W. T. Koiter's Elastic Stability of Solids and Structures

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W. T. KOITER’S ELASTIC STABILITY OF SOLIDS AND STRUCTURES This book deals with the elastic stability of solids and structures, for which Warner Koiter was the world’s leading expert of his time. It begins with fundamental aspects of stability, relating the basic notions of dynamic stability to more traditional quasi-static approaches. The book is concerned not only with buckling, or linear instability, but most importantly with nonlinear postbuckling behavior and imperfection sensitivity. After laying out the general theory, Koiter applies the theory to a number of applications, with a chapter devoted to each. These include a variety of beam, plate, and shell structural problems and some basic continuum elasticity problems. Koiter’s classic results on the nonlinear buckling and imperfection sensitivity of cylindrical and spherical shells are included. The treatments of both the fundamental aspects and the applications are completely self-contained. This book was recorded as a detailed set of notes by Arnold van der Heijden from W. T. Koiter’s last set of lectures on stability theory at TU Delft. Arnold M. A. van der Heijden has his own consultancy, HESTOCON Consultancy B.V. He received his master’s and Ph.D. degrees, with honors, in mechanical engineering and applied mechanics under Professor Koiter. He has been a technical and scientific staff member in the Applied Mechanics Laboratory at Delft University of Technology, an honorary research Fellow at Harvard University, a professor at Eindhoven Technical University, a board member of the Department of Applied Mechanics of the Royal Dutch Institute for Engineers, and co-editor (with J. F. Besseling) of the Koiter symposium book Trends in Solid Mechanics. Dr. van der Heijden has worked on staff and consulted for many corporations, including Royal Dutch Shell (Pernis and The Hague), ABB Lummus Global, and as a project leader of ATEX at General Electric Advanced Materials, SABIC, and Essent. He has done gas explosion calculations for offshore platforms, including structural analysis. He is currently working on improvements in safety management for ProRail.

W. T. Koiter’s Elastic Stability of Solids and Structures Edited by

Arnold M. A. van der Heijden Technische Universiteit Delft

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521515283 © Arnold van der Heijden 2009 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2008

ISBN-13

978-0-511-43674-1

eBook (EBL)

ISBN-13

978-0-521-51528-3

hardback

Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Contents

Preface

page vii

1. Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Discrete systems

1

2. Continuous Elastic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.1 2.2 2.3 2.4

Thermodynamic background Theorems on stability and instability The stability limit Equilibrium states for loads in the neighborhood of the buckling load 2.5 The influence of imperfections 2.6 On the determination of the energy functional for an elastic body

7 11 18 27 41 47

3. Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.1 3.2 3.3 3.4 3.5 3.6

The incompressible bar (the problem of the elastica) Bar with variable cross section and variable load distribution The elastically supported beam Simple two-bar frame Simple two-bar frame loaded symmetrically Bending and torsion of thin-walled cross sections under compression 3.7 Infinite plate between flat smooth stamps 3.8 Helical spring with a small pitch 3.9 Torsion of a shaft 3.10 Torsion of a shaft with a Cardan (Hooke’s) joint 3.11 Lateral buckling of a beam loaded in bending 3.12 Buckling of plates loaded in their plane 3.13 Post-buckling behavior of plates loaded in their plane ´ an-F ´ ¨ 3.14 The “von Karm oppl Equations”

55 59 61 67 72 78 84 101 110 119 126 137 158 166

v

vi

Contents

3.15 Buckling and post-buckling behavior of shells using shallow shell theory 3.16 Buckling behavior of a spherical shell under uniform external pressure using the general theory of shells 3.17 Buckling of circular cylindrical shells 3.18 The influence of more-or-less localized short-wave imperfections on the buckling of circular cylindrical shells under axial compression

169 182 201

221

Selected Publications of W. T. Koiter on Elastic Stability Theory

227

Index

229

Preface

These lecture notes were made after Professor Koiter’s last official course at Delft’s University of Technology, in the academic year 1978–79. Although these notes were prepared in close collaboration with Professor Koiter, they are written in the author’s style. The author is therefore fully responsible for possible errors. This course covers the entire field of elastic stability, although recent developments in the field of stiffened plates and shells are not included. Hopefully, these lecture notes reflect some of the atmosphere of Dr. Koiter’s unique lectures. Delft, June 10, 2008

A. M. A. v. d. Heijden

vii

1

Stability

1.1 Discrete systems Consider a system with a finite number of degrees of freedom. The position of this system is represented by a position vector q(q1 , q2 . . . qn ), where qi (i ∈ 1 . . . n) are n independent coordinates. It is assumed that the system is holonomic, i.e., no relations exist between the derivatives of the coordinates, and scleronomic, i.e., the factor time is not explicitly needed in the description of the system.† Let q˙ i be the generalized velocities. The kinetic energy T is then a homogeneous quadratic function of the generalized velocities, and hence T can be written as T=

1 aij (q)q˙ i q˙ j . 2

(1.1.1)

When the system is non-sclerononic, terms linear in the velocities and a term independent of q˙ i must be added. The coefficient aij (q) is called the inertia matrix. The forces acting upon the system can be expressed by a generalized force vector Q defined by Qi δqi = v.w.,

(1.1.2)

where the right-hand side stands for the virtual work of all the forces acting upon the system. In general, this expression is not a total differential. However, for an important class of problems, it is. Systems for which 1.1.2) is a total differential are called conservative systems. In that case we have Qi δqi = −δP (q) ,

(1.1.3)

where δP (q) is a total differential and P (q) is called the potential energy. In the following, we mainly restrict our attention to conservative systems because for elastic systems, conservative forces play an important role. Introducing a kinetic potential L defined by L = T − P, †

(1.1.4)

This implies that dq = q, k dq k.

1

2

Stability

the Lagrangian equations for a conservative system are d ∂L ∂L − i = 0, dt ∂q˙ i ∂q˙

(i ∈ 1, . . . , n) .

(1.1.5)

Using the expression (1.1.1), we may rewrite this equation to yield d 1 (aij q˙ i ) − ahk,i q˙ h q˙ k + P,i = 0, dt 2

(1.1.6)

where ahk,i =

∂ahk (q) , ∂qi

P,i =

∂P (q) ∂qi

However, it often happens that non-conservative forces are present (e.g., damping forces). It is then advantageous to take these into account separately as follows: Qi δqi = δP (q) + Q∗i δqi ,

(1.1.7)

where Q∗ is the vector of non-conservative forces. The equations of motion then read d ∂L ∂L − i = Q∗i , (i ∈ 1 · · · n) . (1.1.8) i dt ∂q˙ ∂q˙ These are n second-order ordinary differential equations. Let us now consider the stability of discrete systems. For a system to be in equilibrium, the velocities (and hence the kinetic energy) have to vanish. This implies that for a conservative system, we have P,i = 0.

(1.1.9)

In words: The potential energy has a stationary value. By stability we mean that a small disturbance from the state of equilibrium does not cause large deviations from this state of equilibrium. A disturbance from the state of equilibrium implies that the velocities are nonzero or that the position differs from the equilibrium position. We can always choose our coordinate system such that the equilibrium position is given by q = 0. Furthermore, we can always choose the potential energy in such a way that it vanishes in the equilibrium position. Doing so, we may write 1 P,ij (0) qi q j + · · · . (1.1.10) 2 To be able to give a more exact definition of stability, we need a measure to denote the deviation from the state of equilibrium. Remembering that in equilibrium we ˙ is introduced with the following properties: have q = q˙ = 0, a number ρ (q, q) P=

1)

˙ ≤ 0 for ρ (q, q)

q = 0 or

q˙ = 0,

2)

ρ (q1 , q2 , q˙ 1 + q˙ 2 ) ≤ ρ (q1 , q˙ 1 ) ρ (q2 , q˙ 2 ) (triangle inequality),

3)

˙ = |α| ρ (q, q) ˙ ρ (αq, αq)

(1.1.11) α ∈ R.

1.1 Discrete systems

3

We are now in a position to define the following stability criterion. An equilibrium position is stable if and only if for each positive number ε there exists a positive number δ(ε) such that for all disturbances of the equilibrium at the ˙ ˙ time t > 0, with ρ [q(0), q(0)] < δ, the motion for t > 0 satisfies ρ [q(t), q(t)] < ε. Notice that the statement about stability depends on the measure that is used. Different measures yield different criteria for stability. Notice further that different measures may be used for t = 0 and t > 0. This freedom is of great importance for applications. For example, suitable choices for ρ are n 1/2 n i 2 i 2 ρ= (q ) + (q˙ ) , i=1

i=1

ρ = max qi + max q˙ i . For a conservative system, T + P = constant. This well-known result can easily be derived from Lagrange’s equations for a conservative, holonomic, and scleronomic system. Multiplying the equations by q˙ i , we obtain q˙ i or

d ∂L ∂L − q˙ i i = 0 i dt ∂q˙ ∂q

d ∂L ∂L i ∂L q˙ i − q¨ i i − q˙ i i = 0 dt ∂q˙ ∂q˙ ∂q

or

d d i ∂L q˙ i − L = 0. dt ∂q˙ dt

Using Euler’s theorem for homogeneous quadratic functions, we readily obtain d d (2T) − (T − P) = 0, dt dt from which follows T + P = E,

(1.1.12)

where E = T(t = 0) + P(t = 0). This equation enables us to make the following statement about stability. Theorem. The equilibrium is stable provided the potential energy is positive-definite. To prove this theorem, we introduce the following norms: q2 =

n

(qi )2

i=1

˙ 2= q

n i=1

(q˙ i )2 .

4

Stability

Let d(c) denote the minimum of P(q) on the hypersphere q = c . P(q) is positivedefinite when d(c) is a monotonically increasing function of c on the sphere δ ≤ c < R. Proof. T + P = constant = E, T is positive or zero, and P is positive-definite. Restrict the initial disturbance so that q (0) < c1

and

E < d (c1 ) .

This means that T (t = 0) < d (c1 ). Because T + P < d (c1 ) and P is positive-definite, it follows that T < d (c1 ) for all t. On the other hand, because T is positive or zero, it follows that P < d (c1 ) for all t. A similar argument holds for a disturbance ˙ < c 2 . Hence, we may choose an arbitrary (small) disturbance and the dis q(0) placements and velocities will always remain within definable bounds. The converse of this theorem has not yet been proven in all generality. To see some of the difficulties that are encountered, we consider the following example (one degree of freedom): −2

P(q) = e−q cos q−2 . For q = 0, all the derivatives vanish. However, in the immediate vicinity of the origin there are always negative values of P. In spite of this, the system is stable for sufficiently small disturbances. Actual physical systems are never exactly conservative, i.e., there is always some dissipation. The approximation by a conservative system is often a very good approximation. In the presence of damping forces, we need the Lagrangian equations with an additional term for the non-conservative forces. Multiplying by q˙ i , q˙ i

d ∂L ∂L − q˙ i i = Q∗i q˙ i , dt ∂q˙ i ∂q

(i ∈ 1, . . . , n)

from which follows d ˙ . (T + P) = Q∗i q˙ i ≡ −D (q, q) dt

(1.1.13)

Damping implies that the dissipation function D > 0 for q˙ = 0. We now make the following assumptions: 1) The damping forces have the property that Q∗i → 0 for q˙ → 0. ˙ = 0. 2) D(q, q˙ i ) > 0 for q 3) P (q) does not possess stationary values for q˙ < c except at q = 0. Systems satisfying these conditions are called pseudo-conservative. Notice that because of restriction (1), dry friction forces are excluded. Theorem. In the presence of (positive) damping forces, a system with an indefinite potential energy is unstable.

1.1 Discrete systems

5

Proof. If P is indefinite, consider a disturbance of the equilibrium configuration with zero velocity and negative potential energy. The initial total energy is thus negative and, as this configuration cannot be in equilibrium, motion must result, as a result of which energy is dissipated. The total energy must decrease, so the system cannot stay in the vicinity of the origin, which means that the equilibrium configuration is unstable. The great advantage of this stability theorem is that it does not involve the kinetic energy, and hence the inertia matrix aij (q). For a conservative or pseudoconservative system, the stability criterion only depends on the potential energy (a quasi-static criterion). In general, the stability problem is a dynamic problem, and the kinetic energy plays an essential role. An example of such a problem is the behavior of the wings of an airplane in an airflow. In this case, the forces do not depend on only the geometry but also on the velocities. For static loads, it is often sufficient to restrict oneself to conservative loads (e.g., deadweight loads). A more severe restriction for continuous systems is that we must restrict ourselves to elastic systems, i.e., to systems where there is a potential for the internal energy. Such a potential does not exist when plasticity occurs. Let us now have a closer look at the stability problem. As mentioned previously, the stability criterion is fully determined by the potential energy P(q). In the equilibrium position, we have chosen P (0) = 0 and q = 0 so that we may write P (q) =

1 1 P,ij (0) qi qj + · · · ≡ cij qi q j + · · · , 2 2

(1.1.14)

where cij denotes the stiffness matrix in the equilibrium position. It follows that when the stiffness matrix is positive-definite, P (q) is positive-definite and the system is stable. If cij is indefinite (or negative-definite) then the system is unstable. If cij is semi-definite-positive (i.e., non-negative and zero for at least a deflection in one direction), then we must consider higher-order terms in the expansion for P. This case is called a critical case of neutral equilibrium. We shall consider this case in more detail. It is convenient to transform the quadratic form (1.1.14) to a sum of quadratic terms. If the form is positive-definite, then the coefficients in the transformed form are all positive. Applying this transform to (1.1.14) and denoting the transformed coordinates again by qi , we may write 1 ci (qi )2 + · · · + () (q1 )3 + · · · . 2 n

P (q) =

(1.1.15)

i=1

Further, we order the coefficients ci such that c1 ≤ c2 ≤ c3 ≤ · · · ≤ cn . We now consider the case c1 = 0, c2 > 0. Taking all qi = 0 (i > 1), the dominant term will be (q1 )3 . This term can attain negative values, and hence the system will be unstable. A necessary condition for stability is that the coefficient of (q1 )3 is equal

6

Stability y x = y2

x = 2y

f ( x, y) < 0 2

x

Figure 1.1.1

to zero. A further necessary condition for stability is that the coefficient of (q1 )4 is positive. However, this condition is insufficient, as will be shown in the following example. Consider the function P = f (x, y) = (x − y2 )(x − 2y2 ) = x2 − 3xy2 + 2y4 .

(1.1.16)

The graphs of the functions x − y2 = 0 and x − 2y2 = 0 are given in Fig. 1.1.1. The function f (x, y) in an arbitrary small neighborhood of the origin takes on both positive and negative values. In this case, the quadratic form in y vanishes at the origin, and there is no cubic term, but the coefficient in the quartic term is positive. Hence, the necessary conditions for stability are satisfied. However, this system is unstable because in an arbitrarily small neighborhood of the origin, P takes on negative values. The reason that the conditions mentioned here are not sufficient is that we have restricted our investigation to straight lines through the origin (see Fig. 1.1.1). Following these straight lines, we always find only positive values in a sufficiently small neighborhood of the origin. However, if we follow curved lines through the origin (see the dashed lines), we easily find negative values. Once we have recognized the reason why the conditions imposed are insufficient, it is easy to find a remedy. To this end, we consider a line y = constant in the neighborhood of the origin, and we minimize f (x, y) with respect to x, i.e., Min f (x, y) = x2 − 3xy21 + 2y41 . y=y1

(1.1.17)

This yields 2x − 3y21 = 0, and hence x = 3/2 y21 . Substitution of this value into f (x, y) yields min f (x, y) = −1/4 y41 , which means that the function is indefinite. In general, the function P is minimized with respect to qi (i > 1) for fixed q1 . When the coefficient of (q1 )4 is positive-definite, the system is stable.

2

Continuous Elastic Systems

2.1 Thermodynamic background Consider a body that is in a state of equilibrium under conservative loads. Our aim is to investigate this equilibrium state. For an elastic body, the internal energy per unit mass may be represented by U(s, γ), where s denotes the specific entropy and γ is the deformation tensor. Let xi (i = 1, 2, 3) be the components of the position vector x, which describe the position of a material point in the “fundamental state” I, which is to be investigated. Let u(x) be the displacement vector from the fundamental state (u is a small but finite displacement). The corresponding position in the “adjacent state” II is then x + u. The (additional) deformation tensor is now defined by γij =

1 1 (ui,j + uj ,i ) + uh,i uh,j . 2 2

(2.1.1)

The fact that the body has undergone deformations to arrive in the fundamental state is unimportant because the state I is kept fixed. The temperature T is now defined by T=

∂U ∂s

(2.1.2)

(γ is kept constant). From (2.1.2) we obtain ∂T ∂2 U = 2. ∂s ∂s The specific heat of the material is now defined by T

∂s = Cγ , ∂T

(2.1.3)

where Cγ > 0 for a thermodynamically stable material. As ∂2 U/∂s2 is positive (nonzero) we may solve (2.1.2) for s, which yields s = s(T, γ). We now introduce the function F (T, γ), defined by F (T, γ) = U − Ts.

(2.1.4) 7

8


F (T, γ) is called the free energy. Writing (2.1.4) as a total differential, we find (for fixed γ) ∂U ∂F δT = δs − Tδs − sδT. ∂T ∂s

(2.1.5)

Using (2.1.2) we find s=−

∂F . ∂T

(2.1.6)

Let us denote the temperature in the fundamental state (which by virtue of the equilibrium state is equal to the temperature of its surroundings) by TI . A disturbance of the equilibrium state will cause a heat flux in the body. In the following, we will assume that the temperature of the surrounding medium is constant (TI ). Denoting the heat flux by q, the heat flux per unit time through a closed surface is given by A q · n dA, where n denotes the unit normal vector on the surface, positive in the outward direction. According to the second law of thermodynamics,† vheat will flow out of the body when its surface temperature is higher than that of the surrounding medium, i.e., (T − TI )q · n ≥ 0 (on the surface).

