 = = 37.12 MPa σxx = I 30.17 × 103 mm4 The stress tensors of ⎡ 37.12 (4) σH = ⎣ 0 0

subproblem 4 are as follows: ⎤ ⎡ 0 0 0 0 (4) 0 0⎦ MPa; σK = ⎣0 0 0 0 0 0

⎤ 0 0⎦ MPa 0

The fifth subproblem with Mz = 60 Nm is shown in Fig. 13.16(b). In this case, point H lies on the neutral axis HH and, therefore, no stress is developed at point H. At point K, the bending stress developed is (60 × 103 ) × 28 Mz y 2 (Nmm)mm =− = −27.84 MPa σxx = − 3 I 30.17 × 10 mm4

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The stress tensors ⎡ 0 ⎢ (5) σH = ⎣0 0

are as follows: ⎤ ⎡ 0 0 −27.84 ⎥ ⎢ (5) 0 0⎦MPa; σK = ⎣ 0 0 0 0

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435

⎤ 0 0 ⎥ 0 0⎦MPa 0 0

The net stress is the sum of all the five matrices for both points are as follows: ⎡ ⎤ ⎡ ⎤ 37.12 (0.65 + 20.88) 0 37.12 21.53 0 ⎢ ⎥ ⎢ ⎥ 0 0⎦ = ⎣21.53 0 0⎦MPa σH = ⎣(0.65 + 20.88) 0 0 0 0 0 0 ⎡ ⎤ −27.84 0 (−0.87 − 20.88) ⎢ ⎥ 0 0 0 σK = ⎣ ⎦ (−0.87 − 20.88) 0 0 ⎡ ⎤ −27.84 0 −21.75 ⎢ ⎥ 0 0 ⎦MPa =⎣ 0 −21.75 0 0 Interpretation/Comments: In this problem, the combined stress is the superposition of five subproblems involving two kinds of shear forces, one torsional moment and two bending moments. Also, note that in a slender member of this type, the magnitude of shear stress developed by shear forces is usually much smaller than the magnitude of stress developed by the bending moments and the torsional moment. Thus, moments, bending or torsional, develop high stresses in slender members in comparison to stresses developed by shear forces. 13.3

Why Determine Stresses within a Structural Member

Why do we find stress in the internal points of a structural member? How do we use it? Knowing stress and strain fields within a structural member is like earning money. If you have money in your pocket, you can buy things such as a car, groceries, ice cream, and

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movie tickets. Similarly, the field of mechanics of materials is very vast and we can study various different specializations like yielding of a material, fracture mechanics, fatigue failure, buckling, creep, wear, impact damage, stress waves, vibration, friction, and dislocations. Each one of them is analysed with stresses or strains. We shall consider only two of them in this book: • Failure through yielding. • Failure through deflection of a structure member beyond the specifications. Other topics are beyond the scope of this book. Some of them are taught in other courses of undergraduate degree programs. Advanced topics like fracture mechanics are part of postgraduate studies. 13.4

Critical Point and Critical Stress

We have already realized that the developed stress within the member varies from one point to another point except in some special cases like uniform stress distribution in a member with an axial load. In all other three generalized forces, bending moment, shear force, and torsional moment, internal stresses developed vary from one point of the cross-section to another. In Chapter 8, we studied material properties for a uniaxial stress case and found that beyond the yield stress, the strain increases substantially even with a small increase in the stress. For most ductile materials, a structural member does not break into two pieces, when the stress exceeds yield stress. However, most design engineers do not like members to have large deformations. Therefore, they regard the yield stress as the upper limit for designing a member. Thus, the designer ensures that none of the points of a member yields. The location at which the magnitude of the stress is the highest is determined. Some people like to call it the worst loaded point, but it is commonly known as the critical point. If the stress at the critical point is less than the yield stress, the entire member is safe. We therefore determine the location of the critical point and the stress at the critical point. For some cases, it is a simple matter of identifying the location of the critical point. For some problems, it may be difficult to determine the critical point right away. We usually start by guessing the likely

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critical points and then we find stresses at all the guessed points to identify the critical point. We will now elaborate on this point with an example. Example 13.5: A slender member AB of length 1,000 mm is simply supported at its ends. A transverse force P = 5,000 N acts at the midspan, as shown in Fig. 13.17(a). Consider two kinds of cross-section of the member, rectangular shown in Fig. 13.17(b) and T-section shown in Fig. 13.17(c). Determine the critical point/points and the stress at the critical point in each member. Given: Span length L = 1,000 mm; external force P = 5,000 N at x = 500 mm; two kinds of cross-sections, rectangular and T . To find: Location of critical point; stress at the critical point. Strategy: Find the location of the maximum bending moment; guess critical points and determine stress at all the guessed points. Solution: In this simple case, the support reactions are VA = 2,500 N and VB = 2,500 N. The maximum bending moment is under the load at point C and it is equal to Mb =

5,000 × 1,000 PL = N·mm = 1.25 × 106 Nmm 4 4

The magnitude of the bending stress is maximum at the top or at the bottom points of the section where the shear stress is zero. The shear stress is maximum at the points of the neutral axis where the bending stress is zero. Since the bending stresses are much higher than shear stresses for this slender member, the top or bottom points are likely to be critical points. 30

5,000 N

C1 C 1

VA

C

x 500

1,000 mm

(a)

B

50 mm

A

50 mm

y

VB

C2

s

10

s

10

10

(b)

(c)

Fig. 13.17 (a) A simply supported member, (b) rectangular cross-section, and (c) T-section.

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Case (i): The moment of inertia for the rectangular section is (Fig. 13.17(b)) 10 × (50)3 bh3 = = 104.2 × 103 mm4 12 12 In this symmetric rectangular section, the magnitude of compressive bending stress at the top points is the same as that of the tensile stress at the bottom points and, therefore, it is a simple matter of locating critical points; all points of the top edge and all points of the bottom edge of the section at C are critical points. The magnitude of the critical stress is    Mb y  (1.25 × 106 ) × 25 Nmm·mm = = 299.9 MPa σxx = − I  104.2 × 103 mm4 I=

Case (ii): The T-section is a symmetric member as it is symmetric about x–y plane and therefore the bending formula can be invoked. Since this T-section is not symmetric about z-axis, the magnitude of the bending stress at the top points may be different from that at the bottom points. We will determine the magnitude at both locations to identify the critical points. The neutral axis is evaluated with respect to the bottom edge of the section and is given by (Fig. 13.17(c)) s¯ =

(30 × 10) × 45 + (10 × 40) × 20 mm2 ·mm A1 s1 + A2 s2 = A1 + A2 (30 × 10) + (10 × 40) mm2

= 30.71 mm We will now find the moment of inertia. We divide the section into two parts: 30 × 10 rectangle and 10 × 40 rectangle and invoke the parallel axis theorem. Thus,    10 × 403 30 × 103 + 30 × 10 × (45 − 30.71)2 + + 10 × 40 I= 12 12  ×(30.71 − 20)2 = 163.0 × 103 mm4 The bending stress at the top point is t =− σxx

(1.25 × 106 ) × (50 − 30.71) Nmm·mm Mb y =− I 163.0 × 103 mm4 mm4

= −147.9 MPa

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The bending stress at the bottom points is b σxx



1.25 × 106 × (−30.71) Nmm·mm =− = 235.5 MPa 163.0 × 103 mm4 mm4

Thus, the critical points are located at the bottom surface. Interpretation/Comments: In certain cases, we can identify the critical point/points easily without carrying out calculations. In other cases, we guess several likely critical points and then determine stress at those points to identify the critical point. If the yield stress is more than 299.9 MPa, the member of the first case is not yielded at any point. Similarly, if the yield stress is more than 235.5 MPa in the second case, the member is not yielded and all points of the beam are deformed elastically. It is worth noting here that, in the example, only σxx was the nonzero component at the critical points and thus the loading belongs to the case of the uniaxial stress loading. We, therefore, could use the results of uniaxial tension test. However, the more general case of two or more nonzero stress components is tackled with a model. In fact, the model defines an equivalent parameter which accounts for the influence of each nonzero stress component. We then compare the value of this parameter with the maximum allowable limit. 13.5

An Equivalent Parameter

Consider a biaxial problem of a thin plate shown in Fig. 13.18, which is subjected to two nonzero stress components, σxx and σyy . yy

y

x

Fig. 13.18

xx

A plate subjected to two nonzero stress components.

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The stress tensor for this case is ⎡ σxx ⎢ σ=⎣ 0 0

0 σyy 0

⎤ 0 ⎥ 0⎦ 0

In this case, the stress field is uniform, that is, all points are subjected to the same stress and thus all points are critical. Now, our task is to find out which combinations of σxx and σyy yield the plate. There can be an infinite number of combinations. How do we resolve it? The task of a general case with all six nonzero stress components looks still more difficult. We look for an equivalent parameter which can account for all six components of the stress tensor. A model solves this riddle. The model converges the influence of all six parameters into one equivalent parameter. Once we have the single equivalent parameter, we can compare it with the maximum allowable limit. It leads to an equality relation. If the value of this equivalent parameter is less than the critical value, the material does not yield. If it is equal to or greater, the material yields. This equality relation is called the yield criterion. A yield criterion is a model. If it predicts values within a few percent of the actual value and is simple enough to manage mathematically, it is a good model. Since there are many different kinds of materials, no single parameter is appropriate for all materials. For example, there are different failure models for materials like metals, wood, polymers, foams, biomaterials, composite materials, nanomaterials, and amorphous metals. With the development of new kinds of materials, suitable failure models are sought. Before we start discussing them, we will look into the ways of finding principal stresses σ1 , σ2 , and σ3 because it is easier to deal with three parameters rather than work with six parameters. 13.6

Determination of Principal Stresses

We have already discussed how to find principal stresses, σ1 , σ2 , and σ3 , in Chapter 5. But we will review for the sake of convenience. Let us consider a case of a thin plate with x- and y-axes lying on the plane and z-axis normal to the plane of the plate. At all points of the top and bottom surfaces, the three stress components having

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Free surface zz=0 xz=0 yz=0

z

z

Principal direction

Q y

x

xx

yy

x xy

(a)

y

(b)

Fig. 13.19 (a) A thin sheet with top and bottom surfaces as free surfaces and (b) stress element at surface point Q.

z as one of the subscripts is zero, that is, σzz = 0, σxz = 0, and σyz = 0 (Fig. 13.19(a)). These three components may not be zero in the interior points but their values are quite small for thin plates and thus we assume them to be negligible at all points of the plate. Such plates are known as the case of plane stress. Then the stress tensor becomes (Fig. 13.19(b)) ⎤ ⎡ σxx τxy 0 ⎥ ⎢ σ = ⎣ τxy σyy 0⎦ 0 0 0 It is worth noting here that one of the principal stresses of plane stress problems is normal to the surface because σxz = 0 and σyz = 0. However, the magnitude of this principal stress is zero for plane stress problems. The other two principal stresses can easily be obtained through a Mohr circle. In these 2D cases, I personally like to draw a Mohr circle to identify the other two principal stresses. Alternatively, you can use the following relations for this problem:   σxx + σyy σxx − σyy 2 2 + + τxy (13.1a) σ1 = 2 2   σxx + σyy σxx − σyy 2 2 − + τxy (13.1b) σ2 = 2 2 In a general 3D case, the principal stresses can be easily determined as they are the eigenvalues of the stress matrix. Thus, the principal

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stresses are determined by solving the determinant for σ:   (σ − σ) τxy τxz   xx   (σyy − σ) τyz  = 0  τxy    τxz τyz (σzz − σ) Example 13.6: Draw the stress element and determine the principal stresses of the stress tensor: ⎡ ⎤ 50 30 20 ⎢ ⎥ σ = ⎣30 60 10⎦MPa 20 10 20 Given: Stress tensor with six nonzero stress components. To find: Draw stress element and determine σ1 , σ2 , and σ3. Strategy: Through the determinant approach, determine the eigenvalues which are the same as principal stresses. Solution: Figure 13.20(a) shows the stress element. The principal stresses are obtained by solving the determinant:   50 − σ  30 20     60 − σ 10  = 0  30    20 10 20 − σ  The expanded equation is σ 3 − 130σ 2 + 3,800σ − 25,000 = 0 y yy= xz=

60 MPa xy=

20

30

91.4

9.4 29.2

zz=

xx=

20

z

50

x yz= 10

(a)

Fig. 13.20

(b)

(a) The stress element and (b) the resulting Mohr circles.

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The roots of this equation are determined using a scientific calculator as follows: σ1 = 91.4 MPa σ2 = 29.2 MPa σ3 = 9.4 MPa All the three principal stresses are shown in Fig. 13.20(b). From these principal values, all three Mohr circles are drawn, as shown in the figure. Interpretation/Comments: The determinant approach is powerful to evaluate the principal stresses and to draw all three Mohr circles. We will now discuss the following two yield criteria: • von Mises yield criterion, • Tresca yield criterion.

13.7

von Mises Yield Criterion

The equivalent stress σm , known as Mises stress, is defined as  1 [(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 ] (13.2) σm = 2 In this expression, the equivalent parameter σm of the model is determined from the three principal stresses. The von Mises criterion states that if σm is less than the yield stress σys the material is elastic (not yielded) at the point under consideration. If it is equal to or larger, the material yields. Thus, σm ≥ σys : the material yields

(13.3a)

σm < σys : the material does not yield

(13.3b)

Example 13.7: Determine whether the stress of Problem 13.6 yields as per the prediction of von Mises criterion, if σys = 80 MPa. Given: σ1 = 91.4 MPa; σ2 = 29.2 MPa; σys = 80 MPa.

σ3 = 9.4 MPa;

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To find: Whether the material yields or not. Strategy: Determine σm and compare with σys. Solution: For this case, von Mises stress is  1 [(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 ] σm = 2  1 [(91.4 − 29.2)2 + (29.2 − 9.4)2 + (9.4 − 91.4)2 ] = 2 = 74.1 MPa < 80 MPa Since σm is smaller than σys , the material is not yielded at the point where the stress tensor is given. Interpretation/Comments: von Mises criterion is very widely used for ductile metals. It is quite convenient to use as there is only one equation to determine the equivalent parameter σm . In an analysis through a computer package, σm is found out for each and every point of a structural member. The different levels of σm are shown with different colors. In most cases, the red color shows that the portion of the member is subjected to the highest level of σm and green color the lowest level. This facilitates the determination of the critical point and the critical stress. 13.8

Tresca Yield Criterion

The equivalent parameter of the Tresca yield criterion is the maximum shear stress τmax . It is convenient to first evaluate all the three principal stresses and then draw all the three Mohr circles, as shown max

9.4 1

Fig. 13.21

29.2 2

o

91.4 3

All the three Mohr circles and maximum shear stress.

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in Fig. 13.21. It is a common practice to call the maximum principal stress as σ1 and minimum one as σ3 . The maximum shear stress τmax is the radius of the largest Mohr circle, that is, τmax = (σ1 − σ3 )/2. The limiting value of τmax is σys /2. Thus, σys : the material yields 2 σys : the material does not yield < 2

If τmax ≥

(13.4a)

If τmax

(13.4b)

Example 13.8: Determine whether the stress of Example 13.6 yields the material if the yield stress is 80 GPa and Tresca yield criterion is employed. Given: σ1 = 91.4 MPa; σ2 = 29.2 MPa; σ3 = 9.4 MPa; 80 MPa.

