Engineering Mechanics 2: Strength of Materials: An introduction with many examples 3662665891, 9783662665893

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Christian Mittelstedt

Engineering Mechanics 2: Strength of Materials An introduction with many examples

Engineering Mechanics 2: Strength of Materials

Christian Mittelstedt

Engineering Mechanics 2: Strength of Materials An introduction with many examples

Christian Mittelstedt Technical University of Darmstadt Darmstadt, Germany

ISBN 978-3-662-66589-3 https://doi.org/10.1007/978-3-662-66590-9

ISBN 978-3-662-66590-9 (eBook)

Springer Vieweg © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Editor-in-Chief: Michael Kottusch This Springer Vieweg imprint is published by the registered company Springer-Verlag GmbH, DE, part of Springer Nature. The registered company address is: Heidelberger Platz 3, 14197 Berlin, Germany

Preface

This book originated from my lecture notes for the course “Engineering Mechanics 2”, which I hold for students of mechanical engineering in the second semester, but also for students of other disciplines at the Technical University of Darmstadt. The book follows the classical division of engineering mechanics as it is taught at universities in Germany and is dedicated to the determination of stresses and deformations in elastic bodies. Feedback of any kind is welcome anytime.

Darmstadt Autumn 2022

Christian Mittelstedt

v

Contents

1

Introduction to linear elasticity . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . 1.2 State of stress . . . . . . . . . . . . . . . . 1.2.1 Bar unter tension . . . . . . . . . 1.2.2 Stress vector and stress tensor . 1.2.3 Equilibrium conditions . . . . . 1.3 Deformation and strain state . . . . . . . 1.3.1 Bar under tension . . . . . . . . . 1.3.2 Infinitesimal strain tensor . . . . 1.4 The generalized Hooke’s law . . . . . . 1.4.1 Introduction . . . . . . . . . . . . 1.4.2 Three-dimensional material law 1.4.3 Temperature influence . . . . . .

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1 1 2 2 4 7 9 9 10 14 14 16 17

2

Plane stress state . . . . . . 2.1 Introduction . . . . . . 2.2 Stress transformation 2.3 Principal stresses . . . 2.4 Mohr’s stress circle .

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23 23 28 30 33

3

Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Bars . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Bar stresses . . . . . . . . . . . . . . . . 3.1.2 Bar deformations . . . . . . . . . . . . 3.1.3 Statically indeterminate bars . . . . . 3.2 Bar systems . . . . . . . . . . . . . . . . . . . . 3.2.1 Statically determinate bar systems . . 3.2.2 Statically indeterminate bar systems

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45 45 45 50 63 65 65 68

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vii

viii

Contents

4

Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Basic equations for an arbitrary reference frame . . . . 4.3 First cross-sectional normalization: Center of gravity C 4.3.1 Steiner’s theorem . . . . . . . . . . . . . . . . . . . 4.3.2 Selected elementary cases . . . . . . . . . . . . . . 4.3.3 Composite cross-sections . . . . . . . . . . . . . . 4.3.4 Stress analysis . . . . . . . . . . . . . . . . . . . . . 4.4 Second cross-sectional normalization: Principal axes .

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. 73 . 73 . 75 . 88 . 88 . 91 . 97 . 104 . 115

5

Beam deflections . . . . . . . . . . . . . . . . . . . . . 5.1 Basic equations of beam bending . . . . . . . . 5.2 Statically determinate single-span beams . . . 5.3 Statically indeterminate single-span beams . . 5.4 Multi-span beams . . . . . . . . . . . . . . . . . 5.5 Standard bending cases . . . . . . . . . . . . . . 5.5.1 Simply supported beams . . . . . . . . 5.5.2 Cantilever beams . . . . . . . . . . . . . 5.5.3 Statically indeterminate systems . . . . 5.5.4 Multi-span beams and angled systems 5.6 Biaxial bending . . . . . . . . . . . . . . . . . .

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127 127 129 137 140 153 153 155 158 165 169

6

Shear stresses in beams . . . . . 6.1 Introduction . . . . . . . . . . 6.2 Thick-walled cross-sections 6.3 Thin-walled cross-sections . 6.4 Shear center . . . . . . . . . .

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175 175 176 185 201

7

Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Solid bar with circular cylindrical cross-section . . . . . . 7.3 Thin-walled bar with circular cylindrical cross-section . 7.4 Bars with arbitrary thin-walled cylindrical cross-sections 7.5 Bars with open thin-walled cross-sections . . . . . . . . . 7.6 Determination of internal moments . . . . . . . . . . . . .

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207 207 209 211 213 219 224

8

Energy methods . . . . . . . . . . . . . . . . . . . . . . 8.1 Work and energy . . . . . . . . . . . . . . . . . . . 8.1.1 Introduction . . . . . . . . . . . . . . . . . 8.1.2 Internal and external work . . . . . . . . 8.1.3 Principle of work and energy . . . . . . . 8.2 Strain energy and complementary strain energy 8.2.1 The bar . . . . . . . . . . . . . . . . . . . . 8.2.2 The Euler–Bernoulli beam . . . . . . . .

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233 233 233 236 237 237 237 244

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Contents

8.3 8.4

8.5

8.6

8.7 9

ix

8.2.3 Bar under torsion . . . . . . . . . . . . . . . . . . . . . . . . . . 248 8.2.4 Combined load . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 Application of the principle of work and energy to the determination of elastic deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 The principle of virtual forces . . . . . . . . . . . . . . . . . . . . . . . 253 8.4.1 Formulation for the beam . . . . . . . . . . . . . . . . . . . . . 253 8.4.2 The unit load theorem . . . . . . . . . . . . . . . . . . . . . . . 255 8.4.3 Use of integral tables . . . . . . . . . . . . . . . . . . . . . . . . 259 The force method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 8.5.1 Determination of deformations of statically determinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 8.5.2 Statically indeterminate systems . . . . . . . . . . . . . . . . . 270 Reciprocity theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 8.6.1 Betti’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 8.6.2 Maxwell’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 279 Statically indeterminate systems . . . . . . . . . . . . . . . . . . . . . 280

Buckling of bars . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Types of equilibrium . . . . . . . . . . . . . . . . . . 9.3 Determination of critical loads . . . . . . . . . . . . 9.4 Buckling of bars: The four Euler cases . . . . . . . 9.4.1 Introductory example: Euler case II . . . . . 9.4.2 Euler case I . . . . . . . . . . . . . . . . . . . . 9.4.3 Euler case III . . . . . . . . . . . . . . . . . . . 9.4.4 Euler case IV . . . . . . . . . . . . . . . . . . 9.4.5 Summary of the results . . . . . . . . . . . . 9.5 Buckling length . . . . . . . . . . . . . . . . . . . . . 9.6 General form of the buckling differential equation

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289 289 290 292 297 297 300 302 305 306 307 312

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

1

Introduction to linear elasticity

The present chapter deals with the basics of linear elasticity theory and introduces the corresponding state variables, i.e. stresses, strains and displacements. The governing equations are the equilibrium conditions, the kinematic relations and the material law, here in the form of the generalized Hooke’s law. Students are enabled to derive and formulate the basic equations of linear elasticity theory and to formulate and solve simple fundamental problems.

1.1 Introduction In a solid (a so-called continuum) under load, a stress state will develop. Due to the load, deformations occur in the solid, thus displacements of the individual body points occur. Displacements are accompanied by strains, the so-called strain state is formed. For a three-dimensional body, the above quantities – displacements, strains, stresses – are described by the following set of equations under a given load and prescribed boundary conditions:  Kinematic equations: The so-called kinematic equations establish a relationship between the displacements and the resulting strains.  Constitutive equations: The so-called constitutive equations establish a relationship between the stresses and the strains. In the case of linear elasticity, the constitutive equations are described by Hooke’s law.  Equilibrium conditions: Equilibrium must be ensured at every point of the solid under consideration.

© The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_1

1

2

1

Introduction to linear elasticity

1.2 State of stress 1.2.1 Bar unter tension To motivate the concept of the stress state, as an introductory elementary example we consider the straight linear elastic bar unter tension as shown in Fig. 1.1, top. The bar has the length l and the cross-sectional area A and is loaded at its center of gravity by the tensile force F . The applied tensile force F causes forces inside the bar. To investigate this in more detail, we perform a vertical cut at an arbitrary point x of the bar (Fig. 1.1, middle). The forces acting inside the bar must be thought of as being uniformly distributed, and these distributed forces are called stress. It is common to use the symbol  for the stress. Stresses are given in a corresponding unit of a force per unit area, e.g. in the unit [N=m2 ]. It is also possible to use the unit Pascal, where 1 Pa D 1 mN2 . The stresses occurring in the given section are assumed to be constantly distributed over the cross-sectional area and also oriented normal to the cross-section, so that at this point we also speak of the so-called normal stresses  . The normal stresses can be summed up to the normal force N of the bar. Since the normal stress is constant across the cross-section, the following relationship holds between the normal force N D F and the normal stress  : D

N F D : A A

(1.1)

The sign of the normal stress  depends on the sign of the normal force N . If, as in this case, the bar is subjected to tension, the normal force N is positive, and thus, according to (1.1), the normal stress  will also be positive, i.e. it will be a tensile stress. If the normal force N is negative, then  is also negative and thus a compressive stress. We now consider the bar under tension of Fig. 1.1 again and make a cut at an arbitrary location x under an inclination angle  as shown in Fig. 1.2. The crosssectional area to be considered here is called AN and is calculated as AN D cosA  . Such Fig. 1.1 Bar under tension (top), normal stress  for a perpendicular cut at an arbitrary location x (middle), replacement of the normal stress by the normal force N (bottom).

A F

F x

l

F

F x

F

σ

σ N

F

1.2 State of stress

3

Fig. 1.2 Bar under tension (top), cut at an arbitrary location x under the angle , release of the normal stress  and the shear stress  (bottom).

A F

F x

l

F

A

F

x

a cut shows that not only a normal stress  occurs, but also a so-called shear stress , N Again, both stresses are which acts tangentially to the cross-sectional surface A. N We form the sum of assumed to be constantly distributed over the cross-section A. the horizontal forces at the left bar segment and obtain:  AN cos  C  AN sin   F D 0:

(1.2)

Analogously, the sum of the vertical forces results in:

Inserting AN D

A cos 

yields:

 AN sin    AN cos  D 0:

(1.3)

F ; A  tan    D 0:

(1.4)

 C  tan  D

These two equations can be solved for the two stresses  and  as follows: D

1 F ; 2 1 C tan  A

D

tan  F ; 1 C tan2  A

(1.5)

i.e. both stresses depend on the section angle . The shear stress  vanishes for the angle  D 0, as already assumed in Fig. 1.1, and reaches its maximum value F for  D 4 . The normal stress  reaches its maximum value max D FA at max D 2A  D 0. The above results, worked out for a very elementary example, already allow us to draw some important conclusions, which we summarize in a general way below:  The stress state is generally described by the simultaneous occurrence of normal stresses and shear stresses.  The stress state depends on the orientation of the considered cutting direction.  There are special stress states in which only normal stresses occur. There are also conditions (not applicable to the above example of the bar under tension) in which only shear stresses occur.

4

1

Introduction to linear elasticity

1.2.2 Stress vector and stress tensor We now consider an arbitrary solid as shown in Fig. 1.3. Let the body be subjected to arbitrary loads and to arbitrary boundary conditions. In order to describe the state quantities occurring herein, a spatial Cartesian coordinate system x; y; is introduced. We denote the associated displacements as u, v and w. They are generally functions of all three spatial directions: u D u.x; y; z/;

v D v.x; y; z/;

w D w.x; y; z/:

(1.6)

Loads on a solid can occur in various forms. First of all, there are the so called volume forces, which are distributed volumetrically in the body under consideration. They are given in the unit of a force per unit volume, e.g. in the unit [N=m3 ]. Volume forces can occur in all three spatial directions, their components are called fx , fy , fz , and they are summarized in the vector f : 1 fx f D @fy A: fz 0

(1.7)

Loads can also occur in the form of surface loads, which are in the unit of a force per unit area, e.g. in the unit [N=m2 ]. Surface loads can occur on the entire surface of the solid, or only on a part of it. Let the surface loads be given with their components tx , ty , tz , which are arranged in the vector 0 1 tx t D @ty A: tz

(1.8)

Loads can also be in the form of line loads, which have a corresponding unit of force per unit of length, e.g. [N/m]. In addition, point loads may also occur, which are given in the unit [N].

∆F n t

n

t z ,w

∆A

f

t x,u

y,v

Fig. 1.3 Three-dimensional solid under load (left), section through an arbitrary body point (middle), decomposition of the stress vector into normal and shear stress (right).

1.2 State of stress

5

If we now look at a solid under load, forces are generated inside of the solid which can be related to a cut surface. These are called stresses, as already explained at the example of the bar under tension. Stresses in a solid are thus in the unit of a force per unit area, e.g. in the unit [N=m2 ]. For further definition, we consider an arbitrary section through the solid body under consideration and investigate an infinitesimal sectional surface A (Fig. 1.3, middle). The spatial orientation of this surface is denoted by the normal vector n: 0 1 nx (1.9) n D @ny A: nz The force F is now released. The Stress vector is then defined as follows: t D lim

A!0

F : A

(1.10)

The stress vector depends on the orientation of the section and the surface under consideration. The stress vector t is therefore a function of the normal vector n, so that t D t .n/. The stress vector is now decomposed into a component parallel to the normal vector and a component tangential to the cut surface. These components are called normal stress  and shear stress , cf. Fig. 1.3, right. We have: p  D t  n;  D t  t   2 : (1.11) The stress state at a point of a solid is uniquely determined if the stress vector exists in three mutually independent sections through the body point. The normal vectors of these sections must be linearly independent. It is convenient to place these sections so that they are oriented perpendicular to the three coordinate axes x, y, z. In Fig. 1.4 an infinitesimally small cube is shown, which was cut out of the considered solid in this way. Once the stress vectors of these sections are known, they can be used to determine the stress vectors under any other directions. Now let three intersecting surfaces be defined in such a way that their normal vectors are identical with the coordinate axes x, y, z. The three stress vectors t x , t y , t z are then: 0 1 0 1 0 1 xx yx zx t x D @ xy A; t y D @yy A; t z D @ zy A: (1.12) xz yz zz The indexing of the stresses is defined as follows. The first index indicates the direction of the surface normal of the considered section, whereas the second index indicates the direction of action of the considered stress component. Accordingly, stresses with identical indices are normal stresses that are perpendicular to the section under consideration. Stresses with two different indices, on the other hand, are shear stresses, they are tangential to the considered section surface. Obviously, for

6

1

Introduction to linear elasticity

z

tz zz

yz

t

tx

xz

xx

x

xz

zy

zx

f

xx

xy

yx

yy

ty

yz

y dz

yy

yx

xy

zx

zy

zz

dy

dx

Fig. 1.4 Infinitesimal cube and associated stress components.

a complete description of the stress state in a body point, nine stress components are required, namely the three normal stresses xx , yy and zz and the six shear stresses xy , yx , xz , zx , yz and zy . The sign conventions for stresses are analogous to those for forces and moments in bars and beams: a stress is positive if it points in the positive coordinate direction at a positive cutting surface. A cutting surface is positive if the outward normal vector of the considered surface points in positive coordinate direction. In Fig. 1.5 the infinitesimal cube is shown in a top view. By formulating the moment equilibrium around the center of gravity of the cube about all three coordinate axes, it is straightforward to show that mutually associated shear stresses (i.e., shear stresses with identical but interchanged indices) must be identical. We show this by means of Fig. 1.5 for the two shear stresses xy and yx . The sum of the moments about the z-axis gives: 2xy dydz

dx dy  2yx dxdz D 0: 2 2

(1.13)

This yields: xy D yx :

(1.14)

Fig. 1.5 Moment equilibrium at the infinitesimal section element.

xx xy

z yx

y

C

yy dx

yy yx xy

x

xx

dy

1.2 State of stress

7

Analogously, one can form the moment sums around the other two coordinate axes which yield: xz D zx ; yz D zy : (1.15) As a conclusion, it can be stated that mutually associated shear stresses always both point towards or away from the common corner of a section element, as shown in Figs. 1.4 and 1.5. As a result of the above considerations, the number of stress components to be determined is reduced from nine to six. These are the three normal stresses xx , yy , zz and the three shear stresses xy D yx , xz D zx , yz D zy . It is common to summarize the stress components in a symmetric matrix  as follows: 3 2 xx xy xz (1.16)  D 4 xy yy yz 5: xz yz zz This is the so-called Cauchy1 stress tensor. It summarizes the stress vectors t x , t y , t z in columns. Due to the equality of associated shear stresses, the Cauchy stress tensor is always symmetric.

1.2.3 Equilibrium conditions In the following we want to clarify how the stress states of two infinitesimally distant body points are related to each other. For this purpose we consider the so-called local equilibrium conditions. Let us again consider an infinitesimal sectional element of the solid of Fig. 1.3 with the edge dimensions dx, dy, dz, and we apply the respective stress components to the cutting edges (Fig. 1.6). For reasons of clarity, the volume forces fx , fy , fz , which also have to be taken into account, are not shown here. We now assume that the stress components at the positive and negative cutting edges differ by an infinitesimal increment. We denote these increments as dxx , dyy , dzz , and dxy , dxz and dyz , as shown in Fig. 1.6. For the example of the normal stress xx this means that at the negative edge of the cut with respect to the x-direction the normal stress xx .x/ occurs, while at the opposite positive edge of the cut the stress xx .x C dx/ D xx C dxx would be applied. The infinitesimal increments can be developed as Taylor series which we terminate after the first term: @yy @xx @zz dxx D dx; dyy D dy; dzz D dz; @x @y @z @xy @yx @xz dxy D dx; dyx D dy; dxz D dx; @x @y @x @zx @yz @zy dzx D dz; dyz D dy; dzy D dz: (1.17) @z @y @z 1

Augustin-Louis Cauchy, 1789–1857, French mathematician.

8

1

Introduction to linear elasticity

Fig. 1.6 Local equilibrium at the infinitesimal volume element.

z

yx

yy

d

zx

zz

zx

zy

d

zz

d

zy

xx

xy

xz yz xx

xz

d

x

d

xz

xx

xy

d

yx

xy

d

yz

d

yz

yy

d

yy

xy

y

zx

zy

zz

We consider the force equilibrium in the x-direction and obtain by adding up all forces resulting from the stress components:   @xx xx C dx dydz  xx dydz @x   @yx C yx C dy dxdz  yx dxdz @y   @zx C zx C (1.18) dz dxdy  zx dxdy C fx dxdydz D 0: @z Summarizing and observing ij D j i yields: @xx @xy @xz C C C fx D 0: @x @y @z

(1.19)

In the same way, we can proceed with respect to the force equilibria in y- and z-directions, and we obtain the local equilibrium conditions as follows: @xy @xx @xz C C C fx D 0; @x @y @z @xy @yy @yz C C C fy D 0; @x @y @z @yz @xz @zz C C C fz D 0: @x @y @z

(1.20)

The equations (1.20) are three coupled partial differential equations for six unknown stress components. Thus, any solid problem is intrinsically statically indeterminate, and the three equilibrium conditions (1.20) are not sufficient to determine the six stress components. We thus have to use further equations.

1.3 Deformation and strain state

9

1.3 Deformation and strain state 1.3.1 Bar under tension We consider the linear elastic bar under tension of Fig. 1.1 with length l and constant cross-sectional area A and investigate its deformation (Fig. 1.7), which, caused by the tensile force F , consists of a length change l. As a measure of the deformation of the bar, we use the so called normal strain " as the ratio of the length change l to the initial length l. For the special case of a constant cross-sectional area and a constant normal force N D F we obtain: "D

l : l

(1.21)

The normal strain " has no unit. It is positive if the bar becomes longer (i.e. l > 0) and negative if the bar becomes shorter (i.e. l < 0). We will assume in all further explanations that the normal strain is small, so that jlj  l and thus j"j  1 holds. A more general definition of the normal strain of a bar can be obtained by considering the deformation behavior of an infinitesimal bar element (Fig. 1.8). For this purpose, we consider a sectional element of length dx under the displacement u.x/ at its left end at an arbitrary location x. At x C dx, on the other hand, there is the displacement u.x C dx/ D u C du, where the quantity du represents an infinitesimal increase of the displacement u. The length of the infinitesimal bar element in the deformed state is therefore dx C .u C du/  u D dx C du, the length change is therefore du. Thus, the ratio between the length change and the initial length can be expressed as: du : (1.22) "D dx Fig. 1.7 Bar (top), length change under tensile force (bottom).

A x

l

F

F x

Fig. 1.8 Infinitesimal bar element (top), deformation (bottom).

x

l

∆l

dx

x u

x dx+(u+du)-u

u+du

10

1

Introduction to linear elasticity

Thus, once the displacement u is known, the strain " can be obtained from it by differentiating once with respect to x. The displacement u and the strain " are kinematic quantities, and Eq. (1.22) is a so-called kinematic equation.

1.3.2 Infinitesimal strain tensor If a solid body is subjected to load, the body points experience displacements. We want to divide these into the displacement u in the x-direction, the displacement v in the y-direction and the displacement w in the z-direction. Hereby the so called strains are called forth, which we divide into normal strains and shear strains, respectively. This is illustrated at the example of a cube cut out of a solid and subjected to different stress conditions (Fig. 1.9). The material under consideration is isotropic. We first consider the case where the cube of Fig. 1.9 suffers a normal stress xx (Fig. 1.9, left). Then it results that the cube suffers a positive strain "xx in the x-direction (Fig. 1.9, top left), its length in x-direction will increase. At the same time a so-called transverse contraction will occur in the y-direction, i.e. a negative strain "yy , the cube dimension in the y-direction will

Normal strain

xx

Shear strain

z

y xx

z

yz

x

y

x

yz

Normal strain

yy

Shear strain

y

xz

z

xx

x

z

x

y

xz

Normal strain

zz

Shear strain

z xx

x

xy

y y

z

x xy

Fig. 1.9 Normal strains (left) and shear strains (right) on a cube cut out of a solid.

1.3 Deformation and strain state

11

become smaller (Fig. 1.9, middle left). Likewise, a transverse contraction will occur in the z-direction, i.e., a negative strain "zz , associated with a decrease in the cube dimension in the z-direction (Fig. 1.9, bottom left). The three spatial normal strains "xx , "yy and "zz are thus accompanied by a change in the dimensions of the cube, i.e. a volume change, but not by a change in its shape. We now consider the case where the cube is subjected to the shear stress yz (Fig. 1.9, top right). In this case, the right angles of the cube sides will change in the yz-plane as indicated. We refer to this angular change as shear strain yz . Analogously, in the three-dimensional case, the shear strain xz can occur in the xz-plane (Fig. 1.9, center right) as well as the shear strain xy in the xy-plane (Fig. 1.9, bottom right). Thus, for our purposes, a shear strain means that the cube changes its shape, but not its volume. To derive general expressions for normal strains and shear strains in the threedimensional case for the case of small deformations (so called geometric linearity), we consider Fig. 1.10. Shown here is an infinitesimal cuboid with edge lengths dx, dy, dz in top view, in the undeformed state (indicated by dashed lines) and also in two deformed configurations. We first investigate the normal strain "xx in the xy-plane and consider Fig. 1.10, left. Let the cuboid be subjected to a deformation such that its edges are elongated in the xy-plane, but its cuboid shape is preserved. The corners A, B, C , D then undergo various displacements and transform into the points A0 , B 0 , C 0 , D 0 . Let u and v be the displacements of point A in x- and in y-direction, i.e. uA D u, vA D v. Point B will undergo the displacements uB and vB , but here for uB an infinitesimal increment du has to be taken into account, so that the displacement uB of point B amounts to uB D u C du. The increment du can be represented as a Taylor series with respect to the x-direction which we terminate after the first term, i.e., the displacement uB of the point B in the x-direction becomes uB D u C @u @x dx. Its displacement in the y-direction, however, is vB D v as for point A. Quite analogously one can deduce the displacements of point C , and one obtains uC D u C @u @x dx dy, where here the infinitesimal increase for v was developed and vC D v C @v C @y

y

u+du=u+ ∂u ∂ y dy

y

C D

D

C

D

C

v+dv =v+ ∂v ∂ y dy

D

dy

v

C

dy

A

dx

u

v

B

A

B A

A

B

x

u+du=u+ ∂∂ux dx

dx

u

v+dv =v+ ∂v ∂ x dx

B

x

Fig. 1.10 Definition of the infinitesimal normal strains "xx and "yy (left) and the infinitesimal shear strain xy (right).

12

1

Introduction to linear elasticity

in the y-direction. The displacements of the point D follow analogously: uD D u, @v vD D v C @y dy. To obtain a statement about the normal strain "xx , the ratio of the length change with respect to the x-direction to the initial length is formed: "xx D

A0 B 0  AB AB

:

Therein, the edge length in the deformed state A0 B 0 can be represented as:   @u @u A0 B 0 D dx C u C dx  u D dx C dx: @x @x

(1.23)

(1.24)

With AB D dx, it then follows from (1.23):

"xx D

dx C

@u dx  dx @u @x D : dx @x

(1.25)

Quite analogously one can proceed for the strain "yy . We obtain: "yy D

A0 D 0  AD AD

:

(1.26)

With AD D dy and   @v @v A0 D 0 D dy C v C dy  v D dy C dy @y @y

(1.27)

it follows from (1.26): dy C "yy D

@v dy  dy @v @y D : dy @y

(1.28)

A very similar expression can be obtained for the normal strain "zz which, however, is not elaborated on in more detail at this point. The result is: "zz D

@w : @z

(1.29)

We now consider the situation of Fig. 1.10, right. Let the infinitesimal cuboid now be subjected to a deformation such that the cuboid vertices A, B, C , D shift as shown and the original shape of the rectangle formed by A, B, C , D changes into a parallelogram with the vertices A0 , B 0 , C 0 , D 0 . Let the two angles describing the

1.3 Deformation and strain state

13

deviations from the cuboid shape be called ˛ and ˇ. With the help of Fig. 1.10, right, we want to work out a statement about the shear strain xy and consider first the two angles ˛ and ˇ. From Fig. 1.10, right, we obtain: @v dx @x tan ˛ D ; @u dx C dx @x

@u dy @y tan ˇ D : @v dy C dy @y

(1.30)

Because of the smallness of the deformations we assume that the second terms in the denominator of the expressions (1.30) are negligible. Moreover, we can assume small angles with tan ˛  ˛, tan ˇ  ˇ, so that from (1.30): ˛D

@v ; @x

ˇD

@u : @y

(1.31)

The shear strain xy is then the sum of the two angles ˛ and ˇ: xy D

@v @u C : @x @y

(1.32)

The two indices xy indicate that xy is the change of the angle in the xy-plane. The indices x and y are interchangeable, so that xy D yx . Analogous expressions can be derived for the two remaining shear strains yz and xz , but this remains without representation here. In summary, the following definitions are obtained for the spatial normal strains "xx , "yy , "zz and the spatial shear strains xy , xz , yz : @u @v ; "yy D ; @x @y @u @v D C ; @y @x @u @w D C ; @z @x @v @w D C : @z @y

"xx D xy xz yz

"zz D

@w ; @z

(1.33)

They establish a relation between the displacements u, v, w on one side and the normal strains "xx , "yy , "zz and shear strains xy , xz , yz on the other side. Hence they are called kinematic equations. Once the displacements are known, the strains can be determined from them by differentiation. The term strains serves as a generic term for the normal strains and the shear strains.

14

1

Introduction to linear elasticity

It is convenient to summarize the normal strains and the shear strains in a symmetric matrix, the so-called infinitesimal strain tensor ", as follows: 2 3 1 1 "   xy xz 6 xx 2 2 7 6 7 61 1 7 6 (1.34) " D 6 xy "yy yz 7 7: 2 7 62 41 5 1 xz yz "zz 2 2 On its main diagonal the normal strains "xx , "yy , "zz are located. The remaining entries represent the half shear strains.

1.4 The generalized Hooke’s law 1.4.1 Introduction In all explanations of this book we will assume linear elastic material behavior. Accordingly, there is a linear relationship between the stresses and the strains that occur in a solid. In the three-dimensional case, this is described by the so called generalized Hooke’s law2 , which we will consider below. Many materials behave in a linear elastic manner in technically relevant load ranges, as shown in Fig. 1.11, left. A typical stress-strain diagram is shown here, as it would result, for example, in a tensile test for a steel specimen. Typically, there will be a proportional relationship between the applied tensile stress  and the resulting normal strain " until the so-called proportionality limit P is reached. If the stress is increased above this value, an overproportional increase of the normal strain occurs. When the yield stress Y is reached, the strain continues to increase at approximately constant stress. At this point, the material is said to be yielding. Beyond the yield point, it is found that the tensile specimen can take further stresses until at a certain point the specimen breaks. In all further explanations we will restrict ourselves to linear elastic material behavior. Accordingly, there is a linear relationship between the stress  and the normal strain ", as shown in Fig. 1.11, right. In the following, we will derive the

Fig. 1.11 Stress-strain diagram for a tensile specimen made of steel (left) and for an ideal linear elastic material (right).

2

Y P

Robert Hooke, 1635–1703, English mathematician and physicist.

1.4 The generalized Hooke’s law

15

Fig. 1.12 Sheet under tension (left) and under shear (right).

x

F

Cross-sectional area A

x

v

u

T xy

l

l y

b-v b

y

b

corresponding equations describing a linear elastic material behavior. Such equations are also called constitutive equations or material law. For motivation, let us consider the situation of Fig. 1.12 where a linearly elastic very thin sheet-like structure with dimensions b and l is given. Let the sheet be supported at its lower end by a horizontal support in such a way that the strain states considered below can occur unhindered. It is loaded by a tensile force F on the one hand, and by a force T acting tangentially to the surface on the other hand. We first consider the case of Fig. 1.12, left. The applied tensile force F produces the normal stress xx D FA in the sheet, where A is the cross-sectional area. The simplest case of Hooke’s law is in the one-dimensional form xx D E"xx :

(1.35)

Herein, "xx is the normal strain which will be positive in the present case. The quantity E is the so-called Modulus of elasticity, which establishes a relationship between the normal stress xx and the normal strain "xx of the sheet. The modulus of elasticity has the unit of a stress, e.g. [N=m2 ]. From Fig. 1.12, left, the normal strain "xx can be deduced as "xx D ul . Obviously, the modulus of elasticity represents the slope of the straight line in a stress-strain diagram (Fig. 1.11, right). In addition to the normal strain "xx in the x-direction, the sheet will also undergo a so-called transverse contraction, i.e. a normal strain "yy in y-direction, which will be negative (the sheet will shorten in this direction). The two normal strains "xx and "yy are related to each other by the so called Poisson’s ratio :3 "yy D "xx :

(1.36)

The Poisson’s ratio  is unitless and represents a further elementary elasticity constant. For example, for steel the Poisson’s ratio is  D 0:3. We now consider the situation of Fig. 1.12, right, i.e. the sheet under the force T acting tangentially to the surface. The resulting shear stress xy can be calculated as xy D TA if we assume that the shear stress is uniformly distributed over the crosssectional area A. Due to the applied load, the sheet suffers a horizontal deflection, 3

Siméon Denis Poisson, 1781–1840, French mathematician and physicist.

16

1

Introduction to linear elasticity

which assumes its maximum value at the upper edge with the value v. For simplicity, we assume that the vertical edges of the sheet are twisted by an angle xy , but remain straight even in the deformed state. This results in a shear strain xy corresponding to the occurring angle between the unloaded vertical edge of the sheet and the deformed edge, i.e. tan xy D vl . If we again assume small angles, then xy D vl follows. There is a relation between the shear stress xy and the shear strain xy as follows: xy D Gxy : (1.37) Herein, G is the so-called shear modulus. Like the modulus of elasticity, it is in the unit of a stress, e.g. in the unit [N=m2 ]. Isotropic linear elastic material behavior is thus described by the three elastic constants E, , G. They have to be determined experimentally. It can be shown that the following relationship exists between these three material values: GD

E : 2.1 C /

(1.38)

If, for example, the two material constants E and  are available from experiments, e.g. from a tensile test, then the shear modulus G can be determined from (1.38).

1.4.2 Three-dimensional material law With the equations (1.35), (1.36) and (1.37) the constitutive equations are available, with which the material behavior of the sheet of Fig. 1.12 can be described. In this section, we will now extend the considerations to a linear elastic isotropic homogeneous solid described by the material constants E, , G. For this we consider a cuboid taken from a solid body as shown in Fig. 1.13 which is stressed by the normal stress xx . The stress xx causes the normal strain "xx D Exx . Moreover, the transverse normal strain "yy D "xx D  Exx in the y-direction and the transverse normal strain "zz D "xx D  Exx in the z-direction result. Analogously, in the pres ence of the normal stress yy , both the normal strain "yy D Eyy and the two transyy  verse normal strains "xx D "yy D  E and "zz D "yy D  Eyy will occur. Finally, it is necessary to consider the case when the normal stress zz acts. In Fig. 1.13 Cuboid under normal stress xx .

z σxx

σxx

x

y

1.4 The generalized Hooke’s law

17

this case, the normal strain "zz D Ezz results. Moreover, the two transverse normal strains "xx D "zz D  Ezz and "yy D "xx D  Ezz occur. If all three stresses occur simultaneously, then the following expressions for the spatial normal strains "xx , "yy and "zz are obtained by superposition:   1 xx   yy C zz ; E  1 "yy D yy  .xx C zz / ; E   1 zz   xx C yy : "zz D E For the spatial shear strains xy , xz , yz we have: "xx D

xy D

xy ; G

xz D

xz ; G

yz D

yz : G

(1.39)

(1.40)

1.4.3 Temperature influence If the sheet of Fig. 1.12 is not stressed by forces or by the equivalent stresses, but if there is a temperature change T , then corresponding thermal strains are induced. In many cases it can be assumed that there is a linear relationship between the temperature change T and the resulting thermal strain "T as follows: "T D ˛T T:

(1.41)

The material constant ˛T is called thermal expansion coefficient. It is to be determined experimentally and is given in the unit C1ı . In the case where both a mechanical stress  and a temperature change T act together, the individual strain components can be superposed:  (1.42) C ˛T T: "D E Solved for the stress  this relation reads:  D E."  ˛T T /:

(1.43)

For the three-dimensional case we obtain by extension of (1.39) and (1.40):   1 xx   yy C zz C ˛T T; E  1 "yy D yy  .xx C zz / C ˛T T; E   1 zz   xx C yy C ˛T T; "zz D E xy yz xz xy D ; xz D ; yz D : (1.44) G G G Thus, a change in temperature only causes normal strains in the case of isotropic materials, but no shear strains. "xx D

18

1

Introduction to linear elasticity

Example 1.1

A linearly elastic isotropic cube (edge length a, modulus of elasticity E, Poisson’s ratio ) is placed in an ideally rigid enclosure (Fig. 1.14, top). The free surface of the cube is loaded by the constant surface load p0 . The stress state in the cube and the displacement u of the loaded surface in the x-direction are to be determined. In the case where there is no mechanical load but a constant temperature change T is applied, the displacement u is also to be determined. The cube can slide frictionlessly on the walls of the enclosure, which are assumed to be ideally rigid and smooth. Under load, a three-dimensional stress state will form in the cube (Fig. 1.14, bottom), where no shear stresses xy , xz , yz , but only the normal stresses xx , yy , zz occur. Similarly, no shear strains xy , xz , yz arise in the given situation. Due to the rigid wall of the enclosure, both strains "yy and "zz become zero. The normal stress xx corresponds to the negative of the applied load, i.e. xx is a compressive stress: xx D p0 . From Hooke’s law (1.39) follows: 0 D yy  .p0 C zz /;   0 D zz   p0 C yy :

(1.45)

From these two equations, the normal stresses yy and zz can be determined as: yy D zz D p0

 : 1

(1.46)

Both stresses yy and zz are thus compressive stresses, which is an obvious result due to the restrained transverse strain of the cube in y- and z-directions. Fig. 1.14 Cube in enclosure.

p0

z

y

p0

a

y

x

x

a σxx

a σxx

z x

y

σyy

y x

z

a

z

σzz

1.4 The generalized Hooke’s law

19

The normal strain "xx can be determined from the first equation in (1.39) as: "xx D

    p0 1 2 2 xx   yy C zz D  1 : E E 1

(1.47)

The displacement u of the loaded surface of the cube results from the definition of the normal strain "xx D ua as   p0 a 2 2 u D "xx a D  1 : E 1

(1.48)

We now also consider the case where the mechanical load p0 is not present and the cube is subjected to a constant temperature change. Also in this case, a spatial stress state yy , zz will develop, but the normal stress xx will disappear (the cube has a free, unloaded surface whose surface normal coincides with the x-direction). From Hooke’s law (1.44) we obtain from the second and third equations with "yy D "zz D 0:  1 yy  zz C ˛T T; E  1 0D zz  yy C ˛T T: E 0D

(1.49)

From this, both stresses yy and zz can be determined as: yy D zz D 

E˛T T : 1

(1.50)

Also under the present thermal load case, the two normal stresses yy and zz result as compressive stresses, which is an obvious result due to the given geometry and the given thermal load case. Once the stress state is present, then the normal strain "xx can be obtained from the first equation in (1.44): "xx D With "xx D

u a

  1 1C xx   yy C zz C ˛T T D ˛T T : E 1

(1.51)

the displacement u follows as: u D ˛T T

1C a: 1

(1.52) J

20

1

Introduction to linear elasticity

Example 1.2

Two linear elastic isotropic cubes (Fig. 1.15, edge lengths a, moduli of elasticity E1 D E, E2 D 2E, Poisson’s ratios 1 D 2 D , thermal expansion coefficients ˛T;1 D ˛T;2 D ˛T ) are placed in an ideally rigid enclosure as shown. The walls of the enclosure are assumed to be ideally smooth so that the surfaces of the cubes can slide against them without friction. In the case that a constant surface load p0 acts on the surface of cube 1, the stresses and strains in both cubes are to be determined. In the case where there is also a constant temperature change T in the second cube, determine T so that the contact surface between cube 1 and 2 does not shift in the x-direction. If only the mechanical load p0 is present, then the normal stress xx is xx;1 D xx;2 D p0 in both cubes. Moreover, due to the given rigid boundary, the strains "yy and "zz are zero in both cubes: "yy;1 D "yy;2 D 0, "zz;1 D "zz;2 D 0. From Hooke’s law (1.39) we then have for cube 1:  1 yy;1  .p0 C zz;1 / D 0; E   1 "zz;1 D zz;1   p0 C yy;1 D 0: E These two equations can be solved for the stresses yy;1 and zz;1 : "yy;1 D

(1.53)

 : 1 and zz;2 are obtained in cube 2:

yy;1 D zz;1 D p0 Analogously, the stresses yy;2

(1.54)

 : (1.55) 1 Thus, the normal strains "xx;1 and "xx;2 can be determined from the first equation in (1.39). For "xx;1 this results in:     1 p0 2 2 xx;1   yy;1 C zz;1 D  1 : (1.56) "xx;1 D E E 1 yy;2 D zz;2 D p0

Fig. 1.15 Two cubes in enclosure.

p0

1 z x 2

a

p0

a

1

a

y z

y a

x 2

a

a

1.4 The generalized Hooke’s law

21

Analogously, we obtain for "xx;2 : "xx;2 D 

  p0 2 2 1 : 2E 1

(1.57)

The displacement of the contact surface between the two cubes is given as: u D "xx;2 a D 

  p0 a 2 2 1 : 2E 1

(1.58)

We now additionally consider the case where there is a constant temperature increase T in cube 2. In this case the normal stress xx;2 vanishes, and also the two normal strains "yy;2 and "zz;2 are zero. From Hooke’s law (1.44) we obtain the following two relations:  1  yy;2  zz;2 C ˛T T D 0; 2E  1  D zz;2  yy;2 C ˛T T D 0: 2E

"yy D "zz

(1.59)

From this, the two stresses yy;2 and zz;2 can be determined as: yy;2 D zz;2 D 

2E˛T T : 1

(1.60)

Thus, from (1.44), the normal strain "xx;2 can also be calculated: "xx;2 D

1C ˛T T: 1

(1.61)

The displacement u of the contact surface of the two cubes due to the thermal load then follows as: u D "xx;2 a D

1C ˛T T a: 1

(1.62)

To determine the necessary temperature change, the sum of the two displacements (1.58) and (1.62) is set to zero:

This yields:

  p0 a 2 2 1C  1 C ˛T T a D 0: 2E 1 1

(1.63)

  p0 1   2 2 T D 1 : 2E˛T 1 C  1

(1.64) J

22

1

Introduction to linear elasticity

Example 1.3

Let an isotropic cube be positioned in an enclosure as shown in Fig. 1.16. The cube is prevented from expanding in the y-direction, while it can expand unhindered in the x-direction. The cube (edge length a, modulus of elasticity E, Poisson’s ratio , thermal expansion coefficient ˛T ) is stressed by a constant surface load in the z-direction and subjected to a temperature change T . How large must the temperature change T be so that the upper surface of the cube does not move in the z-direction? Let the walls of the enclosure be ideally rigid and smooth, so that the surfaces of the cube can slide along them without friction. For the given situation, it can be immediately concluded that the normal stress zz corresponds to the negative applied load p0 : zz D p0 . Furthermore, xx D 0 since the cube is not prevented from expanding in the x-direction. Therefore, the normal stress yy remains to be determined. Due to the kind of confinement the normal strain "yy becomes zero, so that we can conclude from (1.44):  1 (1.65) 0D yy C p0 C ˛T T: E From this, the normal stress yy can be obtained as: yy D p0  E˛T T:

(1.66)

From the third equation in (1.44) we can thus determine the normal strain "zz : 1 Œp0  .p0  E˛T T / C ˛T T E  p0  D 1   2 C .1 C /˛T T: E

"zz D

(1.67)

We then obtain the displacement u of the surface of the cube loaded by p0 as:  p0 a  (1.68) 1   2 C .1 C /˛T T a: u D "zz a D  E Setting this expression to zero results in T as follows: p0 .1  /: T D E˛T

Fig. 1.16 Cube in enclosure.

(1.69) J

p0

p0

z x

a

z y

a

y

a

x

a

2

Plane stress state

The plane stress state is a very important and practically relevant simplification in the case of thin-walled plane structures loaded in their plane, the so-called disks. First, all the necessary basic equations describing the plane stress state are summarized. After that, the topic of stress transformation is discussed. It turns out that extremal normal stresses, the so-called principal stresses, occur under certain orientations of the coordinate system. Similarly, maximum shear stresses occur under certain directions. A very useful tool for the graphical interpretation of a plane stress state is the so-called Mohr’s stress circle. Students are enabled to formulate and solve simple problems in the plane stress state and to determine stress states under different orientations of the employed Cartesian coordinate system. In addition, students will learn how to analyze plane stress states using Mohr’s stress circle.

2.1 Introduction The previous considerations of the stress state referred to spatial, three-dimensional situations (Chap. 1). In many technical applications, a simplification is possible to the extent that all considerations can be reduced to a two-dimensional structure. This not only allows a considerable reduction of the state quantities to be determined (i.e. displacements, strains and stresses), but also facilitates the analysis immensely, which often makes possible solutions that would not be possible when considering a full three-dimensional problem. One such simplifying situation is the so-called plane stress state, which is an adequate approximation for so-called disks. We use the term disk to refer to a thin plane structure whose thickness t is significantly smaller than its characteristic dimension l in the plane and which is loaded exclusively in its plane (Fig. 2.1, left). For such thin planar structures we can assume in good approximation that those stresses which have the index z, i.e. the stresses zz , xz and yz not only disappear at the disk surfaces at z D  2t and z D  2t , but become zero over the entire thickness t: zz D xz D yz D 0: © The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_2

(2.1) 23

24

2 Plane stress state yy

z y

y

l

xx

xx

x

x t

xy

yy

Fig. 2.1 Disk (left), stresses in a disk (right).

Thus, only the plane stress components xx , yy and xy remain in the plane stress state, which are also independent of the thickness coordinate z (Fig. 2.1, right): xx D xx .x; y/;

yy D yy .x; y/;

xy D xy .x; y/:

(2.2)

For the generalized Hooke’s law (1.39) and (1.40) the following remains in the plane stress state:  1 "xx D xx  yy ; E  1 "yy D yy  xx ; E   "zz D  xx C yy ; E 2.1 C / xy ; xy D E xz D yz D 0: (2.3) In addition to the two plane normal strains "xx and "yy , the normal strain "zz in the thickness direction also occurs in a disk. The two shear strains xz and yz vanish. The nonvanishing strains are exclusively functions of the inplane coordinates x and y. Formulating the constitutive law (2.3) according to the stress components, we obtain:  E  xx D "xx C "yy ; 2 1  E  "yy C "xx ; yy D 2 1 xy D Gxy : (2.4) The kinematic equations (1.33) can be written for the plane stress state as: "xx D

@u ; @x

"yy D

@v ; @y

xy D

@u @v C : @y @x

(2.5)

2.1 Introduction

25

For the equilibrium conditions (1.20) the following remains: @xy @xx C C fx D 0; @x @y @xy @yy C C fy D 0: @x @y

(2.6)

Obviously, the third equation in (1.20) is automatically satisfied. A problem in the plane stress state is described by equations (2.3) and (2.4), (2.5), (2.6) in conjunction with given boundary conditions. Example 2.1

Consider the two disks of Fig. 2.2, which are in plane stress state. Disk 1 (width a, height h1 ) is a triangular disk loaded by the surface load p0 and rests frictionless on disk 2. The rectangular disk 2 has the height h2 and the width a. Both disks are isotropic (modulus of elasticity E, Poisson’s ratio ) and have the thickness t in the z-direction. The stress state in both disks is to be determined. For this purpose, it is assumed that both disks can slide without friction on the walls, which are assumed to be rigid and ideally smooth. Which strain state results in disk 2? What is the maximum permissible surface load p0 if the change in length of disk 2 in the x-direction must not exceed a limit value a? We first cut disk 1 free and consider the force equilibrium in the x-direction and in the y-direction (see Fig. 2.2, right). With l D sinh1˛ the sum of forces in the x-direction yields: xx;1 th1 C p0 t

h1 sin ˛ D 0: sin ˛

(2.7)

From this, the normal stress xx;1 in disk 1 is determined independently of the angle ˛: xx;1 D p0 : (2.8) Fig. 2.2 Triangular disk under surface load p0 and underlying rectangular disk (left), free body image for disk 1 (right).

p0

l

p0

l z y 2

a

1 x

h1

h2

xx

1 yy

a

h1

26

2 Plane stress state

Analogously, the force equilibrium in the y-direction results in: yy;1 ta C p0 t

h cos ˛ D 0: sin ˛

(2.9)

This yields: yy;1 D p0 :

(2.10)

In disk 2, the normal stress xx;2 disappears. Moreover, the normal stress yy;2 of disk 2 is equal to the normal stress yy;1 in disk 1: xx;2 D 0;

yy;2 D p0 :

(2.11)

The strain state of disk 2 with given stresses xx;2 D 0 and yy;2 D p0 can be determined as:  1 p0 "xx;2 D xx;2  yy;2 D ; E E  1 p0 yy;2  xx;2 D  : (2.12) "yy;2 D E E The surface load pN0 that must not be exceeded for the normal strain "xx;2 not to exceed a critical value a is obtained from the following inequality: a  pN0 a  : E a

(2.13)

We solve for pN0 and obtain: pN0 

Ea : a

(2.14) J

Example 2.2

The two isotropic disks shown in Fig. 2.3 (thickness t, width a, height h) are placed in a rigid enclosure of total width b (b > a). There is a gap of width b  a at the sides of both disks where we consider this gap to be small. Both disks are loaded by the surface load p0 as shown. For which surface load p0 D pN0 do the two disks exactly fill the given enclosure? Determine the stress state for the case p0 > pN0 . For disk 1 the elasticity values E1 and 1 are given, disk 2 has the values E2 and 2 . We first determine the normal strain "xx according to "xx D

 1 xx  yy E

(2.15)

in each disk. Since there is a gap on both sides, the two stresses xx;1 and xx;2 are zero. On the other hand, the stresses yy;1 and yy;2 correspond to the negative

2.1 Introduction

27

Fig. 2.3 Two disks in an enclosure.

p0 z y

x

1

2

a b

a b

h

applied load p0 . Then it follows from (2.15):  1  xx;1  1 yy;1 D E1  1  D xx;2  2 yy;2 D E2

"xx;1 D "xx;2

p0 1 ; E1 p0 2 : E2

(2.16)

The two displacements u1 and u2 of the disk edges in x-direction are now required to sum up to twice the gap width: u1 C u2 D 2.b  a/: With "xx;1 D

u1 a

and "xx;2 D

u2 a

(2.17)

it follows:

a"xx;1 C a"xx;1 a D 2.b  a/:

(2.18)

This can be solved for the surface load pN0 , and we obtain:

pN0 D

! b 2  1 E1 E2 a E2 1 C E1 2

:

(2.19)

We now consider the stress state for the case where p0 > pN0 holds. To determine the strains of the disks in the x-direction, we must assume that the normal stresses xx;1 and xx;2 are not zero and are also identical. Therefore xx;1 D xx;2 D xx is valid, and with yy;1 D yy;2 D p0 we obtain for the normal strains "xx;1 and "xx;2 : 1 .xx C 1 p0 /; E1 1 D .xx C 2 p0 /: E2

"xx;1 D "xx;2

(2.20)

28

2 Plane stress state

The condition (2.18) is also valid for the case p0 > pN0 . Substituting (2.20) and solving for xx gives: ! b 2  1 E1 E2  p0 .1 E2 C 2 E1 / a : (2.21) xx D E1 C E2 J

2.2 Stress transformation In the following, we will address the question of how the stress state changes when we perform a coordinate transformation as shown in Fig. 2.4. The new coordinate axes rotated by the angle  are called , , they form an orthogonal coordinate system with the fixed z-axis. An application is shown in Fig. 2.4 in the form of a fibrous material where the fiber direction does not coincide with the reference axes x, y. The stress states with respect to the original axes x, y and with respect to the axes ,  rotated by  are shown in Fig. 2.5. To determine the necessary transformation equations, we consider an infinitesimally small element taken from a disk (Fig. 2.6). We have arranged the element in such a way that there are two cut surfaces parallel to the axes x and y. Furthermore, the cut is made in such a way that a triangular disk is obtained and the remaining cut edge is parallel to the -axis. Let the cut surface that is traversed by the -axis

Fig. 2.4 Axis transformation.

z x y-plane

x

yy

y xx

xx

x

xy

yy

Fig. 2.5 Stress transformation: initial state (left), rotation by the angle  (right).

y

2.2 Stress transformation

29

Fig. 2.6 Infinitesimally small disk element.

y

dAcos

xx

dA=d h dy

x xy

d dx

dAsin

yy

be called dA. It can be calculated as dA D dt, where t is the thickness of the disk. The other two sectional surfaces oriented parallel to x and y with lengths dx and dy then follow as dA sin  and dA cos , respectively. Equilibrium of forces in the direction of the -axis then gives:  dA  xx dA cos  cos   yy dA sin  sin   xy dA cos  sin   xy dA sin  cos  D 0;

(2.22)

where here the equivalence xy D yx was used. After a short transformation we obtain the normal stress  in the coordinate system rotated by the angle  as:  D xx cos2  C yy sin2  C 2xy sin  cos :

(2.23)

In quite the same way, the force equilibrium can be set up in the direction of the -axis, and we obtain the following expression for the shear stress   :     D xx sin  cos  C yy sin  cos  C xy cos2   sin2  : (2.24) On another section, which is not shown here and where the -axis passes through the section surface dA, the following expression for the normal stress  can be obtained:  D xx sin2  C yy cos2   2xy sin  cos : (2.25) It is possible to rearrange the equations (2.23), (2.24) and (2.25) by means of the relations 1 1 .1 C cos 2/; sin2  D .1  cos 2/; 2 2 2 sin  cos  D sin 2; cos2   sin2  D cos 2 cos2  D

(2.26)

as follows:  1  1 xx C yy C xx  yy cos 2 C xy sin 2; 2 2  1  1 D xx C yy  xx  yy cos 2  xy sin 2; 2 2  1 D  xx  yy sin 2 C xy cos 2: 2

 D   

(2.27)

30

2 Plane stress state

These are the transformation equations for the stresses in a disk when the coordinate system is transferred from the reference frame x, y to the reference frame , . Hence, the stresses  ,  and   in any reference frame ,  can be determined from given stresses xx , yy and xy in the reference frame x, y. It can be shown that in the plane state of stress there are so called invariants, i.e., quantities which are invariant to a transformation of the coordinate system as shown above. They are: I1 D xx C yy ;

2 I2 D xx yy  xy :

(2.28)

2.3 Principal stresses In the plane stress state, there are certain coordinate directions under which the stresses assume extreme values. We will first address the question of the angle  at which extremal normal stresses, the so-called principal stresses, occur. For this purpose we solve the following two extreme value problems: d

D 0; d

d D 0: d

(2.29)

The evaluation of both equations leads to the same result, and an equation for the angle p under which the normal stresses become extremal is obtained as: tan 2p D

2xy : xx  yy

(2.30)

The direction marked by the angle p is called principal direction. The corresponding axes and  are denoted as principal axes. Since the tan-function is a periodic function with period , two angles p and p C 2 can always be determined from (2.30) under which the normal stresses become extremal. These two directions are perpendicular to each other and, as we will show, are absolutely equal. Substituting the result (2.30) into the transformation equations (2.27) and using the relations xx  yy 1 cos 2p D p D q ; 2 1 C tan2 2p 2 xx  yy C 4xy tan 2p 2xy sin 2p D p D q 2 1 C tan2 2p 2 xx  yy C 4xy

(2.31)

results in the following equations for calculating the extremal normal stresses, which we will refer to as 1 and 2 : r xx  yy 2 xx C yy 2 : C xy (2.32) 1;2 D ˙ 2 2

2.3 Principal stresses

31

yy 2

2 1

1

y xx

xx

x

xy

1

1 2

yy

p

2

p+ 2

Fig. 2.7 On the equality of the two principal directions p and p C 2 .

These are the so-called principal stresses 1 and 2 , where 1 represents the maximum value and 2 the minimum value, i.e. 1 > 2 . If p or if p C 2 according to (2.30) is inserted into the third transformation equation in (2.27), then it turns out that under p or of p C 2 the shear stress   disappears. Thus, in sections under which the normal stresses become extremal, the shear stress disappears. The two principal directions p and p C 2 are equal as illustrated in Fig. 2.7. This figure shows that the principal stresses occurring in the two principal directions are identical, regardless of whether the principal direction is considered under p or under p C 2 . Only the axes ,  exchange their positions, which, however, is irrelevant for the resulting stress state. The question now also arises under which direction the extremal shear stresses will occur. For this purpose, the following extremal value problem is considered: d  D 0: d

(2.33)

After elementary transformations this leads to the following equation for the determination of the angle p0 : tan 2p0 D 

xx  yy : 2xy

(2.34)

Again, when determining the principal direction for the extremal shear stresses, there are again two completely equal angles, namely the angle p0 and the angle p0 C 2 . Comparing (2.34) with (2.30), we find that the following relation exists: tan 2p0 D 

1 : tan 2p

(2.35)

From this we can conclude that the two directions 2p0 and 2p are oriented perpendicular to each other. Consequently, the direction p0 of the extremal shear stresses

32

2 Plane stress state yy

M 2

max 1

y xx

M

xx

x

xy

M

1 2

yy

p

p

M

4 p

Fig. 2.8 Stress states: given initial state (left), principal stress state (center), principal shear stress state (right).

is rotated by 45ı with respect to the direction p of the extremal normal stresses. We denote the extremal shear stress as principal shear stress max . It is obtained by substituting p0 into the transformation equations (2.27) as: max D ˙

r 

xx

 yy 2 2 : C xy 2

(2.36)

The maximum shear stress max can also be expressed by the two principal stresses 1 and 2 as follows: 1 (2.37) max D ˙ .1  2 /: 2 However, when determining the principal shear stresses, it should be noted that under the angle p0 the two normal stresses do not vanish. Rather, the normal stresses in this case are identical to the value M as follows: M D

 1 xx C yy : 2

(2.38)

Taking into account the invariance (2.28) of the sum of the two normal stresses, which is also valid for the sum of the two principal stresses 1 and 2 , the value M can also be represented as: 1 (2.39) M D .1 C 2 /: 2 Fig. 2.8 shows an overview of the discussed stress states.

2.4 Mohr’s stress circle

33

2.4 Mohr’s stress circle The transformation relations and stress states discussed so far can be treated graphoanalytically in a particularly illustrative way with the help of the so-called Mohr’s stress circle1 which is a graphical representation of the transformation equations (2.27). The respective circle equation is obtained by squaring the first and third equations in the transformation equations (2.27) and adding them together so that the angle  is eliminated. This results in: 

    2  2 1 xx yy 2 2   xx C yy C   D C xy : (2.40) 2 2 Let us assume that the stresses xx , yy and xy are given, so that the right side of the equation (2.40) represents a constant quantity. Let this be abbreviated by r 2 . Thus: 

 2 1 2 C   D r 2; (2.41)   xx C yy 2 wherein r2 D



xx

 yy 2 2 C xy : 2

With the definition (2.38) we get:  2 2   M C   D r 2:

(2.42)

(2.43)

The same result is obtained by using the second and third equations in the transformation equations (2.27). For this reason, we can omit the indices and  for the moment and obtain: (2.44) .  M /2 C  2 D r 2 : This expression is a circle equation in the  -plane. The circle described by it is the so-called Mohr’s stress circle, it has the radius r and the center point .M ; 0/. Mohr’s stress circle can be drawn if the stresses in any initial state are given with the values xx , yy and xy . One first introduces a normal stress axis  and a shear stress axis  and marks the two given normal stresses xx and yy on the  -axis. The shear stress xy is then plotted over the two points thus determined, with the correct sign over xx and with the opposite sign over yy . Thus the two points .xx =xy / and .yy =  xy / are obtained, which can then be connected with a straight line of length 2r (circle diameter). This straight line intersects the  -axis in the point .M =0/, this point is therefore the center of Mohr’s stress circle. With the midpoint present and the two points .xx =xy / and .yy =  xy / determined, Mohr’s stress circle can be drawn, as shown in Fig. 2.9. If a stress state, distinguished by the stresses xx , yy , xy , is given, then the stress state in any direction  can be determined with the help of Mohr’s stress circle. Quite obviously, both the principal normal stresses 1 and 2 and the principal 1

Christian Otto Mohr, 1835–1918, German civil engineer.