(2.1.7)

The heat flux will cause an entropy flux. The entropy flux vector h is given by h=

1 q (per unit time and per unit area). T

For an arbitrary part of the body, the entropy balance is given by ρ˙s dV = h · n dA, V

(2.1.8)

(2.1.9)

A

where ρ is the specific mass. This equation only holds in the absence of irreversible processes in the body. When the state of the body also depends on the deformation rates, irreversible processes will occur, which implies entropy production. In that case, the entropy balance reads ρ˙s dV = h · n dA + ρσ dV, σ ≥ 0, (2.1.10) V

A

V

where σ denotes the entropy production per unit time and mass. This is the more general formulation of the second law of thermodynamics (Clausius-Duhem). The first law of thermodynamics states that the total amount of heat that flows into a body is transformed into internal energy. Let PL [u (x (t))] be the potential energy of the external loads and let 1 ρ u˙ · u˙ dV K [u˙ (x (t))] = 2 V †

This is an early formulation by Clausius (1854).

2.1 Thermodynamic background

be the kinetic energy. The total energy balance is then given by    d ρU (s, γ) dV + K [u˙ (x (t))] + PL [u (x (t))]  dt  V = − q (x (t)) · n dA, (1st law)

9

(2.1.11)

A

where we have a negative sign on the right-hand side of this equation because the heat flux is regarded as positive in the outward direction. To draw conclusions from the first and the second laws, we subtract (2.1.11) from (2.1.10) multiplied by TI . This yields    d ρ [U (s, γ) − TI s] dV + K [u˙ (x (t))] + PL [u (x (t))]  dt  V

=

TI − 1 q · n dA − TI ρσ dV ≤ 0 T

V

(2.1.12)

V

(Duhem, 1911). Here we have made use of the relation d ρ˙s dV = ρs dV dt V

(2.1.13)

V

The first term on the right-hand side of (2.1.12) is negative because the heat flux is in the outward direction when T > TI , and the second term is negative because the entropy production is always positive. The integral on the left-hand side of (2.1.12) may be expressed in terms of the free energy. Using the relation U (s, γ) − TI s = U (s, γ) − Ts + (T − TI ) s = F (T, γ) + (T − TI ) we obtain

    d ∂F dV + PL + K ≤ 0. ρ F (T, γ) + (TI − T)  dt  ∂T

∂F , (2.1.14) ∂T

(2.1.15)

V

Duhem (1911) already discussed the stability of a system on the basis of this equation and came to the conclusion that a system is stable when the form between the braces is positive-definite. In this form, K is a positive-definite function. However, the terms between the square brackets depend on the deformation tensor and the temperature, whereas PL depends on the displacement field. The problem is to separate the influence of the temperature and the displacement field. A straightforward expansion ∂F 1 ∂2 F F (T, γ) = F (TI , γ) + (T − TI ) + (T − TI )2 + · · · ∂T TI 2 ∂T 2 TI

10


does not solve the problem. Following Ericksen (1965), we may write the Taylor expansion of the free energy at constant deformation γ in the form ∂F 1 ∂2 F F (TI , γ) = F (T, γ) + (2.1.16) (TI − T) + (TI − T)2 , ∂T T 2 ∂T 2 ∗ where the first derivative is evaluated at the deformation γ and temperature T, and the second (starred) derivative at the deformation γ and an intermediate temperature T∗ = T + θ (TI − T), where 0 < θ < 1. Using (2.1.16) we may rewrite the term between the square brackets in (2.1.15) as follows: ∂F 1 ∂2 F F (T, γ) + (TI − T) = F (TI , γ) − (TI − T)2 ∂T 2 ∂T2 ∗ (2.1.17) 1 cγ 2 = F (TI , γ) + (TI − T) , 2 T ∗ where we have used the relation cγ ≡ T

∂2 F ∂s = T 2. ∂T ∂T

The first term on the right-hand side of (2.1.17) depends only on the displacement field. The second term is positive-definite. The energy balance may now be written in the form    cγ d 1 ρF (TI , γ) dV + PL [u (x (t))] + K [u˙ (x (t))] + ρ (TI − T)2 dV ≤0.  dt  2 T ∗ V

V

(2.1.18) The last two terms in the left-hand member are positive-definite, and the remaining terms depend only on the displacement field. Our energy balance is not affected when we subtract from the expression between the braces a time-independent quantity, ρF (TI , 0) dV. V

Further, we introduce the notation W (γ) ≡ ρ [F (TI , γ) − F (TI , 0)] ,

(2.1.19)

where W (γ) is the (additional) stored elastic energy in the isothermal (additional) deformation γ at constant temperature TI , from the fundamental state I to the current state. The potential energy functional P is now defined by P [u (x (t))] = W (γ) dV + PL [u (x (t))] . (2.1.20) V

In words: The potential energy is equal to the sum of the increase of the elastic energy for isothermal deformations and the potential energy of the external loads.

2.2 Theorems on stability and instability

11

Hence, stability for the class of problems discussed depends on isothermal constants.† The question of stabilty when TI is not constant is still unsolved (which is important, for example, in problems with thermal stresses). 2.2 Theorems on stability and instability In our discussion on the stability of discrete systems, we have seen that we need measures to be able to specify expressions like “small disturbance” and “not large deviations.” In our discussion of the stability of continuous systems, which is governed by the character of the potential energy, we need a measure for the displacements. A suitable measure is the L2 -norm of the displacement field, defined by u (x (t))2 = 1 M

ρ u (x (t)) · u (x (t)) dV,

(2.2.1)

V

where M is the total mass of the elastic body. We shall employ the same measure for the initial disturbance. We shall assume that the potential energy functional is regular in the following sense. On every ball u = c in the function space of kinematically admissible displacement fields u (x (t)) where the radius c is sufficiently small, the energy functional P [u (x (t))] has a proper minimum that is a continuous function d(c) in the range of c under consideration. The potential energy functional is called positive-definite if the function d(c) is a (positive) increasing function in a range 0 ≤ c < c1 . The functional is called indefinite if the function d(c) is a (negative) decreasing function for 0 ≤ c < c1 . We are now in a position to formulate the stability criterion: The equilibrium in the fundamental state is stable if the potential energy functional P [u (x (t))] is positive-definite. To show this, we introduce the notation V [u (x (t)) , u˙ (x (t)) , T (x (t))] = P [u (x (t))] cγ 1 +K [u˙ (x (t))] + ρ (TI − T)2 dV, 2 T ∗

(2.2.2)

V

where V is the total energy.‡ †

‡

In the literature, one frequently encounters vague and loose statements to the effect that buckling is “rapid” and that it is therefore “reasonable” to assume that the motion is adiabatic. This would imply that elastic stability would be governed by the adiabatic elastic constants rather than by the isothermal elastic properties. This reasoning is erroneous as follows from the foregoing analysis. A simple example is a strut with pinned ends under a compressive load N. For sufficiently small values of the compressive load the straight configuration is stable, and this stability is manifested by a non-vanishing fundamental frequency. This frequency decreases when the critical Euler load N1 is approached, and it vanishes for N1 . The motion at the critical load is thus infinitely slow, and hence isothermal. Duhem called it the “ballistic energy.”

12


According to (2.1.18) we have dV/dt ≤ 0. Let the initial disturbance u (x (0)) satisfy the condition u (x (0)) < α1 (2.2.3) and let the total energy satisfy the condition 0 < V0 < d(α1 ),

(2.2.4)

where V0 = V [u (x (0)) , u˙ (x (0)) , T (x (0))]. Because dV/dt < 0, it follows that P [u (x (t))] < d (α1 ), and hence u(x(t)) < α1 . In words: For a given, sufficiently small initial disturbance and a given total energy, the displacements at t > 0 are bounded by the value of u (x(0)). For an instability criterion, we need the following assumptions: 1. For nonvanishing deformation rates, the entropy production is positive. 2. In a sufficiently small neighborhood of the fundamental state, the potential energy has no stationary values at a different energy level. Under these additional conditions, the following theorem holds. Theorem. If the potential energy functional is indefinite, the system is unstable. To show this, we consider an initial disturbance u (x (0)), as small as we please, in the region in function space where P [u (x)] is negative, and we select initial velocities and temperature variations that are both identically zero. The initial value V0 of the total energy is then negative, and by (2.1.18) this energy will decrease until the motion comes to a final stop. Because no equilibrium will be possible in the range 0 < c < c1 of u, it follows that this norm at some time must approach or exceed the value c1 , no matter how small the initial disturbance has been chosen. It follows that the equilibrium in the fundamental state is unstable. The only case that is not covered by our discussion so far is the case d (c) = 0. However, this case is not important because this condition never occurs in practical problems. The results obtained so far are not as useful as it may seem because so far it has proved to be impossible to show that the elastic energy functional is positivedefinite, even for an elastic body without external loads. To state this problem more clearly, we consider the potential energy in the elastic body P [u (x (t))] = PLu (x (t)) + W (γ) dV, (2.2.5) V

where the elastic potential W (γ) is expanded about its value in the fundamental state 2 ∂W ∂W 1 W (γ) = γij + γij γk + · · · (2.2.6) ∂γij I 2 ∂γij ∂γk I


13

This expansion contains linear and quadratic terms in the strains, which in their turn contain linear and quadratic terms in the derivatives of the displacements. Consider now the displacement fields u = αu1 (x), where u1 (x) is a given (fixed) displacement field and α is a positive number. Stability is now determined by α for α → 0. Now consider all kinematically admissible displacement fields u1 (x). It has been demonstrated that it is impossible to show that W is positive-definite on this basis. We shall also need restrictions on the derivatives of the displacements. Strictly speaking, this condition implies that we cannot use the stability criterion. However, it can be shown that for an indefinite elastic energy functional, the system is unstable. Let us now consider the first term in (2.2.6). By virtue of δW = Sij δγij , where Sij is a symmetric stress tensor, we have 1 ∂W ∂W ∂W Sij = + ≡ . 2 ∂γij ∂γj i I ∂γ(ij ) I

(2.2.7)

(2.2.8)

The potential energy may now be written as 1 Sij (ui,j + uj ,i + uh,i uh,j ) P [u (x (t))] = 2 1 + 2

V 2

∂ W ∂γij ∂γk

γij γk + · · ·

(2.2.9) dV + PL1 [u (x (t))] + PL2 [u (x (t))] + · · · ,

I

where we have expanded the potential of the external loads, PL [u (x (t))] = PL1 [u (x (t))] + PL2 [u (x (t))] + · · · ,

(2.2.10)

where PL1 is linear in u, PL2 is quadratic in u, and so forth. Because the fundamental state I is an equilibrium state, the first variation of P must vanish for all kinematically admissible displacement fields, which implies 1 P1 [u (x (t))] = Sij (ui,j + uj ,i ) dV + PL1 [u (x (t))] = 0. (2.2.11) 2 V

Hence (2.2.9) may be written as 2 1 ∂ W 1 Sij uh,i uh,j + γij γk + · · · dV + PL2 [u (x (t))] + · · · P [u (x (t))] = 2 2 ∂γij ∂γk I V

(2.2.12) In the remaining part of these lectures, we shall restrict ourselves to dead-weight loads, unless mentioned otherwise. Thus, Pd.w.L [u (x (t))] = PL1 [u (x (t))] ,

(2.2.13)

14


so that the discussion of stability is focused on the integral in (2.2.12). We now define a tensor of elastic moduli ∂2 W EIijk ≡ , (2.2.14) ∂γ(ij ) ∂γ(k ) I where W is written symmetrically with respect to γij and γk . Notice that this is not the tensor of elastic moduli that is usually used in the theory of elasticity because that tensor is defined by ∂2 W 0 Eijk ≡ , (2.2.15) ∂γ(ij ) ∂γ(k ) 0 where the index 0 indicates that the second derivatives of W must be evaluated in the undeformed state. Then the tensor of elastic moduli for a homogeneous isotropic material is given by 2ν E0ijk = Gγ δikδj + δi δjk + (2.2.16) δij δk . 1 − 2ν This tensor gives a complete description of the elastic material when the elastic potential is given as W0 =

1 0 0 0 E γ γ , 2 ijk ij k

(2.2.17)

i.e., when W 0 is a homogeneous quadratic form in the strain components. Notice that here we have used Cartesian coordinates in the undeformed state. The deformation tensor is γij =

1 (ui,j + uj ,i + uh,i uh,j ) , 2

(2.2.18)

where u now denotes the displacements with respect to the undeformed configuration and (),i = ∂ () /∂xi , where xi are Cartesian coordinates. The description of W 0 by (2.2.17) is in principle only valid for infinitesimally small strains, and even then only the linear terms in the strain tensor are important. The expression for the elastic potential may now be generalized by assuming that for finite strains (where the quadratic terms in γij may become important), the elastic potential can still be represented by a quadratic function. The fact that quadratic terms in the strain tensor may become important even when the linearized strain tensor θij , 1 (2.2.19) (ui,j + uj ,i ) , 2 is small is immediately clear from the fact that θij 1 does not imply that the linearized rotation tensor θij ≡

ωij ≡ is small.

1 (ui,j − uj ,i ) 2

(2.2.20)


15

The variation of W for a small disturbance from the equilibrium configuration is given by 1 0 0 δW = E0ijk γij0 δγk + E0ijk δγij0 δγk 2

(2.2.21)

(with respect to Cartesian coordinates in the undeformed configuration). Notice that Cartesian coordinates in the undeformed configuration become curvilinear coordinates in the fundamental state I, and vice versa. It is always possible, at least in principle, to find curvilinear coordinates in the undeformed configuration that become Cartesian coordinates in the fundamental state I. In a curvilinear system, the variation of W is given by αβλµ 0 0 γαβ δγλµ

δW = E0

1 αβλµ 0 0 + E0 δγαβ δγλµ , 2

(2.2.22)

where the contravariant components of the tensor of elastic moduli are given by 2ν βµ αβ λµ αµ βλ . (2.2.23a) g g + g g + g Eαβλµ = G gαλ 0 0 0 0 1 − 2ν 0 0 The difference of the metric tensors in the undeformed configuration and in the fundamental state is ij

g0 − δij = O (ε) , where ε is largest principal extension in the fundamental state. Hence it follows that Eαβλµ = E0ijk [1 + O (ε)]

(2.2.23b)

so that 1 αβλµ 0 0 1 E γαβ γλµ = EIijk γij γk [1 + O (ε)] 2 0 2 and hence 1 I 1 Eijk γij γk ≈ Eijk γij γk > 0, 2 2

(2.2.24)

where Eijk is the tensor of elastic moduli that is used in the theory of elasticity. The fact that we have approximated the elastic energy with a relative error of O (ε) does not affect the positive-definite character of the potential energy if the first term in the potential energy (2.2.12) is also multiplied by a factor (1 ± ε). Remarks 1. For large elastic deformations, the approximation (2.2.23b) is not valid because then ε is large. This can occur, e.g., in rubber-like materials. 2. The fact that we have approximated the elastic energy by a quadratic function in the strains with the classical tensor of elastic moduli implies that we may also apply additional approximations used in the theory of elasticity, e.g., beam theory, plate, and shell theories.

16


Example. Consider a simply supported strut, loaded in compression by a force N. ϕ

x N

N

w x u*

Figure 2.2.1

In the engineering approach, the strut is usually assumed to be inextensible. We have the following relations: dw ≡ w dx cos ϕ = 1 − w2 sin ϕ =

cos ϕ

dϕ ≡ w dx

w dϕ =κ= √ dx 1 − w2

u = − 1 − w2 dx. ∗

0

For an inextensible strut, the potential energy is P [w (x)] = 0

w2 1 EI dx + 2 1 − w2

N

1 − w2 − 1 dx.

(2.2.25)

0

Notice that now the potential energy of the external loads is a nonlinear function of the displacement, in contradistinction to what we have used in our theory. This is due to the fact that we have used an auxiliary condition; namely we have assumed that the strut is inextensible, which means 1 + u2 + w2 = 1 or u = 1 − w2 − 1. In this case, we may say that buckling occurs when the energy supplied by the loads is equal to the strain energy in bending. This may not be generalized. Returning to the general theory and restricting ourselves to dead-weight loads and to materials that follow the generalized Hooke’s Law, we may write 1 1 P [u (x)] = (2.2.26) Sij uh,i uh,j + Eijk γij γk dV, 2 2 V

where the strain tensor is given by γij =

1 1 ui,j + uj ,i + uh,i uh,j ≡ θij + uh,i uh,j . 2 2

(2.2.27)


17

The potential energy functional may now be written as (only for dead-weight loads) 1 1 P [u (x)] = Sij uh,i uh,j + Eijk θij θk dV 2 2 V 1 1 Eijk θij um,kum, + Eijk θk uh,i uh,j dV + Eijk uh,i uh,j um,kum, dV, + 2 8 V

V

(2.2.28) where we have arranged the terms so that the integrants contain only terms of second, third, and fourth degree, respectively, in the displacements. Writing P [u (x)] = P2 [u (x)] + P3 [u (x)] + P4 [u (x)] and using the fact that Eijk = Ek ij , we have 1 P2 [u (x)] = Sij uh,i uh,j + Eijk θij θk dV 2 V 1 P3 [u (x)] = Eijk θij um,kum, dV 2 V 1 P4 [u (x)] = Eijk uh,i uh,j um,kum, dV. 8

(2.2.29)

(2.2.30)

A positive-definite energy functional means P2 + P3 + P4 ≥ 0. However, because P3 and P4 contain only higher-order terms, it is usually sufficient to consider only P2 . P2 [u (x)] must be positive-definite for all kinematically admissible displacement fields u (x). When ui and ui,j are sufficiently small, then P2 [u (x)] > 0 is a sufficient condition for stability. (No rigorous proof is offered here.) The limiting case that min P2 [u (x)] = 0 for a nonzero displacement field u1 (x) is called a critical case of neutral equilibrium. P2 is called the second variation, and u1 (x) is called the buckling mode. In the following, we shall argue that this case is only possible in slender constructions. To see this, we first notice that according to (2.2.19) and (2.2.20) we may write ui,j = θij − ωij

(2.2.31)

Sij uh,i uh,j = Sij θhi θhj − 2Sij θhi ωhj + Sij ωhi ωhj .