σys =

To find: Apply Tresca yield criterion to know whether the material yields. Strategy: Draw all three Mohr circles, determine τmax from the largest Mohr circle, and compare it with σys /2. Solution: Figure 13.21 shows all the three Mohr circles. The maximum shear stress is τmax =

80 91.4 − 9.4 = 41 > MPa 2 2

The material yields at the point under consideration. Interpretation/Comments: Note that in the example considered, von Mises yield criterion predicts that material is not yielded (Example 13.7), while the material yields as per Tresca yield criterion. This kind of inconsistency exists due to modelling. A model is not very factual; it approximately predicts values which are close to the actual values. In both cases, the equivalent parameter is close to the allowable limit; in von Mises criterion, σm is marginally smaller than the allowable limit, while τmax of Tresca yield criterion is marginally higher than the maximum allowable limit. How does a designer decide which one to apply? It also depends on the material. Some designers apply the von Mises criterion for ductile metals and the Tresca criterion for materials on the brittle

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side. I personally like to use the Tresca criterion for both kinds of materials because it is slightly on the conservative side. Further, it is convenient to use after I find all the three principal stresses. I just draw all the three Mohr circles and determine the maximum shear stress by finding the radius of the largest circle. We will consider one more example of a 2D case to emphasize the importance of the principal stress in the third direction. Example 13.9: Determine, using the Tresca yield criterion, whether the material is yielded by the stress: ⎡ ⎤ 60 40 0 ⎢ ⎥ σ = ⎣40 120 0⎦MPa 0 0 0 The yield stress of the material is 120 MPa. Given: Three nonzero stress components in x−y plane; σys = 120 MPa. To find: Whether the material yields. Strategy: Determine σ1 , σ2 , and σ3 , draw Mohr circles, find τmax and compare it with σys /2. Solution: Figure 13.22(a) shows the stress element. Since there is no shear stress on z-face, one of the principal stresses is the normal stress on z-face, making σ3 = 0. To determine σ1 and σ2 , we draw a Mohr circle shown in Fig. 13.22(b). The centre of the circle is σo =

60 + 120 σxx + σyy = = 90 MPa 2 2

The radius of the circle is     σxx − σyy 2 60 − 120 2 2 + τxy = + 402 = 50 MPa R= 2 2 Thus, principal stresses are as follows: σ1 = 90 + 50 = 140 MPa σ2 = 90 − 50 = 40 MPa

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y

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447

max

yy=120

MPa xy=40 2

0

xx=60

z

1

40

3

x 60 (a)

90 120 MPa (b)

Fig. 13.22

(a) Stress element and (b) the Mohr circle to determine σ1 and σ2 .

Figure 13.22(b) shows all the three Mohr circles. The maximum shear stress is τmax =

120 140 − 0 = 70 MPa > MPa 2 2

The material yields. Interpretation/Comments: It is worth noting here that σ3 = 0 plays a crucial role in finding τmax . In case we do not account for σ3 = 0, find maximum shear stress as 2D = τmax

140 − 40 = 50 MPa 2

It is less than σys /2 and it predicts that the material does not yield. This may lead to unsafe designs. Thus, it is important to determine all three principal stresses before we start applying a yield criterion or any other failure criterion. 13.9

Other Failure Criteria

I stated in the introduction of this chapter that a structural member may fail in many different ways. Failure through yielding is one of the ways, but it is an important one. Deformation of the member beyond a specified limit is also a failure and we will discuss it in an

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introductory way in the following section. Other ways of failure are fracture, fatigue, vibration, buckling, creep, wear, etc. A structural member is not likely to fail from all of them. I like to compare it with the health of a person who is vulnerable to one or two diseases among diseases like blood pressure, diabetes, arthritis, gastritis, and tuberculosis. He is treated accordingly for the critical one or two diseases. Similarly, the structural member is likely to fail owing to one or two of the many ways. We are not in a position to study some additional failure mechanisms as they are beyond the scope of this book.

13.10

Deformation beyond a Specified Limit

All real-life materials deform but the deformation of a member should not exceed the allowable or the specified limit. A suspension footbridge was constructed over river Ganges in a hilly area. It was not stiff enough and would swing sideways more than a meter or so with wind gusts. The pedestrians were afraid to cross it. It was an unsatisfactory design until it was reinforced to restrict the sideways swing. It is worth noting here that the stresses within the load-bearing suspension cables were much less than the yield stress and the bridge would not fail due to yielding. In earlier times, say before 1930 or so, materials were not welldeveloped and many structures failed due to yielding. However, now the modern materials have been developed with high yield stress and high quality with reliable results. The failure through yielding is no longer as prominent a failure mechanism as it used to be. In fact, some designs are deflection based. One of my friends was involved with the design of a passenger compartment of Indian Railways. The prescribed limit on the deflection of the critical point, which was at the mid-length of the compartment, was 18 mm. It was a stiffnessbased design. Whenever I design a structural member, I calculate the deflection also. If it is excessive, I modify the design. Example 13.10: A L-shaped slender member ABC is made of a round steel tube of 25 mm outside diameter and 1.2 mm wall thickness. It is fixed rigidly at end A and a downward force of P = 850 N is applied at point C, as shown in Fig. 13.23(a). The maximum

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y y 200

A

x

mm 200

A

z

x

B

mm

P=850 N B

z P=850 N 0

15

C

127.5 Nm

(a)

(b)

Fig. 13.23

Example 13.10.

downward deflection should not exceed 8 mm. Determine whether the design is acceptable (modulus: 207 GPa; Poisson’s ratio ν = 0.3). Given: L-shaped steel tube: outside diameter Do = 25 mm, internal diameter Di = 22.6 mm; AB = 200 mm; BC = 150 mm; P = 850 N downward; E = 207 GPa; ν = 0.3. To find: vC . Strategy: We split the problem into two subproblems, Segment AB and Segment BC. (1) Segment AB contributes through the two following kinds of deflection: (i) point B goes down due to bending of AB which makes the point C move down by the same displacement; (ii) the end B rotates due to torsional moment acting at location B which in turn rotates segment BC. (2) Segment BC contributes through the bending moment acting on it. The combination of all the three provides the overall displacement vC . Solution: We first consider Segment AB. Figure 13.23(b) shows the Segment AB with forces acting at point B as Mt = 150 × 850 = 127.5 × 103 Nmm = 127.5 Nm V = −850 N

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450

Since there are two generalized forces, we consider the effect of each one. Segment AB twists due to Mt ; the angle of twist of Segment AB is φB =

Mt L GJ

207 E = = 79.6 GPa and 2(1 + ν) 2(1 + 0.3) π(254 − 22.64 ) π(Do4 − Di4 ) = = Polar moment of inertia J = 32 32 12.74 × 103 mm4 Thus, The shear modulus G =

(1)

φB =

Nmm·mm (127.5 × 103 ) × 200   N (79.6 × 103 ) × (12.74 × 103 ) ·mm4 mm2

= 25.14 × 10−3 Radian Even a small rotation of end B causes a large deflection at point C as it is multiplied with length BC. We like to call it displacement leverage. Thus, point C moves down by

(1) vC = −25.14 × 10−3 × 150 mm = −3.77 mm Segment AB is subjected to transverse force of 850 N acting at point B in downward direction. It is a cantilever problem and the deflection of point B is given by (2)

vB =

P L3 3EI

The moment of inertia I = J/2 = 6.37 × 103 mm4 (2)

N·mm3 (−850) × 2003   N 3 × (207 × 103 ) × (6.37 × 103 ) ·mm4 mm2 = −1.72 mm

Thus, vB =

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Because of the displacement of point B, the entire BC potion moves down, making point C go down by the same displacement, giving (2)

vC = −1.72 mm Now, we focus on portion BC. For its analysis, portion BC is treated as a cantilever with its end fixed at B. Then, its contribution to displacement is (BC)

vC

=

(−850) × 1503 N·mm3 P L3   = N 3EI 3 × (207 × 103 ) × (6.37 × 103 ) ·mm4 mm2

= −0.73 mm Invoking the principle of superposition, we obtain the net displacement of point C as (1)

(2)

(BC)

vc = vC + vC + vC

= −3.77 − 1.72 − 0.73 = −6.22 mm The magnitude of the displacement of point C is less than the prescribed deflection, 8 mm, and thus the design is acceptable as far as the deflection is concerned. Interpretation/Comments: The procedure is similar to the one we adopt to determine the combined stress at a point. In the displacement analysis also, a problem is split into subproblems and each problem is then solved for deflection. However, we should keep track of the displacement leverage, if any. In this problem, the angle of twist of portion AB is multiplied by the length BC, resulting in the largest contribution to the net displacement at point C. In certain cases, this displacement leverage can be very large. Once I designed a frame of a tricycle and there was a small segment which was subjected to torsion. When loaded, the frame moved down more than 100 mm which was not acceptable. After diagnosis, we removed the torsional portion and deflection was reduced to a few millimetres. Further, we have not accounted for displacement caused by shear

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forces acting on both the segments because the effect of shear force is much smaller for slender members and we usually neglect them. We will discuss one more example to illustrate that there can be a large displacement leverage due to rotation caused by a bending moment. Example 13.11: Consider an L-shaped member ABC, made of a square tube of 50 mm × 50 mm outside dimensions and 2 mm wall thickness, is rigidly fixed at end A, as shown in Fig. 13.24(a). A force P = 2,500 N is applied at point C. Determine the displacement of point C in x- and y-directions (E = 207 GPa, Poisson’s ratio ν = 0.3). Given: L-shaped member; AB = 200 mm; BC = 800 mm, P = 2,500; square tube with cross-section of 50 mm×50 mm external dimensions and 2 mm wall thickness; E = 207 GPa; ν = 0.3. To find: Deflection of point C: (i) uC in x-direction and (ii) vC in y-direction. Strategy: Consider portions AB and BC separately; split each case into subproblems; account for displacement leverage while combining the displacements. Solution: Considering portion AB, there are two generalized forces working at point B on portion AB, as shown in Fig. 13.24(b). This is further split into two subproblems, as shown in the figure. C

800 mm

2

P=2,500 N

A

y A

y

x

1 200

x

A B

B

B

A

z

1a

1b

B

(b)

(a)

Fig. 13.24 (a) The member of Example 13.11 and (b) the generalized forces acting on the section at location B and the subproblems.

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453

In the first subproblem with the axial force of 2,500 N, the displacement of point B in x-direction is (2,500) × 200 PL = AE (502 − 462 ) × (207 × 103 )

N·mm = 0.00629 mm N mm2 · mm2 The entire BC portion also moves by the same distance. Thus, its contribution to the x-displacement of point C is (1)

uB =

(1)

uC = 0.00629 mm The second subproblem deals with bending moment Mb = 800 × 2,500 Nmm. Using Figure 11.20(d) of Chapter 11, the displacement of point B in y-direction is (2)

vB = − where I = Thus, (2)

Mb L2 2EI

504 − 464 = 147.7 × 103 mm4 12

Nmm·mm2 (800 × 2,500) × 2002   N 2 × (207 × 103 ) × (147.7 × 103 ) × mm4 mm2 = −1.308 mm

vB = −

The entire portion BC also moves down by this displacement and thus its contribution to the displacement of point C is (2)

vC = −1.308 mm Also, it is important to realize that the end B of Segment AB rotates by angle θB due to the bending moment. This rotation, in turn, rotates the entire segment BC by the same angle, resulting in xdisplacement of point C. Referring to Fig. 11.20(d) of Chapter 11, we obtain the local rotation at point B as θB =

(800 × 2,500) × 200 Nmm·mm Mb L   = 3 3 N EI (207 × 10 ) × (147.7 × 10 ) × mm4 mm2

= 13.08 × 10−3 Radian = 0.750◦

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This rotation causes displacement leverage and contributes to x-displacement of point C as uθC = θB × BC = (13.08 × 10−3 ) × 800 = 10.46 mm Considering portion BC, it works like a cantilever with its end B treated as fixed support. Thus, it contributes to the x-deflection of point C as uBC C =

P L3 (2,500) × 8003 N·mm3   = N 3EI 3 × (207 × 103 ) × (147.7 × 103 ) × mm4 mm2

= 13.96 mm The net displacement of point C in x-direction is (1)

θ BC unet C = uC + uC + uC = 0.0063 + 10.46 + 13.96 = 24.43 mm

The net displacement of point C in y-direction is (2)

net = vC = −1.308 mm vC

Interpretation/Comments: The displacement due to axial force is negligible in this problem. In a large number of problems of this field, deflection due to bending moments and torsional moments is much larger than that of axial force. The local slope of member AB at point B is not large, only 0.75◦ , but it causes a large deflection of 10.46 mm at point C because the long portion BC acts as a displacement leverage. 13.11

Summary

To determine the combined stress at a point, we section a member. Then, all the nonzero generalized forces, acting on the section, are identified and determined. The problem is split into subproblems, each one having only one generalized force acting on its section. The subproblems are solved separately, one at a time, using the analysis presented in Chapters 9–12. In each case, the determined stress at

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455

the point is expressed with a tensor employing a 3 × 3 matrix. The principle of superposition is invoked to obtain the combined stress at the point by summing all the stress matrices of the subproblems. To determine whether the combined stress yields the material, several yield criteria are available. An equivalent parameter is defined for a yield criterion of the commonly used isotropic materials like metals and polymers. The parameter accounts for the influence of all nonzero stress components. If the equivalent parameter is equal to greater than the maximum allowable limit, the material yields. The limit is usually based on the yield stress σys of the material. It is convenient to first determine the principal stresses, σ1 , σ2 , and σ3 . Two yield criteria, von Mises yield criterion and Tresca yield criterion, are widely used for commonly used isotropic materials. The equivalent parameter for von Mises criterion is Mises stress  1 [(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 ]. If σm ≥ σys , the σm = 2 material yields at the point. The equivalent parameter for the Tresca yield criterion is the maximum shear stress τmax . If τmax ≥ σys /2, the material yields. In deflection-based design, the deflection at the point where it is maximum is compared with the allowable deflection. The deflection is also obtained by invoking the principle of superposition. 13.12

Problems

1. A slender bent structural member CAB with a rectangular crosssection (36 mm × 12 mm) is rigidly attached to a wall, as shown in Fig. 13.25(a). A force of 10,000 N is applied at point C in x-direction as shown. Determine the following at location D: (a) Internal generalized forces. (b) Bending stress at the top point H and the bottom point Q at location D and express it in terms of a 3 × 3 matrix. (c) Axial stress at points H and Q. (d) Shear stress at H and Q. (e) Combined stresses at H and Q. Express them with a 3 × 3 matrix and through a stress element. (f) Critical point/points in portion AB and the critical stress.

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456

36 mm

H y

z B

Q 12 Section at D

N

00

80

,0 10

A

C

y 0 20

D

x

mm

0

50

mm

100 mm Dia

p

T=900Nm

z

500 mm

(a)

Fig. 13.25

(b)

(a) Problem 1 and (b) Problem 2.