34

2 Plane stress state

Fig. 2.9 Mohr’s stress circle and associated stress state.

yy

y xx

xx

x xy

(

xy

xx / xy (

yy

r yy M

xx

r xy

(

yy /

xy (

M=

1( + 2 xx

yy(

yy

y xx

xx

x xy

(

xy

xx / xy (

2

yy

r 2

2

yy

( 2 /0 (

2

p r

M

1

( 1/0 (

p xx

1

1 p

xy

(

yy /

2

xy (

Fig. 2.10 Determination of 1 and 2 using Mohr’s stress circle.

shear stress max can be read directly from Mohr’s circle, see Figs. 2.10 and 2.11. It should be noted that, due to the underlying equations, the double angle 2 must always be applied, and in the case of a positive angle, always clockwise. Likewise, any other arbitrary stress state of a disk at the angle  can be read from Mohr’s stress circle (Fig. 2.12). With the help of Mohr’s stress circle, some special cases can be interpreted particularly clearly, which we will discuss briefly below. The first special case is the disk under uniaxial tension xx D 0 , the two stresses yy and xy are zero. The cor-

2.4 Mohr’s stress circle

35

Fig. 2.11 Determination of max using Mohr’s stress circle.

yy

y xx

(

xx

x

M / max (

xy

xy

(

yy

xx / xy (

r

2

p

yy

M

2

xx

p

r xy

(

yy /

xy (

(

M/

max (

M M max p

M M

Fig. 2.12 Determination of the stress state under an arbitrary angle  using Mohr’s stress circle.

yy

y xx

(

/

xy

(

xy

(

xx / xy (

r

2

yy

M

2

xx

r xy

(

xx

x

yy /

xy (

(

/

(

yy

36

2 Plane stress state

( 2/ 2 ( 0

2

p

0

M=

(0/0 (

2

xx = 0

0

2

y

( 0/0 (

p

x

( 2 /- 2 ( 0

= 20 p

0

= 20 =

max =

0

2

Fig. 2.13 Mohr’s stress circle for a disk under uniaxial tension xx D 0 .

responding Mohr’s stress circle is shown in Fig. 2.13. It is tangent to the shear stress axis at its left side, and obviously the initial stress state with xx D 0 , yy D xy D 0 is already the principal stress state with 1 D 0 and 2 D 0, and the principal axis angle p is p D 0ı . The center point of the circle is at .M =0/ D . 20 =0/. The angle p0 is therefore p0 D 45ı , and the corresponding maximum shear stress max is max D 20 . Under the angle p0 the two normal stresses also take the value 20 . Another important special case is the pure shear stress state, as shown in Fig. 2.14. Let the shear stress xy be given the value xy D 0 , and let the two normal stresses xx and yy be zero. The center of Mohr’s stress circle in this case is exactly at the origin of the  -reference frame. Obviously, this stress state is the principal shear stress state, the angle p0 is p0 D 0ı . The given shear stress 0 represents the maximum shear stress max . It follows that the angle p has the value 45ı , and the principal stress state resulting under this angle is shown in Fig. 2.14 as well. The two principal normal stresses occurring under p take the values 1 D 0 and 2 D 0 , and the shear stress vanishes by definition for this condition. A pure shear stress state is thus equivalent to a pure normal stress state that occurs rotated with respect to the reference frame by 45ı and has both a tensile normal stress 1 D 0 and a compressive normal stress 2 D 0 , both of which have identical magnitudes. The last special case we want to discuss at this point is the so-called hydrostatic stress state. It is characterized by the two normal stresses xx and yy , both with the value 0 . A shear stress is not present in this case. The corresponding Mohr’s stress circle is shown in Fig. 2.15. Obviously, Mohr’s stress circle reduces in this case to a point on the  -axis, which is located at the ordinate M D 0 . A transformation of

2.4 Mohr’s stress circle

37

y

xy = 0

xy = 0

x

( 0/ 0 ( =(- 0 /0(

M =0

2

p

2

0

=

0

p

( 0 /0( p

(0/- 0 ( Fig. 2.14 Mohr’s stress circle for a disk under pure shear xy D 0 . Fig. 2.15 Mohr’s stress circle for the hydrostatic stress state xx D yy D 0 .

yy= 0

y

xx = 0

xx

x

yy M= 0

=

0

=

0

the stress state by any angle  then yields the same stress state again. Each angle  is a principal axis angle, and the two given normal stresses xx and yy represent the principal stresses 1 and 2 with the value 0 in any axis system. Example 2.3

Consider the stress state xx D 200 aF2 , yy D 600 aF2 , xy D 300 aF2 , where F is in the unit of a force (e.g. N) and a is in the unit of a length (e.g. m). Find the principal stresses 1 and 2 and the corresponding principal angle p , and deter-

38

2 Plane stress state

mine the principal shear stress max and the corresponding angle p0 . Determine the results both mathematically and graphoanalytically on Mohr’s stress circle. We first determine the principal axis angle p as follows: tan 2p D

2xy 600 3 D D : xx  yy 200  600 4

(2.45)

From this follows 2p D 36:87ı and p D 18:44ı. The principal stresses 1 and 2 can be determined as follows: r xx  yy 2 xx C yy 2 C xy 1;2 D ˙ 2 2 s   200  600 2 200 C 600 D C 3002 D 200 ˙ 500: (2.46) ˙ 2 2 This yields:

F F ; 2 D 300 2 : (2.47) 2 a a We now also determine the maximum shear stress xy and the corresponding angle p0 . For the angle p0 follows: 1 D 700

tan 2p0 D 

xx  yy 200  600 4 D D : 2xy 600 3

F a2 (

(2.48)

M / max ( =(200/500)

500 400

(

xx/ xy ( =(-200/300) 300 200

( 2/0(=(-300/0)

(

M /0( =(200/0)

( 1/0 (=(700/0)

100 -200 -100

100

200

300

400

500

600

2 p =-36.87°

F a2

-100 -200

2 p =53.13°

-300 -400 -500

(

M /- max ( =(200/-500)

Fig. 2.16 Mohr’s stress circle for the given stress state.

(

yy/- xy ( =(600/-300)

2.4 Mohr’s stress circle

39

From this follows 2p0 D 53:13ı as well as the angle p0 as p0 D 26:57ı . The principal shear stress max is given as: r 

 yy 2 2 C xy 2 s   200  600 2 F D˙ C 3002 D ˙500 2 : 2 a

max D ˙

xx

(2.49)

The corresponding stress circle is shown in Fig. 2.16. The value M is given by M D 12 .xx C yy / D 200 aF2 . J Example 2.4

Consider a disk in a plane stress state (Fig. 2.17) with the stress components xx D 1

F ; a2

yy D 7

F ; a2

xy D 3

F : a2

(2.50)

The stresses under an angle of  D 30ı are to be determined. In addition, determine the principal stresses 1 , 2 and the principal shear stress max . We first determine the stresses under an angle  of  D 30ı . For this we use the transformation equations (2.27) and obtain:  1  1 xx C yy C xx  yy cos 2 C xy sin 2 2 2 p 3 1 1 ı ı D .1  7/ C .1 C 7/ cos 2  30  3 sin 2  30 D 1  3 2 2 2 F D 3:60 2 ; a  1   1 D xx C yy  xx  yy cos 2  xy sin 2 2 2 p 3 1 1 ı ı D .1  7/  .1 C 7/ cos 2  30 C 3 sin 2  30 D 5 C 3 2 2 2 F D 2:40 2 ; a  1 D  xx  yy sin 2 C xy cos 2 2 p 3 1 D  .1 C 7/ sin 2  30ı  3 cos 2  30ı D 2 3  2 2 F D 4:96 2 : a

 D



 

(2.51)

40

2 Plane stress state

Fig. 2.17 Given stress state.

F

yy =-7a2

y

F

xx

xx=1a2

x

F

xy =-3a2 yy

F a2 (

/-

(=(-2.4/4.96) 5 4

(

yy /- xy ( =(-7/3) 3 2 1 -8

-7

-6

(

-5

-4

-3

-2

-1

M /0( =(-3/0)

1

F a2

2

-1 -2

2 =60°

-3

(

xx / xy ( =(1/-3)

F

yy =-7a2

-4

(

/ (=(-3.60/-4.96)

-5 y

F

xx

x

xx=1a2

F

xy =-3a2

=-2.40 aF2 =-3.60 aF2

yy

=-4.96 aF2 30° Fig. 2.18 Mohr’s stress circle, determination of stresses under an angle of  D 30ı .

The corresponding stress circle is shown in Fig. 2.18. The principal stress angle p follows from: tan 2p D

2xy 2  .3/ 3 D D : xx  yy 1  .7/ 4

(2.52)

2.4 Mohr’s stress circle

41 F a2 5 4

(

yy /- xy ( =(-7/3) 3

F

2 =-8a2

2

-8

( 2/ 0(=(-8/0)

-7

-6

(

-5

( 1/ 0(=(2/0)

1

2 p =-36.87° -4

M /0( =(-3/0)

-3

-2

-1

1

1

F a2

2

F

1 =2a2

-1 -2 -3

-18.43°

2

(

xx / xy ( =(1/-3)

F

yy =-7a2

-4 -5 y

F

xx

x

xx=1a2

F

xy =-3a2 yy

Fig. 2.19 Mohr’s stress circle, determination of the principal stresses 1 and 2 .

This gives 2p D 36:87ı and p D 18:43ı . The stress components under this angle result in:  D D  D D

 1  1 xx C yy C xx  yy cos 2p C xy sin 2p 2 2 1 F 1 .1  7/ C .1 C 7/ cos.36:87ı /  3 sin.36:87ı / D 2 2 D 1 ; 2 2 a  1  1 xx C yy  xx  yy cos 2p  xy sin 2p 2 2 1 1 .1  7/  .1 C 7/ cos.36:87ı / C 3 sin.36:87ı / D 8F a2 D 2 ; 2 2

  D 

 1 xx  yy sin 2p C xy cos 2p D 0: 2

(2.53)

The corresponding stress circle is shown in Fig. 2.19. Finally, the principal shear stress must be determined. The corresponding angle p0 follows as p0 D p C 45ı D 18:43ı C 45ı D 26:57ı. Then from the

42

2 Plane stress state

(

F a2

M /- max ( =(-3/5) 5 4

(

yy /- xy ( =(-7/3) 3

2 p =53.14°

2 1 -8

-7

-6

(

-5

-4

-3

-2

-1

1

M /0( =(-3/0)

F a2

2

-1 -2 -3

(

xx / xy ( =(1/-3)

F

yy =-7a2

-4 -5

(

y

M / max ( =(-3/-5)

F

xx

x

xx=1a2

F

xy =-3a2

=-3aF2 =-3aF2

yy

F

max =-5a2

26.57° Fig. 2.20 Mohr’s stress circle, determination of the principal shear stress max .

transformation equations (2.27) we obtain:  1  1 xx C yy C xx  yy cos 2p0 C xy sin 2p0 2 2 1 F 1 D .1  7/ C .1 C 7/ cos 53:14ı  3 sin 53:14ı D 3 2 ; 2 2 a  1  1  D xx C yy  xx  yy cos 2p0  xy sin 2p0 2 2 1 F 1 D .1  7/  .1 C 7/ cos 53:14ı C 3 sin 53:14ı D 3 2 ; 2 2 a  1   D  xx  yy sin 2p0 C xy cos 2p0 2 1 F D  .1 C 7/ sin 53:14ı  3 cos 53:14ı D 5 2 : 2 a The corresponding stress circle is shown in Fig. 2.20. J  D

(2.54)

2.4 Mohr’s stress circle

43

Example 2.5

The following data are known from a plane stress state: xx D 6

F ; a2

xy D 2

F ; a2

max D 5:39

F ; a2

M D 1

F : a2

(2.55)

The following quantities are to be determined:  The normal stress yy ,  the principal shear stress angle p0 by using the first transformation equation in (2.27),  the principal stress angle p by using the third transformation equation in (2.27),  the principal stresses 1 and 2 . Draw the corresponding stress circle. At what angle 0 does the normal stress become  D 0 with   > 0? In this case, how large are  and   ? We determine the normal stress yy from the equation for M D 12 .xx C yy /, i.e. F yy D 2M  xx D 2  1  6 D 4 2 : (2.56) a The principal shear stress angle p0 is determined from the first transformation equation in (2.27):  D

 1  1 xx C yy C xx  yy cos 2p0 C xy sin 2p0 : 2 2

(2.57)

Since this is the principal shear stress angle p0 , the normal stress  is present with the value M D 1 aF2 : 1D

1 1 .6  4/ C .6 C 4/ cos 2p0 C 2 sin 2p0 : 2 2

(2.58)

This leads to the following equation for the determination of the angle p0 : 5 tan 2p0 D  : 2

(2.59)

p0 D 34:10ı :

(2.60)

This results in p0 as follows:

We determine the principal stress angle p from the third transformation equation in (2.27):  1 (2.61)   D  xx  yy sin 2p C xy cos 2p : 2

44

2 Plane stress state

F a2 6

(

/

(=(2/5.3)

5 4 3

(

2

xx/ xy ( =(6/2)

1 -5

-4

-3

-2

-1

1

-1

2

(

3

4

5

6

7

F a2

M /0( =(1/0)

-2

(

yy/- xy ( =(-4/-2)

2 0=-57.50°

-3 -4 -5

(

/- (=(0/-5.3)

-6

Fig. 2.21 Mohr’s stress circle, determination of the angle 0 .

In the principal axis system, the shear stress   takes the value zero, and from (2.61) follows: 2 tan 2p D ; (2.62) 5 from which the angle p follows as p D 10:90ı. The principal stresses 1 and 2 can be determined as follows: r xx  yy 2 xx C yy 2 : 1;2 D C xy (2.63) ˙ 2 2 We obtain:

F F ; 2 D 4:39 2 : (2.64) 2 a a The angle 0 under which the normal stress  vanishes and   > 0 holds is determined at Mohr’s stress circle (see Fig. 2.21). We obtain: 1 D 6:39

20 D 57:50ı :

(2.65)

The two stresses  and   follow as:  D 2

F ; a2

  D 5:3

F : a2

(2.66) J

3

Bars

Bars are one-dimensional straight structural elements capable of transmitting tensile and compressive forces. For bars, this chapter introduces the concepts of stress, strain, and displacement and all necessary basic equations, and discusses both statically determinate and statically indeterminate bars and bar systems. Students are enabled to formulate bar problems and solve such problems using the introduced equations that describe the behavior of bars and bar systems.

3.1 Bars 3.1.1 Bar stresses The bar is one of the simplest construction elements. In this chapter we consider exclusively straight bars described by their longitudinal axis. The stress state that arises is particularly simple in the case of the bar. We assume that only normal forces are induced in the member due to the applied loads. The normal forces in turn lead to the normal stress xx (see also Fig. 1.1). The normal stress results from the normal force N divided by the cross-sectional area A: xx D

N : A

(3.1)

If the normal force N is a tensile force, the normal stress xx is also a positive stress (tensile stress). If, on the other hand, the normal force N is a compressive force, then the normal stress xx will be a compressive stress with a corresponding negative sign. If the cross-section A of a bar varies with the bar axis x (i.e. if A D A.x/), then the normal stress xx must be treated as varying over x: xx D xx .x/. In addition, the normal force N can be a function of x, depending on the applied load, i.e. in

© The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_3

45

46

3

Bars

general N D N.x/. Thus, for the normal stress xx .x/ follows: xx .x/ D

N.x/ : A.x/

(3.2)

The prerequisite for this is that the cross-section is only slightly variable over x. It is assumed that the normal stress xx is still uniformly distributed over the crosssection. Example 3.1

Let a conical bar with circular cross-section of length l under compressive force F0 (Fig. 3.1) be given, which has the radius R1 at location x D 0. At the point x D l the radius R2 is given. We are looking for the normal stress xx acting in the bar when R1 D 3R0 and R2 D R0 . How large must the radius R2 be for R1 D 3R0 so that the stress at the point x D l is twice the stress at the point x D 0? We first consider the case R1 D 3R0 and R2 D R0 and determine the normal F0 stress xx .x/ D  A.x/ . The cross-sectional area A.x/ is calculated as: A.x/ D R2 .x/:

(3.3)

Here R.x/ describes the radius of the circular cross section, depending on the position on the bar axis x. To find R.x/, we make a linear approach of the form: R.x/ D ax C b:

(3.4)

The following conditions apply: R.x D 0/ D a  0 C b D R1 ;

R.x D l/ D a  l C b D R2 :

From this, the two constants can be determined as b D R1 and a D we obtain: x R.x/ D .R2  R1 / C R1 : l Fig. 3.1 Conical bar under compressive force F0 .

(3.5) R2 R1 , l

and

(3.6)

F0 R2 l

x

F0 R1

3.1 Bars

47

The cross-sectional area A.x/ then follows for the case R1 D 3R0 and R2 D R0 :  x 2 : A.x/ D R0 3  2 l

(3.7)

The normal stress xx .x/ is then given as: F0

xx .x/ D  R0

x 32 l

!2 :

(3.8)

We now turn to the question of how large the radius R2 must be for R1 D 3R0 so that the stress at the point x D l is twice the stress at the point x D 0. The stress xx at the point x D 0 is: xx .x D 0/ D 

F0 : 9R02

(3.9)

F0 : R22

(3.10)

At the point x D l we obtain: xx .x D l/ D 

We now require xx .x D l/ D 2xx .x D 0/, i.e.:  Solving for R2 yields:

2F0 F0 D : 2 R2 9R02

(3.11)

3 R2 D p R0 : 2

(3.12) J

Example 3.2

Consider a flat bar of length l, where the width 2b.x/ is a parabolic function of x with the initial value b2 and the end value b1 (Fig. 3.2). Let the bar be loaded by a tensile force F0 , and let the thickness of the bar be constant with the value t. We are looking for the normal stress xx .x/ acting in the bar. How large must b2 be so that the stress xx .x D 0/ does not exceed an allowable value 0 ? What is the normal stress xx at the point x D 2l ? Let b.x/ be described by a parabolic function:  x 2 C b2 : b.x/ D .b1  b2 / l

(3.13)

48

3

Fig. 3.2 Flat bar with parabolic boundary under compressive force F0 .

Bars

l F0

x

F0 b2

b1

The normal stress xx .x/ in the bar is then given as: xx .x/ D

F0 D 2b.x/t

2

F0

x 2t 4.b1  b2 / l

!2

3:

(3.14)

C b2 5

The value b2 is now to be determined in such a way that the stress xx at the point x D 0 does not exceed a permissible value 0 . Thus, we require: F0 (3.15)  0 : xx .x D 0/ D 2b2 t Solving for b2 yields: F0 b2  : (3.16) 20 t The stress at the point x D 2l is:   2F0 l D xx x D : (3.17) 2 t.b1 C 3b2 / J Technically relevant are also bars under distributed loads parallel to the bar axis. An example of such a distributed load is the dead weight of a bar. We consider the free body image of Fig. 3.3, in which we investigate an infinitesimal section element of length dx of a bar. With dG D Adx (where  D g is the specific gravity with D density, g D acceleration due to gravity, and A is the cross-sectional area of the bar) the sum of the forces in the x-direction yields: N C dN  N  dG D 0;

(3.18)

dN D Adx:

(3.19)

or

Fig. 3.3 Free body image for determining the normal force N of a bar under dead weight.

N+dN dG

dx N

3.1 Bars

49

Fig. 3.4 Bar under its dead weight.

A,

A, g

l

x

l

x

This yields: dN D A; dx

(3.20)

N D Ax C C1 :

(3.21)

or after integration: Under the dead weight, the normal force is therefore distributed linearly over x. The integration constant C1 is adapted to given boundary conditions. As an example, consider the bar of Fig. 3.4, left, clamped at its lower end which is subjected to its dead weight (specific gravity , cross-sectional area A). The constant C1 is determined from the condition that the normal force N.x/ is zero at the upper end of the bar: N.x D l/ D 0. From this, the constant C1 follows as C1 D Al, and we obtain for the normal force N.x/: N.x/ D Ax  Al D A.x  l/:

(3.22)

Since x  l holds, N is a compressive force for all x. Accordingly, we obtain for the bar stress xx .x/: N.x/ D .x  l/: (3.23) xx .x/ D A If we consider the situation where the bar is clamped at its upper end and subjected to its own weight (Fig. 3.4, right), then the integration constant C1 results from the requirement N.x D 0/ D 0 to C1 D 0 such that: N D Ax:

(3.24)

In this case, the normal force is present with a positive sign for each x, i.e. as a tensile force. We now consider the general case that the bar under consideration is under an arbitrarily distributed axis-parallel line load n.x/. Using the free body image of Fig. 3.5 gives the sum of the forces in the axial direction: N C dN C ndx  N D 0:

(3.25)

50

3

Fig. 3.5 Free body image for determining the normal force N of a bar under an axis-parallel line load n.x/.

Bars

N+dN n(x)

dx N

This can be rearranged as: dN D n: (3.26) dx Accordingly, the applied axis-parallel load n.x/ is the first derivative of the normal force N.x/ with respect to the longitudinal coordinate x, whereby the negative sign must be observed. If the load n.x/ is present, the normal force N.x/ can be determined from this by integration: Z (3.27) N.x/ D  n.x/dx C C1 : Therein, C1 is an integration constant which is adjusted to given boundary conditions.

3.1.2 Bar deformations We will assume linear elastic material behavior in all elaborations of this and all following chapters of this book. Hooke’s law, which describes linear elastic material behavior for a one-dimensional problem, has already been introduced in Chap. 1 with (1.35): xx D E"xx : (3.28) We assume Hooke’s law to be valid also for the bar structures considered here. With xx D NA follows: du N "xx D D u0 D : (3.29) dx EA This equation connects the bar displacement u with the normal force N and thus represents the constitutive law for the bar. In Chap. 1 we have already determined the definition of the normal strain "xx , which is the ratio of length change l and initial length l: "xx D

l : l

(3.30)

A general definition has already been derived with (1.22): "xx D

du : dx

(3.31)

3.1 Bars

51

Fig. 3.6 Bar with constant extensional stiffness EA under force F .

EA F

F

x l

Thus, once the axis-parallel displacement u of the bar is available, the normal strain "xx can be determined from it by derivation. We first consider a homogeneous bar with constant elastic modulus E and constant cross-sectional area A, which is loaded by the force F (see Fig. 3.6). We will refer to the product EA as extensional stiffness in all further elaborations. The normal strain "xx is then given as: l  F D D : l E EA

(3.32)

This expression can be solved for the length change of the bar: l D

Fl : EA

(3.33)

The length change of a homogeneous bar thus depends on the applied force F , the length l and the extensional stiffness EA. If a bar with constant normal force N is treated where the cross section is a function of the longitudinal direction x, i.e. if A D A.x/ holds, then the bar stress follows from: N xx .x/ D : (3.34) A.x/ If we assume that A.x/ is a continuous, integrable function, then the length change l of the bar is the integral over the normal strain "xx over the bar length: Zx l D 0

1 "xx dx D E

Zx xx dx:

(3.35)

0

If we consider a bar where the applied load leads to sectionwise different but constant normal forces Ni (where the index i denotes the section i of the bar) and thus also stresses xx;i D NAi , then the normal strain "xx;i in section i results in  "xx;i D xx;i E . The length change li of section i then follows as: li D

Ni li : EA

(3.36)

The total change of the bar length l is then the sum of all li : l D

n X i D1

when there are n sections in total.

li ;

(3.37)

52

3

Bars

Example 3.3

Consider the bar shown in Fig. 3.7, which has the partial lengths l1 and l2 and is loaded by two single forces as shown. The normal force N1 in section 1 is N1 D 2F0 , while the normal force N2 in section 2 is N2 D 3F0 . The total change in length l results from the partial length changes l1 and l2 as follows: N1 l1 N2 l2 C EA EA 2F0 l1 3F0 l2 F0 D C D .2l1 C 3l2 /: EA EA EA

l D l1 C l2 D

(3.38) J

If the considered bar is not only stressed by mechanical loads, but thermal loads are also present, then the bar elongation can be determined as follows (see also Chap. 1, (1.41)): (3.39) "T D ˛T T: Here ˛T is the coefficient of thermal expansion of the bar material, and T is the temperature change to which the bar is subjected. If the temperature change is a function of the location, that is, if T D T .x/ holds, then it follows: "T .x/ D ˛T T .x/:

(3.40)

If both a mechanical stress  and a temperature change T are present simultaneously, then the individual normal strain components can be superposed to the total normal strain (cf. (1.42)): xx "xx D (3.41) C ˛T T; E or N "xx D u0 D (3.42) C ˛T T: EA Fig. 3.7 Bar under two single forces.

EA

l2 F0 l1

2F0

3.1 Bars

53

One can solve the constitutive expression (3.41) for the stress xx as follows: xx D E."xx  ˛T T /:

(3.43)

The length change l of the bar is given as: Zl  l D 0

 N C ˛T T dx: EA

(3.44)

The special case of the bar with constant extensional stiffness EA and constant temperature change T results from this as: l D

Nl C ˛T T l: EA

(3.45)

If there is exclusively a temperature load, then we obtain: l D ˛T T l:

(3.46)

We now bring the constitutive equation (3.42) and the equilibrium (3.26) into the following form by substituting into each other. Solving (3.42) for the normal force N and substituting into (3.26) yields: 

0 EAu0 D n C .EA˛T T /0 :

(3.47)

For the special case of a constant temperature change T it follows: 0  EAu0 D n:

(3.48)

From this, the bar displacement u.x/ is obtained by twofold integration: 0  EAu0 D n; EAu0 D N D 

Z ndx C C1 ;

“ EAu D 

ndxdx C C1 x C C2 :

(3.49)

The integration constants appearing here are to be determined from given boundary conditions. Fig. 3.8 shows typical cases of boundary conditions for bars. One boundary condition can always be given at each end of the bar, either for u or for N .

54

3

Fig. 3.8 Typical boundary conditions for bars.

N

N

x

u=0

z

N

x

u=0

z

x

EAu =0

z N

Bars

x

u=0

z

Example 3.4

Consider the bar shown in Fig. 3.9. The bar has the length l and the constant extensional stiffness EA. The constant line load n and a single force F at the free end of the member are given as loads. The integration rule (3.49) is as follows for this example: .EAu0 /0 D n;

Z EAu0 D N D  ndx C C1 D nx C C1 ; “ 1 EAu D  ndxdx C C1 x C C2 D  nx 2 C C1 x C C2 : 2

(3.50)

For the given situation, the boundary conditions at the two bar ends are: u.x D 0/ D 0;

N.x D l/ D F:

(3.51)

From this, the integration constants C1 and C2 follow as: C1 D F C nl;

C2 D 0:

Fig. 3.9 Determination of the bar normal force N and the bar displacement u of a clamped straight bar under line load and single force.

(3.52)

EA

x

F

n l

F+nl (+)

F

N(x)

(+)

EA u(x) l

F+ 12 nl

3.1 Bars

55

The normal force distribution N.x/ and the bar displacement u.x/ follow from this as (see Fig. 3.9): N.x/ D F C n.l  x/;

 1 1  nx 2 C .F C nl/x : u.x/ D EA 2

(3.53)

For this example, a linear distribution of the bar normal force N results. The displacement u, on the other hand, is distributed parabolically over the bar length. J For bars where the state variables N and u are not continuous and thus the integration rule (3.49) cannot be carried out over the entire length of the bar, boundary conditions as well as the so-called transition conditions have to be formulated. A simple example is shown in Fig. 3.10. Consider a straight bar with constant extensional stiffness EA and the partial lengths l1 and l2 . At the point x D l1 a single axial force F is applied. We divide the bar into two segments 0  x  l1 and l1  x  l1 C l2 . In segment 0  x  l1 the integration rule is: EAu001 D 0; EAu01 D N1 D C1 ; EAu1 D C1 x C C2 :

(3.54)

For the segment l1  x  l1 C l2 we have: EAu002 D 0; EAu02 D N2 D D1 ; EAu2 D D1 x C D2 :

(3.55)

Fig. 3.10 Clamped bar under single force F .

F

EA

x ll

l2

z

Fl 2 ll+ l2 (+)

N(x)

(-)

Fl1l2 EA(l l + l 2(

u(x)

(+)

- Fl 1 ll+ l2

56

3

Fig. 3.11 Exemplary transition conditions for bars.

Ni

ni

n i+1

Ni+1

Ni

ui = ui+1 EAi u i =EAi+1ui+1

Ni

ni

n i+1

ni

n i+1

Bars

Ni+1

ui = ui+1 EAi u i =EAi+1ui+1

Ni+1

Ni

ni

n i+1

Ni+1 F

ui = ui+1 EAi u i =EAi+1ui+1

ui = ui+1 EAi u i =EAi+1ui+1+F

The boundary and transition conditions for the given bar situation are: u1 .x D 0/ D 0; u1 .x D l1 / D u2 .x D l1 /; EAu01 .x D l1 / D EAu02 .x D l1 / C F; u2 .x D l1 C l2 / D 0:

(3.56)

The integration constants are obtained from this as: F l2 ; C1 D l1 C l2



C2 D 0;

 l2 D1 D F 1 ; l1 C l2

D 2 D F l1 :

(3.57)

The distributions of N.x/ and u.x/ are also shown in Fig. 3.10. In Fig. 3.11 typical transition conditions between two bar segments i and i C 1 are shown. Example 3.5

We reconsider the situation from Example 3.3 and introduce two local reference axes x1 and x2 as shown in Fig. 3.12. The normal force distribution N as well as the displacement u are to be determined. We solve this problem by integration, dividing the bar into two segments. In segment 1 we obtain with n1 D 0: EAu01 D 0; EAu01 D N1 D C1 ; EAu1 D C1 x1 C C2 :

(3.58)

EAu02 D 0; EAu02 D N2 D D1 ; EAu2 D D1 x2 C D2 :

(3.59)

For segment 2 we have:

3.1 Bars

57

Fig. 3.12 Bar under two single forces.

l2

EA x2 F0

l1

x1

2F0

The following boundary conditions apply to the given situation. At the free end of the bar at x1 D 0 the normal force N1 corresponds to the applied force 2F0 : N1 .x1 D 0/ D 2F0 :

(3.60)

From this the integration constant C1 follows as: C1 D 2F0 :

(3.61)

Furthermore, at the clamped end of the bar at x2 D l2 the bar displacement u2 must vanish: u2 .x2 D l2 / D 0: (3.62) This results in the constant D2 as: D2 D 3F0 l2 :

(3.63)

At the transition point x1 D l1 or x2 D 0 between the two segments 1 and 2 the following transition conditions have to be considered. On the one hand, the two displacements u1 and u2 must match at this location: u1 .x1 D l1 / D u2 .x2 D 0/:

(3.64)

From this follows the integration constant C2 as: C2 D F0 .2l1 C 3l2 /:

(3.65)

On the other hand, the equilibrium of forces in the axial direction must be fulfilled at the transition point, i.e.: N1 .x1 D l1 / D N2 .x2 D 0/  F0 :

(3.66)

58

3

Bars

From this the still missing constant D1 results as: D1 D 3F0 :

(3.67)

With the integration constants C1 , C2 , D1 and D2 determined in this way, the normal forces N1 and N2 and the displacements u1 and u2 can be represented as: N1 D 2F0 ; N2 D 3F0 ; F0 .2x1  2l1  3l2 /; u1 D EA   3F0 l2 x2 1 : u2 D EA l2

(3.68)

The bar displacement u1 .x1 D 0/ follows as: u1 D 

F0 .2l1 C 3l2 /: EA

(3.69)

The negative sign indicates that this displacement is opposite to the direction of the x1 -axis. This result thus agrees with (3.38). J Example 3.6

Consider a bar consisting of two segments with constant cross-section A, which is supported at the lower end and held against lateral deflection at the upper end by a horizontal bar (EA ! 1) (Fig. 3.13). The bar is loaded by the single force F0 and the two constant line loads n1 and n2 parallel to the bar axis. The upper section of the bar has the elastic modulus E1 , the lower section has the elastic modulus E2 . We are looking for the distribution of the displacements u.x1 / and u.x2 / as well as the distributions of the normal forces N1 .x1 / and N2 .x2 /. What Fig. 3.13 Bar subjected to single force and normal line load.

F0

EA ∞

l n1

x1

n2

x2

l

E1,A

E2,A

3.1 Bars

59

is the magnitude of the normal force at the base of the column? What is the displacement of the upper point of the bar? The boundary and transition conditions for the given situation are: N1 .x1 D 0/ D F0 ; N1 .x1 D l/ D N2 .x2 D 0/; u1 .x1 D l/ D u2 .x2 D 0/; u2 .x2 D l/ D 0:

(3.70)

This problem is solved by integration: E1 Au001 D n1 ; E1 Au01 D N1 D n1 x1 C C1 ; 1 E1 Au1 D  n1 x12 C C1 x1 C C2 ; 2

(3.71)

and

E2 Au002 D n2 ; E2 Au02 D N2 D n2 x2 C D1 ; 1 (3.72) E2 Au2 D  n2 x22 C D1 x2 C D2 : 2 The integration constants C1 , C2 , D1 and D2 are determined from the boundary and transition conditions. From the condition N1 .x1 D 0/ D F0 follows:  n1  01 C C1 D F0 ;

(3.73)

C1 D F0 :

(3.74)

which can be solved as: The transition condition N1 .x1 D l/ D N2 .x2 D 0/ leads to the expression  n1 l  F0 D D1 :

(3.75)

From this, the constant D1 can be determined as: D1 D n1 l  F0 :

(3.76)

From the requirement u2 .x2 D l/ follows:   1 1  n2 l 2  n1 l 2  F0 l C D2 D 0: E2 A 2

(3.77)

This can be solved for D2 :

  1 D2 D l F0 C l n1 C n2 : 2

(3.78)

60

3

Bars

Finally, the transition condition u1 .x1 D l/ D u2 .x2 D 0/ is considered. It follows:  

  1 1 1 l  n1 l 2  F0 l C C2 D F0 C l n1 C n2 : E1 A 2 E2 A 2

(3.79)

From this, the integration constant C2 follows as:

  E1 1 1 l F0 C l n1 C n2 C n1 l 2 C F0 l: C2 D E2 2 2

(3.80)

Thus the displacements u1 .x1 / and u2 .x2 / can be given as follows:



 1 E1 1 1 1 2 2 l F0 C l n1 C n2 C n1 l C F0 l ;  n1 x1  F0 x1 C u1 .x1 / D E1 A 2 E2 2 2

  1 1 1 u2 .x2 / D  n2 x22  .n1 l C F0 /x2 C l F0 C l n1 C n2 : (3.81) E2 A 2 2 The normal forces N1 .x1 / and N2 .x2 / follow as: N1 .x1 / D n1 x1  F0 ; N2 .x2 / D n2 x2  n1 l  F0 :

(3.82)

The normal force at the lower support is given as: N2 .x2 D l/ D Œ.n1 C n2 /l C F0 :

(3.83)

The displacement at the upper end of the bar follows as:



 1 E1 1 1 2 l F0 C l n1 C n2 C n1 l C F0 l : u1 .x1 D 0/ D E1 A E2 2 2

(3.84) J

Example 3.7

Consider the column shown in Fig. 3.14 under its dead weight and single force F0 . The beam is divided into the segment 0  x  l1 with properties , A, E1 and the segment l1  x  l1 C l2 with , A, E2 , where l1 D 2l2 and E2 D 2E1 . We are looking for the distributions of the normal force and the axial displacement in the two segments. How large must F0 be so that the displacement at the point x D l1 becomes zero? Under the previously calculated force, at what temperature change does the displacement at the location x D l1 C l2 become zero? To answer the last question, we assume a constant temperature expansion coefficient ˛T over the length of the bar.

3.1 Bars

61

Fig. 3.14 Bar consisting of two segments loaded by a single force and its dead weight.

F0 g l2

E2,A,

E1,A, l1 x

The distribution of the normal force N.x/ is continuous over the two segments 1 and 2: N D Ax C C1 : (3.85) The integration constant C1 follows from the boundary condition N.x D l1 C l2 / D F0 as: C1 D .3Al2 C F0 /: (3.86) The normal force distribution results from this: N.x/ D A.x  3l2 /  F0 : From this, the bar stress  .x/ D

N.x/ A

(3.87)

can be determined as:

 .x/ D .x  3l2 / 

F0 : A

(3.88)

If the bar stress  .x/ is given, then the bar normal strain can be determined from Hooke’s law:  .x/ ".x/ D : (3.89) E Since there are different values for E in the two bar segments, the consideration of the bar strains must be done separately for each segment: 

1 F0 "xx;1 .x/ D ; .x  3l2 /  E1 A 

1 F0 "xx;2 .x/ D : (3.90) .x  3l2 /  2E1 A With the bar strain thus given, from the kinematic relation "xx D displacement u can be determined by integration: Z u.x/ D "xx dx C C:

du dx

the bar

(3.91)

62

3

Bars

Thus, in the two bar segments we have: 

 1 2 1 x  3l2 x   E1 2 

 1 2 1 u2 .x/ D x  3l2 x   2E1 2 u1 .x/ D

 F0 x C C2 ; A  F0 x C C3 : A

(3.92)

The integration constants C2 and C3 can be derived from the conditions u1 .x D 0/ D 0; We obtain: C2 D 0;

C3 D

u1 .x D l1 / D u2 .x D l1 /:

(3.93)

  1 F0 4l22  2 l2 : 2E1 A

(3.94)

The displacements are thus:  

 1 2 1 F0 x  3l2 x  x ;  u1 .x/ D E1 2 A 

  1 2 1 F0 2 u2 .x/ D x  3l2 x  4l2  .x C 2l2 / :  2E1 2 A

(3.95)

To determine the force F0 , the displacement is set to zero at the point x D l1 and solved for the force F0 . It follows: F0 D 2Al2 :

(3.96)

Thus, the following state variables occur in the bar: N.x/ D A.x  l2 /;  "xx;1 .x/ D .x  l2 /; E1  .x  l2 /; "xx;2 .x/ D 2E1    1 2 u1 .x/ D x  l2 x ; E1 2   1 2  u2 .x/ D x  l2 x : 2E1 2

(3.97)

To determine the temperature change T at which the displacement of the upper point of the bar becomes zero, the thermal displacement uT is calculated first. It results in: uT .x/ D ˛T T x: (3.98)

3.1 Bars

63

We require: u2 .x D 3l2 / C uT .x D 3l2 / D 0:

(3.99)

The resulting expression can be solved for the temperature change T , and we obtain: 1 l2 T D  : (3.100) 4 ˛T E1 J

3.1.3 Statically indeterminate bars The bar situation of Fig. 3.10 is statically indeterminate. At this point, we take a look at two more possibilities of analysis. Basically, in a statically indeterminate calculation, the bar deformations are to be used in the calculation in addition to the equilibrium conditions. For this purpose, we consider the free body image of Fig. 3.15, bottom. The equilibrium of forces in x-direction results in: AH  BH D F:

(3.101)

This is one equation for the two unknowns AH and BH . Another equation is obtained by considering the displacements of the two bar segments. The left bar segment with length l1 is under constant normal force N1 D AH , so the length change l1 of this segment is given by: l1 D

N1 l1 AH l1 D : EA EA

(3.102)

Similarly, for the length change l2 of the right bar segment we obtain: l2 D

N2 l2 BH l2 D : EA EA

(3.103)

To achieve geometrical compatibility of the deformation state, it is required that the total length change l of the bar becomes zero: l D l1 C l2 D 0: Fig. 3.15 Bar clamped on both sides under single force F (top), free body image (bottom).

(3.104)

F

EA

x ll

l2

z

AH

F

BH

64

3

Bars

This leads to the following equation after substituting (3.102) and (3.103): AH l1 C BH l2 D 0:

(3.105)

With (3.101) and (3.105) two equations for the determination of the support forces AH and BH are available. Solving for AH and BH gives: AH D

F l2 ; l1 C l2

BH D 

F l1 : l1 C l2

(3.106)

With the support reactions available, the normal forces of the bar and thus all other state variables can be determined. Another possibility of analysis is to first make the statically indeterminate member statically determinate and to determine the removed support reaction from the deformation state of the bar. This procedure is known as the so-called force method. For this purpose, we consider again the situation of Fig. 3.10 and make the bar statically determinate by removing the right restraint. The resulting situation is shown in Fig. 3.16, middle. The resulting statically determinate bar is now acted upon by the force F . This is the so-called 0-system. Due to the applied force F , the displacement u0 results at the force application point, whereas the right bar segment with length l2 does not change in its length. Thus: u0 D

F l1 : EA

(3.107)

The support reaction which in reality is present is now applied to the free right end of the bar (Fig. 3.16, bottom), where we use the designation X to make it clear that this is a statically indeterminate quantity that needs to be determined. This results in the displacement u1 at the free bar end, which can be determined as follows: u1 D 

Xl1 Xl2  : EA EA

(3.108)

The negative sign indicates that this displacement runs against the direction of the x-axis. Fig. 3.16 Bar clamped on both sides under single force F (top), 0-system (middle) and 1-system (bottom).

F

EA

x ll

l2

z

0-system

F

u0

u0

1-system u1

X

3.2 Bar systems

65

The total displacement u results from the superposition of the two displacements u0 and u1 : F l1 Xl1 Xl2 u D u0 C u1 D   : (3.109) EA EA EA However, since in reality there is a restraint at the right end of the bar, the displacement u must become zero: F l1 Xl1 Xl2   D 0: EA EA EA

(3.110)

This expression can be solved for X, and we obtain: XD

F l1 : l1 C l2

(3.111)

The statically indeterminate quantity X is identical to the support reaction BH , so that: F l1 BH D : (3.112) l1 C l2 Thus, the result (3.106) is confirmed, where the different signs can be attributed to the different directions of action of the considered forces.

3.2 Bar systems 3.2.1 Statically determinate bar systems In this section, we want to study the deformations of statically determinate bar systems and, for introduction, consider the system shown in Fig. 3.17, left. Given are the two bars 1 and 2 of equal extensional stiffness EA. The vertically acting single force F acts at the point of intersection where an ideal hinge is assumed. The nodal displacements u and w are to be determined as shown in Fig. 3.17, right.

F EA 1

w

45°

2l

l EA

2

Fig. 3.17 Truss under single force F (left), deformed system (right).

u

66

3

Bars

Since this is a statically determinate system, the bar forces N1 and N2 can be determined from the equilibrium conditions. They follow as: N1 D F;

p N2 D  2F:

(3.113)

From this, we can then determine the length changes l1 and l2 of the bars. They are: N1 l1 Fl l1 D D ; EA EA N2 l2 2F l D ; (3.114) EA EA where we find that both bars become shorter. We can now use the length changes of the bars determined in this way to determine the node displacements we are looking for. The resulting deformation state is shown in Fig. 3.17, right. The calculation of the two nodal displacements u and w can then be performed as follows (see Fig. 3.18). The two bars will move on a circular path around their support points when loaded (Fig. 3.18, left), where we have to consider here the changes in length l1 and l2 of the bars due to the bar forces N1 and N2 . Accordingly, the circular path forpbar 1 has the radius l C l1 , whereas for bar 2 the radius of the circular path is 2l C l2 . The point where the two circular arcs intersect is the hinge position in the deformed state. Since we assume that the deformations remain small, we can replace the circular arcs by their tangents, as shown in Fig. 3.18, middle. From elementary trigonometric considerations we can then determine the two displacements u and w we are looking for (Fig. 3.18, right). We have: l2 D

u D jl1 j D

Fl ; EA

p 2 Fl  1C2 2 : w D jl1 j C p jl2 j D EA 2 ∆l1

∆l1

∆l2

∆l1

∆l2

w

w

(3.115)

1 ∆l 2 2

w

45°

∆l ∆l1+ 22 u

u u Fig. 3.18 Determination of the node displacements u and w.

∆l2

∆l2 2

3.2 Bar systems

67

Example 3.8

Consider the truss of Fig. 3.19, which consists of two bars with extensional stiffness EA at an angle of inclination ˛ to the horizontal and is loaded at its nodal point by a horizontal force F . The displacement u is to be determined. We first determine the two bar forces N1 and N2 , which are identical for the given example: F N1 D N2 D N D  : (3.116) 2 cos ˛ From this, the length changes of the bars with the length l1 D l2 D determined as: Fl : l1 D l2 D l D  2EA cos2 ˛

l cos ˛

can be (3.117)

From the displacement plan of Fig. 3.19, right, we thus determine: uD

Fl jlj D : cos ˛ 2EA cos3 ˛

(3.118) J

Example 3.9

Consider the truss of Fig. 3.20, which consists of two bars of extensional stiffness EA and is loaded at its nodal point by the vertical force F . The two nodal displacements u and w are to be determined. We first determine the two bar forces N1 and N2 from nodal equilibrium and obtain: F F N1 D ; N2 D  : (3.119) tan ˛ sin ˛ Thus, the length changes l1 and l2 of the bars with the lengths l1 D l and l2 D cosl ˛ can be given as: l1 D

N1 l1 Fl D ; EA EA tan ˛

l2 D 

Fl : EA sin ˛ cos ˛

(3.120)

From the displacement plan of Fig. 3.20, right, we can determine the displacements u and w as: u D l1 D

Fl ; EA tan ˛

  1 1 u Fl jl2 j C C D : wD sin ˛ tan ˛ EA sin2 ˛ cos ˛ tan2 ˛

(3.121) J

68

3

Fig. 3.19 Truss (left), deformed system and displacement plan (right).

Bars

α 2

EA

F u 1

EA

∆l1 ∆l2 u

l

Fig. 3.20 Truss (left), deformed system and displacement plan (right)

F 1 w

α

EA

u

EA

2

∆l1 ∆l2 ∆l2 sinα

l

w

u tanα

α

u

3.2.2 Statically indeterminate bar systems If a statically indeterminate system is given, then we must consider the material law and the kinematics of the given system in addition to the equilibrium conditions. To illustrate this, we consider again the truss from Example 3.8, which now includes an additional horizontal bar of length l2 (Fig. 3.21). The two bars under the angle ˛ have the extensional stiffness EA1 , the horizontal bar has the extensional stiffness EA2 . We want to determine the nodal displacement u and the bar forces of the system. We first consider the nodal section of Fig. 3.21, top right, where we can already assume that the bar forces of the two diagonal bars are identical due to the symmetry of the system. The sum of the horizontal forces gives: 2N1 cos ˛ C N2 C F D 0:

(3.122)

The length changes l1 and l2 of the bars can be given as: N1 l1 N1 l D ; EA1 EA1 cos ˛ N2 l2 l2 D : EA2 l1 D

(3.123)

3.2 Bar systems

69

Fig. 3.21 Truss (left), nodal section and displacement plan (right)

N1

α

N2 EA1

α α

F

N1

EA2

F

EA1

∆ l2

∆ l1 ∆ l1

u l2 l

From the displacement plan of Fig. 3.21, bottom right, we can conclude: l1 D u cos ˛;

l2 D u:

(3.124)

Substituting (3.124) into (3.123), the resulting expressions can be solved for the bar forces N1 and N2 as follows: N1 D

EA1 u cos2 ˛ ; l

N2 D

EA2 u : l2

(3.125)

Substituting the bar forces into the equilibrium condition (3.122), we obtain the displacement u as: F : (3.126) uD 2EA1 cos3 ˛ EA2 C l l2 With the now known nodal displacement u the bar forces N1 and N2 can be determined from (3.125): F cos2 ˛ ; N1 D  EA2 l 3 2 cos ˛ C EA1 l2 EA2 l F EA1 l2 : N2 D  EA2 l 2 cos3 ˛ C EA1 l2

(3.127)

70

3

Bars

Example 3.10

Consider a rigid bar of length 2a supported by two linear elastic bars (stiffness EA) and loaded by a single force F as shown in Fig. 3.22. The bar forces N1 and N2 , the length changes l1 and l2 of the two elastic bars, the deflection w of the force application point and the support reactions at the left support are to be determined. Using the free body image of Fig. 3.22, middle, we first form the equilibrium conditions. The two force sums in horizontal and vertical direction as well as the moment sum around the left support result in: 1 AH C p N2 D 0; 2 1 AV C N1 C p N2  F D 0; 2 p N1 C 2N2  2F D 0:

(3.128)

Obviously, these three equations are not sufficient to determine the four forces AH , AV , N1 , N2 occurring here. We must therefore include the deformations of the system in the calculations. The length changes l1 and l2 of the two elastic bars can be formulated as: N1 l1 N1 l l1 D D ; EA EA

N2 l2 l2 D D EA

Fig. 3.22 Considered system (top), free body image (middle), displacement plan (bottom).

p

2N2 l : EA

(3.129)

rigid

F EA

1

l

EA

2

l

l

AH AV

F

N1 N2 ∆l1

∆ l2

w

3.2 Bar systems

71

Using the displacement plan of Fig. 3.22, bottom, we can derive the following relationship: 1 l2 ; (3.130) cos 45ı D p D w 2 from which: wD

p 2l2 :

(3.131)

Moreover, we have the following relation between l1 and the deflection w: l1 D

w : 2

(3.132)

Combining the two expressions (3.131) and (3.132), it follows that there is the following relationship between the length changes l1 and l2 : 1 l1 D p l2 : 2

(3.133)

With (3.128), (3.129) and (3.133) there are six equations for the unknown quantities AH , AV , N1 , N2 , l1 , l2 . These can be solved as follows. First, we put the expressions (3.129) for the length changes l1 and l2 into the kinematic relation (3.133), which leads to N1 D N2 :

(3.134)

Thus, the equilibrium conditions (3.128) can be solved for the unknown forces as follows: p  p 2 2 p F; AV D 1  2 F; N1 D N2 D p F: (3.135) AH D  1C 2 1C 2 With the bar forces thus determined, the length changes l1 and l2 as well as the deflection w can finally be determined: l1 D 

2F l ; p 1 C 2 EA

l2 D 

p 2 2F l ; p 1 C 2 EA

4F l wD : (3.136) p 1 C 2 EA J

4

Beams

A beam is a straight one-dimensional structural element subjected to bending. The constitutive law for the beam under normal force and bending is first derived for an arbitrary coordinate system, where different recurring area integrals are also introduced. Special attention is paid to the so-called moments of inertia and the so-called deviation moment. The two cross-sectional normalizations, i.e. the reference of the considerations to the center of gravity of the considered cross-section and the transfer to the so-called principal axis system, simplify the considerations enormously. Finally, the bending stresses occurring in beams are to be determined. Students are enabled to apply the basic equations for beam structures and to determine normal stresses in arbitrary cross-sections.

4.1 Introduction This chapter is dedicated to the treatment of straight beams under normal force and bending. The loading of the beam structures that we will discuss in this chapter cause bending moments and transverse shear forces where we want to consider normal forces in this chapter as well. Torsional action is excluded at this point, this topic will be discussed in Chap. 7. The load may occur in the form of the two line loads qy and qz (shown in Fig. 4.1 as constant along the longitudinal axis x), which can be arbitrary functions of x. Similarly, there can be any number of single forces Fy and Fz at arbitrary locations on the beam. Single bending moments are also allowed as loads, but are not shown in Fig. 4.1. The load of the beam produces internal internal forces and moments, which we can interpret as resultants of the internal beam stresses. In the beam structures considered here, the normal force N , the two bending moments My and Mz , and the two transverse shear forces Vy and Vz are present as internal internal forces and moments (Fig. 4.2). The corresponding displacements and rotations are the three displacements u, v and w in x-, y- and z-directions and the two rotations 'y and 'z , the so-called bending angles. The displacments and rotations are shown in Fig. 4.1, © The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_4

73

74

4 Beams Fz qz

Fy

y

y qy

Vy

z

N

My

Vz

z

φy

v

u w

x

Mz

x φz

Fig. 4.1 Loads, forces and moments of the beam (left), displacements (right).

Fz qz

Fy

y

y qy

Vy

z

xx

N

My

Vz

z x

Mz

N

x

Fz

Fz

qz

qz

y

y

xx z

My

z x

Vz

x

Fy

y

Fy

y

qy

qy Vy

z

x

xx

z x

Mz

Fig. 4.2 Internal forces and moments of the beam and the associated stress components.

4.2 Basic equations for an arbitrary reference frame

75

right. A rotation of the cross-section about the x-axis, as it occurs in torsion, is excluded here. Fig. 4.2 shows to what extent the forces and moments N , My , Mz , Vy and Vz are related to the stress components occurring in the beam. The stresses in the beam are the normal stress xx and the shear stress . The loads qz and Fz cause the bending moment My and the transverse shear force Vz , whereas qy and Fy are responsible for the occurrence of the bending moment Mz and the transverse shear force Vy . Furthermore, a normal force N may also be present. From Fig. 4.2 it can be seen that the normal force N gives rise to the normal stress xx which, under certain conditions yet to be discussed, is constantly distributed over the crosssection. We will show in this chapter that the bending moment My also causes the normal stress xx , which is then linearly distributed over the cross-section as shown in Fig. 4.2, center left. The transverse shear force Vz is the cause of the shear stress  as shown in Fig. 4.2, center right. The determination of shear stresses due to shear forces is the subject of Chap. 6 and will not be discussed in the present chapter. It can be shown that, for the specific case discussed here, the shear stress is linear in the flanges of the I-section and has a parabolic distribution in the web of the crosssection. The direction of action of the shear stress is indicated by arrows, as shown in Fig. 4.2, center right. Because of the circumferential action of , indexing of this stress component is often omitted. Like My , the bending moment Mz evokes the normal stress xx , which also varies linearly across the cross-section, as shown in Fig. 4.2, bottom left. The transverse shear force Vy , on the other hand, causes the shear stress  (Fig. 4.2, bottom right). In practice, the main task of engineers is to determine the stresses in beam structures and to design components in such a way that the existing stresses do not exceed certain limits. Likewise, in many cases, the deformations of such structures have to be considered and verified accordingly. In this chapter, we will consider beam structures and the determination of the normal stress xx . The determination of shear stresses due to transverse shear forcess and torsion will be considered in the Chaps. 6 and 7. The analysis of deformations of beams is dealt with in Chap. 5.

4.2 Basic equations for an arbitrary reference frame The beam theory we will discuss in this chapter goes back to Leonhard Euler and Jakob Bernoulli1 and is appropriately referred to as Euler–Bernoulli beam theory. This elementarily important theory, widely used in engineering practice, is based on the following assumptions:  We assume that the hypothesis of flat cross-sections as well as the normal hypothesis are valid. Cross-sections do not warp when the beam deflects and are orthogonal to the longitudinal axis of the beam even in the deformed state (Fig. 4.3). 1

Leonhard Euler, 1707–1783, and Jakob Bernoulli, 1654–1705, both Swiss mathematicians and physicists.

76

4 Beams

Fig. 4.3 Hypothesis of flat cross-sections and normal hypothesis.

q

Center line

    

As a consequence, shear strains in the beam are explicitly excluded. However, we will see later that this assumption leads to contradictions, but we use it as a suitable basis for all further considerations. We assume linear-elastic material behavior, Hooke’s law is valid. The cross-sectional dimensions are significantly smaller than the length of the beam. An idealization within the framework of a beam theory is thus justified. The cross-sectional shape is completely preserved during the deformation of the beam under load. We assume geometric linearity, i.e. the deformations are assumed to be small. Only straight beam structures are considered.

Hooke’s law for the beam situation considered here is as follows: xx D E"xx ;

 D G:

(4.1)

We will start the explanations with Fig. 4.4, where we use NxN for the normal force N . At an arbitrary point of the cross-section, which does not necessarily have

y y My

Center of gravity C

x z Mz

v Nx φy vP(x,y,z) uP(x,y,z)

x P

wP(x,y,z)

u

z w

φz Fig. 4.4 Forces and displacements in an arbitrary reference system x; N y; N zN and displacements uN P , vNP and wN P of the point P.

4.2 Basic equations for an arbitrary reference frame Fig. 4.5 Kinematics of the Euler–Bernoulli beam theory.

77

h

x zP P

z Undeformed

u uP

w wP

-φy zP ∆u

Deformed

h

-φy

P

∆u

dw dx

to be the center of gravity C, a Cartesian coordinate system is introduced, where we denote the axes as x, N y, N zN . The overline indicates that this is a reference frame at an arbitrary point of the cross-section. In the following we want to determine the deflection wN P as well as the longitudinal displacement uN P of an arbitrary point P, which is located at an arbitrary position of the cross-section. The displacement vNP in the direction of the reference axis y, N which generally also occurs, is taken into account at a later point. The resulting displacements are u, N vN and zN as well as the two bending angles 'yN and 'zN . We now consider the kinematics of the Euler–Bernoulli beam shown in Fig. 4.5 by means of a sectional element of the beam in the undeformed and in the deformed state. Let the height of the cross-section be h. Due to the bending moment and normal force, the section element will undergo both a deflection wN and a longitudinal displacement u. N Due to the deformation, the inclination ddwxNN D wN 0 is also called forth. The displacements concerning point P are denoted as wN P and uN P . Due to the rotation 'yN the coordinate origin will suffer a longitudinal displacement uN which is different from the displacement uN P of point P. Let the difference of these two displacement values be called u. N The difference uN can be obtained from the triangle shown in Fig. 4.5, right: sin.'yN / D

uN : zNP

(4.2)

Since in the context of the present beam theory we assume small deformations and thus also small rotations 'yN and 'zN , sin.'yN / ' 'yN can be assumed so that: uN D NzP 'yN :

(4.3)

Based on the hypothesis that no cross-sectional distortions occur and the normal hypothesis is valid, the slope ddwxNN D wN 0 of the beam deflection wN can be equated with the bending angle 'yN : (4.4) uN D zNP wN 0 :

78

4 Beams

The longitudinal displacement uN P of the point P is thus: uN P D uN  uN D uN  zNP wN 0 :

(4.5)

This relation is valid exclusively for the case of the so-called uniaxial bending, i.e. for the case that only the deformations uN and wN occur in the xN z-plane. N If the given load leads to deformations uN and wN as well as the displacement v, N then we speak of biaxial bending. In this case, (4.5) is extended by the displacement vN as follows: uN P D uN  zNP wN 0  yNP vN 0 : (4.6) Here yNP is the distance of the point P from the origin of the reference frame with respect to the y-direction. N Since we assume that the cross-sectional dimensions do not change due to the beam deformation, we can equate the deflection wN of the coordinate origin and the deflection wN P of the point P: wN P D w: N

(4.7)

This is also true for the displacements in the y-direction: N vNP D v: N

(4.8)

The equations (4.6), (4.7) and (4.8) describe the displacement field of the Euler– Bernoulli beam. From this, the strain field of the beam can be determined. Since we assume a slender beam where the length is significantly larger than its crosssectional dimensions and we have also excluded any shear strains from the outset, the only remaining strain component is the normal strain "xN xP N . It is calculated from the kinematic equation (1.33) as: "xN xP N D

duN P D uN P0 D uN 0  zNP wN 00  yNP vN 00 : dxN

(4.9)

Since the relations derived so far have validity for arbitrary points of the crosssection, we will omit the index P in the following. From Hooke’s law (4.1), the normal stress xN xN of the beam is obtained as follows:   xN xN D E"xN xN D E uN 0  zN wN 00  yN vN 00 :

(4.10)

The normal stress xN xN leads to the following normal force and bending moments of the beam: Z Z Z NxN D xN xN dA; MyN D xN xN zdA; N MzN D  xN xN ydA: N (4.11) A

A

A

4.2 Basic equations for an arbitrary reference frame

79

Inserting (4.10) yields: Z NxN D E A

D E uN 0 Z MyN D E A

D E uN 0

 Z

 uN 0  zN wN 00  yN vN 00 dA dA  E wN 00

A

MzN D E A

D E uN 0

Z ydA; N A

 uN 0  zN wN 00  yN vN 00 zdA N zN dA  E wN 00

A

Z

zN dA  E vN 00

A

 Z

Z

Z

zN 2 dA  E vN 00

A

 Z

Z yN zN dA; A

 uN 0  zN wN 00  yN vN 00 ydA N ydA N C E wN 00

A

Z

Z

yN zdA N C E vN 00

A

yN 2 dA:

(4.12)

A

A series of recurring area integrals that are typical for beam structures occurs at this point. The cross-sectional area A of the beam in units of a length to the second power is defined as follows: Z AD

dA:

(4.13)

A

Furthermore, in (4.12) first-order area integrals occur, namely the so-called static moments SyN and SzN : Z Z N SyN D zN dA: (4.14) SzN D ydA; A

A

The static moments are in the unit of a length to the third power. Finally, the second-order area integrals occurring in (4.12) are the so called moments of inertia IyN yN , IzN zN as well as the so-called deviation moment IyN zN : Z IyN yN D A

zN 2 dA;

Z IzN zN D A

yN 2 dA;

Z IyN zN D

yN zN dA:

(4.15)

A

The two moments of inertia IyN yN and IzN zN always take values greater than zero. The deviation moment IyN zN , on the other hand, can have both positive and negative values or even become zero in special cases as we will discuss later. Both the moments of inertia and the deviation moment have the unit of a length to the fourth power.