(2.2.32)

so that

The components of the stress tensor Sij are small compared to those of the tensor of elastic moduli Eijk , which are of O(G) where G is the shear modulus. In the elastic range (for engineering materials) the strains must be small so that only terms involving the rotations might compete with terms with Eijk . This is only possible for large rotations, so the last term in (2.2.32) is the principle term. To show that the

18


construction under consideration must be slender, we write 1 1 1 (ui,j − uj ,i )k = (ui,jk − uj ,ik) = (uj ,k − uk,j )i 2 2 2 1 − (ui,k + uk,i )j = θjk,i − θik,j . 2 For a sufficiently supported construction, ω ωij ,k = O , ωij ,k =

(2.2.33)

where is a characteristic length of the construction. Large values of one or more components of the rotation tensor are only possible if one or more strains are of order (ω/ ) ∗ where ∗ / 1. This condition means that the construction has a second characteristic length ∗ that is considerably smaller than . This implies that the construction is slender (e.g., beams, plates, shells). A rigorous mathematical proof of this scenario was given by Fritz John. He showed that when γij = O(ε), ε 1, and the dimensions of a body are all of the same order of magnitude, the rotation vector is given by ω = ω0 + O (ε) where ω0 is a constant rotation vector. For an adequately supported construction, this means that ω = O(ε), i.e., the rotations and the strains are of the same order of magnitude. 2.3 The stability limit For conservative dead-weight loads, and under the assumption that the material follows the generalized Hooke’s Law, the potential energy is given by 1 1 P [u (x)] = (2.3.1) Sij uh,i uh,j + Eijk γij γk dV. 2 2 As discussed in Section 2.2, stability is primarily determined by the character of P2 , given by 1 1 P2 [u (x)] = (2.3.2) Sij uh,i uh,j + Eijk θij θk dV. 2 2 If the second variation is positive-definite, the equilibrium is stable, and if the second variation is indefinite (or negative-definite), the equilibrium is unstable. The second variation is unable to give a valid decision on the stability of instability in the critical case that it is semi-definite positive. Let u (x) be a minimizing displacement field. Then P2 [u (x) + εζ(x)] ≥ P2 [u (x)]

(2.3.3)

for all kinematically admissible displacement fields ζ(x) and for sufficiently small values of ε ∈ R. Here and in the following it will be assumed that ζ is continuously

2.3 The stability limit

19

differentiable. Expanding the left-hand side in (2.3.3), we obtain P2 [u] + P11 [u, εζ] + P2 [εζ] ≥ P2 [u] ,

(2.3.4)

where we have written u instead of u (x) and so on, from which εP11 [u, ζ] + ε2 P2 [ζ] ≥ 0 for ∀ζ and

∀ε ∈ R,

|ε| 1.

(2.3.5)

Because this expression must hold for all sufficiently small values of |ε|, it follows that P11 [u, ζ] = 0.

(2.3.6)

This equation is the variational equation for neutral equilibrium. From the functional (2.3.2), we now obtain for the bilinear term 1 (2.3.7) Sij uh,i ζh,j + Eijk θij (ζk, + ζ ,k) dV, P11 [u, ζ] = 2 where ζk are the Cartesian components of ζ. Due to the symmetry of Eijk in the indices k , we may write Eijk θij ·

1 (ζk, + ζ ,k) = Eijk θij ζk, . 2

Further, we introduce the notation Eijk θij = σk .

(2.3.8)

Here σk are the stresses corresponding to the linearized strain tensor in the absence of prestresses. Thus (2.3.7) may now be rewritten as (2.3.9) (Sij uh,j + σhj ) ζh,j dV = 0, and using the divergence theorem we obtain (Sij uh,i + σhj ) ζh nj dA − (Sij uh,i + σhj ),j ζh dV = 0 AP

for ∀ζ.

(2.3.10)

V

According to the principal theorem in the calculus of variations, we then must have (Sij uh,i + σhj ),j = 0 in V,† (Sij uh,j + σhj ) nj = 0 ‡

on AP ,

ui = 0,

(2.3.11) on Au .

(2.3.12)

Suppose (2.3.11) does not hold in a point x∗ , say, Sij uh,i + σhj j > 0. Then choose  2   xi − x∗i xi − x∗i − R2 for xi − x∗i xi − x∗i ≤ R2 ∗ ∗ ζ= 0 for xi − xi xi − xi ≥ R2   and thus, the surface integral in (2.3.10) vanishes. However, the volume integral is positive, so (2.3.10) is violated, and hence (2.3.11) must hold.

20


σy

Sy Syx

dy dx I Fundamental State

ψ/ 2

Sxy

Sx σx

y

Sx

τyx τxy

(1 + ε

Sxy

) dy

rxy

Sy S yx

) dx (1 + ε x ψ/ 2

ψ/ 2

rxy

ψ/ 2

II No Rotation

ω III Final State

Figure 2.3.1

Notice that in the absence of prestresses Sij , these equations reduce to the equations from the classical theory of elasticity. Performing differentiation by parts, we obtain from (2.3.11) Sij ,j uh,i + Sij uh,ij + σhj ,j = 0.

(2.3.13)

The equilibrium equations and boundary conditions in the fundamental state I are given by Sij ,j + Xi = 0 in V , (2.3.14) Sij nj = Pi on Ap where Xi are the mass forces and Pi are prescribed tractions on Ap . Using these expressions, we can rewrite (2.3.12) and (2.3.13) as − Xi uh,i + Sij uh,ij + σhj ,j = 0 p i uh,i + σhj nj = 0 on Ap ,

in V

ui = 0

on Au .

(2.3.15) (2.3.16)

These equations and boundary conditions were derived for the first time by Trefftz (1930, 1933). Different but equivalent equations were derived earlier by Biezeno and Hencky (cf. C.B. Biezeno and R. Grammel, Engineering Dynamics, Vol. I). We shall reproduce here their derivation for the two-dimensional case (to simplify the analysis). Consider a rectangular material element with dimensions dx and dy in the fundamental state, loaded by stresses Sx , Sy , and Sxy (see Figure 2.3.1). The final state is reached in two steps: First, a deformation without a rotation of the deformed element (state II), and then a rotation of the deformed element (state III). In state II, the element will not be in moment equilibrium under the forces Sx , Sy , and Sxy acting on the deformed element. To restore equilibrium (to a first approximation), we add additional (small) forces σx , σy , and τxy (τxy = τyx ). These additional forces do not enable us to satisfy the equilibrium of moments exactly.


21

To reach this goal, skew-symmetric shear forces rxy (ryx = −rxy ) must be added. It is obvious that the final rotation does not disturb the equilibrium of moments. The equilibrium of moments requires 1 Sx dy dx ψ+ Syx dx (1 + εy ) dy − Sxy dy (1 + εx ) dx 2 1 − Sy dx ψdy − rxy dy dx − rxy dx dy = 0, 2 from which follows 2rxy =

1 ψ(Sx + Sy ) + Sxy (εy − εx ) . 2

(2.3.17)

The equilibrium of forces yields (Sx + σx − Sxy ω) , x + (Syx + τyx − rxy − Sy ω) , y + X = 0 (Sy + σy − Syx ω) , y + (Sxy + τxy − rxy − Sx ω) , x + Y = 0.

(2.3.18)

In the fundamental state, we have Sx,x + Syx,y + X = 0

(2.3.19)

Sy,y + Sxy,x + Y = 0.

Substitution of these equations into (2.3.18) and using the relation (2.3.17) yields 1 1 σx,x + τyx,y − (Sxy ω) ,x − (Sy ω) ,y − [ψ(Sx − Sy )] ,y − [Sxy (εy − εx )] ,y = 0 4 2 (2.3.20) 1 1 τxy,x + σy,y + (Syx ω) ,y + (Sx ω) ,x + [ψ(Sx − Sy )] ,x + [Sxy (εy − εx )] ,x = 0. 4 2 With ω = 12 (v,x −u,y ) and ψ = v,x +u,y , for the first of the equations we finally obtain (2.3.20) 1 1 σx,x + τyx,y − Sxy (v,x −u,y ) − Sy (v,x −u,y ) ,y 2 2 ,x (2.3.21) 1 1 − [(v,x + u,y ) (Sx − Sy )],y − [Sxy (v,y − u,x )],y = 0. 4 2 The general result from Biezeno and Hencky may be written in the form 1 1 σij + Shi θhj − Shj θhi + Sih ωhj =0 2 2 ,i or rewritten as

or

(2.3.22)

1 1 σij + Shi (θhj + ωhj ) − Shi θhj − Shj θhi = 0 2 2 ,i

1 1 σij + Shi uj ,h − Shi θhj − Shj θhi 2 2

= 0. ,i

(2.3.23)

22


(Equation 2.3.21) is the two-dimensional form of (2.3.22) for j = 1. Our earlier result (2.3.11) was (σij + Shi uj ,h ),i = 0.

(2.3.24)

The additional terms in (2.3.23) can easily be derived from the variational approach by adding to the energy density the term 12 Shi θhj θij . This is small, of order O(ε) compared to 12 Eijk θij θk , in which we had already admitted a relative error of O(ε). Hence it follows that the equations for neutral equilibrium derived by Biezeno and Hencky and those derived by Trefftz are equivalent within the scope of our theory. We now continue with our general discussion of neutral equilibrium, and we consider the case that P2 [u1 (x)] = 0,

P2 [u (x)] ≥ 0,

where u1 (x) is the buckling mode. The question now arises whether u1 is the only kinematically admissible displacement field for which P2 [u (x)] vanishes. To investigate this condition we might look for additional solutions of the equations of neutral equilibrium, but it proves to be more useful to proceed as follows. Consider the set of orthogonal (kinematically admissible) displacement fields. To define orthogonality, we introduce the positive-definite auxiliary functional T2 [u (x)] ≥ 0.

(2.3.25)

This functional defines the measure in the energy space. Applying similar arguments as in the discussion of P2 , we find that the bilinear term must vanish, T11 [u (x) , v (x)] = 0, which defines the orthogonality of u and v. A possible choice for T2 is 1 T2 [u] = Gui,j ui,j dV, 2

(2.3.26)

(2.3.27)

V

where the integrant is a positive-definite quadratic function of the displacement gradients. The factor G has been added to give T2 the dimension of energy. The vanishing of the bilinear term yields G ui,j vi,j dV = 0. (2.3.28) V

Another suitable choice is T2 [u] =

1 2

ρui ui dV, V

(2.3.29)


23

where ρ is the mass density. This choice is motivated by Rayleigh’s principle for the determination of the lowest eigenfrequency of small vibrations about an equilibrium configuration, which states ω2 = Min V

P2 [u (x)] 1 ρui ui 2

dV

.

(2.3.30)

In the critical case of neutral equilibrium, ω vanishes, which means that the motion is infinitely slow. It will turn out that the results are independent of the particular choice for T2 .† An arbitrary displacement field can always be written in the form u (x) = a u1 (x) + u (x) ,

T11 [u1 , u] = 0.

(2.3.31)

Namely, T11 [u1 , u] = a T11 [u1 , u1 ] + T11 [u1 , u] , where the last term vanishes by virtue of (2.3.31). It follows that a=

T11 [u1 , u] T11 [u1 , u] = , T11 [u1 , u1 ] 2T2 [u1 ]

(2.3.32a)

where we have used the relation T2 [u1 + u1 ] = T2 [2u1 ] = 4T2 [u1 ] = T2 [u1 ] + T11 [u1 , u1 ] + T2 [u1 ] . The second variation for an arbitrary displacement field can now be written as P2 [u] = P2 [au1 + u] = a2 P2 [u1 ] + aP11 [u1 , u] + P2 [u] ,

(2.3.32b)

where the first term on the right-hand side vanishes because u1 is the buckling mode, and the second term vanishes because P11 vanishes for all displacement fields, and hence also for displacement fields orthogonal to u1 . In other words, it follows that when P2 [u] = 0 for u = u1 , P2 [u] also vanishes for a displacement field orthogonal to u1 . We may thus restrict ourselves to displacement fields that are orthogonal to u1 . As P2 [αu] → 0 for α → 0, it is more suitable to consider the following minimum problem: Min

T11 [u1 ,u]=0

P2 [u (x)] = λ2 , T2 [u (x)]

(2.3.33)

where λ2 is the minimum of the left-hand member. If λ2 = 0, then we have a second buckling mode, and when λ2 > 0, u1 is then the unique buckling mode. Let u2 be the second buckling mode; then P2 [u2 + εη] P2 [u2 ] ≥ = λ2 T2 [u2 + εη] T2 [u2 ]

(2.3.34)

T11 [u2 , u1 ] = T11 [η, u1 ] = 0.

(2.3.35)

and

†

See following equation (3.3.55).

24


Rewriting (2.3.34), we find P2 [u2 ] + εP11 [u2 , η] + ε2 P2 [η] ≥ λ2 T2 [u2 ] + εT11 [u2 , η] + ε2 T2 [η] where P2 [u2 ] − λ2 T2 [u2 ] = 0, so that ε P11 [u2 , η] − λ2 T2 [u2 , η] + ε2 P2 [η] − λ2 T2 [η] ≥ 0.

(2.3.36)

Because this equation must hold for arbitrary small values of ε, the term linear in ε must vanish, i.e., P11 [u2 , η] − λ2 T2 [u2 , η] = 0

(2.3.37)

for all displacement fields with T11 [u1 , η] = 0. In this condition, the displacement field η is orthogonal to u1 . In our original condition of P11 [u, ζ] = 0, ζ was not submitted to this requirement. We shall now show that (2.3.37) also holds for arbitrary displacement fields ζ. To show this, we write ζ = tu1 + η, whereη satisfies the orthogonality condition. Replacing η in (2.3.37) by ζ, we find P11 [u2 , tu1 + η] − λ2 T11 [u2 , tu1 + η] = tP11 [u2 , u1 ] + P11 [u2 , η] − λ2 tT11 [u2 , u1 ] − λ2 T11 [u2 , η] = t P11 [u2 , u1 ] − λ2 T11 [u2 , u1 ] + P11 [u2 , η] − λ2 T11 [u2 , η]

(2.3.38)

= P11 [u2 , η] − λ2 T11 [u2 , η] = 0 It follows that without a loss of generality we may restrict ourselves to displacement fields that are orthogonal to the ones that are already known. Suppose we have found m linearly independent solutions. The solution of P11 [u, ζ] = 0 can then be written as u (x) = ah uh ,

h ∈ (1, . . . m)

with T11 [uh , uk] = 0

for h = k.

Further, we may normalize the buckling modes by requiring T11 [u1 , u1 ] = · · · = T11 [um, um] = 1.

(2.3.39)

To investigate the stability of the critical case of neutral equilibrium, we write for an arbitrary displacement field u (x) u (x) = ah uh (x) + u (x) ,

T11 [u, uk] = 0

k ∈ (1, 2, . . . , m) .

(2.3.40)

This is always possible, namely, T11 [uk, u] = T11 [uk, ah , uh + u] = ah T11 [uk, uh ] + T11 [uk, u] = ak.

(2.3.41)

A substitution of (2.3.40) into the energy functional yields P [u (x)] = P2 [ah uh + u] + P3 [ah uh + u] + P4 [ah uh + u] + · · · = P2 [ah uh ] + P11 [ah uh , u] + P2 [uh ] + P3 [ah uh ] + P21 [ah uh , u] + P12 [ah uh , u] + P3 [u]

(2.3.42)

+ P4 [ah uh ] + P31 [ah uh , u] + P22 [ah uh , u] + P13 [ah uh , u] + P4 [u] + · · ·.


25

When the material follows the generalized Hooke’s Law, the expansion terminates after P4 [u]; in other cases, higher-order terms follow. Because uh are buckling modes, the first term on the right-hand side vanishes, P11 [ah uh , ζ] = 0, for all displacement fields ζ, and hence also for u. Because we have already found m buckling modes, P2 [u] must satisfy the relation P2 [u] ≥ λm+1 T2 [u] ,

λm+1 > 0.

(2.3.43)

Let us now first consider the case that u = ≡ 0. We then only have to deal with the terms P3 [ah uh ] and P4 [ah uh ]. A necessary condition for stability is that P3 [ah uh ] vanishes and that P4 [ah uh ] ≥ 0. Hence it follows that from the mathematical point of view, systems will generally be unstable because functions for which P2 = P3 = 0 and P4 ≥ 0 are exceptions. However, in applications this often happens due to the symmetry of the structure. The conditions mentioned are necessary conditions; to obtain sufficient conditions, we consider small values of u. In fact, the stability conditions must be satisfied for small deviations from the fundamental state. Suppose now that the necessary conditions are satisfied; thus the most important terms in (2.3.42) are P2 [u] ,

P21 [ah uh , u] ,

P4 [ah uh ] ,

namely, for u = O(a(h) , u(h) 2 ) these terms are of the same order of magnitude, whereas all other terms have integrands of order O(a(h) , u(h) n ), n ≥ 5. We now first consider the minimum problem, Min

T11 [uk,u]=0 ak const. k∈(1,...,m)

(P2 [u] + P21 [ah uh , u]) .