2. Within a machine, a circular tube of 100 mm outside diameter with closed end is pressurized with an air of pressure 2 MPa. The tube is made of steel with 2 mm wall thickness. The tube is also loaded with a torque of 900 Nm, as shown in Fig. 13.25(b). Determine the following: (i) stress at a point of the tube in the cylindrical coordinate system, (ii) principal stresses and maximum shear stress, and (iii) invoke Tresca criterion to explore whether the design of the tube is safe if maximum allowable uniaxial tensile stress is 100 MPa. 3. A T-shaped structural member is rigidly fixed at location B, as shown in Fig. 13.26. Three forces, 800, 6,000, and 800 N, are applied, as shown in the figure. Portion AB is made of a round tube with an outside diameter of 30 mm and a wall thickness of 3 mm. Determine the following: (a) Internal generalized forces at location D. (b) Bending stresses at points H and Q. (c) Stress due to axial force at points H and Q. (d) Combined stresses at points H and Q and show them through a stress element. (e) Critical points and critical stress in AB. 4. A structural member of size 320 mm × 40 mm × 20 mm is fixed rigidly to a machine at its bottom face, as shown in Fig. 13.27(a). A compressive force of 10,000 N is applied at point H as shown which develops the bending moment as well as axial compression

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Combined Stresses and Yield Criteria

800

457

B

N

y 0

200

500

x

20

mm

00 N

0

z 6,000 N

D

30 mm

20

y

A

8

H

x

Q

Section of D (b)

(a)

Fig. 13.26

x P =10,000 N 10

y

H

1.6 m

H

z

320 mm

320 mm

z

P2=10,000 N P1=5,000 N 10

1m

y

x

Problem 3.

0.1m 100 Kg 3.2 m

4 mm H

Q

x 40

20

(a)

Fig. 13.27

40

20

(b)

y

A

60 mm (c)

Section at A

(a) Problem 4, (b) Problem 5 and (c) Problem 6.

in the member. Determine the following: (i) the critical point and (ii) the stress at the critical point. 5. A structural member of size 320 × 40 × 20 is fixed rigidly to a machine at its bottom face, as shown in Fig. 13.27(b). Two compressive forces, 10,000 N and 5,000 N, are applied at the top face. Looking at the member, the critical point is expected

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9.

l

T

T

H

H

B

y

5,000 N

x A

10

N

E

z D 90

C

l

y

y

10

8.

page 458

at one of the four vertical edges. Determine the location and the stress tensor of the critical point (note that two kinds of bending moments, My and Mz and an axial force are developed). A signboard of Joy Pizza is of size 1.6 m × 1.0 m and is attached to a vertical pole, as shown in Fig. 13.27(c). It is designed for a maximum wind load of 0.25 kPa, acting normal to the face of the signboard. The cross-section of the cylindrical pole is shown in the figure. The internal torsional moment developed in the pole is uniform through the length of the pole, while the maximum bending moment is at location A. Determine the critical point and stress at the critical point. Determine principal stresses of the critical point of Q6 and invoke Tresca and von Mises failure criteria. The yield stress of the material is 200 MPa and a factor of safety of 2 is chosen, that is, the maximum allowable stress of a uniaxial tensile test is 100 MPa. Is the design safe? A beam AB of rectangular cross-section is fixed to a wall at location B, as shown in Fig. 13.28(a). At location A, a force of 5,000 N is applied at the centroid of the face as shown. Consider a cross-section at location D shown in Fig. 13.28(b). Determine stresses at points T, H, and N. Also, consider a section at location E (Fig. 13.28(c)) and find stress at points T , H , and N. The cross-section of a C-clamp is T-section, as shown in Fig. 13.29(a). If the clamp is designed for a maximum force P = 800 N, determine the stress at one of the two likely critical points, H and Q.

50

7.

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6.

9in x 6in

z

Nl

C

m

0m

20

15

(a)

Fig. 13.28

Problem 8.

At D

At E

(b)

(c)

z

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459

200 dia A

P=800 N

15

3

32 mm

H

y 2.5

80 mm

Q

C

y C 600 N x E z 1,000 N

x 1,500 N

Section at C

P=800 N

15

F

0 35

400 N

300 dia G

0m

m

B

20

0

(a)

Fig. 13.29

(b)

(a) Problem 9 and (b) Problem 10.

10. Power is transmitted from belt pulley drive at location C of shaft AB to another belt drive at location G, as shown in Fig. 13.29(b). The tension of the belt on one side of the pulley is different from the belt on the other, which in turn, applies torque to the shaft. The diameter of the shaft is 30 mm. The shaft is supported on bearings at locations E and F which do not resist rotational motion. Plot the bending moment diagram and the torsional moment diagram. Determine the critical point on the shaft and the stress tensor of the critical point. Neglect the stress developed by internal shear forces. 11. The yield stress of the material of a structural member is 150 MPa. The stress at the critical point of the member is ⎡ ⎤ 100 30 20 ⎢ ⎥ σ = ⎣ 30 50 40 ⎦MPa 20 40 −60 (a) Determine the cubic equation to obtain eigenvalues of the stress tensor. (b) Use a scientific calculator to determine all the three principal stresses. (c) Draw all the three Mohr circles. (d) Determine maximum shear stress and invoke the Tresca yield criterion to check whether the loading is safe.

B1948

Governing Asia

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Chapter 14

Energy Methods 14.1

Introduction

Powerful vehicles like cars, buses, locomotives, and ships were developed to go from one place to another. These machines transported us much faster than the earlier means like horse carts, bullock carts, and camels. However, we always try to achieve more. Thus, a much faster way, that is, air travel, was developed. We are now ambitious enough to wish to go from London to New York in less than an hour through space vehicles using rockets. Similarly, the development of stress and strain parameters and other associated relations was a big breakthrough in our field of mechanics of materials. Many bridges, complex machines, tunnels, high-rise buildings, etc., have been designed using the analysis. However, the analysis was limited either to simple cases or simplified approximate analyses because the rigorous analysis was difficult. One reason was that stress and strain are second-order symmetric tensors, each having six independent components making the analysis complex. In the last 70–80 years, energy methods were developed which are quite effective in solving the complex problems of the mechanics of materials. In fact, energy is a scalar and, therefore, it often makes it easier for us to solve complex problems. With the help of energy methods, several powerful computer packages have been developed that enable us to quickly analyze a proposed structure with high accuracy. Many sophisticated energy methods have been developed, but adequate background has not been provided in this book to study them in depth. In this chapter, we have restricted our discussion to methods dealing only with strain energy applied to linear elastic materials. 461

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462

When we do some work on a structure that deforms elastically, it is stored as strain energy. Using strain energy, we will explore how to find the deflection of a point employing a very handy theorem known as Castigliano’s theorem. The theorem also enables us to determine reaction components of an indeterminate structure. The energy approach is so powerful that some people go to the extent of saying that energy methods can be seen as an alternative analysis to the equilibrium approach. 14.2

Strain Energy through Axial Forces

The concepts of strain energy are explained in this section through a relatively simpler analysis of axial forces so that one can appreciate the effectiveness of the energy approach. In the following sections, we will consider general cases involving bending moments, shear forces, and torsional moments. It is worth noting here that strain energy in a member is usually determined using internal generalized forces. We first determine internally developed generalized forces in terms of external forces, and then the internal forces are employed to determine the strain energy. Of course, the eventual expression of the strain energy is in terms of the external forces. Consider a straight member shown in Fig. 14.1(a). We find internal force in the member taking a section at distance x as shown in Fig. 14.1(b). In this simple case, the internal axial force F turns out to be equal to P irrespective of the location. In fact, the axial force (a)

(d)

(b) (c)

Fig. 14.1 (a) A slender member, (b) internal axial force F, (c) axial force diagram, and (d) area under the curve is strain energy.

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463

diagram (AFD) is a constant line as F = P as shown in Fig. 14.1(c). Under the internal axial force F , the length of the member changes by Δu. The increase in strain energy ΔU becomes ΔU = F Δu

(14.1)

Integrating,  U=

F du

(14.2)

Thus, strain energy U is the area under the curve relating the linear relationship between F and u shown in Fig. 14.1(d) and it is given by 1 U = Fu 2

(14.3)

But, the displacement u is expressed as (Eq. (9.4)) u=

FL AE

(14.4)

where L is the length of the member, A is the area of cross-section, and E the modulus. Substituting u in Eq. (14.3), we obtain U=

F 2L 2AE

(14.5)

This relation can be expressed in integral form if F , A, and E vary with distance as  L 1 F 2 dx (14.6) U= 0 2 AE In this expression, the principle of superposition does not work because U does not depend on F linearly. Thus, we should determine internal generalized forces from all the external forces before we determine the strain energy. Further, the integral form of strain energy is general enough to account for variation in the crosssectional area or modulus along the length. Usually, the modulus does not vary in a member, but in certain cases it may vary. For example,

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(a)

(b)

Fig. 14.2

(a) Axial member with two segments and (b) axial force diagram.

two metals having different moduli can be brazed or fused together. These days, graded materials are being developed, in which the properties vary with distance. We will now discuss several solved examples to look into various facets of strain energy. Let us consider a case involving three forces shown in Fig. 14.2(a). The axial force diagram is shown in Fig. 14.2(b). This problem has two segments, AC and CB. Using Eq. (14.5), we obtain strain energy in segment 1 as 1 U1 = (3P )2 2

  9 L /AE = P 2 L/AE 2 4

and in segment 2 as 1 U2 = (5P )2 2

  25 L /AE = P 2 L/AE 2 4

So, the total strain energy becomes U = U1 + U2 =

9 2 25 17 2 P L/AE + P 2 L/AE = P L/AE 4 4 2

A question may be raised here: How is strain energy stored in the material? The distances between the neighboring atoms are altered. This altered state is not the natural state and the atoms try to come back to their original positions. Thus, the altered state is capable of doing work which is equal to the strain energy stored for a conservative system.

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What is a conservative system? If a portion of the member deforms plastically, heat is generated, resulting in an increase in temperature. Then, energy as heat is lost to the surroundings. The system is no longer conservative because a part of the energy, which has been provided to the member by the external forces, is consumed. Another example of a nonconservative system is that the member loses energy if sliding friction is involved during deformation. In a conservative system, the entire energy is recoverable. In this chapter, the analysis through strain energy is applicable only for conservative systems. Another interesting parameter is strain energy density (SED ), which is defined as strain energy per unit volume. Consider the prismatic member of Fig. 14.1(a) with cross-sectional area A and length L. All points of this member are subjected to the same state of stress. The strain energy density then becomes   1 P 2L  2 2 AE 1 P U = = SED = AL AL 2E A Noting that P/A is the axial normal stress σ, we obtain SED =

σ 2 1 = σ 2E 2

(14.7)

Several aspects of SED are worth discussing. First, its unit is stress as strain has no unit. Second, SED is a point function and it varies, in general, from one point to another. Third, its expression can be used for a general case of normal stress. For example, axial stress is developed when a member is subjected to a bending moment; the axial stress so developed varies from point to point. Thus, the expression of Eq. (14.7) can be used to determine the total strain energy of a bent member. How do we use strain energy to solve problems? We make use of the Castigliano theorem which is stated with the help of Fig. 14.3. In a general case, there are several external forces P1 , P2 , P3 , etc., acting on the member and there are several supports. According to the Castigliano theorem, the partial derivative of the strain energy with respect to an applied force is equal to the displacement at the point on which the force is applied and the displacement is in the

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Fig. 14.3

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Displacement of point C is determined using Castigliano’s theorem.

Fig. 14.4

Axial member of Example 14.1.

direction of the force. For example, the displacement δc of point C, where the external force P3 is applied, is given by δc =

∂U ∂P3

(14.8)

where δc is in the direction of force P3 . We will discuss the proof of Castigliano’s theorem later in this chapter. Let us solve a problem with two methods: (i) using a conventional force displacement equation and (ii) a strain energy approach using Castigliano’s theorem. Example 14.1: A slender member is made of two segments of diameters d1 and d2 and lengths L1 and L2 as shown in Fig. 14.4. An external axial force P is applied at location C. All the three supports are shown in the figure. Determine the displacement δc of point C if the modulus of the material is E. Given: Segment 1: diameter d1 , length L1 ; segment 2: diameter d2 , length L2 ; hinge support at A; roller support at B; roller support at C; external axial force = P applied at C; modulus E.

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To find: δc by (i) Method 1: conventional force displacement relation; (ii) Method 2: strain energy method Strategy: Invoke Eq. (14.4) for Method 1 and Eq. (14.8) for Method 2 Solution: Method 1: The displacement at point C is given by δc = δ1 + δ2

(14.9)

The internal force F (=P ) is the same in both segments. Equation (14.4) gives δ1 = Thus, δc = δ1+ δ2 =

P E



L1 A1

P L1 ; A1 E  L2 +A 2

δ2 =

P L2 A2 E

Method 2: The strain energies in the two segments are U1 =

1 2

U2 =

1 2



L

P 2 L1 F 2 dx = 2AE 2A1 E

L

P 2 L2 F 2 dx = 2AE 2A2 E

o

 o

The overall strain energy becomes P2 U = U1 + U2 = 2E



L1 L2 + A1 A2



The Castigliano theorem (Eq. (14.8)) yields   P L1 ∂U L2 = + δc = ∂P E A1 A2 We thus obtain the same results from both the methods. Interpretation/Comments: The energy method is quite effective for the problems with many segments and other features.

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Fig. 14.5

Axial member of Example 14.2.

Example 14.2: Consider an axial member BCD shown in Fig. 14.5. The member is made of a graded material with modulus varying as E = Eo eax where Eo and a are known constants. An axial force P1 acts at location C and another axial force P2 at location D. Both supports are shown in the figure. Determine the displacement of points C and D if the area of cross-section is A. Given: Segment 1: length = L/2, area cross-section = A; Segment 2: length = L/2, area of cross-section = A; modulus of material E = Eo eax ; axial force P1 acts at location C; axial force P2 acts at location D; hinge support at B; roller support at D To find: (i) δC and (ii) δD Strategy: Determine strain energy in each segment to obtain total strain energy; differentiate it with respect to P1 to determine δc Solution: In segment 1: The internal axial force F = P1 + P2 . U1 then becomes 



(P1 + P2 )2  −ax L/2 (P1 + P2 )2 dx e = 0 2AEo eax 2AEo (−a) 0  (P1 + P2 )2  1 − e−aL/2 = 2AEo a

U1 =

F 2 dx = 2AE

L/2

In segment 2: The internal axial force F = P2 and thus  U2 = =

F 2 dx = 2AE



L L/2

(P2 )2  −ax L (P2 )2 dx e = L/2 2AEo eax 2AEo (−a)

 (P2 )2  −aL/2 e − e−aL 2AEo a

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The total strain is   (P1 + P2 )2  (P2 )2  −aL/2 −aL/2 −aL 1−e e −e + U = U1 + U2 = 2AEo a 2AEo a The displacement of location C becomes δC = ∂U/∂P1 =

 (P1 + P2 )  1 − e−aL/2 AEo a

And, the displacement of point D is δD = ∂U/∂P2 =

  (P1 + P2 )  P2  −aL/2 1 − e−aL/2 + e − e−aL AEo a AEo a

Interpretation/Comments: Since the principle of superposition does not work on strain energy, the strain energy in segment 1 is evaluated by taking the total force (P1 + P2 ). In structures with multiple members, such as a truss, the strain energy is determined by the expression for n number of elements: i=n  Fi2 Li U= 2Ai Ei

(14.10)

i=1

In the expression, the integral form is not adopted. In a truss, the members are two-force members carrying only axial forces. In the expression, it is implicitly assumed that a member of the truss is prismatic, that is, it is straight and the cross-section is uniform along its length. However, if material property E or cross-section of a member varies along its length, the integral form should be invoked. Now, we take up a solved example to show the application of Castigliano’s theorem to a structure made of axial members. Example 14.3: Links 1 and 2 are attached to a rigid wall with pin joints shown in Fig. 14.6(a). Two external forces P1 and P2 are applied at pin D. Determine the displacement of point D using the strain energy approach. The links are made of the same material of modulus E and the same cross-sectional area A.

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(a)

Fig. 14.6

(b)

(a) Linkages 1 and 2 and (b) forces acting at point D.