80

4 Beams

A further parameter which can be derived from this is the so-called polar moment of inertia Ip , which results from the sum of the two moments of inertia IyN yN and IzN zN : Z Z  2  2 (4.16) Ip D r dA D yN C zN 2 dA D IyN yN C IzN zN : A

A

With (4.13), (4.14) and (4.15) the equation system (4.12) can be represented as follows: NxN D EAuN 0  ESyN wN 00  ESzN vN 00 ; MyN D ESyN uN 0  EIyN yN wN 00  EIyN zN vN 00 ; MzN D ESzN uN 0 C EIyN zN wN 00 C EIzN zN vN 00 : (4.17) This can be represented in a vector-matrix-notation as follows: 0 1 2 30 0 1 NxN A SyN SzN uN @ MyN A D E 4SyN IyN yN IyN zN 5@wN 00 A: MzN SzN IyN zN IzN zN vN 00

(4.18)

It follows that given the choice of an arbitrary coordinate system x, N y, N z, N the normal N force NxN and the bending moments MyN , MzN are coupled with all displacements u, v, N w. N The set of equations (4.17) or (4.18) is also called the constitutive law of the beam formulated here for the arbitrarily chosen reference system x, N y, N zN . The products EIyN yN and EIzN zN are called bending stiffnesses. Example 4.1

We demonstrate the determination of the are integrals occurring in (4.12) at the example of a rectangular cross-section (height h, width b) as shown in Fig. 4.6. We first consider the area integral (4.13), i.e. the cross-sectional area A. For such an elementary simple rectangular cross-section, the area can be determined immediately as A D bh. However, we want to formally evaluate the area integral (4.13) in order to learn how to deal with such integral expressions. We use the infinitesimal area element dA for the calculation as shown in Fig. 4.6, bottom right, i.e. dA D dydN N z . Then it follows from (4.13): Z AD

Zh Zb dA D

A

0

dydN N z D yj N b0 zN jh0 D bh:

(4.19)

0

Next, we determine the static moments SyN and SzN . To calculate SzN we use the infinitesimal area element of Fig. 4.6, bottom center, i.e. we set dA D hdyN and integrate from yN D 0 to yN D b: Z SzN D

Zb ydA N D

A

0

ˇb hyN 2 ˇˇ hb 2 hyd N yN D D : ˇ 2 0 2

(4.20)

4.2 Basic equations for an arbitrary reference frame

81

b y

h

z y

y

y dA

z

z dA y dz dy

y

dA dy dz

z

z

z

Fig. 4.6 Rectangular cross-section (top), different types of infinitesimal area elements dA (bottom).

Analogously we can proceed for the determination of SyN and obtain with the infinitesimal area element dA D bdNz : Z SyN D

Zh zdA N D

A

0

ˇh b zN 2 ˇˇ bh2 b zN dNz D D : ˇ 2 0 2

(4.21)

Finally, we determine the moments of inertia IyN yN and IyN zN as well as the deviation moment IyN zN . For the moment of inertia IyN yN it follows from (4.15) with the infinitesimal element dA D bdNz (Fig. 4.6, bottom left): Z IyN yN D

2

Zh

zN dA D A

0

ˇh b zN 3 ˇˇ bh3 b zN dNz D D : 3 ˇ0 3 2

(4.22)

Analogously, for IzN zN we obtain using dA D hdyN Z IzN zN D

2

Zb

yN dA D A

0

hyN 2 dyN D

ˇb hyN 3 ˇˇ hb 3 D : ˇ 3 0 3

(4.23)

82

4 Beams

Finally, we determine the deviation moment IyN zN using the infinitesimal area element dA of Fig. 4.6, bottom right, i.e. dA D dydN N z . Then: Z IyN zN D

Zh Zb yN zdA N D

A

0

0

ˇb ˇh yN 2 ˇˇ zN 2 ˇˇ b 2 h2 yN zN dydN N zD D : 2 ˇ0 2 ˇ0 4

(4.24)

The constitutive law (4.18) can then be represented for the current example as: 2

bh2 2 bh3 3 b 2 h2 4

6 bh 1 6 NxN 6 2 @ MyN A D E 6 bh 6 6 2 MzN 4 hb 2 2 0

3 hb 2 0 1 2 7 7 uN 0 b 2 h2 7 7@wN 00 A: 7 4 7 vN 00 35 hb 3

(4.25)

J

In many technically relevant cases, cross-sections of beams are composed of a number of segments whose area integrals are easy to determine. If such a case exists, then the area integrals of (4.13), (4.14) and (4.15) can be determined as the sum of the partial integrals over the segments of the cross-section. If a cross-section is given with n segments, we obtain: Z dA D

AD A

n Z X i D1 A

ydA N D

n Z X

A

i D1 A

Z

n Z X

SyN D Z IyN yN D

ydA N iD

zN dAi D

i D1 A

n Z X

A

Z

n Z X

i

yN 2 dA D

A

Z

n Z X

A

Sy;i N ;

n X

yN 2 dAi D

IyN y;i N ;

i D1 A

i

n X

IzN zN ;i ;

i D1

i

yN zN dA D

Sz;i N ;

i D1

i D1 A

IyN zN D

n X

zN 2 dAi D

i D1 A

IzN zN D

n X

i D1

i

zN 2 dA D

Ai ;

i D1

i

zdA N D A

n X i D1

i

Z SzN D

dAi D

yN zN dAi D

n X i D1

IyN z;i N :

(4.26)

4.2 Basic equations for an arbitrary reference frame

83

Example 4.2

Consider the thin-walled I-section of Fig. 4.7, which has the constant wall thickness t, the web height h and the flange width 2b. The area integrals (4.26) are to be evaluated, where the cross-section is to be divided into n D 3 segments. The wall thickness of the cross-section is halved at each point by the so called skeleton line. We start our considerations by determining the cross-sectional area. For the partial areas we obtain: A1 D 2bt;

A2 D .h  t/t;

A3 D 2bt:

(4.27)

The total area then follows as: A D A1 C A2 C A3 D 4bt C ht  t 2 :

(4.28)

If we assume a very thin-walled cross-section with t  b; h, then the last term of order t 2 is negligible, and the following expression remains: A D t.4b C h/:

(4.29)

Furthermore, we consider the static moment SzN . We have: SzN D

3 Z X

ydA N iD

i D1 A

i

3 X

Sz;i N :

(4.30)

i D1

With the area elements dA1 D tdyN (upper flange), dA2 D hdyN (web), dA3 D tdyN (lower flange) the following static moments Sz;i N of the partial areas result: Z2b Sz;1 N D 0

ˇ2b t yN 2 ˇˇ t yd N yN D D 2b 2 t; 2 ˇ0

bC 2t

Z

Sz;2 N D b 2t

ˇbC t hyN 2 ˇˇ 2 hyd N yN D D bht; 2 ˇb t 2

2 Sz;3 N D Sz;1 N D 2b t:

Fig. 4.7 Thin-walled I-section.

(4.31)

t

1

y t

2

h t

b

z 3

b

84

4 Beams

The static moment SzN of the cross-section then follows as: SzN D

3 X

Sz;i N D bt.4b C h/:

(4.32)

i D1

We can proceed quite analogously for the static moment SyN and obtain:   h SyN D th C 2b : 2

(4.33)

When determining the static moments, it turns out that the calculation can be simplified quite considerably. From the integration rule (4.26) it follows that in the case of a cross-section consisting of n segments, it is sufficient to form the sum over the partial areas Ai , multiplied by the locations yNi or zNi of their center of gravity: n n X X Ai yNi ; SyN D Ai zN i : (4.34) SzN D i D1

i D1

This is shown for the static moment SzN . With A1 D 2tb;

A2 D t.h  t/;

A3 D 2tb;

yN1 D b;

yN2 D b;

yN3 D b

(4.35)

we obtain: SzN D

3 X

Ai yNi D 2tb 2 C t.h  t/b C 2tb 2 D 4tb 2 C thb  t 2 b:

(4.36)

i D1

Assuming a very thin-walled cross-section for which t  b; h holds, the last term in (4.36) can be neglected, and we have: SzN D bt.4b C h/:

(4.37)

Apparently, this result is identical to (4.32). We now determine the moment of inertia IyN yN . For this purpose the infinitesimal area elements dA1 D 2bdNz , dA2 D tdNz , dA3 D 2bdNz are used. We obtain

4.2 Basic equations for an arbitrary reference frame

85

for the partial moments of inertia: Z IyN y;1 N D

t

Z2

2

zN dA1 D Z

IyN y;2 N D

2

 2t

A1

Zh

2

ˇt 2b zN 3 ˇˇ 2 bt 3 2b zN dNz D D ; ˇ 3 t 6 2

zN dA2 D

ˇh t zN 3 ˇˇ th3 t zN dNz D D ; 3 ˇ0 3

A2

0

Z

hC 2t

IyN y;3 N D

zN 2 dA3 D

Z

2

ˇhC t   2b zN 3 ˇˇ 2 2bt t2 2 D C 3h : 3 ˇh t 3 4

2b zN 2 dNz D

2

h 2t

A3

(4.38)

The moment of inertia then follows from the sum of the partial moments of inertia as follows:   bt 3 t2 th3 2bt 2 IyN yN D C C 3h C : (4.39) 6 3 3 4 If we assume a very thin-walled cross-section with t  b; h, then in (4.39) terms of order t 3 are negligible, and the following remains: IyN yN D th2



 h C 2b : 3

(4.40)

Furthermore, we determine the moment of inertia IzN zN according to (4.26) and first determine the partial moments of inertia IzN z;1 N , IzN z;2 N , IzN z;3 N . With the area elements dA1 D tdy, N dA2 D hdy, N dA3 D tdyN we have: Z IzN z;1 N D

2

Z2b

yN dA1 D

ˇ2b t yN 3 ˇˇ 8tb 3 t yN dyN D D ; 3 ˇ0 3

A1

0

Z

bC 2t

IzN z;2 N D

2

Z

yN dA2 D A2

Z

IzN z;3 N D A3

b 2t

2

ˇbC t   hyN 3 ˇˇ 2 ht t2 2 hyN dyN D D C 3b ; 3 ˇb t 3 4 2

yN 2 dA3 D IzN z;1 N D

2

8tb 3 : 3

(4.41)

86

4 Beams

Thus, the moment of inertia can be determined from the sum of the partial moments of inertia:   16 3 ht t2 2 3b C : (4.42) IyN yN D tb C 3 3 4 If we again neglect terms of order t 3 under the assumption of a thin-walled crosssection, we obtain:   2 16 IyN yN D tb bCh : (4.43) 3 Finally, the deviation moment IyN zN must be determined. For the partial moments we have: t

Z IyN z;1 N D

Z 2 Z2b yN zN dA1 D

t

Zh bC Z 2

Z IyN z;2 N D

2

 2t 0

A1

ˇ2b ˇ t yN 2 ˇˇ zN 2 ˇˇ 2 yN zd N ydN N zD D 0; 2 ˇ0 2 ˇ t

yN zN dA2 D A2

0 b t 2

Z

hC 2t 2b Z Z

IyN z;3 N D

yN zN dA3 D h 2t

A3

0

ˇbC t ˇh yN 2 ˇˇ 2 zN 2 ˇˇ bh2 t yN zN dydN N zD D ; 2 ˇb t 2 ˇ0 2 2

ˇ2b ˇhC t yN 2 ˇˇ zN 2 ˇˇ 2 yN zd N ydN N zD D 2b 2 ht: 2 ˇ0 2 ˇh t

(4.44)

2

Thus the deviation moment results as:   h IyN zN D tbh C 2b : 2

(4.45)

The constitutive law (4.18) then takes the following form: 2

h th C 2b 2

!

3

bt.4b C h/ 7 6 t.4b C h/ 7 6 6 ! ! ! 70 uN 0 1 NxN 7 6 h h h 7 @ MyN A D E 6 C 2b th2 C 2b tbh C 2b 7@wN 00 A: 6th 7 6 2 3 2 MzN 6 ! !7 vN 00 7 6 h 16 5 4 bt.4b C h/ tbh C 2b tb 2 bCh 2 3 (4.46) J 0

1

4.2 Basic equations for an arbitrary reference frame

87

Example 4.3

Consider the Z-section shown in Fig. 4.8 with constant wall thickness t. We want to determine the area integrals according to (4.26). In the following, only the results are presented without going into details concerning the analysis itself: A D t.2b C h/;   h Cb ; SyN D th 2 SzN D tb.2b C h/;   h Cb ; IyN yN D th2 3   8b IzN zN D tb 2 Ch ; 3   h 3b IyN zN D tbh C : 2 2

(4.47)

The constitutive law (4.18) is then as follows for this cross-sectional form: 2

h th Cb 2

!

3

tb.2b C h/ 7 6 t.2b C h/ 7 6 6 ! ! !70 0 1 uN NxN 6 h 3b 7 h h 7@ 00 A @ MyN A D E 6 Cb th2 Cb tbh C 7 wN : (4.48) 6th 6 2 3 2 2 7 vN 00 MzN 6 ! !7 7 6 h 3b 8b 5 4 tb.2b C h/ tbh C tb 2 Ch 2 2 3 J 0

1

Fig. 4.8 Thin-walled Z-section.

t

y h

t

z

t

b

b

88

4.3

4 Beams

First cross-sectional normalization: Center of gravity C

4.3.1 Steiner’s theorem With equations (4.17) and (4.18), the constitutive law for the Euler–Bernoulli beam is given for an arbitrary choice of the reference frame. However, it turns out that here every force and moment quantity is coupled with every displacement quantity. This can be simplified quite substantially if the reference frame is shifted to a particular point where the two static moments Sz and Sy vanish. We will show below that this is the center of gravity C of the cross-section. The reference of the considerations to the center of gravity C of the cross-section is called first cross-sectional normalization. The shift of the reference frame is shown in Fig. 4.9. The center of gravity C is at the position .yN D yNC ; zN D zN C /. The reference system referred to the center of gravity C is designated as x, y, z. In the same way, let the related normal force, bending moments and displacements be denoted as N , My , Mz and u, v, w, whereas in the following we want to omit the indexing of the normal force N . The centroid axes x, y, z are related to the reference axes x, N y, N zN as follows: y D yN  yNC ;

x D x; N

z D zN  zNC :

(4.49)

The reference of the axis system to the center of gravity C thus corresponds to a pure translation of the reference system in the yN zN -plane. A prerequisite for the center of gravity coordinate system is that the two static moments Sz and Sy vanish: Z Z Z Sz D ydA D ydA N  yNC dA D 0; A

A

Z

A

Z

Sy D

zdA D A

Z zN dA  zN C

A

dA D 0:

(4.50)

A

yC zC My

y

C

y

x

y

Mz Mz

x

y v

z

N

My

z

Nx

x

z

C

v

u

x u

z

w

w

Fig. 4.9 First cross-sectional normalization: Reference to the center of gravity C with the center of gravity coordinate system x, y, z and the associated force and displacement quantities.

4.3 First cross-sectional normalization: Center of gravity C

89

We can immediately solve this requirement for the centroid coordinates yNC , zN C : Z Z ydA N zN dA SzN SyN A A D ; zN C D Z D : yNC D Z (4.51) A A dA dA A

A

Furthermore, the moments of inertia Iyy , Izz and the deviation moment Iyz are determined. For Iyy we obtain: Z Z 2 Iyy D z dA D .Nz  zNC /2 dA A

A

Z

2

Z

zN dA  2NzC

D A

zN dA C zNC2

A

Using SyN D zN C A yields:

Z

dA D IyN yN  2NzC SyN C zNC2 A:

(4.52)

A

Iyy D IyN yN  zN C2 A:

(4.53)

Similar expressions are obtained for Izz and Iyz , and we obtain Steiner’s theorem2 as follows: Iyy D IyN yN  zN C2 A; Izz D IzN zN  yNC2 A; Iyz D IyN zN  yNC zNC A:

(4.54)

Steiner’s theorem, often also addressed as the so-called parallel-axis theorem, describes the change of the moments of inertia and of the deviation moment when any Cartesian reference system is shifted to the center of gravity C of a cross-section. We will show later that the sign of those parts in (4.54) which contain the crosssectional area A is reversed if the reference system is moved from the center of gravity to any other point of the cross-section. From Steiner’s theorem it also follows immediately that the moments of inertia and the deviation moment related to the center of gravity take the smallest possible values. Example 4.4

We again consider the cross-section of Fig. 4.6 and determine the area integrals with respect to the centroid coordinate system x, y, z (see Fig. 4.10). The centroid coordinates yNC , zNC follow as: yNC D

2

SzN hb 2 b D D ; A 2hb 2

zNC D

Jakob Steiner (1796–1863), Swiss mathematician.

SyN bh2 h D D : A 2bh 2

(4.55)

90

4 Beams

Fig. 4.10 Rectangular cross-section with centroid coordinate system x, y, z.

b y

C

y

z

h

z

The moments of inertia Iyy and Izz and the deviation moment Iyz result from Steiner’s theorem (4.54) as follows: bh3 h2 bh3  bh D ; 3 4 12 hb 3 b 2 hb 3 Izz D IzN zN  yNC2 A D  bh D ; 3 4 12 b 2 h2 b h Iyz D IyN zN  yNC zN C A D  bh D 0: (4.56) 4 22 Obviously, for the considered rectangular cross-section, besides the two static moments Sz and Sy , the deviation moment Iyz also disappears. This is due to the fact that this is a doubly symmetric cross-section where the reference axes coincide with the symmetry axes. We will come back to this issue at a later point of this chapter. J Iyy D IyN yN  zN C2 A D

Example 4.5

We consider the I-cross-section of Fig. 4.7. The center of gravity C is located at the following coordinates yNC and zN C : SzN bt.4b C h/ D D b; A t.4b C h/   SyN th h2 C 2b h D D : zN C D A t.4b C h/ 2

yNC D

(4.57)

The moments of inertia and the deviation moment can then be determined using Steiner’s theorem as follows: 1 3 Iyy D IyN yN  zN C2 A D th C tbh2 ; 12 4 Izz D IzN zN  yNC2 A D tb 3 ; 3 Iyz D IyN zN  yNC zNC A D 0: (4.58) For the given double-symmetric I-cross-section, the deviation moment Iyz vanishes. J

4.3 First cross-sectional normalization: Center of gravity C

91

Example 4.6

For the Z-section of Fig. 4.8, the following centroid coordinates yNC and zNC result: SzN tb.2b C h/ D D b; A t.2b C h/  h SyN th 2 C b h D D : zN C D A t.2b C h/ 2

yNC D

(4.59)

The moments of inertia IyO yO and IzO zO as well as the deviation moment IyO zO result from Steiner’s theorem as: 1 3 1 th C tbh2 ; 12 2 2 3 2 D IzN zN  yNC A D tb ; 3 1 D IyN zN  yNC zN C A D tb 2 h: 2

Iyy D IyN yN  zN C2 A D Izz Iyz

(4.60)

For this cross-section, the deviation moment Iyz does not disappear. We will come back to this issue at a later point. J

4.3.2 Selected elementary cases In many technical applications, cross-sections consist of elementary simple segments for which the moments of inertia and the deviation moment can be determined easily and from which the entire cross-section can then be composed. In this section we will consider such elementary cases and provide some examples, always referring to the center of gravity C of the cross-section. As a first example, consider again the rectangular cross-section of width b and height h (Fig. 4.11). The two moments of inertia Iyy and Izz and the moment of deviation Iyz with respect to the center of gravity of the cross-section are to be determined. The moment of inertia Iyy Z Iyy D

z 2 dA

(4.61)

A

can be determined with the area element dA D bdz (Fig. 4.11, middle) as: h

Z2 Iyy D  h2

ˇh bz 3 ˇˇ 2 bh3 z bdz D D : 3 ˇ h 12 2

2

(4.62)

92

4 Beams

b

h

C

y

C

y

z

y

dA

z

z

C

z

y dA z

Fig. 4.11 Determination of the moments of inertia and of the deviation moment of a rectangular cross-section.

Similarly, the moment of inertia Izz is obtained as: Izz D

hb 3 : 12

For the determination of the deviation moment Z Iyz D yzdA

(4.63)

(4.64)

A

the area element dA D dydz (Fig. 4.11, right) is used. We obtain: h

b

ZC 2 ZC 2 Iyz D

yzdydz D 0:

(4.65)

 h2  b2

These results apparently agree with the results (4.56) that we have previously obtained using Steiner’s theorem. Example 4.7

Consider the box cross-section of Fig. 4.12 with height h, width b and constant wall thickness t. We introduce the two auxiliary quantities bN D b  2t and hN D h  2t . This box section can be treated by subtracting from the moments of inertia of the solid section of height h and width b the moments of inertia of the N We obtain: inscribed section of width bN and height h. Iyy D

bh3 bN hN 3 1  3 N N3   D bh  b h : 12 12 12

(4.66)

4.3 First cross-sectional normalization: Center of gravity C

93

Fig. 4.12 Determination of the moments of inertia for a box cross-section.

b t t

h

C

y

t

t

z

For the moment of inertia Izz the result is analogous: Izz D

hb 3 hN bN 3 1  3 N N 3  D hb  hb : 12 12 12

(4.67)

The deviation moment Iyz vanishes for this cross-section. J Example 4.8

Consider the circular cross-section (radius R) of Fig. 4.13. The two moments of inertia Iyy and Izz as well as the deviation moment Iyz are to be determined. In addition, these area integrals are to be determined for a circular ring crosssection. We first determine the polar moment of inertia Ip : Z Ip D

r 2 dA:

(4.68)

A

We can interpret the area element dA as a circular ring of thickness dr, where dA D 2 rdr. Thus: ZR R4 Ip D 2 r 3 dr D : (4.69) 2 0

y

R C

y

rC dr

dA z

z

Ri C Rm

y

t

y

C t Rm

Ra

z

z

Fig. 4.13 Determination of the moments of inertia for different kinds of circular cross-sections.

94

4 Beams

The polar moment of inertia is the sum of Iyy and Izz . Since the circular crosssection is also double symmetric (therefore Iyy D Izz ), it follows: Iyy D Izz D

R4 : 4

(4.70)

The deviation moment Iyz is zero for this cross-section. Now consider a circular ring cross-section with the wall thickness t, the inner radius Ri and the outer radius Ra . We obtain the required area integrals by subtraction: Iyy D Izz D

 Ra4 Ri4  4  D Ra  Ri4 : 4 4 4

(4.71)

Furthermore, let the mean radius Rm be defined as an auxiliary quantity: Rm D

1 .Ra C Ri /: 2

(4.72) 2

3 Furthermore, the relation Ra4  Ri4 D 4tRm .1 C 4Rt 2 / is valid. If we consider m a very thin-walled circular ring cross-section with t  Rm , then we can write in good approximation: 3 Iyy D Izz D  tRm : (4.73) J

Example 4.9

Consider the elliptical cross-section of Fig. 4.14 with the two radii a and b as shown. The two moments of inertia Iyy and Izz and the deviation moment Iyz are to be determined. The boundary of the cross-section is given by the elliptic equation  y 2  z 2 C D 1: (4.74) a b We first determine the moment of inertia Iyy using the area element dA as shown in Fig. 4.14. From the ellipse equation (4.74), the half-width y.z/ of the area element dA can be derived as: r  z 2 y.z/ D a 1  : (4.75) b Fig. 4.14 Elliptical crosssection.

b y

z

C

a dA

y(z) z

4.3 First cross-sectional normalization: Center of gravity C

95

For dA we then have: r dA D 2y.z/dz D 2a 1 

 z 2 b

dz:

(4.76)

The moment of inertia Iyy can then be determined as: Z Iyy D A

ZCb r  z 2 z dA D 2a z 2 1  dz: b 2

(4.77)

b

With the substition z D b sin

' 2

(4.78)

the expression (4.77) transfers into: Iyy

ZC ' ' D ab sin2 cos2 d': 2 2 3

(4.79)



The integral occurring here takes the value Iyy D

 4

after partial integration, so that:

ab 3 : 4

(4.80)

The moment of inertia Izz can be deduced from this by swapping a and b: Izz D

ba3 : 4

(4.81)

Since the cross-section is doubly symmetrical, the deviation moment Iyz disappears: Iyz D 0: (4.82) J Example 4.10

Consider the triangular cross-section of Fig. 4.15. The triangle given here has the dimensions b and h, and the moments of inertia Iyy and Izz as well as the deviation moment Iyz are to be determined. We start the considerations with the moment of inertia Iyy and use the area element dA D bA .z/dz shown in Fig. 4.15. We can express the width bA .z/ as: bA .z/ D y.z/ C

b ; 3

(4.83)

96

4 Beams

Fig. 4.15 Triangular crosssection.

b b 3 y

h 3

C yA

z dA

h

bA(z) z

where y.z/ is the coordinate y depending on z and can be determined as: y.z/ D  Thus:

bz b C : h 3 2h

Z

2

Iyy D

C3 Z

z dA D A

z

2



 h3 2h

C3 Z

D  h3

(4.84)

 b dz y.z/ C 3

  2b bh3 bz z2  C dz D : h 3 36

(4.85)

The moment of inertia Izz can then be obtained by swapping b and h: Izz D

hb 3 : 36

(4.86)

For the determination of the deviation moment Iyz the area element of Fig. 4.15 is used. The position of the center of gravity yA .z/ can be given as a function of z: bz yA .z/ D  : (4.87) 2h Thus, the deviation moment Iyz can be determined as: 2h

C3 Z

Z Iyz D

yzdA D  A

 h3 2h

C3 Z

D  h3

  bz 2 b y.z/ C 2h 3

  bz 2 2b bz b 2 h2  C D : 2h h 3 72

(4.88) J

4.3 First cross-sectional normalization: Center of gravity C

97

Example 4.11

For the semi-circular cross-section shown in Fig. 4.16, the moment of inertia Iyy is to be determined. With the distance z D Rm cos ' and the infinitesimal area element dA D tRm d' we obtain the moment of inertia Iyy by integration: Z Iyy D

2

Z

z dA D A

2

.Rm cos '/ tRm d' D 0

3 Rm t

Z

cos2 'd' D

0

3 t Rm : (4.89) 2

J

4.3.3 Composite cross-sections Steiner’s theorem is particularly useful for cross-sections whose center of gravity has already been determined and which are composed of elementary areas whose center of gravity positions and moments of inertia are known. We already know from (4.26) that the area integrals of cross-sections consisting of several segments result from the sum of the individual partial integrals. Accordingly, this is also true for the moments of inertia and for the deviation moment also with respect to the centroidal coordinate system y; z, i.e.: Z Iyy D

2

z dA D

Izz D

y 2 dA D

n Z X

y 2 dAi D

i D1 A

yzdA D

Iyy;i ;

n Z X i D1 A

i

n X

Izz;i ;

i D1

i

Z

A

n X i D1

i

A

Iyz D

z 2 dAi D

i D1 A

A

Z

n Z X

yzdAi D

n X

Iyz;i :

(4.90)

i D1

Here the partial moments of inertia Iyy;i , Izz;i , Iyz;i are related to the center of gravity axis system y, z. In the following we want to work out how these partial Fig. 4.16 Thin-walled semicircular cross-section.

dA=tRmd Rm cos y

Rm

z

98

4 Beams

moments of inertia can be determined as expediently as possible. For this purpose, we consider the axis translation of Fig. 4.9 again and now shift the centroid coordinate system y, z by yNC and zN C from the centroid C to any arbitrary other point. From the relation yN D y C yNC ; zN D z C zN C (4.91) the following alternative representation of Steiner’s theorem results: IyN yN D Iyy C zN C2 A; IzN zN D Izz C yNC2 A; IyN zN D Iyz C yNC zN C A;

(4.92)

which describes the change of the second order area integrals with respect to the center of gravity in the course of an axis translation. Accordingly, the area integrals Iyy , Izz , Iyz of a cross-section change by zN C2 A, yNC2 A, yNC zN C A, respectively, when the reference frame is translated from the center of gravity, and pass into the expressions IyN yN , IzN zN , IyN zN . From (4.92) one learns again that the area integrals Iyy , Izz , Iyz are minimal with respect to the center of gravity C. The representation (4.92) of Steiner’s theorem has therefore a very essential meaning, since it is based on the area integrals with respect to the center of gravity C of the cross-section. In many practical applications, the first step is to determine the center of gravity C of the cross-section under consideration, and the next step is to calculate the area integrals from there or, in the case of segmented cross-sections (see Fig. 4.17 for such a cross-section, where we assume that the cross-section is thin-walled and that all dimensions are related to the skeleton line) by summing up the contributions of the individual cross-section segments. For a cross-section composed of n elementary segments, the cross-sectional area A is the sum of the partial areas Ai : Z n X A D dA D Ai : (4.93) A

Fig. 4.17 Segmented crosssection.

i D1

y zC y

C

zC,i zC,i

Ci

yi Ai

zi

z

z

yC,i

yC yC,i

4.3 First cross-sectional normalization: Center of gravity C

99

For the static moments with respect to the arbitrarily chosen reference frame y, N zN one obtains: Z SyN D

zN dA D

n X

Z Ai zNC;i ;

SzN D

i D1

A

ydA N D

n X

Ai yNC;i :

(4.94)

i D1

A

Herein, yNC;i and zN C;i are the centroid coordinates of segment i. The coordinates of the center of gravity are then given as: Z

n X

ydA N A

yNC D Z

D dA

A

Z

i D1 n X

n X

zN dA

Ai yNC;i ;

A

zNC D Z

D dA

Ai A

i D1

Ai zN C;i

i D1 n X

:

(4.95)

Ai

i D1

The moments of inertia Iyy , Izz and the deviation moment Iyz , related to the center of gravity C, result as the sum of all moments of inertia IyN y;i N , IzN z;i N , IyN z;i N and those portions which include the area Ai and the coordinates yC;i , zC;i : Iyy D

n X

IyN y;i N C

i D1

Izz D

n X

n X i D1

2 zC;i Ai ;

i D1

IzN z;i N C

i D1

Iyz D

n X

n X

2 yC;i Ai ;

i D1

IyN z;i N C

n X

yC;i zC;i Ai :

(4.96)

i D1

The quantities IyN y;i N , IzN z;i N , IyN z;i N are the moments of inertia and the deviation moment of segment i with respect to its local center of gravity axes yNi , zN i . The quantities yC;i and zC;i are the distances of the center of gravity Ci of segment i from the center of gravity C of the cross-section. Example 4.12

Consider the I-cross-section of Fig. 4.18, left. The moments of inertia Iyy and Izz as well as the deviation moment Iyz are to be determined with the help of Steiner’s theorem concerning the center of gravity C of the cross-section. The cross-section is divided into segments 1 (top flange), 2 (web), and 3 (bottom flange), and local reference frames yN1 , zN 1 , yN2 , zN2 , and yN3 , zN3 are assigned whose origins are located at the centroids of each segment (Fig. 4.18, right). We assume a very thin-walled cross-section in all further considerations, and we are

100

4 Beams

Fig. 4.18 Application of Steiner’s theorem to an Icross-section.

t

y1 t h y

C

y2

t

z1

zC,1

z2

zC,3

y3 b

z3

b

z

guided by the skeleton line in determining all geometric relationships. The centroids of the segments are located at yC;1 D 0, zC;1 D  h2 , yC;2 D 0, zC;2 D 0, yC;3 D 0, zC;3 D h2 . The moments of inertia and the deviation moments of the individual segments can be given as follows: IyN y;1 N D IyN y;3 N D

2bt 3 ' 0; 12

IyN y;2 N D

th3 ; 12

IzN z;1 N D IzN z;3 N D

t.2b/3 2tb 3 D ; 12 3

IzN z;2 N D

ht 3 ' 0; 12

IyN z;1 N D IyN z;2 N D IyN z;3 N D 0:

(4.97)

The cross-sectional areas Ai are as follows when oriented along the skeleton line: A1 D A3 D 2tb; A2 D th: (4.98) Steiner’s theorem (4.96) then gives: Iyy D

 2   h th3 th3 h 2 2tb C 2tb D C  C tbh2 ; 12 2 2 12

Izz D

2tb 3 2tb 3 4tb 3 C D ; 3 3 3

Iyz D 0:

(4.99)

Apparently, these results correspond to the values previously determined in Example 4.5. J Example 4.13

Consider the Z-cross-section of Fig. 4.19, left. The moments of inertia Iyy and Izz as well as the deviation moment Iyz with respect to the center of gravity C of the cross-section are to be determined with the help of Steiner’s theorem. We

4.3 First cross-sectional normalization: Center of gravity C Fig. 4.19 Application of Steiner’s theorem to a Zcross-section.

101

yC,1 t

y1 t h y

z1

y2

C t

z2

zC,1 zC,3

y3 b

z3

b

yC,3

z

assume that the cross-section is thin-walled and that we can relate all considerations to the skeleton line which bisects the wall thickness at each point of the cross-section. The cross-section is divided into three segments as shown in Fig. 4.19, right. The centroid coordinates of the individual segments are yC;1 D  b2 , zC;1 D  h2 , yC;2 D 0, zC;2 D 0 and yC;3 D b2 , zC;3 D h2 , with respect to the centroid C of the cross-section. The moments of inertia and the deviation moments of the segments are: IyN y;1 N D IyN y;3 N D

bt 3 ' 0; 12

IyN y;2 N D

th3 ; 12

IzN zN ;1 D IzN z;3 N D

tb 3 ; 12

IzN z;2 N D

ht 3 ' 0; 12

IyN zN ;1 D IyN z;2 N D IyN z;3 N D 0:

(4.100)

The areas Ai of the segments are: A1 D A3 D tb;

A2 D th:

(4.101)

Thus, from Steiner’s theorem we obtain:  2   h th3 th3 h 2 tbh2 tb C tb D C  C ; 12 2 2 12 2  2   b tb 3 2tb 3 tb 3 b 2 D tb C tb D C C  ; 12 12 2 2 3    b h bh tb 2 h D   tb C tb D : 2 2 22 2

Iyy D Izz Iyz

These results agree with the values already determined. J

(4.102)

102

4 Beams

Example 4.14

Consider the cross-section of Fig. 4.20, left. The two moments of inertia Iyy and Izz as well as the deviation moment Iyz with respect to the center of gravity of the cross-section are to be determined. The reference system y, N zN is to be used. To determine the center of gravity C, the cross-section is divided into four sub-areas as shown in Fig. 4.20, right. Due to the symmetry of the cross-section with respect to the zN -axis, the centroid coordinate yNC is yNC D 0. The centroid positions zN C;i of the partial areas A1 D 9t 2 , A2 D 9t 2 , A3 D 8t 2 , A4 D 10t 2 result as: zN C;1 D 4:5t;

zN C;2 D 4:5t;

zN C;3 D 9:5t;

zN C;4 D 12:5t:

(4.103)

The position zNC of the center of gravity can then be determined as: P4 zNC D

i D1

P4 2

Ai zNC;i

i D1

Ai

9t  4:5t C 9t 2  4:5t C 8t 2  9:5t C 10t 2  12:5t 9t 2 C 9t 2 C 8t 2 C 10t 2 47 D t: 6 D

(4.104)

With the centroid coordinates yNC , zN C thus determined, the centroid positions of the partial areas can be specified with respect to the centroid C as (Fig. 4.21): yC;1 D 3:5t; 10 zC;1 D  t; 3

yC;2 D 3:5t; yC;3 D 0; 10 5 zC;2 D  t; zC;3 D t; 3 3

y

yC;4 D 0; 14 zC;4 D t: 3

(4.105)

y zC,1= zC,2 t

t z

10t

C1

C2

z C3

t

C4

5t

3t 2t 3t Fig. 4.20 Considered cross-section.

zC,3

zC,4

4.3 First cross-sectional normalization: Center of gravity C Fig. 4.21 Center of gravity position as well as centers of gravity of the partial areas.

yC,1

103

yC,2

C2 C C3

C1

y

zC,1= zC,2 zC,3

y1

y2 z1

zC,4

C4

y3 y4

z2 z3

z4 z

We first determine the moment of inertia Iyy . The moments of inertia IyN yN of the partial areas, related to the local axes y, N zN , are obtained as follows: IyN y;1 N D

t  .9t/3 243 4 D t ; 12 4

IyN y;2 N D

243 4 t ; 4

IyN y;3 N D

8t  t 3 2 D t 4; 12 3

IyN y;4 N D

2t.5t/3 125 4 D t : 12 6

Steiner’s theorem Iyy D

4 X i D1

IyN y;i N C

4 X

2 zC;i Ai

(4.106)

(4.107)

i D1

then yields the following moment of inertia Iyy of the cross-section with respect to the center of gravity C: Iyy D

243 4 243 4 2 4 125 4 t C t C t C t 4 4 3 6    2  2 2  5 14 2 10 10 2 2 2 C  t  9t C  t  9t C t  8t C t  10t 2 3 3 3 3

D 583t 4 : (4.108) To determine the moment of inertia Izz , the moments of inertia IzN zN of the partial areas are required in relation to their respective centers of gravity. They can be

104

4 Beams

determined as: IzN z;1 N D

9t  t 3 3 D t 4; 12 4

IzN z;2 N D

3 4 t ; 4

IzN z;3 N D

t.8t/3 128 4 D t ; 12 3

5t.2t/3 10 4 D t : 12 3 can then be determined as follows:

IzN z;4 N D The moment of inertia Izz Izz D

(4.109)

3 4 3 4 128 4 10 4 t C t C t C t 4 4 3 3 2 2 C .3:5t/  9t C .3:5t/2  9t 2 C 0  8t 2 C 0  10t 2

D 268t 4 :

(4.110)

The deviation moment in this example is zero due to the symmetry of the crosssection with respect to the z-axis: Iyz D 0:

(4.111) J

4.3.4 Stress analysis In order to determine the normal stress xx in the cross-section, the normal force and bending moments are first related to the center of gravity C. On the basis of Fig. 4.9, the following relationships can be determined: N D NxN ;

My D MyN  NxN zNC ;

Mz D MzN C NxN yNC :

(4.112)

The constitutive law (4.17) then takes the following form: N D EAu0 ; My D EIyy w 00  EIyz v 00 ; Mz D EIyz w 00  EIzz v 00 : In vector-matrix notation we obtain: 1 2 0 A 0 N @ My A D E 4 0 Iyy Mz 0 Iyz

30 0 1 0 u Iyz 5@w 00 A: Izz v 00

(4.113)

(4.114)

4.3 First cross-sectional normalization: Center of gravity C

105

This can be solved for the displacement derivatives as follows: N ; EA 1 Iyz My C Iyy Mz ; v 00 D 2 E Iyy Izz  Iyz 1 Izz My C Iyz Mz : w 00 D  2 E Iyy Izz  Iyz u0 D

(4.115)

The first cross-sectional normalization thus obviously leads to a decoupling of bar action and bending action. Therefore, in practical applications, one will always try to relate the considerations to the center of gravity C of the cross-section under consideration. However, even with reference to the center of gravity C, the two bending actions, expressed by the two displacements v and w as well as the two bending moments My and Mz , are still coupled due to the generally non-vanishing deviation moment Iyz . Substitution of (4.115) into Hooke’s law   xx D E u0  zw 00  yv 00 (4.116) yields the following expression for the normal stress xx , which can be used to calculate the normal stress for given normal force N and bending moments My , Mz at each point y and z of the cross-section: xx D

Iyz My C Iyy Mz Izz My C Iyz Mz N z y: C 2 2 A Iyy Izz  Iyz Iyy Izz  Iyz

(4.117)

For the special case of a vanishing deviation moment Iyz , the expression (4.117) for the normal stress xx simplifies considerably, and we obtain: xx D

My N Mz z y: C A Iyy Izz

(4.118)

Obviously, the fraction of xx due to the normal force N is constant over the whole cross-section, as also discussed for the bar in Chap. 3. However, due to the bending moment My , the fraction of xx is linear over the coordinate z. In quite the same way, due to the bending moment Mz , we obtain a linear distribution of the normal stress xx over the coordinate y (Fig. 4.2). The superposition principle applies at this point, so we can additively compose the normal stress xx from these expressions. This is shown in Fig. 4.22 for a rectangular cross-section. A very important parameter in the calculation of the normal stress xx is the so-called neutral axis. Equations (4.117) and (4.118), respectively, represent plane equations describing the surface of the stress distribution xx by means of the two coordinates y and z. Setting (4.117) or (4.118) to zero, one can express one of the two coordinates y or z by the other. The resultant expression describes a straight line which mathematically describes at which points the normal stress xx shows zero values.

106

4 Beams

y

C

x z

σ xx N A N

σ xx

My z Iyy

My σ xx

Mz y Izz

Mz

My Mz σ xx N z y A Iyy Izz

Neutral axis Fig. 4.22 Normal stress due to normal force and bending moments.

Example 4.15

Let us consider the cross-section from Example 4.12, where we want to investigate here the special case h D 2b (Fig. 4.23). The normal stress distribution xx over the cross-section for the two load cases N D 0, My D M0 , Mz D 0 and N D 0, My D M0 , Mz D M0 is to be determined. For the first load case, it

4.3 First cross-sectional normalization: Center of gravity C

107

Fig. 4.23 Considered Icross-section.

t t h= 2b y

C t

b

b

z should also be determined how large the normal force N must be so that no compressive normal stresses xx occur in the cross-section. To solve the problem, we first determine the moments of inertia Iyy and Izz (the deviation moment Iyz is zero for the considered cross-section). For the special case h D 2b we obtain from Example 4.12: Iyy D

th3 14 3 C tbh2 D tb ; 12 3

Izz D

4 3 tb ; 3

Iyz D 0:

(4.119)

Since the deviation moment becomes zero in this case, the expression (4.118) can be used to determine the normal stress distribution. For load case 1 with N D 0, My D M0 , Mz D 0, (4.118) leads to the following expression: My 3M0 xx D zD z: (4.120) Iyy 14tb 3 Apparently, the normal stress xx is linear over z and constant over y. The maximum values of the stresses occur at the upper and lower edges of the crosssection: ! t 3M0 b C   2 t ; D xx z D b  3 2 14tb ! t 3M0 b C   2 t xx z D b C : (4.121) D 3 2 14tb At the top of the cross-section at b  2t , xx is present as compressive stress, whereas at the bottom at b C 2t , xx is present as tensile stress (Fig. 4.24, left). The neutral axis follows from the requirement xx D 0, which leads to z D 0. Accordingly, the neutral axis is identical to the y-axis (Fig. 4.24, left).

108

4 Beams

y

y

σxx

σxx

z My = M0

z

Neutral axis

My = M0

x

x

Fig. 4.24 Stress distribution at the I-cross-section for N D 0, My D M0 , Mz D 0 (left), stress distribution assuming a thin-walled cross-section with t  b (right).

If we assume a very thin-walled cross-section, i.e., t  b, then the considerations can be related to the skeleton line of the cross-section, and the stress values (4.121) are obtainedas (Fig. 4.24, right): xx .z D b/ D 

3M0 ; 14tb 2

xx .z D b/ D

3M0 : 14tb 2

(4.122)

For load case 2 with N D 0, My D M0 , Mz D M0 , the stress calculation (4.118) yields: My Mz 3M0  z y : (4.123) z yD  xx D Iyy Izz 2tb 3 7 2 We evaluate this expression at the four corner points of the profile:   3M0 t xx y D b; z D b  .9b C t/; D 2 28tb 3   3M0 t .5b  t/; xx y D b; z D b  D 2 28tb 3   3M0 t .9b C t/; xx y D b; z D b C D 2 28tb 3   3M0 t .5b  t/: xx y D b; z D b C D 2 28tb 3

(4.124)

The neutral axis follows from the requirement xx D 0 as zD

7 y: 2

(4.125)

A graphical representation can be found in Fig. 4.25, left. Assuming again at this point that the cross-section is thin-walled with t  b, the fractions due to the

4.3 First cross-sectional normalization: Center of gravity C

y

109

y

σxx z My = M0

σxx z

My = M0

x

Mz = M0

Mz = M0

x

Neutral axis Fig. 4.25 Stress distribution at the I-cross-section for N D 0, My D M0 , Mz D M0 (left), stress distribution assuming a thin-walled cross-section with t  b (right).

wall thickness t in (4.124) can be neglected, and it follows: 27M0 ; 28tb 2 15M0 ; xx .y D b; z D b/ D 28tb 2 27M0 ; xx .y D b; z D b/ D 28tb 2 15M0 : xx .y D b; z D b/ D  28tb 2 xx .y D b; z D b/ D 

(4.126)

This is shown in Fig. 4.25, right. An alternative, two-dimensional representation of the stress distribution is shown in Fig. 4.26, where we assume a thin-walled cross-section and refer to the skeleton line. We now consider again the case My D M0 , Mz D 0 and determine the normal force N D N0 for which no compressive stress occurs in the cross-section. We start from the stress formulation My N z; C A Iyy

(4.127)

N0 3M0 z: C t.4b C h  t/ 14tb 3

(4.128)

xx D which can be expressed as follows: xx D

The highest compressive stress in the cross-section occurs at the top edge at z D b  2t , so we require:     N0 3M0 t t D  D 0: xx z D b  b C 2 t.4b C h  t/ 14tb 3 2

(4.129)

110

4 Beams

Fig. 4.26 Two-dimensional representation of the stress distribution of the I-crosssection for N D 0, My D M0 , Mz D M 0 .

σxx

My M0

2

y 7

z

Neutral axis

Mz M0

This can be solved for the normal force N0 as follows:   3M0 t N0 D .4b C h  t/: b C 14b 3 2

(4.130)

The maximum tensile stress then occurs at the lower edge of the cross-section and amounts to:     3M0 t t D : (4.131) xx z D b C bC 2 7tb 3 2 The resulting stress distribution is shown in Fig. 4.27. J

y Neutral axis N = N0

z My = M0

σ xx

x Fig. 4.27 Stress distribution on the I-cross-section for N D N0 , My D M0 , Mz D 0.

4.3 First cross-sectional normalization: Center of gravity C

111

Example 4.16

Consider the cantilever beam of Fig. 4.28, left, under the two forces F and P and under the boundary moment M D F l. For the case P D 0, the normal stress distribution is to be determined at the point of highest bending moment. In addition, it is to be clarified how large the compressive force P acting at the center of gravity must be so that tensile stresses do not occur at any point in the cross-section. The cross-section is a circular cross-section with radius R2 , which contains a circular hole with radius R1 . We assume that R2 D 2R1 . The highest bending moment occurs in the range 2l  x  l with the value My D F l. We first determine the center of gravity of the cross-section and consider the circular hole with radius R1 as the partial area A1 , where we assume the area A1 with a negative sign, i.e. A1 D R12 . The partial area A2 is A2 D R22 D 4R12 . The ordinates of the centroids of the partial areas, related to the coordinate system y, N zN are: R1 (4.132) ; zN C;2 D 0: 2 The centroidal coordinate yNC is found to be zero due to the symmetry of the cross-section with respect to the zN -axis. The coordinate zNC follows as: yNC;1 D 0;

yNC;2 D 0;

zN C D

zN C;1 D

A1 zNC;1 C A2 zN C;2 R1 D : A1 C A2 6

(4.133)

Thus the centers of gravity of the partial areas, related to the center of gravity coordinate system y; z, result as: 2 1 (4.134) R1 ; zC;2 D R1 : 3 6 In order to determine the stress distribution xx , the moment of inertia Iyy of the cross-section is required. It is calculated from Steiner’s theorem (4.96) as: zC;1 D

2 2 Iyy D IyN y;1 N C IyN y;2 N C zC;1 A1 C zC;2 A2   2 2 2 1 R4 D  1 C 4R14  R1 R12 C R1 4R12 4 3 6 41 D R4 : 12 1

F

M0 = Fl

R2

R2

P y

x l 2

R1

l 2 z

z

Fig. 4.28 Cantilever beam under load (left), cross-section (right).

(4.135)

R1 2

112

4 Beams

22 Fl 41 R13 11 6 R1

R2

y

R1

R1 6

xx

Neutral axis

13 6 R1 26 Fl 41 R13

z

Fig. 4.29 Stress distribution under bending moment M and force F .

The stress distribution over the cross-section can be determined as: xx D

My z: Iyy

(4.136)

The ordinates at the upper and lower edge of the cross-section follow as:   11 22 F l ; xx z D  R1 D  6 41 R13   13 26 F l : (4.137) xx z D R1 D 6 41 R13 The stress distribution is shown in Fig. 4.29. We determine the compressive normal force P necessary to prevent tensile stresses from developing in the entire cross-section from the requirement that R: the boundary stress becomes zero at the point z D 13 6 1   13 xx z D R1 D 0: (4.138) 6 From the stress formulation xx D

My N z C A Iyy

after inserting N D P , My D F l, Iyy D 41 R14 and z D 12 expression for the compressive force P follows: P D

78 F l : 41 R1

(4.139) 13 R, 6 1

the following

(4.140)

R is calculated as: The normal stress xx thus arising at the upper end at z D  11 6 1   11 48 F l xx z D  R1 D  : (4.141) 6 41 R13 The resulting stress distribution is shown in Fig. 4.30. J

4.3 First cross-sectional normalization: Center of gravity C

113

48 Fl 41 R13 11 6 R1

R2

y

R1

R1 6

xx

13 6 R1

Neutral axis z Fig. 4.30 Stress distribution under bending moment M , force F and compressive normal force P .

Example 4.17

Consider the cross-section of Fig. 4.31, left. The load is given in the form of a bending moment M0 D 50;000F t oriented under 60ı to the horizontal, where t is the wall thickness of the cross-section. The distribution of the normal stress xx over the cross-section is to be determined. To determine the center of gravity, the reference system y, N zN as shown in Fig. 4.31, right, is employed, and the cross-section is divided into three partial areas. The partial areas amount to A1 D 5t 2 , A2 D 6t 2 , A3 D 5t 2 , their center of gravity coordinates are yNC;1 D 3:5t, yNC;2 D 0, yNC;3 D 3:5t, zNC;1 D 2:5t, zNC;2 D 0:5t, zN C;3 D 2:5t. The centroid coordinates of the total cross-section are then obtained as yNC D 0 due to the symmetry of the cross-section and zNC D

5t 2  2:5t C 6t 2  0:5t C 5t 2  2:5t D 1:75t: 5t 2 C 6t 2 C 5t 2

(4.142)

Thus, the center of gravity positions of the partial areas with respect to the center of gravity C can be given as: yC;1 D 3:5t; yC;2 D 0; yC;3 D 3:5t; zC;1 D 0:75t; zC;2 D 1:25t; zC;3 D 0:75t:

M0

yC,1

t y

60°

C

yC,3

y

zC,2

C2

zC,1= zC,3 C1

5t

C3

z z

t

t 8t

Fig. 4.31 Considered cross-section and load case.

(4.143)

114

4 Beams

The moments of inertia of the partial areas are as follows: IyN y;1 N D

1t  .5t/3 D 10:42t 4 ; 12

IyN y;2 N D

6t  .1t/3 D 0:5t 4 ; 12

4 IyN y;3 N D 10:42t ;

IzN z;1 N D

5t  .1t/3 D 0:42t 4 ; 12

IzN z;2 N D

1t  .6t/3 D 18t 4 ; 12

4 IzN z;3 N D 0:42t ;

IyN z;1 N D IyN z;2 N D IyN zN ;3 D 0:

(4.144)

The two moments of inertia Iyy and Izz of the cross-section with respect to the center of gravity C can then be determined using Steiner’s theorem as follows: 2 2 2 Iyy D IyN y;1 N C IyN y;2 N C IyN y;3 N C zC;1 A1 C zC;2 A2 C zC;3 A3

D 10:42t 4 C 0:5t 4 C 10:42t 4 C .0:75t/2  5t 2 C .1:25t/2  6t 2 C .0:75t/2  5t 2 D 36:34t 4 ; 2 2 2 Izz D IzN z;1 N C IzN zN ;2 C IzN z;3 N C yC;1 A1 C yC;2 A2 C yC;3 A3

D 0:42t 4 C 18t 4 C 0:42t 4 C .3:5t/2  5t 2 C 02  6t 2 C .3:5t/2  5t 2 D 141:34t 4 :

(4.145)

The applied bending moment M0 is now decomposed into its two components My and Mz : My D M0 cos 60ı D 25;000F t; Mz D M0 sin 60ı D 43;301:27F t: (4.146) The stress distribution xx is then obtained as: xx D

My Mz Fz Fy z y D 687:95 3 C 306:36 3 : Iyy Izz t t

(4.147)

The equation of the neutral axis can be determined from the requirement xx D 0 as: z D 0:45y: (4.148)

4.4 Second cross-sectional normalization: Principal axes Fig. 4.32 Neutral axis and stress distribution.

115

B y

-2429.35 F2 t

C

Neutral axis

z

A

3461.28 F2 t

It is shown in Fig. 4.32. The two points farthest from the neutral axis are points A and B, as shown in Fig. 4.32. Their coordinates are: yA D 4t;

yB D 4t;

zA D 3:25t;

zB D 1:75t:

(4.149)

The stresses at these two points are: xx;A D 3461:28

F ; t2

xx;B D 2429:35

F : t2

(4.150) J

4.4 Second cross-sectional normalization: Principal axes With the help of the first cross-sectional normalization, a decoupling of the bar action from the bending action is achieved. However, after performing the first crosssectional normalization, the deviation moment Iyz generally still occurs, so that the two bending effects, characterized by the two bending moments My and My as well as the deflections w and v, are coupled with each other. A very significant simplification is achieved by determining a particular coordinate system yp , zp , which is obtained from the reference system y; z by rotation about the x-axis by an angle '0 yet to be determined, and in which the deviation moment vanishes. We will also show that the two momenta of inertia attain extremal values in this special reference frame, which we will refer to as the so-called principal axes. The disappearance of the deviation moment in the principal axis system then decouples the two bending effects from each other, as can be easily seen from (4.113). The coordinate rotation by the angle '0 is called the second cross-sectional normalization. The rotation of the reference axes y; z around the x-axis by an arbitrary angle ' is shown in Fig. 4.33. Let the rotated axes be denoted as , . The transformation

116

4 Beams

Fig. 4.33 Rotation of the coordinate system x, y, z by the angle ' into the reference system x, , .

C

y

x

My

N M

z M Mz Fig. 4.34 Relation between y, z and , 

y P

zP

P

P

yP

z

of the coordinates is derived using Fig. 4.34 by considering a point P with coordinates yP , zP and P , P , respectively. It follows:

D y cos ' C z sin ';

 D y sin ' C z cos ':

(4.151)

The x-axis remains unchanged under this pure rotation of the reference frame. We now determine the deviation moment I  in the rotated coordinate system as follows: Z Z I  D dA D .y cos ' C z sin '/.y sin ' C z cos '/dA A

A

Z

2

2

Z

y dA C cos '

D  sin ' cos ' A

Z

2

Z

yzdA  sin ' A

yzdA A

2

z dA

C sin ' cos ' A

  D Izz sin ' cos ' C Iyz cos2 '  sin2 ' C Iyy sin ' cos '     D Iyz cos2 '  sin2 ' C Iyy  Izz sin ' cos ':

(4.152)

4.4 Second cross-sectional normalization: Principal axes

117

Using cos2 ' D 12 .1 C cos 2'/, sin2 ' D 12 .1  cos 2'/ and 2 sin ' cos ' D sin 2', we obtain:  1 I  D Iyy  Izz sin 2' C Iyz cos 2': (4.153) 2 To determine the principal axis angle '0 we set the expression (4.153) for the deviation moment I  to zero and obtain: tan 2'0 D

2Iyz : Izz  Iyy

(4.154)

Thus, an equation for the principal axis angle '0 under which the deviation moment I  vanishes and the two moments of inertia I and I become extremal is found. Since the tan-function in (4.154) is periodic, in addition to the angle '0 , one can determine another angle '0 C 2 under which the deviation moment Iyz becomes zero and the moments of inertia become extremal. At this point, one also speaks of the so called principal moments of inertia. Since the two principal axis systems are oriented orthogonally to each other, no further insight results from considering ' D '0 C 2 , but only the values of I and I are interchanged. We denote the reference axes under the principal axis angle '0 as yp , zp . Analogous expressions to (4.153) can also be found for the transformation of the moments of inertia, and the following expressions for arbitrary angles ' are obtained:  1  1 Iyy C Izz C Iyy  Izz cos 2'  Iyz sin 2'; 2 2  1  1 D Iyy C Izz  Iyy  Izz cos 2' C Iyz sin 2'; 2 2  1 D Iyy  Izz sin 2' C Iyz cos 2': 2

I D I I 

(4.155)

We determine the values for the principal moments of inertia by means of the expressions Izz  Iyy 1 cos 2'0 D p D q ; 2 2 1 C tan 2'0 2 Izz  Iyy C 4Iyz 2Iyz tan 2'0 sin 2'0 D p D q ; 2 2 1 C tan 2'0 2 Izz  Iyy C 4Iyz

(4.156)

and it follows from (4.155) for the two principal moments of inertia with (4.154): I1;2

Iyy C Izz D ˙ 2

s 

Izz  Iyy 2

2 2 : C Iyz

(4.157)

Herein, I1 is the maximum moment of inertia, and I2 denotes the minimum value.

118

Iyz I yz

(Izz , Iyz (

I2 (I2,0( I yz

2 Izz

Iyy

0

2

0

I1 I ,I (I1,0 yy zz (

Fig. 4.35 Inertia circle.

4 Beams

(Iyy , Iyz (

Analogous to Mohr’s stress circle (Chap. 2), one can also create an analogous circle for the moments of inertia, which can be derived from Mohr’s stress circle by replacing the normal stresses xx and yy by the moments of inertia Iyy and Izz and the shear stress xy by the deviation moment Iyz . The inertia circle is shown in Fig. 4.35. It should be noted here that equation (4.154) for the principal axis angle '0 can also be obtained from the requirement that the two moments of inertia I and I have extreme values: dIyy dIzz D 0; D 0: (4.158) d' d' For the treatment of cross-sections, some general rules can be established, which are also shown in Fig. 4.36:  If there is one axis of symmetry for a cross-section, then one of the two principal axes yp , zp coincides with this axis of symmetry. The principal axis angle '0 is zero if there is at least one axis of symmetry and one of the two reference axes is parallel to this axis of symmetry. However, in general the center of gravity C has to be determined. Cross-sections to which this applies are shown in Fig. 4.36, top row.  If a cross-section has two axes of symmetry, then these represent the two principal axes. The deviation moment Iyz vanishes in this case, so that the principal axis angle '0 also becomes zero. Thus, if such a cross-section is present, then the second cross-sectional normalization can be omitted. Also, the first crosssectional normalization is not necessary since the center of gravity C of the cross-section is always located at the intersection of the two axes of symmetry. Examples of such cross-sections are shown in Fig. 4.36, middle row.  If the cross-section has no symmetry properties, both the first and the second cross-sectional normalization must be performed. In Fig. 4.36, bottom row, two examples are shown.

4.4 Second cross-sectional normalization: Principal axes

119 Symmetry axis

Symmetry axis

t1

t1

t1

t2

y

y

C

t2

C

y

C

Symmetry axis

t2 t1

t3

z

z

z Symmetry axis

Symmetry axis

Symmetry axis

t1

t1 t2

y

C Symmetry axis

t2

C

y

t2

Symmetry axis

C t

y

Symetry axis

t1

t1

z

yp

z

t1

z t1 t2

t2

y yp

φ0

φ0

y

C

C

φ0 t3

φ0

zp z

zp

z

Fig. 4.36 Selected cross-sections.

It can be shown that the sum of the two moments of inertia Iyy and Izz or of I and I has always the same value, independent of the choice of the reference frame, i.e. it represents an invariant: Iyy C Izz D I C I D Ip :

(4.159)

Here, Ip is the polar moment of inertia, i.e. the sum of the two moments of inertia. 2 It can also be shown that the expression Π21 .Iyy  Izz / 2 C Ixy is another invariant. In addition, the bending moments My and Mz have to be related to the reference system , . This is done analogously to the transformation (4.151): M D My cos ' C Mz sin ';

M D My sin ' C Mz cos ':

(4.160)

If the principal axis system x, D yp ,  D zp under the angle '0 is given, then the deviation moment I  vanishes. The constitutive equations are then: N D EAu0 ;

M D EI w 00 ;

M D EI v 00 :

(4.161)

120

4 Beams

Herein, for simplicity, we have used the terms u, v, w for the displacements in x-,

-, and -direction. In vector-matrix-notation, these equations read: 0

1 2 N A 0 @ M A D E 4 0 I

0 0 M

30 0 1 u 0 0 5@w 00 A: I v 00

(4.162)

One can invert these relations as follows: u0 D

N ; EA

w 00 D

M

; EI

v 00 D

M : EI

(4.163)

The constitutive equations are significantly simplified when the considerations are related to the principal axis system. In addition to the already existing decoupling of bar action and bending action, the two bending actions are now also decoupled from each other. Using Hooke’s law   (4.164) xx D E u0  w 00  v 00 ; then (4.163) can be used to derive the following expression for the normal stress xx : M M

N 

: (4.165) xx D C A I

I There is a considerable simplification compared to the expression (4.117), which was based on the consideration of the center of gravity axes, which are not the principal axes. Again, an expression for the neutral axis can be obtained by setting (4.165) to zero. If a beam situation is given where there is exclusively bending about one of the two principal axes y or z, then the maximum normal stresses xx;max necessary for a stress justification are obtained as: j jxx;max D

My ; Wy

or j jxx;max D 

Mz : Wz

(4.166)

(4.167)

The quantities Wy and Wz are the so called resistance moments or section moduli of the considered cross-section. They are obtained as follows: Wy D

Iyy ; jzjmax

Wz D

Izz : jyjmax

(4.168)

Here zmax and ymax are the coordinates of the point of the cross-section farthest from the center of gravity axis C.