(2.3.44)

This is a meaningful minimum problem because P2 is a positive-definite functional and P21 is a functional linear in u. Let u = v be the solution to this problem; thus for a variation of this field v + εη we have P2 [v] + εP11 [v, η] + ε2 P2 [η] + P21 [ah uh , v] + εP21 [ah uh , η] ≥ P2 [v] + P21 [ah uh , v] or ε (P11 [v, η] + P21 [ah uh , η]) + ε2 P2 [η] ≥ 0.

(2.3.45)

Because this inequality must hold for all sufficiently small values of ε, it follows that P11 [v, η] + P21 [ah uh , η] = 0,

(2.3.46)

which is an equation for v. To answer the question of whether the solution to (2.3.44) is unique, we consider a second solution v∗ . Subtracting the equation obtained from (2.3.46) by replacing v with v∗ , from (2.3.46) we find P11 [v − v∗ , η] = 0.

(2.3.47)

26


By arguments similar to those following (2.3.37), we may also replace η by ζ, where ζ is an arbitrary displacement field.† Further, we have T11 [ah uh , v − v∗ ] = 0,

(2.3.48)

i.e., v − v∗ is orthogonal with respect to the linear combination of buckling modes, and due to (2.3.47), P2 [v − v∗ ] = 0.

(2.3.49)

However, this result contradicts our assumption (2.3.43) that λm+1 > 0. This completes our proof that v is unique. Equation (2.3.46) is quadratic in the buckling modes but linear in v, so the solution can be written as v (x) = ah akvhk (x) ,

vhk = vkh .

(2.3.50)

Substitution of this expression into (2.3.46) yields 1 P11 [vhk, ζ] + P111 [uh , uk, ζ] = 0, 2

(2.3.51)

where we have used the relation P21 [ah uh ; ζ] =

1 ah akP111 [uh , uk, ζ] . 2

For a known field v = ah akvhk, we may now determine the minimum value of the energy functional (2.3.42). Only retaining the most important terms, we find Min P11 [u] = P4 [ah uh ] + P2 [ah akvhk] + P21 [ah uh , aka vk ] + · · · .

ak=const.

(2.3.52)

Using (2.3.46) with ζ = ah akvhk and making use of the relation P11 [ah akvhk, a amv m] = 2P2 [ah akvhk] , we find 2P2 [ah akvhk] + P21 [ah uh , aka vk ] = 0

(2.3.53)

so that (2.3.52) may be rewritten to yield Min

(ak=const.)

P [u] = P4 [ah uh ] − P2 [ah akvhk] + · · · 1 = P4 [ah uh ] + P21 [ah uh , aka vk ] + · · · . 2

(2.3.54)

It follows that the (necessary) condition P4 [ah uh ] ≥ 0 is not a sufficient condition, because a positive-definite term P2 [ah akvhk] is subtracted. Introducing the notation P4 [ah uh ] − P2 [ah akvhk] ≡ Aijk ai aj ak a , †

(2.3.55)

The equation with ζ is preferable because then we do not have the side condition of orthogonality.

2.4 Equilibrium states for loads in the neighborhood of the buckling load

27

our results may be summarized as follows: r Aijk ai aj aka ≥ 0 is a necessary condition for stability, and r A a a a a as positive-definite (i.e., zero only for a displacement field u = 0 ) is ijk i j k a sufficient condition for stability. We shall now show that the results obtained are independent of the particular choice of the auxiliary functional T2 . Let T2∗ be a positive-definite functional (T2∗ = T2 ), so then we may construct a minimizing displacement field u∗Min = ah akv∗hk (x) .

(2.3.56)

This field satisfies the same equations as the field v = ah akvhk(x) but is subjected to a different orthogonality condition. The difference of these fields must satisfy the homogeneous variational equation P11 [vhk − v∗hk, ζ] = 0,

(2.3.57)

which means that vhk − v∗hk is a linear combination of buckling modes, say vhk − v∗hk = chk u ,

(2.3.58)

which does not satisfy one of the orthogonality conditions implied by T2 and T2∗ . The minimum of the potential energy functional is now given by Min

(ak=const.)

P (u) = P4 [ah uh ] − P2 [ah akv∗hk] + · · · ,

(2.3.59)

where P2 ah akv∗hk may be written as P2 [ah ak (vhk − chkl ul )] = P2 [ah akvhk] − P11 ah akvhk, ap aq cpqr ur + P2 [ah ak chk u ] .

(2.3.60)

The second term on the right-hand side of (2.3.60) vanishes because P11 [ah uh , ζ] = 0 for all fields ζ, and hence also for ζ = v, and the last term vanishes because the argument is a linear combination of buckling modes. Hence it follows that the minimum value of P [u] is independent of the particular choice of the auxiliary functional. 2.4 Equilibrium states for loads in the neighborhood of the buckling load In the foregoing analysis, we employed the fundamental state as a reference. Because we want to investigate equilibrium states for loads in the neighborhood of the buckling load, we must investigate the behavior of the structure for small but finite deflections. The fundamental state then depends on the load, so it cannot be used as a reference state. We need a fixed reference state, and it is suitable to choose the undeformed (stress-free) state as the reference state.

28


0

I

x

x +U

Undeformed (stress-free) State

Fundamental State

II

x +U + u Adjacent State

Figure 2.4.1

The strain tensor in state II is now given by 1 1 1 (Ui + ui ),j + (Uj + uj ),i + (Uh + uh ),i (Uh + uh ),j 2 2 2 1 1 = (Ui,j + Uj ,i ) + Uh,i , Uh,j 2 2 1 1 1 1 + (ui,j + uj ,i ) + uh,i uh,j + Uh,i , uh,j + Uh,j , uh,i ≡ ij + γij 2 2 2 2

(2.4.1)

where 1 1 (2.4.2) (Ui,j + Uj ,i ) + Uh,i Uh,j 2 2 is the strain tensor going from the undeformed state to the fundamental state, and ij ≡

1 1 1 1 (2.4.3) (ui.j + uj ,i ) + uh,i uh,j + Uh,i uh,j + Uh,j uh,i 2 2 2 2 describes the strains going from the fundamental state to the adjacent state. Notice that this tensor also depends on U. The elastic energy density in state II is given by γij ≡

1 Eijkl (ij + γij ) (k + γk ) . (2.4.4) 2 The increment of elastic energy density going from state I to state II is given by 1 1 1 Eijk (ij + γij ) (k + γk ) − Eijk ij k = Eijk ij γk + Eijk γij γk 2 2 2 1 = sk γk + Eijk γij γk . (2.4.5) 2 This expression is of the same form as our original energy functional but the strain tensor is different. We consider loads that can be written as a unit load multiplied by a load factor λ. This means that the displacement field in the fundamental state also depends on λ, i.e., U = U (x; λ). The increment of the potential energy going from state I to state II is now given by PII − PI = P [u (x) ; λ] = P2 [u (x) ; λ] + P3 [u (x) ; λ] + · · · .

(2.4.6)

Notice that this expression starts with quadratic terms because the fundamental state is an equilibrium configuration, and also that this expression depends on U, i.e., it refers to the fundamental state.


29

For sufficiently small loads, the potential energy will be positive-definite and the equilibrium configuration will be stable and unique (Kirchhoff’s uniqueness theorem). Now let λ ≥ 0 be monotonically increasing; then for a certain value of λ, say λ = λ1 , the potential energy will become semi-definite-positive. (The case λ < 0 is a different stability problem that may be treated similarly.) We shall now expand the potential energy with respect to the load parameter λ in the vicinity of λ = λ1 (assuming that such an expansion is possible), e.g., P2 [u (x) ; λ] = P2 [u (x) ; λ1 ] + (λ − λ1 ) P2 [u (x) ; λ1 ] + · · ·

(2.4.7)

where () ≡ ∂ () /∂λ. As mentioned previously, the potential energy depends on U (x, λ). In the following, we shall assume that the fundamental state is known and the question is whether there exist equilibrium configurations in the vicinity of the fundamental state. Therefore stationary values of P [u (x) ; λ] are considered for nonzero displacement increments from the fundamental state. A necessary condition for equilibrium is that P11 [u (x) , ζ(x) ; λ] + P21 [u (x) , ζ(x) ; λ] + · · · = 0.

(2.4.8)

Notice that this equation is satisfied for u(x) = 0, the fundamental state, as it should be. At the critical load, the infinitesimal displacement field is a linear combination of buckling modes, i.e., u = ah uh . Assuming that for small but finite displacements the field can also (approximately) be represented by this expression, we write u (x) = ah uh (x) + u (x) ,

T11 [uk, u] = 0,

(2.4.9)

which is always possible. The energy functional now becomes P [u (x) ; λ] = P2 [ah uh ; λ1 ] + P11 [ah uh , u; λ1 ] + P2 [u; λ1 ] + (λ − λ1 ) P2 [ah uh ; λ1 ] + (λ − λ1 ) P11 [ah uh , u; λ1 ]

+ (λ − λ1 ) P2 [u; λ1 ] + · · · + P3 [ah uh ; λ1 ] + P21 [ah uh , u; λ1 ] + · · · + (λ − λ1 ) P3 [ah uh ; λ1 ] + (λ − λ1 ) P21 [ah uh , u; λ1 ] + · · ·

+ P4 [ah uh ; λ1 ] + · · · .

(2.4.10)

By known arguments, the first two terms on the right-hand side of (2.4.10) vanish. We shall now first consider the terms depending on u. First of all, we have the positive-definite term P2 [u; λ1 ], thus the term (λ − λ1 ) P11 [ah uh , u; λ1 ] is important because it contains u linearly, and is also linear in ah uh and λ − λ1 . The term (λ − λ1 ) P2 [u; λ1 ] may be neglected because it is quadratic in u and is multiplied by λ − λ1 . The next important term depending on u is P21 [ah uh , u; λ1 ], which is linear in u and quadratic in ah uh . Other terms depending on u may be neglected because they are either of higher order in u or they contain higher-order terms in ah uh or (λ − λ1 ). We now determine the minimum with respect to u of the three terms mentioned, Min P2 [u; λ1 ] + (λ − λ1 ) P11 [ah uh , u; λ1 ] + P21 [ah uh , u; λ1 ] .

w.r.t. u

(2.4.11)

30


The solution to this minimum problem is unique. Suppose that u∗ is a second solution. Then the following equation must holds, P11 [u − u∗ , η] = 0 for ∀η| T11 [u, η] = T11 [u∗ , η] = 0.

(2.4.12)

The solutions of this equation are linear combinations of the buckling modes, but u and u∗ are orthogonal with respect to these buckling modes, so we must have u − u∗ = 0. Guided by the structure of (2.4.11), we try a solution of the form uMin (x) = (λ − λ1 ) ah vh (x) + ah akvhk (x) .

(2.4.13)

Because uMin (x) is the solution to (2.4.11), we have for a variation of this field, say uMin + εη, P2 [uMin + εη; λ1 ] + (λ − λ1 ) P11 [ah uh , uMin + εη; λ1 ] + P21 [ah uh , uMin + εη; λ1 ] ≥ P2 [uMin ; λ1 ] + (λ − λ1 ) P11 [ah uh , uMin ; λ1 ] ,

for ∀ε.

(2.4.14)

This inequality must be satisfied for all sufficiently small values of ε ∈ R, which implies P11 [uMin , η; λ1 ] + (λ − λ1 ) P11 [ah uh , η; λ1 ] + P21 [ah uh , η; λ1 ] = 0.

(2.4.15)

Substitution of (2.4.13) into (2.4.15) yields (λ − λ1 ) P11 [ah vh , η; λ1 ] + P11 [ah akvhk, η; λ1 ] + (λ − λ1 ) P11 [ah uh , η] + P21 [ah uh , η; λ1 ] = 0.

(2.4.16)

Because this equation must hold for all (λ − λ1 ), we must have P11 [vh , η; λ1 ] + P11 [uh , η; λ1 ] = 0

(2.4.17)

1 P11 [vhk, η; λ1 ] + P111 [uh , uk; λ1 ] = 0, 2

(2.4.18)

and

where for (2.4.18) we have used the relation P21 [ah uh , η; λ1 ] =

1 ah akP111 [uh , uk, η; λ1 ] . 2

From these linear equations, vh and vhk can be determined. In the following, we shall need the following property: Min(F2 [u] + F1 [u]) = −F2 [u∗ ],

(2.4.19)

where F2 [u] is a positive-definite functional and u∗ is the minimizing vector field. To show this, we first notice that the minimizing displacement field satisfies the equation F11 [u∗ , ζ] + F1 [ζ] = 0 for ∀ζ. Now choose ζ = u∗ , then F1 [u∗ ] = −F11 [u∗ , u∗ ] .

(2.4.20)


With

F2 [u] = F2

31

1 1 1 1 u + u = F2 [u] + F11 [u, u] 2 2 2 4

or F11 [u, u] = 2F2 [u] we find F1 [u∗ ] = −2F2 [u∗ ] , so that F2 [u∗ ] + F1 [u∗ ] = −F2 [u∗ ] . Using this property, we find that the minimum of (2.4.11) is Min P2 [u; λ1 ] + (λ − λ1 ) P11 [ah uh , u; λ1 ] + P21 [ah uh , u; λ1 ]

w.r.t .u

= −P2 [(λ − λ1 ) ah vh (x) + ah akvhk (x)] = − (λ − λ1 )2 P2 [ah vh (x) ; λ1 ] − (λ − λ1 ) P11 [ah vh (x) , ah akvhk (x) ; λ1 ] −P2 [ah akvhk (x) ; λ1 ] .

(2.4.21)

Let us now first consider the case that P3 [ah uh ; λ1 ] ≡ 0. In this case, the term P4 [ah uh ; λ1 ] is small compared to the cubic term, and may thus be neglected in (2.4.10). The minimum of the remaining terms with u is given in (2.4.21), and here the term P2 [ah akvhk (x) ; λ1 ] may be neglected because it is of fourth degree in the amplitudes of the buckling modes. Further, the term (λ − λ1 ) P11 ah vh (x), ah ak vh (x) ; λ1 ] may be neglected because it is of third degree in the amplitudes of the buckling modes and multiplied by (λ − λ1 ). Finally, we may also neglect the first term on the right-hand side of (2.4.21) because it is of order O[(λ − λ1 )2 a2 ], where a represents the order of magnitude of the amplitudes of the buckling mode. The only remaining term in (2.4.10) that still is to be discussed is (λ − λ1 )P2 [ah uh ; λ1 ]. This term is of order O[(λ − λ1 )a2 ] and may thus be important. Our final result is now given by Min P(u; λ) = (λ − λ1 )P2 [ah uh ; λ1 ]

(ah =const)

+ P3 [ah uh ; λ1 ] + O(a4 , (λ − λ1 )a3 , (λ − λ1 )2 a2 ).

(2.4.22)

In other words, to a first approximation we might have neglected the terms containing u to obtain the same result. Let us now consider the case P3 [ah uh ; λ1 ] ≡ 0. Now we must take into account the term P4 [ah uh ; λ1 ], and hence also the last term in (2.4.21). Further, the term (λ − λ1 ) P2 [ah uh ; λ1 ] is important, whereas the first two terms on the right-hand side of (2.4.21) may be neglected compared to this term (under the assumption that P2 [ah uh ; λ1 ] ≡ 0). In this case, our result is Min

(ah =const.)

P (u; λ) = (λ − λ1 ) P2 [ah uh ; λ1 ] + P4 [ah uh ; λ1 ] − P2 [ah akvhk; λ1 ] + O[a5 , (λ − λ1 ) a3 , (λ − λ1 )2 a2 ].

(2.4.23)

32

Continuous Elastic Systems λ

λ

λ1 λ1

u

u Equilibrium possible for λ > λ1

No equilibrium possible for λ > λ1 , no linear terms in λ − λ1

Figure 2.4.2

Notice that the last two terms in this expression are the terms that decide on the stability of neutral equilibrium (cf. 2.3.55). In our discussion, we have restricted ourselves to cases where P2 [ah uh ; λ1 ] ≡ 0, i.e., we have assumed that the derivative of P2 [ah uh ; λ] with respect to the load parameter λ exists for λ − λ1 (the fundamental state). This means that we have assumed that there are equilibrium states in the vicinity of the fundamental state. This is not always the case (see Figure 2.4.2). In the following, we shall see that the present theory only enables us to treat branch points. Finally, we notice that when P2 [ah uh ; λ1 ] ≡ 0, this term must be negative because for λ − λ1 < 0, the equilibrium in the fundamental state is stable, i.e., P2 [ah uh ; λ] = P2 [ah uh ; λ1 ] + (λ − λ1 ) P2 [ah uh ; λ1 ] + · · · ≥ 0

for λ − λ1 < 0.

The first term on the right-hand side is zero, and hence P2 [ah uh , λ1 ] < 0.

(2.4.24)

Let us now consider the simple case that m = 1, i.e., there is only one buckling mode, say u (x) = au1 (x). Let us further introduce the notation P3 [u1 ; λ1 ] = A3 ,

P2 [u1 ; λ1 ] = A2 .

(2.4.25)

First, consider the case A3 ≡ 0, A2 < 0. Then we obtain from (2.4.22) Min P [u (x) , λ] = (λ − λ1 ) A2 a2 + A3 a3 .

ak=const.

(2.4.26)

The condition for equilibrium is obtained by differentiating this expression with respect to a and equating this result to zero, 2 (λ − λ1 ) A2 a + 3A3 a2 = 0.