Given: BD= L; ∠BDC =45 o ; hinge at B; hinge at C; pin joint at D; external force P1 acting at D in x-direction; external force P2 at D in negative y−direction; modulus E; cross-sectional area A To find: δxD ; δyD Strategy: Determine axial forces in the links by invoking equilibrium at pin D; determine strain energies in links; invoke Castigliano’s theorem Solution: Since links 1 and 2 are two-force members, they carry only axial forces. Equilibrium of pin D, shown in Fig. 14.6(b), gives F2 − F1 − √ + P1 = 0 2 F2 √ − P2 = 0 2

ΣFx : ΣFy : On solving, we obtain F2 =

√

2P2 ;

F1 = P1 − P2

The strain energies are obtained using Eq. (14.10) as √

2 √ √ 2 2P2 ( 2L) (P1 − P2 )2 L 2P2 L ; U2 = = U1 = 2AE 2AE AE

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The total strain energy is 

(P1 − P2 )2 √ 2 L + 2P2 U = U1 + U2 = 2 AE The displacement of point D is obtained from Castigliano’s theorem as ∂U (P1 − P2 ) L = ∂P1 AE

 L √ ∂U = = − (P1 − P2 ) + 2 2P2 ∂P2 AE  L

 √  = 1 + 2 2 P2 − P1 AE

δxD = δyD

Interpretation/Comments: In this problem, the determination of δxD and δyD is quite involved if we find them by displacement–force relations. The strain energy approach is relatively much simpler. 14.3

Strain Energy Density

For normal stress, we have already obtained stain energy density 2 as 12 σE or 12 σ where σ and  are normal strain and corresponding normal strain. Similarly, if a member is subjected to shear stress τ , the strain energy density is given by SED =

1 1 τ2 = τγ 2G 2

(14.11)

For a general case, SED =

1 [σxx xx + σyy yy + σzz zz + τxy γxy + τxz γxz + τyz γyz ] 2 (14.12)

Using the stress–strain relations like xx = E1 [σxx − ν (σyy + σzz )] and γxy = τxy /G, the above relation can be expressed in terms of

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stress components as SED =

 1  2 2 σxx + σyy +σ 2zz − 2ν (σxx σyy + σxx σzz + σyy σzz ) 2E  1  2 2 2 τxy + τxz + τyz (14.13) + 2G

If the axes coincide with the principal axes, the relation simplifies to SED =

1 2 [σ + σ22 + σ32 − 2ν(σ1 σ2 + σ1 σ3 + σ2 σ3 )] 2E 1

(14.14)

It is important to realize that the expressions for SED involve quadratic terms, and the equations are no longer linear. Thus, the principal of superposition does not work. 14.4

Strain Energy in Slender Members

In this book, our focus is mainly on slender members. As we have been discussing, there are four kinds of internal generalized forces on a cross-section: (i) axial force, (ii) bending moment, (iii) torsional moment, and (iv) shear force. We will develop relations for each case in this section. 14.4.1

Strain energy due to an internal axial force

We have already developed the expression for it in Eq. (14.6) as 1 Ua = 2



L o

F 2 dx AE

This integral form is convenient and accounts for variation of internal force F , cross-sectional area A, and modulus E with x. 14.4.2

Strain energy due to a bending moment

We will prove that the strain energy due to bending moment Mb on a slender member is  1 L Mb2 dx (14.15) Ub = 2 o EI

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(a)

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473

(b)

(c)

Fig. 14.7 (a) Bending moment acting on a small length dx, (b) resulting curved shape, and (c) an infinitesimal element of length dx and area dA.

The form of this expression is the same as that of the axial case. The integral also allows variation of modulus E and moment of inertia I along the length of the member. We can deduce this expression in two ways. In the first method, we consider a small element dx of the member which bends with curvature ρ under the bending moment Mb as shown in Fig. 14.7(a). The work done by Mb is the internal energy of the element and is expressed as 1 Mb Δθ 2

dUb =

The factor 1/2 is present as moment increases linearly with θ. Since Δθ = dx/ρ (Fig. 14.7(b)), dUb =

1 Mb dx 2 ρ

In Section 11.2, we have shown that curvature κ = the above equation changes to   1 Mb dx dU b = Mb 2 EI

1 ρ

= Mb /EI and

leading to Ub =

1 2



L o

Mb2 dx EI 2

xx . In the second method, we make use of the expression of SED = σ2E Consider a small volume dV which is a product of area dA on the

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cross-section A and small length dx as shown in Fig. 14.7(c). Then, 2 xx dA dx. Thus, the strain energy the strain energy in this volume is σ2E in the slender member is   2  σxx dA dx Ub = 2E We now substitute the equation of bending stress σxx = − MIb y to have     Mb y 2 1 − dA dx Ub = 2E I This is rearranged as 1 Ub = 2 Since



Ay

2 dA



Mb2 EI 2

 y dA dx



2

A

= I , the expression simplifies to Ub =

1 2



L o

Mb2 dx EI

This is the same as Eq. (14.15). 14.4.3

Strain energy due to an internal torsional moment

We will show that the strain energy Ut in a slender member due to internal torsional moment Mt is 1 Ut = 2

 o

L

Mt2 dx GJ

(14.16)

The shear modulus G or polar moment of inertial J can vary with x. The form of this expression is also similar to those of the axial and bending cases. This expression can also be obtained through two methods. In the first method, a small length element dx is subjected to torsional moment Mt as shown in Fig. 14.8. With respect to the

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Fig. 14.8 Mt .

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475

A small length portion of a slender bar subjected to torsional moment

face at A, the face B rotates by the angle Δφ. The strain energy dU t then is given by 1 dUt = Mt Δφ 2 Δφ is obtained from Mt using the angle of twist equation Δφ = Mt Δx/GJ. Substituting in the above equation, we have dUt =

1 Mt2 Δx 2 GJ

The integral form is Ut =

1 2



L

o

Mt2 dx GJ

In this case as well, the polar moment of inertia J and shear modulus can vary with x. 2 We make use of SED = 12 τG in the second method. We multiply it with the infinitesimal volume Δx. ΔA, similar to the way we did it for the bending case, to have strain energy as  2 τ dA dx dUt = 2G Then,    Ut =

τ2 2G

 dA dx

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Substituting τ =

Mt r J ,

 Ut = But,

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we have

Mt2 r 2 dA dx = 2GJ 2



Mt2 2GJ 2



 r dA dx 2

r 2 dA = J, and the above expression is simplified to Ut =

1 2



Mt2 dx GJ

This expression is the same as that of Eq. (14.16). 14.4.4

Strain energy due to a shear force

The expression for strain energy developed by a shear force V in a slender member by making use of SED is dUs =

1 τ2 1 τ2 × Volume = dA dx 2G 2G

We found τ in Section 10.32 as τ = VbIQ in which the shear stress is evaluated at distance y from the neutral axis. The width of the crosssection is b at distance y and Q is the moment of the area beyond distance y. Substituting τ in the above equation and manipulating the expression, we obtain  

    V 2 Q2 V2 1 Q2 dA dx dA dx = 2 GI 2 b2 Gb2 I 2     V2 A Q2 1 dA dx = 2 GA I 2 b2

1 Us = 2

We define shape factor fs as fs =

A I2



Q2 dA b2

(14.17)

The shape factor depends on the shape of the cross-section. Its value is 6/5 for a rectangular cross-section as shown in Section 14.9,

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Energy Methods Table 14.1 S. No. 1 2 3 4

*fs =

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477

Strain energy in slender members.

Generalized force

Strain energy

Axial

Ua = Ub =

Bending Torsional Shear

Ut = Us∗ =

1 2 1 2 1 2 1 2

   

L

F 2 dx AE

o L

Mb2 dx EI

L

Mt2 dx GJ

o

o

fs

V2 dx GA

A  Q2 dA (Eq. 14.17) I2 b2

Appendix A. The shape factors for other cross-sections can be similarly determined using Eq. (14.17). Then, the expression for strain energy becomes  1 V2 fs dx (14.18) Us = 2 GA The expressions of strain energy of a slender member are summarized in Table 14.1. 14.5

Castigliano’s Theorem and its Applications

We have already seen the statement of Castigliano’s theorem in Section 14.2 and discussed its application to axial cases to determine the displacement of the point at which an external force is applied. The theorem is applicable for strain energy developed through all the four internal forces, that is, axial, bending moment, torsional moment, and shear force. However, the contribution of shear stress is usually much smaller for slender members (with length at least eight times larger than the largest dimension of the cross-section) and therefore we neglect its effect for most cases of slender members. Example 14.4 illustrates this point. Figure 14.9 shows a slender member with an axial external force P1 , two external transverse forces, P2 and P3 , one external torque T ,

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Fig. 14.9

External forces acting on a slender member.

and one external moment Mo . These external forces develop internal generalized forces, axial force F , bending moment Mb , torsional moment Mt , and shear force V, within the body of the member. In general, these internal generalized forces vary with distance in x-direction. It is helpful to draw all diagrams, namely, axial force diagram (AFD), bending moment diagram (BMD), torsional moment diagram (TMD), and shear force diagram (SFD). To calculate strain energy in the entire member, we determine strain energy in each segment of the member (BC, CD, DE, EG, GH, and HK) in the case of Fig. 14.9. Then, we determine the total strain energy by adding the strain energy of all segments. The effect of all external forces should be considered to determine internal generalized forces, F, Mb , Mt , and V. If the member remains linear elastic at all points of the member, we can invoke Castigliano’s theorem to obtain the following values: • • • • •

∂U Displacement of point B in x-direction as ∂P 1 ∂U Deflection of point D in negative y-direction as ∂P 2 ∂U Deflection of point G in y-direction as ∂P 3 Angle of twist (φx) of the member at location E as ∂U ∂T Rotation (θz ) of member about z-axis at location K as

∂U ∂Mo

The proof of the Castigliano theorem is conceptually involved and we would not want to get stuck in its proof and lose sight of its strength in solving problems. The proof is presented in Section 14.10, Appendix B. Now, we take up a solved example to show that the effect of shear force in determining the strain energy is negligibly small. Example 14.4: A transverse force P acts on a simply supported slender member (Fig. 14.10(a)) with it rectangular cross-section (Fig. 14.10(b)). Determine: (i) strain energy Ub developed by the bending moment, (ii) strain energy Us developed by the shear force,

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(c)

(f)

(e)

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479

(a)

(d)

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(g)

Fig. 14.10 (a) The member, (b) cross-section, (c) FBD of the member, (d) section at distance s, (e) section at distance r from end B, (f) SFD, and (g) BMD.

(iii) Us /Ub , (iv) deflection δc of point C, and (v) value of δc if P = 2000 N, L = 400 mm, h = 40 mm, b = 20 mm, E = 70 GPa, and ν = 0.3. Given: Length AB= L; hinged at A; roller supported at B; transverse force P at x =L/3 To find: (i) Ub , (ii) Us , (iii) Us /Ub , (iv) δc , and (v) δc for given values of L, P , h, b, E, and ν Strategy: Draw SFD and BMD, determine strain energy in each segment, differentiate U with P to obtain δc Solution: In this simple case, FBD of the member yields (Fig. 14.10(c)) 2 RA = P 3

and

1 RB = P 3

In segment 1 (Fig. 14.10(d)), 2 Mb = P s 3

2 and V = − P 3

In segment 2 (Fig. 14.10(e)), Mb = 13 P r and V = 13 P

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Figure 14.10(f) shows the SFD and Fig. 14.10(g) BMD. Strain energy due to bending moment in segment 1 is (1) Ub

2  L/3  Mb2 1 2 dx = P x dx EI 2EI 0 3    1 L 3 P 2 L3 1 4 P2   = 0.0329 = × 2 9 E bh3 3 3 Ebh3 1 = 2



12

Invoking Eq. (14.18) with fs =6/5 for the rectangular cross-section beam, we obtain Us(1)

2  L/3    fs V 2 1 2 6 dx = P dx GA 2GA 0 5 3

0.267P 2 L3 P 2L = 0.089 = Gbh Gbh 1 = 2



Strain energies in segment 2 are (2) Ub

2  2L/3  Mb2 1 1 dr = P r dr EI 2EI 0 3    1 2L 3 P 2 L3 1 1 P2   = 0.0658 = × 2 9 E bh3 3 3 Ebh3 1 = 2



12

Us(2)

 2L/3    2 fs V 2 1 1 6 dr = P dr GA 2GA 0 5 3

0.0667P 2 2L P 2L 3 = 0.0444 = Gbh Gbh 1 = 2



The total strain energy due to bending moment adds to (1)

Ub = Ub

(2)

+ Ub

= 0.0987

= (0.0329 + 0.0658)

P 2 L3 Ebh3

P 2 L3 N 2 m 3

→ Nm → J Ebh3 mN2 mm3

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481

The total strain energy due to shear force adds to Us = Us(1) + Us(2) = (0.089 + 0.0444) = 0.1334

P 2L Gbh

P 2 L N 2 .m

→ Nm → J Gbh mN2 .m.m

The ratio of the two strain energies is 2

0.1334 PGbhL h2 E Us = = 1.352 2 3 Ub L2 G 0.0987 PEbhL3 But,

E G

=

E

E 2(1+ν)

= 2(1 + ν).

For ν =0.3, the above expression gives E/G =2.6 and then h2 h2 Us = 1.352 × 2.6 2 = 3.52 2 Ub L L For h/L =1/8, Us = 0.055 Ub The contribution of shear force is only about 5%. The ratio becomes still smaller for higher values of L/h. The total strain energy becomes U = Ub + Us = 0.0987

P 2 L3 P 2L +0.1334 Ebh3 Gbh

The Castigliano theorem yields P L3 ∂U PL = 0.198 (14.19) +0.267 ∂P Ebh3 Gbh For P = 2000 N, L = 400 mm, h = 40 mm, b = 20 mm, E = 70 GPa, and G = 70/(2(1 + 0.3) = 26.9 GPa, δc =

0.198 × 2000× (400)3 N .mm3

δc = N 3 70×103 ×20×403 mm2 .mm.mm +

N .mm 0.267 × 2000 × 400 N

3 26.9×10 ×20 × 40 mm2 .mm.mm

= 0.283 mm + 0.0099 mm = 0.293 mm

(14.20)

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Interpretation/Comments: In this case, L/h = 10 which is greater than 8; the effect of shear force is only 3.5 % and, therefore, it can be neglected. The algebra to determine deflection at the load point can be simplified. The general expression for determining deflection at the load point of external force Pi is      Mb2 1 Mt2 1 F2 ∂U ∂ 1 dx + dx + dx = δi = ∂Pi ∂Pi 2 EI 2 GJ 2 AE We now can take the differential to the integrands of all the three terms to have ∂U δi = = ∂Pi



Mb ∂Mb dx + EI ∂Pi



Mt ∂Mt dx + GJ ∂Pi



 F ∂F dx AE ∂Pi (14.21)

If we deal with a truss consisting of only axial members, U =Σ

Fj2 Lj 2Aj Ej

where Fj is the internal axial force in member j. The displacement of the point where Pi is applied is given by δi =

∂U Fj Lj ∂Fj =Σ ∂Pi Aj Ej ∂Pi

(14.22)

We will now solve a few problems to appreciate the usefulness of simplified Eqs. (14.21) and (14.22). Example 14.5: Determine the deflection of point C of Example 14.4 using the simplified formulas. Given: Same as for Example 14.4 To find: δc (Fig. 14.10(a)) Strategy: Invoke Eq. (14.21)

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483

Solution: In segments 1 and 2, the bending moments are (1)

Mb

2 P s; 3

=

(2)

Mb

=

1 Pr 3

The differentiation leads to (1)

∂Mb ∂P

=

2 s; 3

(2)

∂Mb ∂P

=

1 r 3

Eq. (14.21) yields        L  2L  3 3 2 1 1 1 2 1 Ps s ds + Pr r ds δc = EI 0 3 3 EI 0 3 3  L  2L 3 3 P 1 2 2   4s ds + r dr = 9 E bh3 0 0 12      4 P 4 L 3 1 2L 3 + = 3 bh3 3 3 3 3 =

P L3 16 P L3 = 0.198 81 bh3 bh3

The result is the same as obtained in Eq. (14.19) for the contribution of bending alone. Interpretation/Comments: The form of Eq. (14.21) is simple which can help reach the right answer because the differentiation involved in Castigliano’s relation is merged with the integration needed in obtaining the strain energy. Example 14.6: Consider a cantilever with a triangular distributed load and an external force P acting at x = L/2 as shown in Fig. 14.11(a). Given: Length = L; rigidly fixed at C; free end at A; triangular distributed load with qo at C and zero at A; force −P at x = L/2 ; modulus E; moment of inertia = I To find: δB Strategy: Invoke Eq. (14.21)

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484 (a)

(b)

(c)

Fig. 14.11

(a) The member, (b) Mb in segment 1, and (c) Mb in segment 2.