4.4 Second cross-sectional normalization: Principal axes

121

Fig. 4.37 Thin-Walled rectangular cross-section.

t y

h

z

Example 4.18

Consider the thin-walled rectangular cross-section (height h, thickness t, t  h) as given in Fig. 4.37. Determine the moments of inertia I , I and the deviation moment I  for a coordinate system rotated by the angle '. For the moments of inertia Iyy , Izz and the deviation moment Iyz we have: Iyy D

th3 ; 12

Izz D

ht 3 ' 0; 12

Iyz D 0:

(4.169)

With the help of the transformation equations (4.155) the moments of inertia and the deviation moment result in: I D

 1  1 th3 Iyy C Izz C Iyy  Izz cos 2'  Iyz sin 2' D .1 C cos 2'/; 2 2 24

I D

 1  1 th3 Iyy C Izz  Iyy  Izz cos 2' C Iyz sin 2' D .1  cos 2'/; 2 2 24

I  D

 1 th3 Iyy  Izz sin 2' C Iyz cos 2' D sin 2': 2 24

(4.170) J

Example 4.19

Consider the Z-cross-section of Fig. 4.38, left. The cross-section has the flange width b and the web height h D 2b. The center of gravity coordinate system y, z is given, the load consists of the two bending moments My D M0 and Mz D M0 . The cross-section has a constant wall thickness t. We want to determine the stress distribution xx . The cross-section can be treated as very thin-walled (t  b), so that all considerations can be related to the skeleton line. We first determine the moments of inertia Iyy , Izz and the deviation moment Iyz with respect to the center of gravity coordinate system y, z. For this purpose, the cross-section is divided into three segments (Fig. 4.38, right), namely the upper flange (segment 1, A1 D tb), the web (segment 2, A2 D 2tb) and the lower flange (segment 3, A3 D tb). The moments of inertia of the segments with

122

4 Beams

Fig. 4.38 Z-cross-section.

t

y1 M0

t 2b y

b

C t

z1

y2

b

y3 b

z2

zb z3

M0

b

b

respect to their local axes are as follows: IyN y;1 N D

bt 3 t.2b/3 bt 3 2 ' 0; IyN y;2 D tb 3 ; IyN y;3 ' 0; N D N D 12 12 3 12

IzN z;1 N D

tb 3 ; 12

IzN z;2 N D

2bt 3 ' 0; 12

IzN z;3 N D

tb 3 ; 12

IyN z;1 N D IyN z;2 N D IyN z;3 N D 0:

(4.171)

Steiner’s theorem then yields: Iyy D

3 X

IyN y;i N C

i D1

3 X

2 zC;i Ai

i D1

2 8 D tb 3 C .b/2 tb C b 2 tb D tb 3 ; 3 3 Izz D

3 X

IzN zN ;i C

i D1

3 X

2 yC;i Ai

i D1

  2  b tb b 2 2 D2 tb C  tb D tb 3 ; C 12 2 2 3 3

Iyz D

3 X i D1

IyN zN ;i C

3 X

yC;i zC;i Ai

i D1

  b b D .b/tb C  btb D tb 3 : 2 2

(4.172)

The principal axis angle '0 follows from (4.154): tan 2'0 D

2Iyz D 1: Izz  Iyy

(4.173)

4.4 Second cross-sectional normalization: Principal axes

We obtain

2'0 D 45ı ;

123

(4.174)

ı

i.e. the principal axis angle '0 is '0 D 22:5 . The moments of inertia in the principal axis system can be determined using the transformation equations (4.155):  1  1 Iyy C Izz C Iyy  Izz cos 2'0  Iyz sin 2'0 2 2   5 p D C 2 tb 3 D I1 ; 3

I D

 1  1 Iyy C Izz  Iyy  Izz cos 2'0 C Iyz sin 2'0 2 2   5 p D  2 tb 3 D I2 ; 3

I D

I  D

 1 Iyy  Izz sin 2'0 C Iyz cos 2'0 D 0: 2

(4.175)

Alternatively, (4.157) can be used, which leads to the same result. However, this representation does not allow to assign the two principal moments of inertia I1 and I2 uniquely to their coordinate directions. The transformation of the moments of inertia can also be done graphoanalytically using Mohr’s circle. This is shown in Fig. 4.39. There are two possibilities to determine the distribution of the normal stress xx . The first possibility is to perform the calculation with respect to the axes y and z, i.e. with respect to the center of gravity C of the cross-section, without having performed the second cross-sectional normalization. Here, due to the non-vanishing deviation moment Iyz , equation (4.117) must be used. We obtain: xx D

Izz M0 C Iyz M0 Iyz M0 C Iyy M0 3 M0 z yD .z C 5y/: 2 2 Iyy Izz  Iyz Iyy Izz  Iyz 7 tb 3

(4.176)

The neutral axis follows from the requirement xx D 0 and results in: z D 5y:

(4.177)

It is shown in Fig. 4.40. The two points of the cross-section farthest from the neutral axis are points A and B as shown in Fig. 4.40. Their coordinates are yA D b, zA D b, yB D b, zB D b, and from the stress equation (4.176) the following values are obtained: xx;A D 

12 M0 M0 D 1:71 2 ; 2 7 tb tb

xx;B D

12 M0 M0 D 1:71 2 : 2 7 tb tb

(4.178)

The second way of determining the stress distribution is to perform the second cross-sectional normalization and use the stress equation (4.165). Here, the

124

4 Beams

Iyz y tb 3 2tb3 3 tb3 3

3

– tb 3 3 – 2tb 3 – tb

z tb3 I2 3

2tb3 tb 3 3

4tb3 3

2φ 0

2φ 0 5tb3 2tb3 7tb3 8tb3 3tb3 I1 3 3 3

Iyy ,Izz

φ0

yp

zp

3

Fig. 4.39 Mohr’s circle for the Z-cross-section. Fig. 4.40 Stress distribution and neutral axis for the Zcross-section.

σxx

12 M0 7tb2

12 M0 7tb2

A

y

5 1

B

z

Neutral axis

transformed bending moments are to be applied, which result from (4.160) as follows: M D M0 cos '0 C M0 sin '0 D 1:30M0; M D M0 sin '0 C M0 cos '0 D 0:54M0 :

(4.179)

The expression (4.165) then gives: xx D 0:42

M0  M0

 2:16 3 : 3 tb tb

(4.180)

4.4 Second cross-sectional normalization: Principal axes

125

Fig. 4.41 Stress distribution and neutral axis for the Zcross-section in the principal axis system.

xx

12 M0 7tb2

12 M0 7tb2

A 1

yp

5.14

B

zp Neutral axis

The neutral axis follows as:  D 5:14 :

(4.181)

It is shown in Fig. 4.41. The two relevant points for stress determination are again the two points A and B when using the principal axis system. Their coordinates A , A , B , B follow from the transformation equations (4.151) and result as:

A D 0:54b;

A D 1:30b;

B D 0:54b;

B D 1:30b:

(4.182)

The stresses in these points then follow as: xx;A D 1:71

M0 ; tb 2

xx;B D 1:71

M0 : tb 2

(4.183)

These results are identical to (4.178). J Example 4.20

Consider the cross-section shown in Fig. 4.42, left, which is loaded by the bending moment My D 250F t. Let the wall thickness be t. The position of the center of gravity is already shown in Fig. 4.42, left, and both the moments of inertia and the deviation moment with respect to the center of gravity C are given as follows: Iyy D 155:65t 4 ;

Izz D 52:18t 4 ;

Iyz D 51:90t 4 :

(4.184)

The stress distribution xx is to be determined. The principal axis angle follows from (4.154): tan 2'0 D

2Iyz D 1:003: Izz  Iyy

(4.185)

126

4 Beams

Fig. 4.42 Cross-section under consideration (left), stress distribution (right).

12.12 F2 t xx

1.65t

A yp y

3.4t

y

t

10t

1

2.41

14.28 F2 t

z t 6.5t

This yields:

Neutral axis

zp z B

'0 D 22:55ı :

(4.186)

The transformation of the moments of inertia results is performed according to (4.155):  1  1 Iyy C Izz C Iyy  Izz cos 2'0  Iyz sin 2'0 D 177:19t 4 D I1 ; 2 2  1  1 D Iyy C Izz  Iyy  Izz cos 2'0 C Iyz sin 2'0 D 30:65t 4 D I2 ; 2 2  1 D Iyy  Izz sin 2'0 C Iyz cos 2'0 D 0: (4.187) 2

I D I I 

The moment transformation (4.160) leads to: M D 230:89F t;

M D 95:87F t:

(4.188)

The stress equation (4.165) then leads to the following expression: xx D 1:30

F t F t

C 3:13 4 : t4 t

(4.189)

From this, the neutral axis can be determined as:  D 2:41 :

(4.190)

It is shown in Fig. 4.42, right. The points relevant for the stress calculation are points A and B marked in Fig. 4.42, right. Their coordinates follow from (4.151) as:

A D 2:83t;

A D 2:51t;

B D 1:93t;

B D 6:34t:

(4.191)

Thus, the following stresses at points A and B can be determined from (4.189): xx;A D 12:12

F ; t2

xx;B D 14:28

F : t2

(4.192) J

5

Beam deflections

In many applications, it is of technical relevance to determine deflections of beam structures. In this chapter, the necessary basic equations of beam bending are provided and applied to the analysis of statically determinate and statically indeterminate single-span and multi-span beams. This is followed by a presentation of elementary bending cases, which are then used for the analysis of statically indeterminate systems and of multispan and angled beams. Finally, biaxial bending is considered, i.e. bending about two axes. Students are enabled to formulate beam bending problems and apply the basic equations to determine the deflections of beam structures.

5.1

Basic equations of beam bending

In addition to determining the stresses in beams, it is also important in many applications to make statements about the deformations of beam structures. This field of activity concerns the determination of the so-called elastic line of a beam. In general, the three displacements u, v, w occur in a beam structure, where the displacement u is the displacement in axial direction. The determination of the longitudinal displacement u of a bar was already the subject of Chap. 3 and will not be treated further here. Hence, in the following we will consider the determination of the displacements v and w of beam structures. We assume in all cases a principal axis system, which we will denote as x, y, z. Starting point of the considerations are the equilibrium conditions for the beam with variable bending stiffness EIyy .x/, which is loaded by the line load qz , see Fig. 5.1, left. We assume here that the reference frame x, y, z is a principal axis system and consider the infinitesimal beam element of Fig. 5.1, right (length dx). The equilibrium conditions result in: dVz D Vz0 D qz ; dx

dMy D My0 D Vz : dx

© The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_5

(5.1)

127

128

5 Beam deflections

Vz

qz(x)

x

EIyy(x)

qz

dMy My+ dx dx

My dx

dV Vz + dxzdx

l

z Fig. 5.1 Equilibrium at the infinitesimal beam element.

If we use the constitutive law (4.161) at this point, where for simplicity we want to denote the reference axes and  as y and z, but clearly assume that these are the principal axes of the cross-section, then we obtain with My0 D Vz and Vz0 D qz : My D EIyy w 00 ;

 0 Vz D  EIyy w 00 ;

 00 EIyy w 00 D qz :

(5.2)

If the beam under consideration has constant properties with respect to x and thus has a constant bending stiffness EIyy , then we have: My D EIyy w 00 ;

Vz D EIyy w 000 ;

EIyy w 0000 D qz :

(5.3)

Regarding the bending moment Mz , the transverse shear force Vy and the line load qy , the equilibrium conditions are as follows: dVy D Vy0 D qy ; dx

dMz D Mz0 D Vy : dx

(5.4)

With the constitutive law (4.161) follows: Mz D EIzz v 00 ;

 0 Vy D  EIzz v 00 ;



EIzz v 00

00

D qy :

(5.5)

If the beam has a constant bending stiffness EIzz , then the following expressions result: (5.6) Mz D EIzz v 00 ; Vy D EIzz v 000 ; EIzz v 0000 D qy : In the following, we will first restrict ourselves to the case of uniaxial bending in the xz-plane and employ equations (5.2) and (5.3), respectively.

5.2 Statically determinate single-span beams

129

5.2 Statically determinate single-span beams From the equations (5.2) and (5.3), respectively, it can be seen that the determination of the bending deformations or the so-called elastic line w.x/ can be accomplished by integration of these equations. In principle, there are several ways to do this. One possibility is to first determine the bending moment My .x/ from equilibrium considerations for statically determinate systems and then to determine both the inclination w 0 .x/ and the elastic line w.x/ by single and double integration, respectively. Thus, for a given moment distribution My .x/ we obtain: EIyy w 00 D My ; Z EIyy w 0 D  My dx C C1 ; “ EIyy w D  My dxdx C C1 x C C2 :

(5.7)

Here, the two integration constants C1 and C2 are to be determined from given geometric boundary conditions. Likewise, we can start from the applied load qz .x/. This is especially the case for statically indeterminate systems where the moment distribution My cannot be readily determined from equilibrium considerations. EIyy w 0000 D qz ; 000

Z

EIyy w D Vz D

qz dx C C1 ; “ EIyy w 00 D My D qz dxdx C C1 x C C2 ; • 1 qz dxdxdx C C1 x 2 C C2 x C C3 ; EIyy w 0 D 2 ZZZZ 1 1 EIyy w D qz dxdxdxdx C C1 x 3 C C2 x 2 C C3 x C C4 : 6 2

(5.8)

Here, too, the integration constants are to be determined from given boundary conditions, whereby a distinction is to be made here between geometric boundary conditions and static boundary conditions. Geometric boundary conditions, in the case of a beam structure, refer to such conditions that concern the deflection w and its first derivative/the inclination of the beam axis w 0 . Static boundary conditions are conditions related to the transverse shear force Vz and the bending moment My . A selection of typical boundary conditions is shown in Fig. 5.2. In detail the depicted boundary conditions are:  At a simply supported beam end, both the deflection w and the bending moment My D EIyy w 00 vanish (Fig. 5.2, top left).

130

5 Beam deflections

Fig. 5.2 Typical boundary conditions for beams.

x

My Vz

Vz

w=0 EIw =0

z

Vz

x

My Vz

w=0 w =0

z

EIw =0 EIw =0

z

x

My

x

My

z

w =0 EIw =0

 If there is a free unloaded beam end without any support, then both the transverse shear force Vz D EIyy w 000 and the bending moment My D EIyy w 00 vanish (Fig. 5.2, top right).  At a clamped beam end, both the deflection w and the inclination w 0 of the elastic line become zero (Fig. 5.2, bottom left).  If there is a parallel guidance at one end of the beam (Fig. 5.2, bottom right), then both the inclination w 0 of the elastic line and the transverse shear force Vz D EIyy w 000 become zero at this beam end. In this section we restrict ourselves to single-span beams, i.e., beams for which both the state variables Vz , My , w 0 , and w and the load qz are continuous and differentiable functions, and illustrate the procedure at an elementary example. We consider the cantilever beam of Fig. 5.3 of length l, which is clamped at its left end and loaded at its free end by a force F . The beam has a constant bending stiffness EIyy . We determine the elastic line of the beam by integrating the beam differential equation (5.3) four times: EIyy w 0000 D qz .x/ D 0; EIyy w 000 D Vz .x/ D C1 ; EIyy w 00 D My .x/ D C1 x C C2 ; 1 EIyy w 0 D C1 x 2 C C2 x C C3 ; 2 1 1 EIyy w D C1 x 3 C C2 x 2 C C3 x C C4 : 6 2

(5.9)

Fig. 5.3 Cantilever beam.

F x

w(x)

l z

wmax wmax

5.2 Statically determinate single-span beams

131

The integration constants C1 ; : : : ; C4 are determined from the boundary conditions, which can be stated here as follows: w.x D 0/ D 0;

w 0 .x D 0/ D 0;

Vz .x D l/ D F;

My .x D l/ D 0: (5.10)

From the requirement w.x D 0/ D 0 we immediately obtain C4 D 0:

(5.11)

Similarly, the condition w 0 .x D 0/ D 0 leads to C3 D 0:

(5.12)

The requirement Vz .x D l/ D F results in: C1 D F:

(5.13)

Finally, the last condition My .x D l/ D 0 yields: C1 l C C2 D 0;

(5.14)

C2 D F l:

(5.15)

and with (5.13) this leads to Thus, the state variables of the beam can be specified as: Vz .x/ D F; My .x/ D F l

x

1 ;

l F l2 x  x w .x/ D 2 ; 2EIyy l l

  x 2  x 3 F l3  3 : w.x/ D 6EIyy l l 0

(5.16)

0 of the elastic line w.x/ as well as the maximum The maximum inclination wmax deflection wmax occur at the free end of the beam with the values 0 D wmax

F l2 ; 2EIyy

wmax D

F l3 : 3EIyy

(5.17)

The second possibility to determine the elastic line w.x/ is to start from the bending moment My .x/, which can be determined at this statically determinate beam from elementary equilibrium considerations. We have (see also (5.16)): EIyy w 00 D My D F .x  l/;  2  x EIyy w 0 D F  lx C C1 ; 2  3  x lx 2 EIyy w D F  C C1 x C C2 : 6 2

(5.18)

132

5 Beam deflections

Fig. 5.4 Superposition principle

F q0

=

w(x)

q0

wq(x)

+ F

wF(x)

The two integration constants occurring here are determined from the two geometric boundary conditions w.x D 0/ D 0 and w 0 .x D 0/ D 0 and are obtained as: C1 D 0;

C2 D 0:

(5.19)

Thus, the inclination w 0 .x/ and the elastic line w.x/ of the cantilever beam are as follows: x F l2 x  2 ; w 0 .x/ D 2EIyy l l

  F l3 x 2  x 3 w.x/ D  3 : (5.20) 6EIyy l l Apparently, these results are identical to (5.16). For the determination of beam deflections, the superposition principle applies since the beam differential equation EIyy w 0000 D qz is a linear differential equation. This means that the deformations resulting from individual load cases may be added together to give total deformations of a beam under combined load. This is illustrated in Fig. 5.4 at the example of a simply supported beam, which is under the uniform line load q0 and the single force F . Here the elastic line w.x/ is composed of the two parts wq .x/ and wF .x/, i.e.: w.x/ D wq .x/ C wF .x/:

(5.21)

Example 5.1

Consider the beam of Fig. 5.5 (length l, constant bending stiffness EIyy ), which is loaded by the uniform line load q0 . We want to determine the elastic line w.x/ of the beam.

5.2 Statically determinate single-span beams

133

Fig. 5.5 Simply supported beam under line load q0 .

q0 x

wmax w(x)

l z

We solve the problem by starting from the beam differential equation EIyy w 0000 D q0 : EIyy w 0000 D q0 ; EIyy w 000 D Vz D q0 x C C1 ; 1 EIyy w 00 D My D q0 x 2 C C1 x C C2 ; 2 1 1 0 3 EIyy w D q0 x C C1 x 2 C C2 x C C3 ; 6 2 1 1 1 EIyy w D q0 x 4 C C1 x 3 C C2 x 2 C C3 x C C4 : 24 6 2

(5.22)

The boundary conditions to be applied here to determine the integration constants C1 , C2 , C3 and C4 are: w.x D 0/ D 0; My .x D 0/ D 0; w.x D l/ D 0; My .x D l/ D 0:

(5.23)

From the condition w.x D 0/ D 0 we obtain the constant C4 as C4 D 0:

(5.24)

From the condition My .x D 0/ D 0 follows immediately: C2 D 0:

(5.25)

The condition My .x D l/ D 0 gives: 1 C1 D  q0 l: 2

(5.26)

Finally, from the condition w.x D l/ D 0 we obtain: C3 D

1 q0 l 3 : 24

(5.27)

134

5 Beam deflections

The state variables of the beam then follow as: q0 l  x 2 1 ; 2 l q0 lx  x 1 ; My .x/ D  2 l 

  x 2 q0 l 3 x 3 0 w .x/ D 6 C1 ; 4 24EIyy l l   4  4 x x 3 x q0 l : 2 C w.x/ D 24EIyy l l l Vz .x/ D 

(5.28)

The maximum bending moment occurs at the center of the beam at x D 2l and amounts to:   q0 l 2 l Mmax D My x D D : (5.29) 2 8 The rotations of the two supports follow from the inclination w 0 as: w 0 .x D 0/ D

q0 l 3 ; 24EIyy

w 0 .x D l/ D 

q0 l 3 : 24EIyy

(5.30)

In order to check the correctness of the computations one can use the criterion that the slope w 0 of the elastic line w must become zero at the center of the beam at x D 2l . One can easily convince oneself that this is the case. The maximum deflection occurs at the center of the beam at x D 2l and amounts to:   l 5q0 l 4 : (5.31) wmax D w x D D 2 384EIyy J Example 5.2

Consider the beam shown in Fig. 5.6 with constant bending stiffness EIyy , which is loaded by a sinusoidal line load qz .x/ D q0 sin. x /. We want to determine the l elastic line w.x/. Fig. 5.6 Simply supported beam loaded by sinusoidal line load qz .x/.

q0

x

wmax w(x)

l z

5.2 Statically determinate single-span beams

135

To solve the problem, we start from the beam differential equation EIyy w 0000 D qz : EIyy w 0000 D q0 sin

 x l

;

 x l C C1 ; cos  l  2  l x C C1 x C C2 ; EIyy w 00 D My D q0 sin  l  3  x 1 l cos C C1 x 2 C C2 x C C3 ; EIyy w 0 D q0  l 2  4  l x 1 1 sin C C1 x 3 C C2 x 2 C C3 x C C4 : EIyy w D q0  l 6 2

EIyy w 000 D Vz D q0

(5.32)

The evaluation of the boundary conditions w.x D 0/ D 0; My .x D 0/ D 0; w.x D l/ D 0; My .x D l/ D 0

(5.33)

leads to C1 D C2 D C3 D C4 D 0, so that the state variables of the beam can be specified as follows:  x l ; cos  l  2  l x My D q0 sin ;  l  3  x l q0 cos ; w0 D EIyy  l  4  l q0 x : wD sin EIyy  l Vz D q0

(5.34)

The maximum deflection wmax is obtained at the center of the beam and amounts to    4 l l q0 : (5.35) D wmax D w x D 2 EIyy  The maximum bending moment Mmax is also present at x D value    2 l l : D q0 Mmax D My x D 2 

l 2

and takes the (5.36) J

136

5 Beam deflections

Example 5.3

Consider the cantilever beam shown in Fig. 5.7 which has a constant bending stiffness EIyy and is loaded by a linear line load qz .x/ D q0 xl . The elastic line of the beam is to be determined. We employ the beam differential equation EIyy w 0000 D qz as follows: x EIyy w 0000 D q0 ; l q0 x 2 C C1 ; 2 l 3 q0 x EIyy w 00 D My D C C1 x C C2 ; 6 l q0 x 4 1 EIyy w 0 D C C1 x 2 C C2 x C C3 ; 24 l 2 q0 x 5 1 1 EIyy w D C C1 x 3 C C2 x 2 C C3 x C C4 : 120 l 6 2

EIyy w 000 D Vz D

(5.37)

The boundary conditions to be applied here are: w.x D 0/ D 0; w 0 .x D 0/ D 0; Vz .x D l/ D F; My .x D l/ D 0:

(5.38)

This leads to the following expressions for the integration constants: C1 D 

q0 l ; 2

C2 D

q0 l 2 ; 3

C3 D 0;

C4 D 0:

(5.39)

Thus, the state variables of the beam can be given as:

 x 2  q0 l Vz D 1 ; 2 l

 q0 l 2  x 3 x C3 2 ;  My D 6 l l   x 2 3  4 x q0 l x 6 C8 ; w0 D 24EIyy l l l

   x 2  4 x 5 x 3 q0 l  10 C 20 : wD 120EIyy l l l

(5.40) J

Fig. 5.7 Cantilever beam under linear line load qz .

q0

x l z

5.3 Statically indeterminate single-span beams

5.3

137

Statically indeterminate single-span beams

The beam situations considered so far were statically determinate; both the support reactions as well as the transverse shear force and the bending moment could be determined from equilibrium conditions. We now extend the considerations to statically indeterminate single-span beams. The procedure for the determination of the state quantities Vz , My , w 0 , w is identical to the explanations for statically determinate systems, but for statically indeterminate beams we must perform the analysis using the beam differential equation EIyy w 0000 D qz in all cases since the moment distribution My cannot be obtained exclusively from equilibrium considerations in such systems. As an example we consider the beam of Fig. 5.8, which has the length l and the constant bending stiffness EIyy and which is loaded by the uniform line load q0 and the edge moment M0 . The beam is clamped at its left end at x D 0 and simply supported at its right end at x D l. We solve the given statically indeterminate problem by quadruple integration of the beam differential equation: EIyy w 0000 D q0 ; EIyy w 000 D Vz D q0 x C C1 ; EIyy w 00 D My D

1 q0 x 2 C C1 x C C2 ; 2

EIyy w 0 D

1 1 q0 x 3 C C1 x 2 C C2 x C C3 ; 6 2

EIyy w D

1 1 1 q0 x 4 C C1 x 3 C C2 x 2 C C3 x C C4 : 24 6 2

(5.41)

The boundary conditions to be considered here are: w.x D 0/ D 0;

w 0 .x D 0/ D 0;

w.x D l/ D 0;

My .x D l/ D M0 : (5.42)

From the condition w.x D 0/ D 0 follows immediately: C4 D 0:

(5.43)

Fig. 5.8 Statically indeterminate beam under uniform line load q0 and edge moment M0 .

q0 x l z

M0

138

5 Beam deflections

The condition w 0 .x D 0/ D 0 gives: C3 D 0:

(5.44)

From the condition w.x D l/ D 0 the following expression can be derived: q0 l 4 1 1 C C1 l 3 C C2 l 2 D 0: 24 6 2

(5.45)

Finally, from the condition My .x D l/ D M0 we obtain: q0 l 2 C C1 l C C2 D M0 : 2

(5.46)

The equations (5.45) and (5.46) represent two equations for the two still unknown constants C1 and C2 . The solution is: C1 D 

5q0 l 3M0 C ; 8 2l

C2 D

q0 l 2 M0  : 8 2

(5.47)

With the constants C1 , C2 , C3 and C4 thus determined, the state variables of the statically indeterminate beam can be given as follows: i 3M q0 l h x 0 8 5  ; 8 l 2l

 q0 l 2  x 2 x M0  x 3 1 ; My D  5 C1  4 8 l l 2 l  

 q0 l 3 4  x 3 5  x 2 x x M0 l x 2  C 2 ; C 3 w0 D 8EIyy 3 l 2 l l 4EIyy l l



 M0 l 2  x 3  x 2 q0 l 4 1  x 4 5  x 3 1  x 2  C  C : (5.48) wD 8EIyy 3 l 6 l 2 l 4EIyy l l Vz D 

For the special case M0 D 0, the elastic line of the statically indeterminate beam is shown in Fig. 5.9 qualitatively. The support reactions MA , AV and BV (Fig. 5.10) can be determined as: MA D My .x D 0/ D  AV D Vz .x D 0/ D

q0 l 2 M0 C ; 8 2

5q0 l 3M0  ; 8 2l

BV D Vz .x D l/ D

3q0 l 3M0 C : 8 2l

(5.49)

5.3 Statically indeterminate single-span beams

139

Fig. 5.9 Elastic line w.x/ of the statically indeterminate beam under uniform line load q0 .

q0 x

w(x)

l z Fig. 5.10 Determination of the support reactions of the statically indeterminate beam.

q0

MA

M0

AV

BV

Example 5.4

Consider the beam of Fig. 5.11 (length l, constant bending stiffness EIyy ), which is clamped at its left end at x D 0 and has a vertically guided support at its right end. The load consists of a uniform line load q0 . The state variables Vz , My , w 0 and w of the beam are to be determined. The problem is solved by quadruple integration of the beam differential equation EIyy w 0000 D q0 : EIyy w 0000 D q0 ; EIyy w 000 D Vz D q0 x C C1 ; 1 EIyy w 00 D My D q0 x 2 C C1 x C C2 ; 2 1 1 0 3 EIyy w D q0 x C C1 x 2 C C2 x C C3 ; 6 2 1 1 1 4 q0 x C C1 x 3 C C2 x 2 C C3 x C C4 : EIyy w D 24 6 2

(5.50)

The boundary conditions are: w.x D 0/ D 0;

w 0 .x D 0/ D 0;

Vz .x D l/ D 0;

Fig. 5.11 Statically indeterminate beam with one clamped end and one end with a vertically guided support.

w 0 .x D l/ D 0: q0 x l

z

(5.51)

140

5 Beam deflections

Fig. 5.12 Support reactions of the statically indeterminate beam with one clamped end and one end with a vertically guided support.

q0

MA

MB

AV

From this, the following integration constants C1 , C2 , C3 and C4 can be obtained: C1 D q0 l;

C2 D

q0 l 2 ; 3

C3 D 0;

C4 D 0:

The state variables of the beam then follow as: x 1 ; Vz D q0 l l 

 1 x 2 x 1 2 ; C  My D q0 l  2 l l 3 

q0 l 3 1  x 3 1  x 2 1 x w0 D  C ; EIyy 6 l 2 l 3l

 q0 l 4 1  x 4  x 3  x 2  C : wD 6EIyy 4 l l l

(5.52)

(5.53)

The support reactions MA , AV and MB (Fig. 5.12) can be obtained as: q0 l 2 ; 3 AV D Vz .x D 0/ D q0 l;

MA D My .x D 0/ D 

MB D My .x D l/ D

q0 l 2 : 6

(5.54) J

5.4 Multi-span beams The explanations so far assumed beam structures where the quantities qz , Vz , My , w 0 and w are continuous and differentiable over the whole length of the beam. Technically very relevant is also the case of the so-called multi-span beams. If the considered beam is supported by an intermediate support or interrupted by a joint, or if a beam is to be analyzed where single forces or moments act at any point, or where discontinuities of the bending stiffness EIyy are present, then the beam has to be divided mathematically into regions within which both the load qz and the state variables Vz , My , w 0 , w are continuous. An example for such a multi-span beam

5.4 Multi-span beams

141

Fig. 5.13 Exemplary multispan beam.

x z

is shown in Fig. 5.13. If we assume the beam differential equation EIyy w 0000 D qz in each area when considering this example, then the fourfold integration to be performed here to determine the elastic line must be performed separately in each subdomain. For the example of Fig. 5.13, this results in four subdomains with a total of 4  4 D 16 integration constants. In order to determine these, in addition to the boundary conditions, so-called transition conditions must be formulated at the boundaries between the subdomains. A selection of typical transition conditions for the transition between two subdomains i and i C 1 which are loaded by the line loads qi and qi C1 and have the bending stiffnesses EIi and EIi C1 , respectively, is shown in Fig. 5.14. At each transition point, four transition conditions can be formulated. For illustration we consider the example of Fig. 5.13 and formulate the boundary and transition conditions. For this purpose, the subdomain axes xi (i D 1; 2; 3; 4)

qz,i

qz,i+1

My,i

qz,i

qz,i+1

My,i+1 My,i Vz,i

wi =0 wi+1=0 wi = wi+1 EIi wi =EIi+1wi+1

qz,i

qz,i+1

My,i

F qz,i+1

M y,i

My,i+1 wi = wi+1 Vz,i+1 wi = wi+1 EIi wi = EIi+1wi+1 EIi wi =EIi+1wi+1 -F

qz,i+1 M

wi = wi+1 wi = wi+1 EIi wi =EIi+1wi+1 EIi wi =EIi+1wi+1

Vz,i

wi = wi+1 Vz,i+1 EIi wi =0 EIi+1wi+1=0 EIi wi =EIi+1wi+1

Vz,i

My,i

Vz,i+1

qz,i My,i+1

qz,i

My,i+1 Vz,i

Vz,i+1

My,i+1

wi = wi+1 Vz,i+1 wi = wi+1 EIi wi = EIi+1wi+1-M EIi wi =EIi+1wi+1

Vz,i

Fig. 5.14 Exemplary transition conditions.

142 Fig. 5.15 Formulation of boundary and transition conditions for the multi-span beam.

5 Beam deflections

q0

M0

x1

F0

x2

x3

x4

z l1

l2

l3

l4

are introduced as shown in Fig. 5.15. Let the bending stiffness EIyy of the beam be constant over its entire length. At the left support point at x1 D 0 the following boundary conditions have to be fulfilled: w1 .x1 D 0/ D 0; M1 .x1 D 0/ D M0 :

(5.55)

Herein M1 is the bending moment of subdomain 1. The second boundary condition in (5.55) can also be represented as follows: w100 .x1 D 0/ D 

M0 : EIyy

(5.56)

At the joint at the position x1 D l1 or x2 D 0 the following transition conditions have to be fulfilled: w1 .x1 D l1 / D w2 .x2 D 0/; V1 .x1 D l1 / D V2 .x2 D 0/; M1 .x1 D l1 / D 0; M2 .x2 D 0/ D 0:

(5.57)

Herein, the moment and transverse shear force conditions can be specified as: w1000 .x1 D l1 / D w2000 .x2 D 0/; w100 .x1 D l1 / D 0; w200 .x2 D 0/ D 0:

(5.58)

At the support point x2 D l2 or x3 D 0 the following transition conditions have to be fulfilled: w2 .x2 D l2 / D 0; w3 .x3 D 0/ D 0; w20 .x2 D l2 / D w30 .x3 D 0/; M2 .x2 D l2 / D M3 .x3 D 0/;

(5.59)

5.4 Multi-span beams

143

where the last condition occurring here can also be formulated as follows: w200 .x2 D l2 / D w300 .x3 D 0/:

(5.60)

At the force application point x3 D l3 or x4 D 0 the transition conditions are: w3 .x3 D l3 / D w4 .x4 D 0/; w30 .x3 D l3 / D w40 .x4 D 0/; V3 .x3 D l3 /  V4 .x4 D 0/ D F0 ; M3 .x3 D l3 / D M4 .x4 D 0/:

(5.61)

The latter two conditions can be formulated as: w3000 .x3 D l3 /  w4000 .x4 D 0/ D 

F0 ; EIyy

w300 .x3 D l3 / D w400 .x4 D 0/:

(5.62)

Finally, at the clamped edge at x4 D l4 , the following boundary conditions apply: w4 .x4 D l4 / D 0; w40 .x4 D l4 / D 0:

(5.63)

Hence, a total of 16 conditions are available for the determination of the 16 integration constants. Example 5.5

For the beam shown in Fig. 5.16, the state variables Vz , My , w 0 and w are to be determined. The beam has the bending stiffness EIyy in the area of the length l1 between the two supports. In the cantilevered area with length l2 , the bending stiffness is 2EIyy . For convenience, the two reference axes x1 and x2 are introduced for the two areas. There are again two ways to determine the state variables. On the one hand, we can determine the moment distribution in advance and, starting from this, determine the inclination w 0 and the elastic line w by simple and twofold integration, respectively. On the other hand, we can determine the state variables by quadruple integration of the beam differential equation. Both ways are described below. Fig. 5.16 Two-span beam under edge moment.

EIyy

x1 l1

z

2EIyy x2

l2

M0

144

5 Beam deflections

Fig. 5.17 Moment distribution for the two-span beam.

M0

x1

x2

AV

My

BV

We first determine the moment distribution My .xi / (i D 1; 2), it is shown in Fig. 5.17. It can be formulated as follows: My .x1 / D 

M0 x 1 ; l1

My .x2 / D M0 :

(5.64)

The support reactions AV and BV are obtained with the directions of action shown in Fig. 5.17: M0 : (5.65) AV D BV D l1 We now determine the inclination w 0 .xi / and the elastic line w.xi / of the beam by single and twofold integration of the bending moment My .xi /, respectively, dividing the considerations here into two subdomains. In subdomain 1 we have: EIyy w100 .x1 / D My .x1 / D

M0 x 1 ; l1

EIyy w10 .x1 / D

M0 x12 C C1 ; 2l1

EIyy w1 .x1 / D

M0 x13 C C1 x 1 C C2 : 6l1

(5.66)

In subdomain 2 we obtain: 2EIyy w200 .x2 / D My .x2 / D M0 ; 2EIyy w20 .x2 / D M0 x2 C D1 ; 2EIyy w2 .x2 / D

M0 x22 C D1 x2 C D2 : 2

(5.67)

The integration constants C1 , C2 , D1 and D2 are determined from the given geometric boundary and transition conditions. At the left support point at x1 D 0, the deflection w1 must disappear: w1 .x1 D 0/ D 0:

(5.68)

5.4 Multi-span beams

145

From this we obtain directly: C2 D 0:

(5.69)

Analogously, at the right support point at the location x2 D 0, it is required that the deflection w2 becomes zero, which leads to: D2 D 0:

(5.70)

The deflection w1 becomes zero at the right support point at x1 D l1 : w1 .x1 D l1 / D 0:

(5.71)

This results in:

M0 l 1 : (5.72) 6 Finally, we need to impose the requirement that on both sides of the right support point the slopes w10 and w20 of the two subdomains are identical: C1 D 

w10 .x1 D l1 / D w20 .x2 D 0/: From this we obtain: D1 D

2 M0 l 1 : 3

(5.73)

(5.74)

With the integration constants thus determined, the inclination w 0 and the elastic line w of the beam can be specified. In area 1 we obtain: # "   x1 2 M0 l 1 D 1 ; 3 6EIyy l1 "  # x1 3 x1 M0 l12  : w1 .x1 / D 6EIyy l1 l1 w10 .x1 /

(5.75)

In subdomain 2 we have:   M0 l 1 3 x 2 D C1 ; 3EIyy 2 l1   M0 l 1 x 2 1 x 2 2 C w2 .x2 / D : 2EIyy 2 l1 3

w20 .x2 /

(5.76)

The maximum deflection wmax between the two supports in area 1 is obtained at the point where the slope w10 becomes zero: # "   x1 2 M0 l 1  1 D 0: 3 6EIyy l1

(5.77)

146

5 Beam deflections

This can be solved for x1 as follows: l1 x1 D p : 3

(5.78)

Thus, the maximum deflection wmax in subdomain 1 is given as:  wmax D w1

l1 x1 D p 3



M0 l 2 D p 1 : 9 3EIyy

(5.79)

The inclination of the beam as well as its deflection at the cantilever end at x2 D l2 are given as:   M0 l 1 3 l 2 0 C1 ; w2 .x2 D l2 / D 3EIyy 2 l1   M0 l 1 l 2 1 l 2 2 C w2 .x2 D l2 / D : (5.80) 2EIyy 2 l1 3 We now show how the determination of the state variables Vz , My , w 0 , w can be performed if we start from the beam differential equation EIyy w 0000 D qz in both subdomains and integrate fourfold. In domain 1 we obtain: EIyy w10000 D q0 D 0; EIyy w1000 D Vz D C1 ; EIyy w100 D My D C1 x1 C C2 ; EIyy w10 D

1 C1 x12 C C2 x1 C C3 ; 2

EIyy w1 D

1 1 C1 x13 C C2 x12 C C3 x1 C C4 : 6 2

(5.81)

In subdomain 2 we have: 2EIyy w20000 D q0 D 0; 2EIyy w2000 D Vz D D1 ; 2EIyy w200 D My D D1 x2 C D2 ; 2EIyy w20 D

1 D1 x22 C D2 x2 C D3 ; 2

2EIyy w2 D

1 1 D1 x23 C D2 x22 C D3 x2 C D4 : 6 2

(5.82)

The boundary and transition conditions to be imposed for the given beam situation are as follows. At the left support at x1 D 0, both the deflection w1 and the

5.4 Multi-span beams

147

bending moment My vanish: w1 .x1 D 0/ D 0;

My .x1 D 0/ D 0:

(5.83)

At the right support point at x1 D l1 or x2 D 0 the following transition conditions apply. First of all, it must be required that the displacements of the subdomains 1 and 2 disappear at this support, i.e.: w1 .x1 D l1 / D 0;

w2 .x2 D 0/ D 0:

(5.84)

In addition, the inclinations of the beam axis must be identical on both sides, so that we have: w10 .x1 D l1 / D w20 .x2 D 0/: (5.85) Finally, the bending moment on both sides of the support must have identical values in both subdomains, i.e.: My .x1 D l1 / D My .x2 D 0/:

(5.86)

At the cantilever end at x2 D l2 , the following static boundary conditions apply: My .x2 D l2 / D M0 ;

Vz .x2 D l2 / D 0:

(5.87)

From the boundary and transition conditions, the integration constants C1 , C2 , C3 , C4 and D1 , D2 , D3 , D4 can be determined as: C1 D

M0 ; l1

D1 D 0;

M0 l 1 ; 6

C4 D 0;

2 M0 l 1 ; 3

D4 D 0:

C2 D 0;

C3 D 

D 2 D M0 ;

D3 D

(5.88)

With this, the state variables Vz , My , w 0 , w of the two subdomains can then be given as follows. In subdomain 1 we obtain: Vz .x1 / D 

M0 ; l1

M0 x1 ; l1 # "   2 x l M 0 1 1 w10 .x1 / D 1 ; 3 6EIyy l1 "  # x1 3 x1 M0 l12 w1 .x1 / D  : 6EIyy l1 l1

My .x1 / D 

(5.89)

148

5 Beam deflections

For subdomain 2, the result is: Vz .x2 / D 0; My .x2 / D M0 ;

 3 x2 C1 ; 2 l1   M0 l 1 x 2 1 x 2 2 w2 .x2 / D C : 2EIyy 2 l1 3 w20 .x2 / D

M0 l 1 3EIyy



(5.90)

The support reactions AV and BV can be obtained from this with the directions of action as shown in Fig. 5.17: AV D Vz .x1 D 0/ D

M0 ; l1

BV D Vz .x1 D l1 / C Vz .x2 D 0/ D

M0 : l1

(5.91) J

Example 5.6

For the cantilever beam shown in Fig. 5.18, the elastic line w and the inclination w 0 of the elastic line are to be determined. The beam has the constant bending stiffness EIyy , the two local axes x1 and x2 are introduced as shown. For the solution of this problem we start from the differential equation EIyy w 0000 D qz . The beam differential equation EIyy w 0000 D qz is solved in the two subdomains by quadruple integration. In the left subdomain we obtain: EIyy w10000 D q1 ; EIyy w1000 D Vz;1 D q1 x1 C C1 ; 1 q1 x12 C C1 x1 C C2 ; 2 1 1 EIyy w10 D q1 x13 C C1 x12 C C2 x1 C C3 ; 6 2 1 1 1 EIyy w1 D q1 x14 C C1 x13 C C2 x12 C C3 x1 C C4 : 24 6 2

EIyy w100 D My;1 D

Fig. 5.18 Beam under line load.

q1 x1

q2 x2

l z

(5.92)

l

5.4 Multi-span beams

149

In subdomain 2 we have: EIyy w20000 D q2 ; EIyy w2000 D Vz;2 D q2 x2 C D1 ; 1 q2 x22 C D1 x2 C D2 ; 2 1 1 EIyy w20 D q2 x23 C D1 x22 C D2 x2 C D3 ; 6 2 1 1 1 EIyy w2 D q2 x24 C D1 x23 C D2 x22 C D3 x2 C D4 : 24 6 2

EIyy w200 D My;2 D

(5.93)

The integration constants are determined from the boundary and transition conditions of the system. The boundary conditions at the clamped end x1 D 0 are: w1 .x1 D 0/ D 0;

w10 .x1 D 0/ D 0:

(5.94)

From this, the two integration constants C3 and C4 follow as: C3 D C4 D 0:

(5.95)

The boundary conditions at the free cantilever end at x2 D l are: Vz;2 .x2 D l/ D 0;

My;2 .x2 D l/ D 0:

(5.96)

From this, D1 and D2 can be determined as: D1 D q2 l;

D2 D

1 2 q2 l : 2

(5.97)

We now consider the transition conditions at the point x1 D l or x2 D 0. There, the equality of the two transverse forces has to be required: Vz;1 .x1 D l/ D Vz;2 .x2 D 0/:

(5.98)

From this, the integration constant C1 results as: C1 D l.q1 C q2 /:

(5.99)

Likewise, the bending moments must be identical at the transition point between the two subdomains: My;1 .x1 D l/ D My;2 .x2 D 0/: It follows: C2 D

1 2 l .q1 C 3q2 /: 2

(5.100)

(5.101)

150

5 Beam deflections

At the transition point, the slope of the elastic line is also identical in both subdomains: w10 .x1 D l/ D w20 .x2 D 0/: (5.102) From this, the integration constant D3 results as: D3 D l

3



 1 q1 C q2 : 6

(5.103)

Finally, it is required that the deflections w1 and w2 at the transition point between the two subdomains match: w1 .x1 D l/ D w2 .x2 D 0/:

(5.104)

From this the integration constant D4 follows as: D4 D l

4



 1 7 q1 C q2 : 8 12

(5.105)

With the integration constants determined in this way, the expressions (5.92) and (5.93) can be evaluated. For w10 .x1 / and w20 .x2 / we obtain:

  x 2  x  x1 3 l3 1 1 D  3.q1 C q2 / C 3.q1 C 3q2 / ; q1 6EIyy l l l

   x 2 x x2 3 l3 2 2 C q1 C 6q2 :  3q2 C 3q2 q2 (5.106) w20 .x2 / D 6EIyy l l l w10 .x1 /

The values at the transition point and at the free cantilever end are: w10 .x1 D l/ D

l3 .q1 C 6q2 /; 6EIyy

w20 .x2 D l/ D

l3 .q1 C 7q2 /: 6EIyy

(5.107)

For the deflections w1 .x1 / and w2 .x2 / of the subdomains we obtain:

  x 3  x 2  x1 4 l4 1 1 w1 .x1 / D  4.q1 C q2 / C 6.q1 C 3q2 / q1 ; 24EIyy l l l

  x 3  x 2 x2 4 l4 2 2  4q2 C 6q2 q2 w2 .x2 / D 24EIyy l l l  x 2 C 3q1 C 14q2 : C 4.q1 C 6q2 / (5.108) l

5.4 Multi-span beams

151

The deflections at the transition point x1 D l or x2 D 0 and at the cantilever end x2 D l follow as: w1 .x1 D l/ D

l4 .3q1 C 14q2 /; 24EIyy

w2 .x2 D l/ D

l4 .7q1 C 41q2 /: 24EIyy

(5.109) J

Example 5.7

Consider the beam shown in Fig. 5.19, which is loaded by a force F . The beam of length l is divided by the point of application of the force into two subdomains of lengths l1 and l2 . The beam has a constant bending stiffness EIyy . The two reference axes x1 and x2 are introduced as shown. The resulting moment diagram is also shown in Fig. 5.19. The slope w 0 and the elastic line w of the beam are to be determined. We solve this problem by integrating the bending moment in each subdomain twice. The moment distributions can be formulated as follows: My .x1 / D

F l2 x 1 ; l

My .x2 / D

F l1 .l2  x2 /: l

(5.110)

The twofold integration in subdomain 1 then gives: EIyy w100 D 

F l2 x 1 ; l

EIyy w10 D 

F l2 x12 C C1 ; 2l

EIyy w1 D 

F l2 x13 C C1 x 1 C C2 : 6l

Fig. 5.19 Simply supported beam loaded by a force F (top), moment distribution (bottom).

(5.111)

F x1

x2 l2

l1 l z

My

AV

Fl2 l

Fl1l2 l

BV

Fl1 l

152

5 Beam deflections

In subdomain 2 follows: F l1 .l2  x2 /; l F l1 .l2  x2 /2 C D1 ; EIyy w20 D 2l F l1 EIyy w2 D  .l2  x2 /3  D1 .l2  x2 / C D2 : 6l

EIyy w200 D 

(5.112)

Here the coordinate .l2  x2 / was used. The integration constants C1 , C2 , D1 and D2 are determined from the given boundary and transition conditions. At the beam end x1 D 0, the deflection w1 must vanish. This leads to: C2 D 0: (5.113) Likewise, it is required that at the right end of the beam the deflection w2 disappears, which leads to D2 D 0: (5.114) The transition condition w1 .x1 D l1 / D w2 .x2 D 0/ gives the following expression: F l1 l23 F l 3 l2 (5.115)  1 C C1 l 1 D   D 1 l2 : 6l 6l Analogously, the following condition can be derived from the requirement w10 .x1 D l1 / D w20 .x2 D 0/: 

F l12 l2 F l1 l22 C C1 D C D1 : 2l 2l

(5.116)

From (5.115) and (5.116) the following expressions for the integration constants C1 and D1 can be obtained: C1 D

F l1 l2 .l C l2 /; 6l

D1 D 

F l1 l2 .l C l1 /: 6l

(5.117)

The slope w10 or w20 and the elastic line w1 or w2 can then be represented as:   F l1 l22 l x12 0 w1 .x1 / D C 2 C1 ; 3 6lEIyy l1 l2 l   F l1 l22 l C l2 x13 C x1 ; w1 .x1 / D  6lEIyy l1 l2 l2 ! 2 2 F l1 l2 l C l1 .l2  x2 / 0  3 ; w2 .x2 / D 6lEIyy l1 l2 l1 ! F l12 l2 l C l1 .l2  x2 /3 C .l2  x2 / : (5.118)  w2 .x2 / D 6lEIyy l1 l2 l1 J

5.5 Standard bending cases

153

5.5 Standard bending cases 5.5.1

Simply supported beams

A number of frequently recurring standard bending cases are summarized in the following. A constant bending stiffness EIyy is assumed for all subsequent cases. Unless otherwise noted, the elastic line w.x/, the maximum deflection wmax and its location, and the edge rotations of the considered beams are given. For the simply supported beam under constant line load q0 (Fig. 5.20), we obtain: w.x/ D

 x 2 x  q0 l 4  x 4 2 C 24EIyy l l l

wmax D

5q0 l 4 384EIyy

wA0 D wB0 D

at

xD

l 2

q0 l 3 24EIyy

(5.119)

For the simply supported beam under linear line load with the maximum value q0 (Fig. 5.21) follows: 

  x 3 q0 l 4 x x 5  10 C7 3 360EIyy l l l s r 8 q0 l 4 D 0:006522 at x D l 1  EIyy 15

w.x/ D wmax

wA0 D

Fig. 5.20 Simply supported beam under constant line load q0 .

7q0 l 3 360EIyy

wB0 D 

q0 l 3 45EIyy

(5.120)

q0 A

x

B

wmax w(x)

l z Fig. 5.21 Simply supported beam under linear line load.

A

wmax

x

w(x)

l z

q0 B

154

5 Beam deflections

Fig. 5.22 Simply supported beam under single force at an arbitrary position.

F A

x1 wmax l1

wF

x2

B

l2 l

z Fig. 5.23 Simply supported beam under centric single force.

F A

B

x wmax l 2

l 2

z

For the beam simply supported on both sides under single force (subdomain lengths l1 and l2 , Fig. 5.22), the following results: 

  F l1 l22 l x13 w1 .x1 / D C 1C  x1 6lEIyy l1 l2 l2 # "   l F l12 l2 .l2  x2 /3 C C 1 .l2  x2 /  w2 .x2 / D 6lEIyy l1 l2 l1 wF D

F l12 l22 3lEIyy

s l 2  l22 F l2 lN3 lN D wmax D l1  l2 3lEIyy 3 s 3  F l1 l2  lN l 2  l12 lN D l2  wmax D l1  l2 3lEIyy 3 F l1 l2 .l C l2 / F l1 l2 .l C l1 / wA D wB D  (5.121) 6lEIyy 6lEIyy For the simply supported beam under a centric single force follows (Fig. 5.23) 

 x F l3 l x 3 C3 w.x/ D 0x 4 48EIyy l l 2 # " 3 2 l Fl .l  x/ C 3.l  x/ w.x/ D xl 4 2 48EIyy l 2 wmax D

F l3 48EIyy

wA0 D wB0 D

F l2 16EIyy

(5.122)

5.5 Standard bending cases

155

Fig. 5.24 Simply supported beam under edge moment M0 .

M0 A

B

x wmax l z

Fig. 5.25 Simply supported beam under centric moment M0 .

M0 A

wmax

wmax

x l 2

B l 2

z

In the case of the beam simply supported on both sides under the edge moment M0 (Fig. 5.24) we obtain: 

 x 2 M0 l 2  x 3 x w.x/ D 3 C2 6EIyy l l l p   2 3M0 l 1 wmax D at x D l 1  p 27EIyy 3 l l M M 0 0 wB0 D  wA0 D 3EIyy 6EIyy

(5.123)

For the simply supported beam under the centric bending moment M0 (Fig. 5.25) the following results: 

 M0 l 2 l x 3 x w.x/ D C 0x 4 24EIyy l l 2 

  x 2 M0 l 2 x x 3 w.x/ D C 12  11 C 3 4 24EIyy l l l M0 l 2 p at 72 3EIyy M0 l wA0 D wB0 D 24EIyy

wmax D

5.5.2

l xD p 2 3

l xl 2   1 and x D l 1  p 2 3 (5.124)

Cantilever beams

For the cantilever beams of length l with constant bending stiffness EIyy considered below, both the elastic line w.x/ and the maximum deflection wmax as well as 0 the maximum slope wmax are given.

156

5 Beam deflections

Fig. 5.26 Cantilever beam under single force F .

F x

w(x)

wmax wmax

l z Fig. 5.27 Cantilever beam under single force F .

F wmax

x l1

w(x)

wmax

l2 l

z

For a cantilever beam with a force F at the free end of the beam (Fig. 5.26) we have:

  x 2  x 3 F l3 3  w.x/ D 6EIyy l l wmax D

F l3 ; 3EIyy

0 wmax D

F l2 2EIyy

(5.125)

If we consider a cantilever beam with a single force F acting at the point x D l1 (Fig. 5.27), then we obtain: "    3 # x x 2 F l13  w.x/ D 3 0  x  l1 6EIyy l1 l1   F l13 x l1  x  l 3 1 6EIyy l1 "    3 # l1 l1 2 F l3 F l12 0 D  D 3 ; wmax 6EIyy l l 2EIyy

w.x/ D wmax

(5.126)

For a cantilever beam with an edge moment M0 (Fig. 5.28) the following results: w.x/ D

M0 x 2 2EIyy

wmax D

M0 l 2 ; 2EIyy

0 wmax D

M0 l EIyy

(5.127)

5.5 Standard bending cases

157

Fig. 5.28 Cantilever beam under edge moment M0 .

M0 x

w(x)

wmax wmax

l z Fig. 5.29 Cantilever beam under line load q0 .

q0 x

w(x)

wmax wmax

l z Fig. 5.30 Cantilever beam under linear line load with maximum value q0 .

q0

x

w(x)

l

wmax wmax

z

If a cantilever beam under the uniform line load q0 (Fig. 5.29) is considered, then one obtains:

  x 3  x 4  q0 l 4 x 2 w.x/ D 4 C 6 24EIyy l l l wmax D

q0 l 4 ; 8EIyy

0 wmax D

q0 l 3 6EIyy

(5.128)

For the case where the considered cantilever beam is under a linear line load with the maximum value q0 (Fig. 5.30) follows: w.x/ D

  x 3  x 2  x 5 q0 l 4  10 C 20 120EIyy l l l

wmax D

11q0 l 4 ; 120EIyy

0 wmax D

q0 l 3 8EIyy

(5.129)

158

5.5.3

5 Beam deflections

Statically indeterminate systems

The presented standard bending cases can also be used to analyze statically indeterminate systems in a very straightforward way. Consider again the beam of Fig. 5.8, which is clamped at its left end and simply supported at its right end. Let the load be given in the form of a uniform line load q0 and an edge moment M0 (Fig. 5.31, top). We want to determine the support reactions of this beam, taking advantage of the fact that we can superpose the beam deformations due to different load cases. We first make the beam statically determinate by removing the right support (so-called 0-system). This results in the elastic line w0 .x/ as shown in Fig. 5.31, middle. In particular, we are interested here in the deflection w0 .x D l/ at the right end, which is now thought to be free, resulting in w0 .x D l/ D

q0 l 4 M0 l 2 C ; 8EIyy 2EIyy

(5.130)

as can be easily deduced from the standard bending cases of Figs. 5.28 and 5.29. In the next step, the support reaction BV is applied (so-called 1-system), which is of course present in reality. It causes the elastic line w1 , and the displacement w1 .x D l/ at the right end of the beam is (cf. the standard bending case of Fig. 5.26): w1 .x D l/ D 

BV l 3 : 3EIyy

(5.131)

One can interpret the actually statically indeterminate system as the superposition of the two described load cases of the 0-system and the 1-system. At this point we can use the fact that the displacement at the support point x D l has to vanish. Since w.x/ D w0 .x/ C w1 .x/ holds, we can therefore require: w0 .x D l/ C w1 .x D l/ D 0: Fig. 5.31 Statically indeterminate beam under constant line load q0 and edge moment M0 (top), statically determinate beam and elastic line (middle), elastic line due to the statically indeterminate support force BV (bottom).

(5.132)

q0

M0

x l z

q0

M0 w0(x) w1(x)

w0

w1 BV

5.5 Standard bending cases

159

This leads to the expression q0 l 4 M0 l 2 BV l 3 C  D 0; 8EIyy 2EIyy 3EIyy

(5.133)

which can be solved for the unknown support force BV as follows: BV D

3q0 l 3M0 C : 8 2l

(5.134)

This result agrees with (5.49). The elastic line w.x/ D w0 .x/ C w1 .x/, but also the slope w 0 .x/ D w00 C w10 as well as the bending moment M.x/ D M0 .x/ C M1 .x/ and the transverse shear force V .x/ D V0 .x/ C V1 .x/ of the statically indeterminate system can be determined from superposition. Example 5.8

Consider the two-span beam shown in Fig. 5.32, top, with two subdomains of length l. The load consists of the constant line load q0 . We obtain the statically determinate 0-system by removing the middle support and thus making the beam statically determinate. The resulting deflection w0 .x/ of the beam is shown in Fig. 5.32, middle. Of particular interest here is the deflection in the center at the support point B, which amounts to: w0 .x D l/ D

5q0 .2l/4 5q0 l 4 D : 384EIyy 24EIyy

(5.135)

In the next step, we apply the statically indeterminate support force BV to the system (1-system, Fig. 5.32, bottom) and thus obtain the displacement w1 .x D l/ Fig. 5.32 Statically indeterminate beam under uniform line load q0 (top), statically determinate beam and elastic line (middle), elastic line due to the statically indeterminate support reaction BV (bottom).

q0 x

l

l

z

q0

w0(x) w1(x)

BV

160

5 Beam deflections

at the support point at the location x D l: w1 .x D l/ D 

BV .2l/3 BV l 3 D : 48EIyy 6EIyy

(5.136)

The compatibility condition to be applied here requires that the two displacements w0 .x D l/ and w1 .x D l/ in sum become zero: w0 .x D l/ C w1 .x D l/ D 0;

(5.137)

BV l 3 5q0 l 4  D 0: 24EIyy 6EIyy

(5.138)

i.e.:

From this, the statically indeterminate support reaction can be determined as: BV D

5 q0 l: 4

(5.139)

An alternative way to make the given static system statically determinate is to introduce a full joint at the support point B (Fig. 5.33, top). The resulting rotations 0 0 w0;L .x D l/ and w0;R .x D l/ at the support point B then amount to: 0 .x D l/ D  w0;L

q0 l 3 ; 24EIyy

0 w0;R .x D l/ D

q0 l 3 : 24EIyy

(5.140)

In the next step, the statically determinate system is subjected to the statically 0 indeterminate bending moment MB D My .x D l/. The two rotations w1;L .x D l/ 0 and w1;R .x D l/ on both sides of the support point are then: 0 .x D l/ D w1;L

MB l ; 3EIyy

0 w1;R .x D l/ D 

MB l : 3EIyy

(5.141)

The compatibility condition to be raised here requires equality of rotations on both sides of the support B, i.e.: 0 0 0 0 w0;L .x D l/ C w1;L .x D l/ D w0;R .x D l/ C w1;R .x D l/:

Fig. 5.33 Statically determinate beam and elastic line (top), elastic line due to the statically indeterminate bending moment MB (bottom).