(2.4.27)


33

λ A3 > 0

A3 < 0

unstable

λ1

stable

a Figure 2.4.3

The solutions are a1 = 0 A 2 a2 = − (λ − λ1 ) 2 3 A3

(the fundamental state)

(2.4.28)

(branched equilibrium state).

The stability condition is that the second derivative of (2.4.26) is non-negative, 2 (λ − λ1 ) A2 + 6A3 a ≥ 0.

(2.4.29)

It follows that the fundamental state is stable for λ < λ1 and unstable for λ > λ1 and λ = λ1 (because A3 = 0 ). Substitution of a2 into (2.4.29) yields − 2 (λ − λ1 ) A2 ≥ 0,

(2.4.30)

which implies that the branched solution is stable for λ > λ1 and unstable for λ < λ1 and λ = λ1 (because A3 = 0 ). These results are plotted in Figure 2.4.3. The curved line is the behavior of the exact solution. An example of a structure with such a behavior is the two-bar structure shown in Figure 2.4.4 (to be discussed later). For λ sufficiently close to λ1 , our approximate solution will be a good approximation to the actual solution. However, it is not possible to assess the accuracy of the approximation. F

Figure 2.4.4

34


λ

λ1 λ1

a

A3 = 0

a

A4 > 0

A3 = 0

(I)

A4 < 0 (II)

Figure 2.4.5

Let us now consider the case A3 = 0. From (2.4.23) we obtain Min P [u, λ] = (λ − λ1 ) A2 a2 + A4 a4 ,

ak=const.

(2.4.31)

where A4 ≡ P4 [uh , λ1 ] − P2 [vhk; λ1 ] . The branch point is stable when A4 > 0 and unstable when A4 < 0. The equilibrium condition reads 2 (λ − λ1 ) A2 a + 4A4 a3 = 0,

(2.4.32)

from which a1 = 0 (fundamental state) A 1 a22 = − (λ − λ1 ) 2 . 2 A4

(2.4.33)

Because for real values of a2 the left-hand side is positive, the right-hand side must be positive, i.e., λ ≷ λ1

for A4 ≷ 0.

(2.4.34)

The stability condition reads 2 (λ − λ1 ) A2 + 12A4 a2 ≥ 0,

(2.4.35)

so the fundamental state is stable for (λ < λ1 ) and unstable for λ > λ1 . Substitution of a22 into this condition yields − 4 (λ − λ1 ) A2 ≥ 0,

(2.4.36)

from which follows that the branched equilibrium state is stable for λ > λ1 and A4 > 0, and unstable for λ < λ1 and A4 < 0. These results are plotted in Figure 2.4.5 and Figure 2.4.6.


35

F

rigid rod

Figure 2.4.6

Examples of structures with the behavior of (I) are the Euler column and plates loaded in their plane. An example of a structure with the behavior of (II) is given in Figure 2.4.6. From Figure 2.4.3 and Figure 2.4.5, we see that branched equilibrium states for loads below the critical load are unstable. Later, we shall prove that this is always the case.† The converse that branched equilibrium states would always be stable for loads exceeding the critical load is not true (except for the single mode case). Let us now proceed with the discussion of Minah =const P[u, λ], which is given by (2.4.32) when P3 [ah uh ; λ1 ] ≡ 0 and by (2.4.23) when this term is identically equal to zero. The term P2 [ah uh ; λ1 ] is quadratic in the buckling modes and may be transformed to a form containing only quadratic terms, which will be normalized so that the coefficients are all equal to −1 (remember P2 [ah u; λ1 ] ≤ 0), so that we may write P2 [ah uh ; λ1 ] = −ai ai .

(2.4.37)

Further, we introduce the notation P3 [ah uh ; λ1 ] ≡ Aijkai aj ak, P4 [ah uh ; λ1 ] − P2 [ah akvhk; λ1 ] ≡ Aijkl ai aj akal .

(2.4.38)

We may normalize the load factor λ so that at the critical load, λ = λ1 = 1. Denoting Min P [u; λ] by F [ai ; λ], our discussion of stability is reduced to the discussion of the quadratic form Aijkai aj ak F (ai ; λ) = (1 − λ) ai ai + (2.4.39) Aijk ai aj aka where the term on top is important when Aijkai aj ak0, else the quartic term must be used. †

W. T. Koiter, Some properties of (completely) symmetric multilinear forms . . . . Rep. Lab. For Appl. Mech. Delft, 587 (1975).

36


Let δai be a variation of ai , F (ai + δai ; λ) = (1 − λ) ai ai + 2 (1 − λ) ai δai + (1 − λ) δai δai Aijk (ai aj ak + 3aj akδai + 3akδai δaj + δai δaj δak) + Aijkl (ai aj akal + 4aj akal δai + 6akal δai δaj + 4al δai δaj δak + δai δaj δakδal ) = F (ai ; λ) + δF (ai ; λ) + δ2 F (ai ; λ) + · · · so that

(2.4.40)

3Aijkaj ak δF (ai ; λ) = 2 (1 − λ) ai + δai , 4Aijk aj aka 3A a ijk k δ2 F (ai ; λ) = 1 − λ + δai δaj . 6Aijk aka

(2.4.41)

(2.4.42)

The equilibrium equations are obtained from δF = 0, for all variations δai , so that 3Aijkaj ak = 0. (2.4.43) 2 (1 − λ) ai + 4Aijk aj aka The condition for stability is that δ2 F ≥ 0, 3Aijkakδai δaj ≥ 0. (1 − λ) δai δaj + 6Aijk aka δai δaj

(2.4.44)

Let δah = bh be a unit vector, i.e., b = bi bi = 1, and let ah = aeh , where e is a unit vector and a is the amplitude of a. The equilibrium equations now read 3aAijkej ek 2 (1 − λ) ei + = 0, (2.4.45) 4a2 Aijk ej eke and the stability condition is now given by 3aAijkei bi bj ≥0 1−λ+ 6a2 Aijk eke bi bj

(∀ b | bi bi = 1) .

(2.4.46)

We are now in a position to prove the statement that the branched equilibrium states are unstable for loads below the critical load, i.e., for λ < 1. To show this, we first notice that (2.4.46) must hold for all unit vectors b, and hence also for b = e, so that 3aAijkei ej ej 1−λ+ ≥ 0. (2.4.47) 6a2 Aijk ei ej eke Multiplying (2.4.45) by ei , we obtain 2 (1 − λ) +

3aAijkei ej ek = 0, 4a2 Aijk ei ej eke

(2.4.48)

from which 3aAijkei ej ek = −2 (1 − λ) ,

(2.4.49)


37

or when the cubic term is missing, 1 a2 Aijk ei ej eke = − (1 − λ) . 2

(2.4.50)

Substitution of these expressions into the corresponding form of the stability condition (2.4.47) yields − (1 − λ) ≥ 0

resp. − 2 (1 − λ) ≥ 0,

and hence λ ≥ 1. It follows that for λ < 1 the equilibrium state is unstable. Let us now consider stationary values of the multilinear forms Aijkti tj tk resp. Aijk ti tj tkt on the unit sphere t = 1. Introducing a Lagrangean multiplier µ, we look for stationary values of Aijkti tj tj + µ (ti ti − 1)

resp. Aijk ti tj tkt + µ (ti ti − 1) ,

(2.4.51)

i.e., the first variation of these forms must vanish. This yields 3Aijktj tk − 2µti = 0

resp. 4Aijk tj tkt − 2µti = 0.

(2.4.52)

These equations are of the same form as our equilibrium equations – in other words, the equilibrium equations yield stationary directions on the unit sphere e = 1. Among these stationary directions, there are minimizing directions. Let t = t∗ be a minimizing direction, and let Aijkti∗ tj∗ tk∗ = A3 (t∗ ) = A∗3 ≤ 0

and

Aijk ti∗ tj∗ tk∗ t ∗ = A4 (t∗ ) = A∗4 .

(2.4.53)

For A∗3 ≡ 0, i.e., A∗3 < 0, the equilibrium state is unstable. For A∗3 ≡ 0 and A∗4 > 0, the equilibrium state is stable. To show this, we need the following property† : Min

e=b=1

Aijkl ei ej bkbl = Min Aijk ti tj tkt . t=1

(2.4.54)

Using this property, we obtain from the stability condition (2.4.46) 1 − λ + 6a2 A∗4 ≥ 0.

(2.4.55)

From the equilibrium equation (2.4.45) for e = t∗ and multiplied by t∗i , we obtain 2 (1 − λ) + 4a2 A∗4 = 0,

(2.4.56)

so we may rewrite (2.4.56) to yield − 2 (1 − λ) ≥ 0.

(2.4.57)

Hence the branched equilibrium state is stable for (λ > 1). This property has been proved for a minimizing direction t∗ . For other solutions of the equilibrium equations, the question of stability cannot be established in general. †

W. T. Koiter, Some properties of (completely) symmetric multi-linear forms . . . . Rep. Lab. For Appl. Mech. Delft, 587 (1975).

38

Continuous Elastic Systems − A* > 0 (steepest rise) 3

other solutions

λ

stable

1

unstable

A*3 < 0 (steepest descent)

0 0 a

λ

λ

A*4 > 0 (smallest rise)

1

1 A*3 = 0

A*3 = 0 A*4 < 0 (steepest descent)

0

0 0

0 a

a

Figure 2.4.7

From the equilibrium equations (2.4.45) with e = t∗ and multiplied by t∗i , we find 21−λ 3 A∗3 11−λ a2 = − 2 A∗4 a=−

(A∗3 < 0) (A∗3 = 0) .

(2.4.58)

Our results are shown in Figure 2.4.7. When A∗3 < 0, the dashed line corresponding to the minimizing solution is a line of steepest descent because the amplitude a for fixed λ is monotonically increasing for increasing values of A3 ∈ [A∗3 , 0) ⊂ R− . Similarly, the dashed curve for A∗4 < 0 is a curve of steepest descent. The solid curve for the minimizing solution when A∗4 > 0 is a curve of smallest rise because for fixed λ, the amplitude a is monotonically decreasing with increasing values of A4 ∈ [A∗4 , ∞) ⊂ R+ . We emphasize here once again that our positive verdict about stability only holds for curves of smallest rise, and that for other rising solutions stability must be examined in each particular case. Descending solutions are unstable. The behavior of the structure for minimizing solutions may be presented more clearly. To this end, we consider the generalized displacement corresponding to the external load. This generalized displacement is such that the product of the load and the displacement is equal to the increase of the elastic energy in the structure. For example, consider an elastic bar (Figure 2.4.3),


39

εl

N

Figure 2.4.8

where ε is the generalized displacement. The behavior of this structure is plotted in Figure 2.4.9. Here the curve I corresponds to the fundamental state, which is stable up to the critical load N1 and unstable for loads exceeding N1 . The branched solution II is stable. In the fundamental state, for N = NB the elastic energy is given by the area of ODB, and the energy of the load is given by NB(ε )B, the area of the square ODBNB. The potential energy in B is thus given by the negative of the area OBNB. In the branched state for N = NB the elastic energy is given by the area of OACE, and the energy of the load is given by NB(ε )C, the area of the square OECNB. The potential energy is then given by the negative of the area OACNB. Hence the increment of the potential energy going from the fundamental state B to the branched state C is given by P [ueq ; N] = − area ABC,

(2.4.59)

where ueq indicates that u is the displacement from the equilibrium state B to the equilibrium state C. The first variation of P [ueq ; λ] is given by δP [ueq ; λ] = − δ (area ABC) = − (εl) δN

(2.4.60)

or ε = −

∂P [ueq ; N] . ∂N

(2.4.61)

I

N B

NB N1

A1

A

}

O

C

II δN

D ∆εl E Figure 2.4.9

εl

40

Continuous Elastic Systems I − A3*

λ

1

∆ε = a 2 =

unstable

4 (1 − λ ) 9 A3* 2

2

( )

other solutions

A*3 < 0

A3*

0 0

ε

I

λ

A4*

I

λ

1

1

unstable

other solutions A*4

0

0 0 A*3 = 0, A*4 > 0

other solutions 0

ε

A*3 = 0, A*4 < 0

ε

Figure 2.4.10

In words: The increment of the generalized displacement is equal to the negative of the derivative of the potential energy with respect to the external load, evaluated in the branched equilibrium state. In general, denoting the generalized displacement by ε, we may now write ε = −

d eq ∂F eq F ai ; λ = − a ; λ = ai ai = a2 . dλ ∂λ i

(2.4.62)

It follows that a2 is a natural measure for the increment of the generalized displacement ε, and this justifies our earlier notation a = ae. We can now make the graphs shown in Figure 2.4.10. When A∗3 = 0 and A∗4 = 0, we have from (2.4.58) ε = a2 = −

11−λ . 2 A∗4

(2.4.63)

Let us make some final remarks on the results obtained so far: i. Our results are only valid in the immediate vicinity of the branch point. It is not possible to assess the accuracy of the results for finite deflections from the fundamental state. ii. A structure will (unless prohibited) follow the path of steepest descent or weakest ascent. This suggests that for a structure with geometrical imperfections,

2.5 The influence of imperfections

41

the imperfections coinciding with minimizing directions are the most dangerous imperfections. It is therefore meaningful to consider these imperfections in particular. iii. The fact that the slope of the branched equilibrium path is always smaller than that of the fundamental path implies that the stiffness of the structure after buckling is always decreased. 2.5 The influence of imperfections In the foregoing discussions, it was assumed that the geometry of the structure and the distribution of the loads were known exactly. Further, it was assumed that the material was elastic and followed the generalized Hooke’s Law. However, in practice the quantities are not known exactly, e.g., it is impossible to manufacture a perfectly straight column of constant cross section, or a perfect cylindrical shell. The deviations from the “idealized structure” in the absence of loads are called initial imperfections. Some structures are extremely sensitive to initial imperfection, e.g., cylindrical shells under axial compressive loads. The primary effect of initial imperfection is that the fundamental state of the idealized perfect structure, described by the displacement vector U from the undeformed state, does not represent a configuration of equilibrium of the actual imperfect structure. The bifurcation point of the perfect structure can also not be a bifurcation point of the structure with imperfections. In our previous discussions, we have seen that for structures under a dead– weight load in an equilibrium configuration, the expansion of the potential energy functional starts with terms quadratic in the displacements. However, in the presence of imperfections the fundamental state is not an equilibrium configuration, and hence the expansion of the potential energy functional will start with a term linear in the displacements. Furthermore, if we restrict ourselves to small initial imperfections, this term will also be linear in these imperfections. This implies that our energy functional F (ai ; λ) is replaced by the functional F ∗ (ai ; λ), defined by ∗

F (ai ; λ) = (1 − λ) ai ai +

Aijkai aj ak + µBi ai , Aijk ai aj aka

(2.5.1)

where µ is a measure for the magnitude of the imperfections and B is determined by the shape of the imperfections. It should be noted that our approach is valid under the restrictions |ai | 1,

|1 − λ| 1,

|µ| 1.

(2.5.2)

The equilibrium equations now follow from δF ∗ (ai ; λ) = 0, which yields 2 (1 − λ) ai +

3Aijk aj ak + µBi = 0. 4Aijk aj ak a

(2.5.3)

42


The stability condition is not affected by the linear term µBi ai , and is still given by (2.4.44). Writing ai = aei , |e| = 1, the equilibrium equations become 3a2 Aijk ej ek + µBi = 0, (2.5.4) 2 (1 − λ) aei + 4a3 Aijk ej ek e and the stability condition becomes 1−λ+

3aAijkek bi bj ≥ 0, 6a2 Aijk ek e bi bj

(2.5.5)

where we have put δai = bi , b = 1. The equations (2.5.4) are m equations for ei , i ∈ (1, m) ⊂ N, and the amplitude a., i.e., m − 1 unknowns because e = 1. In general, this problem is extremely complicated, but for the most dangerous imperfections, the ones with directions corresponding to the minimizing directions t∗ , the problem is simplified enormously. We shall now treat this problem in more detail. First, choose the imperfections in the directions of stationary values on the unit ball, i.e., Bi = ti , and choose ei = ti . Let us now first consider the case P3 [ah uh ; λ] 0. The equilibrium equation now reads 2 (1 − λ) ati + µti + 3a2 Aijk tj tk = 0.

(2.5.6)

Multiplying this equation by ti , summing over i, and using the notation Aijkti tj tk = A3 (t), we obtain 2 (1 − λ) a + µ + 3a2 A3 (t) = 0.

(2.5.7)

This equation is to be solved under the side condition a ∈ R+ . The stability condition reads 1 − λ + 3aAijkbi bj tk ≥ 0

for ∀ b| b = 1.

(2.5.8)

Now consider the particular case that the stationary directions are minimizing directions, i.e., t = t∗ . Using the property Min

a=b=c=1.

Aijk ai bj ck = Aijk ti∗ tj∗ tk∗ = A3 (t∗ ) = A∗3 < 0,

(2.5.9† )

1 − λ + 3aA∗3 ≥ 0,

⇒ λ < 1.

(2.5.10)

⇒ λ∗ < 1.

(2.5.11)

we obtain from (2.5.8)

The stability limit follows from 1 − λ∗ + 3a∗ A∗3 = 0, †

W. T. Koiter, Some properties of (completely) symmetric multi-linear forms . . . Rep. Lab. For Appl. Mech. Delft, 587 (1975).

2.5 The influence of imperfections λ

λ

µ0

1

λ* A3* < 0

0

0 0

a

*

0

a

a

Figure 2.5.1

For t = t∗ , we obtain from (2.5.7) a∗ [2 (1 − λ∗ ) + 3a∗ A∗3 ] + µ = 0.