Solution: The intensity of the distributed force at distance x is (Figs. 14.11(b) and 14.11(c)) q=

x L

qo



In segment 1, the equivalent total force is 12 Lx qo x which acts at distance x/3 from the sectional plane. Thus, the bending moment is (Fig. 14.11(b)) (1)

Mb ∂M

=

   x3 1 x  x qo x = qo 2 L 3 6L

(1)

Then, ∂Pb = 0. In segment 2, the bending moment is (Fig. 14.11(c)) (2) Mb

       L x3 1 x  L x qo x −P x− = qo − P x − = 2 L 3 2 6L 2 (2)

Then,

∂Mb ∂P



= − x − L2 .

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Invoking Eq. (14.21), we obtain δB as    L/2 3  L  3 x 1 x 1 L qo − P x − ( qo ) × (0)dx + EI 0 6L EI L/2 6L 2    L dx × − x− 2 ⎡ ⎤ ⎡ ⎤ L   L   ⎥ ⎢ ⎥ x4 x3 L2 qo ⎢ 2 ⎢ ⎥+ P ⎢ − x + dx dx⎥ − Lx + = ⎣ ⎦ ⎣ ⎦ EI 6L 12 EI 4

δB =

L 2

=−

L 2

P L3 49 qo L4 + 3840 EI 24EI

Interpretation/Comments: The direction of δB is the same as that of external force P , that is, in the downward direction. Of course, the contribution of force P is in the direction of P , while the contribution of the distributed force opposes the deflection because the distributed force is applied in the opposite direction. Example 14.7: A slender member AD is rigidly fixed at point D and is made of three segments with different diameters as shown in Fig. 14.12(a). A force P is applied at the free end A. Determine the deflection δA of point A if the modulus is E. Given: AD fixed at D; segment 1: AB = L, dia. d; segment 2: BC= L, dia. =1.5d; segment 3: CD= L, dia. 2d; external force P at location A; modulus= E (a)

(b)

(c)

(d)

Fig. 14.12 (a) The member, (b) section in segment 1, (c) section in segment 2, and (d) section in segment 3.

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To find: δA Strategy: Determine bending moment in each segment and use Eq. (14.21) Solution: In segment 1 (Fig. 14.12(b)), (1)

Mb

= −P x

0≤x≤L

and, (1)

∂Mb ∂P In segment 2 (Fig. 14.12(c)), (1)

Mb

= −P x

= −x.

L ≤ x ≤ 2L

and, (2)

∂Mb ∂P In segment 3 (Fig. 14.12(d)), (3)

Mb

= −P x

= −x.

2L ≤ x ≤ 3L

and, (3)

∂Mb = −x. ∂P Invoking Eq. (14.21), we obtain  L  2L 1 (−P x) (−x) (−P x) (−x) dx + dx δA = E 0 I1 I2 L   3L (−P x) (−x) dx + I3 2L   2L  3L L x2 x2 x2 P dx + dx + = 4 dx π(1.5d)4 E 0 (πd4 ) L 2L π(2d) 64 64 64  24.24P L3 64P L3 (2L)3 − L3 (3L)3 − (2L)3 + + = = πEd4 3 3 × 5.063 3 × 16 Ed4 Interpretation/Comments: Energy methods are handy to solve problems of several segments.

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(a)

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487

(b)

Fig. 14.13 (a) The tensile coil spring and (b) the torsional moment at a section of its wire.

Example 14.8: Determine spring constant k of a tensile coil spring shown in Fig. 14.13(a). The wire diameter of the spring is d and the diameter of the coils is D. There are n number of coils and the modulus of the spring material is E. Given: Coil dia.= D; wire dia.= d; number of coils= n; modulus E To find: k Strategy: We will determine displacement δ of spring for axial force of P and then evaluate k using the relation k = P/δ. We cut the spring at any place and realize that the cross-section of the cut wire is subjected to torsional moment Mt =PD/2 (Fig. 14.13(b)). The torsional moment is the same at all points of the coils of the spring. We will neglect the effect of shear force. We will then determine the strain energy due to this internal torsional moment and then differentiate to find the displacement δ caused by force P . Solution: The torsional moment at a section of the wire is Mt = P D/2 Its differential form is ∂Mt = D/2 ∂P The total length of the wire of the spring is πDn. We neglect the deformation of the wire hook through which the force is applied.

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Thus, displacement δ is determined using Eq. (14.21) as   L  πDn  1 8P D 3 n ∂Mt PD D 1 ds =  4  ds = Mt δ= GJ o ∂P 2 2 Gd4 o G πd 32 The spring constant becomes k = P/δ =

P 8P D 3 n Gd4

=

Gd4 8D 3 n

It is always advisable to check the dimension of a resulting expression. (N/mm2 ).mm4 N → mm . For this case, dimensions are mm3 Interpretation/Comments: The spring constant can also be determined using the angle of twist relation. The spring can be modeled as a long wire of length πDn with torque of PD /2. The angle of twist is φ=

(P D/2) πDn 16P D 2 n Mt L   = = 4 GJ Gd4 G πd 32

To obtain the deflection of the load point, we multiply the angle of rotation by the distance between the wire of the spring and the load line. Thus,   8P D 3 n 16P D 2 n D = δ = φ × (D/2) = Gd4 2 Gd4 This is the same as obtained through energy approach. 14.6

Fictitious Force to Determine Deflection at Any Point

We have so far learnt how to find deflection only of a point at which an external force is applied. In a first glance, it seems that the scope of the application of Castigliano’s theorem is limited because we are not able to find deflection of any point of the structural member. However, we can extend the scope of Castigliano’s theorem by applying a fictitious force. We thus apply a fictitious force Q at the point

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Fig. 14.14

(a)

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489

Fictitious force Q at point D in the desired direction q .

(b)

(c)

(d)

Fig. 14.15 (a) Cantilever, (b) with fictitious load Q, (c) section in segment 1, and (d) section in segment 2.

of our interest and along the direction in which we intend to find the deflection. For example, we are interested in finding deflection at point D of the member shown in Fig. 14.14 in the direction q . We, therefore, apply force Q in the direction q as shown. We then determine the strain energy considering all the applied forces and Q. The strain energy thus obtained is differentiated with respect to Q. The resulting expression still has terms with Q. We then make Q = 0 to have the displacement of point D in direction q due to external forces P1 , P2 , and P3 of Fig. 14.14. The procedure is justified as force Q can be as small as we want and still the Castigliano theorem works. We will now solve two problems to demonstrate the powerfulness of Castigliano’s theorem. Example 14.9: On a cantilever BC, a force P is applied at its free end B as shown in Fig. 14.15(a). Determine the deflection of point

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490

D (x = L/2 ) in the downward direction if the moment of inertia is I and modulus E of the member. Given: BC= L; fixed at C; free at B; external force P at B; modulus= E; moment of inertia= I To find: Deflection δD of point D Strategy: Apply a fictitious force Q at location D in the downward direction (Fig. 14.15(b)); determine bending moment in both segments and apply Eq. (14.21) to determine δD. Solution: In segment 1, the bending moment and its differentiation with respect to Q are (Fig. 14.15(c)) (1)

Mb

(1)

∂Mb ∂Q

= −P x;

=0

In segment 2, the bending moment and its differentiation with respect to Q are (Fig. 14.15(d)) (2) Mb



L = −P x − Q x − 2



(2)

;

∂Mb ∂Q



L =− x− 2



Equation (14.21) yields   L (1) (2) L/2 1 (1) ∂Mb (2) ∂Mb dx + dx Mb Mb δD = EI 0 ∂Q ∂Q L/2  L    L 2 L 1 (−P x × 0) dx + −P x − Q x − = L EI 0 2 2     L dx × − x− 2  L

2

 1 2 2 P x − Lx/2 + Q x − Lx + L /4 dx = L EI 2

Making Q = 0, we obtain  L 2

1 5 P L3 P x − Lx/2 dx = δD = L EI 48 EI 2

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491

(a)

(b)

Fig. 14.16 (a) Links 1 and 2, (b) fictitious force Q at D, and (c) force equilibrium at point D.

Interpretation/Comments: The positive value of δD means point D moves in the direction of the fictitious force Q we have chosen. If we had chosen our fictitious force in the upward direction, δD will turn out to be negative. It means that the displacement is in the direction opposite to the direction of the fictitious force. Example 14.10: Consider a structure with two links as shown in Fig. 14.16(a). Both links, 1 and 2, are made of the same modulus E, but the area of cross-section is different, A/2 for link 1 and A for link 2. Determine the displacement of pin D in x-direction. Given: BC = L; ∠CBD = 60o ; ∠BCD = 30o ; pin joints at B, C, and D; modulus = E; area of cross-section of link 1=A/2 ; area of cross-section of link 2 = A. x To find: Displacement of pin D, δD

Strategy: Apply a fictitious force Q at point D in x−direction; use F L ∂F δi = Σ Ajj Ejj ∂Pji Solution: Force equilibrium at pin D yields (Fig. 14.16(b)) ΣFx :

F1 cos30o − F2 cos60o + Q = 0

ΣFy :

F1 sin30o + F2 sin60o − P = 0

On solving, we obtain

√ P 3 Q; F1 = − 2 2

√ F2 =

Q 3 P+ 2 2

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492

Then,

√ 3 ∂F1 =− ; ∂Q 2

Thus,

⎡

Fj Lj ∂Fj 1 x =Σ = ⎣ δD Aj Ej ∂Pi E √ +

3 2 P

+

Q 2

A



×

L 2

P 2

∂F2 1 = ∂Q 2

−

√

3 2 Q



×

√

3 2 L

A/2

 √  3 × − 2

⎤ 1 N.mm × ⎦ N

2 mm2 mm2

Substituting Q = 0, we obtain  √ 3 PL 1 3 x − P+ P × L = −0.533 δD = AE 4 8 AE Interpretation/Comments: The negative sign means that the point D moves in the direction which is opposite to that of the fictitious force Q. One can justify the negative sign because Link 1 elongates more due to its smaller cross-sectional area and longer length.

14.7

Solution of Indeterminate Problems using Castigliano’s Theorem

Castigliano’s theorem is very effectively used to determine reactions in indeterminate problems. We recall that a problem is known as indeterminate if the number of unknown reaction components exceeds the number of equilibrium equations. Since in most problems, the support points do not move, we are able to develop additional equations by determining the deflection of the support point and equating it to zero. We will now illustrate this point by solving an example. Example 14.11: A slender member BC is rigidly fixed at location C and is roller supported at location B as shown in Fig. 14.17(a). Determine all reaction components if modulus is E and moment of inertia I.

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(a)

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493

(b)

(c)

(d)

Fig. 14.17 (a) The slender member, (b) FBD, (c) section in segment 1, and (d) section in segment 2.

Given: BC = L; fixed at C; roller supported at B; force P applied at x = L/2 ; modulus = E; moment of inertia = I To find: All reaction components Strategy: There are three unknown reaction, MC , VC , and VB (Fig. 14.17(b)). In addition, there are two equilibrium equations: ΣFy and ΣMz . We need to develop an additional equation. We can treat one of the unknown reaction components as external force, say VB . Determine strain energy and differentiate it with respect to VB . The resulting deflection is equated to zero. Solution: In segment 1, the bending moment is (Fig. 14.17(c)) (1)

Mb

= xVB

yielding, (1)

∂Mb =x ∂VB In segment 2, the bending moment is (Fig. 14.17(d)) (2)

Mb (2)

yielding,

∂Mb ∂VB

=x

  L P = xVB − x − 2

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494

Then, δB is obtained as   L (1) (2) L/2 1 (1) ∂Mb (2) ∂Mb Mb dx + Mb dx = 0 δB = EI 0 ∂VB ∂VB L/2 or, L

L 

2

(xVB ) x dx + 0

or,

L 2



L xVB − x − 2

  P x dx = 0

    3 ! 1 L 3 1 L 3 L − + VB 3 2 3 2   3 !  2 ! 1 L L L 3 2 −P L − L − − =0 3 2 4 2

or, VB =

5 P 16

To determine the other two reaction components, we invoke equilibrium equations to have ΣFy :

VB + VC = P

or, 11 5 VC = P − VC = P− P = P 16  16 5 L − PL = 0 ΣMzC : M C + P 2 16 or, MC = −

3 PL 16

Interpretation/Comments: If the unknown reactions are more than the equilibrium equations, the technique of Castigliano’s theorem is handy.

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495

Summary

Energy theorems are quite powerful in solving problems of the mechanics of materials, especially for nonlinear cases such as of plastic deformation. The energy theorems are often convenient to use because, unlike stress or strain tensor, energy is scalar. In this chapter, we have discussed only one kind of energy, strain energy, due to the limited exposure so far in this field. Also, the entire discussion is confined to linear elastic materials. During loading, distances between molecules of the material change. However, the molecules tend to come back to their natural state. Thus, strain energy accounts for the internally stored energy. The unit of strain energy is joule (J). The strain energy expressions for a slender member are  1 L F 2 dx Axial: U a = 2 o AE  1 L Mb2 dx Bending: U b = 2 o EI  L 2 1 Mt dx Torsional: U t = 2 o GJ  1 V2 dx (fs : shape factor) fs Shear: Us = 2 GA Strain energy density (SED) is strain energy per unit volume. Its unit is the same as that of stress, that is, MPa. It is a point function which, in general, varies from one point to another. In a slender member loaded with only axial stress σxx , the SED is equal to 12 σxx xx . In a case where the member is subjected to only shear stress τxy , the SED is equal to 12 τxy xy . Castigliano’s theorem is very effective in solving problems through the strain energy. The theorem states that the partial derivative of the total strain energy with respect to an externally applied force provides the displacement at which the force is applied and the direction ∂U ). The of the displacement is the same as that of the force (δi = ∂P i theorem is handy in finding displacement of a point. In case we want to find displacement at a point where no external force is present, we apply a fictitious force Q at the point and determine strain energy of the member in terms of Q. The strain energy thus obtained is

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partially differentiated with respect to Q. We then make Q equal to zero to obtain the required displacement. Castigliano’s theorem is also convenient for developing additional equations for solving the cases of indeterminate structure. Quite commonly, the strain energy is partially differentiated with respect to a reaction of a support point (which does not allow displacement) and is equated to zero to obtain the required additional equation.