(5.142)

q0

w0,L(x=l)

w0,R(x=l)

w1,L(x=l)

w1,R(x=l)

MB

MB

5.5 Standard bending cases

161

From this, the bending moment MB can be determined as: MB D

q0 l 2 : 8

(5.143)

Since in the present case the slope of the elastic line on both sides of the support is zero (both the beam and its load are symmetrical), the compatibility requirement can be given as follows: 0 0 w0;L .x D l/ C w1;L .x D l/ D 0; 0 0 .x D l/ C w1;R .x D l/ D 0: w0;R

(5.144)

This again gives the result (5.143). With the bending moment MB thus determined, all other internal forces and moments of the statically indeterminate system can also be determined. J Example 5.9

We consider the beam shown in Fig. 5.34, top, that is clamped on both sides. The beam has the length l and the constant bending stiffness EIyy and is loaded by the uniform line load q0 . This beam situation is statically indeterminate to the second degree. We first solve the problem by integrating the beam differential equation EIyy w 0000 D q0 . We obtain: EIyy w 0000 D q0 ; EIyy w 000 D Vz D

Z

q0 dx C C1 ; “ EIyy w 00 D My D q0 dxdx C C1 x C C2 ; • 1 q0 dxdxdx C C1 x 2 C C2 x C C3 ; EIyy w 0 D 2 ZZZZ 1 1 EIyy w D q0 dxdxdxdx C C1 x 3 C C2 x 2 C C3 x C C4 : 6 2

(5.145)

We obtain the four integration constants C1 , C2 , C3 and C4 from the underlying boundary conditions. The condition w.x D 0/ D 0 results in: C4 D 0:

(5.146)

From the condition w 0 .x D 0/ D 0 we can conclude: C3 D 0:

(5.147)

162

5 Beam deflections

Fig. 5.34 Statically indeterminate beam (top), support reactions and elastic line (bottom).

q0 x l z 2

q0l 2 12

q0l 12

w(x)

q0l 2

q0l 2

Similarly, at the right beam end at x D l, the deflection w must vanish, i.e. w.x D l/ D 0. We obtain the following expression from this: 1 1 1 q0 l 4 C C1 l 3 C C2 l 2 D 0: 24 6 2

(5.148)

Finally the requirement w 0 .x D l/ D 0 leads to 1 3 1 q0 l C C1 l 2 C C2 l D 0: 6 2

(5.149)

From the two conditions (5.148) and (5.149), the two integration constants C1 and C2 can be determined as: 1 C1 D  q0 l; 2

C2 D

1 q0 l 2 : 12

Thus, the state variables of the beam can be given as: x q0 l  12 ; Vz D 2 l

  q0 l 2 x x 2 My D C6 1 ; 6 12 l l  x 2 x  3  3 l q x 0 3 C ; w0 D 2 12EIyy l l l

 x 3  x 2  q0 l 4  x 4 2 C : wD 24EIyy l l l

(5.150)

(5.151)

The support reactions of the statically indeterminate beam follow from this as: q0 l 2 q0 l 2 ; MB D My .x D l/ D  ; 12 12 q0 l q0 l ; BV D Vz .x D l/ D : AV D Vz .x D 0/ D 2 2

MA D My .x D 0/ D 

(5.152)

5.5 Standard bending cases

163

Fig. 5.35 Statically determinate beam (0-system, top), 1-system (middle) and 2-system (bottom).

q0 x

w0(x)

w0(x=l) w0(x=l )

l

w1(x=l )

w1(x)

w1(x=l)

BV MB w2(x) w2(x=l)

w2(x=l )

Next, we will show how we can determine the support reactions of the statically indeterminate beam using standard bending cases. We choose the statically determinate 0-system in such a way that we remove the right restraint, thereby transforming the static system into a statically determinate beam clamped on one side (Fig. 5.35, top), which has the elastic line w0 .x/ under the given line load q0 . The kinematic quantities of interest here are the end displacement w0 .x D l/ and the end slope w00 .x D l/, which are as follows: w0 .x D l/ D

q0 l 4 ; 8EIyy

w00 .x D l/ D

q0 l 3 : 6EIyy

(5.153)

We now apply at the cantilever end the support force BV acting in reality (1-system, see Fig. 5.35, center) and determine the end displacement w1 .x D l/ and the end slope w10 .x D l/ as follows: w1 .x D l/ D 

BV l 3 ; 3EIyy

w10 .x D l/ D 

BV l 2 : 2EIyy

(5.154)

Finally, we apply the restraining moment MB existing in reality at this point (2-system, see Fig. 5.35, bottom) and determine the end displacement w2 .x D l/ and the end slope w20 .x D l/ as: w2 .x D l/ D

MB l 2 ; 2EIyy

w20 .x D l/ D

MB l : EIyy

(5.155)

In order to maintain compatibility and to take into account the rigid clamping actually present at the right end of the beam, it is necessary to require, on the one hand, that the deflection w disappears at the point x D l and, on the other hand, that the slope w 0 of the elastic line also becomes zero at this point. We ensure

164

5 Beam deflections

this by superposing the individual components: w0 .x D l/ C w1 .x D l/ C w2 .x D l/ D 0; w00 .x D l/ C w10 .x D l/ C w20 .x D l/ D 0:

(5.156)

This leads to the following two expressions: q0 l 2 BV l MB  C D 0; 8 3 2 q0 l 2 BV l  C MB D 0: 6 2

(5.157)

From this, the two support reactions BV and MB can be determined as: BV D

q0 l ; 2

MB D

q0 l 2 : 12

(5.158)

An alternative solution can be found by selecting the 0-system in such a way that the two clamped ends of the statically indeterminate beam are transformed into hinged supports (Fig. 5.36, top). The two resulting end rotations are calculated as: q0 l 3 q0 l 3 w00 .x D 0/ D ; w00 .x D l/ D  : (5.159) 24EIyy 24EIyy In the next step, we apply the still unknown moment MA (1-system, Fig. 5.36, middle) and determine the resulting edge rotations as: w10 .x D 0/ D 

MA l ; 3EIyy

w10 .x D l/ D

MA l : 6EIyy

(5.160)

Finally, the restraining moment MB is applied to the right end of the beam (2-system, Fig. 5.36, bottom). We obtain the resulting edge rotations as follows: w20 .x D 0/ D 

MB l ; 6EIyy

w20 .x D l/ D

MB l : 3EIyy

Fig. 5.36 Statically determinate beam (0-system, top), 1-system (middle) and 2-system (bottom).

(5.161)

q0 x w0(x=0)

MA w1(x=0)

w2(x=0)

l

w0(x) w0(x=l)

w1(x)

w2(x)

w1(x=l)

w2(x=l)

MB

5.5 Standard bending cases

165

The compatibility constraints to be imposed here require that the superposition of the individual edge rotations w00 , w10 , w20 becomes zero at both ends of the beam, i.e.: w00 .x D 0/ C w10 .x D 0/ C w20 .x D 0/ D 0; w00 .x D l/ C w10 .x D l/ C w20 .x D l/ D 0:

(5.162)

This leads to the following expressions: q0 l 2 MB  MA  D 0; 8 2 MA q0 l 2 C C MB D 0:  8 2

(5.163)

From this, the two support moments MA and MB can be determined as: MA D MB D

5.5.4

q0 l 2 : 12

(5.164) J

Multi-span beams and angled systems

The standard bending cases provided previously can also be applied advantageously to the analysis of multi-span beams and angled systems. As an introduction, we consider again the beam from Example 5.6, shown in Fig. 5.37, top, and in which we have marked the points A (clamping point), B (transition between the two subregions), and C (free end) for better clarity. We are looking for the deflection wC of point C and the slope wC0 of the elastic line at this point. We can take advantage of the fact that we can superpose deflections from different loads and decompose the given static system into several subsystems as shown in Fig. 5.37, bottom. We first consider the right half of the system, which we may take to be a cantilever beam with its clamped end at the transition point at x D l. This cantilever beam is under the uniform line load q2 , and we determine the deflection wC q and the slope wC0 q as follows: wC q D

q2 l 4 ; 8EIyy

wC0 q D

q2 l 3 : 6EIyy

(5.165)

Furthermore, we consider the left half of the system, which we may also consider as a cantilever beam of length l, where we have to take into account not only the line load q1 but also the bending moment MB D 12 q2 l 2 and the transverse shear force FB D q2 l, which we can interpret as statically equivalent applied loads at this location. Due to q2 , MB and FB , the displacements wBq , wBM , wBF and the slopes

166

5 Beam deflections

Fig. 5.37 Beam under line load (top), decomposition into subsystems (bottom).

q2

q1 A

C

x

B

wC

l

wC

l q2

z

C

B

wCq

q1

wCq

B

A

wBq

wBq

MB B

A

wBM

wBM

FB B

A

wBF

wBF

0 0 0 wBq , wBM , wBF are induced, which result as follows:

wBq D

q2 l 4 ; 8EIyy

wBM D

q2 l 4 ; 4EIyy

wBF D

q2 l 4 ; 3EIyy

0 wBq D

q2 l 3 ; 6EIyy

0 wBM D

q2 l 3 ; 2EIyy

0 wBF D

q2 l 3 : 2EIyy

(5.166)

From these deformations, the deflection wB of point B and the rotation wB0 at this point can be determined as: wB D wBq C wBM C wBF D

q1 l 4 q2 l 4 q2 l 4 C C 8EIyy 4EIyy 3EIyy

D

l4 .3q1 C 14q2 /; 24EIyy

0 0 0 wB0 D wBq C wBM C wBF

D

q1 l 3 q2 l 3 q2 l 3 C C 6EIyy 2EIyy 2EIyy

D

l3 .q1 C 6q2 /: 6EIyy

(5.167)

5.5 Standard bending cases

167

These values are consistent with the results from Example 5.6. To determine the deflection wC of point C, we add the two deflections wB and wC q . However, we must take into account that the given load leads to a rotation wB0 at point B as determined, so that the additional portion wB0 l of the deflection wC results due to this rotation. Hence: wC D wB C wB0 l C wC q D

l4 l4 q2 l 4 .3q1 C 14q2 / C .q1 C 6q2 / C 24EIyy 6EIyy 8EIyy

D

l4 .7q1 C 41q2 /: 24EIyy

(5.168)

To determine the slope wC0 , the two rotations wB0 and wC0 q are added together: wC0 D wB0 C wC0 q D

l3 q2 l 3 .q1 C 6q2 / C 6EIyy 6EIyy

D

l3 .q1 C 7q2 /: 6EIyy

(5.169)

Again, this is in agreement with the results from Example 5.6. Example 5.10

We consider the beam from Example 5.5 (see Fig. 5.38) and want to determine the deflection of point C by using standard bending cases. The given static system is decomposed as shown in Fig. 5.38, bottom, into two statically equivalent subsystems. One is the cantilever beam of the subsection between points B and C, and the simply supported beam between points A and B which is loaded at its right support by the moment M0 . A

EIyy

B

2EIyy

C

M0

x l1

l2

z C

B

wCM

wBM A

B

M0 wCM

M0

Fig. 5.38 Beam under edge moment (top), decomposition into subsystems (bottom).

168

5 Beam deflections

We first determine the deflection WCM at the free end on the cantilever beam between points B and C. It results as: wCM D

M0 l22 : 4EIyy

(5.170)

0 at the support point B of the simply Furthermore, we calculate the slope wBM supported beam between the points A and B, which results as: 0 D wBM

M0 l 1 : 3EIyy

(5.171)

The deflection wC at the free end of the given beam is then obtained by adding 0 up the deflection wCM and the slope wBM , multiplied by the length l2 : 0 wC D wCM C wBM l2 :

(5.172)

This leads to the following expression: wC D

  M0 l22 M0 l 1 l 2 M0 l 1 l 2 1 l 2 2 C D C : 4EIyy 3EIyy 2EIyy 2 l1 3

(5.173) J

Example 5.11

Consider the angled beam of Fig. 5.39 (length l, height h, constant bending stiffness EIyy and extensional stiffness EA) loaded by the uniform line load q0 . The deflection of point C is to be determined by superposition of standard bending cases. To solve the problem, we decompose the angled beam into the two statically equivalent subsystems of Fig. 5.39, right. The region between the two points B Fig. 5.39 Angled beam under line load (left), decomposition into subsystems (right).

q0 B

q0 wC

C

C

B

wCq

FB

h

MB

wBM B

wBM A

l

A

wCq

5.6 Biaxial bending

169

and C is represented by a cantilever beam of length l under the line load q0 , whereas the vertical segment between the two points A and B is represented by a straight beam clamped at the bottom under the statically equivalent load FB D q0 l and bending moment MB D 12 q0 l 2 . We first determine the deflection wC q of the cantilever due to the uniform line load q0 . It results as: q0 l 4 wC q D : (5.174) 8EIyy Furthermore, we determine the length change uBF of the vertical beam segment due to the statically equivalent force FB D q0 l as follows: uBF D

FB h q0 lh D : EA EA

(5.175)

0 Likewise, the rotation wBM at point B due to the statically equivalent bending 1 2 moment MB D 2 q0 l must be determined: 0 wBM D

MB h q0 l 2 h D : EIyy 2EIyy

(5.176)

The required deflection wC is then the sum of the two displacements wC q 0 and uBF . In addition, there is also the portion resulting from the rotation wBM multiplied by the length l: 0 wC D wC q C uBF C wBM l:

(5.177)

This leads to the following expression for the deflection wC : wC D

q0 lh q0 l 3 h q0 l 4 C : C EA 8EIyy 2EIyy

(5.178) J

5.6 Biaxial bending If a given beam situation not only leads to the deflection w.x/ in the z-direction, but also causes a deflection v.x/ in the y-direction, then these two deflections are to be superposed to the total deflection of the beam. This case is called biaxial bending. For illustration, let us consider the beam of Fig. 5.40, which is clamped at one end and loaded at its free end by a single force F oriented at an angle of 30ı to the y-axis. We first decompose the force F into its components Fy and Fz : p 3 Fy D F cos 30 D F; 2 ı

Fz D F sin 30ı D

F : 2

(5.179)

170

5 Beam deflections

Fig. 5.40 Cantilever beam under single force.

y

z 30°

x

F

We can then determine the deflections v.x/ and w.x/ as: p

 3F l 3  x 2  x 3 v.x/ D  3 ; 12EIzz l l

  x 2  x 3 F l3  3 : w.x/ D 12EIyy l l

(5.180)

The deflections at the cantilever end amount to: p v.x D l/ D

3F l 3 ; 6EIzz

w.x D l/ D

F l3 : 6EIyy

(5.181)

The resulting deflection f at the cantilever end is then determined as: p F l3 f D v2 C w2 D 6

s

3 .EIzz /2

C

1 EIyy

2 :

(5.182)

The procedure is largely analogous also in the case that an axis system is considered which is not the principal axis system. Then the constitutive relations (4.115) are to be employed, which we give here once more for the better clarity. Ev 00 D

Iyz My C Iyy Mz ; 2 Iyy Izz  Iyz

Ew 00 D 

Izz My C Iyz Mz : 2 Iyy Izz  Iyz

(5.183)

As an example, we consider the cantilever beam of Fig. 5.41, which is loaded at its free end by a force F . We consider a Z-cross-section which has already been treated

5.6 Biaxial bending

171

Fig. 5.41 Cantilever beam (left), cross-section (right).

t

F t

y

x

h t

l b

z

b

z

in Example 4.13. With h D 2b the following moments of inertia and deviation moment result for this cross-section: th3 tbh2 8 C D tb 3 ; 12 2 3 2 D tb 3 ; 3 tb 2 h D D tb 3 : 2

Iyy D Izz Iyz

(5.184)

At the given cantilever beam, the two bending moments My .x/ and Mz .x/ are obtained as follows: My D F .x  l/; Mz D 0: (5.185) From the constitutive law (5.183) then follows: Iyz My 9F D .x  l/; 2 Iyy Izz  Iyz 7tb 3 9F .x  l/2 C C1 ; Ev 0 D 14tb 3 3F .x  l/3 C C1 x C C2 : Ev D 14tb 3

Ev 00 D

(5.186)

Analogously we have: Izz My 6F D .x  l/; 2 Iyy Izz  Iyz 7tb 3 3F .x  l/2 C C3 ; Ew 0 D  7tb 3 F .x  l/3 C C3 x C C4 : Ew D  7tb 3

Ew 00 D 

(5.187)

Evaluating the boundary conditions v.x D 0/ D 0, v 0 .x D 0/ D 0 and w.x D 0/ D 0, w 0 .x D 0/ D 0 leads to the following integration constants: C1 D 

9F l 2 ; 14tb 3

C2 D

3F l 3 ; 14tb 3

C3 D

3F l 2 ; 7tb 3

C4 D 

F l3 : 7tb 3

(5.188)

172

5 Beam deflections

Thus the displacements v.x/ and w.x/ in the y-direction and the z-direction can be given as: # " x 3F l 3 .x  l/3 3 C1 ; v.x/ D 14Etb 3 l3 l # " x .x  l/3 F l3 C3 1 : (5.189)  w.x/ D 7Etb 3 l3 l The deflections at the cantilever end x D l follow as: v.x D l/ D 

3F l 3 ; 7Etb 3

w.x D l/ D

2F l 3 : 7Etb 3

(5.190)

Obviously, deflections in both coordinate directions y and z result under the bending moment My . We obtain the total deflection f at the cantilever end as: p p 13F l 3 2 2 f D v Cw D : (5.191) 7Etb 3 Example 5.12

Consider the simply supported beam of length l as shown in Fig. 5.42. We consider a rectangular cross-section of width b and height h D 2b, installed at an angle of 45ı . Let the load consist of a uniform line load q0 . The deflection of the beam is to be determined. We first determine the moments of inertia IyN yN and IzN zN of the cross-section (the deviation moment IyN zN is zero). We obtain: IyN yN D

b  .2b/3 2b 4 D ; 12 3

IzN zN D

2b  b 3 b4 D : 12 6

(5.192)

From this, the moments of inertia Iyy and Izz as well as the deviation moment Iyz can be determined by transformation:  1  1 5b 4 IyN yN C IzN zN C IyN yN  IzN zN cos 2'  IyN zN sin 2' D ; 2 2 12  1  1 5b 4 D IyN yN C IzN zN  IyN yN  IzN zN cos 2' C IyN zN sin 2' D ; 2 2 12  1 b4 D IyN yN  IzN zN sin 2' C IyN zN cos 2' D  : 2 4

Iyy D Izz Iyz

Fig. 5.42 Simply supported beam under uniform line load (left), cross-section (right).

q0

(5.193)

b 2b

y

x

45°

l

y

z z

5.6 Biaxial bending

173

Fig. 5.43 Free body image.

q0 q0l 2

My

x

In order to calculate the beam deformations, we need to determine the bending moment My .x/ which can be obtained by considering the free body image of Fig. 5.43. We obtain by forming the moment equilibrium around a point at an arbitrary location x:

 q0 l 2 x  x 2 My .x/ D  : 2 l l

(5.194)

We obtain the deflection v.x/ from the constitutive law (5.183) by integration:

 Iyz My 9q0 l 2 x  x 2 D  ; Ev D 2 Iyy Izz  Iyz 8b 4 l l

 9q0 l 3 1  x 2 1  x 3  Ev 0 D  C C1 ; 8b 4 2 l 3 l

 3q0 l 4  x 3 1  x 4  Ev D  C C1 x C C2 : 16b 4 l 2 l 00

(5.195)

The two integration constants C1 and C2 can be determined from the boundary conditions v.x D 0/ D 0 and v.x D l/ D 0 and are as follows: C1 D

3q0 l 3 ; 32b 4

C2 D 0:

(5.196)

Thus, the deflection v.x/ can be given as:

 x 3  x  3q0 l 4  x 4 v.x/ D 2 C : 32Eb 4 l l l

(5.197)

We proceed analogously for the determination of the deflection w.x/:

 Izz My 15q0 l 2 x  x 2 D   ; 2 Iyy Izz  Iyz 8b 4 l l

 15q0 l 3 1  x 2 1  x 3  Ew 0 D  C C3 ; 8b 4 2 l 3 l

 5q0 l 4  x 3 1  x 4  Ew D  C C3 x C C4 : 16b 4 l 2 l

Ew 00 D 

(5.198)

174

5 Beam deflections

From the boundary conditions w.x D 0/ D 0 and w.x D l/ D 0 the integration constants C3 and C4 can be determined: C3 D

5q0 l 3 ; 32b 4

C4 D 0:

(5.199)

For the deflection w.x/ we thus obtain:

 x 3  x  5q0 l 4  x 4 : w.x/ D 2 C 32Eb 4 l l l

(5.200) J

6

Shear stresses in beams

Beams under bending generally do not only develop bending stresses, i.e. normal stresses. Rather, it turns out that shear stresses in the plane of the cross-section also occur, which are related to transverse shear forces. All the necessary equations for the analytical determination of shear stresses are derived for both thick-walled and thin-walled cross-sections, and it is shown how to carry out the analysis of shear stresses in beams due to transverse shear forces. An important geometric quantity is the so-called shear center of a cross-section, where transverse shear forces have to act in order to avoid (generally unwanted) torsion in the considered beam. Students are enabled to determine the shear stresses in thick- and thin-walled cross-sections and to determine the location of the shear center of a cross-section.

6.1 Introduction So far we have considered straight bars and beams under normal force and bending, where the normal stress xx occurs as a consequence of the normal force N and the two bending moments My and Mz . However, the equilibrium conditions for the beam (see (5.1) and (5.4)) show that the transverse shear forces Vy and Vz must also be present in order to ensure equilibrium. In the following we will assume that only the transverse shear force Vz is present and that the transverse shear force Vy is zero. From Fig. 4.2 we can conclude that Vz is related to a shear stress , and in this chapter we will deal with the question of how we can determine shear stresses  due to a transverse shear force Vz . The explanations can easily be applied to the case of the transverse shear force Vy which we will, however, not address in this chapter. In the following explanations, we distinguish between thick-walled and thin-walled cross-sections. The fundamental problem we face in determining the shear stresses due to transverse shear force is the fact that the Euler–Bernoulli beam theory assumes the normal hypothesis and the hypothesis that the cross-sections remain plane also in the deformed state. This is equivalent to the fact that we do not have any shear strains  in the beam. However, the shear stress  is connected to the shear strain  © The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_6

175

176

6

Shear stresses in beams

by Hooke’s law  D G:

(6.1)

Since as a result of the assumptions of the underlying beam theory the shear strains  disappear, the shear stresses  also would have to disappear. This is a contradiction of the Euler–Bernoulli beam theory which results directly from the assumptions of this theory and which is unavoidable. Accordingly, no constitutive law is available for the determination of the shear stress , and we must obtain statements about the shear stress  from equilibrium considerations. In the following explanations, let x, y, z be again the principal axes of the considered cross-section, and given loads act exclusively in the direction of the principal axis z and result in bending about the y-axis. In addition, a normal force may occur which is assumed to be constant along the beam length or only slightly variable. The considerations are also limited to straight beams which have constant properties or only slightly varying properties along the longitudinal axis x. In the following, we will assume linear elasticity and small deformations, with the beam material being isotropic and homogeneous.

6.2 Thick-walled cross-sections We consider a beam with an arbitrary cross-section (shown as a simple rectangular cross-section in Fig. 6.1) subjected to load that leads to the transverse shear force Vz and the bending moment My as shown in Fig. 6.1, left. We now consider an infinitesimal element of length dx, as indicated in Fig. 6.1, left, and cut this element free (Fig. 6.1, middle). Here, at the negative edge, we have released the transverse shear force Vz and the bending moment My as well as the bending stress xx , whereas at the positive edge, these quantities occur with an infinitesimal dMy dxx z increment, i.e. Vz C dV dx dx, My C dx dx and xx C dx dx. Similarly, the shear stress xz is released on the vertical surfaces of the infinitesimal element. We further consider an element of infinitesimal height dz, as indicated in Fig. 6.1, middle, and cut it free (Fig. 6.1, top right). The shear stress xz (upper surface) and xz C ddzxz dz (lower surface) occurs on the two horizontal surfaces of the element of size bdx. The sum of the forces in x-direction gives:   dxx dx bdz  xx bdz xx C dx   dxz xz C bdx  xz bdx D 0: dz

(6.2)

We introduce at this point the shear flow T .z/ D xz .z/b.z/, which can be interpreted as the mean value of the shear stress over the width b.z/. Furthermore, let the resulting normal force flow nxx of the normal stress xx be introduced as nxx .z/ D xx .z/b.z/. The equilibrium condition (6.2) then leads to the following

6.2 Thick-walled cross-sections

177

Vz

xx

xz

My xx

y

d xx xx + dx dx d xz xz + dz dz

qz

z

xx +

xz

qz

d xx dx dx

dz dx

y dx My

z

x

Vz

d My My+ dx dx

x dV Vz + dxz dx

Fig. 6.1 On the determination of the shear stress xz .

expression: dT dnxx D : dz dx

(6.3)

We perform an integration with respect to z on both sides of the equation in the limits z D z0 and z and obtain: Zz z0

dT dOz D T .z/  T .z D z0 / D  dOz

Zz z0

dnxx dOz : dx

(6.4)

Therein, T .z/ is the shear flow at an arbitrary point z, and T .z D z0 / is the initial value of the shear flow at the starting point of integration at z D z0 . We assume that the reference axes x, y, z are the principal axes of the crosssection and that the applied load causes the bending moment My . In addition, there is a normal force N acting at the center of gravity of the cross-section. The normal force flow nxx then follows as:  nxx D

 My N z b: C A Iyy

(6.5)

In (6.4) we have the first derivative of nxx with respect to x, which we obtain as follows:   0 My0 N dnxx z b: (6.6) D C dx A Iyy

178

6

Shear stresses in beams

In the following we want to assume that the first derivative N 0 of the normal force N vanishes. Furthermore, My0 D Vz , so that (6.6) transitions into: dnxx Vz bz : D dx Iyy

(6.7)

Thus it follows from (6.4): Vz T .z/  T .z D z0 / D  Iyy

Zz zbdO O z:

(6.8)

z0

The integral expression appearing herein is the static moment Sy .z/ with respect to a point z, i.e. Zz Sy .z/ D zO bdOz : (6.9) z0

Thus, we obtain for the shear flow T .z/: T .z/  T .z D z0 / D 

Vz Sy .z/ : Iyy

(6.10)

.z/ can be Once the shear flow has been determined, the shear stress xz .z/ D Tb.z/ obtained as: Vz Sy .z/ xy .z/  xy .z D z0 / D  : (6.11) Iyy b.z/ As an elementary example, consider the rectangular cross-section (width b, height h) of Fig. 6.2, left. Let the shear force Vz be given. We first determine the static moment of the partial area A, which can be understood as the sum of the infinitesimal areas dA D bdOz between z D  h2 and the arbitrary location z, multiplied by the distance z: O # "  z ˇ Z 2z 2 1 2 ˇˇz bh2 Sy .z/ D zbdO O z D b zO ˇ D 1 : (6.12) 2 8 h h 2

 h2

The static moment takes the value zero for z D ˙ h2 , for z D 0 it amounts to 2 Sy .z D 0/ D  bh8 . b

b h 2

y

h 2 z

y

z

A dA=bdz^ dz^

z^

z

Fig. 6.2 Determination of the shear stress xz on a rectangular cross-section.

τ (z ) τmax

6.2 Thick-walled cross-sections

179 3

With the moment of inertia Iyy D bh 12 , the constant section width b and the initial value xz .z D z0 / D 0 (at the point z D  h2 , i.e. at the upper edge of the section, no shear stresses are introduced into the cross-section), the shear stress xz .z/ is obtained as: "  2 # 2z 3Vz xz .z/ D 1 : (6.13) 2bh h The shear stress is distributed parabolically over z (Fig. 6.2, right), vanishes at the two cross-section edges z D ˙ h2 and takes its maximum value at the location z D 0 3Vz . The shear at the height of the center of gravity of the cross-section with max D 2bh stress xz .z/ is constant over the section width b and is not a function of y for the considered example. The static moment can be interpreted as the area A.z/ distinguished by the point z under consideration multiplied by the centroidal distance zA of this area to the total center of gravity C of the cross-section. As an example we again consider the rectangular cross-section of Fig. 6.2 and want to determine the static moment Sy .z D 0/ for the coordinate z D 0 (Fig. 6.3). The considered area A.z D 0/ is , the distance zA is zA D  h4 . Thus, the static moment Sy .z D 0/ is A.z D 0/ D bh 2 given as: bh h bh2 Sy .z D 0/ D A.z D 0/zA D  D : (6.14) 2 4 8 3Vz . With (6.10) this gives the maximum value of the shear stress as max D 2bh It should be noted that the static moment Sy .z/ for a partial area distinguished by the coordinate z is identical to the negative static moment of the remaining residual N i.e. Sy .A/ D Sy .A/ N holds (see Fig. 6.4). area A,

Fig. 6.3 Determination of the static moment on a rectangular cross-section.

b A(z) zA

y

z Fig. 6.4 Determination of the static moment for a partial area and the remaining residual area.

A y

z A

z

180

6

Shear stresses in beams

Example 6.1

Consider the triangular cross-section of Fig. 6.5. The static moment Sy .z/ and the shear stress xz at the point z D 0 due to a given transverse shear force Vz are to be determined. Let the moment of inertia Iyy be given with the value 3 ca3 Iyy D 2ca 36 D 18 . The static moment Sy .z/ can be determined from the infinitesimal surface element dA as: Zz b.Oz /Oz dOz : (6.15) Sy .z/ D  13 a

We formulate the width b.z/ as a function of z as follows: b.z/ D 

4c 2c zC : a 3

(6.16)

Substitution into (6.15) yields:   ˇ Zz  2c 3 2c 2 ˇˇz 2c 4c zdO O z D  zO C  zO C Sy .z/ D zO ˇ a 3 3a 3 1a 3

 13 a

D

2c 3 2c 2 8ca2 z C z  : 3a 3 81

(6.17)

The static moment Sy at the center of gravity at z D 0 results from this as: Sy .z D 0/ D 

8ca2 : 81

(6.18)

Thus, the shear stress xz at the location z D 0 with width b.z D 0/ D determined as: ! 8ca2 Vz  81 4Vz xz .z D 0/ D  : D 3 3ac ca 4c

4c 3

can be

(6.19)

18 3

J

Fig. 6.5 Triangular crosssection.

2c b(z) y

dz^

z

^

dA=bdz^ z

a 3 2a 3

6.2 Thick-walled cross-sections

181

Example 6.2

Consider the trapezoidal cross-section shown in Fig. 6.6. Let the moment of inertia Iyy be given with the value Iyy D 78a4 . The static moment Sy .z/ as well as the shear stress xz at the height of the center of gravity at z D 0 are to be determined. The static moment as a function of the coordinate z can be specified as: Zz Sy .z/ D

b.Oz /Oz dOz :

(6.20)

 10 3 a

The width b.z/ can be formulated as a function of z as follows: b.z/ D

z 14a C : 2 3

(6.21)

Then it follows from (6.20): Zz  Sy .z/ D  10 3 a

  3 ˇz zO zO 14a 7Oz 2 a ˇˇ C zO dOz D C ˇ 10a 2 3 6 3  3

z3 7az 2 1600a3 C  : (6.22) 6 3 81 The static moment Sy at the location of the center of gravity at z D 0 is given as: D

1600a3 : 81 Thus, the shear stress at z D 0 can be determined as: ! 1600a3 Vz  81 400Vz : D xz .z D 0/ D  7371a2 14a 4 78a 3 Sy .z D 0/ D 

Fig. 6.6 Trapezoidal crosssection.

(6.23)

(6.24) J

3a y

z^

10 a 3 6a dz^ 8a 3 dA=bdz^

z b(z) 6a

182

6

Shear stresses in beams

Example 6.3

Consider the I-shaped cross-section of Fig. 6.7, top left. We want to determine the shear stress distribution xz due to a given transverse shear force Vz . We first determine the moment of inertia Iyy of the cross-section as follows: Iyy

  4a  6a3 12a  .6a/3 D2 C C 2  6a  12a  .6a/2 D 5688a4: 12 12

(6.25)

To determine the static moment Sy .z/ and the shear stress xz .z/, the three local axes s1 , s2 and s3 are introduced as shown in Fig. 6.7, top left. We first consider the upper segment 0  s1  6a (Fig. 6.7, top right). The area A1 of height s1 , highlighted in dark gray, is calculated as: A1 D 12as1 :

(6.26)

The center of gravity coordinate zC1 of the considered partial area A1 is then: zC1 D 9a C

s1 : 2

(6.27)

s1 6a y

s2 6a 18a

z 4a

s1

s1 2 s1 9a

y

s3 6a

z

4a 4a 12a

s2 2 y

s2

s2 3a

y

3a s3

z Fig. 6.7 I-shaped cross-section.

z

s3 s3 2

6.2 Thick-walled cross-sections

183

Thus, the static moment Sy;1 .s1 / of the partial area A1 can be given as: Sy;1 .s1 / D zC1 A1 D 6as12  108a2s1 :

(6.28)

The static moment Sy;1 .s1 / takes the value zero at s1 D 0. At s1 D 6a the value Sy;1 .s1 D 6a/ D 432a3 results. We now consider the area 0  s2  6a. The partial area A2 is then: A2 D 4as2 :

(6.29)

The center of gravity coordinate zC 2 is given here as: zC 2 D 3a C

s2 : 2

(6.30)

The static moment Sy;2 .s2 / is then the product of zC 2 and A2 , where here the static moment of the entire partial area A1 must be added. We obtain: Sy;2 .s2 / D 2as22  12a2 s2  432a3 :

(6.31)

Evaluated at s2 D 0, s2 D 3a (center of gravity of the cross-section) and s2 D 6a, we get: Sy;2 .s2 D 0/ D 432a3 ; Sy;2 .s2 D 3a/ D 450a3 ; Sy;2 .s2 D 6a/ D 432a3 :

(6.32)

Finally, the lower section in the range 0  s3  6a is considered. The partial area A3 and its centroid coordinate zC 3 are: A3 D 12as3 ;

zC 3 D 3a C

s3 : 2

(6.33)

This gives the static moment in this area as: Sy;3 .s3 / D zC 3 A3 C Sy .s2 D 6a/ D 6as32 C 36a2 s3  432a3 :

(6.34)

Evaluation yields: Sy;3 .s3 D 0/ D 432a3 ;

Sy;3 .s3 D 6a/ D 0:

(6.35)

184

6

Shear stresses in beams

Fig. 6.8 Shear stress distribution in the I-cross-section.

V 0.0063 a2z

V 0.019 a2z V τmax =0.020 2z a

y

V 0.019 a2z V 0.0063 a2z

z

With the static moment thus determined, the shear stress xz can be determined as follows: xz .s1 D 0/ D 0;

  Vz  432a3 xz .s1 D 6a/ D  4 5688a   12a  Vz  432a3 xz .s2 D 0/ D  4  4a 5688a   Vz  450a3 xz .s2 D 3a/ D  4  4a 5688a   Vz  432a3 xz .s2 D 6a/ D  4  4a 5688a   Vz  432a3 xz .s3 D 0/ D  5688a4  12a xz .s3 D 6a/ D 0:

Vz ; a2 Vz D 0:019 2 ; a Vz D 0:020 2 ; a Vz D 0:019 2 ; a Vz D 0:0063 2 ; a D 0:0063

(6.36)

A graphical representation of the shear stress distribution is given in Fig. 6.8. J

Example 6.4

Consider the circular cross-section of Fig. 6.9, left, under a transverse shear force Vz . We want to determine the static moment Sy .z/ and the shear stress xz distribution as functions of z. Fig. 6.9 Circular crosssection.

b(z) R y

y

z

R

dz^

z

dA=bdz^ z z^

6.3

Thin-walled cross-sections

185

We first formulate the width b of an infinitesimal element as a function of z. From the Pythagorean theorem it follows on the basis of Fig. 6.9, right: p (6.37) b.z/ D 2 R2  z 2 : The static moment is then obtained from integration in the limits of z D R up to an arbitrary point z: Zz Sy D

Zz p zO b.Oz /dOz D 2 zO R2  zO 2 dOz

R

2 D 3

R

ˇz q ˇ 3ˇ 2 2 .R  zO / ˇ

R

D

2 3

q .R2  z 2 /3 :

Thus, the shear stress in the given circular cross-section with Iyy D A D R2 can be determined as: ! q 2 3 Vz   .R2  z 2 /   3 z2 4Vz 1 2 : D xz D  3A R R4 p 2  2 R  z2 4

(6.38) R4 4

and

(6.39)

Its maximum value at the height of the center of gravity at z D 0 is given by max D

6.3

4Vz : 3A

(6.40) J

Thin-walled cross-sections

In this section, we now consider straight beams that can be considered to be very thin-walled. In this case, we can assume that the shear stresses  are constant over the thickness of the individual cross-section segments and, moreover, always run in the tangential direction. To derive the basic equations for such thin-walled beams, we consider Fig. 6.10, which shows a thin-walled beam loaded, for example, by a line load qz . However, also single forces in the z-direction and bending moments around the y-axis are possible. Again, y and z are the principal axes of the crosssection. The applied load is assumed to cause the transverse shear force Vz .x/ and the bending moment My .x/. We now examine an infinitesimal section element of length dx and cut it out of the beam. The bending moment My causes the linearly distributed normal stress xx over z, as shown in the free body image of Fig. 6.10, middle. Note that the normal stress xx is also a function of the longitudinal coordinate x, so that we apply the normal stress xx to the negative edge of the infinitesimal element, whereas at the positive edge we have the normal stress xx

186

6

Shear stresses in beams

Vz My

n xx σ xx

qz

y

Ts s

dT Ts+ dxsdx

dT Ts+ dssds s

s

z

s t(s)

qz dx

y

My

Vz

d σ xx dx dx

Ts (s)

My+

x

σ xx +

s

s

dx z

ds

n xx +d n xx dx

d My dx dx

x dV Vz + zdx dx

Fig. 6.10 Determination of the shear stress  .s/ on a thin-walled cross-section.

together with an infinitesimal increment ddxxx dx. We proceed in the same way with the transverse shear force Vz and the bending moment My . We now want to determine the distribution of the shear stress  over the thin-walled cross-section in the presence of the normal stress xx . For this purpose, we introduce a circumferential coordinate s as indicated in the infinitesimal element. Since the shear stress is always parallel to the local coordinate s, we will refer to it as s . We will show later that we can appropriately describe the shear stress as a function of s, i.e. s D s .s/ holds. Let the wall thickness also be a function of s, i.e. t D t.s/. At this point, we again introduce the shear flow Ts .s/, which is the resultant of the shear stress s over the section thickness t: Ts .s/ D s .s/t.s/:

(6.41)

The shear flow thus has the unit of a force per unit length. With (6.41) we assume that the shear stress s .s/ at each location s is constantly distributed over the wall thickness t.s/ of the cross-section. This is a useful and valid assumption for thinwalled cross-sections. In the free body image of Fig. 6.10, right, an infinitesimal sectional element of width ds is given, which we have cut out from the infinitesimal sectional element of Fig. 6.10, middle. Here, on the one hand, the shear force flow Ts .s/ with its respective infinitesimal increments is shown. In addition, we have the resulting normal force flow nxx , which represents the resultant of the normal stress xx over the cross-section thickness t: nxx .s/ D xx t.s/: (6.42) We form the equilibrium of forces in the x-direction and obtain:     dnxx dTs dx ds  nxx ds C Ts C ds dx  Ts dx D 0: nxx C dx ds

(6.43)

6.3

Thin-walled cross-sections

187

This results in:

dnxx dTs dxds C dsdx D 0: (6.44) dx ds We divide by dx and perform an integration over s from a starting point s D sA to an arbitrary point s. It follows: Zs sA

dTs dOs D Ts .s/  Ts .s D sA / D  dOs

Zs sA

dnxx dOs : dx

(6.45)

The force flow nxx follows from the normal stress xx due to uniaxial bending with additional normal force according to My N z: C A Iyy

(6.46)

 My N z t; C A Iyy

(6.47)

xx D 

Thus: nxx D so that

dnxx D n0xx D dx



 My0 N0 z t: C A Iyy

(6.48)

We assume that there is a constant normal force N over x, so we can assume N 0 D 0. With the relation between My and Vz dMy D My0 D Vz dx follows: n0xx D

(6.49)

Vz zt: Iyy

(6.50)

Then we obtain from (6.45): Zs Ts .s/  Ts .s D sA / D  sA

Vz Vz tzdOs D  Iyy Iyy

Zs tzdOs :

(6.51)

sA

The integral appearing in this expression is the static moment Sy .s/ at an arbitrary point s of the section element under consideration. Thus: Ts .s/  Ts .s D sA / D 

Vz Sy .s/: Iyy

(6.52)

With (6.52) an expression for the determination of the shear force flow Ts .s/ at an arbitrary location s is found, which depends on the acting shear force Vz and also

188

6

Fig. 6.11 Double symmetric I-cross-section with local circumferential axes si (i D 1; 2; : : : ; 5).

Shear stresses in beams t s1

t h y

s2

s3

C t

s4

b

s5

b

z

requires the knowledge of the moment of inertia Iyy and the static moment Sy .s/. The term Ts .s D sA / is the initial value of the shear force flow at a point s D sA . Once the shear flow Ts .s/ for a thin-walled cross-section is known, then the shear stress s .s/ at each location s can be determined as: s .s/ D

Ts .s/ : t.s/

(6.53)

In this section, we restrict our considerations to open, thin-walled cross-sections in which no shear loads are applied at the free ends. Closed cross-sections are not discussed here. In this case, we can set the initial value Ts .s D sA / to zero if we place the initial value of the integration in (6.45) at a free end. For the shear flow Ts .s/ this leaves the following expression: Ts .s/ D 

Vz Sy .s/: Iyy

(6.54)

To illustrate the determination of the shear flow Ts .s/ and the shear stresses s .s/, we consider the I-cross-section of Fig. 6.11 (web height h, flange width 2b, constant wall thickness t). We assume that this is a very thin-walled cross-section, so we will again refer exclusively to the skeleton line in all considerations. The coordinate system is a principal axis system. For the calculation, the local circumferential axes si (i D 1; 2; : : : ; 5) are introduced as shown. It should be noted that the choice of the starting points and directions of the local axes are arbitrary. The moment of inertia Iyy can be taken directly from Example 4.12: Iyy D

th3 C tbh2 : 12

(6.55)

We will first determine the static moment Sy .s/, which is given by the following equation: Zs Sy .s/ D tzdOs : (6.56) sA

We first examine segment 1 at z D  h2 , which is described by the local axis s1 with 0  s1  b (Fig. 6.12, left). From the integration rule (6.56) we can conclude that

6.3

Thin-walled cross-sections

Fig. 6.12 Determination of the static moment for the double symmetric I-crosssection.

189 s3 2

s1 h 2

h 2

s1

y

h 2

s3 s3

y

h 2 b

b

z

b

b

z

the distribution of the static moment in this segment will be linear with s1 since the coordinate z is constant with the value z D  h2 . It is therefore sufficient to determine the static moment for s1 D 0 and for s1 D b. Between these two edge values the static moment is linear, as can be seen from the integration up to an arbitrary value s1 : ˇ  Zs1  h thOs1 ˇˇs1 ths1 D : Sy .s1 / D t  dOs1 D  2 2 ˇ0 2

(6.57)

0

At the point s1 D 0 the static moment takes the value zero, whereas at the point s1 D b we obtain: tbh Sy .s1 D b/ D  : (6.58) 2 We would also obtain the result (6.58) if we multiplied the centroid coordinate zC;1 D  h2 with the area A1 D tb. We proceed analogously in segment 2, where a linear distribution of Sy .s/ is also obtained which results in Sy .s2 D 0/ D 0 and Sy .s2 D b/ D  t bh 2 . The web is considered by use of the local reference axis s3 (Fig. 6.12, right), which has its origin at the intersection point of the two upper flange segments. The two values of the static moments Sy .s1 D b/ and Sy .s2 D b/ at the point s3 D 0 add up, and we obtain: Sy .s3 D 0/ D Sy .s1 D b/ C Sy .s2 D b/ D tbh:

(6.59)

The course of the static moment Sy .s3 / over s3 can again be obtained by integration: Zs3 Sy .s3 / D tbh C

tzdOs3 :

(6.60)

0

With z D  h2 C s3 this yields:  Zs3  h ths3 ts 2 Sy .s3 / D tbh C t  C sO3 dOs3 D tbh  C 3: 2 2 2 0

(6.61)

190

6

Shear stresses in beams

At the location of the center of gravity of the cross-section at s3 D h2 , the static moment Sy .s3 / results in:     h h D th b C : Sy s3 D 2 8

(6.62)

This result is also obtained if the considered partial area of the web t  h2 is multiplied by its centroid coordinate  h4 and the value Sy .s3 D 0/tbh is added. At the bottom of the web at the location s3 D h we obtain from (6.61): Sy .s3 D h/ D tbh:

(6.63)

It should be noted here that when considering the web, the function s to be integrated is a linear function, as can be seen from (6.61). Therefore, the static moment Sy over s3 is parabolic in the web, with the maximum value at the center of gravity of the cross-section. Consideration of the two areas 4 and 5, i.e. the two lower flange segments, yields the following values for the static moment Sy : tbh ; 2 tbh : Sy .s5 D 0/ D 0; Sy .s5 D b/ D 2

Sy .s4 D 0/ D 0; Sy .s4 D b/ D

(6.64)

The variation of the static moment over the cross-section is shown in Fig. 6.13, left. With the static moment thus determined, both the shear flow Ts .s/ and the shear

tbh 2

s

s

(– )

s1

s3 s4

s2 (– )

s5

(+)

s

tbh 2 S y (s)

s

s

tbhVz 2 Iyy

tbh s1 s2 s3 th b+ h8 s4 tbh s

s

(+)

(–)

s2 (+)

s2 V th b+ h8 I z yy

tbhVz 2 Iyy Vz Sy (s) Ts(s) = I yy

s1

(+)

s

s3

(+)

s2

s4

s5 s

s

tbhVz Iyy

bhVz 2 Iyy

tbhVz Iyy

s

s2 V h b+ h8 I z yy

s5

(–)

bhVz 2 Iyy

τ s(s) = -

bhVz Iyy

s

bhVz Iyy

Vz Sy (s) I yy t(s)

Fig. 6.13 Static moment Sy .s/, shear flow Ts .s/ and shear stress s .s/ at the I-cross-section under the transverse shear force Vz .

6.3

Thin-walled cross-sections

191

stress s .s/ can be determined. They are also shown in Fig. 6.13. Here, the sign indicates the direction of action of Ts .s/ and s .s/. If a positive sign is present, then the direction of action corresponds to the direction of the local axis si . If, on the other hand, there is a negative sign, then the direction of action of Ts .s/ and s .s/ is opposite to the direction of si . The direction of action of shear flow and shear stress is indicated by arrows. Some general rules can be derived from the previous considerations:  At a free profile end, the shear flow is always zero, unless tangential shear loads are applied there.  If z D 0, then Ts and s show constant values. In the case of a constant value for z not equal to zero (as in the above example in the case of the flange segments) the course of Ts and s is linear over the local reference axis s. If, on the other hand, a linearly varying z-coordinate is present in a segment under consideration (as was the case for the web in the above example), then Ts and s follow as parabolic functions of s.  The shear flow Ts due to a transverse shear force Vz assumes extreme values exactly where the y-axis intersects the skeleton line of the cross-section.  If the cross-section under consideration has an axis of symmetry, then the course of Ts and s is symmetrical due to the shear force acting in the axis of symmetry (in the above example, the shear flow Ts and shear stress s are symmetrical to the z-axis). The shear flow Ts and the shear stress s due to Vz show zero values at intersections of the z-axis with the skeleton line, if z is a symmetry axis.  A very simple verification of the analysis can be performed by considering that the resultants of the shear stress s of all segments must be equal in sum to the acting transverse shear force. If only one shear force Vz is acting, then the z-components of the resultants of the shear flow Ts must add up to the shear force Vz .  At such points where several segments intersect, the sum of all resulting shear forces must be zero. For this purpose, we consider again the I-cross-section of Fig. 6.11. In Fig. 6.14, the section element of Fig. 6.10 of length dx is shown again. We cut out another section element exactly at the intersection point of the two flanges and the web and investigate the equilibrium of the resultants of the shear flows acting at this point. The sum of all forces in x-direction then leads to .Ts .s1 D b/ C Ts .s2 D b/  Ts .s3 D 0//dx D 0. On the basis of Fig. 6.13 it can be seen that this requirement is fulfilled here.

192

6

Shear stresses in beams

s1 s2 s3

qz s4

y

s4

dx

Ts (s) Ts(s2= b)

dx

Ts(s1= b) My

z

x

Ts(s3= 0)

Vz

Fig. 6.14 Verification of the analysis at the example of the I-cross-section.

Example 6.5

Consider the C-cross-section of Fig. 6.15. Let the arbitrary reference frame y, N zN be defined as indicated, and we first determine the centroid coordinates yNC , zN C . We obtain: Z ydA N b 2b 2 A 2 D b yNC D Z D ; t.h C 2b/ 2b C h dA A

Z

zN dA A

zN C D Z

D

h Cbt h h 2 D : t.h C 2b/ 2

ht 

dA

(6.65)

A

Fig. 6.15 C-cross-section with reference axes y, N zN (left), center of gravity position and center of gravity coordinate system y, z (right).

b b

b 4

y h 2

t t

h

y

h

C

h 2 t

z

z

6.3

Thin-walled cross-sections

193

For simplicity, let us assume that the web height h is exactly twice the flange width b, i.e. h D 2b. For the coordinates of the center of gravity then follows: yNC D

b ; 4

zNC D

h D b: 2

(6.66)

The moments of inertia IyN yN , IzN zN and the deviation moment IyN zN follow as: Z IyN yN D

zN 2 dA D

20tb 3 ; 3

yN 2 dA D

2tb 3 ; 3

A

Z IzN zN D A

Z

yN zN dA D tb 3 :

IyN zN D

(6.67)

A

Steiner’s theorem then gives the following values for Iyy , Izz , Iyz with respect to the center of gravity C: 8tb 3 ; 3 5tb 3 D IzN zN  yNC2 A D ; 12 D IyN zN  yNC zN C A D 0:

Iyy D IyN yN  zN C2 A D Izz Iyz

(6.68)

The deviation moment becomes zero with respect to the center of gravity, i.e. y and z constitute a principal axis system. We again determine the course of the static moment Sy .s/ segment by segment, and for this purpose we introduce the circumferential local reference axes s1 , s2 and s3 as shown in Fig. 6.16. We first determine the static moments as Fig. 6.16 Determination of the static moment Sy .s/ for the C-cross-section.

s2 2

s1 b

s1

s2

b

b

s2

b

b b

s3 s3

b

b

b

194

6

Shear stresses in beams

functions of the circumferential coordinates s1 , s2 and s3 . We obtain: Sy .s1 / D .ts1 /  .b/ D tbs1 ;  1 s2 D ts22  tbs2  tb 2 ; Sy .s2 / D Sy .s1 D b/ C ts2 b C 2 2 Sy .s3 / D Sy .s2 D 2b/ C ts3 b D tb 2 C tbs3 :

(6.69)

Evaluation at relevant points yields: Sy .s1 D 0/ D 0; Sy .s1 D b/ D t  b  .b/ D tb 2 ; Sy .s2 D 0/ D Sy .s1 D b/ D tb 2 ;   b 3tb 2 Sy .s2 D b/ D tb 2 C bt  D ; 2 2 Sy .s2 D 2b/ D tb 2 C 2bt  0 D tb 2 ; Sy .s3 D 0/ D Sy .s2 D 2b/ D tb 2 ; Sy .s3 D b/ D 0:

(6.70)

The distribution of Sy .s/ is shown in Fig. 6.17, left. With this, both the shear flow Ts .s/ (Fig. 6.17, middle) and the shear stress s .s/ (Fig. 6.17, right) can be calculated. The shear flow can be given with (6.69) as a section-wise function as: 3 Vz s1 Ts .s1 / D ; 8 b b  2  3 Vz 1 s2 s2 Ts .s2 / D    1 ; 8 b 2 b2 b 3 Vz  s3 Ts .s3 / D  1 : (6.71) 8 b b J Fig. 6.17 Static moment Sy .s/, shear flow Ts .s/ and shear stress s .s/ for the C-cross-section under transverse shear force Vz .

s

s ( )

s1

( )

2

tb

s2

s2 2 3tb 2

( )

s3

2

tb

( )

s

3Vz 8b

2

tb

S y (s)

s1

s2

( )

3Vz 8b s2 9Vz 16b

s3 ( )

s

3Vz 8tb

s

( )

s1

s2

( )

3Vz 8tb s2 9Vz 16tb

s3 3Vz 8b

3Vz 8b Vz Sy (s) Ts(s) = I yy

( )

s

3Vz 8tb

3Vz 8tb τ s(s) = -

Vz Sy (s) I yy t(s)

6.3

Thin-walled cross-sections

195

Example 6.6

Consider a thin-walled box-cross-section (width b, height h D b, constant wall thickness t, Fig. 6.18, top left), which is loaded by a transverse shear force Vz and which has an opening in the center at its upper edge. The distribution of the shear flow Ts .s/ is to be determined. How does the distribution of the shear flow change if the opening is placed in the lower left corner instead (Fig. 6.18, middle left)? What is the shear flow distribution for a closed box-cross-section 3 (Fig. 6.18, bottom left)? The moment of inertia Iyy is Iyy D 2t3b . We first consider the cross-section of Fig. 6.18, top left, and introduce the local circumferential axes si as indicated. The variation of the static moment as

t

s5

s1

s2 h=b

y

tb 4

t

t

tb2 4

2

s2

C t s4 s3

(–)

(–)

s

(–) 2

s

s

tb 4

z b

y

s2 tb2 ( 8

C

)

2

tb 2

z t

2

tb 4 t

s2

C

3tb2 8 tb2 4

z

(+)

(+)

s

t

tb2 4

(+)

s2

t

(+)

Sy(s)

s1

y

(+)

tb2 2

s

s3 s4

(+)

Sy(s)

3tb2 8 tb2 4

s

tb2 4

tb2 4 ( )

( )

s

s

tb2 4 (+)

(+)

Sy(s) ( ) 2

tb 4

s

s

(+)

tb2 4

3Vz 8b s

3Vz 8b s 3Vz ( ) 3Vz (-) 8b 8b s2 s2 (+) s2 (-) Ts(s) 9Vz 3tb2 9Vz 16b 8 16b 3Vz ( ) 3Vz (-) tb2 s s 8b 8b 4 3Vz 3Vz 8b 8b tb2 4

s

2

tb 2 s2 5tb2 8

3Vz 4b (-)

s2 3Vz 16b

( )

(-)

Ts (s)

tb2 2

s

(-)

3Vz 4b

3Vz 4b s2 15Vz 16b 3Vz 4b

3Vz 8b s 3Vz ( ) 3Vz (-) 8b 8b s2 s2 (+) s2 (-) Ts(s) 9Vz 3tb2 9Vz 8 16b 16b 3Vz ( ) 3Vz (-) tb2 s s 8b 8b 4 3Vz 3Vz 8b 8b tb2 4

3Vz 8b s

Fig. 6.18 Considered cross-sections (left), static moments (middle), shear flow distributions (right).

196

6

Shear stresses in beams

a function of si can be given as: Sy .s1 D 0/ D 0;     b b b tb 2 Sy s1 D Dt   D ; 2 2 2 4   b tb 2 D ; Sy .s2 D 0/ D Sy s1 D 2 4     b tb 2 b b 3tb 2 D Ct    D ; Sy s2 D 2 4 2 4 8 tb 2 tb 2 Ct b0D ; Sy .s2 D b/ D  4 4 tb 2 Sy .s3 D 0/ D Sy .s2 D b/ D  ; 4   b tb 2 b b Sy s3 D D C t   D 0; 2 4 2 2 2 tb b tb 2 Ct b D ; Sy .s3 D b/ D  4 2 4 tb 2 Sy .s4 D 0/ D Sy .s3 D b/ D ; 4   b tb 2 b b 3tb 2 Sy s4 D D Ct   D ; 2 4 2 4 8 tb 2 tb 2 Ct b0D ; Sy .s4 D b/ D 4 4 tb 2 Sy .s5 D 0/ D Sy .s4 D b/ D ; 4     b tb 2 b b Sy s5 D D Ct    D 0: 2 4 2 2

(6.72)

The distribution of the static moment Sy .s/ for the cross-section with the centric opening in the upper flange is shown in Fig. 6.18, top center. From this, the shear flow Ts .s/ can then be determined, which is shown in Fig. 6.18, top right. If the cross-section has its opening in the lower left corner as shown in Fig. 6.18, middle left, then the course of the static moment Sy .s/ results as shown in Fig. 6.18, middle. The step-by-step representation of the calculation is omitted here. The resulting shear flow Ts .s/ is shown in Fig. 6.18, center right. We now consider the closed box-cross-section of Fig. 6.18, bottom left. For closed cross-sections the problem arises that the initial value Ts .s D sA / (see (6.52)) cannot be neglected a priori. The analysis of closed cross-sections requires more advanced computational methods, which we will not explain here. However, for the specific case of Fig. 6.18, bottom left, we can take advantage of the fact that in the two horizontal segments of this doubly symmetric cross-

6.3

Thin-walled cross-sections

197

section, at those points where the z-axis intersects the skeleton line, the static moment and hence the shear flow Ts become zero. Therefore, for this case, it is sufficient to consider one of the two symmetry halves and to set the static moment to zero at the resulting free ends. This results in the distributions for the static moment Sy .s/ and the shear flow Ts as shown in Fig. 6.18, bottom center and bottom right, respectively. Apparently, identical distributions are obtained here as those already determined for the cross-section with a centric opening in its upper flange (Fig. 6.18, top). J Example 6.7

Consider the circular cross-section of constant wall thickness t as shown in Fig. 6.19, left. We are looking for the distribution of the shear flow Ts . How does the shear flow change when the cross-section has a centric opening at its right edge (Fig. 6.19, right)? We first consider the closed circular cross-section of Fig. 6.19, left, taking advantage of the fact that it is a symmetric cross-section. Under the shear force Vz , the shear flow will exhibit zero values at those points where the z-axis intersects the cross-section respectively the skeleton line. It is therefore sufficient to cut the cross-section into two halves and to consider one of these two halves (Fig. 6.20, top left). To describe the static moment and the shear flow, the angle ' is introduced as shown. The infinitesimal element dA can then be determined as dA D tRm d'. The static moment Sy at any point ' D ˛ is then calculated with z D Rm cos ' as: Z˛ Sy D

ˇ˛ 2 2 .Rm cos '/tRm d' D Rm t sin ' ˇ0 D Rm t sin ˛:

(6.73)

0

The distribution of the static moment over half the cross-section is shown in 2 Fig. 6.20, top center. At the point ' D 2 the maximum value with Sy D Rm t 3 occurs. The resulting shear flow curve T with Iyy D  tRm is shown in Fig. 6.20, top right. The maximum value of the shear flow occurs at the height of the center Vz of gravity at ' D 2 and amounts to max D R . m The course of the shear flow T shown in Fig. 6.20, top right, refers only to the left half of the actually closed, thin-walled circular cross-section. The course in the right half is obtained by mirroring the left half, the distribution for the closed cross-section is as shown in Fig. 6.20, bottom. Fig. 6.19 Thin-walled circular cross-section (left), circular cross-section with an opening (right).

Rm

Rm

y

y

t

z

t

z

198

6

Fig. 6.20 Determination of the static moment Sy and the shear flow T at the thinwalled circular cross-section.

Shear stresses in beams

dA=tRmd Rm cos y

Rm

Rm2 t

(-)

Vz Rm

Sy

(+)

T

z

Vz Rm

(+)

)+(

T

Vz Rm

We now consider the cross-section of Fig. 6.19, right, with a centric opening at its right edge, and determine the static moment Sy as shown in Fig. 6.21, left. With z D Rm sin ' the static moment follows as: Z˛ Sy D

ˇ˛ 2 2 .Rm sin '/tRm d' D Rm t cos ' ˇ0 D Rm t.cos ˛  1/:

(6.74)

0

Thus, the shear flow T can be determined as: T D

Vz .1  cos ˛/: Rm

(6.75)

The maximum value results at the point ' D  and amounts to: Tmax D

2Vz : Rm

(6.76)

This value is exactly two times higher than the maximum value of the closed profile. J

dA=tRmd Rm

y

Rm sin

2Vz Rm

(+)

T

z

Fig. 6.21 Determination of the static moment Sy at the thin-walled circular cross-section with a centric opening at its right edge (left), distribution of the shear flow T (right).