(2.5.12)

Because the term between the square brackets is positive, this equation can only be satisfied for µ < 0. From (2.5.11), we find for the amplitude at the stability limit a∗ = −

1 − λ∗ , 3A∗3

(2.5.13)

and substituting this expression into (2.5.12), we find for the load factor at the stability limit (1 − λ∗ )2 − 3µA∗3 .

(2.5.14)

The general expression for the amplitude of the buckling modes in the presence of imperfections in the minimizing directions follows from (2.5.7) with t = t∗ , 1 2 ∗ − − λ) ± (2.5.15) − 3µA a1,2 = − λ) (1 (1 3 . 3A∗3 These amplitudes are real for (1 − λ)2 > 3µA∗3 , and they are both positive when µ < 0. When µ > 0, only the negative sign can be used because a ∈ R+ . These results are shown in Figure 2.5.1. One easily shows that for µ < 0, the path is stable for a < a∗ and unstable for a > a∗ , and that for µ > 0 the path is stable. Notice that due to the presence of imperfections, the branch point is not reached. The stability limit is now a limit point and not a branch point, as was the case for the perfect structure. In the foregoing discussion, we have made it plausible that imperfections in the minimizing directions are the most harmful ones. To show this, we return to (2.5.4). Multiplying this equation by ei and summing over i, we find 2 (1 − λ) a + 3a2 A3 (e) + µBi ei = 0.

(2.5.16)

The stability condition (2.5.5) reads 1 − λ + 3aAijkbi bj ek ≥ 0 for

∀ b| b = 1.

(2.5.17)

44


Let us now assume that e is known, and let the minimum of the trilinear form in (2.5.17) for fixed e be C3 (e). We then have the following inequalities; A∗3 = A3 (t∗ ) ≤ C3 (e) ≤ A3 (e) .

(2.5.18)

The stability limit now follows from 1 − λ + 3aC3 (e) = 0;

(2.5.19)

hence, a=−

1−λ . 3C3 (e)

(2.5.20)

Because we are interested in the reduction of the critical load due to imperfections (1 < λ), it suffices to consider only negative values of C3 (e). Substitution of (2.5.20) into (2.5.16) yields A3 (e) = 3µBi ei C3 (e) . (2.5.21) (1 − λ)2 2 − C3 (e) For the right-hand side of (2.5.21), we have the estimate 3µBi ei C3 (e) ≤ 3µC3 (e) ≤ 3µA∗ . 3

(2.5.22)

The term between the square brackets in (2.5.21) is always larger than 1, for A3 (e) > 0 it is larger than 2, and for A3 (e) < 0 we have A3 (e) /C3 (e) ≤ 1 as then A3 (e) ∈ [C3 (e) , 0) ⊂ R− . The critical load occurs for µ < 0, so we may write (1 − λ)2 ≥ 3µA∗3 = (1 − λ∗ )2 ,

(2.5.23)

which implies λ∗ < λ, i.e., λ∗ is a lower bound for the critical load. Let us now consider the case P3 [ah uh ; λ] ≡ 0 and Aijk ai aj aka indefinite. The equilibrium equation now reads 2 (1 − λ) aei + 4Aijk ej eke a3 + µBi = 0

(2.5.24)

and the stability condition is given by 1 − λ + ba2 Aijk bi bj eke ≥ 0 for

∀ b| b = 1.

(2.5.25)

Because the fourth degree form is indefinite, we have Min Aijk ti tj tkt = A4 (t∗ ) = A∗4 < 0.

t=1

(2.5.26)

Now we consider imperfections in the direction of the minimizing direction, i.e., Bi = ti∗ . From (2.5.24) and with e = t∗ , we now obtain [2 (1 − λ) a + µ] ti∗ + 4Aijk ti∗ tk∗ t ∗ a3 = 0.

(2.5.27)

Multiplying by ti∗ and summing over i, we obtain for the amplitude 2 (1 − λ) a + 4A∗4 a3 + µ = 0.

(2.5.28)

2.5 The influence of imperfections

45

The stability condition (2.5.25) must hold for all b with b = 1, and hence also for b = t∗ , so 1 − λ + 6A∗4 a2 ≥ 0.

(2.5.29)

The stability limit follows from 1 − λ∗ + 6A∗4 a∗2 = 0,

(2.5.30)

from which follows a∗1,2 = ± −

1 − λ∗ 6A∗4

(λ∗ < 1, because A∗4 < 0) .

(2.5.31)

Substitution into (2.5.28) yields 3 (1 − λ∗ )3/2 = ∓ µ −6A∗4 . 4

(2.5.32)

As λ∗ < 1, the left-hand side of (2.5.32) is positive and hence we must choose the negative sign in (2.5.31) when µ < 0. This implies that (2.5.32) can be rewritten to yield 3 (2.5.33) (1 − λ∗ )3/2 = |µ| −6A∗4 . 4 Equation (2.5.33) is a formula for the reduction of the stability limit due to imperfections in the direction of the minimizing direction, corresponding to the line of steepest descent. We shall now show that this reduction is the largest possible reduction. Consider imperfections in an arbitrary direction. Multiplying the equilibrium (equation 2.5.24) by ei and summing over i, we obtain 2 (1 − λ) a + 4A4 (e) a3 + µBi ei = 0.

(2.5.34)

Introducing C4 (e) defined by Min

w.r.t. b, b=1 e fixed

Aijk bi bj eke = C4 (e) ,

(2.5.35)

we obtain from the stability condition (2.5.25) 1 − λ + 6a2 C4 (e) ≥ 0.

(2.5.36)

Further, we note the inequalities A∗4 ≤ C4 (e) ≤ A4 (e) .

(2.5.37)

Because we are interested in the reduction of the critical load due to imperfections, it suffices to restrict ourselves to values of λ for which λ < 1. From (2.5.36), it then follows that C4 (e) < 0. With the stability limit, we have 1 − λ + 6a2 C4 (e) = 0,

(2.5.38)

46


λ

1

1

other imperfections

other imperfections

λ*

*

λ

A4* < 0

A3* < 0

a

A3* = 0

a

Figure 2.5.2

so that a1,2 = ± −

1−λ 6C4 (e)

.

Substitution into (2.5.34) yields A4 (e) 2 (1 − λ)3/2 1 − = ±µBi ei −6C4 (e), 3C4 (e)

(2.5.39)

(2.5.40)

which determines the reduction of the critical load at the stability limit in the presence of imperfections in arbitrary directions. Let us start the discussion of this expression with the right-hand member. Because B and e are unit vectors, we have ∗ − Bi ei ≤ 1. Because C −6C4 (e) ≤ , 0 ⊂ R we have, by virtue of (2.5.37), ∈ A (e) 4 4 ∗ ∗ −6A4 , so that Bi ei −6C4 (e) ≤ −6A4 . On the left-hand side of (2.5.40), the term between the square brackets is positive and greater than 1 when A4 (e) is positive, and greater than or equal to 2/3 for A4 (e) < 0, so that we can write 3 (2.5.41) (1 − λ)3/2 ≥ ∓ µ −6A∗4 . 4 Because the left-hand member is positive, we must choose the negative sign in (2.5.39) when µ < 0 and the positive sign when µ > 0. Hence, we may write 3 (2.5.42) (1 − λ)3/2 ≥ |µ| −6A∗4 . 4 Comparing this expression with (2.5.33), we find that λ∗ yields the largest possible reduction of the critical load, and is thus a lower bound for the critical load. Summarizing our results, we conclude that imperfections in the direction of the minimizing directions always cause the largest reduction of the critical load.† The results obtained so far are shown in Figure 2.5.2. Notice that for imperfections in the direction of the minimizing direction, the critical load is obtained as a limit point, whereas for other imperfections the critical load λ > λ∗ may be obtained as either a bifurcation point or a limit point. †

This theorem was first proved by D. Ho (Int. J. Sol. Struct. 10, 1315–1330, 1974) for cubic systems.

2.6 On the determination of the energy functional for an elastic body λ*

47

λ* A3* < 0

1

0

0

µ

0

A*3 = 0, A*4 < 0

1

µ

0

Figure 2.5.3

Let us finally consider the influence of the magnitude of the imperfections on the reduction of the critical load for imperfections in the direction of the minimizing direction. When P3 [ah uh ; λ] ≡ 0, the stability limit is given by (2.5.14), from which 3A∗3 dλ∗ = d |µ| 1 − λ∗

(< 0) ,

(2.5.43)

i.e., for λ∗ = 1 the tangent to the λ∗ − |µ| curve is vertical. Similarly, for the case that the cubic terms are identically equal to zero, we obtain from (2.5.33) −6A∗3 dλ∗ , (2.5.44) =− √ d |µ| 2 1 − λ∗ which implies that also in this case the tangent is vertical for λ∗ = 1. This explains why very small imperfections may lead to a considerable reduction of the critical load. 2.6 On the determination of the energy functional for an elastic body In the foregoing section we have given the general theory while assuming that the elastic energy functional is known. In this section, we shall determine the energy functional for a new class of problems. Our line of thought is depicted in Figure 2.6.1: O

U

I

u

II

u0 U O*

u I* Figure 2.6.1

II*

48


Let O be the undeformed (stress-free) state for the idealized (perfect) structure. The position of a material point in this configuration is given by the Cartesian components of the position vector x. The position of this point in the fundamental state I (equilibrium configuration) is given by x + U, and the position in an adjacent state II (not necessarily an equilibrium configuration) by x + U + u. Let O∗ be the undeformed (stress-free) state for the actual (imperfect) structure, obtained by superposition of a displacement field u0 on the state O, without introducing stresses. The position of a material point is now given by the Cartesian components of the position vector x + u0 , where u0 = u0 (x). In the state I∗ (not an equilibrium configuration), the position of the material point is x + u0 + U. In the adjacent state II∗ , the position is x + u0 + U + u, and we shall try to determine u so that the state II∗ is an equilibrium state. In our discussion, we shall need the metric tensors in the various states, g0ij = δij

(2.6.1)

0 0 0 0 g0∗ ij = δij + ui,j + uj ,i + uh,i uh,j

(2.6.2)

gIij = δij + Ui,j + Uj ,i + Uh,i Uh,j ≡ δij + 2ij

(2.6.3)

0 0 0 0 gI∗ ij = δij + Ui,j + Uj ,i + Uh,i Uh,j + ui,j + uj ,i + uh,i uh,j

+ Uh,i u0h,j + Uh,j u0h,i ≡ δij + u0i,j + u0j ,i + u0h,i u0h,j + 2∗ij gII ij = δij + Ui,j + Uj ,i + Uh,i Uh,j + ui,j + uj ,i + uh,i uh,j + Uh,i uh,j + Uh,j uh,i ≡ δij + 2ij + 2γij + Uh,i uh,j + Uh,j uh,i

(2.6.4)

(2.6.5)

gII∗ ij = δij + Ui,j + Uj ,i + Uh,i Uh,j + ui,j + uj ,i + uh,i uh,j + u0i,j + u0j ,i + u0h,i u0h,j + Uh,i uh,j + Uh,j uh,i + Uh,i u0h,j + Uh,j u0h,i + uh,i u0h,j + uh,j uh,i

(2.6.6)

≡ δij + 2∗ij + 2γij∗ + u0i,j + u0j ,i + u0h,i u0h,j . For convenience, we note the relations 1 (U + Uj ,i + Uh,i Uh,j ) 2 i,j 1 ∗ij = ij + (Uh,i u0h,j + Uh,j u0h,i ) 2 1 γij = (ui,j + uj ,i + uh,i uh,j ) 2 1 1 ∗ γij = γij + uh,i u0h,j + uh,j u0h,i + (Uh,i uh,j + Uh,j uh,i ). 2 2 ij =

(2.6.7)

2.6 On the determination of the energy functional for an elastic body

49

Further, we introduce the following notation for the linearized strain tensors and the linearized rotation tensors: 1 (ui,j + uj ,i ) 2 1 ωij = (uj ,i − ui,j ) 2 θij =

1 (Ui,j + Uj ,i ) 2 1 ij = (Uj ,i − Ui,j ) . 2 ij =

(2.6.8)

We shall now derive an approximate expression for the energy functional based on the following assumptions: 1. 2. 3. 4. 5.

The elastic energy density follows the generalized Hooke’s law. The imperfections are (infinitesimally) small. The displacements u are small (but finite). The loads are dead-weight loads. The fundamental state is (approximately) linear, i.e., , , and are of the same order of magnitude.

Let us now first consider the idealized (perfect) structure. The increment of the elastic energy density going from the fundamental state I to the adjacent state II is 1 1 WII − WI = Eijkl ij + γij + (Uh,i uh,j + Uh,j uh,i ) 2 2 1 1 × k + γk + (Um,kum, + Um, um,k) − Eijk ij k 2 2 (2.6.9) 1 = Eijk k γij + (Uh,i uh,j + Uh,j uh,i ) 2 1 1 1 + Eijk γij + (Uh,i uh,j + Uh,j uh,i ) γk + (Um,kum, + Um, um,k) . 2 2 2 Because we are dealing with dead-weight loads, the terms linear in u are balanced by the energy of the external loads, so the potential energy is given by ! 1 1 1 P [u] = Eijk k uh,i uh,j + Eijk γij + (Uh,i uh,j + Uh,j uh,i ) 2 2 2 V " (2.6.10) 1 dV. × γk + (Um,kum, + Um, um,k) 2 This expression is fully exact for dead-weight loads and a material that follows the generalized Hooke’s Law. To simplify this functional, we need estimates for the various terms. Using the relations uh,i = θih + ωih

and

Uh,i = ih + ih ,

(2.6.11)

we have the following estimates: γij = O(θ, ω2 )

1 (Uh,i uh,j + Uh,j uh,i ) = O (ω, θ) = O (ω) 2

(2.6.12)

50


where θ = θij max ,

ω = ωij max ,

and so on.

(2.6.13)

Notice that for the estimate of the second term in (2.6.12), we have made use of the fact that the fundamental state is linear. We now have the following estimates: 1 γij γk = O(θ2 , θω2 , ω4 )γij (Uh,i uh,j + Uh,j uh,i ) 2 3 1 = O(θω, ω ) (Uh,i uh,j + Uh,j uh,i )(Um,kum, + Um, um,k) 4 1 = O(2 ω2 ) kl uh,i uh,j = O(ω2 ). 2

(2.6.14)

Because buckling in the elastic range only occurs in slender structures, we have ω θ. Now consider the linear terms in (2.6.9), Eijk k (ui,j + Uh,i uh,j ) = Eijk k (δih + Uh,i ) uh,j . Because in the energy functional these terms are balanced by the energy of the external loads, we find for the equilibrium equations [Eijk k (δhi + Uh,i )],j + ρXh = 0.

(2.6.15)

Because Uh,i = O(), we can say that Eijk k = Sij

(2.6.16)

is proportional to the external loads. Here Sij is the symmetric stress tensor of Kirchhoff-Trefftz.† It follows that we may write (1)

Sij = λSij ,

(2.6.17)

(1)

where Sij is the stress tensor for a unit load. Further, we shall normalize λ so that at the critical load, λ = 1. The first term in (2.6.10) is now proportional to ω2 , so that of the other terms in (2.6.14), the term of O(θ2 ) may be comparable. Competition of the terms of O(ω2 ) √ and O(θ2 ) is only possible when θ = O(ω ). It follows that the term O(θ ω) is of order O(3/2 ω2 ) and may thus be neglected compared to O(ω2 ). Terms of O(ω3 ) and O(2 ω2 ) in (2.6.14) may also be neglected. This means that the energy functional (2.6.10) can be simplified to yield 1 (1) 1 P [u] = (2.6.18) λS uh,i uh,j + Eijk γij γk dV 2 ij 2 V

√ with a relative error of O( ). Notice that this expression is linear in the load factor λ. †

Notice that this is not the same stress tensor as the one introduced in (2.2.7).