14.9

Appendix A: Shape Factor of Rectangular Cross-Sections

Consider a rectangular cross-section shown in Fig. 14.18. We will show that the shape factor for this cross-section is 6/5. The shape factor is defined as fs =

A I2

  A

Q2 b2

 dA

where A =bh, I = bh3 /12 and Q at distance y is given by Q = (Area above BD) × (distance of its centroid)         h b h2 h −y + y /2 = − y2 = b 2 2 2 4

Fig. 14.18

A rectangular cross-section of a slender member.

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Substitution of A, I, and Q in the expression of fs yields

fs = 

(bh) 2 bh3 12

36 = 5 h



h 2

− h2



h 2

− h2



 b 2

h2 4

− y2 b2

2 (bdy)

 h4 h2 y 2 4 − + y dy = 6/5 16 2

Thus, the shape factor of a rectangular section is 6/5.

14.10

Appendix B: Proof of Castigliano’s Theorem

We find that the Castigliano theorem is very effective in solving problems. We deliberately avoided the proof in the main text of the chapter to emphasize the usefulness of the theorem by solving different kinds of problems. Consider a structural member with forces P1 , P2 , . . . , Pi , . . . , Pn acting on its surface as shown in Fig. 14.19. The strain energy due to all these forces is U . Our interest is to find displacement of any one of the forces. Let us focus our attention on force Pi acting at point D. The Castigliano theorem states that the displacement of point D in

Fig. 14.19

A structural member subjected to various forces.

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the direction of force Pi is given by δi =

∂U ∂Pi

Proof : Let us vary force Pi by a small amount of dP i which changes the strain energy by dU. U + dU = U +

∂U dPi ∂Pi

(14.23)

For linear elastic materials, U does not depend on the sequence of applying forces, that is, if P2 is applied first and then P1 or vice versa, the strain energy developed in the member is the same. We make use of this realization. Consider the member with no external loads on it. Now, we first apply only dPi at point D in the direction of force Pi . Due to the application of only dP i , point D moves by a small displacement Δ which is also infinitesimal. The strain energy developed in the structural member is (1/2)dP i Δ. This is the product of two infinitesimal quantities and therefore can be neglected. Now, we apply all the forces, P1 , P2 , . . . , Pi , . . . ,Pn , on the member while dP i continues to act on the member. It is worth noting that dP i is also being displaced and is moved by displacement δi . This δi is much larger than Δ because it is caused by finite magnitude forces. The total strain energy is given by U + dU = U + dP i δi

(14.24)

The right-side expressions of Eqs. (14.23) and (14.24) are equivalent and therefore ∂U dPi = dP i δi ∂Pi or, δi =

∂U . ∂Pi

It is worth remembering that displacement δi is in the direction of Pi .

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499

Problems 1. A tube of outside diameter 25 mm and wall thickness 1.5 mm is made from four different materials, namely, alloy steel, aluminum alloy, glass fiber reinforced plastic (GFRP), and carbon fiber reinforced plastic (CFRP). It is loaded in cantilever mode as shown in Fig. 14.20. Determine displacement v in y-direction for all the four cases. Plot P − v curves for all the four cases on a single figure so that the area under the curves can be compared. Determine stored strain energy for each case. Why does the GFRP tube store the highest strain energy? Note that GFRP is used for making the poles for the pole vaulting athletic event and diving boards at swimming pools. The modulus of the materials is as follows: alloy steel = 207 GPa; alloy aluminum = 70 GPa; GFRP = 38 GPa; CFRP = 127 GPa. 2. Consider two configurations of a cantilever which are of the same length and same material as shown in Figs. 14.21(a) and 14.21(b). In Config. (a), the cantilever is made thicker in segment AB. In Config. (b), the segment AB is made wider in such a way that the volume of the two cantilevers is the same. Determine the strain energy of both configurations (Modulus E = 207 GPa). Why does Config. (b) store more strain energy? The leaf springs of the suspension of automobiles or trucks are made on this realization. The Config. (c) is equivalent to Config. (b) as the additional portion in segment AB is placed at the top in such a way that the two plates are free to slide during loading. 3. Consider three cantilevers of the same length and same material, alloy steel of modulus E =207 GPa: (i) a circular bar of diameter 10 mm as shown in Fig. 14.22(a); (ii) a tube of the same length with outside diameter 25 mm and wall thickness is chosen in such

Fig. 14.20

Problem 1.

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500

(a)

(b)

(c)

Fig. 14.21 Problem 2: (a) with thicker AB segment, (b) with wider segment AB, and (c) configuration equivalent to the wider configuration.

(a)

(b)

(c)

Fig. 14.22 Problem 3: (a) circular cross-section of diameter 10 mm, (b) tube of 25 mm OD, and (c) flat of 2 mm thickness.

a way that the area of cross-section is the same as of the circular bar (Fig. 14.22(b)); and (iii) a flat bar of the same length and of 2 mm thickness having the same area of cross-section (Fig. 14.22(c)). The maximum stress of the critical point should not exceed 400 MPa in all the three cases. Determine maximum load P for each case and the corresponding strain energy. Compare the strain energy through a table. Can we deduce something out of the resulting data? 4. A circular bar AB of diameter d is fixed at end B as shown in Fig. 14.23(a). At the free end A, axial force P and torque T are applied. Determine displacement and rotation at the free end A using Castigliano’s theorem. 5. A slender member AB of diameter d is supported at A through a hinge while it is supported with a roller at B. At the midpoint C, a moment Mo acts as shown in Fig. 14.23(b). Determine the

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(b)

(a) Problem 4 and (b) Problem 5.

(a)

Fig. 14.24

page 501

501

(a)

Fig. 14.23

b4632-ch14

(b)

(a) Problem 6 and (b) Problem 7.

rotation of the member at point C (Modulus = E; Moment of inertia = I). 6. Member AB is rigidly fixed at end B and a moment Mo acts on its free end as shown in Fig. 14.24(a). A force P also acts at location C. Determine the rotation at A and downward deflection of point C. 7. A structure is made to mount an advertisement board at the side of a road. The entire structure is in an L shape and is made of a steel tube of 100 mm outside diameter and 5 mm wall thickness. The board applies a load of 500 N as shown in Fig. 14.24(b). Determine: (a) Vertical deflection of point C (b) Rotation of the free end at C. 8. A slender member AB is supported on a hinge at location A and roller supported at B as shown in Fig. 14.25(a). A distributed load q N/m is applied from the midpoint C to point B. Determine

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502

(a)

Fig. 14.25

(a)

Fig. 14.26

(b)

(a) Problem 8 and (b) Problem 9.

(b)

(a) Problem 10 and (b) Problem 11.

the downward deflection of point C (Modulus = E; Moment of inertia = I). 9. A slender member is made thicker in the center portion as shown in Fig. 14.25(b). A force P acts at the mid-length. Determine deflection under the load P . The moment of inertia of thin and thick portions is I1 and I2 , respectively. 10. Around a steel circular rod of 10 mm diameter, a copper tube of 15 mm outside diameter and 10 mm inside diameter is bonded as shown in Fig. 14.26(a). This composite member is rigidly fixed at location A and is being pulled at location B by axial force of 9000 N. Determine the extension of the composite member and strain energy stored in each material (modulus of copper = 117 GPa; modulus of steel = 207 GPa). Note that the axial strain in both members is the same. 11. A slender member AB is hinged at both ends as shown in Fig. 14.26(b). It is made in two segments AC and BC in such a way that the cross-sectional area of segment AC is twice that of segment CB. An external axial force P acts at location C as shown. Determine the reactions at the support and the strain energy of the member.

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Chapter 15

Historical Perspective 15.1

Introduction

The field of mechanics of materials has been of interest to mankind right from the Stone Age because products such as huts, boats, homes, bridges, agricultural implements, and kitchen utensils have to be used for a long time. With a lot of experience, our ancestors had some idea about the strength and deflection of materials and developed broad guidelines and constructed large structures, such as forts, mosques, temples, and churches. Sometimes, the designed structures were not strong enough and failed. But in most cases, they were made very conservatively, that is, they were too strong for their needs. For example, in my ancestral house in a remote village of northern India, the walls are about 600 mm thick, made of small high strength baked bricks. It has already survived the weather of 240 years and in my assessment, it will last at least for another 500 years. The builders probably had no quantitative knowledge about the strength of walls and made them too conservatively. The modern philosophy of product design is that we analyze and design a structure that is safe and economical. In fact, we use just enough materials and resources required to meet the specifications. However, there are always some unexpected additional forces. We thus apply an appropriate factor of safety to consider unexpected situations. Many scientists and engineers have been working over the years to develop appropriate material testing equipment and to develop analytical tools which can predict the failure conditions quite well. The first documented evidence of quantitative work 503

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in this field was carried out by famous Italian artist and scientist, Leonardo da Vinci (1452–1519). Since then, the field has developed gradually to a very sophisticated analysis of modern times. The research and exploration into an untapped territory is a slow process as it usually takes a roundabout route to reach an elegant analysis. This point will be elaborately discussed in Section 15.3 on the bending problem which evolved very gradually. The field of mechanics of materials is very vast and a lot has been developed since Leonardo’s work. 15.2

Axial Force and Uniaxial Testing

Leonardo was very creative in developing a test machine to determine the strength of a wire. He suspended the specimen wire from a high point, as shown in Fig. 15.1(a). A basket was tied to the lower end of the wire. Dry sand was poured slowly into the basket from a hopper using a mechanism attached to its pouring mouth. As soon as enough sand was collected in the basket to break the wire, the mechanism would close the mouth of the hopper. On the floor, a resting pit was made just under the basket to avoid excessive tilt and spilling of the sand collected inside. The strength of the wire was the combined weight of the collected sand and the basket. Leonardo was a scientist in its true sense. He repeated the experiment several times, every time with a fresh wire, to ascertain that the variation was within a small limit. Leonardo was wonderfully talented. In fact, he designed many machines in his lifetime. I visited an exhibition in Venice, Italy displaying the models of the machines based on the design of Leonardo. I was amazed to see his vast work in engineering. He did not publish his work but his notes are available. He is now regarded as the father of experimental mechanics by some people. Another famous Italian, Galileo Galilei (1564–1642), who did pathbreaking work in physics and astronomy, also contributed a lot to the mechanics of materials. Galilei published a book “Two New Sciences” in 1638 in which a portion was devoted to the mechanics of materials. His work is considered to be the first publication in this field. He designed a test set-up to determine the tensile strength

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Specimenwire

Hopper Opening mechanism Basket

(a)

(b)

Fig. 15.1 (a) Leonardo’s test set-up to determine the strength of a wire and (b) determination of tensile strength of bars by Galilei.

of bars, as shown in Fig. 15.1(b). Galilei then used this strength to determine the strength of a beam; we will come back to it in Section 15.3. Further, Galilei correctly figured out that the strength of a bar was proportional to its area of cross-section and did not depend on the length of the bar. Robert Hooke (1635–1703), an English scientist, developed watches based on springs. Until then, watches were pendulum based using Earth’s gravitational force. Hooke’s watches not only were compact but also could be used in any orientation. While studying the deformation behavior of springs, he conducted tests on metal wires and found that the stretching was proportional to the applied axial force. His contribution, involving deformation of the material, was very significant because most investigators until that time were mainly concerned with the strength of a material. His finding is now considered as the beginning of the elasticity. To honour him, stress– strain relations are now known as Hooke’s law although the definitions of stress and strain tensors were formulated much later. Edme Mariotte (1620–1684), a Frenchman, carried out tensile tests on several materials like wood and paper. He realized that the linear behavior between force and stretching was valid up to a certain limit and thus developed the concept of proportional limit. He was in charge of supplying water to the Palace of Versailles through piping. He developed a testing procedure to determine the

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strength of a thin walled cylindrical drum. High internal pressure within a circular drum was generated using a high-water column, as shown in Fig. 15.2(a) and the bursting pressure was determined. He inferred that the bursting wall thickness t required for a pipe was proportional to the applied pressure p and the diameter of the pipe d. His inferences were correct as we know now (Section 9.7.1) that the hoop stress is governed by the relation σθθ = pd/2t. Rephrasing it to t = pd/(2σ θθ ) gives t ∝ p and t ∝ d. Petrus van Musschenbroek (1692–1761), a Dutch scientist, developed several test machines to determine tensile, compressive, and flexural strength using a lever system to magnify the applied load considerably (Fig. 15.2(b)). His results were widely used by the practicing engineers. C. A. Coulomb (1736–1806), who contributed enormously to many fields of science, also contributed immensely to our field. He developed a test machine to determine the compressive strength of a brick, an important result in those days for making brick buildings. He also experimented with the elastic limit of a material and found that heat treatment changed the elastic limit considerably. In addition, he found that the linear relation between force and the resulting displacement, now known as modulus, did not change with

Water column

Fulcrum link

Drum

Dead weight

Specimen

(a)

(b)

Fig. 15.2 (a) Determination of bursting strength of a thin wall cylindrical drum and (b) Muschenbroek’s test setup to determine tensile strength.

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heat treatment. He argued correctly that the relation was a characteristic of the arrangement of molecules of the material. The atomic theory was not well-developed in his time and, therefore, his inference was quite remarkable. His contribution toward obtaining the right formula for determining the bending strength was amazing and we will elaborate on it in the following section. Thomas Young (1773–1829), an English man, was a very intelligent scientist, but lived in his own world and did not care much about whether his lectures and writings were understood by his contemporaries. Although he worked on several important topics of mechanics of materials, he was not given due credit. He was the one who advocated that the applied force should be divided by the cross-sectional area and the change in length by the original length (ΔL/L). This realization of Young was a milestone. We no longer need to specify the details of the dimensions of the tested specimen and the results can be used in a widespread way to design various structures. This led to the definition of stress and strain which is an important realization to carry out the calculations of elasticity. The ratio of these two processed numbers, stress and strain, was called modulus. Since it was an important formulation, many engineers in modern times still call it Young’s modulus. Lagerhjelm, a Swedish physicist, worked extensively with all kinds of iron specimens and published his work in 1828. He showed that the modulus of a material was independent of the technological process employed such as rolling, hammering, and heat treatment, but the ultimate strength and elastic limit could be changed radically by these processes. For example, EN 24 is a very widely used alloy steel these days because its yield stress can be changed from 200 MPa to 1,700 MPa through various heat treatments. The modulus and the strength are very different and independent material properties. The modulus characterizes the molecular structure and tells us how much a structure will deform under given loading conditions. On the other hand, the yield stress or ultimate strength indicates how strong the material is, whether it is likely to fail or not. I find that many students confuse between these two properties, probably because a stiffer material (high modulus) is usually stronger too. It need not be so. For example, some polymers can be made almost as strong as mild steel but have a much smaller modulus.