6.3

Thin-walled cross-sections

199

Example 6.8

Consider the triangular cross-section with an opening at its right edge as shown in Fig. 6.22, left. The cross-section is subjected to the transverse shear force Vz . The distribution of the shear flow Ts is to be determined. The dimension a is given, the wall thickness is constant with the value t. The local circumferential axes s1 , s2 , s3 are introduced as indicated in Fig. 6.22. We determine the static moment Sy as a function of the circumferential coordinates s1 , s2 , s3 as follows: ˇs  Zs1 sO12 ˇˇ 1 s2 sO1 Sy .s1 / D tdOs1 D  t ˇ D t 1 ;  2 4 0 4 0

 ˇs Zs2 a taOs2 t sO22 ˇˇ 2 ta2  C sO2 tdOs2 C Sy .s1 D a/ D   C Sy .s2 / D 2 2 2 ˇ0 4 0

ts22 tas2 ta2   ; 2 2 4   ˇs Zs3 taOs3 t sO32 ˇˇ 3 ta2 a sO3  Sy .s3 / D  tdOs3 C Sy .s2 D a/ D  2 2 2 4 ˇ0 4 D

0

D

ts32 tas3 ta2 C  : 4 2 4

a

s2 a y

z

ta 2 4

ta 2 4

s1

t

s3

(6.77)

a

3ta2 8 ta 2 4

(-) (-)

Sy(s) (-)

ta 2 4

1 Vz ta 2 4 Iyy 3 Vz ta 2 8 Iyy

(+)

1 Vz ta 2 4 Iyy

1 Vz ta 2 4 Iyy (+)

Ts(s) (+)

1 Vz ta 2 4 Iyy

Fig. 6.22 Triangular cross-section with an opening at its right edge (left), static moment Sy (middle), shear flow Ts (right).

200

6

Shear stresses in beams

We calculate the static moment at relevant points of the cross-section: Sy .s1 D 0/ D 0;  a ta2 D Sy .s1 D a/ D t  a   ; 4 4 ta2 ; 4  ta2 3 a a  a D D  ta2 ; Sy s2 D Ct    2 4 2 4 8 Sy .s2 D 0/ D Sy .s1 D a/ D 

Sy .s2 D a/ D 

ta2 ; 4

Sy .s3 D 0/ D Sy .s2 D a/ D  Sy .s3 D a/ D 

ta2 ; 4

ta2 a C t  a  D 0: 4 4

(6.78)

The distribution of the static moment Sy .s/ is shown in Fig. 6.22, middle. The resulting shear flow is shown in Fig. 6.22, right. The shear flow can be represented using (6.77) as follows: Vz Vz ts12 Sy .s1 / D ; Iyy Iyy 4   ts 2 Vz Vz tas2 ta2 Sy .s2 / D C ;  2 C Ts .s2 / D  Iyy Iyy 2 2 4   Vz Vz ts32 tas3 ta2 Sy .s3 / D  C : Ts .s3 / D  Iyy Iyy 4 2 4 Ts .s1 / D 

(6.79)

We also determine the forces F1 , F2 , F3 resulting from Ts .si / by integrating (6.79): Za F1 D 0

Vz t Ts .s1 /ds1 D Iyy 4

Za F2 D

Ts .s2 /ds2 D 0

Za F3 D 0

Vz Iyy

Vz Ts .s3 /ds3 D Iyy

Za 0

s12 ds1 D

Vz ta3 ; Iyy 12

 Za  Vz ta3 tas2 ta2 ts 2 C ds2 D ;  2 C 2 2 4 Iyy 3 0

Za  0

 ts32 tas3 Vz ta3 ta2  C ds3 D : 4 2 4 Iyy 12

(6.80)

6.4 Shear center

201

Fig. 6.23 Resulting forces on the triangular cross-section with an opening at its right edge.

F1 F2 F3

To verify our calculation we form the horizontal and vertical sum of forces (Fig. 6.23). The sum of horizontal forces results in F1 cos 30ı F3 cos 30ı , which in sum gives the value zero. This is an obvious result because there is no shear force component in the horizontal direction. The vertical force sum leads to F2  F1 cos 60ı  F3 cos 60ı D

1 Vz 3 ta : 4 Iyy

(6.81)

We now determine the moment of inertia Iyy of the cross-section. The part of t a3 the vertical web results in IyN y;2 N D 12 . The moments of inertia IyN y;1 N and IyN y;3 N of the segments running diagonally are obtained with the result from Example 4.18 with ' D 60ı as IyN y;1 N D IyN y;3 N D

ta3 ta3 .1 C cos 120ı / D : 24 48

(6.82)

Steiner’s theorem then yields: 2 2 Iyy D IyN y;1 N C IyN y;2 N C IyN y;3 N C zS;1 A1 C zS;3 A3

D

 a 2 ta3 ta3 ta3  a 2 ta C ta C C C  48 12 48 4 4

D

ta3 : 4

(6.83)

Thus, the vertical force sum yields the value Vz . This is also an obvious result – the sum of vertical forces must be equal to the applied transverse shear force Vz . J

6.4 Shear center In a number of practically relevant cross-sections, the application of the shear force at the center of gravity C leads to undesired torsion of the beam. There is a specific point where a transverse shear force has to be applied if this (generally unplanned) torsion is to be avoided. This point is called the shear center M, and we devote this section to determining the location of this specific point. For motivation, we

202

6

Fig. 6.24 Resulting forces on the C-cross-section under the transverse shear force Vz .

Shear stresses in beams

3Vz 8tb

s

( +)

s1

s2

( +)

3Vz 8tb s2 9Vz 16tb

3V F1 = 16z

F2 =Vz

s3 ( +)

s

3Vz 8tb

3Vz 8tb τ s(s) = -

3V F1 = 16z

Vz Sy (s) I yy t(s)

consider the C-cross-section of Fig. 6.24 (flange width b, web height h D 2b), which is subjected to the transverse shear force Vz . If the line of action of Vz passes through the center of gravity C of the cross-section, then a torsional effect occurs in this cross-section – the cross-section twists around the x-axis. To show this, we first form the forces F1 and F2 resulting from the shear flow Ts according to (6.71) acting in the flanges and web, respectively. It follows by integration: 3 Vz F1 D 8 b

Zb 0

3 Vz F2 D  8 b

s1 3 ds1 D Vz ; b 16

Z2b 0

 1 s22 s2   1 ds2 D Vz : 2 b2 b

(6.84)

It turns out that the resulting force F2 corresponds exactly to the shear force Vz , which is an obvious result for equilibrium reasons. In addition, the two partial forces F1 acting in the flanges of the cross-section are equal in magnitude, but have different directions of action, so that they cancel each other out. This is also an obvious result for reasons of equilibrium since no shear force Vy is acting. From Fig. 6.24 it can be concluded that the resulting forces F1 and F2 produce a torsional moment Mx about the center of gravity C: Mx D 2  F1  b  F2 

b 5Vz b D : 4 8

(6.85)

Obviously, when the shear force acts at the center of gravity C, there is an additional and usually unwanted torsional effect in addition to a bending effect. This can be

6.4 Shear center

203

Fig. 6.25 Position of the shear center M for the Ccross-section.

b b 3b 4 8 Vz

h 2

y

h

C

M

h 2

z

avoided only if the shear force acts at the so-called shear center M. Let yM and zM be the coordinates of the shear center M. We then require: Mx D 

5Vz b Vz yM : 8

(6.86)

The y-coordinate yM of the shear center M then follows as: yM D 

5b : 8

(6.87)

The shear center M is marked in Fig. 6.25. It is located outside the cross-section. If the additional torsional effect due to the shear force Vz is to be avoided, it must be ensured by design that the line of action of Vz passes through the shear center M at yM . The determination of the shear center coordinate zM can be omitted in this example. In the presence of a symmetry axis, as given for this example, the shear center is always located on this symmetry axis. Some general rules can be derived for the determination of the shear center M of thin-walled cross-sections:  If there is an axis of symmetry for a cross-section, then the shear center M is always located on this axis of symmetry.  If the cross-section under consideration is a point-symmetric or double-symmetric cross-section, the shear center M is always located in the center of gravity C.  If a cross-section consists of a number of thin-walled straight segments all intersecting at one point, then the shear center M is located exactly at the intersection of the segments. Fig. 6.26 shows the shear center positions for selected thin-walled cross-sections.

204

6

Shear stresses in beams

b1 Symmetry axis

Symmetry axis

h I3 I1+ I3 e b31t1 h I1 = 12 y b33t3 I3 = 12

t1

t1

e=

t2 M

C

t2

C

h y

e=

t1

b2h2t1 4Iyy

M

y

C

Symmetry axis

t2

M

t1

t3

z

z

b3

b

Symmetry axis

z e Symmetry axis

Symmetry axis

t1

t1 t2

y

C

M Symmetry axis

t2

C

y

M

t2

Symmetry axis

t

y

C

M

Symmetry axis

t1

t1 t

z

yp

z t

y

C

C

y

M

z t1

M

t2

yp z

zp

t

zp z

Fig. 6.26 Location of the shear center M for selected cross-sections.

Example 6.9

For the triangular cross-section discussed in Example 6.8, the location of the shear center is to be determined. For convenience, we use a reference frame y, Q zQ as shown in Fig. 6.27. The equivalence of moments about the x-axis then gives: p Vz yQM D F2 Solving for yQM gives with F2 D

Vz t a3 Iyy 3

yQM

3 a: 2

and Iyy D

p 2 3 D a: 3

(6.88) t a3 : 4

(6.89)

6.4 Shear center

205

Fig. 6.27 Position of the shear center M for the triangular cross-section.

F1 Vz F2

M

y~ F3 ~ z

y~M

The position of the shear center with respect to the z-direction Q can be directly inferred as zQM D 0. The cross-section is simply symmetric, and the shear center is always positioned on the symmetry axis. J Example 6.10

For the thin-walled circular cross-section with an opening at its right edge as discussed in Example 6.7, we want to determine the position of the shear center (Fig. 6.28). For this purpose, we consider the infinitesimal element dA D tRm d˛ on which the shear stress .˛/ acts. The resulting moment about the x-axis is then given by the circumferential integral of the shear stress .˛/ with the constant lever arm Rm : Z2 Mx D

Z2 .˛/Rm d˛ D

0

0

Vz Rm .1  cos ˛/d˛ D 2Vz Rm : 

(6.90)

If we equate this with the resulting moment Vz yM of the shear force Vz , we obtain: yM D 2Rm : (6.91) Again, the determination of the shear center coordinate zM can be omitted since we are dealing with a simply symmetrical cross-section, so that zM D 0 results. J Fig. 6.28 Position of the shear center M for the circular cross-section with an opening.

( ) dA=tRmd Vz

M

Rm

y

yM

z

7

Torsion

This chapter is devoted to the consideration of straight bars under torsion. First, the constitutive law of St. Venant’s torsion is derived, and members with circular cross-sections as well as with open and closed cross-sections with thin walls are discussed. Finally, the determination of the state variables for statically determinate and statically indeterminate single-span and multi-span members is shown. Students are enabled to solve simple torsion problems of straight bars analytically with the help of the provided equations.

7.1 Introduction So far, straight bars and beams under normal forces and bending have been considered. However, torsion is an equally important aspect of analysis and design for bars, and this chapter is devoted to the consideration of torsion of straight bars, here the so-called St.-Venant’s torsion1 . In particular, we assume that the twisting of the cross-section causes shear stresses in the plane of the cross-section, but that no cross-section distortions occur and thus no normal stresses develop in the direction of the bar axis (so-called warping stresses). The latter aspects are part of the so-called non-uniform torsion or warping torsion which we will not discuss in this book. Consider the beam shown in Fig. 7.1, which is subjected to an eccentric force F with the eccentricity e. The beam is supported at both ends against deflection as well as against torsion of the cross-section. The statically equivalent condition is to shift the force to the beam axis (system bending) and at the same time to apply a moment acting around the beam longitudinal axis, the torsional moment MT D F e (system torsion). In this case, the system concerning bending consists of a beam simply supported at both ends under the single force F , in which bending moments My and shear forces Qz are generated (if the reference axes are a principal axis system). The stress state in this beam will then be composed of the normal 1

Adhémar Jean Claude Barré de Saint-Venant, 1797–1886, French scientist.

© The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_7

207

208

7

Torsion

Fork restraint y

F

y

y

F

z z

z

e

MT =Fe

x

x

x

Bending Torsion

Fig. 7.1 Beam under eccentric load (left) and the resulting analysis models: system bending (middle), idealized as a beam simply supported at its edges, and system torsion (right), idealized as a beam with fork restraints at its both ends. Fig. 7.2 Bar supported on both sides by fork restraints under torsional moment mT distributed along the length of the bar.

y

mT z

x

stress xx due to bending and the shear stress  due to shear, which we can treat using the means of the previous chapters. However, due to the applied torsional moment MT D F e, an internal torsional moment Mx will develop, which in turn leads to shear stresses  in the cross-sectional plane. Torsional moments can also occur in a distributed form mT (see Fig. 7.2), which in turn result in internal torsional moments Mx . The special feature of the so-called fork restraint for members under torsion, as shown in Fig. 7.1, right, will be discussed later. However, it should be noted at this point that a fork restraint supports the bar in such a way that rotation of the bar about its longitudinal axis x is prevented at this point. In this chapter we will assume that torsion and bending can be considered independently from each other. In this basic introduction to the subject of torsion, we cannot fully explain when this assumption is actually valid, but we will assume that beam bending and torsion can be treated independently. St.-Venant torsion of straight bars is based on the following assumptions:  Due to torsion of the bar, a twisting of the cross-section will occur. We assume that the axis of rotation can adjust itself freely and is not forced by any external constructive measures.  We assume that only a rotation respectively an angle of twist #.x/ about the member axis x results.

7.2 Solid bar with circular cylindrical cross-section

209

 We assume that torsion and bending can be treated independently and that the superposition principle is valid.  We assume that although the cross-section of the bar will rotate about the bar axis and the angle of twist #.x/ about the longitudinal axis x will result, the cross-section will retain its shape and dimensions also in the twisted state. This assumption is called the hypothesis of straight radii.  We assume warping-free cross-sections. This means that due to the angle of twist # around the bar axis x there are no displacements u in the longitudinal direction.  Geometric and material linearity is assumed, i.e. the deformations are small and Hooke’s law is valid.

7.2 Solid bar with circular cylindrical cross-section As a first example we consider the clamped solid bar with circular cross-section (radius R, Fig. 7.3). The bar is loaded at its free end by the torsional moment MT , which causes the constant internal torsional moment Mx in the bar. We now consider an infinitesimal section element of length dx, which we cut out of the bar (Fig. 7.3, right). The torsional loading results in the shear strain  of the surface and the incremental angle of twist d# of the cross-section. The radial coordinate, measured from the center of gravity of the cross-section, is denoted by r. The shear strain  and the angle of twist of the cross-section # are related as follows: dx D rd#: (7.1) This expression can be solved for the shear strain :  Dr

d# D r# 0 : dx

(7.2)

We call the first derivative of the angle of twist # occurring in this expression the twist # 0 of the bar. Mx x

r

MT l

x

dx (+)

R

Mx

d

Mx

MT Fig. 7.3 Solid bar with circular cross-section under torsional moment MT and corresponding internal torsional moment Mx (left), infinitesimal element (right).

210

7

Torsion

If the shear strain  is given, then the shear stress distribution over the bar crosssection can be calculated. We use the constitutive law  D G, where G is the shear modulus of the material:  D G D Gr# 0 : (7.3) Obviously, there is a linear shear stress distribution over the radial coordinate r. At the center of gravity of the cross-section, the shear stress becomes zero, and it reaches its yet to be determined maximum value at the edge of the cross-section at r D R. Once the shear stress distribution has been determined, we can calculate the resulting torsional moment Mx : Z Z Z (7.4) Mx D  rdA D Gr 2 # 0 dA D G# 0 r 2 dA: A

A

A

The area integral appearing in (7.4) can be subdivided into two partial integrals with the area element dA D rdrd' (Fig. 7.4, left) as follows: Mx D G# 0

Z2 ZR 0

r 3 drd' D G# 0

0

R4 4

Z2

d' D G# 0

0

R4 : 2

(7.5)

At this point, the torsional moment of inertia IT is introduced as follows: IT D We then obtain:

R4 : 2

(7.6)

Mx D GIT # 0 :

(7.7)

The product GIT of shear modulus and torsional moment of inertia is called torsional stiffness. Equation (7.7) represents the constitutive law for St.-Venant torsion. Thus (7.3) can be represented as:  D Gr# 0 D

Mx r: IT

(7.8)

The maximum value of the torsional stress  at the section edge r D R can then be given as (Fig. 7.4, right): Mx R : (7.9) D IT Fig. 7.4 Infinitesimal area element dA (left), shear stress profile  .r/ over the circular cross-section (right).

dA=rdrd d dr r

(r)

Mx R IT

7.3 Thin-walled bar with circular cylindrical cross-section

211

From equation (7.8) in conjunction with Fig. 7.4, right, it can be concluded that the shear stress  is linear along the radial coordinate. Thus, in the region of the center of gravity of the considered cross-section, there is a poor utilization of the material.

7.3

Thin-walled bar with circular cylindrical cross-section

As another elementary case, we consider the thin-walled bar with circular cylindrical cross-section of constant wall thickness t (see Fig. 7.5). We assume that the wall thickness t is significantly smaller than the two radii Ri and Ra , where r D Ri and r D Ra are the locations of the inner and outer surface of the cylinder. Let the wall of the circular cylindrical cross-section be bisected at each point of the circumferential coordinate s by the skeleton line at the point r D Rm . From the consideration of the solid circular cross-section we already know that the shear stress according to (7.8) is distributed linearly over the radial coordinate r. Since we assume that the wall thickness t is very small for the example of a thinwalled circular cylindrical cross-section, we can assume that the shear stress  is constant over the thickness t of the cross-section. This is shown in Fig. 7.5, bottom right. Since we will refer all considerations to the skeleton line with the circumferential axis s, we will refer to the shear stress as s accordingly. Let us now consider an infinitesimal element which we cut out of the cylinder (Fig. 7.6) and where we apply the shear stress s . At the infinitesimal element of

Fig. 7.5 Bar segment with circular cylindrical crosssection under torsional moment Mx (top), dimensions (bottom left), shear stress distribution (bottom right).

Mx r

x

t

s

Fig. 7.6 Infinitesimal element cut out of the bar with circular cylindrical crosssection.

t

Mx

dx

Ri Rm Ra

τs (s)

dx

∂τs ds ∂s ∂τs τs+ dx ∂x

τs +

τs

ds

τs

212

7

Torsion

Fig. 7.6, right, the force equilibrium with respect to the x-direction is obtained as:   @s  ts dx  s C ds tdx D 0: (7.10) @s This leads to the following expression: @s D 0: @s

(7.11)

From this it can be concluded that the shear stress s does not undergo any change over the circumferential coordinate s, and thus it is constant over the entire circumference of the circular cylindrical cross-section. The internal torsional moment Mx Z follows as: Z (7.12) Mx D  rs dA D s rdA: A

A

Since the radius is constant with r D Rm it can be drawn in front of the integral: Z Mx D Rm s dA: (7.13) A

If we use the area element dA D Rm td', we obtain: Mx D

2 s Rm t

Z2

2 d' D 2Rm ts :

(7.14)

0

This can be solved for s : s D

Mx : 2t 2Rm

(7.15)

Since the shear stress s was assumed to be constant over the wall thickness, this also represents the maximum value: max D

Mx : WT

(7.16)

Herein, WT is the section modulus of torsion or the torsional resistance moment: 2 t: WT D 2Rm

(7.17)

At this point we use the relation (7.2) as well as the constitutive law  D G, and it follows: (7.18) s D Gr# 0 : Solving for # 0 with r D Rm and inserting (7.15) gives: #0 D

s Mx D : 3 Gt GRm 2Rm

(7.19)

7.4 Bars with arbitrary thin-walled cylindrical cross-sections

213

With the torsional moment of inertia 3 IT D 2Rm t

we obtain: #0 D

(7.20)

Mx : GIT

(7.21)

This can be brought into the form Mx D GIT # 0 ;

(7.22)

and again the constitutive law of St. Venant’s torsion is obtained, as already shown with (7.7). The mutual angle of twist of the two bar ends # is obtained by integrating the twist # 0 over the bar length: Zl # D

0

Zl

# dx D 0

0

Mx dx: GIT

(7.23)

In case of a constant torsional moment Mx and a torsional moment of inertia IT which is constant over the bar length, the mutual angle of twist of the bar ends is given as: Mx l : (7.24) # D GIT If one is interested in designing a circular cylindrical cross-section that is as stiff as possible, then one will try to maximize the torsional stiffness GIT . This is best done by controlling the radius Rm , which enters IT to the third power.

7.4 Bars with arbitrary thin-walled cylindrical cross-sections We now extend the considerations to the case of a bar having an arbitrary thinwalled cylindrical cross-section with arbitrarily varying wall thickness t D t.s/ along the circumference (Fig. 7.7). The axis about which the cross-section twists is called axis of rotation D. At this point we introduce the shear flow Ts as follows: Ts D s .s/t.s/:

(7.25)

The shear flow is the resultant of the shear stress s .s/ over the section thickness t.s/, and it has a corresponding unit of force per unit length, e.g. [N/m]. Using the free body image of Fig. 7.7, right, the force equilibrium concerning the x-direction gives the following result: @Ts D 0: (7.26) @s

214

7

t(s)

ds dx

Torsion

Ts

Ts

τs (s)

s Ts +

D

∂T Ts + s ds ∂

τs (s)

τ s (s)

x

∂Ts dx ∂x

Mx

τ s (s)

Fig. 7.7 Arbitrary thin-walled cylindrical cross-section with arbitrarily varying wall thickness along the circumference.

This means that the shear flow Ts takes the same value at every point s on the skeleton line, i.e. it is constant over the circumference of the cross-section. Here, however, it is important to note that this does not hold for the shear stress s .s/ D T .s/ , which is shown in Fig. 7.7 in a qualitatively manner. The smaller the wall t .s/ thickness, the higher the shear stress will be. The maximum shear stress max therefore occurs at the location of the smallest wall thickness tmin . We now want to establish a relationship between the shear flow Ts and the torsional moment Mx . For this purpose, we consider the free body image of Fig. 7.8, in which a sectional element of the cross-section of length ds is shown. The shear flow Ts acting in this sectional element has the lever arm r t (measured from the tangent of the skeleton line) with respect to the axis of rotation D as shown. The sectional element has the distance r.s/ from the axis of rotation. The shear flow causes the partial torsional moment dMx as follows: dMx D Ts dsr t :

(7.27)

The torsional moment Mx is obtained by integration over the circumference over the cross-section, represented by a ring integral: I Mx D Ts r t ds: (7.28) With dAm D

1 r t ds 2

(7.29)

Fig. 7.8 Free body image of a cross-sectional element ds.

Tangent of the skeleton line

ds

rt

D

dAm

Ts

7.4 Bars with arbitrary thin-walled cylindrical cross-sections

we can solve for ds: ds D 2

dAm : rt

215

(7.30)

Substituting into (7.28) gives: I Mx D 2Ts

dAm

(7.31)

Am

or

Mx D 2Ts Am :

(7.32)

Herein, Am is the area of the cross-section enclosed by the skeleton line. With this, the shear flow Ts due to the torsional moment Mx can be determined as: Ts D

Mx : 2Am

(7.33)

The shear flow Ts as a consequence of the torsional moment Mx is thus calculated for an arbitrary thin-walled cross-section as the torsional moment Mx divided by twice the area Am enclosed by the skeleton line. Equation (7.33) is the so-called First Bredt formula2 . Hence, the shear stress can also be determined, and it follows: s .s/ D

Ts Mx D : t.s/ 2Am t.s/

(7.34)

It should be noted that s , in contrast to Ts , is generally a function of the circumferential coordinate s. It reaches its maximum at the point of the cross-section with the smallest wall thickness t D tmin : max D

Mx : 2Am tmin

(7.35)

Consequently, the smallest wall thickness tmin is decisive for the design of a thinwalled hollow section under torsion. At this point, the torsional resistance moment WT according to WT D 2Am tmin

(7.36)

can be introduced. Then we can write for (7.35): max D

Mx : WT

(7.37)

We will now consider the determination of the torsional moment of inertia IT for the thin-walled closed cross-section considered here. For this purpose, we examine the cross-section twisted about the axis of rotation D and consider Fig. 7.9. Since we assume that the cross-section remains unchanged in shape and dimensions, the 2

Rudolf Bredt, 1842–1900, German engineer.

216

7

Fig. 7.9 Displacement of a point P on the skeleton line.

Torsion

P rt

dv r

D

d

dr

P

Skeleton line

Tangent

rotation of the cross-section around the axis of rotation D is a pure rigid body rotation. We now examine the displacements of the point P according to Fig. 7.9, where P is displaced by the measure dr perpendicular to the radius r. Let the displacement projected onto the tangent of the skeleton line be dv. Since we can assume small deformations and hence small angles, we can deduce from Fig. 7.9: dr D rd#: We also find:

(7.38)

dv dv D : dr rd#

(7.39)

rt : r

(7.40)

dv D r t d#:

(7.41)

cos ˛ D Also, the following relation is valid:

cos ˛ D Equating (7.39) and (7.40), we obtain:

We derive (7.41) once, and assuming that r t is independent of x for a prismatic bar, we arrive at: dv d# (7.42) D rt D r t d# 0 : dx dx Now the shear strain of an infinitesimal element of the bar is considered (see Fig. 7.10). The shear strain is composed of the angles ˛ and ˇ, which can be determined as follows, assuming small angles: ˛D

dv ; dx

Thus:  D˛Cˇ D

ˇD

du : ds

du dv C : dx ds

(7.43)

(7.44)

Using Hooke’s law  D G yields: D

s dv du D C : G dx ds

(7.45)

7.4 Bars with arbitrary thin-walled cylindrical cross-sections

217

x dx

t(s)

ds dx

dv ds

s D

s

du Mx

x

Fig. 7.10 Shear strain of an infinitesimal element.

With the shear flow Ts D s .s/t.s/ as well as (7.41) we obtain: Ts du D rt # 0 C : Gt.s/ ds

(7.46)

represents the change of the displacement u of the cross-section The expression du ds over the circumferential coordinate s and thus the warping of the cross-section. In this context warping refers to displacements u of the cross-section which occur out of the cross-section plane in the x-direction. We form the circumferential integral over the expression (7.46): I I I du Ts ds D r t # 0 ds C ds: (7.47) Gt.s/ ds We first consider the second integral on the right side. We can consider the circumferential integral as an integral from a starting point s D sA to the end point s D sE . Hence: I ZsE du du (7.48) ds D ds D u.s D sE /  u.s D sA /: ds ds sA

However, since we are currently considering a closed cross-section where the points s D sE and s D sA coincide, this integral becomes zero: I du ds D 0: (7.49) ds We thus obtain:

Ts G

I

1 ds D # 0 t.s/

I r t ds:

The integral of the right side gives the value 2Am : I 1 Ts ds D 2Am # 0 : G t.s/

(7.50)

(7.51)

218

7

Using the constitutive law (7.21) we obtain: I 1 Ts Mx : ds D 2Am G t.s/ GIT From (7.32) we can conclude Mx D 2Ts Am so that: I 1 Ts ds D 4A2m : Ts t.s/ IT

Torsion

(7.52)

(7.53)

This expression can be solved for the torsional moment of inertia IT : IT D I

4A2m : ds t.s/

(7.54)

This is the so-called Second Bredt formula. In many engineering applications, one is dealing with cross-sections composed of n rectilinear segments of lengths li and whose wall thicknesses ti are constant. One can then transform equation (7.54) into the following form: IT D

4A2m n Z X ds ti i D1 li

0

D

4A2m : n X li

(7.55)

t i D1 i

In the special case that the cross-section has the constant wall thickness t, then this expression becomes: 4A2m t : (7.56) IT D U Here U is the circumference of the cross-section described by the skeleton line. A discussion of the result for the torsional stiffness GIT of a bar shows that for a given material it can only be controlled by the torsional moment of inertia IT . If one is interested in the highest possible torsional stiffness, then according to (7.54) and (7.56) one will strive to maximize the area Am enclosed by the skeleton line, which enters here in second power. If it is a cross-section with constant wall thickness t, then it turns out that the circular cross-section is the optimal shape. Example 7.1

Consider the thin-walled box cross-section of Fig. 7.11 (width b, height h, wall thicknesses t1 , t2 , t3 , t4 as indicated). From (7.54) and (7.55), respectively, with Am D bh, the torsional moment of inertia IT for the given cross-section follows as: 4b 2 h2 : (7.57) IT D b h b h C C C t1 t2 t3 t4

7.5 Bars with open thin-walled cross-sections

219

Fig. 7.11 Thin-walled box cross-section.

t1 t2

t4

h

t3

b

We consider the case t1 D t, t2 D 2t, t3 D 5t, t4 D 2t and b D 100t, h D 200t. For this we obtain: IT D

4.20;000t 2 /2 100t 200t 100t 200t C C C t 2t 5t 2t

D 5  106 t 4 :

(7.58)

If we consider the special case that all wall thicknesses have the identical value t, then (7.56) can be used to calculate IT . With U D 600t, the result is: IT D

4.20;000t 2 /2 t 8 D  106 t 4 : 600t 3

(7.59) J

7.5 Bars with open thin-walled cross-sections In this section we consider thin-walled open cross-sections, where we want to study cross-sections composed of segments with constant thickness t. Some technically relevant examples are shown in Fig. 7.12. In Fig. 7.13, left, a segment of such a thin-walled cross-section is given. Let this segment be a rectangular cross-section with height h and thickness t, assuming t  h. We can interpret this cross-section as a number of nested hollow crosssections, as indicated in Fig. 7.13, left. The wall thickness of such a hollow crosssection is dy.

t1 t2

t3

t1

t1

t1 t2

t2

t3

t2

t1 t2

t3

Fig. 7.12 Thin-walled open cross-sections consisting of segments of constant thickness.

220

7

Torsion

t dy Am dy

τ 2y τ (y)= τmax t

y

h

z Fig. 7.13 Cross-sectional segment and its idealization.

Since we assume that the thickness t of the considered segment is sufficiently small, we can assume (similarly to the solid circular cross-section) that the shear stress is linearly distributed across the thickness and runs exclusively in the z-direction in the present case, reaching its maximum value max at the edge of the cross-section. This is shown in Fig. 7.13, middle. The circulating effect of the shear stress due to torsion, which actually occurs at the upper and lower ends of the segment, is not considered due to the assumption t  h. Thus, we assume a number of partial cross-sections as shown in Fig. 7.13, right, which run completely over the height h of the segment. If we assume sufficiently small values for dy, then we can assume that the shear stress is constant in each of these subsections, as shown in Fig. 7.13, right. Each of these partial cross-sections encloses the partial surface Am D 2yh. Using the first Bredt formula (7.33), the partial shear flow dTs D .y/dy gives the corresponding partial torsional moment, which is transmitted over the considered partial cross-section: dMx D 2Am dTs : With .y/ D max

(7.60)

2y t

(7.61)

and Am D 2yh we obtain: dMx D

8y 2 hmax dy: t

Integration over the cross-section from y D 0 to y D Mx D

1 max ht 2 : 3

(7.62) t 2

provides: (7.63)

7.5 Bars with open thin-walled cross-sections

221

Using the torsional resistance moment WT D

1 2 ht 3

(7.64)

leads to the maximum shear stress as follows: max D

Mx : WT

(7.65)

An expression for the torsional moment of inertia can be found in the same way. We use the second Bredt formula (7.54) and apply it to the partial cross-section of Fig. 7.13. For the partial moment of inertia dIT we obtain: dIT D 8hy 2 dy:

(7.66)

Integrating over the entire cross-section leads to the torsional moment of inertia IT : t

Z2 IT D

IT dy D 0

1 3 ht : 3

(7.67)

Obviously, there is a significant difference between an open cross-section and a closed cross-section in terms of stress distributions across the cross-section thickness. This is shown in Fig. 7.14. Here, the arbitrary cross-section of Fig. 7.7 (shown again in Fig. 7.14, left) is compared with an identical cross-section, but converted into an open cross-section by a cut in the longitudinal direction (Fig. 7.14, right). In an open cross-section, the shear stresses s .s/ due to torsion run linearly over the wall thickness and assume their maximum values at the edge of the wall. They therefore reach their maximum at the point of the cross-section where the wall thickness has its maximum, i.e. at t D tmax . This is in contrast to the closed

t(s)

τ s (s)

t(s)

τ s (s)

s τ s (s)

s τ s (s)

τ s (s)

x τ s (s)

Mx

τ s (s)

x

Mx

τ s (s)

Fig. 7.14 Shear stress distribution in an arbitrary closed cross-section and a similar open crosssection.

222

7

Torsion

cross-section, where the maximum shear stress is reached where the smallest wall thickness t D tmin is present. We now consider thin-walled open cross-sections as shown in Fig. 7.12. Based on the assumption that the cross-sections do not change in shape or dimensions due to torsion, we can assume that all segments of the cross-section undergo the same angle of twist # and thus also the same twist # 0 . We can write for the segment i of the cross-section: Mx;i #i0 D D # 0: (7.68) GIT;i The same applies to the total cross-section: #0 D or rearranged:

Mx ; GIT

(7.69)

Mx D GIT # 0 :

(7.70)

The total torsional moment Mx is composed of the partial moments of the segments of the cross-section: Mx D

n X

Mx;i D

n X

i D1

# 0 GIT;i D # 0

i D1

n X

GIT;i :

(7.71)

i D1

Equating (7.70) and (7.71), we obtain: GIT # 0 D # 0

n X

GIT;i ;

(7.72)

i D1

or after eliminating the twist # 0 : GIT D

n X

GIT;i :

(7.73)

i D1

The torsional stiffness GIT of a cross-section composed of several segments is thus obtained by adding the individual contributions GIT;i of the segments of the crosssection. Example 7.2

An I-cross-section as shown in Fig. 7.15 is considered in which the torsional moment of inertia IT results from the sum of the individual contributions of the segments of the cross-section: IT D

n X i D1

n

IT;i D

1X 3 li t : 3 i D1 i

(7.74)

7.5 Bars with open thin-walled cross-sections

223

Fig. 7.15 Shear stresses due to torsion in the I-crosssection.

t

τ s (s)

t h

Mx

t

b

b

Here li is the length of the segment i, ti is the corresponding wall thickness. The following torsional moment of inertia results: IT D

1 .4b C h/t 3 : 3

(7.75)

The resulting distribution of the shear stresses due to torsion at this I-crosssection is shown in Fig. 7.15, right, qualitatively. J Example 7.3

Consider the circular ring profile of Fig. 7.16 with constant thickness t and radius Rm with respect to the skeleton line. The profile is considered once as a closed cross-section and once as an open cross-section, and essential differences in the load-bearing behavior with respect to torsion are to be studied on the basis of this example. First, the torsional moments of inertia IT are calculated for both crosssections. For the closed cross-section, using (7.56) yields:  2 2 t 4 Rm 4A2m t 3 IT D D 2Rm t: (7.76) D U 2Rm For the open cross-section follows with (7.67): IT D

1 3 1 2 ht D .2Rm /t 3 D Rm t 3 : 3 3 3

(7.77)

We now form the ratio of these values and obtain:   3 2Rm t Rm 2 D3 : t 2 Rm t 3 3

(7.78)

Fig. 7.16 Closed and open circular ring cross-section.

t

Mx

τ s (s)

Rm

t

Mx

Rm

τ s (s)

224

7

Torsion

It can be seen that the torsional moment of inertia IT of the closed cross-section is higher than the torsional moment of inertia of the open cross-section by a factor of 3. Rtm /2 . Many cross-sections in practical applications will be thin-walled where the wall thickness t is significantly smaller than the radius Rm . For this reason, the difference between the two torsional moments of inertia will be significant in many technically relevant cases. In a similar way, we want to investigate the difference in the shear stresses due to torsion. In both cases, the maximum shear stress is calculated using the x formula max D M WT . The torsional resistance moment WT is obtained for the closed cross-section as: 2 WT D 2Am tmin D 2Rm t:

(7.79)

For the open cross-section we obtain with (7.64): WT D

1 2 2 ht D Rm t 2 : 3 3

(7.80)

If we compare the maximum shear stresses, we obtain the following result: 2 Rm t 2 max;c 1 t 3 D : D 2t max;o 2Rm 3 Rm

(7.81)

The maximum shear stresses at the open cross-section are thus higher than in the case of the closed cross-section by a factor of 3 Rtm . J

7.6 Determination of internal moments We consider the constitutive law (7.22): Mx D GIT # 0 :

(7.82)

Equilibrium at an infinitesimal element of the bar (Fig. 7.17) yields the following x equilibrium condition with dMx D dM dx: dx dMx D Mx0 D mT : dx mT

Mx

(7.83)

mT

Mx + dM x

x

dx l

Fig. 7.17 Free body image for the infinitesimal element of the bar under torsion.

7.6 Determination of internal moments

225

The first derivative of the torsional moment Mx corresponds to the negative distributed torsional load mT . Substituting this into the constitutive law (7.82), we obtain: .GIT # 0 /0 D mT : (7.84) In the case of a prismatic bar with constant torsional stiffness GIT , it follows: GIT # 00 D mT :

(7.85)

From (7.84) and (7.85), respectively, one can then determine the twist # 0 and the torsional moment Mx as well as the angle of twist # by single and double integration of the load mT , respectively: .GIT # 0 /0 D mT ; GIT # 0 D Mx D 

Z mT dx C C1 ;

“ GIT # D 

mT dxdx C C1 x C C2 :

(7.86)

The constants C1 and C2 are determined from given boundary conditions. A selection of typical boundary conditions is shown in Fig. 7.18. In the case of a fork restraint at x D 0 (Fig. 7.18, left), the cross-section is prevented from twisting. Thus: #.x D 0/ D 0: (7.87) If the bar is not subject to any support at the point x D 0 (Fig. 7.18, middle), then the torsional moment Mx becomes zero at this location: Mx .x D 0/ D 0:

(7.88)

Because of the relation (7.82) this is equivalent to the fact that at this point the twist # 0 disappears, i.e.: # 0 .x D 0/ D 0: (7.89) At a free bar end where a torsional moment Mx;0 is applied, the torsional moment Mx must be equal to the applied torsional moment Mx;0 : Mx .x D 0/ D Mx;0 : Mx x

Mx x

(7.90)

Mx,0

Mx x

Fig. 7.18 Typical boundary conditions of a bar under torsion at the point x D 0: Fork restraint (left), free end (center), free end with end moment (right).

226

7

Torsion

Example 7.4

We consider the bar of constant torsional stiffness GIT shown in Fig. 7.19 under a constant torsional load mT . The bar is subject to fork restraints at both ends. The torsional moment Mx .x/ and the angle of twist #.x/ are to be determined. To determine the torsional moment Mx and the angle of twist # we employ (7.86). The following boundary conditions are given: #.x D 0/ D 0;

#.x D l/ D 0:

(7.91)

The first condition in (7.91) results in: C2 D 0:

(7.92)

From the second condition in (7.91) we obtain: C1 D

1 mT l: 2

(7.93)

The angle of twist thus follows as: #.x/ D

mT x .l  x/: 2GIT

(7.94)

The torsional moment Mx results as:  Mx .x/ D mT

 l x : 2

(7.95)

The state variables #.x/ and Mx .x/ are shown in Fig. 7.19. Under a constant load mT , the torsional moment Mx is linear and the angle of twist is quadratically distributed over x. J Fig. 7.19 Bar supported by fork restraints at both ends under uniform load mT (top), resulting state variables #.x/ and Mx .x/ (bottom).

mT x

GIT l

ϑ (x) (+)

ϑ max=

mT l 2

(+)

mT l

2

8GIT (-)

Mx (x) mT l 2

7.6 Determination of internal moments

227

Example 7.5

We consider the bar under fork restraints shown in Fig. 7.20 under the single moment MT;0 . The bar has the constant torsional stiffnesses GIT;1 and GIT;2 in the sections of lengths l1 and l2 , respectively. We want to determine the distribution of the torsional moment Mx . We introduce the two longitudinal axes x1 and x2 as shown in Fig. 7.20. From (7.86), we then obtain for the section 0  x1  l1 : GIT;1 #100 D mT;1 ; GIT;1 #10 D Mx D mT;1 x1 C C1 ; 1 GIT;1 #1 D  mT;1 x12 C C1 x1 C C2 : 2

(7.96)

In the section 0  x2  l2 we have: GIT;2 #200 D mT;2 ; GIT;2 #20 D Mx D mT;2 x2 C C3 ; 1 GIT;2 #2 D  mT;2 x22 C C3 x2 C C4 : 2

(7.97)

In (7.96) and (7.97) the terms mT;1 and mT;2 are zero. From the boundary and transition conditions the constants C1 ; : : : ; C4 can be determined: #1 .x1 D 0/ D 0; #1 .x1 D l1 / D #2 .x2 D 0/; Mx;1 .x1 D l1 / D Mx;2 .x2 D 0/ C MT;0 ; #2 .x2 D l2 / D 0:

Fig. 7.20 Bar supported by fork restraints at both ends under single moment MT;0 (top), distribution of the torsional moment Mx .x/ (bottom).

(7.98)

MT,0 x1

x2

GIT,1

GIT,2

l1 Mx (x1) =MT,0 -

l2 MT,0 l1 l1+l2GIT,1 GIT,2

(+)

Mx

(-)

Mx (x2) = -

MT,0 l1 l1+l2GIT,1 GIT,2

228

7

Torsion

The resulting system of equations leads to the constants C1 ; : : : ; C4 as follows: 0 B B C1 D MT;0 B1  @

C3 D 

1 C C C; C2 D 0; GIT;1 A l1 C l2 GIT;2 l1

MT;0 l1 GIT;1 l1 C l2 GIT;2

;

C4 D

MT;0 l1 l2

:

(7.99)

:

(7.100)

GIT;1 l1 C l2 GIT;2

The moment distribution Mx can thus be determined as: 0 1 B B Mx;1 D MT;0 B1  @

C C C; GIT;1 A l1 C l2 GIT;2 l1

Mx;2 D 

MT;0 l1 GIT;1 l1 C l2 GIT;2

The distribution of Mx is shown in Fig. 7.20. J

Example 7.6

Consider the cantilever of Fig. 7.21 (length 2l, constant torsional stiffness GIT ), which is loaded by two single torsional moments MT;0 . The maximum shear stress max in the given thin-walled rectangular cross-section is to be determined. What is the angle of twist at x D 2l? In addition, determine how large the ratio Ram must be if the rectangular cross-section is replaced by a circular cross-section with radius Rm and constant wall thickness t and the maximum shear stress  is MT,0

t

x

MT,0

l

t

2t

l

2a (+)

Rm

Mx

MT,0 2MT,0 a Fig. 7.21 Cantilever under two single moments MT;0 (top), distribution of the torsional moment Mx .x/ (bottom), considered cross-sections (right).

7.6 Determination of internal moments

229

to remain the same. How large is Ram if the angle of twist at the free cantilever end #.x D 2l/ is to remain the same? For the given rectangular cross-section, the maximum shear stress max is given as: max D

2MT;0 2MT;0 2MT;0 MT;0 D D : D WT 2Am tmin 2  .2a  a/  t 2ta2

(7.101)

To determine the angle of twist of the bar end at x D 2l, the torsional moment of inertia IT is required. It can be determined from Bredt’s second formula as: 4A2m 4.2a  a/2 ! D 4a3 t: IT D I D ds a 2a 2 C t.s/ t 2t

(7.102)

The required angle of twist is obtained by adding the angles of twist of the two segments 0  x  l and l  x  2l of the bar: #.x D 2l/ D

MT;0 l 3MT;0 l 2MT;0 l C D : GIT GIT 4Ga3 t

(7.103)

In order to determine the ratio Ram which must be maintained if the same maximum shear stress is to occur for the circular cross-section, the torsional resistance moment WT of the circular cross-section is first determined: 2 t: WT D 2Am tmin D 2Rm

(7.104)

This gives the maximum shear stress at the circular cross-section as: max D

2MT;0 MT;0 D : 2 2t 2Rm t Rm

(7.105)

Equating with (7.101) gives: Rm D a

r

2 : 

(7.106)

To determine the ratio Ram so that the same angle of twist #.x D 2l/ occurs for the circular cross-section as for the rectangular cross-section, the torsional moment of inertia of the circular cross-section is first calculated. It follows as: 3 t: IT D 2Rm

(7.107)

230

7

Torsion

Thus, the angle of twist #.x D 2l/ can be determined as: #.x D 2l/ D

3MT;0 l 3MT;0 l D : 3t GIT 2GRm

(7.108)

Equating with (7.103) gives: Rm D a

r 3

2 : 

(7.109) J

Example 7.7

A bar of length 2l supported on both sides by fork restraints is considered which is loaded by the torsional moment flow mT (Fig. 7.22). In the range 0  x1  l the torsional stiffness GIT;1 is given, in the range 0  x2  l we have the torsional stiffness GIT;2 . The torsional moment Mx and the angle of twist # are to be determined. We solve the given problem by integration. For 0  x  l follows: GIT;1 #100 D mT ; GIT;1 #10 D Mx;1 D mT x1 C C1 ; 1 GIT;1 #1 D  mT x12 C C1 x1 C C2 : 2

(7.110)

In the section 0  x2  l2 we have: GIT;2 #200 D mT ; GIT;2 #20 D Mx;2 D mT x2 C D1 ; 1 GIT;2 #2 D  mT x22 C D1 x2 C D2 : 2

(7.111)

The boundary and transition conditions to be applied here are: #.x1 D 0/ D 0;

#.x2 D l/ D 0;

Mx;1 .x1 D l/ D Mx;2 .x2 D 0/; #1 .x1 D l/ D #2 .x2 D 0/:

Fig. 7.22 Bar under fork restraitns at both ends under torsional moment flow mT .

(7.112)

mT x1

GIT,1 l

x2

GIT,2 l

7.6 Determination of internal moments

231

From the first condition we obtain C2 D 0. The three remaining conditions lead to the following linear system of equations for the constants C1 , D1 and D2 : 1  mT l 2 C D1 l C D2 D 0; 2 mT l C C1 D D1 ;   1 1 1 2 D2 :  m T l C C1 l D GIT;1 2 GIT;2

(7.113)

This yields: GIT;2 GIT;1 ; GIT;2 1C GIT;1 1 0 GIT;2 3C C mT l B GIT;1 C B D1 D  2 C; B A 2 @ GIT;2 1C GIT;1 0 1 GIT;2 3 C mT l 2 B GIT;1 C B C D2 D B3  C: 2 @ GIT;2 A 1C GIT;1 mT l C1 D 2

3C

(7.114)

Thus, the torsional moments in the two segments can be formulated as: 0

Mx;1

Mx;2

1 GIT;2 3C B l GIT;1 C B C D mT Bx1  C; @ 2 GIT;2 A 1C GIT;1 1 0 GIT;2 3C C mT l B GIT;1 C B x2 D C 2 C: B2  A 2 @ l GIT;2 1C GIT;1

(7.115)

232

7

Torsion

The angles of twist in the two segments follow as: 2

3 GIT;2 3C  mT l 2 6 GIT;1 7 6 x1 2 x1 7 #1 D   6 7; 2GIT;1 4 l l GIT;2 5 1C GIT;1 1 0 2 3 GIT;2 GIT;2 3C C B 3 C GI  mT l 2 6 GIT;1 7 C 6 x2 2 x2 B 7 T;1  B  2C  3 C #2 D  6 7: A 2GIT;2 4 l l @ GIT;2 GIT;2 5 1C 1C GIT;1 GIT;1

(7.116) J

8

Energy methods

Energy methods are well suited to determine both the deformations and the static state quantities such as forces and moments in elastic structures. After a brief introduction to the concepts of work and energy, strain energy and complementary strain energy are introduced for bars and beams. Then, the principle of work and energy is used to determine deformations in beam and bar structures. Another essential aspect of energy methods in structural mechanics concerns the so-called principle of virtual forces, from which the unit load theorem can be derived for the determination of deformations of structures. In formalized form, this procedure is also known as the so-called force method, and it is shown how the force method can be used for the analysis of deformations of beam and bar structures, but also of statically indeterminate structures. When applied to multiply statically indeterminate systems, the so-called reciprocity theorems, here in the form of Betti’s and Maxwell’s theorems, are also used. Students are enabled to apply the provided energy methods to the analysis of statically determinate and statically indeterminate bar and beam structures.

8.1 Work and energy 8.1.1 Introduction In the following, we always assume so-called conservative forces. These are forces for which the work W done by them depends only on the starting and end point of their motion, but not on the distance covered. All following explanations assume elastic material behavior. Fig. 8.1 shows a particle under a force F which is displaced by the action of the force by the distance u in the direction of action of the force. The work W corresponds to the force F multiplied by the distance u: W D F u: © The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_8

(8.1) 233

234

8 Energy methods

Fig. 8.1 Displacement of a particle under the force F by the distance u in the direction of action of the force.

F

F

u

This very simple result follows from the fact that we have tacitly agreed that the force F remains constant over the distance u covered and that the displacement u points exactly in the direction of action of the force F . We now want to make these considerations more general and consider the situation of Fig. 8.2. A particle under the force F D .Fx ; Fy ; Fz /T is shown, which moves on a spatial trajectory with the position vector r D .rx ; ry ; rz /T from a point A to a point B. Let the position vectors associated with points A and B be rA and r B , respectively. Let the force F be location dependent, which means that F can have different magnitudes and directions of action at different positions on the trajectory. The work increment dW , which is driven by the force F D .Fx ; Fy ; Fz /T along a displacement increment du D .du; dv; dw/T can be expressed as the scalar product of F and du: dW D F du D Fx du C Fy dv C Fz dw:

(8.2)

The total work done by F on the trajectory between points A and B then follows as the sum of all work increments. Thus, the integral over the scalar product F du is to be formed: ZB W D F du: (8.3) A

Work is a scalar quantity. It is expressed in Newton meters1 [Nm] or in Joules2 [J], where 1 Nm D 1 J. The Joule is the SI unit for work and energy: 1J D 1

kg  m2 : s2

(8.4)

If we consider a moment M D .Mx ; My ; Mz /T subjected to a rotation ' D .'x ; 'y ; 'z /T , the procedure is quite analogous. The work increment dW performed Fig. 8.2 Particle under force F on a spatial trajectory between points A and B.

F

z A rA

dr=du

r r+dr

rB

x 1 2

Isaac Newton, 1642–1726, English scientist. James Prescott Joule, 1818–1889, English physicist.

y

B

8.1

Work and energy

235

by the moment M along the rotation d' D .d'x ; d'y ; d'z /T then is: dW D M d' D Mx d'x C My d'y C Mz d'z :

(8.5)

The total work performed between an initial rotation 'A and a final rotation 'B is given as: Z'B W D M d': (8.6) 'A

In the following, some types of mechanical work are briefly considered. First of all, we will consider the so-called lifting work, which is done when an object of mass m is lifted against the gravitational field of the earth by the amount h. This is shown in Fig. 8.3. The lifting work performed in this case can be derived from (8.3) by replacing the integral limits A and B with the values 0 and h and defining the force vector F as F D .mg; 0; 0/T D const. It follows: W D mgh:

(8.7)

For this simple example, the work performed is the weight mg multiplied by the distance h. Another type of work is the so-called friction work, which is performed when a body is displaced by a distance s on a rough surface with friction coefficient (Fig. 8.4). Here, the frictional force FR must be overcome, which corresponds to the force F multiplied by the coefficient of friction , which the body exerts on the surface. The friction work performed is then given as: W D FR s D F s;

(8.8)

as one can easily convince oneself by means of (8.3). As a final example, consider the work that is performed when applying a tensile force F on a linear elastic spring. In Fig. 8.5a linear elastic spring (spring constant k) is shown, which is loaded by a tensile force F . The work performed when the spring is tensioned from its unloaded state to the length u and thus the force F Fig. 8.3 Lifting work.

g

h

m Fig. 8.4 Friction work.

s

FR= μF F

236

8 Energy methods

Fig. 8.5 Length change u of a spring with stiffness k loaded by force F .

F k u

F

W u

is increased from the value 0 to its final value F during the deformation process follows with the linear elastic spring law F D ku from (8.3): Zu W D

Zu F duO D

0

k ud O uO D 0

1 2 1 ku D F u: 2 2

(8.9)

Thus, the performed work corresponds to the area below the line in the force-displacement diagram shown in Fig. 8.5, right.

8.1.2 Internal and external work For a solid body problem in elastostatics, a distinction must be made between the external work We on the one hand and the internal work Wi on the other. External work We is work performed by the external loads acting on a solid body along the displacements of the body. Considering the example of the linear elastic spring of Fig. 8.5, the work done in tension W according to (8.9) is an external work. The loads acting on a body cause a state of stress inside the body. Internal work Wi is thus the work which the internal forces perform along the displacements of the body points. This is particularly easy to illustrate with the example of the linear elastic spring already considered. Here the tensile force F will cause an inner spring force with the magnitude F , and the inner work increment dWi is thus the product of inner force F and displacement increment du: dWi D F du:

(8.10)

The total internal work Wi is then the sum of all work increments dWi , i.e.: Zu Wi D

F du: O

(8.11)

0

If a linear-elastic spring with the spring law F D ku is given, then the following results: Zu 1 1 O uO D ku2 D F u: (8.12) Wi D k ud 2 2 0

8.2 Strain energy and complementary strain energy

237

Thus Wi D We holds, i.e. internal work and external work are identical. At this point we introduce the so-called strain energy or internal energy ˘i . If we consider an elastic structure, or even, as already shown in the example of the spring of Fig. 8.5a linear elastic structure, then the internal work Wi is completely stored in the solid as internal energy or as strain energy ˘i and can be completely recovered when the load is removed from the body. The term potential ˘i is also common at this point.

8.1.3 Principle of work and energy The principle of work and energy of elastostatics states that the work We done by the given loads on a solid body is fully converted into internal work Wi in a conservative system. Hence: Wi D We : (8.13) At the same time, it follows that the internal work Wi is stored as internal energy or as strain energy in the solid under consideration and can be completely recovered when the load is removed from the body. Thus, we have: ˘i D We :

(8.14)

This relation is also called the principle of work and energy of elastostatics and has many applications. The principle of work and energy is valid for any frictionless elastic self-contained system. It also states that no energy can be lost in such a system and is thus also called the principle of conservation of energy.

8.2 Strain energy and complementary strain energy 8.2.1 The bar We consider the homogeneous bar of Fig. 8.6 (cross-sectional area A, modulus of elasticity E) under a tensile normal force N . The normal stress xx D NA acts in the bar, it is uniformly distributed over the cross-section. Let the bar be elastic, but not necessarily linearly elastic. We cut an infinitesimal section element of length dx out of the bar and investigate its deformation behavior. At the negative cutting edge the displacement u is present, while at the positive cutting edge the displacement u C du dx D u C u0 dx D u C "xx dx occurs. The normal stress xx dx performs a work increment along the infinitesimal strain d"xx of xx Ad"xx dx. With dV D Adx this can be rewritten as xx d"xx dV . At this point, let us introduce the abbreviation dU0 D xx d"xx , so that dU0 dV is obtained. The quantity dU0 is the so-called incremental strain energy density, it is shown in the stress-strain diagram of Fig. 8.7, top left.

238

8 Energy methods

Fig. 8.6 Bar (top), infinitesimal element (bottom) in its undeformed and its deformed state.

Cross-sectional area A

N

N

x

xx

xx

u u+du=u+ du dx dx =u+ xx dx

dx

Fig. 8.7 Stress-strain diagram with strain energy density U0 and complementary strain energy density UN 0 (left), force-displacement diagram with strain energy ˘i and complementary strain energy ˘N i (right) using the example of the bar under tension; elastic material behavior (top) and the special case of linear elasticity (bottom).

F Wi = Π i U0 U0 Wi = Π i

dU0

u

d F Wi = Π i U0 U0 dU0

d

Wi = Π i

u

The strain energy density U0 is obtained by integrating dU0 over the entire strain history up to the desired strain "xx : Z"xx U0 D

xx dO"xx :

(8.15)

0

The strain energy density U0 is the area below the graph of Fig. 8.7, top left. If the special case of linear elasticity is given, then with Hooke’s law xx D E"xx and with the kinematic relation "xx D u0 the following expression for the normal stress xx follows: xx D E"xx D Eu0 : (8.16) The normal force N of the bar follows from the integration of the normal stress xx over the cross-sectional area A: Z Z Z 0 0 N D xx dA D Eu dA D Eu dA D EAu0 : (8.17) A

A

A

Solving for u0 yields: u0 D

N : EA

(8.18)

8.2 Strain energy and complementary strain energy

239

For the normal stress xx the following well-known relation follows from (8.16): xx D Eu0 D

N : A

(8.19)

The strain energy density U0 then follows as: Z"xx U0 D

Z"Oxx xx dO"xx D E

0

"xx dO"xx D 0

1 2 1 1 E"xx D xx "xx D Eu02 : 2 2 2

(8.20)

The strain energy density is the energy per unit volume introduced into the bar as a result of the strain "xx . In the case of an elastic bar, where the deformation state is fully reversible, it is possible to fully recover this energy and convert it into work. The term density, which appears here, indicates that the strain energy is related to the volume of the considered element. The work increment dWi at the infinitesimal element follows when the strain energy density U0 is multiplied by the volume dV : Z"xx dWi D

xx dO"xx dV D U0 dV:

(8.21)

0

The internal work done Wi and thus the strain energy ˘i stored in the bar follows from the integration over the bar volume. With dV D Adx it follows: Z l Z"xx Wi D

Zl xx dO"xx Adx D A

0

0

U0 dx D ˘i :

(8.22)

0

The strain energy ˘i is shown in the force-displacement diagram of Fig. 8.7, right, for the general case of elasticity and for the special case of linear elasticity. In both cases, it is the surface under the displayed graphs. In the case of linear elasticity we obtain: Z Wi D V

1 U0 dx D E 2

Using u02 D "2xx D

2 xx E2

D

Zl Z 0 A

N2 , E 2 A2

1 Wi D EA 2

Zl 0

1 u dAdx D EA 2 02

Zl

u02 dx D ˘i :

(8.23)

0

this can be written as: N2 1 dx D 2 2 E A 2

Zl 0

N2 dx D ˘i : EA

(8.24)

240

8 Energy methods

If both the normal force N and the extensional stiffness EA are constant, then the following expression results: Wi D ˘i D

N 2l : 2EA

(8.25)

From similar considerations, the complementary strain energy density can be derived: Zxx U0 D "xx dO xx : (8.26) 0

From Fig. 8.7, left, it can be concluded that U0 C U 0 D xx "xx

(8.27)

holds. For linear elasticity we obtain: U0 D

1 xx "xx : 2

(8.28)

One can also deduce this result directly from Fig. 8.7, bottom left. For the case of linear elasticity, the stress-strain diagram is a straight line and the area above this line is exactly the value of U 0 . The inner complementary work W i or the complementary strain energy ˘ i can be determined as: Zxx "xx dO xx dV D U 0 dV: (8.29) dW i D 0

Integration over the bar volume results in: Z l Zxx Wi D

Zl "xx dO xx Adx D A

0

0

U 0 dx D ˘ i :

(8.30)

0

For the special case of linear elasticity we obtain: 1 Wi D 2

Zl 0

N2 dx D ˘ i : EA

(8.31)

Obviously, the expressions for Wi and W i are identical for linear elasticity (cf. Fig. 8.7, bottom).