51

Let us now consider the actual (imperfect) structure. The increment in the elastic energy density going from state I∗ to state II∗ is 1 ∗ 1 ∗ ) − E∗ijk ∗ij ∗k E (∗ + γij∗ )(∗k + γk 2 ijk ij 2 1 1 ∗ Um,ku0m, + Um,ku0m, = E∗ijk ∗k γij∗ + E∗ijk γij∗ γk = E∗ijk k + 2 2 1 1 (2.6.19) uh,i u0h.j + uh,j u0h.i + (Uh,i uh,j + Uh,j uh,i ) × γij + 2 2 1 1 1 + E∗ijk γij + uh,i u0h.j + uh,j u0h.i + (Uh,i uh,j + Uh,j uh,i ) 2 2 2 1 1 × γk + um,ku0m, + um,l u0m,k + (Um,kum, + Um, um,k) , 2 2

WII∗ ∗ − WI∗∗ =

where E∗ijk is the tensor of elastic moduli in O∗ , E∗ijk = Eijk + O(γ 0 E),

(2.6.20)

so we may replace E∗ijk by Eijk . For our discussion of the terms with the imperfections, we shall need the following expressions, θ0 =

1 0 u + u0j ,i max 2 i,j

and

ω0 =

1 0 u − u0i,j max . 2 j ,i

(2.6.21)

We now have the following estimates, kl (uh,i u0h,j + uh,j u0h,i ) = O(ωω0 , θω0 , ωθ0 , θθ0 ) γij (u0h,i uh,j + u0h,j uh,i ) = O(θω0 ω, ω3 ω0 ) γij (Um,ku0m,

+

Um, u0m,k)

(2.6.22)

= O(θω , ω ω ). 0

2

0

Further, (2.6.19) contains terms that are quadratic in the imperfections, which can be neglected. At the critical load, we have ω θ. We know that is small but finite, and we can always choose deviations from the fundamental state as small as we need so that θ . The leading term in (2.6.22) is of order ωω0 . The only other term that may be important is the term O (ωθ0 ), and all other terms are small compared to these terms. Using (2.6.20) and (2.6.17) and the fact that we have dead-weight loads, we can simplify the energy functional corresponding to (2.6.19) to yield (1) ∗ P [u] = P [u] + λSij u0h,i uh,j dV, (2.6.23) V

where P [u] is the functional for the perfect structure defined in (2.6.18). Notice that the term with the imperfections depends linearly on the load factor λ. Let us now consider the effect of eccentric loads. In this case, only the functional of the external loads must be modified. Let x denote the position of a material point, and let ρX be the force per unit volume at this point. The potential energy of the dead-weight load is then −ρX · u (x), where u denotes the displacement of the point

52


λ 1

1

λ

λ*

*

A*3 = 0, A*4 < 0

A3* < 0

0

0 0

a

0 a

Figure 2.6.2

in question. When the load is applied eccentrically, say, in a point y close to x, the potential energy becomes − ρX · u (y) = −ρX · [u (x) + (yi − xi ) u,i (x) + · · ·] ,

(2.6.24)

where yi − xi determines the eccentricity of the loads. To a first approximation, the correction of the functional of the external loads is linear in the eccentricity of the loads and linear in the loads. It follows that for both the case of imperfection and the case of eccentric loads, the additional term in the energy functional is of the form λQ [u]. It is now convenient to slightly modify our function F ∗ (ai , λ), Aijkai aj ak ∗ F (ai , λ) = (1 − λ) ai ai + + λµBi ai (2.6.25) Aijk ai aj aka so that it explicitly shows the linear dependence on the load factor in the last term. This expression is valid for λ ≥ 0, e.g., the equilibrium equations, when P3 (ah uh , λ) ≡ 0, now read 2 (1 − λ) ai + 3Aijkaj ak + λµBi = 0.

(2.6.26)

For λ = 0, the solution is ai = 0, and for λ 1, we have ai = −

λµBi . 2 (1 − λ)

(2.6.27)

Our figures relating λ to a now are shown in Figure 2.6.2. The accuracy of these curves increases with decreasing values of λ. In the case that we only have to deal with geometrical imperfections, we can slightly modify the functional (2.6.25). To this end, the imperfections are represented by u0 = a0h uh + u0

(2.6.28)

and the displacement field is written as u = ah uh + u.

(2.6.29)


We now return to (2.6.23), which fully written out, reads 1 (1) 1 P∗ [u] = λSij uh,i uh,j + u0h,i uh,j + Eijk γij γk dV. 2 2

53

(2.6.30)

V

The second variation is P2 [u, λ] =

1 (1) 1 λSij uh,i uh,j + Eijk θij θk dV. 2 2

(2.6.31)

V

Expanding (2.6.31) with respect to λ about λ = 1, and using the fact that P2 [u, 1] = 0, we obtain (1) 1 P2 [u, λ] = (1 − λ) Sij uh,i uh,j dV + · · · ≡ (1 − λ) P2 [u, 1] + · · · (2.6.32) 2 V

Substitution of (2.6.29) into (2.6.32) yields for the integral 1 (1) Sij akal ukh,i u h,i + akukh,i uh,j + al u h,j uh,i + uh,i uh,j dV. 2

(2.6.33)

V

The orthogonality of u with respect to the buckling modes is now defined by 1 (1) S akukh,i uh,j dV = 0. (2.6.34) 2 ij V

The last term in (2.6.33) may be neglected because it is small compared to the first term. The buckling modes are now normalized so that 1 (1) (1) S akal ukh,i u h,i dV = akal Sij ukh,i u h,j dV = akal δk = akak. (2.6.35) 2 ij V

V

Using (2.6.28) and (2.6.29), we find for the term with the imperfections in (2.6.30), (1) (2.6.36) λSij a0kal ukh,i u h,j + a0kukh,i uh,j + a u h,j u0h,i + u0h,i u0h,j dV. V

The second term vanishes by virtue of (2.6.34), and the third term vanishes by virtue of the orthogonality condition, (1) Sij a u h u0h,i dV = 0. (2.6.37) V

The last term in (2.6.36) may be neglected because it is small compared to the first term, so we have (1) 0 λakal Sij ukh,i u h,j dV = 2λa0kal δk = 2λa0kak, (2.6.38) V

where we have made use of (2.6.35).

54


In (2.6.25), the influence of the imperfections is accounted for by the term λµBi ai , where µ < 0, and is equivalent to (2.6.38). Hence, we may modify (2.6.25) to yield Aijkai aj ak ∗ − 2λa0i ai . (2.6.39) F (ah , λ) = (1 − λ) ai ai + Aijk ai aj aka It is this form of F ∗ (ah , λ) that is most suitable for application.

3

Applications

3.1 The incompressible bar (the problem of the elastica) We consider a simply supported incompressible bar loaded by compressive forces N at its ends, as shown in Figure 3.1.1. Its length is denoted by . Let ϕ denote the angle between the horizontal and the tangent at a point of the deflected bar. The following relations then hold: dw = w dx cos ϕ · ϕ = w sin ϕ =

ϕ =

(3.1.1)

w w =√ , cos ϕ 1 − w2

where ϕ denotes the curvature of the bar at the corresponding point. The displacement of the moveable end is now given by − = −

1 − 1 − w2 dx. cos ϕ dx =

0

(3.1.2)

0

Here we have made use of the fact that the bar is incompressible, i.e., inextensional. An explicit condition for incompressibility reads [(1 + u ) + w2 ](dx)2 = (dx)2 2

or u =

1 − w2 − 1.

(3.1.4) −∆ l

x N

(3.1.3)

w(x)

N

l Figure 3.1.1 55

56

Applications

The potential energy of this system is now given by w2 1 1 2 2 EIϕ dx + N = EJ P [w; N] = +N 1 − w − 1 dx. 2 2 1 − w2 0

0

(3.1.5) The expression (3.1.5) was already given in (2.2.25), and it was discussed there that the potential energy of the external load is a nonlinear function of the displacement due to the incompressibility condition imposed. Using the expansion 1 = 1 + w2 + w4 + · · · 1 − w2 1 1 1 − w2 = 1 − w2 − w4 − · · · , 2 8

(3.1.6a)

we may rewrite (3.1.5) to yield P [w; N] =

1 1 2 1 4 EIw2 (1 + w2 + · · ·) − N w + w + · · · dx, 2 2 8

(3.1.6b)

0

and hence we have P2 [w; N] =

1 1 2 2 EIw − Nw dx 2 2

(3.1.7)

0

P3 [w; N] = 0 P4 [w; N] =

1 1 2 2 4 EIw w − Nw dx. 2 8

(3.1.8)

(3.1.9)

0

The variational equation for neutral equilibrium now follows from (3.1.7), P11 [w, ζ; N] =

[EIw ζ − Nw ζ ] dx = 0.

(3.1.10)

0

Integration by parts yields

EIw ζ 0

− (EIw + Nw ) ζ 0 +

(EIw + Nw ) ζdx = 0.

(3.1.11)

0

Because this equation must hold for all kinematically admissible displacement fields, we have EIw + Nw = 0

(3.1.12)

EIw ( ) = EIw (0) = 0,

(3.1.13)

3.1 The incompressible bar (the problem of the elastica)

57

while the kinematic conditions at the supports lead to the requirement w( ) = w(0) = 0.

(3.1.14)

The general solution to (3.1.12) reads w = c1 + c2 x + c3 cos αx + c4 sin αx,

(3.1.15)

where we have introduced the parameter α2 =

N . EI

(3.1.16)

The boundary conditions (3.1.13) and (3.1.14) admit a nonzero value for c4 only. This value exists if and only if sin α = 0,

(3.1.17)

so α = kπ for k = 1, 2, . . . . The constant c4 remains undetermined. The critical load now follows from the smallest root, N1 =

π2 EI . 2

(3.1.18)

We shall now investigate whether this equilibrium state is stable. The necessary condition for stability P3 [w; N] = 0 is satisfied. The displacement field for small deflections from the equilibrium configuration is now written as w(x) = aw1 (x) + w(x),

(3.1.19)

where w1 (x) is the buckling mode and w(x) is the orthogonal remainder. We must now consider the minimum problem, as per (2.3.44), Min

T11 [w,w]=0

P2 [w] + P21 [w1 , w]

(3.1.20)

but the last term vanishes because P3 [w; N] ≡ 0. Conseqently, the quantity that governs stability in the critical state is P4 [w1 ; N1 ] = 0 (see 2.3.54). Taking w1 (x) = 1 · sin πx/ , we find P4 [w1 ; N1 ] =

1 π6 πx πx 1 π4 πx EI 6 cos2 sin2 − N1 4 cos4 dx 2 8

0

=

1 π4 N1 2 4

1 2πx 1 1 2πx sin2 − · 1 + cos2 4 4 4

(3.1.21)

2 dx =

1 π4 N1 , 64 4

0

which is positive, and hence equilibrium at the buckling load is stable. Let us now consider the behavior of the structure for loads slightly exceeding N1 . The increment of the second variation is 1 P2 [w1 ; N] − P2 [w1 ; N1 ] = 2

0

(N1 − N)w2 1 dx =

π4 (N1 − N ) . 4 2

(3.1.22)

58

Applications

The function F (a; N) is now given by F (a; N) =

π4 1 π4 π2 N1 π2 4 2 4 2 , (3.1.23) − N) a + N a = + a − λ) a (N (1 1 1 4 2 64 4 4 2 16 2

where λ = N/N1 . The equilibrium condition follows from π2 N1 ∂F π2 2 =0= a (1 − λ) + 2 a . ∂a 2 2 8

(3.1.24)

Besides the fundamental solution a = 0, (3.1.23) has the solution √ † 2 2 a= λ − 1. π

(3.1.25)

Let us now consider the displacement − , 1 − = 2

1 2 2 w dx 1 − 1 − w dx ≈ 2 0

1 = a2 2

0

(3.1.26)

π2 a2 π2 2 πx dx = cos = 2 (λ − 1) , 2 2

0

where we have made use of (3.1.25). The effective relative shortening is N − = 2 (λ − 1) = 2 −1 . N1 Further, we have

2 1 2 2 ∂ 2 2 − = = = 2 , ∂N N1 π EI EA π2 i 2

(3.1.27)

(3.1.28)

√ where i = I/A is the radius of gyration of the cross section. Our results are now plotted in Figure 3.1.2. N

Et = E

N1

E 1 0 0 − Figure 3.1.2 †

λ = 1.01 already gives a deflection a ≈ 0.1 .

∆l l

π 2i 2 2l2

3.2 Bar with variable cross section and variable load distribution

59

It should be noted that these results are only valid for small deflections. Comparison with the exact solution shows that the results are valid up to deflections of the middle of the bar of about 20% of its length.† 3.2 Bar with variable cross section and variable load distribution Consider a bar with variable cross section loaded in compression by a normal force N(x) and a load distribution P(x), clamped at one end (see Figure 3.2.1). The bar is considered to be incompressible. Consider an infinitesimal part of the bar in the deflected configuration (see Figure 3.2.2). Due to the deflection of the bar, the load P (x) dx will move over a distance x dx [1 − cos ϕ (x)], and hence the contribution to the potential energy is given by 0

x −P (x) dx

0

[1 − cos ϕ (ξ)] dξ, 0

so the total potential energy is given by P [ϕ (x) ; λ] =

1 B (x) ϕ2 (x) dx + λ 2

0

P (x)



0

+ λN

 x 

0

 

[cos ϕ (ξ) − 1] dξ dx  (3.2.1)

[cos ϕ (x) − 1] dx, 0

where B (x) denotes the bending stiffness. The normal force in the bar is given by N (x) = N +

P (ξ) dξ,

(3.2.2)

x

A

P ( x)

N

l Figure 3.2.1

[

]

Figure 3.2.2

†

Cf. W. T. Koiter, Stability of Elastic Equilibrium, Sect. 6 (Thesis, Delft).

+x

60

Applications

and hence we have N (x) = −P (x). Using this relation to rewrite the second term in (3.2.1), we find

x

−N (x) 0

[cos ϕ (ξ) − 1] dξ dx 0

x

= −N (x) 0

[cos ϕ (ξ) − 1] dξ + N (x) [cos ϕ (x) − 1] dx 0

= N ( )

0

N (x) [cos ϕ (x) − 1] dx,

[cos ϕ (x) − 1] dx + 0

0

so that (3.2.1) can be rewritten to yield ! P [ϕ (x) ; λ] =

" 1 2 B (x) ϕ (x) + λN (x) [cos ϕ (x) − 1] dx. 2

(3.2.3)

0

The second, third, and fourth variations are given by 1 1 B (x) ϕ2 (x) − λN (x) ϕ2 (x) dx 2 2

(3.2.4)

P3 [ϕ (x) ; λ] ≡ 0

(3.2.5)

P2 [ϕ (x) ; λ] = 0

P4 [ϕ (x) ; λ] =

1 λN (x) ϕ4 (x) dx. 24

(3.2.6)

0

Because P3 [ϕ (x) ; λ] ≡ 0 at the critical load factor, and where P2 [ϕ (x) ; λ1 ] is semidefinite-positive, stability is governed by the sign of P4 [ϕ (x) ; λ1 ]. The condition P11 [ϕ (x) , ψ(x) ; λ1 ] = 0 yields P11 [ϕ (x) , ψ(x) ; λ] =

[B (x) ϕ (x) ψ (x) − λN (x) ϕ (x) ψ(x)] dx = 0, (3.2.7)

0

where ψ(x) is a kinematically admissible rotation field. Integrating by parts, we obtain B (x) ϕ (x) ψ(x)0 −

[B (x) ϕ (x)] + λN (x) ϕ (x) ψ(x) dx = 0.

(3.2.8)

0

The first term vanishes when the ends of the bar are hinged or clamped. From the second term, we obtain

[B (x) ϕ (x)] + λN (x) ϕ (x) = 0.

(3.2.9)

3.3 The elastically supported beam

61

The boundary conditions are

ϕ (0) = 0 B(0)ϕ (0) = 0

(when A is clamped) (when A is hinged)

B( )ϕ ( ) = 0.

(3.2.10)

From the equation for neutral equilibrium (3.2.9) we obtain at the buckling load

λ1 N (x) ϕ1 (x) = − [B (x) ϕ1 (x]) ,

(3.2.11)

so that −

P4 [ϕ1 (x) ; λ1 ] =

1 [B (x) ϕ1 (x)] ϕ31 (x) dx 24

0

1 = − B (x) ϕ1 (x) ϕ31 (x)0 + 24

1 B (x) ϕ1 (x) · 3ϕ21 (x) ϕ1 (x) dx. 24

0

The first term vanishes when the ends are either clamped or hinged, so 1 P4 [ϕ1 (x) ; λ1 ] = 8

2 B (x)ϕ2 1 (x) ϕ1 (x) dx > 0,

(3.2.12)

0

and hence the equilibrium at the buckling mode is stable. Notice that N (x) may also be a locally tensile force. In the case that A is an elastic hinge, we must add an additional term to the potential energy given by 12 Cϕ2 (0). The boundary condition for x = 0 then becomes B (0) ϕ (0) = Cϕ (0)

(3.2.13)

and in the expression for P4 [ϕ1 (x) ; λ1 ], from the stock term we now have a contribution at x = 0, 1 B (0) ϕ1 (0) ϕ31 (0) > 0, 24

(3.2.14)

so that for a elastically hinged bar the equilibrium at buckling is also stable. 3.3 The elastically supported beam† Consider a uniform elastically supported hinged-hinged beam loaded in compression by a normal force N (see Figure 3.3.1). †

Cf., J. G. Lekkerkerker, On the stability of an elastically supported beam . . . (Proc. Kon. Ned. Ak. Wet. B65, no. 2, 190–197 (1962).

62

Applications x w (x)

N

Figure 3.3.1

The reaction of the springs per unit length of the beam is cw, and the corresponding energy is 12 cw2 (x). The potential energy of this system is now given by P [w (x) ; N] = 0

1 w2 1 2 2 − 1 B cw + + N 1 − w dx, 2 1 − w2 2

(3.3.1)

and hence P2 [w; N] =

1 1 1 Bw2 + cw2 − Nw2 dx 2 2 2

(3.3.2)

P3 [w; N] ≡ 0

(3.3.3)

0

P4 [w; N] =

1 1 Bw2 w2 − Nw4 dx. 2 8

(3.3.4)

0

The variational equation for neutral equilibrium follows from (3.3.2), P11 [w, ζ; N] =

(Bw2 ζ + cwζ − Nw ζ ) dx = 0.

(3.3.5)

0

From integration by parts, we obtain

Bw ζ 0

+ 0

=

Bw ζ 0

(−Bw ζ + cwζ + Nw ζ) dx − Nw ζ 0 − Bw ζ 0 − Nw ζ 0 +

(3.3.6) (Bw

+ Nw + cw) ζdx = 0,

0

from which follows Bw + Nw + cw = 0,

(3.3.7)

with the dynamic boundary conditions Bw ( ) = Bw (0) = 0.

(3.3.8)

3.3 The elastically supported beam

63

Further, we have the kinematic boundary conditions w(0) = w( ) = 0.

(3.3.9)

The deflection w(x) is now written as a Fourier sine series, w(x) =

∞ k=1

ak sin

kπx .