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Navier, a French scientist, contributed immensely toward the development of the field of mechanics of materials. He emphasized, in his book published in 1826 on strength of materials, that it was important to know the limit up to which a material behaves perfectly elastic and suffers no permanent deformation. Beyond the elastic limit, the relations between force and deformation become very complicated and no simple formula could be formulated to estimate the ultimate strength. Navier suggested that structures should be designed to have only elastic deformation. This practice is being followed even today to design structural components. However, in the last 60–70 years, some designers have started allowing a limited plastic deformation in a structural member. S. D. Poisson (1781–1840) was interested in the theory of elasticity based on molecular structure. He found that there was a lateral contraction when a prismatic bar was pulled. He suggested lateral contraction yy as μxx where xx is the strain in the pulling direction. He found μ = −0.25. With better instrumentation and more accurate test machines, we can now determine it more accurately and is found to be between 0.27 and 0.38 for metals. In a uniaxial tensile test, we yy in which the negative sign now define Poisson’s ratio as ν = − xx has been introduced to have positive value of ν. Thus, in an isotropic material, the stress–strain relations are usually expressed with two material constants, Young’s modulus E and Poisson’s ratio ν. Professional working in this field felt that the test methods should be standardized so that results of one laboratory could be compared with those of other laboratories and designers need not do their own testing. They can just pick up the data from the standardized tests. J. Bauschinger (1833–1893) arranged a conference in Munich in 1884 which was attended by experts coming from many countries. Several committees were formed to draft the standards of various tests. In the follow-up meeting in 1886, the specifications of standardized tests were approved. 15.3

Bending of a Beam

As already stated several times in this book, a bending moment is the most dangerous force in comparison to other generalized forces in many field applications. Thus, people realized the importance of

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analysing the effects of bending moment even in the era much earlier than the Industrial Revolution. However, its formulation was quite difficult and it took a long slow evolutionary process of three centuries to come up with the right formulas. Leonardo da Vinci naturally got interested to look into the important and useful problem of bending around 1500 CE. In fact, the documentation on the quantitative approach started from his work. He considered a beam simply supported at its ends (Fig. 15.3). He realized that the strength of the beam was inversely proportional to its length L and also directly proportional to width b. Let us check whether he was right by solving it in terms of the modern way. The maximum bending moment is at the load point with bending moment PL/4. The stress at the outer fibre is σxx

PL h   Mb h2 × 3 PL = − 4  32 = − =− h I 2 bh2 b 12

The critical load Pc depends on the critical value σc as Pc =

2 bh2 σc 3 L

Thus, Leonardo’s inferences were correct. The critical load is proportional to the width of the beam and inversely proportional to the length for members of the rectangular cross-section. However, his notes are silent on the dependence of strength on the depth h. The early investigators applied their analysis on a test problem, a simple cantilever. We now are going to discuss it in detail. P b h

L/2 L

Fig. 15.3

A beam, supported at the ends.

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Test problem

The analysis was made on a cantilever beam of rectangular crosssection, as shown in Fig. 15.4. Rectangular sections of wood or stones were commonly used in those days. A force P was applied in downward directions at the free end as shown. The concepts of stress, moment of inertia, etc., were still not developed. Galilei tried an interesting way of finding bending strength. The only result available in those days was the strength S of a prismatic bar loaded in tension, as shown in Fig. 15.5(a). The same prismatic axial member is made to act as a cantilever beam with transverse end load P (Fig. 15.5(b)). What is the strength P of the beam in terms of S? In other words, express P in terms of S. y

z

x

b

h

L Fig. 15.4

P

The test problem.

L L

A B

P

S (a)

(b)

Fig. 15.5 (a) A prismatic member loaded in tension and (b) same member loaded in bending with end load P.

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In modern terms, the strength of the beam is evaluated by the following expression: σu = −

(−P L) × Mb y =− bh3 I

h 2

12

Substituting σu =

S bh

=6

PL bh2

(15.1)

and rearranging, we obtain  P =

S bh

 ×

bh2 6L

or, P =

Sh 6L

(15.2)

This relation expresses the maximum load P in terms of tensile strength S of the same member. Our focus will be on the factor (1/6) in the relation. Galilei obtained it as (1/2). Later on, other investigators improved it to (1/3). Only Coulomb in 1773 found it correctly as (1/6). We will obtain the same result by adopting the technique of the early investigators. The approach is indirect and may look strange to you. But we put ourselves in the shoes of the early investigators to study the evolution and appreciate it. The member of Fig. 15.4 is analyzed through Fig. 15.6. At section AB, the fibre at point A would be loaded to the same extent as in the case of axial strength S. It cannot be higher; otherwise, failure will be initiated at this point and would start growing toward point B. For this symmetric rectangular cross-section, the neutral axis is at half the height. Due to the linear elastic behavior of the material, the distribution of the internal force in the upper half is triangular. Thus, the upper half of the cross-section is able to take force equal to S × 1 1 S 2 × 2 = 4 ; first (1/2) is due to the fact that only half of the total area is taking the force and the second (1/2) is because the distribution is triangular and the average value is half. The resultant force in the upper half acts at the point which is at the distance of  two-thirds  . h/2. Thus, the moment of the upper portion is S4 × 23 × h2 = Sh 12 For this symmetric cross-section, the lower half applies the same moment Sh 12 and in the same direction. Thus, the total moment due

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L

S A 2 xh = h 3 2 3

F=S/4

h/2 h

F=S/4 B

P Fig. 15.6

Solving the test problem using the technique employed in those days.

Sh to internal force is 2 × Sh 12 = 6 . The moment at the cross-section AB due to load P is PL. Equating the two moments, we obtain

PL =

Sh 6.

→

P =

Sh 6L

This result is the same as obtained through the modern formula. 15.3.2

The evolution

Galilei, in his book “Two New Sciences”, published in 1638 CE, made the first attempt to come up with a bending formula. He correctly argued that the bending moment was largest at base AB (Fig. 15.7(a)). He also argued correctly that an internal bending moment is developed at AB, which was equal to an external bending moment. Further, he argued that the internal bending moment was developed through tensile loading at AB. They were important realizations. It is worth noting here that in the case of the axial loading, the internal loads are also in the direction of pulling force and it was S

L

S

A

F=S

L A

h F=S/2

h/2 B

P (a)

h

2h/3 B (b)

P

Fig. 15.7 (a) Uniform distribution of internal forces taken by Galilei and (b) triangular distribution taken by Mariotte in his earlier work.

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intuitively obvious. However, in the bending problem, the external load is in y-direction, which in turn, generates a bending moment. The bending moment is then resisted by tensile internal forces in x-direction. The question was: How could one determine the internal moment? Galilei did not take the correct distribution of internal tensile force at AB. He assumed the distribution of internal force to be uniform on the entire cross-section, as shown in Fig. 15.7(a). Also, Galilei argued that the cantilever rotates about point B, probably based on observing the failing beams in which the fibres start breaking at A and then the entire beam rotates about point B as the failure propagates. For the uniform internal force distribution, the internal moment is force times half the distance (h/2). Thus, the internal bending  moment becomes S × h2 . It is equal to the external moment PL. Thus, PL =

Sh 2

or, Sh 2L The form of this bending formula was correct, except that the factor (1/2) was three times higher than the correct value. For another 135 years, scientists worked on improving this factor. Galilei laid the right foundation for future investigators. However, Galilei did not balance the forces on the free body diagram of the cantilever beam. He did not realize that the net internal force on section AB should be zero in x-direction; the only net internal force on section AB is P in y-direction which is accounted for by the shear stresses developed on section AB. However, the analysis of Galilei was pathbreaking. In spite of the very limited development of mechanics of materials during his days, he came up with such a clever and creative scheme to assess the bending strength. I salute him. Robert Hooke (1635–1703) realized that fibres on the convex side were extended and those on the concave side were compressed. However, he did not do anything to improve the result on the bending strength. P =

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Mariotte (1620–1684) modified Galilei’s expression. He also took a rotational point at B but argued that the fibres at point A were stretched most and thus they took higher internal force. Since the material was linear elastic, he assumed the distribution of internal forces as triangular, as shown in Fig. 15.7(b). In his analysis, the net internal force was S/2 and it acted at a distance (2/3)h from point B. Thus,     2 Sh S × h = Internal moment = 2 3 3 External moment = PL Equating these two relations, we obtain P =

Sh 3L

This result was an improvement on Galilei’s expression, P = Sh/2L. However, P was still two times larger than the correct value. Mariotte also did not have the right concept of balancing the forces in the horizontal direction at cross-section AB on the free body diagram. Mariotte also realised later that the upper fibres were in tension and the bottom fibres in compression. He got the correct distribution of internal forces, as shown in Fig. 15.8. Due to the symmetric rectangular cross-section, the upper half takes the tension and the lower half compression. He argued that the upper half took force which  2 S/2 h was away from the centre point C by the distance 3 × 2 . Thus, the     upper portion contributed moment equal to S2 × 23 × h2 = Sh 6 . The L

S A 2 xh = h 3 2 3

F=S/2 Incorrect

h/2 h

F=S/2 B

P Fig. 15.8

Triangular distribution by Mariotte in his later work.

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lower half of the cross-sectional area applies the same moment with the overall moment of the internal force as Sh/3 which, in turn, gives P = Sh 3L . Mariotte was on the right path but made an error in the above calculations. The net force in the upper half was S/4 not S/2 because it was acting on half the total cross-sectional area. The incorrect result thus obtained by Mariotte survived for a long time and was used by practising engineers till Coulomb came up with the correct result in 1773. However, Antoine Parent (1666–1716), a French scientist, came up with the correct analysis in his memoirs of 1705 A. He took the correct expression for internal moment on the tensile side as well as on the compression side. He was not an experimentalist and was not aware that the material elastic behavior (modulus in modern terminology) was the same in tension and compression. He did, however, resolve that the net force on tension should be equal to the net force in compression. In other words, he found out the location of the neutral axis. Parent also pointed out that there should be a shear force acting on a plane equal to the external force. He thus had a fairly good idea of the free body diagram. Parent solved the problem correctly and completely. However, his work was not taken seriously and the practicing engineers did not have access to the correct analysis for 68 more years. The right analysis was accepted only after Coulomb analysed the problem correctly. Why was Parent not taken seriously? The readability of Parent’s writings was not good and he did not publish his work in reputed journals. Instead, he published his own memoirs which were poorly edited with many mistakes. Further, Parent was too critical about the works of his contemporaries and thus antagonized them. Even in a technical field, one should learn how to market his original work. About 68 years later, Coulomb (1736–1806) published the correct analysis of the bending problem in his paper of 1773 and obtained P = Sh/6L for beams with rectangular cross-sections. Most probably, Coulomb was not aware of Parent’s work as he did not refer him. Coulomb clearly stated that the sum of compressive internal forces should be equal to the sum of tensile forces. This facilitated the determination of the neutral axis; it should pass through the centroid of the cross-section and the moments should be taken about the neutral axis. He also stated that the vertical force on section AB should be equal to external force P . He noted correctly that the

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effect of shear force at section AB was negligible while determining the bending strength. Navier (1785–1836) was probably not aware of the works of Parent and Coulomb. In his 1819 paper, he advocated that the neutral axis should be found out by equating the moment caused by internal tensile forces with that of compressive forces. This model worked well on simple and symmetrical cross-sections but would not work on other cross-sections. However, Navier soon realized the mistake in his model. In his publication of 1826, he stated that the neutral axis must pass through the centroid of the section. Navier found out that a plane section, which is normal to the neutral axis in the undeformed condition, remains plane and normal after deformation when a pure bending moment was applied. This assumption works well and all introductory modern books on mechanics of materials invoke it to derive the bending formula. If we recall the development of the bending formula in Chapter 10, we also used the assumption that a plane normal to the neutral axis remains plane and normal after deformation. I find that most students remember the bending formula to find bending stresses, but do not remember this crucial assumption. It is important to remember the assumptions made in formulating a problem. In this fast-changing world, the same assumptions and the resulting formulas may not be valid for some newly developed materials. Then, we need to seek another set of assumptions appropriate for a new material. In the same paper of 1826, Navier found that the curvature of the deformed beam was proportional to a bending moment. He expressed the curvature κ of the beam as Mb κ= EI where I is the moment of inertia and E the modulus. The curvature κ was known in terms of lateral deflection v as d2 v dx2

κ=   dv 2 3/2 1 + dx Navier argued that if the slope

dv dx

κ=

 1, one can write d2 v dx2

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Then, the deformation of the beam is given by the relation Mb d2 v = 2 dx EI We used this equation in Chapter 11 to determine the deflection of a beam. Navier was the first one to evolve the method of analysing statically indeterminate problems. In these structures, the equilibrium equations alone cannot solve the problem. But Navier realized that indeterminate structures have constraints on deformation. Knowing the relations between stresses and deformation, the geometrical constraints yield additional equations. For example, a beam fixed at one end and simply supported at the other end (Fig. 15.9) cannot be solved using only the equilibrium equations. But at point B, the deflection in y-direction is zero, yielding an additional equation and the problem can be solved completely. 15.3.3

Summary on the evolution of bending analysis

The serious and scientific analysis to determine internal forces and deflection of a bending beam was started by Leonardo in 1500 CE or so. He obtained partial results stating that the strength of a beam was proportional to its width and inversely proportional to its length. Galilei published an analysis in his book “Two New Sciences” in 1638. He gets the credit of finding the correct form of the strength P = Sh 2L for a cantilever beam of rectangular cross-section. However, his assumption on the internal distribution of tensile forces at the base plane of the cantilever beam was incorrect. Thus, he obtained factor (1/2) in place of the correct factor (1/6). The analysis was further improved by Mariotte with better distribution of internal P

y A

B

x L/3 L

Fig. 15.9

A problem of indeterminate structure.

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distribution, although it was still incorrect. The factor was improved from 1/2 to 1/3. Parent in 1705 came up with the correct analysis, but he was not taken seriously. Mariotte’s incorrect results were continued to be used by professionals till 1773 when Coulomb came up with the correct analysis and stated how to find the neutral axis. Navier in his publication of 1826 derived the bending formula to determine internal stresses the way we have used in this book by stating that a plane and normal section remains plane and normal after deformation. Navier also showed how to find the deflection of a beam subjected to a pure bending moment. We have used the same methodology in Chapter 11. Navier was the first one to develop the technique to solve statically indeterminate structures. 15.4

Shear Stress and Stress Tensor

To find internal forces at an internal point, say H, we take an imaginary section passing through the point. We then save one of the two portions and determine forces applied by the other portion which has not been saved. In Fig. 15.10, we focus on the left potion. The traction (force per unit area) at point H is not normal, in general, to the cut plane. Cauchy (1789–1857), a French mathematician born in the year French Revolution started, pointed out in 1823 that the traction is made of both normal and shear components. Until his realization, the investigators were heavily influenced by the fluid mechanics where the internal pressure is normal to a surface and the shear component is negligible. Thus, Cauchy realized the presence of six shear components, τxy , τyx , τxz , τzx, τyz , and τzy . Then, Cauchy also proved that the shear components were symmetric. He then P2

y C B

P1

H

C

B

t

x

x

M2

t z

Fig. 15.10

M1

A

P3

H

A

A member is sectioned by a plane passing through point H.