8.2 Strain energy and complementary strain energy

241

Example 8.1

Consider the truss of Fig. 8.8 loaded by the single force F . We want to determine the strain energy ˘i and the complementary strain energy ˘ i . The horizontal members 1, 2, 4 and 5 have the cross-sectional areas A1 and the elastic modulus E1 . The vertical bars 3, 6 and 8, on the other hand, have the values A2 and E2 . The diagonal bars 7 and 9 have the cross-sectional area A3 and the modulus of elasticity E3 . Hooke’s law is assumed to be valid. To solve the problem we first determine the bar forces Ni (i D 1; 2; : : : ; 9). Obviously, the forces 1, 2 and 8 are zero, so that: N1 D N2 D N8 D 0:

(8.32)

At the left support (see free body image of Fig. 8.8, top right), the bar force N6 is given as: F (8.33) N6 D  : 2 For reasons of symmetry, this result is also valid for bar 3: N3 D 

F : 2

(8.34)

At the free body image of the upper left node (Fig. 8.8, lower right), the vertical sum of forces is obtained as follows: N7 sin 45ı 

F D 0: 2

(8.35)

This results in the following bar force N7 : F N7 D p : 2

(8.36)

For reasons of symmetry, the same result is obtained for the bar force N9 : F N9 D p : 2 Fig. 8.8 Truss under single force F (left), free body images (right).

(8.37)

4

5 l 6

7

N6 9

8

3

0

F N1 2

2

1

F 45°

l

l

F 2

N5 N7

242

8 Energy methods

From the horizontal sum of forces follows: N7 cos 45ı C N5 D 0:

(8.38)

We can determine the bar force N5 as: N5 D 

F : 2

(8.39)

For reasons of symmetry, this result is also valid for the bar force N4 : N4 D 

F : 2

(8.40)

At this point, all bar forces are available, from which we can determine the bar stresses as: Nj : (8.41) xx;j D Aj We have: xx;1 D xx;2 D 0; xx;4 D xx;5 D 

F ; 2A2 F Dp ; xx;8 D 0: 2A3

xx;3 D xx;6 D  F ; xx;7 D xx;9 2A1

(8.42)

From this we can determine the bar strains "xx;j as: "xx;j D

xx;j : Ej

(8.43)

We obtain: "xx;1 D "xx;2 D 0; "xx;4 D "xx;5 D 

F ; 2E2 A2 F Dp ; "xx;8 D 0: 2E3 A3

"xx;3 D "xx;6 D  F ; 2E1 A1

"xx;7 D "xx;9

(8.44)

Thus, the strain energy densities U0;j of the bars can be determined as: U0;j D

1 xx;j "xx;j ; 2

(8.45)

i.e.: U0;1 D U0;2 D 0; U0;4 D U0;5 D

U0;3 D U0;6 D

F2 ; 8E2 A22

F2 F2 ; U D U D ; U0;8 D 0: 0;7 0;9 8E1 A21 4E3 A23

(8.46)

8.2 Strain energy and complementary strain energy

243

From this, the strain energies of the individual bars follow as: Zlj Z

Z ˘i;j D

U0;j dVj D Vj

U0;j dAj dx D U0;j Aj lj ;

(8.47)

0 Aj

thus: ˘i;1 D ˘i;2 D 0; ˘i;4 D ˘i;5 D

F 2l ; 8E2 A2 p 2 2F l D ; 4E3 A3

˘i;3 D ˘i;6 D

F 2l ; 8E1 A1

˘i;7 D ˘i;9

˘i;8 D 0:

(8.48)

The total strain energy stored in the truss then follows from the sum of the contributions of the individual members: p ! 9 X 1 2 F 2l 1 ˘i D ˘i;j D C C : (8.49) 2 2E1 A1 2E2 A2 E3 A3 j D1 For the special case where all bars have identical extensional stiffnesses EA, we get: p F 2l ˘i D .1 C 2/ : (8.50) 2EA To determine the complementary strain energy, we first determine the complementary strain energy densities U 0;j of the individual bars: U 0;j D

1 xx;j "xx;j ; 2

(8.51)

which leads to the following expressions: U 0;1 D U 0;2 D 0; U 0;4 D U 0;5 D

U 0;3 D U 0;6 D

F2 ; 8E2 A22

F2 F2 ; U D U D ; U 0;8 D 0: 0;7 0;9 8E1 A21 4E3 A23

(8.52)

Integration over the bar volume leads to the complementary strain energy: Zlj Z

Z ˘ i;j D

U 0;j dVj D Vj

U 0;j dAj dx D U 0;j Aj lj : 0 Aj

(8.53)

244

8 Energy methods

Thus: ˘ i;1 D ˘ i;2 D 0; ˘ i;4 D ˘ i;5 D

F 2l ; 8E2 A2 p 2 2F l D ; ˘ i;8 D 0: 4E3 A3

˘ i;3 D ˘ i;6 D

F 2l ; ˘ i;7 D ˘ i;9 8E1 A1

(8.54)

The complementary strain energy of the truss is the sum of the fractions of the individual members: p ! 9 X 2 F 2l 1 1 ˘i D ˘ i;j D C C : (8.55) 2 2E A 2E A E A 1 1 2 2 3 3 j D1 If identical extensional stiffnesses EA are given in all bars, then we obtain: ˘ i D .1 C

p F 2l 2/ : 2EA

(8.56)

Since we have assumed linear elasticity for this example, this expression is identical to the strain energy ˘i . J

8.2.2 The Euler–Bernoulli beam We now extend our considerations to the Euler–Bernoulli beam under uniaxial bending about the y-axis with simultaneous normal force action, i.e. the normal force N , the transverse shear force Qz and the bending moment My occur. A prerequisite for uniaxial bending is that one of the cross-sectional principal axes coincides with the z-axis and thus no biaxial bending occurs. The beam has the extensional stiffness EA and the flexural stiffness EIyy . The area integrals occurring in the following are the moment of inertia Iyy , the static moment Sy and the cross-sectional surface A: Z

Z AD

dA; A

Sy D

Z zdA;

Iyy D

A

z 2 dA:

(8.57)

A

The displacement field of the Euler–Bernoulli beam with simultaneous bar action is: uP D u  zP w 0 ;

wP D w:

(8.58)

Herein, P is an arbitrary point of the cross-section at the location zP , and u and w are the displacements of the centroid axis in the x-direction and the z-direction. The normal strain "xx can be determined as follows: "xx D

duP D uP0 D u0  zP w 00 : dx

(8.59)

8.2 Strain energy and complementary strain energy

245

The normal stress xx can be determined from Hooke’s law xx D E"xx as:   xx D E u0  zw 00 : (8.60) In the sense of generality of the explanations we want to drop the indexing concerning the point P from this point on. The normal force N follows as: Z Z Z N D xx dA D Eu0 dA  Ezw 00 dA A

A

D Eu

0

A

Z

dA  Ew

00

A

Z zdA:

(8.61)

A

Since the cross-sectional axes are assumed to be the principal axes (i.e., Sy D 0), the second term in (8.61) vanishes, and the following expression remains: N D EAu0 :

(8.62)

The bending moment My can be determined in the same way: Z Z Z My D xx zdA D Eu0 zdA  Ez 2 w 00 dA A

A

D Eu

0

A

Z

zdA  Ew A

00

Z

z 2 dA:

(8.63)

A

With Sy D 0 and the moment of inertia Iyy follows: My D EIyy w 00 :

(8.64)

Substituting (8.62) and (8.64) into (8.60) gives the following expression for the normal stress xx : My N z: (8.65) C xx D A Iyy The strain energy density U0 for the linear elastic Euler–Bernoulli beam follows as: Z"xx U0 D

Z"xx xx dO"xx D E

0

"Oxx dO"xx D 0

2 1 2 1  E"xx D E u0  zw 00 : 2 2

(8.66)

The strain energy ˘i follows from integration over the beam volume dV D dAdx: Z ˘i D V

1 U0 dV D 2

Zl Z 0 A

 E u02  2z u0w 00 C z 2 w 00 2 dAdx;

(8.67)

246

8 Energy methods

and taking into account Sy D 0: 1 ˘i D 2

Zl 0

1 EAu dx C 2 02

Zl

2

EIyy w 00 dx:

(8.68)

0

This can be expressed by the normal force N and the bending moment My as: 1 ˘i D 2

Zl 0

N2 1 dx C EA 2

Zl

My2 EIyy

0

dx:

(8.69)

Similarly, a shear component due to shear stress xz and shear strain xz can be formulated for the inner potential ˘i : 1 ˘i D 2

Zl Z 0 A

1 xz xz dAdx D 2

Zl Z 0 A

2 xz dAdx: G

(8.70)

This part must not be neglected a priori, since even if the shear strain xz vanishes due to the kinematic assumptions of the Euler–Bernoulli beam (see Chap. 4), the shear stress xz must occur in order to maintain equilibrium (see Chap. 6). SubstiVz Sy tuting the relation xz D  Iyy b for a rectangular cross section of width b in (8.70), we obtain:   Zl Z Vz Sy 2 1 1 dAdx; (8.71)  ˘i D 2 G Iyy b 0 A

or after integration: 1 ˘i D 2

Zl 0

Vz2 dx: GAeff

(8.72)

Herein, Aeff is the so-called effective cross-sectional area, which is defined as 1 1 D 2 Aeff Iyy

Z A

Sy2 b2

dA:

(8.73)

This is the effective area of a cross-section under transverse shear that is available for load transfer. It can be calculated with the help of the so-called shear correction factor K and the cross-sectional area A: Aeff D KA:

(8.74)

8.2 Strain energy and complementary strain energy

Hence:

Zl

1 ˘i D 2

247

Vz2 dx: KGA

0

(8.75)

In combination with (8.69) we obtain: 1 ˘i D 2

Zl 0

N2 1 dx C EA 2

Zl 0

My2

1 dx C EIyy 2

Zl 0

Vz2 dx: KGA

(8.76)

The complementary strain energy density can be determined in a similar manner. We have: Zxx Zxz U0 D "xx dO xx C xz dOxz : (8.77) 0

0

The second term appearing here again must not be neglected a priori. With Hooke’s law xx D E"xx and xz D Gxz then follows: Zxx U0 D 0

Zxz

xx dO xx C E

0

xz 2 2 dOxz D xx C xz : G 2E 2G

(8.78) V S

z y The complementary strain energy ˘ i can be determined with xz D  Iyy b e.g. for a rectangular cross-section of width b by integration over the beam volume:

Zl Z 

Z ˘i D

U 0 dV D V

Zl Z "

D 0 A

0 A

 2 xx 2 C xz dAdx 2E 2G



My 1 N z C 2E A Iyy

2

 # Vz Sy 2 1 C  dAdx; 2G Iyy b

(8.79)

or after integration: 1 ˘i D 2

Zl 0

N2 1 dx C EA 2

Zl 0

My2

1 dx C EIyy 2

Zl 0

Vz2 dx: KGA

(8.80)

The comparison between (8.80) and (8.76) shows that both expressions are identical. The shear correction factor K is K D 56 for a rectangular cross-section, independent of the concrete dimensions of the cross-section. However, the general determination of the shear correction factor for arbitrary cross sections cannot be dealt with in this book.

248

8 Energy methods

8.2.3 Bar under torsion We consider a straight bar of length l subjected to the torsional moment Mx . The torsional stiffness GIT is given. The work increment dWi is obtained with the angle of twist # as: 1 (8.81) dWi D Mx d#: 2 From the constitutive law (7.7) for torsion GIT # 0 D Mx follows: Mx dx : GIT

(8.82)

 2 1 Mx2 1 dx D GIT # 0 dx: 2 GIT 2

(8.83)

d# D Equation (8.81) then transforms into: dWi D

Integrating over the bar length l, we obtain the strain energy of the bar due to torsion: Zl Zl Zl  2 Mx2 1 1 1 0 dx D Mx # dx D GIT # 0 dx: (8.84) ˘i D 2 GIT 2 2 0

0

0

8.2.4 Combined load If there is a situation with simultaneous axial strain, bending about the two principal axes y and z, transverse shear forces in both principal axial directions, and torsion, then the strain energy or inner potential can be formed by superposition of the individual components as follows: 1 ˘i D 2 C

Zl

1 EAu dx C 2 02

0

1 2

Zl

Zl 0

1 EIyy w dx C 2 00 2

Zl

2

EIzz v 00 dx

0

 2 GIT # 0 dx;

(8.85)

0

or formulated in the occurring forces and moments: 1 ˘i D 2

Zl 0

1 C 2

N2 1 dx C EA 2 Zl 0

Vy2

Zl 0

My2

1 dx C EIyy 2

1 dx C Ky GA 2

Zl 0

Zl 0

Mz2 dx EIzz

Vz2 1 dx C Kz GA 2

Zl 0

Mx2 dx: GIT

(8.86)

8.3 Application of the principle of work and energy

249

The complementary strain energy ˘ i is: 1 ˘i D 2

Zl 0

1 C 2

N2 1 dx C EA 2 Zl 0

Zl 0

Vy2

My2

1 dx C EIyy 2

1 dx C Ky GA 2

Zl 0

Zl 0

Mz2 dx EIzz

Vz2 1 dx C Kz GA 2

Zl 0

Mx2 dx: GIT

(8.87)

Here Ky and Kz are the shear correction factors assigned to the two transverse shear forces Vy and Vz . Further details are not given here.

8.3 Application of the principle of work and energy to the determination of elastic deformations We want to use the principle of work and energy (8.14), i.e. the equality of strain energy and external work ˘i D We , to calculate deformations of bar and beam structures under single forces and moments. For this purpose, we consider the truss of Fig. 8.9 consisting of two members. Both members have the same extensional stiffness EA. The vertical displacement w of the force application point is to be determined. For this purpose, the principle of work and energy ˘i D We is employed. From the free body image of Fig. 8.9, the bar forces can be determined F F as N1 D tan.˛/ and N2 D  sin.˛/ . The strain energy ˘i can then be determined as follows: Zl

Z

(8.88)

  F 2l 1 1 C : 2EA tan2 .˛/ cos3 .˛/

(8.89)

0

N12 1 dx C EA 2

l cos.˛/

N22 dx: EA

1 ˘i D 2

0

If we insert the bar forces N1 and N2 , we obtain: ˘i D

Fig. 8.9 Truss under single force F .

F 1

α

w

N1

α

2

l

N2

F

250

8 Energy methods

The external work We is obtained as: We D

1 F w: 2

(8.90)

If we evaluate the principle of work of energy ˘i D We , we obtain:   F 2l 1 1 1C D F w: 2EA tan2 .˛/ cos3 .˛/ 2

(8.91)

This expression can be solved directly for the displacement w: wD

  1 Fl 1 C : EA tan2 .˛/ cos3 .˛/

(8.92)

The principle of work and energy can be used for the simple determination of deformations in structures under single forces and moments. However, this procedure is limited in scope, since it can only be used to determine displacements or rotations at the points of application of the individual forces and moments, and furthermore only those deformations whose direction also corresponds to the direction of action of the applied force or moment. Example 8.2

Consider the linear elastic cantilever beam of Fig. 8.10, which is loaded by the single force F . Let the constant bending stiffness EIyy be given. The vertical displacement of the force application point is to be determined. When determining the displacement, only the bending moment My is to be taken into account in the strain energy. First, the bending moment My .x/ is determined, it is shown in Fig. 8.10. The strain energy follows as: 1 ˘i D 2

Zl 0

My2 EIyy

dx D

F 2l 3 : 6EIyy

(8.93)

The external work We is given as: We D

1 F w: 2

(8.94)

The deflection w then follows from the principle of work and energy ˘i D We as: F l3 : (8.95) wD 3EIyy J

8.3 Application of the principle of work and energy

F

Fl

F

x

251

0

w

F

l

Fl

z

My(x)

Fl

F

(-) My

x

Fig. 8.10 Cantilever beam under single force F .

Example 8.3

We consider the simply supported linear elastic beam (constant bending stiffness EIyy ) of Fig. 8.11, which is loaded at its left support by the edge moment M0 . The rotation of the left support is to be determined. The strain energy of the beam reads: 1 ˘i D 2

Zl 0

My2 EIyy

The external work We is: We D

dx D

M02 l : 6EIyy

(8.96)

1 M': 2

(8.97)

From ˘i D We the rotation ' of the left support follows as: 'D

M0 l : 3EIyy

(8.98) J

Fig. 8.11 Beam under edge moment M0 .

M0

l

M0

(+)

252

8 Energy methods

Fig. 8.12 Truss under single force, free body images to determine support reactions and bar forces.

l

l

4 45°

5

N5 N2

3

l 1

45°

2

F 2

N4

F F

F 2

0 F 2

F 2 F

F F 2

F 2

45°

N1

Example 8.4

Consider the truss of Fig. 8.12. The horizontal deflection w of the point of application of the force F is to be determined using the principle of work and energy. All members have identical extensional stiffnesses EA. The determination of the support reactions and the bar forces is left to the reader. Here, only the result for the strain energy ˘i is reported: 9

1X ˘i D 2 j D1

Zlj 0

Nj2 Ej Aj

9

dxj D

p F 2l 1 X Nj lj D .1 C 2 2/ : 2 j D1 EA 4EA 2

(8.99)

The external work is:

1 F w: (8.100) 2 The principle of work and energy ˘i D We then leads to the horizontal deflection w of the force application point as follows: We D

p Fl : w D .1 C 2 2/ 2EA

(8.101) J

Example 8.5

Consider the angled beam shown in Fig. 8.13, left, with constant extensional stiffness EA and constant bending stiffness EIyy . The deflection of the force application point in the direction of the force F is to be determined using the principle of work and energy ˘i D We .

8.4 The principle of virtual forces

253

Fig. 8.13 Static system (left), normal force and bending moment diagrams (right).

w

F

h

EA, EIyy

0

F

N(x)

(+)

x

My(x)

l

(-) (-)

Fh

Fh

The distributions of the normal force N.x/ and the bending moment M.x/ are shown in Fig. 8.13, right. The strain energy ˘i can be determined as: 1 ˘i D 2

Zl 0

N2 1 dx C EA 2

Zh 0

N2 1 dx C EA 2

resulting in: ˘i D F 2



Zl 0

My2

1 dx C EIyy 2

l h2 C 2EA 2EIyy



h Cl 3

Zh 0

My2 EIyy

dx;

(8.102)

 :

(8.103)

Herein, we have neglected the influence of the transverse shear force Vz . With the external work We D 12 F w, the deflection w follows from ˘i D We :

l h2 wDF C EA EIyy



h Cl 3

 :

(8.104) J

8.4 The principle of virtual forces There are work and energy principles that are based on the consideration of socalled virtual forces. For this purpose, we first introduce the concepts of virtual forces and the so-called complementary virtual work.

8.4.1 Formulation for the beam We consider a body in its equilibrium state. The body is subjected to a virtual equilibrium group F , which performs the complementary virtual work ıW along the actual displacements: ıW D ıF u:

(8.105)

254

8 Energy methods

Fig. 8.14 Cantilever beam under virtual forces.

δFV l δFH

x δFV

δFH EA, EIyy , KGA

δFV

l

z

As an example, consider the cantilever beam of Fig. 8.14 (length l, stiffnesses EA, EIyy , KGA). We assume that the given axis system is a principal axis system, and only uniaxial bending with respect to the y-axis is considered here. Let the beam be clamped at its left end, and at its free end the virtual forces ıFV and ıFH are applied as shown. The virtual forces will cause the virtual horizontal support force ıFH and the virtual vertical support force ıFV . In addition, the virtual support moment ıFV l at the clamping point occurs. These virtual force quantities thus form an equilibrium group, all equilibrium conditions are satisfied. However, the virtual complementary work ıW a is only caused by the two virtual forces ıFH and ıFV at the cantilever end, because no displacements and thus no virtual work are possible at the clamping point. Thus: ıW a D ıFV w.x D l/ C ıFH u.x D l/: (8.106) The virtual complementary internal work ıW i follows from the virtual normal stress ıxx and the virtual shear stress ıxz and the associated real strains: Z ıW i D ."xx ıxx C xz ıxz /dV : (8.107) V

We divide the strain "xx into the two parts "0xx and "1xx . Here "0xx is related to the center of gravity of the beam and thus describes a bar action. The term "1xx describes the linear variability of the strain over z and thus describes a curvature of the beam due to a bending moment. The volume integral is decomposed into an integral over the cross-sectional area A and an integral over the longitudinal axis x: Zl Z ıW i D



  "0xx C z"1xx ıxx C xz ıxz dAdx:

(8.108)

0 A

Integration over the cross-sectional area A results in: Zl ıW i D 0



 "0xx ıN C "1xx ıMy C xz ıVz dx:

(8.109)

8.4 The principle of virtual forces

255

Herein, ıN , ıMy and ıVz are the virtual forces and moment induced by the virtual external forces. They perform virtual work along the real strains. If we assume linear elasticity, then we obtain for "0xx and "1xx : "0xx D Thus:

Zl  ıW i D 0

N ; EA

"1xx D

M : EIyy

 My ıMy N ıN Vz ıVz C C dx: EA EIyy KGA

(8.110)

(8.111)

The virtual force quantities have the following properties:  The virtual force quantities are virtual and do not exist in reality.  The virtual force quantities are infinitesimally small.  The virtual force quantities must satisfy equilibrium conditions and obey all given stress boundary conditions. The principle of virtual forces requires the equality of the virtual internal complementary energy and the virtual external complementary work: ıW i D ıW a :

(8.112)

This principle applies to arbitrary virtual force quantities satisfying given stress boundary conditions. The principle of virtual forces applies to any material behavior.

8.4.2 The unit load theorem A very useful theorem that can be derived from the principle of virtual forces is the so-called unit load theorem. We consider an arbitrary beam loaded by the two virtual forces ıFH and ıFV and the virtual single moment ıM . The corresponding real displacements and rotation are u, w and '. The principle of virtual forces ıW i D ıW a is then: Zl  0

 My ıMy N ıN Vz ıVz C C dx D uıFH C wıFV C 'ıM: EA EIyy KGA

(8.113)

In the case of a truss with m bars loaded exclusively by single forces at the nodes of the members, the following expression remains due to the constancy of the member forces: m X Ni ıNi li D uıFH C wıFV : (8.114) .EA/i i D1 Here li and .EA/i are the length and the extensional stiffness of the bar i.

256

8 Energy methods

δFV

F EA 1

δFH w

45°

2l

l EA

u

2

Fig. 8.15 Truss under single force F (left), deformed structure under virtual forces with real nodal displacements (right).

The practical value of the unit load theorem proves itself in the determination of deformations of static systems. For illustration, we consider the truss of Fig. 8.15. Let the load consist of the vertically acting single force F , and we want to determine the two nodal displacements u and w as indicated. Both members have identical constant extensional stiffnesses EA. To determine the two unknown nodal displacements u and w, we apply the two virtual forces ıFV and ıFH at the node. The principle of virtual forces is then: m X Ni ıNi li i D1

.EA/i

p N1 ıN1 l N2 ıN2 2l D C D ıFH u C ıFV w: EA EA

(8.115)

To determine the displacement u, we set the virtual force ıFV to zero: p N1 ıN1 l N2 ıN2 2l C D ıFH u: EA EA

(8.116)

Virtual forces may be arbitrary within the limits of the static system under consideration. Consequently, for ıFH a unit force ıFH D 1 can be assumed, and we obtain: p N1 ıN1 l N2 ıN2 2l C D u: (8.117) EA EA Obviously, the nodal displacement u can be read directly from (8.117). Here the forces N1 and N2 are the real bar forces due to the single force F . The forces ıN1 and ıN2 are the bar forces due to the virtual unit force ıFH D 1. We have: p N1 D F; N2 D  2F; ıN1 D 1; ıN2 D 0: (8.118) For the nodal displacement u thus follows from (8.117): uD

F  .1/  l Fl D : EA EA

(8.119)

8.4 The principle of virtual forces

257

The same procedure can be followed to determine the displacement w. In (8.115) we set the virtual single force ıFH to zero, and the virtual single force ıFV is assumed as a unit force ıFV D 1. It follows: p N1 ıN1 l N2 ıN2 2l C D w: (8.120) EA EA The real bar forces N1 and N2 due to F are already given in (8.118). The virtual bar forces ıN1 and ıN2 caused by the virtual unit force ıFV D 1 are determined as: p ıN1 D 1; ıN2 D  2: (8.121) We can thus calculate the nodal displacement w as follows: p Fl w D .1 C 2 2/ : EA

(8.122)

Obviously, the unit load theorem allows us to determine deformations in static systems quite easily. In formalized form, this procedure is also called the force method, which we will discuss later. Example 8.6

Consider the beam shown in Fig. 8.16. The beam has the length l and the constant bending stiffness EIyy . The load is given in the form of a single force F acting at the center of the beam, and the deflection w in the center of the beam is to be determined by means of the principle of virtual forces, wherein we want to assume the contribution of the transverse shear force Vz as being negligible. We first determine the moment distribution of the beam due to the given single force F , it is shown in Fig. 8.16, left. To calculate the deflection w, we also apply a virtual single force ıF at the center of the beam and determine the corresponding moment diagram, shown in Fig. 8.16, right. The principle of virtual

F x1

δF

EIyy

l/2

x2

x1

EIyy

l/2

l

M0

l

M1

(+) Fx 2 1

x2

Fl 4

- F x2 + Fl 2 4

(+) 1x 2 1

l 4

- 1 x2 + l 2 4

Fig. 8.16 Beam under single force F (top left), beam under virtual force ıF D 1 (top right), corresponding moment diagrams (bottom).

258

8 Energy methods

forces is:

Zl 0

My ıMy dx D wıF: EIyy

(8.123)

Herein, My is the bending moment due to the given force F , ıMy is the bending moment due to the virtual force ıF . If we set ıF as unit load ıF D 1, we are left with: Zl My ıMy dx D w: (8.124) EIyy 0

Thus, the deflection we are looking for can be read from (8.124) in a simple and straightforward way. It is now our task to perform the integration of the two presented moment functions prescribed in (8.124). Since both moment lines My and ıMy have discontinuities at the center of the beam, the two reference axes x1 and x2 are introduced as shown in Fig. 8.16 and the integral in (8.124) is decomposed into two partial integrals. It follows: l

Z2 wD 0

l

M.x1 /ıM.x1 / dx1 C EIyy

Z2 0

M.x2 /ıM.x2 / dx2 : EIyy

(8.125)

A very common notation is to not denote the virtual quantities with the symbol ı, but rather to denote the moment distributions as M0 (bending moment due to the given load) and M1 (bending moment due to the virtual unit force): l

Z2 wD 0

l

M0 .x1 /M1 .x1 / dx1 C EIyy

Z2 0

M0 .x2 /M1 .x2 / dx2 : EIyy

(8.126)

If we perform the prescribed integrations, we obtain: l

Z2 wD 0

l

1 F 1 x1 x1 dx1 C EIyy 2 2

Z2 0

   1 Fl l 1 F  x2 C dx2 :  x2 C EIyy 2 4 2 4 (8.127)

Evaluation yields: wD

F l3 : 48EIyy

(8.128) J

8.4 The principle of virtual forces

259

8.4.3 Use of integral tables The preceding example has shown that the determination of deflections with the help of the unit load theorem is done by integration of products of state variables, here the products of the moment distributions M0 .x1 / and M1 .x1 / as well as M0 .x2 / and M1 .x2 /. For more complex systems, however, this approach quickly becomes impractical due to the necessity of solving integrals. One can make use of the fact that diagrams of the state variables of static systems are in most cases built up of elementary standard cases, for which the results of the integrations can be summarized in so-called integral tables. This is shown below at the example of two linear functions f .x/ and g.x/ defined in the limits of x D 0 and x D l. Let the boundary values of f .x/ be f .x D 0/ D 0 and f .x D l/ D F , and let those of g.x/ be g.x D 0/ D 0 and g.x D l/ D G. The two functions f .x/ and g.x/ can then be formulated as follows: f .x/ D The integral

F x; l

g.x/ D

G x: l

(8.129)

Zl f .x/g.x/dx

(8.130)

0

then gives the value 13 F Gl. One can always reuse this result when one has to perform the integration of the product of two linear functions as defined in (8.129). Furthermore, for two linear functions f .x/ and g.x/ with boundary values f .x D 0/ D F and f .x D l/ D 0 as well as g.x D 0/ D G and g.x D l/ D 0, the value of the integral (8.130) is the same. Thus, for Example 8.6, using the result provided above, one can immediately evaluate the two integrals without first multiplying the two functions and actually solving the integrals. For the deflection of the beam in Example 8.6 follows: l

Z2

l

Z2 M0 .x1 /M1 .x1 / M0 .x2 /M1 .x2 / dx1 C dx2 wD EIyy EIyy 0 0   1 Fl l l 1 1 Fl l l D    C    EIyy 3 4 4 2 3 4 4 2 D

F l3 : 48EIyy

(8.131)

Results for the integrals of two multiplied functions are summarized in so called integral tables. A selection of results for frequently occurring functions f .x/ and g.x/ is shown in Fig. 8.17.

260

8 Energy methods

g(x)

G

f(x)

G l

G

G

G

l

l

l

FGl

1 FGl 2

1 FGl 2

2 FGl 3

1 FGl 2

1 FGl 3

1 FGl 6

1 FGl 3

1 FGl 2

1 FGl 6

1 FGl 3

1 FGl 3

1 FGl 2

1 FGl(1+ ) 6

1 FGl(1+ ) 6

l 2 (F1 +F2 )G

l 6 (F1 +2F2 )G

l 6 (2F1 +F2 )G

F l

F l

F l

F l

l

1 FGl(1+ 3

)

l

F1 F2 F2

F1

l 3 (F1 +F2 )G

l

F

2 FGl 3

1 FGl 3

1 FGl 3

8 15FGl

F

1 FGl 3

1 FGl 4

1 12FGl

1 5 FGl

G

1 FGl 3

1 12FGl

1 FGl 4

1 5 FGl

Gl(F +4F +F ) 3 2 6 1

Gl(2F +F ) 6 2 3

Gl(F +2F ) 2 6 1

Gl 15(F1 +8F2 +F3 )

Quadratic functions

l

G l

F l

F1

F2

F3 l

Fig. 8.17 Integral table

Rl 0

f .x/g.x/dx for a selection of typical cases.

8.5 The force method

261

8.5 The force method The analysis of static systems by means of the principle of virtual forces in connection with the unit load theorem is often summarized under the term force method. In this section, we will formalize this procedure and look at it in more depth, highlighting other applications besides the determination of deformations.

8.5.1 Determination of deformations of statically determinate systems We start our considerations with the analysis of deformations of statically determinate truss structures. For this purpose, consider a truss consisting of m members. The truss is loaded in such a way that constant normal forces Ni (i D 1; 2; : : : ; m) occur in the members. Let the length of the bar i be li , and let the constant extensional stiffness EIi be present in bar i. Then, using the force method, one can calculate a deformation w at any point of the truss as follows:  For the given truss, the bar forces Ni (i D 1; 2; : : : ; m) are determined as a result of the given real loads.  Furthermore, the bar forces ıNi D N i are determined due to a virtual single force F D 1 at the considered location in the direction of the displacement w to be determined.  The displacement w can be calculated according to the following formula: wD

m X Ni N i li : .EA/i i D1

(8.132)

The procedure is illustrated at the example of the truss shown in Fig. 8.18, on which the indicated deflection w is to be determined by means of the force method.

F 5

6 l 1

7

8

4

1

7

8

9 2 l

w

5

6

2F

9 3 l

3

2 F =1

Fig. 8.18 Truss under point forces (left), truss under virtual point load (right).

4

262

8 Energy methods

We first determine the bar forces Ni (i D 1; 2; : : : ; 9) due to the two point forces and obtain: F 3F F N1 D ; N2 D 2F; N3 D 0; N4 D  ; N5 D ; 2 2 2 F F 3F (8.133) N6 D ; N7 D  p ; N8 D p ; N9 D F: 2 2 2 The bar forces N i due to the virtual unit force F D 1 are obtained as: 1 1 1 N 1 D  ; N 2 D 0; N 3 D 0; N 4 D  ; N 5 D  ; 2 2 2 1 1 1 N 6 D  ; N 7 D p ; N 8 D p ; N 9 D 0: (8.134) 2 2 2 The deflection w can then be determined as follows: 9 X Ni N i li wD .EA/i i D1 

1 3F 1 p F 1 3F 1 F 1 F 1 F 1 p D 2l C p p 2l :  lC l l lp p EA 2 2 2 2 2 2 2 2 2 2 2 2 (8.135) This results in: wD

p Fl 2 : EA

(8.136)

Example 8.7

Consider the truss shown in Fig. 8.19. The deformed truss is to be determined by means of the force method. Fig. 8.19 Truss under point load F (top), deformed structure (bottom, displacements Fl in unit EA ).

VII

l

6

VI

5

V

4 7

8

I

9 1

10

11 2

II

l

F

III

IV

l

l 2.00

3

3.00 17.66

6.83

F 3.00 5.83

5.00 16.66

6.00 29.48

8.5 The force method

263

Table 8.1 Bar forces for the individual load cases. Bar li Ni N i;I;v N i;II;h N i;II;v N i;III;h N i;III;v N i;IV;h N i;IV;v N i;V;h N i;V;v N i;VI;h N i;VI;v

1

2

3

l 3F 0 1 1 1 2 1 3 0 2 0 1

l 2F 0 0 0 1 1 1 2 0 1 0 0

l F 0 0 0 0 0 1 1 0 0 0 0

4 p 2l p 2F 0 0 0 0 0 0 p 2 0 0 0 0

5 l F 0 0 0 0 0 0 1 1 0 0 0

6 l 2F 0 0 0 0 1 0 2 1 1 1 0

7 l 0 F 0 0 0 0 0 0 0 0 0 0

8 p 2l p 2F 0 0 p 2 0 p 2 0 p 2 0 p 2 0 p 2

9 l F 0 0 0 0 1 0 1 0 1 0 1

10 p 2l p 2F 0 0 0 0 p 2 0 p 2 0 p 2 0 0

11 l F 0 0 0 0 0 0 1 0 1 0 0

To determine the deformed truss, only the horizontal and vertical displacements of the truss nodes are required, between the nodes the deformation figure will consist of straight lines. To determine the nodal displacements, first the member forces Ni (i D 1; 2; : : : ; 11) due to the applied single force F are determined. After that, horizontal and vertical virtual single forces F D 1 are applied at each node, and the virtual bar forces N i are calculated, which we want to identify with Roman numbers for the node number and with h and v (horizontal and vertical), respectively. The virtual forces are applied acting to the right and downward, respectively, in all cases. The determined bar forces are summarized in Table 8.1. The required displacements can be determined as follows: m X Ni N i li : (8.137) ıD .EA/i i D1 We obtain: Fl ıI;v D 0; ıII;h D 3 ıI;h D 0; ; EA  p Fl Fl Fl D 5:83 ; ıIII;h D 5 ; ıII;v D 3 C 2 2 EA EA EA  p Fl Fl Fl D 16:66 ; ıIV;h D 6 ; ıIII;v D 11 C 4 2 EA EA EA  p Fl Fl Fl D 29:49 ; ıV;h D 3 ; ıIV;v D 21 C 6 2 EA EA EA  p Fl Fl Fl D 17:66 ; ıVI;h D 2 ; ıV;v D 12 C 4 2 EA EA EA  p Fl Fl D 6:83 : ıVI;v D 4 C 2 2 EA EA The deformed truss is shown in Fig. 8.19, bottom. J

(8.138)

264

8 Energy methods

q

F

F F =1

x

x

w

l

l

l

Fl (-)

2Fl Fl

M0 due to F(x=2l) M0 due to F(x=l)

Fl

(-)

=

Fl (-)

(-)

Fl (-)

M0 due to q

ql 2 8

0

2ql 2

1 ql 2 (-)

2

=

(+)

1 ql 2 2 (-) (-)

l (-)

l (-)

ql 2 8

(+)

2ql 2 (-)

2l M1 due to F =1

l

=

l (-)

l (-)

Fig. 8.20 Beam under line load q and point loads F (top left), beam under virtual point load F D 1 (top right), corresponding moment diagrams (bottom).

The force method can also be applied to the determination of deformations of beams and beam systems. As an introductory example, we consider the cantilever beam of Fig. 8.20. The cantilever beam has the length 2l and the constant bending stiffness EIyy . The load consists of the constant line load q and the two point forces F . The deflection w at the end of the cantilever beam is to be determined. We first determine the moment distribution M0 due to the applied load and the moment diagram M1 due to the virtual single force F D 1 at the end of the cantilever wherein we decompose the moment diagrams into elementary segments as shown in Fig. 8.20. We also use the superposition principle and determine the moment line M0 for the respective partial loads separately. The evaluation is performed section by section on the intervals 0  x  l and l  x  2l. The deflection w1 due to the single force F at x D 2l follows as: w1 D

1 EIyy



 1 3 1 3 1 3 1 8 F l3 : F l C F l C F l C F l3 C F l3 D 3 2 2 3 3 EIyy

(8.139)

For the deflection w2 due to the force F at x D l we obtain: w2 D

1 EIyy



 1 3 1 3 5 F l3 : Fl C Fl D 3 2 6 EIyy

(8.140)

8.5 The force method

265

The deflection w3 due to the line load q follows as: w3 D

1 EIyy

D2



1 4 2 4 1 1 1 1 1 ql C ql  ql 4 C ql 4 C ql 4  ql 4 C ql 4  ql 4 12 3 24 4 12 6 24

ql 4 : EIyy



(8.141)

The total deflection w is then the sum of the partial deflections w1 , w2 , w3 : w D w1 C w2 C w3 D

8 F l3 5 F l3 ql 4 1 C C2 D 3 EIyy 6 EIyy EIyy EIyy



 21 3 F l C 2ql 4 : 6 (8.142)

Example 8.8

Consider the frame of Fig. 8.21, top left. The frame is subjected to a constant line load q and has the constant bending stiffness EIyy . We want to determine the mutual displacement w of the vertical frame members at their half height. We apply opposing virtual loads to determine the mutual displacement w of the vertical frame members, as shown in Fig. 8.21, top right, and determine the moment diagrams for both the given load and the virtual load case. They are

q

h 2

F=1

F=1

h 2 l 2

M0

(-) 1ql 2 (-) 8 1 ql =16

l 2

(-)

(-)

M1 1 2 (-) 8ql

2

(-)

0 1h 4

(-)

0 1h 4 (-)

(-)

Fig. 8.21 Frame structure under line load q, determination of the mutual horizontal displacement at half height of the vertical frame members.

266

8 Energy methods

shown in Fig. 8.21, bottom. Superposition of the moment lines M0 and M1 then gives the mutual displacement of the considered points as follows: wD

ql 2 h2 : 64EIyy

(8.143) J

Example 8.9

For the beam shown in Fig. 8.22, the rotation of the left support is to be determined. The beam is loaded by the line load q and has the length l and the constant bending stiffness EIyy . Superposition of the two moment lines M0 and M1 yields the rotation ' as follows: 1 ql 2 1 ql 3   : (8.144) 'D 1l D EIyy 3 8 24EIyy J Example 8.10

For the beam structure of Fig. 8.23, the mutual rotation ' of the beam segments intersecting in the joint is to be determined. The beam has the constant bending stiffness EIyy . The superposition of the two moment diagrams M0 and M1 leads to the desired mutual rotation ' as follows: 'D

 q 3 l2  2.3l2 C 2l1 /l12 : 24EIyy

(8.145) J

Example 8.11

Consider the beam of Fig. 8.24 (length 2l), which is loaded by two edge moments MA and MB . The rotations 'A and 'B are to be determined. We also want to clarify how large the moment MB must be for a given MA so that 'A D 'B . In the left half, the beam has the bending stiffness 2EIyy , in the right half the value EIyy is given. In order to apply the force method, we first determine the moment line M0 due to the two edge moments MA and MB , as shown in Fig. 8.25. The bending stiffness EIyy is not constant along the beam length. Therefore, it is convenient to decompose the moment diagram as shown in Fig. 8.25 into triangles and rectangles of length l. In this way, the later superposition of the moment diagrams is facilitated. To determine the rotation 'A , a virtual single moment is applied to the left support (Fig. 8.26) and the corresponding moment diagram is determined. In or-

8.5 The force method

267

q

M=1

x l

(+)

1

M0

1ql 2 8

(+)

M1

Fig. 8.22 Determination of the rotation of the left support of a simply supported beam under line load q.

q

M=1

x l1

1 ql l 2 12 (-)

l2 l

1+ l 1

M0

(+)

1

2

M1

1 ql 2 8 2 Fig. 8.23 Determination of the mutual rotation of the beam segments at the joint. Fig. 8.24 Beam under two edge moments MA and MB .

MA

2EIyy

EIyy

φB

MB

x φ A l

l

der to determine the rotation 'A the moment lines M0 and M1 are superimposed: Z

M0 M1 dx EIyy   1 MA 1 1 MA 1 1 MA 1 MA 1 l   C   C   C  D 2EIyy 3 2 2 2 2 2 2 2 2 2 2     l 1 MB 1 1 MB 1 1 MB 1 1 MB 1 l    C      C   2EIyy 6 2 2 2 2 2 EIyy 6 2 2 2 2 2 1 MA 1 MB l 3MA l l    : (8.146)  D C EIyy 3 2 2 8EIyy 4EIyy

'A D

In quite the same way, the rotation 'B can be determined (Fig. 8.27). The superposition of the moment diagram M0 and the moment diagram M1 results in the

268 Fig. 8.25 Moment diagram M0 due to the two edge moments MA and MB .

8 Energy methods

MB

MA x l

MA

l

(+) (-)

=

MA

MB

(+)

+ (-)

MB

=

MA 2 (+)

+

MB 2

(-)

MB 2 Fig. 8.26 Moment diagram M1 due to a virtual single moment at the left support.

1 x l

1

l

(+)

=

1 2

1 2

(+)

Fig. 8.27 Moment diagram M1 due to a virtual single moment at the right support.

1 x l

l

(-)

1 2

=

1 1 2

(-)

8.5 The force method

269

rotation 'B : 'B D 

MA l 5MB l C : 4EIyy 8EIyy

(8.147)

By equating the two expressions (8.146) and (8.147), one can determine the moment MB as a function of MA such that the two rotations 'A and 'B are identical. It follows: 3MA l MB l MA l 5MB l  D C : (8.148) 8EIyy 4EIyy 4EIyy 8EIyy This expression can be solved for MB : MB D

5 MA : 7

(8.149) J

Example 8.12

Consider the beam of Fig. 8.28 loaded by a single force F , which exhibits different bending stiffnesses EI1 D 3EIyy , EI2 D 2EIyy , EI3 D EIyy in its segments. The displacements w and wF as well as the mutual rotation ' of the two beam segments intersecting at the joint are to be determined using the force method. The moment diagram due to the given load is shown in Fig. 8.28, bottom. We apply the virtual forces and moments corresponding to the displacements and the mutual rotation at the relevant points of the beam and determine the corresponding moment distributions. They are shown in Fig. 8.29. For the displacement w follows: Zl wD 0

M0 M1 1 1 F l3 dx D  l F l l D : 3EIyy 3EIyy 3 9EIyy

Fig. 8.28 Beam under single force F , moment diagram M0 (bottom).

(8.150)

w F

φ

x l

EI1

EI2

EI3

l

l

Fl (-) (+)

Fl

wF

F

270

8 Energy methods

Fig. 8.29 Beam under virtual forces and moments, resulting moment diagrams M1 .

l 1

(+)

0

0 l

1

(-) (+)

l 2 (+)

0 1

1

The displacement wF results as: Zl wF D 0

M0 M1 dx C 3EIyy

Zl 0

M0 M1 dx C 2EIyy

Zl 0

M0 M1 dx EIyy

1 1 1 1 1 1  l F l l C  l F l l C  l F l l D 3EIyy 3 2EIyy 3 EIyy 3 D

11F l 3 : 18EIyy

(8.151)

The mutual rotation ' is determined as follows: Zl 'D 0

M0 M1 dx C 3EIyy

Zl 0

M0 M1 dx 2EIyy

1 1 1 1 1 1   l  1  .F l/ C  l F l 2C  l F l 1 D 2EIyy 6 3EIyy 3 3EIyy 6 D

8.5.2

7F l 2 : 36EIyy

(8.152) J

Statically indeterminate systems

The force method also proves to be very useful in the analysis of statically indeterminate systems. At this point we will restrict ourselves to system which are statically indeterminate to the first degree. As an introductory example, consider the system shown in Fig. 8.30, top. Let the beam of length l and constant bending stiffness EIyy be clamped at its left end and simply supported at its right end. The load is given in the form of a uniform line load q. The moment distribution M.x/ is to be determined.

8.5 The force method

271

q B

x l

q

1 ql 2

2

δ0

F=1

ql 2 8

1 ql 2 2

=

(-) (+)

M0 due to q

ql 2 8

l δ0

B=1

M1 due to F1=1 l

δ1

M1 due to B =1

Final moment distribution: 1 ql 2 8

Fig. 8.30 Statically indeterminate beam under line load q, representation of the calculation steps for determining the moment line M.x/.

In order to determine the state variables of the statically indeterminate beam, an arbitrary support reaction is released by removing the corresponding kinematic bond. In this example, we remove the support B. As a consequence, the support reaction B is released and the beam becomes statically determinate. The resulting statically determinate system is called the 0-system. The choice of the statically determinate 0-system is arbitrary, but it is imperative to ensure that the system does not become displaceable and thus unusable by removing the kinematic bond. The moment diagram M0 due to the given load is now determined at the statically determinate 0-system. It is shown in Fig. 8.30, middle. The deflection ı0 at the support point B can then be determined by means of the force method. We apply a virtual single force F D 1 at the right end of the beam and determine the moment diagram M1 . In order to perform the superposition of the two moment lines M0 and M1 appropriately, we have decomposed the moment diagram M0 into a quadratic ql 2 parabola with the maximum value and a triangular area with the maximum 8 2 value  ql2 . Then, for the deflection ı0 we obtain: Zl ı0 D 0

M0 M1 1 dx D EIyy EIyy



 1 1 1 1 3ql 4 2 2 : l l  ql  l l  ql D 3 2 3 8 24EIyy (8.153)

272

8 Energy methods

In the next step, we determine the deflection ı1 that would result from the support force B. At this point, the support force B is still unknown, so we assume B to be a unit force B D 1. The deflection ı1 due to B D 1 (so-called 1-system) then results as: Zl M1 M1 l3 ı1 D dx D : (8.154) EIyy 3EIyy 0

The two deflections ı0 and ı1 cannot occur in reality due to the support at point B. Therefore the following compatibility condition applies: ı0 C Bı1 D 0:

(8.155)

This means that both the deflection ı0 and the deflection ı1 multiplied by B must cancel each other out. From this requirement, the support force B can be determined as: 3ql 4 3 24EIyy ı0 BD D D  ql: (8.156) ı1 8 l3 3EIyy The negative sign indicates that B actually acts in the opposite direction than shown in Fig. 8.30. Once the support force B is given, then all further support reactions and consequently all state variables can be determined. The final moment line M.x/ on the statically indeterminate system is shown in Fig. 8.30, bottom. In general, support reactions and internal forces and moments of the system (here symbolically denoted by S) can be determined from the superposition of the individual calculations: (8.157) S D S0 C BS1 : Here S0 are the internal forces and moments on the statically determinate 0-system due to the given load. The quantities S1 are the internal forces and moments at the statically determinate 0-system due to the unit load B D 1, which must be multiplied in (8.157) by the determined value for B. The moment distribution M for the statically indeterminate system follows as: M D M0 C BM1 :

(8.158)

It is common to denote a statically indeterminate support reaction as X. The compatibility condition is then in general form: ı0 C Xı1 D 0:

(8.159)

The solution is:

ı0 : (8.160) ı1 The internal forces and moments in the statically indeterminate system are then: X D

S D S0 C XS1 :

(8.161)

8.5 The force method

273

Example 8.13

Consider the statically indeterminate beam shown in Fig. 8.31 under a single force F . The support reactions A, B and MA of this statically indeterminate beam are to be determined using the force method. The statically determinate 0-system is generated by releasing a statically intermediate quantity. Here, we will treat the support moment MA as statically indeterminate quantity by replacing the clamped end with a hinged support. The statically determinate 0-system is shown in Fig. 8.32, top. Fig. 8.31 Statically indeterminate beam (top), support reactions (bottom).

F x ll

l2 l

F

MA A Fig. 8.32 Statically determinate 0-system (top), moment diagram M0 (middle) due to the given load, moment diagram M1 (bottom) due to the statically indeterminate quantity X D 1 (1-system).

B F

X=1 x ll

l2 l

F

l

F l +2l l

l

(+)

F l +ll

2

l

ll

F l +l l2 l

X=1

1 l l+ l 2

l2 l l+ l 2

1

2

(-)

ll l l+ l 2 (-)

=

l2 l l+ l 2

1 l l+ l 2

2

274

8 Energy methods

We determine the moment diagram M0 for the statically determinate 0-system. It is given in Fig. 8.32, middle. Moreover, the moment diagram M1 is required as a result of the statically indeterminate moment X D 1, it is shown in Fig. 8.32, bottom. The statically indeterminate moment X can then be determined using the rotations ı0 and ı1 of the support point: Zl

M0 M1 dx EIyy 0   1 l1 l2 l2 l1 l2 1 l1 l2 1 l1 l2 1 D  l1 F  l2 F  l1 F EIyy 6 l1 C l2 l1 C l2 2 l1 C l2 l1 C l2 3 l1 C l2 l1 C l2  2 2 l1 F l1 l2 l1 l2 l2 D C  (8.162) 2 EIyy .l1 C l2 / 6 2 3

ı0 D

and Zl

M1 M1 dx EIyy 0   l12 l22 l22 1 l1 l2 1 1 D C l C l C l1 l 1 1 2 EIyy 3 .l1 C l2 /2 l1 C l2 l1 C l2 .l1 C l2 /2 3 .l1 C l2 /2   3 l1 1 l23 2 2 D l C l l C C l : (8.163) 2 1 1 2 EIyy .l1 C l2 /2 3 3

ı1 D

The statically indeterminate support moment X D MA then follows from the compatibility equation ı0 (8.164) X D ; ı1 thus: XD

  F l1 l2 l2 1C D MA : 2.l1 C l2 / l1 C l2

(8.165)

The support force B can be determined from this as: B D B0 C XB1 D

F l12 Œl2 C 2.l1 C l2 / : 2.l1 C l2 /3

(8.166)

The support force A follows as: A D F  B:

(8.167) J

8.5 The force method

275

Example 8.14

Consider the statically indeterminate beam shown in Fig. 8.33 which is loaded by the line load q. The moment diagram of the beam that is statically indeterminate to the first degree is to be determined by means of the force method. The beam has a constant bending stiffness EIyy . We make the beam statically determinate by releasing the moment above the middle support by inserting a full joint at this location. The moment diagrams M0 and M1 are shown in Fig. 8.34. The compatibility equation to be applied here is: ı0 X D : (8.168) ı1 The two rotations ı0 and ı1 are: Z

M0 M1 1 1 ql 2 ql 3 dx D 2   l  ; 1D EIyy EIyy 3 8 12EIyy Z M1 M1 1 1 2l dx D 2   l 1D : ı1 D EIyy EIyy 3 3EIyy ı0 D

(8.169)

Thus, the statically indeterminate bending moment can be determined as: X D

ql 2 : 8

(8.170)

The reaction forces A of the outer supports follow from this as: A D A0 C XA1 D

3 ql: 8

(8.171)

Fig. 8.33 Static system and load.

q x l

l

q

X =1 x

x

ql 2

M0

ql 2

ql (+)

M1

(+)

1ql 2 8

1 l

1ql 2 8

Fig. 8.34 Moment diagrams M0 (left) and M1 (right).

2 l (+)

1

1 l

276

8 Energy methods

Fig. 8.35 Moment diagram for the statically indeterminate beam.

1ql 2 8

M (+)

(+)

3ql 8

3ql 8

5ql 4

The support force B of the middle support is: B D B0 C XB1 D

5 ql: 4

(8.172)

The final moment distribution in the statically indeterminate system follows from the condition (8.173) M D M0 C XM1 : J

The moment diagram is shown in Fig. 8.35. Example 8.15

Consider the frame that is shown in Fig. 8.36, which is statically indeterminate to the first degree. The bending stiffness EIV is given in the vertical bars, in the horizontal beam the value EIH is given. The frame is loaded by a single force as shown. The system is made statically determinate by introducing a moment joint, and the static quantity thus released is to be determined. In addition, the support reactions and the moment diagram of the statically indeterminate system for the special case EIV D EIH and h D l are to be determined. We use the force method for the solution of this problem, taking into account only the contribution due to the bending moments in the deformation calculations. The statically determinate 0-system is generated by introducing a moment joint at the force application point and releasing the bending moment at this Fig. 8.36 Statically indeterminate frame under single force.

F EIH h

EIV

EIV

l 2

l 2 l

8.5 The force method

Fl 4 Fl 4

277

Fl 4

F (-)

(-)

(-)

(-)

M0

Fl 4h F 2

1 Fl 4

1

(-)

1

1 (-)

Fl 4h

M1

1 h

F 2

(-)

X =1

0

1 h 0

Fig. 8.37 Moment diagrams M0 (left) and M1 (right).

point. The resulting moment lines M0 and M1 at the 0-system and the 1-system are shown in Fig. 8.37. From the compatibility equation we determine the statically indeterminate bending moment X D M as: X D

ı0 : ı1

(8.174)

The quantities ı0 and ı1 result from the superposition of the moment diagrams as:   Z M0 M1 h Fl l ı0 D dx D C ; EIyy 2 3EIV 4EIH Z M1 M1 2h l dx D C : (8.175) ı1 D EIyy 3EIH EIV This leads to the statically indeterminate bending moment as: ! h l Fl C 3EIV 4EIH ! : X D 2h l 2 C 3EIH EIV

(8.176)

If the special case EIV D EIH and h D l is considered, we obtain: X D

7 F l: 40

(8.177)

The support reactions and the moment distribution are calculated from the condition (8.178) S D S0 C XS1 : They are shown in Fig. 8.38. J

278 Fig. 8.38 Support reactions and moment diagram of the statically indeterminate frame for the special case EIV D EIH and h D l.

8 Energy methods

3 40 Fl 3 40 Fl

3 40 Fl

F

(-)

3 40 Fl

(-) (+)

(-)

(-)

7 40 Fl 3 40 F

3 40 F

M F 2

F 2

8.6 Reciprocity theorems 8.6.1 Betti’s theorem We want to motivate Betti’s theorem3 using the situation of Fig. 8.39. Consider a linear elastic solid loaded by two point forces F1 and F2 . Let us denote the displacements of the two force application points as u1 and u2 , respectively. We now perform the following thought experiment. We first consider the situation that only the force F1 acts on the solid and the force F2 is not yet present. The force F1 then performs work along the displacement u1 as follows: 1 .1/ (8.179) We D F1 u1 : 2 .1/

Here the superscript .1/ indicates that the displacement u1 is caused by the force F1 . In the next step, the force F2 is applied to the solid body, and we want to assume that the force F1 still acts at full magnitude. The total work done is then: 1 1 .1/ .2/ .2/ (8.180) We D F1 u1 C F2 u2 C F1 u1 : 2 2 Fig. 8.39 On Betti’s theorem.

F2

F1 u2 u1

3

Enrico Betti, 1823–1892, Italian mathematician.

8.6 Reciprocity theorems

279 .1/

In addition to the work 12 F1 u1 already performed, another share results from the .2/ force F2 performed along the displacement u2 . However, additionally the part of the force F1 has to be taken into account, which performs additional work along the .2/ displacement u1 caused by F2 (so-called passive work). The thought experiment is now repeated in reverse order, and we first apply the force F2 . It follows: 1 .2/ (8.181) We D F2 u2 : 2 The force F2 now acts at full magnitude, and the force F1 is additionally applied. The total work done is then: 1 1 .2/ .1/ .1/ (8.182) We D F2 u2 C F1 u1 C F2 u2 : 2 2 The total work must be identical in both cases. If we equate (8.180) and (8.182), we obtain: 1 1 1 1 .1/ .2/ .2/ .2/ .1/ .1/ (8.183) F1 u1 C F2 u2 C F1 u1 D F2 u2 C F1 u1 C F2 u2 : 2 2 2 2 .1/

.2/

The terms 12 F1 u1 and 12 F2 u2 cancel out, and the following expression remains: .2/

.1/

F1 u1 D F2 u2 :

(8.184)

This is Betti’s theorem. It states that the work done by a force F1 along a dis.2/ placement u1 caused by a force F2 acting at another point is identical to the work .1/ performed by the force F2 along the displacement u2 caused by F1 . In symbolic form we can also write: W12 D W21 ; (8.185) or in general form: Wij D Wj i ;

i ¤ j:

(8.186)

Herein, the index i indicates the location and the index j indicates the cause of the performed work.

8.6.2 Maxwell’s theorem Maxwell’s theorem4 is obtained by taking the forces F1 and F2 in Betti’s theorem as unit loads. Maxwell’s theorem then reads: .2/

.1/

u1 D u2 :

(8.187)

The following notation will prove to be useful at a later point: ıij D ıj i ; with i ¤ j . 4

James Clerk Maxwell, 1831–1879, Scottish physicist.

(8.188)

280

8 Energy methods

8.7 Statically indeterminate systems Betti’s theorem and Maxwell’s theorem can be applied to the analysis of systems that are statically indeterminate to a higher degree using the force method. For this purpose, the introductory example of Fig. 8.40, top left, is considered. The beam under consideration has the total length 2l and the constant bending stiffness EIyy . The beam is supported in such a way that at its left end it is rigidly clamped. In the middle of the beam and at the right end, the beam is simply supported. This beam is thus statically indeterminate to the second degree. In order to be able to apply the force method, we make the beam statically determinate by releasing a sufficient number of kinematic bonds and releasing the statically indeterminate reaction forces without rendering the system unusable. In the example, this is done by removing the two simple supports, see Fig. 8.40, top right. Due to this removal, the displacements ı10 and ı20 arise at the two support points due to the given load (0-system), which of course cannot occur at the actually statically indeterminate system. Therefore, we now apply the still unknown support force X1 (Fig. 8.40, 1-system, bottom left) at the center support point, which we consider as a unit force. It invokes the two displacements ı11 and ı21 on the 1-system. From the indexing of these displacements it is clear that these are displacements at points 1 and 2 (indicated by the first index in each case) caused by X1 (indicated by the second index). In the same way, we can proceed with the unknown support force at the right support by applying the unit load X2 D 1 (2-system) and considering the two displacements ı12 and ı22 caused by it (Fig. 8.40, bottom right). All displacements mentioned above cannot occur in the actually statically indeterminate system. Therefore, there are two compatibility conditions for a system that is statically indeterminate to the second degree. Regarding ı10 , ı11 and ı12 at the support point in the beam center, it follows: ı10 C X1 ı11 C X2 ı12 D 0: q

(8.189)

q

x l

δ10

δ20

δ12

δ22

l

δ11

X1 =1

δ21

X2 =1

Fig. 8.40 Statically indeterminate beam, determination of the support reactions.

8.7

Statically indeterminate systems

281

At the support point at the right end of the beam, on the other hand, the compatibility condition to be fulfilled is: ı20 C X1 ı21 C X2 ı22 D 0:

(8.190)

These are two equations for the two unknown support reactions X1 and X2 . From Maxwell’s theorem the equality of the two displacements ı12 and ı21 can be concluded: (8.191) ı12 D ı21 : In general terms, we can write: ıij D ıj i ;

i ¤ j:

(8.192)

The two compatibility conditions then take the following form: ı10 C X1 ı11 C X2 ı12 D 0; ı20 C X1 ı12 C X2 ı22 D 0:

(8.193)

Thus, if a system is considered that is statically indeterminate to the second degree, which is to be treated by means of the force method, then a total of five displacements are to be determined. If only the influence of the bending moment M is taken into account, then these displacements result as: Z ıij D

Mi Mj dx: EIyy

(8.194)

The compatibility conditions (8.193) can be solved for the statically indeterminte quantities X1 and X2 as follows: X1 D

ı10 ı22 C ı20 ı12 ; 2 ı11 ı22  ı12

X2 D

ı20 ı11 C ı10 ı12 : 2 ı11 ı22  ı12

(8.195)

Once the support reactions of the beam have been determined, the further support reactions as well as the internal forces and moments of the statically indeterminate system can be determined: S D S 0 C X1 S 1 C X2 S 2 :

(8.196)

Here S1 and S2 are the forces and moments due to the statically indeterminate quantities X1 and X2 .