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formulated three normal strain components and three shear strain components. A shear strain component was defined as the change in an original right angle. He proceeded further to obtain elastic stress– strain relations. If we look up the achievements of Cauchy, we find that he contributed immensely to mathematics. Sometimes, historians mentioned briefly that he also contributed to the mechanics of materials. Some other historians do not even mention his contribution to the mechanics of materials. To them, he contributed enormously to the field of mathematics and the contribution to mechanics of materials was his minor work. To us, the minor job is a huge contribution to our field in laying the foundation of the mechanics of materials. Cauchy’s formulation of shear stress components and strain components led to the development of stress and strain tensors, discussed in Chapter 5. 15.5

Shear Force and Shear Stress Distribution

Galilei did not realize that a shear force should act on the base plane AB (Fig. 15.7(a)). Parent (1705) realized it, for the first time, that a shear force on the base plane existed whose magnitude is equal to external force P. Coulomb (1773) conducted experiments to determine shear strength and reported that it was equal to the ultimate tensile strength. Although the results were not correct, he gets the credit for realizing that shear load was also important and a structural member may fail under a shear load. He also found out correctly that the effect of shear force on bending analysis was negligible if the depth of the beam is much smaller than the length of the beam. It is worth noting here that the formulation of shear stress was not developed well until Cauchy defined the shear stresses properly in 1823. Vicat (1786–1861), a French engineer, tested several metals to find their shear strength. He showed that shear force becomes important for beams of short length (1833). His work drew attention toward the importance of shear forces. Saint-Venant (1797–1886) showed a plane section remains plane only for beams subjected to pure bending moment. When shear force exists, the cross-section does not remain plane after deformation. Thus, a plane section warps. He further showed that warping is the

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y M Ml Nl

N

V x

z

Fig. 15.11

A section of a beam subjected to shear stress.

same for any two cross-sections. Thus, it does not produce any change in length of a fibre, as shown in Fig. 15.11. Line MN moves to M N with no change in length due to shear deformation. Jourawski (1821–1891), a Russian engineer, worked extensively with wooden beams and built-up wooden beams made by bonding planks of wood together. He observed that beams were failing due to shearing stress close to the neutral axis. He made calculations in 1844 to determine shear stress at the neutral surface by considering the free body diagram of the upper half of a cantilever made of a rectangular cross-section (Fig. 15.12). In fact, he considered the cantilever, similar to the one used by Galilei. Carrying out the force balance in x-direction, one obtains shear force T acting on the bottom face as 1 T = σmax (bh/2) 2

(15.3)

where factor (1/2) is for the average stress on face CA, b is the width of the beam, and (bh/2) is the area of cross-section. If τ is the average y

P

L

b

A h/2

C

x

T

C

h/2

B

Fig. 15.12 section.

Considering the top half of a cantilever beam of rectangular cross-

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shear stress on the bottom face CC , the total force T becomes T = τ bL. Substituting it in Eq. (15.1), one obtains   h 1 τ = σmax 2 2L

(15.4)

σmax is determined using the bending formula as:     P L h2 Mb h2 6P L =  3 = σmax = bh I bh2 12

Substituting σmax in Eq. (15.2), we have     3P h 1 6P L = × τ= 2 bh2 2L 2 bh Thus, Jourawski obtained the correct expression for shear stress at the neutral surface. However, the analysis was not general enough, but it turned out to be a breakthrough and drew the attention of experts in the field. The rigorous analysis for determining shear stress was developed by Saint-Venant in 1856. The relation between shear force and bending moment, V = dMb − dx , was developed in 1851 by Schwester (1823–1894), a German civil engineer who designed many public buildings and bridges. Using this relation, Schwester showed that a bending moment reaches its maximum value at the location where the shearing force changes its sign. The determination of shear stress distribution was relatively more difficult and it took about 200 years after the first attempt of Galilei in analyzing the internal forces in a cantilever beam. 15.6

Torsion of a Prismatic Bar

Coulomb, as early as 1784, realized the importance of the torsion of a structural member. He performed an interesting experiment by

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supporting a metal wire of circular cross-section from a high point. He attached a cylindrical shaped weight at the lower end. Studying the torsional vibration, he found out that the torsional moment Mt was proportional to the angle of twist φ and developed the correct relation: Mt =

μd4 φ L

where d is the diameter of the wire and L is the length. By his time, the shear modulus was not developed and, therefore, he called μ a material elastic consultant. Thomas Young (1773–1829) pointed out that, in bars with circular cross-section, applied torque is mainly balanced by shearing stresses acting on the cross-sectional plane and the shearing stresses are proportional to the distance from the axis of the member. Saint-Venant (1797–1886), the famous French mathematician, carried the work forward and solved the torsional problem for circular and noncircular cross-sections in 1853. 15.7

Rigorous Analysis of Advanced Levels

There are two basic approaches to solve problems of this field: 1. Mechanics of materials (also known as strength of materials), 2. Mechanics of solids (also known as elasticity; continuum mechanics). In the first approach of mechanics of materials, a relatively simpler but approximate process is adopted. In the analysis, the deformed shape of the structural member is assumed. Based on these assumptions, we evaluate strain at a point in terms of an unknown. For example, in Chapter 11, for bending problems, we assumed that an original plane section, which is normal to the axis of the slender member, remains plane and normal after it is deformed. This immediately

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Historical Perspective

enables us to determine axial strain ⎡ y −ρ ⎢ xx = ⎢ ⎣ 0 0

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at all points. It becomes ⎤ 0 0 ⎥ 0 0⎥ ⎦ 0 0

where ρ is the radius of curvature of the deformed beam. It is to be noted here that ρ is still unknown and will be evaluated later. Knowing the strain, we evaluate the stress tensor at all points using elastic stress–strain relations. Then, we invoke the equilibrium conditions which gives the location of the neutral axis and relation between the bending moment and ρ. The problem is now completely solved. In such an approach, the complexity of mathematics is not high but visualization is required. The problem of torsion of circular cross-section was solved the same way in Chapter 12. Again, it was assumed that a plane section normal to the axis remains plane and normal after deformation. The strain field is then expressed in terms of an unknown, the angle of rotation per unit length, α. The strain field is then converted to stress field and the equilibrium equations yield the value of α in terms of the torsional moment. The simpler approach is capable of solving only simple problems. It cannot even solve the torsion of the square bar because the plane section does not remain plane after deformation. Even the simpler approach seems to be difficult to some beginners. The concept of stress with six independent components is abstract. The students have problems, especially with shear stress components. They find visualization difficult and some of them are not able to distinguish τxz from τyz easily. Probably because of this, almost all books on the introductory level teach this subject in 2D dealing only with three in-plane stress components σxx , σyy , and τxy . However, the world is 3D and many real-life problems require 3D consideration. The level of analysis and sophistication is going up, especially when we make use of commercial software packages. Since the numerical analysis is capable of solving even complex problems in 3D, we should learn to visualize in 3D so as to make good use of the packages. The concept of drawing a stress element in 3D facilitates the visualization. In the second approach of mechanics of solids, no assumptions on the shape of the deformed body are made. Instead, there are field

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equations which are solved using appropriate boundary conditions. Two sets of field equations have already been developed in this book. They are as follows: 1. Strain–Displacement relations xx =

∂ux ∂x

yy =

∂uy ∂y

zz =

∂uz ∂z

γxy =

∂ux ∂uy + ∂y ∂x

γxz =

∂ux ∂uz + ∂z ∂x

γyz =

∂uy ∂uz + ∂z ∂y

2. Stress–Strain relations 1 [σxx − ν(σyy + σzz )] E 1 = [σyy − ν(σxx + σzz )] E 1 = [σzz − ν(σxx + σyy )] E = μγxy

xx = yy zz τxy

τxz = μγxz τyz = μγyz There is another set of field equations, equilibrium equations, which have not been developed in this book because they are hardly needed to solve problems through the mechanics of materials

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approach. The equilibrium equations are as follows: ∂τxz ∂σxx ∂τxy + + =0 ∂x ∂y ∂z ∂σyy ∂τyz ∂σxy + + =0 ∂x ∂y ∂z ∂τyz ∂σzz ∂τxz + + =0 ∂x ∂y ∂z and, τxy = τyx

τxz = τzx

τyz = τzy

There is another set of field equations which ensure that a continuous body remains continuous after deformation. For example, if there is a plate with no hole, it will not have holes after deformation. These field equations are called compatibility relations. In fact, there are many field equations to solve. In the advanced analysis, all the field equations and the boundary conditions are accounted for. Various solving techniques have been evolved in the last 200 years of work by many researchers. Further, handbooks have been developed which list the solutions to many different kinds of real-life problems. I personally consult the book “Roark’s Formulas for Stress & Strain by W. C. Young”. If the closed-form solution is not available for a case under consideration, we can always use a numerical technique like finite element analysis. These days, excellent software packages are available to determine stress and strain fields in a structural component. The basic knowledge of solid mechanics is useful in solving problems of many advanced topics, including fracture mechanics, fatigue, wear, creep, buckling, finite element method, stress waves in solids, plasticity, viscoelasticity, dislocation behavior, and vibration.

B1948

Governing Asia

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Index

A

cold working, 201–202 cold-rolled materials, 202 combined stress, 413, 415, 435, 454 compatibility, 219–220, 227, 236 compatibility relation, 228, 233, 237, 240, 358 compliance matrix, 177 compliant member, 26, 168 compression test, 212 conservative system, 465 constitutive relations, 209 Coulomb, C. A., 506, 515, 518, 521 critical point, 95, 436–437 critical stress, 436 cup and cone, 197 curvature, 261, 267, 322 cylindrical coordinate system, 114 cylindrical coordinates, 379, 385 cylindrical storage tank, 242

action and reaction, 7 anelastic deformation, 186 angle of twist, 381, 384, 387, 389, 392, 406 angular momentum, 14 axial force diagram, 62 B ball and socket joint, 29–30, 47 Bauschinger, J., 508 bending moment diagram, 69 bending stress, 259, 439 body forces, 30–31 brittle materials, 197, 210 buckling, 200 bulk modulus, 203, 206 C

D

case hardening, 202 Castigliano’s theorem, 462, 465–467, 477, 489, 494–495, 497 Cauchy, Augustin-Louis, 115–116, 151–152, 518–519 Celsius, Anders, 8 center of mass, 10 centroid, 222, 266, 424 centroid of a distributed force, 82 coefficient of thermal expansion, 167–168, 234, 251

da Vinci, Leonardo, 504, 509, 517 deformation, 3, 147, 149 degree of indeterminacy, 365 determinant approach, 443 direct integration method, 324 dislocations, 190–192, 195, 201 displacement leverage, 451 distance leverage, 320 distributed force, 82, 324 dog bone shape, 184 527

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ductile material, 196–198, 210 ductility, 199

gravitational force, 50 green function approach, 354–355

E

H

eigenvalues, 136–137, 139, 442 eigenvectors, 136, 139, 160 elastic curve, 321, 327, 329, 354 elastic deformation, 175 elastic modulus, 187 elastic recovery, 192–193 elongation, 199 engineering shear strains, 158 engineering strain, 151–152, 170 engineering strain components, 177 equilibrium equations, 15, 27 equivalent parameter, 439–440, 443, 455 Euler’s first law, 14 Euler’s second law, 14 Euler, Leonhard, 13 extensometer, 185 external forces, 50

heterogeneous materials, 202 homogeneous material, 202, 380 Hooke’s law, 505 Hooke, Robert, 203, 505 hoop stress, 243, 247, 251, 506

F

Joule, James Prescott, 8 Jourawski, D. J., 520–521

fictitious force, 488 first moment of the area, 432 flexural rigidity, 267, 322 force balance equations, 15 fourth-order tensor, 98 free body diagram, 23, 37–38 free surface, 111–112, 117, 163 G Galilei, Galileo, 504–505, 510–513, 517, 519 gauge length, 185 generalized forces, 11, 13, 23 generalized Hooke’s law, 176–177, 203 generalized internal forces, 419 geometrical compatibility, 230, 237, 251, 339, 356, 358 geometrical constraints conditions, 341

I in-plane tensile stresses, 245 indeterminate structure, 46–48, 227–228, 231, 251, 355–356 internal forces, 1–2, 55–56, 58, 95 internal generalized forces, 219, 377, 413, 431, 462 internal torsional moment, 377, 379 isotropic elastic materials, 176 isotropic material, 178–180, 188–189, 203, 205, 209, 234, 380 J

L Lagerhjelm, Per, 507 Lame’s constants, 203, 206 lateral forces, 258 laws of motion, 8–9 linear momentum, 8, 14 lower yield point, 195–196 M Mariotte, Edme, 505, 514 maximum shear stress, 129–130, 133–134, 138 method of superposition, 324 modulus, 186, 188, 192, 203, 205, 214, 330, 338 Mohr circle, 127–129, 131, 136, 140, 159, 170, 214 Mohr, Christian Otto, 127

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Index moment balance equation, 16–17 moment of inertia, 267, 330, 426 N Navier, Claude-Louis, 508, 516–518 neck, 197–199, 210, 212 negative shear stress, 108 neutral axis, 260, 437 neutral surface, 260–261, 321 Newton’s law, 7 Newton, Issac, 8 nominal strain, 185, 200 nominal stress, 185, 200, 211 nonconservative system, 465 O offset line, 190, 193 P Parent, Antoine, 515–516, 518 Pascal, Blaise, 9, 106 physical quantity, 121, 140 plane of symmetry, 112, 117 plastic deformation, 175 Poisson’s effect, 268, 321 Poisson’s ratio, 188, 203, 205, 210, 214, 508 Poisson, S. D., 188, 508 polar moment of inertia, 381, 383, 397, 426 polymer composite material, 178–179, 200, 203, 209, 268 positive shear stress, 108 principal directions, 135–137, 160, 162 principal strains, 160, 162 principal stresses, 128, 130, 134, 136, 139, 440, 442–443 proportional limit, 189 R radius of curvature, 261–262, 264 recrystallization temperature, 201–202 redundant force, 356, 358–359, 361, 363

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529 right-hand rule, 61 rigid body rotation, 148 rigid body translation, 148 rosette strain gauge, 165–166 S Saint Venant’s principle, 183, 392 Saint-Venant, 519, 521–522 scalar, 96–97 Schwester, J. W., 521 shape factor, 476–477, 496 shear blocks, 168 shear force diagram, 69 shear modulus, 178, 203, 205, 209–210, 214 sign convention, 59 significant digits, 40 spherical shell, 250 spring constant, 487 statically determinate, 46 statically indeterminate, 338–339, 345, 361, 369, 403, 415 stiffness matrix, 177 stiffness-based design, 448 strain energy, 320, 462–463, 465, 469, 472, 474, 476 strain energy density, 471, 495 strain gauges, 163–164, 189 strain hardening, 191–192, 210 strain tensor, 160 stress element, 102, 116 stress–strain relations, 175–177, 203, 209 superposition, 203, 223, 345–347, 349, 352, 355, 358, 369, 406, 413–414, 416, 435, 451, 455, 463 surface forces, 30–31 symmetric plane, 114 T tensorial shear strains, 158 tensorial strain, 151–152, 157, 170 tensorial strain components, 177 tensors, 96–97

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Mechanics of Materials: A Friendly Approach

thermal strains, 166–167, 170, 219–220, 233–236 thermal stresses, 220, 249 thin-walled cylindrical members, 241, 244 thin-walled spherical shell, 249 third-order tensor, 98 torsional moment diagram, 65, 395 torsional rigidity, 389 traction vector, 100, 105, 116 transformation of strain components, 158 transformation of stress components, 121 transformation relations, 124, 158 transition point, 323–325, 345 transverse forces, 258 Tresca yield criterion, 444–446, 455 true strain, 200–201, 210, 212 true stress, 200, 210, 212 truss, 36 Two New Sciences, 512 two-force member, 33–34, 43, 225 U ultimate tensile strength, 196–198, 210

uniaxial strain, 180 uniaxial stress, 179, 198–199 uniaxial stress test, 182 upper yield point, 195–196 V van Musschenbroek, Petrus, 506 vectors, 96–97 Vicat, Louis, 519 von Mises stress, 444 von Mises yield criterion, 443–444, 455 W Watt, James, 8 Y yield criterion, 440 yield point, 191, 195 yield points in mild steel, 195 yield stress, 189–190 Young’s modulus, 186 Young, Thomas, 186, 507 Young, W. C., 525

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