282

8 Energy methods

Example 8.16

Consider the statically indeterminate system shown in Fig. 8.41, left. The given frame structure is statically indeterminate to the second degree. The support forces and the moment distribution are to be determined. We want to use the force method for the analysis, normal force and transverse shear force influences are to be neglected. We choose the statically determinate 0-system as shown in Fig. 8.41, right. Here, a moment joint was introduced at the upper left corner, and the clamped end is transformed into a hinged support. The moment diagram M0 is shown in Fig. 8.42, top. The corresponding bending moment diagrams as a consequence of the statically indeterminate quantities X1 D 1 and X2 D 1 are shown in Fig. 8.42, middle. The compatibility equations to be fulfilled here are: ı10 C X1 ı11 C X2 ı12 D 0; ı20 C X1 ı12 C X2 ı22 D 0:

(8.197)

The quantities ıij (i; j; D 0; 1; 2) currently amount to: ı10 D

F l2 ; 3EIyy

ı20 D 

l ; EIyy

ı11 D

ı12 D 

2l ; 3EIyy

F l2 5l ; ı22 D : 2EIyy 3EIyy

(8.198)

The statically indeterminate quantities X1 and X2 can thus be determined as: X1 D 

2 F l; 11

X2 D

5 F l: 22

(8.199)

The support reactions and the resulting moment distribution of the statically indeterminate system are shown in Fig. 8.42, bottom. J Statically indeterminate system

0-system

F

l

l

l

Fig. 8.41 Statically indeterminate system (left), statically determinate 0-system (right).

F

8.7

Statically indeterminate systems

283

0-system

Fl (-)

0

M0

0

0

0

F

2F

1-system

2-system

X 2=1 (-)

1

1 (+)

0

1

1

(+)

(-)

(+)

M1

0

M2

X1 =1 1 l

(+)

1

1 l

1 l

1 l 1 l

0

1 l 0

Support reactions and bending moment 13 Fl Fl 22

5 22 Fl

(-) (-) (+) (+)

(-)

5 22 Fl

(-)

2 11 Fl 9 22 F

9 22 Fl

M 9 22 F

2 11 Fl 9 11 F

20 F 11

Fig. 8.42 Bending moment distributions for the 0-system, the 1-system and the 2-system; support reactions and bending moment diagram for the statically indeterminate system.

Systems with higher degrees of static indeterminacy can be treated quite analogously using the force method. If a system is considered that is statically indeterminate to the third degree, then the following compatibility conditions are to be considered: ı10 C X1 ı11 C X2 ı12 C X3 ı13 D 0; ı20 C X1 ı21 C X2 ı22 C X3 ı23 D 0; ı30 C X1 ı31 C X2 ı32 C X3 ı33 D 0; (8.200)

284

8 Energy methods

or after application of Maxwell’s theorem: ı10 C X1 ı11 C X2 ı12 C X3 ı13 D 0; ı20 C X1 ı12 C X2 ı22 C X3 ı23 D 0; ı30 C X1 ı13 C X2 ı23 C X3 ı33 D 0:

(8.201)

In vector-matrix notation, we have: 2

ı11 4ı12 ı13

ı12 ı22 ı23

30 1 0 1 ı13 X1 ı10 ı23 5@X2 A D @ı20 A: ı33 X3 ı30

(8.202)

Thus there are three equations for the determination of the three unknown quantities X1 , X2 and X3 . The coefficient matrix is symmetric as a consequence of Maxwell’s theorem. We now consider a static system with the degree of static indeterminacy n. The n compatibility conditions to be considered are then: ı10 C X1 ı11 C X2 ı12 C X3 ı13 C    C Xn ı1n D 0; ı20 C X1 ı12 C X2 ı22 C X3 ı23 C    C Xn ı2n D 0; ı30 C X1 ı13 C X2 ı23 C X3 ı33 C    C Xn ı3n D 0; :: : ın0 C X1 ı1n C X2 ı2n C X3 ı3n C    C Xn ınn D 0:

(8.203)

In vector-matrix notation we can write: 2

ı11 6ı12 6 6ı13 6 6 :: 4 :

ı12 ı22 ı23 :: :

ı13 ı23 ı33 :: :

   :: :

ı1n

ı2n

ı3n



30 1 0 1 ı1n ı10 X1 B X2 C Bı20 C ı2n 7 7B C B C B C B C ı3n 7 7BX3 C D Bı30 C; B :: C :: 7B :: C @ : A : 5@ : A Xn ın0 ınn

(8.204)

or in symbolic form: ıX D ı 0 : The matrix ı is symmetric as a consequence of Maxwell’s theorem.

(8.205)

8.7

Statically indeterminate systems

285

Example 8.17

Consider the frame structure shown in Fig. 8.43 which is statically indeterminate to the third degree. We want to determine the support reactions and the moment distribution. The depicted statically determinate 0-system is to be used. Currently, the following compatibility conditions have to be considered for the determination of the three statically indeterminate quantities X1 , X2 and X3 : ı10 C X1 ı11 C X2 ı12 C X3 ı13 D 0; ı20 C X1 ı12 C X2 ı22 C X3 ı23 D 0; ı30 C X1 ı13 C X2 ı23 C X3 ı33 D 0:

(8.206)

The displacements and rotations ıij are obtained as: 1 F h.l C h/; 3EIyy   1 1 1 D lC h ; EIyy 2 6   1 1 1 D Fh l C h ; EIyy 2 3 1 D .l C h/; 2EIyy   1 1 D l Ch : EIyy 3

1 .l C 2h/; 3EIyy   1 1 1 D lC h ; EIyy 3 2   1 2 D lC h ; EIyy 3   1 1 1 D Fh l C h ; EIyy 3 2

ı10 D 

ı11 D

ı12

ı13

ı20 ı23 ı33

ı22 ı30

(8.207)

Statically indeterminate system

Statically determinate 0-system

F

h

l Fig. 8.43 Statically indeterminate frame (left), statically determinate 0-system (right).

286

8 Energy methods

Equation (8.206) takes the following form: ! !3 1 0 1 1 1 1 1 1  lC h  lC h 7 6 .l C 2h/ F h.l C h/ C 6 3 2 6 3 2 70 1 B B 3 !C 7 6 ! ! B 7 X1 6 1 C 1 C B 1 1 2 1 7 6 F h l C h C: lC h .l C h/ 7@X2 A D B 6 l C h C B 2 3 7 6 2 6 3 2 B !C 6 ! ! 7 X3 C B 7 6 @F h 1l C 1h A 1 1 1 1 5 4 .l C h/ l Ch  lC h 3 2 3 2 2 3 (8.208) This results in the following statically indeterminate quantities X1 , X2 , X3 : 2

X1 D

h.3h C l/ F; 2.6h C l/

X2 D 

3h2 F; 2.6h C l/

X3 D 

h.3h C l/ F: 2.6h C l/

(8.209)

The support reactions of the statically indeterminate system then result in: 3h2 F; l.6h C l/ F AH D Ah0 C X1 AH1 C X2 AH 2 C X3 AH 3 D ; 2 3h2 B V D B V 0 C X1 B V 1 C X2 B V 2 C X3 B V 3 D  F; l.6h C l/ F BH D BH 0 C X1 BH1 C X2 BH 2 C X3 BH 3 D  : 2 AV D AV 0 C X1 AV 1 C X2 AV 2 C X3 AV 3 D

(8.210)

The moment distribution M of the statically indeterminate system then follows from superposition of the individual partial moment lines: M D M0 C X1 M1 C X2 M2 C X3 M3 : It is shown in Fig. 8.44, bottom. J

(8.211)

8.7

Statically indeterminate systems

287

Fh F (+) Fh

(+)

0

(-)

1 (-)

0-system

1-system

1 h

F

0 Fh l

1 l

X 1=1

1 l (+)

1 1

1

1

(+)

(+)

(+)

0

2-system

1 h

1 h 0

(-)

(-)

F

2

(+)

3h 2(6h+l) F (+)

h(3h+l) 2(6h+l) F

2

Fig. 8.44 Moment diagrams.

2

3h 2(6h+l) F

(-)

F 2

h(3h+l) h(3h+l) 2(6h+l) F 2(6h+l) F 3h l(6h+l) F

2

0 1 l

Fh (+)

3-system

0

0

Support reactions and bending moment

F 2

1

1

(+)

2

1 h

(+)

Fh l X 2=1

3h 2(6h+l) F

1

3h l(6h+l) F

1 l

X 3=1

9

Buckling of bars

This chapter deals with the fundamentals of the buckling of bars under compressive load. After a brief introduction to the stability of equilibrium and the determination of critical loads (i.e. the load under which a previously straight member changes to a deflected position), as necessary for the understanding of this chapter, we discuss the determination of buckling loads of bars under compressive load under various boundary conditions. The purpose of this chapter is to enable students to treat stability problems of compressed members and to determine buckling loads for elementary cases.

9.1 Introduction The aim of the design of a structure is to provide sufficient stiffness and strength, taking into account specific requirements. Engineers in practice are therefore often concerned with providing the corresponding verification, usually in the form of a strength verification, i.e. the verification that the stresses in a structure do not exceed permissible values. Considering the tensile member of Fig. 9.1, it must be verified that the tensile stress  D FA does not exceed an allowable value allow (taking safety factors into account if necessary). Thus, the bar under tension requires the consideration of a strength problem. A completely different situation arises when the considered bar is subjected to a compressive load F (Fig. 9.2). In many technically relevant cases there is no strength problem (i.e. the corresponding strength of the material is not reached by the applied load), but rather, depending on the magnitude of the applied load, several equilibrium states can occur that describe different configurations of the bar. Thus, Fig. 9.1 Bar under tensile load.

F l

© The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9_9

289

290

9

undeformed

Buckling of bars

F

deformed

l Fig. 9.2 Bar under compressive load.

there is not necessarily an unambiguous relationship between the applied load and the resulting bar response. In addition to the straight configuration, configurations are also possible that are accompanied by a bending of the bar. This phenomenon, i.e. the bending of a compression member when a certain load is reached, is referred to as buckling. Also, the term of the so-called flexural buckling is commonly used to make it clear that, although we are dealing with a bar, it reacts to the applied load like a beam by bending. This deflection generally occurs abruptly when a defined load level F D Fcrit is reached, which we refer to as critical load or buckling load. This field of knowledge is the so-called stability theory, and it is the objective of a stability analysis to determine the critical load or buckling load Fcrit at which the previously straight member changes into a deflected position.

9.2 Types of equilibrium In what follows we want to clarify the term stability and the related stability of equilibrium in more detail. For this purpose we consider the so-called sphere analogy (see Fig. 9.3). A rigid sphere of mass m is considered, which is located on differently shaped surfaces. Gravity acts as indicated. First, we consider the case that the sphere lies on a concave surface (Fig. 9.3, top left). In this case, the sphere will come to rest by itself in its rest position at the lowest point of the surface and will find its natural equilibrium position there. If the position of the sphere is changed by deflecting it infinitesimally and then leaving it to itself (i.e. if an infinitesimal perturbation of the state of equilibrium is applied), then the sphere will always return to its original position under the influence of the gravitational field of the earth and through the influence of sliding friction and will resume its state of rest at the lowest point of the concave surface. Such an equilibrium position is called stable equilibrium. This behavior of the rigid sphere on a concave surface can be directly transferred to the elastic bar under compressive axial load of Fig. 9.2 (see also Fig. 9.3, bottom left). The bar has the length l and the bending stiffness EI . The force F is below the buckling load, i.e. F < Fcrit . If an infinitesimal lateral deflection is applied to the bar (i.e. if the straight equilibrium position is subjected to a disturbance) and the bar is then left to itself, the bar will return to the straight undeformed position by itself. Thus, the bar under compression is in stable equilibrium for all loads F below the buckling load Fcrit .

9.2 Types of equilibrium

Stable equilibrium Infinitesimally adjacent position

g

291

Indifferent equilibrium

Unstable equilibrium Rest position

Infinitesimally adjacent position

Infinitesimally adjacent position

Rest position

Rest position m

FFcrit

undeformed bar

undeformed bar

undeformed bar

Infinitesimally adjacent position

Infinitesimally adjacent position

Infinitesimally adjacent position

EI

EI

returns by itself to initial position

EI

both positions are possible

l

x

l

x

remains in deflected position, deflections increase with load increase

l

x

Fig. 9.3 Sphere analogy and its transfer to the straight bar under axial compression.

We will now consider the case where the sphere is in a state of equilibrium on a horizontal plane (Fig. 9.3, top center). Now, we will apply a perturbation in such a way that we place the sphere at another position on the plane and then leave it there to itself. In this case, the sphere will not return to its initial equilibrium position, but rather remains in its new rest position. Accordingly, in this case both mentioned positions are possible equilibrium positions, and the equilibrium in this case is no longer unambiguous – several equilibrium positions exist simultaneously. This type of equilibrium is called indifferent equilibrium. This part of the sphere analogy can also be applied to the bar under compression. It can be seen that when a certain load level is reached and an infinitesimal lateral perturbation is applied, the bar will no longer return to the straight initial position, but will rather remain in the deflected position. Obviously, several equilibrium positions exist at the same time, so that indifferent equilibrium exists at this point. This state can only occur at a certain magnitude of the load F , namely exactly when the force F reaches the critical load: F D Fcrit . At this point, we also speak of bifurcation of equilibrium. The occurrence of indifferent equilibrium is often also referred to as onset of buckling. The last part of the sphere analogy concerns the case where the sphere is at rest on a convex surface (Fig. 9.3, top right). If the sphere is subjected to an infinitesimal perturbation and then left to itself, the sphere will not return to its initial rest position. The motion of the sphere follows the direction of the applied perturbation. Such an equilibrium position is called unstable equilibrium. Unstable equilibrium

292

9

Buckling of bars

is characterized by the fact that the rest position is left at the slightest disturbance and is not resumed by itself. Considering the bar under compression, unstable equilibrium means that the case exists where the force F has already exceeded the buckling load and thus F > Fcrit and the bar is still in its straight configuration. Even though everyday experience shows us that this case will not occur in any practically relevant system, this is, as we will show, a mathematically possible solution. In this condition at this load level, even the smallest perturbation in the form of an infinitesimal lateral deflection is sufficient to move the bar from its straight position to an adjacent deflected position (i.e. a buckled position), and thus the straight position is unstable for F > Fcrit . The greater the load F , the greater the lateral deflection will be. The laterally deflected position, however, is stable. The distinction whether an equilibrium position is stable, indifferent or unstable can be made particularly advantageously by energy considerations. The example of the sphere analogy shows that in the case of the sphere which takes its rest position at the lowest point of a concave surface, the potential energy is a minimum – any other position is accompanied by an increase of the potential energy. This means that energy must be supplied to move the sphere from its rest position. Stable equilibrium is therefore accompanied by a minimum of energy. Conversely, unstable equilibrium means that energy is released when the sphere is moved from its rest position. Unstable equilibrium is therefore accompanied by a maximum of potential energy.

9.3

Determination of critical loads

The determination of critical loads can be motivated particularly clearly by systems of rigid bars. As an introductory example, we consider the elastically supported beam of Fig. 9.4, left. We consider a straight, ideally rigid and massless bar of length l, which is simply supported at its lower end. In addition, the support is reinforced by an elastic rotational spring with stiffness k (linear-elastic spring law M D k'). The bar is loaded at its upper end by the compressive force F . We assume that the applied force retains its direction even when the bar is deflected from the equilibrium position, i.e. we consider a conservative force. There are basically two ways to determine critical loads. One is the determination via equilibrium considerations (also referred to as equilibrium method), and the other is the consideration of the energy of the system. We start the elaborations with the consideration of the equilibrium method. For this purpose, we deflect the bar from its straight configuration by the angle ', Fig. 9.4, middle. The angle ' is required to be j'j < 2 , but we do not impose the requirement that it must be a small angle. The consideration of equilibrium of the deformed system is common in stability theory and in fact mandatory to obtain statements about critical loads. The rotation of the bar by the angle ' around its support point leads to a restraining moment of the magnitude k'. If the sum of the moments about the support point

9.3 Determination of critical loads

F

293

F F lsinφ

l(1-cosφ) Postbuckling range φ =0

rigid

Postbuckling range φ =0

l φ

k

kφ

l

lcosφ Bifurcation

Fcrit = k l

–φ

Asymptote for linearized analysis Prebuckling range φ =0

φ

Fig. 9.4 Elastically clamped massless ideally rigid bar under compressive load F (left), deflected position (middle), force-rotation diagram (right).

is calculated, the result is: k'  F l sin ' D 0:

(9.1)

This equation has two solutions. On the one hand, (9.1) can be solved by setting ' D 0, independent of the acting force F : ' D 0:

(9.2)

This result means that the straight undeformed configuration with ' D 0 is an equilibrium position. This is shown in Fig. 9.4, right, indicated as a vertical line in the force-rotation diagram. The second solution is obtained by discussing (9.1) in more detail. Since '  sin ' holds for all angles ', the solution ' ¤ 0 is only possible if the force F takes values F > kl . The resulting force-rotation curve is also shown in Fig. 9.4, right. As a consequence, for all F < kl only the solution ' D 0 exists. Accordingly, for the force F D kl both the straight and the deflected configurations are possible, and it is said that a bifurcation of equilibrium occurs at this point. The force F D kl apparently characterizes the transition from the straight state to the deflected position and thus represents the critical load Fcrit of the rigid elastically clamped bar: k Fcrit D : (9.3) l It is common to speak of the so-called prebuckling range at load levels below the critical load. The postbuckling range, on the other hand, is reached when the force F assumes values above the critical load Fcrit .

294

9

Buckling of bars

It is important to note at this point that the described procedure applies to the discussed example of the elastically clamped rigid bar for arbitrary angles with j'j < 2 . Thus, it is a solution for a so-called geometrically nonlinear problem, i.e. a problem with large deformations. It should be noted, however, that finding such geometrically nonlinear solutions is very difficult, and considering the postbuckling range is also not relevant for a good number of engineering applications – in many engineering situations, one will design a bar-like structure against the critical load, but not beyond it. For this reason, we want to linearize the present geometrically nonlinear problem, which means that we assume small angles '. Then we have: sin ' ':

(9.4)

The equilibrium condition (9.1) then takes the following form: .k  F l/' D 0:

(9.5)

If the trivial solution ' D 0 is not considered further (which would correspond to the bar remaining straight) and the bracket term in (9.5) is set to zero, then this expression can be solved directly for the force F . The solution then corresponds to the critical load Fcrit : k Fcrit D : (9.6) l Thus, from the linearization of the problem, the critical load Fcrit can be immediately determined. In many practical applications, linearized considerations are used to determine critical loads. As a drawback, it must be accepted that when considering a linearized problem, no statements about the postbuckling range with F > Fcrit are possible (see Fig. 9.5). If the equilibrium method is used, statements can be made about critical loads, but no statements about the type of equilibrium (stable, indifferent, unstable) are possible in this way. Such statements are only possible by means of an energetic consideration, which we will briefly discuss using the example of the elastically clamped rigid bar. For this purpose, we consider the total potential of the bar, again in the deflected state. The total potential is denoted here as ˘ , it is composed of the inner potential ˘i of the elastic spring and the external potential ˘e of the applied Fig. 9.5 Force-rotation diagram for linearized analysis.

F

Bifurcation

Asymptote for linearized analysis

Fcrit = k l

–φ

φ

9.3 Determination of critical loads

295

force F . The bar is assumed to be ideally rigid, so that it does not deform and undergoes a pure rigid body rotation '. Consequently, no energy is stored in the bar itself. The total potential ˘ is therefore: ˘ D ˘i C ˘e D

1 2 k'  F l.1  cos '/: 2

(9.7)

This means that the elastic spring stores potential energy, whereas the applied force F has lost potential energy, which explains the negative sign of the corresponding term in (9.7). The condition for the existence of an equilibrium state is that the first variation ı˘ of the total potential ˘ vanishes: ı˘ D 0:

(9.8)

Since we are dealing with a system with one degree of freedom ', we may write (9.8) in the following form: @˘ D k'  F l sin ' D 0: @'

(9.9)

It turns out that this expression corresponds exactly to the equilibrium condition (9.1) that we have already derived using the equilibrium method. The nature of the equilibrium can be determined by considering the second variation ı 2 ˘ , which for the present example reads: @2 ˘ D k  F l cos ': @' 2

(9.10)

We first consider the case of the undeflected bar and set ' D 0: @2 ˘ D k  F l: @' 2

(9.11)

There are three cases to be distinguished here. 2

 If F < kl D Fcrit , then @@'˘2 > 0 holds. This is equivalent to the total potential for 0  F < Fcrit assuming a minimum in the equilibrium position ' D 0. Obviously, energy must be expended to change into any other possible configuration. Thus, the equilibrium position ' D 0 is a stable equilibrium position for 0  F < Fcrit . 2  If F > kl D Fcrit , then @@'˘2 < 0 holds. The change into any other possible adjacent equilibrium position is therefore accompanied by a loss of energy. Consequently, the equilibrium position ' D 0 is unstable for F > kl D Fcrit . Thus, the bar will immediately move from the straight position to an adjacent position and remain in this position, no matter how small the perturbation. 2  If F D kl D Fcrit , then @@'˘2 D 0 holds. Thus, a change to a possible adjacent configuration is neither associated with energy loss, nor is energy input necessary. For ' D 0 at F D kl D Fcrit we thus have indifferent equilibrium.

296

9

Buckling of bars

Fig. 9.6 Functions f .'/ D ' and f .'/ D tan.'/.

tan f( ) tan

/2

3 /2

These conclusions can be stated more generally as:  ı 2 ˘ D 0: Indifferent equilibrium  ı 2 ˘ > 0: Stable equilibrium  ı 2 ˘ < 0: Unstable equilibrium Finally, we want to consider which equilibrium state arises when ' ¤ 0 holds. As we already know, this state can only occur for forces F > kl D Fcrit . From the condition 2 @˘ D k'  F l sin ' D 0 we obtain F l D k sin' ' , which after substituting into @@'˘2 @' transforms into the following expression:   @2 ˘ ' D k 1  : (9.12) @' 2 tan ' 2

Since tan' ' < 1 (Fig. 9.6), provided that j'j < 2 , we can conclude @@'˘2 > 0 for any angle ' ¤ 0. Thus, the equilibrium position ' ¤ 0 is a stable equilibrium position. The types of equilibrium that arise are shown in Fig. 9.7. Fig. 9.7 Stable, unstable and indifferent equilibrium states.

F

Postbuckling range φ =0 (unstable)

Postbuckling range φ =0 (stable) Bifurcation

Fcrit = k l (indifferent)

–φ

Prebuckling range φ =0 (stable)

φ

9.4 Buckling of bars: The four Euler cases

297

9.4 Buckling of bars: The four Euler cases 9.4.1 Introductory example: Euler case II In this section we will discuss a deformable linear elastic bar under compressive load (Fig. 9.8). The bar configurations considered in this section are also commonly referred to as the so-called four Euler cases1 . Consider an elastic (modulus of elasticity E) and ideally straight bar of length l, which has the moment of inertia I . The bar is simply supported at its left end and is loaded at this point by the compressive force F while the right end is also simply supported. The aim of the following explanations is to determine the buckling load Fcrit for this bar. The force is applied at the center of gravity of the beam’s cross-section. This special situation of a straight bar under compression is commonly referred to as Euler case II. As already shown for the example of the rigid bar of Sect. 9.3 we have to investigate the equilibrium at the deformed system (deflection w.x/) to determine the buckling load Fcrit . Here it is assumed that these deformations remain infinitesimally small, which is commonly referred to as the so-called second-order theory. We consider the free body image of Fig. 9.8, right, in which we investigate the deformed bar cut at the arbitrary location x. We form the moment equilibrium around the point S in the deflected configuration and obtain (note that both the normal force N and the transverse shear force V have no lever arm here and accordingly do not appear in the moment sum, and that furthermore there is no vertical support force): M.x/  F w.x/ D 0: (9.13) Here it was implicitly assumed that length change of the bar due to the applied compressive load can be neglected. Application of the constitutive law

then yields:

EI w 00 .x/ D M.x/

(9.14)

EI w 00 .x/ C F w.x/ D 0:

(9.15)

EI

F x

w(x)

F

S l

0 x

Fig. 9.8 Euler case II.

1

V

w(x)

Leonhard Euler, 1707–1783, Swiss mathematician and physicist.

N

M

298

9

Buckling of bars

It is useful at this point to use the abbreviation 2 D

F ; EI

(9.16)

so that (9.15) takes the following form: w 00 .x/ C 2 w.x/ D 0:

(9.17)

This is a second-order ordinary linear homogeneous differential equation with constant coefficients. It is commonly referred to as buckling differential equation. Its general solution is: w.x/ D C1 sin x C C2 cos x; (9.18) where C1 and C2 are constants. This means that the so-called buckling shape or buckling mode is composed of a sine function and a cosine function. The boundary conditions to be applied here require that the deflection w.x/ vanishes at the two support points x D 0 and x D l: w.x D 0/ D 0;

w.x D l/ D 0:

(9.19)

Evaluating the first boundary condition immediately gives C2 D 0, leaving the following expression for the buckling shape: w.x/ D C1 sin x:

(9.20)

Apparently, the buckling shape of this simply supported beam is described by a single sinusoidal function. Evaluating the second boundary condition in (9.19) yields: C1 sin l D 0:

(9.21)

Obviously, there are two possible solutions. The first solution is C1 D 0. However, since C2 has already been determined to be zero, this solution would mean that the buckling shape (9.18) becomes zero overall and the bar does not buckle. This is a mathematically correct solution, but it has no technical relevance, so it is often referred to as trivial solution. We therefore require that the sine function in (9.21) vanishes: sin l D 0: (9.22) This can be fulfilled only if the argument l of the sine function corresponds to a multiple of , i.e. if: n l D n: (9.23) Herein, n is an integer greater than zero. We have to exclude the value n D 0, because it leads again to the trivial solution. Also negative values for n are excluded. We have added the additional index n to the value n to make clear that it depends

9.4 Buckling of bars: The four Euler cases

299

on n. The quantities n are also called the eigenvalues of the buckling problem under consideration. Since the number n is not bounded above, there is an infinite number of eigenvalues n . We substitute (9.16) into (9.23) and obtain: r Fn l D n: (9.24) EI This expression can be solved for the buckling load, and it follows: Fn D

n2  2 EI : l2

(9.25)

Thus, depending on the value n, there is an infinite number of solutions for the buckling load Fn . Technically relevant from the set of all possible buckling loads is always the smallest load, which results here with n D 1. Thus it follows: F1 D

12  2 EI  2 EI D D Fcrit : l2 l2

(9.26)

At this point it is of interest to determine the concrete form of the buckling shape w.x/. With 1 according to (9.16) we have: r F1  D : (9.27) 1 D EI l With (9.20) we obtain: w.x/ D C1 sin

 x

; (9.28) l i.e. the buckling shape is described by a sine function which describes one half wave along the entire length of the bar. Note, however, that the constant C1 remains undetermined. For its determination, investigations in the framework of a so-called geometrically nonlinear theory have to be carried out, which, however, is not part of this chapter. Accordingly, we know at this point that the buckling shape is described by the sine function given in (9.28), but we cannot make any statements about its amplitude. A calculation within the framework of the second order theory is obviously limited to the onset of buckling when the critical load is reached. Fig. 9.9 shows the first three buckling modes w1 , w2 , w3 of the bar together with the corresponding buckling loads F1 , F2 , F3 . Obviously, the first and technically relevant buckling mode w1 is described by one sine half-wave over the length of the bar, whereas the second buckling shape w2 describes two sine half-waves. Analogously, the third eigenmode w3 is already characterized by three sine half-waves, and quite analogous conclusions can be drawn for higher buckling modes. Also on the basis of this representation it can be concluded that only the buckling load F1 together with the corresponding buckling mode w1 is of interest – it is obvious that such a bar always buckles in a single sine half-wave over its entire length, but never with several half-waves.

300

9

Buckling of bars

2 2 F1 = 1 2 EI l

w1(x)=C1sin 1 x l 2 2 F2 = 2 2 EI l

2 2 F3 = 3 2 EI l

w2(x)=C1sin 2 x l

w3(x)=C1sin 3 x l

Fig. 9.9 The first three buckling modes for Euler case II.

9.4.2 Euler case I We consider the bar of Fig. 9.10, top. Let an elastic ideally straight bar of length l with the bending stiffness EI be given, which is rigidly clamped at its left end and free at its right end. At the free end x D l the compressive force F is applied. This special situation is commonly referred to as Euler case I. Again, we consider the bar in its deflected position (Fig. 9.10, middle). The bar exhibits the deflection w.x D l/ D ı at its free end x D l. To determine the buckling load, we consider the free body diagram of Fig. 9.10, bottom, and form the sum of all moments about the point S at an arbitrary location x. We obtain: M.x/ C F .ı  w.x// D 0;

(9.29)

Fig. 9.10 Euler case I.

F

x

EI l

w(x) δ

F x

l-x

M N

S δ-w(x)

V

F

l-x

9.4 Buckling of bars: The four Euler cases

which can be brought into the following form with 2 D

301 F EI

:

w 00 .x/ C 2 w.x/ D 2 ı:

(9.30)

This is a second order ordinary linear inhomogeneous differential equation with constant coefficients. Its total solution w.x/ is composed of a homogeneous solution wh .x/ and a particular solution wp .x/, where the homogeneous solution wh can be given as: wh .x/ D C1 sin x C C2 cos x: (9.31) Since the term 2 ı is a constant term, an approach of the form wp .x/ D C D const: for the particular solution is used. Substituting in (9.30) then immediately gives C D ı, so that: wp .x/ D ı: (9.32) The total solution of the buckling differential equation for Euler case I is then: w.x/ D wh .x/ C wp .x/ D C1 sin x C C2 cos x C ı:

(9.33)

We first consider the boundary condition that at the clamping point x D 0 the deflection of the bar w.x D 0/ vanishes, i.e. we require w.x D 0/ D 0. With (9.33) we obtain: w.x D 0/ D C2 C ı D 0; (9.34) from which C2 D ı:

(9.35)

The solution (9.33) then takes the following form: w.x/ D C1 sin x  ı cos x C ı:

(9.36)

The second boundary condition requires that at the clamped end x D 0 the inclination w 0 .x D 0/ of the member axis becomes zero: w 0 .x D 0/ D 0. With (9.33) this results in: C1 D 0: (9.37) If we neglect the solution  D 0 (which would be equivalent to a vanishing buckling load), only the solution C1 D 0 remains at this point. For (9.36) thus remains: w.x/ D ı.1  cos x/:

(9.38)

We now also require that at the free end x D l the deflection w.x/ must take the value ı: w.x D l/ D ı. This leads to: ı.1  cos l/ D ı:

(9.39)

cos l D 0:

(9.40)

Thus:

302

9

Buckling of bars

This condition can only be satisfied if the argument l of the cosine function is a multiple of , again using the notation n : n l D

2n  1 : 2

Herein n is again an integer greater than zero. With 2n D the buckling load Fn as follows: Fn D

(9.41) Fn EI

this can be solved for

.2n  1/2  2 EI : 4l 2

(9.42)

From this we obtain the technically relevant buckling load F1 with n D 1, it represents at the same time the critical load Fcrit : F1 D

 2 EI D Fcrit : 4l 2

(9.43)

 : 2l

(9.44)

The corresponding value 1 is: 1 D

Thus, the buckling shape w.x/ according to (9.38) can be given as: h  x i : w.x/ D ı 1  cos 2l

(9.45)

It is shown in Fig. 9.10. Again, the amplitude of the buckling shape, represented by the value ı, remains undetermined as already discussed for Euler case II.

9.4.3 Euler case III As a further elementary case of bar buckling (the so-called Euler case III) we consider the bar of Fig. 9.11, top. Given is a linear elastic and ideally straight bar of length l, which is clamped at its left end, whereas the right end is simply supported and is loaded by the compressive force F . We cut the bar at an arbitrary point x and form the moment equilibrium around the point S, taking into account the support reaction B at the right support. The result is: M.x/ C B.l  x/  F w.x/ D 0: (9.46) F This equation can be put into the following form with 2 D EI using the constitutive law M.x/ D EI w 00 .x/:

w 00 .x/ C 2 w.x/ D

B .l  x/: EI

(9.47)

9.4 Buckling of bars: The four Euler cases

303

Fig. 9.11 Euler case III.

F

x

EI l w(x)

F

x

l-x

M N

V w(x)

F

S

B l-x

Also in this case, the total solution of this inhomogeneous ordinary linear differential equation with constant coefficients is composed of a homogeneous solution wh .x/ and a particular solution wp .x/, where the homogeneous solution can be given as (cf. also (9.31)) wh .x/ D C1 sin x C C2 cos x:

(9.48)

For the particular solution wp .x/ we use an approach as follows: wp .x/ D C C Dx;

(9.49)

with the two constants C and D to be determined. Inserting (9.49) into the differential equation (9.47) results in the following constants C and D after coefficient comparison: B Bl ; DD : (9.50) C D F F Thus, the total solution of the differential equation (9.47) can be given as: w.x/ D wh .x/ C wp .x/ D C1 sin x C C2 cos x C

B .l  x/: F

(9.51)

The following boundary conditions apply at x D 0: w.x D 0/ D 0;

w 0 .x D 0/ D 0:

(9.52)

This leads to the constants C1 and C2 as follows: C1 D

B ; F

C2 D 

Bl : F

(9.53)

304

9

Buckling of bars

Thus, the buckling shape w.x/ can be represented according to (9.51) as: w.x/ D

  x Bl sin x  cos x C 1  : F l l

(9.54)

As is common for stability problems in the framework of second-order theory the amplitude of the buckling shape remains undetermined. Evaluating the boundary condition w.x D l/ D 0 at the right end of the bar results in:   Bl sin l  cos l D 0: (9.55) F l At this point only the zeroing of the parenthesis term leads to a meaningful result, which eventually yields: tan l D l: (9.56) This is an implicit equation for the determination of the buckling load Fcrit , which generally requires a numerical-iterative solution. Equations of this type occur frequently in the analysis of stability problems. Such an equation is called buckling condition. The solution can be determined straightforwardly by plotting both the tangent function tan l and the function l and finding the first intersection point greater than zero (Fig. 9.12). The intersection point is found at l D 4:49, which can also be represented as a multiple of  as l D 1:43. Thus, using (9.16), the buckling load Fcrit can be given as: Fcrit D

1:432  2 EI 2:04 2 EI D : l2 l2

(9.57)

Fig. 9.12 Functions f .l/ D l and f .l/ D tan.l/.

tan( l)

l tan( l)

/2 l

4.49

3 /2

9.4 Buckling of bars: The four Euler cases

305

9.4.4 Euler case IV As the fourth and last Euler case (the so-called Euler case IV) we want to consider the bar of Fig. 9.13 of length l which has the constant bending stiffness EI and is rigidly clamped at its left end at x D 0. At the right end, let the bar be guided in such a way that the deflection w and the slope w 0 are prevented, but a displacement in the axial direction and thus also an introduction of the compressive force F are possible. We cut the bar free at an arbitrary point x and consider the free body image shown in Fig. 9.13, bottom. Here, a support moment MB must be taken into account. We form the sum of all moments around the point S and obtain the following buckling differential equation: w 00 .x/ C 2 w.x/ D

MB : EI

(9.58)

This is a second-order ordinary linear inhomogeneous differential equation with constant coefficients. Its solution, consisting of homogeneous and particular solution, can be given as: w.x/ D C1 sin x C C2 cos x C

MB : F

(9.59)

The two constants C1 and C2 can be determined from the boundary conditions w.x D 0/ D 0 and w 0 .x D 0/ D 0 as: C1 D 0;

C2 D 

MB : F

(9.60)

The buckling form w.x/ is then: w.x/ D

MB .1  cos x/: F

(9.61)

Again, the amplitude of the buckling shape remains undetermined. Fig. 9.13 Euler case IV.

F

x

EI l w(x)

F x

l-x

M V N

w(x)

F S

MB l-x

306

9

Buckling of bars

The two boundary conditions w.x D l/ D 0 and w 0 .x D l/ D 0 give rise to the following two conditions: cos l D 1;

sin l D 0:

(9.62)

Both conditions are to be fulfilled simultaneously. Obviously, the first condition is satisfied if the argument of the cosine function is an integer positive even multiple of . The second condition, on the other hand, is satisfied if the argument of the sine function is an integer positive multiple of . Thus, both conditions can only be satisfied for l D 2; 4; 6; : : :. The technically relevant solution is again the smallest possible value, i.e. l D 2. From this then follows the critical load Fcrit as: 4 2 EI : (9.63) Fcrit D l2

9.4.5 Summary of the results The four Euler cases are summarized in Fig. 9.14. The following conclusions can be drawn from these results:  The buckling load depends on the modulus of elasticity of the material of the bar. Consequently, a high modulus of elasticity increases the buckling load Fcrit .  A high moment of inertia I results in a high buckling load Fcrit .  The higher the length of the bar, the smaller the buckling load Fcrit becomes. Thus, a long bar buckles at smaller loads than a short bar.

Euler case I

Euler case II

Euler case III

F

F

F

l

l

x

x

2

Fcrit = EI2 4l Fig. 9.14 Euler cases I-IV.

F

l

x

2

Fcrit = EI 2 l

Euler case IV

l

x

2

Fcrit = 2.04 EI 2 l

2 Fcrit = 4 2EI l

9.5 Buckling length

307

 The buckling load Fcrit is particularly strongly dependent on the boundary conditions of the bar. The lowest buckling load is obtained for Euler case I (Fig. 9.14, left), which can be explained by the free end at the location x D l. Obviously, free ends lead to a reduction of the buckling resistance. The buckling load for Euler case II is already four times higher than for Euler case IV, and there is even a factor of 16 between Euler cases I and IV. The correct consideration of the boundary conditions in the analysis of the stability behavior of bars is therefore particularly important.

9.5 Buckling length The buckling loads Fcrit for Euler cases I-IV can be expressed in the form Fcrit D

 2 EI  2 EI D ; .ˇl/2 s2

(9.64)

where we want to call the term s D ˇl the so-called buckling length. This is shown in Fig. 9.15. The buckling length is the distance between two inflection points of

F

F

F

F

l=0.7l l=l

l=0.5l

l=2l

2 Fcrit= EI2 4l

=2

2 Fcrit= EI 2 l

2 Fcrit= 2.04 EI 2 l

=1

Fig. 9.15 Definition of the buckling length s D ˇl.

=0.7

2 Fcrit= 4 2EI l

=0.5

308

9

Buckling of bars

the buckling shape. It amounts to 2l for Euler case I, l for Euler case II, 0:7l for Euler case III and finally 0:5l for Euler case IV. Thus, all four Euler cases can be represented in the form (9.64), and only the correct buckling length s D ˇl or the correct buckling coefficient ˇ has to be used. Example 9.1

Consider the static system of Fig. 9.16, top. We want to determine how high the moment of inertia I of the column has to be for a triple safety against buckling. Furthermore, how does the result change if the column is supported at mid-height by an additional support (Fig. 9.16, bottom)? The column has the elastic modulus E, the lengths l and h and the angle ˛ are given. We determine the compressive force N acting on the column. In order to do this, we first determine the forces S of the diagonal struts. From the free body image of Fig. 9.16, right, we obtain: SD

F : 2 sin ˛

(9.65)

The compressive force N then follows as: N D 2S cos ˛ D F cot ˛:

(9.66)

As shown in Fig. 9.16, top, the buckling shape of the column corresponds to Euler case II. The critical load of the column is thus: Fcrit D

l

 2 EI ; s2

(9.67)

l α h

F

S

F

F

α S

h

α

S α h F

α α N

F

α

h

Fig. 9.16 Static system (top), system with additional support of the column (bottom).

S

9.5 Buckling length

309

where the buckling length is s D 2h, thus: Fcrit D

 2 EI  2 EI D : 2 .2h/ 4h2

(9.68)

The criterion for a safety factor of 3 against buckling to be applied here is as follows: 1 jN j  Fcrit ; (9.69) 3 thus:  2 EI : (9.70) F cot ˛  12h2 This can be solved for the moment of inertia I , and we obtain: I 

12F h2 cot ˛: E 2

(9.71)

If the column is supported by an additional centric support, the buckling length is s D h, and the critical load Fcrit is then: Fcrit D

 2 EI : h2

(9.72)

From the criterion jN j  13 Fcrit then follows: I 

3F h2 cot ˛: E 2

(9.73) J

Example 9.2

Consider the lever of Fig. 9.17, top, which is supported by the two columns 1 and 2 (moduli of elasticity E1 , E2 , moments of inertia I1 , I2 ). The two moments of inertia I1 , I2 of the columns are to be determined such that there is a triple safety against buckling. The compressive forces N1 and N2 can be determined from the free body image of Fig. 9.17 as: N1 D F; N2 D 2F: (9.74) We first consider column 1, and if this column is the cause of the buckling failure of the structure, a buckling figure will occur as shown in 9.17, bottom left. This corresponds to Euler case II with the buckling length s D 2l. The buckling load Fcrit;1 is thus:  2 E1 I1  2 E1 I1 Fcrit;1 D D : (9.75) .2l/2 4l 2

310

9

Buckling of bars

The criterion for triple buckling safety of column 1 is: N1 

1 Fcrit;1 : 3

(9.76)

This can be solved for I1 using (9.74) and (9.75), and we obtain: I1  12

F l2 : E1  2

(9.77)

The buckling shape for column 2 is shown in Fig. 9.17, bottom right. This corresponds to Euler case I, and the buckling length s is s D 2  3l D 6l. The buckling load can thus be given as: Fcrit;2 D

 2 E2 I2  2 E2 I2 D : .6l/2 36l 2

(9.78)

For a triple buckling safety we require: N2 

1 Fcrit;2 : 3

(9.79)

Solving this expression for I2 with (9.74) and (9.78), we obtain: I2  216

F l2 : E2  2

(9.80) J

Fig. 9.17 Static system (top), buckling figures (bottom).

E1,I1 F

E2,I2

2l

2l

2l

F

N1 N2

3l

9.5 Buckling length

311

Example 9.3

The horizontal beam of length 2l shown in Fig. 9.18, top, is loaded by the constant line load q0 and is simply supported at its left end. At its right end, the beam is supported by a vertical column of length h. The column has the bending stiffness EI . We want to determine the bending stiffness EI so that there is n-fold safety against buckling for the column. Furthermore, we want to determine how the result changes if the support at the left end is replaced by a fixed hinged support as shown. Due to the applied load in the form of the constant line load q0 , the column experiences the compressive force N D q0 l. The resulting buckling figure is shown in Fig. 9.18, middle, and corresponds to Euler case I with the buckling length 2h. The critical load then follows as: Fcrit D

 2 EI  2 EI D : .2h/2 4h2

(9.81)

For an n-fold buckling safety we require N 

Fig. 9.18 Static system (top), buckling figure (middle), buckling figure with a fixed hinged support at the left end of beam (bottom).

1 Fcrit : n

(9.82)

q0

EI

2l q0

q0

h

312

9

Buckling of bars

This can be solved for the bending stiffness EI as follows: EI 

4nq0 h2 l : 2

(9.83)

If the support at the left end of the beam is now exchanged for a fixed hinged support, the buckling figure results as shown in Fig. 9.18, bottom. This corresponds to Euler case III, and the corresponding buckling load is: Fcrit D 2:04

 2 EI : h2

(9.84)

With N D q0 l and the criterion (9.82) the bending stiffness EI follows as: EI 

nq0 h2 l : 2:04 2

(9.85) J

9.6 General form of the buckling differential equation The buckling differential equations (9.17), (9.30), (9.47) and (9.58) derived so far have validity only for the specific Euler cases under consideration and are therefore not generally applicable to other buckling problems. We therefore want to extend our considerations at this point and derive a general form of the buckling differential equation, employing very basic equilibrium considerations and using the free body image of Fig. 9.19. Shown here is a bar under a compressive force F . No boundary conditions are provided at this point for the bar in order to make the considerations as general as possible. For the sake of simplicity, we restrict ourselves to the case of a constant bending stiffness EI over the entire length l of the beam. Using the free body image, we now form the force and moment balances. The equilibrium of forces in the direction of the normal force N results in: N  V d'  N  dN D 0;

(9.86)

dN C V d' D 0:

(9.87)

or: Expansion of this expression as follows

yields:

d' dN dx C V dx D 0 dx dx

(9.88)

N 0 C V ' 0 D 0:

(9.89)

In quite the same way we can form the equilibrium of forces in the direction of the transverse shear force: (9.90) V 0  N' 0 D 0:

9.6 General form of the buckling differential equation

F

F

313

(V+dV)sin(d φ) ~ Vd φ

x

V+dV dφ

R

M

dφ

(V+dV)cos(d φ) ~ V+dV

N V

R

N+dN dφ

(N+dN)sin(d φ) ~ Nd φ (N+dN)cos(d φ) ~ N+dN

V+dV

M+dM N+dN Fig. 9.19 Free body image.

Finally, we form the moment equilibrium around the positive cutting edge of the free body image of Fig. 9.19. This results in the well-known relationship between bending moment and transverse shear force: M 0 D V:

(9.91)

We first consider equation (9.89). With the constitutive law M D EI ' 0 and V D EI ' 00 we obtain: N 0 C EI ' 00 ' 0 D 0: (9.92) The expression ' 00 ' 0 can be considered to be very small, so we can neglect the second term in (9.92). Thus: N 0 D 0: (9.93) This means that the normal force N in the considered bar is constant over the bar length and its first derivative N 0 disappears, which is immediately obvious due to the considered bar situation. For equation (9.90) we obtain with V 0 D EI ' 000 and N D F : EI ' 000 C F ' 0 D 0;

(9.94)

or with ' D w 0 (assumption of small angles): EI w 0000 C F w 00 D 0: With the abbreviation 2 D

F EI

(9.95)

(9.96)

314

9

Buckling of bars

Equation (9.95) can be written as: w 0000 C 2 w 00 D 0:

(9.97)

This is the general buckling differential equation, which can be used to treat arbitrary bar buckling problems. It is a fourth order ordinary linear homogeneous differential equation. Its general solution is: w D C1 sin.x/ C C2 cos.x/ C C3 x C C4 ;

(9.98)

with the integration constants C1 , C2 , C3 and C4 . We demonstrate the procedure for determining the buckling load using (9.97) and (9.98) on the basis of Euler case II (see Fig. 9.8). The boundary conditions for this buckling situation can be given as follows: w.x D 0/ D 0; M.x D 0/ D EI w 00 .x D 0/ D 0; w.x D l/ D 0; M.x D l/ D EI w 00 .x D l/ D 0:

(9.99)

Evaluation of the boundary conditions employing (9.98) gives the following linear system of equations: C2 C C4 D 0; C2 D 0; C1 sin.l/ C C2 cos.l/ C C3 l C C4 D 0; C1 sin.l/ C C2 cos.l/ D 0: In vector-matrix notation, this can be represented as: 30 1 0 1 2 0 0 1 0 1 C1 7BC2 C B0C 6 0 1 0 0 7B C B C 6 4sin.l/ cos.l/ l 15@C3 A D @0A: 0 sin.l/ cos.l/ 0 0 C4

(9.100)

(9.101)

In order to avoid the trivial solution we determine the coefficient determinant and equate it to zero: ˇ ˇ ˇ 0 ˇ ˇ 1 0 1ˇˇ ˇ ˇ 0 1 0ˇˇ ˇ 0 ˇ ˇ 1 0 0 ˇ ˇ ˇ ˇ ˇsin.l/ cos.l/ l 1ˇ D ˇsin.l/ cos.l/ l ˇ ˇ ˇ ˇsin.l/ cos.l/ 0ˇ ˇsin.l/ cos.l/ 0 0ˇ ˇ ˇ ˇsin.l/ l ˇ ˇ ˇ D l sin.l/ D 0: Dˇ (9.102) sin.l/ 0ˇ The buckling condition thus determined can be satisfied only if l is an integer multiple of : l D n; for n D 1; 2; 3; : : : (9.103)

9.6 General form of the buckling differential equation

315

Using  according to (9.96), we immediately obtain the buckling load as follows: Fn D

n2  2 EI : l2

(9.104)

Technically relevant is the smallest value with n D 1, so that: Fcrit D

 2 EI : l2

(9.105)

The buckling load determined in this way agrees with the expression (9.26). We also determine the corresponding buckling shape from the system of equations (9.101). It is immediately obvious that we can set the constants C2 and C4 to zero. From the third equation in (9.101) we obtain C1 sin.l/ C C3 l D 0;

(9.106)

which we can solve for C3 : C3 D C1

sin.l/ ; l

(9.107)

so that the solutiong (9.98) takes the following form: w.x/ D C1 sin.x/  C1

sin.l/ x: l

(9.108)

Since sin.l/ must be zero according to the eigenvalue equation (9.102), we have: w.x/ D C1 sin.x/:

(9.109)

With  D n l the buckling shape w.x/ for the technically relevant case n D 1 is obtained as follows:  x w.x/ D C1 sin : (9.110) l Again, the constant C1 and thus the amplitude of the buckling mode remains undetermined. Example 9.4

We consider Euler case I (Figs. 9.10 and 9.20) and want to use the differential equation (9.97) together with its solution (9.98) to determine the critical load for this situation. The boundary conditions to be applied here are as follows. At the clamped end of the bar, both the deflection w and its derivative w 0 must vanish: w.x D 0/ D 0;

w 0 .x D 0/ D 0:

(9.111)

316

9

Fig. 9.20 Euler case I.

F

Buckling of bars

F

F EI l φ=w

x

V

The bending moment M D EI w 00 must become zero at the top of the beam: EI w 00 .x D l/ D 0;

(9.112)

which can also be formulated more simply as: w 00 .x D l/ D 0:

(9.113)

As a final condition, the equilibrium between the transverse shear force V and the compressive force F must be considered (see Fig. 9.20, right). The sum of forces in the direction of the transverse shear force results in: V .x D l/  F sin.'.x D l// D 0:

(9.114)

Assuming small angles, i.e. sin.'/ D ' D w 0 , gives: V .x D l/  F w 0 .x D l/ D 0:

(9.115)

With the constitutive relation V D EI w 00 it follows:

or

 EI w 000 .x D l/  F w 0 .x D l/ D 0;

(9.116)

w 000 .x D l/ C 2 w 0 .x D l/ D 0:

(9.117)

This last boundary condition states that the transverse shear force in the buckled state has a component in the direction of the applied compressive force F . Evaluating the four boundary conditions yields the following linear homogeneous system of equations for the constants C1 ; : : : ; C4 : C2 C C4 D 0; C1  C C3 D 0; C1 sin.l/ C C2 cos.l/ D 0; C3 D 0:

(9.118)

9.6 General form of the buckling differential equation

In vector-matrix notation this can be written as: 30 1 0 1 2 0 0 1 0 1 C1 7BC2 C B0C 6  0 1 0 7B C B C 6 4sin.l/ cos.l/ 0 05@C3 A D @0A: 0 0 0 1 0 C4

317

(9.119)

This is a homogeneous linear system of equations with four equations for the four constants C1 ; : : : ; C4 . To avoid the trivial solution we require the vanishing of the coefficient determinant: ˇ ˇ ˇ 0 1 0 1ˇˇ ˇ ˇ  0 1 0ˇˇ ˇ (9.120) ˇsin.l/ cos.l/ 0 0ˇ D 0: ˇ ˇ ˇ 0 0 1 0ˇ This leads to: ˇ ˇ 0 1 ˇ ˇ  0 ˇ ˇsin.l/ cos.l/ ˇ ˇ 0 0

0 1 0 1

ˇ ˇ ˇ 1ˇˇ ˇ  0 1ˇˇ ˇ 0ˇˇ D ˇˇsin.l/ cos.l/ 0ˇˇ 0ˇˇ ˇ 0 0 1ˇ 0ˇ ˇ ˇ ˇ  0 ˇˇ ˇ D  cos.l/ D 0: (9.121) D ˇ sin.l/ cos.l/ˇ

The expression cos.l/ D 0 is the buckling condition required to determine the buckling load Fcrit . It is fulfilled if the following holds: l D

2n  1 ; 2

for n D 1; 2; 3; : : :

(9.122)

From this, the buckling load Fn follows as a function of n: Fn D

 2 EI .2n  1/2 : 4l 2

(9.123)

The technically relevant case follows with n D 1 F1 D

 2 EI D Fcrit : 4l 2

This expression is in agreement with (9.43). J

(9.124)

318

9

Buckling of bars

Example 9.5

Consider the bar shown in Fig. 9.21. The bar has the constant bending stiffness EI and length l and is clamped at its lower end at x D 0. At its upper end at x D l, the bar is guided horizontally in such a way that a bending moment can be absorbed, but no transverse shear force can occur at this point. Starting from the solution (9.98) of the buckling differential equation (9.97), the evaluation of the boundary conditions w.x D 0/ D 0; w 0 .x D 0/ D 0; w 0 .x D l/ D 0; w 000 .x D l/ D 0

(9.125)

yields the following homogeneous linear system of equations: C2 C C4 C1  C C3 C1  cos l  C2  sin l C C3 C1 cos l C C2 sin l

D 0; D 0; D 0; D 0:

(9.126)

To avoid the trivial solution, the coefficient determinant is set to zero: ˇ ˇ ˇ ˇ ˇ 0 1 0 1ˇˇ ˇ ˇ  0 1ˇˇ ˇ ˇ ˇ  0 1 0 ˇ ˇ ˇ ˇ ˇ cos.l/  sin.l/ 1 0ˇ D ˇ cos.l/  sin.l/ 1ˇ ˇ ˇ  cos l ˇ sin l 0ˇ ˇ  cos l sin l 0 0ˇ D  sin l D 0:

(9.127)

The resultant buckling condition is therefore sin l D 0:

(9.128)

Fig. 9.21 Bar under compressive load F clamped at the lower end with a horizontally guided support at the upper end.

F

F

EI l

x

9.6 General form of the buckling differential equation

319

This condition apparently agrees with the buckling condition of Euler case II (see (9.22)), so that the technically relevant buckling load is obtained as: Fcrit D

 2 EI : l2

(9.129) J

Example 9.6

Consider a straight elastic bar of length l with constant bending stiffness EI which is simply supported at its left end at x D 0 and loaded by the compressive force F . The right end at x D l is also simply supported and is reinforced by a torsional spring with the stiffness k' (see Fig. 9.22). We want to determine the buckling condition for this bar. For this situation, the boundary conditions are: w.x D 0/ D 0; M.x D 0/ D EI w 00 .x D 0/ D 0; w.x D l/ D 0; M.x D l/ C k' w 0 .x D l/ D EI w 00 .x D l/ C k' w 0 .x D l/ D 0:

(9.130)

Evaluating these conditions with the general solution (9.98) of the buckling differential equation (9.97), we obtain the following linear homogeneous equation system: C2 C C4 D 0; C2 D 0; C1 sin.l/ C C2 cos.l/ C C3 l C C4 D 0; .EI 2 sin.l/ C k'  cos.l//C1 C.EI 2 cos.l/  k'  sin.l//C2 C C3 k' D 0:

(9.131)

To avoid the trivial solution, we set the coefficient determinant to zero and, after a short calculation, we obtain the following buckling condition which describes the buckling of the elastically clamped bar: k' sin.l/  l.EI  sin.l/ C k' cos.l// D 0:

(9.132)

Obviously, this is an implicit buckling condition that requires a numerical evaluation. This will not be discussed here. Fig. 9.22 Bar with elastic clamping.

kφ

F x l

320

9

Buckling of bars

The present buckling situation includes two limit cases. If the spring stiffness k' tends to zero, then the buckling condition (9.132) reduces to: sin.l/ D 0:

(9.133)

This corresponds to the buckling condition for Euler case II. For the limit case k' ! 1, which describes a rigid clamping at x D l, the following buckling condition is obtained from (9.132): tan.l/  l D 0:

(9.134)

This buckling condition has already been derived for Euler case III, which is an equally obvious result. J

Index

A Angle of twist, 208

B Bar, 45 Beam theory, 75 Beams, 73 Bending angle, 73 Bending stiffness, 80 Betti’s theorem, 278 Biaxial bending, 78, 169 Bifurcational buckling, 291 Bredt’s formulas First Bredt formula, 215 Second Bredt formula, 218 Buckling Energy method, 294 Influence of boundary conditions, 307 Linearization, 294 Buckling condition, 304 Buckling differential equation, 298, 314 Buckling length, 307 Buckling load, 290

C Cauchy stress tensor, 7 Compatibility condition, 272 Complementary strain energy, 240 Complementary strain energy density, 240 Complementary virtual work, 253 Constitutive equations, 1, 15

Constitutive law, 50 Euler–Bernoulli beam, 80 Torsion, 210 Critical load, 290

D Deformation, 9 Deviation moment, 79 Disk, 23 Displacements, 1, 10

E Effective cross-sectional area, 246 Elastic line, 127 Equilibrium Bifurcation, 291 Indifferent, 291 Stable, 290 Unstable, 291 Equilibrium conditions, 1, 127 Local, 7, 8 Euler cases, 297 Euler–Bernoulli beam theory, 75 Extensional stiffness, 51 External work, 236

F First cross-sectional normalization, 88 Flexural buckling, 290 Force equilibrium, 8 Force method, 64, 261 Fork restraint, 225

© The Author(s), under exclusive license to Springer-Verlag GmbH, DE, part of Springer Nature 2023 C. Mittelstedt, Engineering Mechanics 2: Strength of Materials, https://doi.org/10.1007/978-3-662-66590-9

321

322 G Generalized Hooke’s law, 14 Geometric boundary conditions, 129 Geometric linearity, 11

H Hooke’s law, 14 Hydrostatic stress state, 36 Hypothesis Flat cross-section, 75 Normal hypothesis, 75 Straight radii, 209

I Indifferent equilibrium, 291, 296 Inertia circle, 118 Infinitesimal strain tensor, 14 Integral table, 259 Internal energy, 237 Internal work, 236 Invariant, 30, 119

J Joule, 234

K Kinematic equations, 1, 10, 13 Kinematics Euler–Bernoulli beam, 77

L Line loads, 4 Linear elasticity, 14

M Material law, 15 Maxwell’s theorem, 279 Modulus of elasticity, 15 Mohr’s stress circle, 33 Moment equilibrium, 6 Moments of inertia, 79 Multi-span beam, 140

N Neutral axis, 105 Normal strain, 9, 10 Normal stress, 2, 5

Index P Parallel-axis theorem, 89 Plane stress state, 23 Point load, 4 Poisson’s ratio, 15 Polar moment of inertia, 80 Potential, 237 Principal axes Moments of inertia, 115 Normal stresses, 30 Principal direction Normal stresses, 30 Shear stresses, 31 Principal moments of inertia, 117 Principal shear stress, 32 Principal stresses, 30 Principle of conservation of energy, 237 Principle of virtual forces, 255 Principle of work and energy, 237

R Resistance moment, 120

S Second cross-sectional normalization, 115 Second-order theory, 297 Section modulus, 120 Shear center, 201 Shear correction factor, 246 Shear flow, 176, 186 Torsion, 213 Shear modulus, 16 Shear strains, 10 Shear stress, 3, 5 Torsion, 210 Transverse shear force, 175 Skeleton line, 83, 211 Stability theory, 290 Stable equilibrium, 290, 296 Standard bending cases, 153 Static boundary conditions, 129 Static moment, 79 Statically indeterminate systems, 270 Steiner’s theorem, 89 Strain energy, 237 Strain energy density, 238 Strain state, 1 Strain tensor, 14 Strains, 10 Stress, 2 Stress state, 1, 5 Stress vector, 5

Index St.-Venant’s torsion, 207 Superposition principle, 132 Surface loads, 4

T Thermal expansion coefficient, 17 Torsion, 207 Axis of rotation, 213 Constitutive law, 210 First Bredt formula, 215 Second Bredt formula, 218 Shear flow, 213 Shear stress, 210 Twist, 209 Torsional moment, 209 Torsional moment of inertia, 210 Torsional resistance moment, 212, 215 Torsional stiffness, 210 Transformation equations Moments of inertia, 117 Plane stress state, 30

323 Transition conditions, 55, 141 Transverse contraction, 10, 15 Twist, 209

U Uniaxial bending, 78 Unit load theorem, 255 Unstable equilibrium, 291, 296

V Virtual forces, 254 Volume forces, 4

W Warping, 217 Warping-free cross-section, 209 Work, 233