(3.3.10)

Notice that each term in this series satisfies the kinematic boundary conditions. To determine the stability limit, we might substitute this series into (3.3.7); however, then we need the fourth derivative of this series, and it is unknown beforehand whether the series obtained is convergent. It is therefore more suitable to return to the second variation as this only requires the second derivative of this series, which is expected to be convergent in view of the fact that the dynamic boundary conditions also contain a second derivative. Substitution of (3.3.10) into (3.3.2) yields  %∞ %∞ &2 &2 k2 π2 1 kπx 1 kπx P2 [w; N] =  B − 2 ak sin + c ak sin 2 2 k=1 k=1 0 %∞ &2  ∞ kπ k4 π4 1 kπx  k2 π2 2 B 4 + c − N 2 ak. ak cos − N dx = 2 4 k=1

k=1

(3.3.11) This expression is positive when all its coefficients are positive. The coefficient of ak becomes zero for N1 =

k2 π2 2 B + C. 2 k2 π2

(3.3.12)

This is the stability limit. To obtain the lowest (critical) value of N, we must minimize (3.3.12) with respect to k, dN1 kπ 2 2 = 2 2 B − 3 2 C = 0, dk kπ

(3.3.13)

from which kπ =

1/4 C . B

(3.3.14)

However, here we have assumed that k ∈ R+ , but k ∈ N+ , so k cannot generally take on the value in (3.3.14). For k ∈ R+ , we have from (3.3.14) and (3.3.12) √ NMin = 2 BC. (3.3.15) √ It is now convenient to plot N/2 BC versus (C/B)1/4 /π for various values of k ∈ N+ (see Figure 3.3.2).

64

Applications

N 2 BC

k =1

k=2

k=3

1

STABLE 0 0

1

2

3

(C B )1 4

l π

Figure 3.3.2

Suppose now that we have /π (C/B)1/4 = 3.5. We then investigate what the value of N is for k = 3 and k = 4. From (3.3.12), we obtain for k = 3 √ √ 9 √ 1 BC + (3.5)2 BC = 2.096 BC, 2 (3.5) 9 √ whereas the actual minimizing value is N = 2 BC. Similarly, we obtain for k = 4, √ N = 2.072 BC. In general, for a sufficiently high wave number k we may assume that k ∈ R+ , and then NMin is given by (3.3.15). As indicated in Figure 3.3.2, equilibrium is stable in the shaded region but not necessarily at the stability limit, i.e., on the curves bounding the shaded region. Because P3 [w; N] ≡ 0, stability is governed by the sign of P4 [w1 ; N1 ]. With N1 given by (3.3.12) and N=

w1 (x) = ak sin

kπx ,

k ∈ N+ ,

(3.3.16)

we obtain from (3.3.4) 2 2 2 1 kπ kπ kπx kπx 2 1 kπ kπx 2 dx P4 [w1 ; N1] = − N1 ak cos B − 2 ak sin ak cos 2 8 0

=

1 k6 π6 4 2 2kπx 1 k4 π4 4 4 kπx dx a sin B − N1 4 ak cos 8 6 k 8

(3.3.17)

0

1 k6 π6 4 k6 π6 4 3 1 k2 π2 4 k4 π4 = a − N1 3 ak = B − 3C . B a 16 5 k 64 64 k 4

The sign of this expression depends on C, and the critical value is C=

k4 π4 B. 3 4

(3.3.18)

3.3 The elastically supported beam N 2 BC

k =1

k=2

65

stable k=3 unstable

1 0

0 0

1

2

3

(C B )1 4

l π

Figure 3.3.3

For the minimizing value of kπ/ given in (3.3.14), we obtain ) 1 4 C C < 0, P4 [w1 ; NMin ] = − ak 32 B

(3.3.19)

which means that at the minimizing values of kπ/ , the equilibrium is unstable. We now plot our results in Figure 3.3.3. For values of N exceeding N1 , the increment of the second variation is given by P2 [w1 ; N] =

1 k2 π2 kπx − (N − N1 ) a2k 2 cos2 dx 2 0

=

1 k2 π2 (N1 − N) a2k 4

=

1 k2 π2 , (1 − λ) a2kN1 4

(3.3.20)

and for N1 = Nmin and kπ/ given by (3.3.14), we obtain P2 [w1 ; N] =

1 (1 − λ) a2kc . 2

(3.3.21)

The function F (ak; λ) is now given by ) C 1 1 1 k2 π2 4 1 2 4 2 a = c (1 − λ) ak − a , (3.3.22) F (ak; λ) = c (1 − λ) ak − 2 16 B k 2 16 2 k Let us now consider the influence of imperfections on the behavior of this structure. In (2.6.39), we saw that for imperfections of the shape of the buckling mode, we must add to (3.3.22) a term −2λak0 ak, hence 1 1 k2 π2 4 ∗ 2 0 (3.3.23) F (ak; λ) = c (1 − λ) ak − a − 2λakak . 2 16 2 k

66

Applications

The equilibrium condition is

1 1 k2 π2 3 ∂F ∗ 0 a − 2λak (ak; λ) = 0 = c 2 (1 − λ) ak − ∂ak 2 4 2 k

(3.3.24)

and the stability condition is

∂2 F ∗ 3 k2 π2 2 1 ≥ 0. c 2 − λ) − ; λ) = a (1 (a k 2 4 2 k ∂a2k

(3.3.25)

The stability limit is reached for 8 2 (3.3.26) (1 − λ∗ ) 2 2 . 3 kπ Substitution into the equilibrium condition yields √ 4√ 4 − − 2λ∗ a0k = 0 6 (1 − λ∗ )3/2 (1 − λ∗ )3/2 6 3 kπ 9 kπ or 3 √ kπ ∗ 0 6 λ ak. (3.3.27) (1 − λ∗ )3/2 = 8 This result shows that for small imperfections the critical load is reduced appreciably, e.g., ka0k/ ≈ 0.01 yields λ∗ ≈ 0.91, i.e., a reduction of about 10% in the critical load. Summarizing, we may say that the elastic foundation increases the buckling load but it also increases the imperfection sensitivity. Finally, let us calculate the tangent modulus, which determines the stiffness of the structure after buckling. The extra shortening after buckling is a2k =

1 2 2 − = w dx 1 − w − 1 dx ≈ 2 0

1 = 2

0

k2 π2 a2k 2

1 k2 π2 2 kπx 1 dx = ak = cos 4 4

)

2

C 2 a . B k

(3.3.28)

0

With (3.3.26), we find 1 − = 4 or 2 − = 3

)

)

C8 2 (1 − λ∗ ) 2 2 B3 kπ

C (1 − λ∗ ) B

)

B 2 = (1 − λ∗ ) , C 3

(3.3.29)

so that ∂ (− / ) 1 2 = =− . ∂N 3Nmin Et A

(3.3.30)

Hence,

√ BC 3 Nmin = −3 . Et = − 2 A A This result is shown in Figure 3.3.4.

(3.3.31)

3.4 Simple two-bar frame

67

N

N1 Et = −3 E

I A

C E

−∆ l l Figure 3.3.4

3.4. Simple two-bar frame We consider a structure consisting of two incompressible bars of length , rigidly attached to each other on one end and hinged on the other end (see Figure 3.4.1), loaded by a force N. Let () denote the derivative with respect to x and ()· the derivative with respect to y. At point A, the following kinematic relations then hold: w ( ) = v· ( ) = − sin θ

(3.4.1)

1 − v·2 − 1 dy

(3.4.2)

1 − 1 − w2 dx. v ( ) =

(3.4.3)

w ( ) =

0

0

l

N

v ( y) −θ

A

l – w (x)

B

x

Figure 3.4.1

C y

68

Applications N

Figure 3.4.2

The potential energy is now given by P [v (y) , w (x) ; N] = 0

1 1 w2 v··2 2 + N 1 − w − 1 dx + dy, B B 2 1 − w2 2 1 − v·2 0

(3.4.4) where B is the bending stiffness of the bars. However, it is inconvenient to formulate the problem this way due to the nonlinear kinematic side conditions (3.4.2) and (3.4.3). It is more convenient to replace the rigid connection in A by the elastic element of Figure 3.4.2, where both springs have a stiffness c. The corresponding additional energy (penalty function) is given by 2 2   1  1 1 − 1 − w2 dx . (3.4.5) 1 − v·2 − 1 dy + c w ( ) − c w ( ) − 2 2 0

0

Using the appropriate expansions, we find the following expressions for the second and third variation: P2 [v, w; N, c] =

1 1 ··2 1 1 2 2 Bw − Nw dx + Bv dy + cw2 ( ) + cv2 ( ) 2 2 2 2

0

(3.4.6)

0

1 P3 [v, w; N, c] = cw ( ) 2

1 v dy − cv ( ) 2 ·2

0

w2 dy.

(3.4.7)

0

In this case, we have P3 ≡ 0. The condition for neutral equilibrium follows from (3.4.6), P11 [v, η, w, ζ; N, c] =

1 1 Bw ζ − Nw ζ dx 2 2

0

+ 0

1 ·· ·· Bv η dy+ cw ( ) ζ( ) + cv ( ) η( ) = 0. (3.4.8) 2


69

From integration by parts, we obtain 0=

Bw ζ 0

− Bw ζ 0 − Nw ζ 0 +

+ Bv·· η· 0 − Bv··· η 0 +

(Bw + Nw ) ζdx

0

(3.4.9)

····

Bv η dy + cw ( ) ζ( ) + cv ( ) η( ) , 0

from which the following differential equations are obtained: Bw + Nw = 0,

Bv···· = 0.

(3.4.10)

For the evaluation of the stock terms, we must bear in mind that η and ζ are kinematically admissible functions, and that they must satisfy the kinematic condition (3.4.1), i.e., ζ ( ) = η· ( ) .

(3.4.11)

Bw ( ) + Bv·· ( ) = 0

(3.4.12)

We then obtain

Bw (0) = Bv·· (0) = 0 − Bw ( ) − Nw ( ) + cw ( ) = 0

(3.4.13)

···

− Bv ( ) + cv ( ) = 0. In addition to these dynamic boundary conditions, we have the kinematic boundary conditions w (0) = v (0) = 0 w ( ) = v· ( ) = − sin θ.

(3.4.14)

These boundary conditions and the differential equations are also homogeneous and linear, so they always have the trivial solution v = w = 0. We now introduce the parameter N/B = α2 .

(3.4.15)

Choosing solutions for u and v so that the first two conditions of (3.4.13) and the first two conditions of (3.4.14) are satisfied, we obtain w(x) = A1 sin αx + A2 x v(x) = A3 y + A4 y3 .

(3.4.16)

From the remaining conditions, we obtain the following set of equations: −α2 A1 sin α + 6A4 = 0 Bα3 A1 cos α − N (αA1 cos α + A2 ) + c (A1 sin α + A2 ) = 0 −6BA4 + c A3 + A4 3 = 0 αA1 cos α + A2 = A3 + 3A4 2 = − sin θ.

(3.4.17)

70

Applications

The second of these conditions can be simplified because the first two terms cancel, so that N A1 sin α + A2 1 − = 0. (3.4.18) c For c → ∞, we obtain from this equation and the fourth equation of (3.4.17), A1 =

sin θ , sin α − α cos α

A2 =

− sin α sin θ . sin α − α cos α

(3.4.19)

From the third equation of (3.4.17) and the fourth equation of (3.4.17), for c → ∞ we obtain A3 =

1 sin θ, 2

1 A4 2 = − sin θ. 2

(3.4.20)

From the first of the equations of (3.4.17), we now obtain the bifurcation condition α2 2 sin α +3=0 sin α − α cos α

(3.4.21)

or tan α =

α . 1 1 + (α )2 3

(3.4.22)

This bifurcation equation is valid for c → ∞, i.e., for our actual structure. We now determine the value of P3 at the critical load,   1 1  w2 (3.4.23) P3 [v1 , w1 ; N1 , c → ∞] = lim  cw1 ( ) v·2 1 dy − cv1 ( ) 1 dx . c→∞ 2 2 0

0

From (3.4.16) and (3.4.18), we obtain lim cw1 ( ) = A2 N =

c→∞

3N sin θ 2

(α )

=

3B sin θ ( )2

,

(3.4.24)

where we have used (3.4.19), (3.4.22), and (3.4.15). From (3.4.16) and the third of equation of (3.4.17), we obtain lim cv1 ( ) = 6BA4 = −

c→∞

We now obtain

3B sin θ ( )2

.

(3.4.25)

  3 B sin θ  2 A3 + 3A4 y2 dy + (α1 A1 cos α1 x + A2 )2 dx P3 [v1 , w1 ; N1 , c → ∞] = 2 2 0 0 3 B sin θ 9 = A3 + 2A3 A4 3 + A24 5 (3.4.26) 2 2 5 1 1 sin 2α1 + 2A1 A2 sin α1 + A22 . + αA21 + 2 2α


71

Using (3.4.20) and A1 =

3 sin θ , −(α )2 sin α

A2 =

3 sin θ , (α )2

(3.4.27)

which follow from (3.4.19) and (34.22), after some algebra we find 3N1 sin3 θ 7 9 P3 [v1 , w1 ; N1 , c → ∞] = . + 20 2 (α1 )2 (α1 )2

(3.4.28)

The smallest positive root of (3.4.22) is α1 = 1.1861π,

(3.4.29)

and using this value we finally find P3 [v1 , w1 ; N1 , c → ∞] = 0.14565N1 sin3 θ.

(3.4.30)

This means that we have the behavior shown in Figure 3.4.3. The equilibrium path is stable for positive values of wmax and unstable for negative values of wmax . λ

Wmax

N

N

stable

unstable

Figure 3.4.3

72

Applications

For a further discussion of this problem, we refer the reader to W. T. Koiter, Postbuckling analysis of a simple two-bar frame, Recent Progress in Applied Mech., The Folke Odquist Volume (337–354). 3.5 Simple two-bar frame loaded symmetrically We consider again the simple two-bar frame, but now loaded symmetrically by two compressive forces N (see Figure 3.5.1). The stiff joint is again replaced by an elastic element, and the potential energy is now given by P [v (y) , w (x) ; N] = 0

1 w2 2 B +N 1 − w − 1 dx 2 1 − w2

1 v··2 ·2 − 1 B + N 1 − v dy 2 1 − v·2 0  2 1  1 − v·2 − 1 dy + c w ( ) − 2

+

(3.5.1)

0

 1 + c v ( ) − 2

2 1 − 1 − w2 dx . 0

The second variation is now given by P2 [v, w; N, c] = 0

1 1 Bw2 − Nw2 dx 2 2

+

1 ··2 1 ·2 1 1 Bv − Nv dy + cw2 ( ) + cv2 ( ) . 2 2 2 2

0

N

l

N

l

Figure 3.5.1

(3.5.2)

3.5 Simple two-bar frame loaded symmetrically

73

The third variation is still given by (3.4.7) but vanishes due to the symmetry of the structure and its loading. We will also need the fourth variation,

1 1 ··2 ·2 1 ·4 1 2 2 4 P4 [v, w; N, c] = Bw w − Nw dx + Bv v − v 2 8 2 8 0 0 (3.5.3)   2 2 1 ·2  1 2  1 1 v dy + c  w dx . + c 2 2 2 2 0

0

From (3.5.2), similar to the previous example we now obtain Bw + Nw = 0,

Bv··· + Nv··· = 0.

(3.5.4)

The kinematic condition (3.4.12) is now automatically satisfied because the bending moments vanish at x = y = . The dynamic boundary conditions are Bw (0) = Bv·· (0) = 0 −Bw ( ) − Nw ( ) + cw ( ) = 0 ···

(3.5.5)

·

−Bv ( ) − Nv ( ) + cv ( ) = 0. Further, we have the kinematic boundary conditions w (0) = v (0) = 0 w ( ) = v· ( ) = − sin θ.

(3.5.6)

The differential equations and the boundary conditions are satisfied by w1 = a sin

πx ,

v1 = a sin

πy ,

N1 π2 = 2, B

(3.5.7)

and we have cw1 ( ) = cv1 ( ) = 0 a=

(3.5.8)

sin θ. π

(3.5.9)

To investigate the stability of this neutral equilibrium state, for small displacements from this configuration we write w (x) = a sin

πx + w (x) ,

v (x) = a sin

πy + v (y) ,

(3.5.10)

where w and v are orthogonal to w1 and v1 , respectively. We now consider (see 2.4.11) Min P2 [v, w; N1 , c → ∞] + (N − N1 ) P11 [v1 , w1 , v, w; N1 , c → ∞] w.r.t. v, w + P21 [v1 , w1 , v, w; N1 , c → ∞] . (3.5.11)

74

Applications

Now P11

·

[v1 , w1, v, w ; N1 , c → ∞] = −

w1 w

dx −

0

v·1 v· dy = 0

(3.5.12)

0

due to the orthogonality conditions. From (3.5.2) we obtain P2 [v, w· ; N1 , c] =

1 1 ··2 1 ·2 1 Bw2 − Nw2 dx + Bv − Nv dy 2 2 2 2 0 0 (3.5.13) 1 2 1 2 + cw ( ) + cv ( ) . 2 2

From (3.4.7), we obtain 1 P3 [v1 + v, w1 + w; N1 , c] = c [w1 ( ) + w ( )] 2

·2 v1 + 2v·1 v· + v2 dy

0

1 − c [v1 ( ) + v ( )] 2

2 w1 + 2w1 w + w2 dx

0

1 = cw1 ( ) 2

v·2 1

1 dy − cv1 ( ) 2

0

+ cw1 ( )

0

v·1 v· dy − cv1 ( )

0

=

1 cw ( ) 2

w2 1 dx (3.5.14) w1 w dx + · · ·

0

1 v·2 1 dy − cv ( ) 2

0

w2 1 dx + · · ·

0

= P21 [v1 , w1 , v, w· ; N1 , c] + · · · . Let v, w be the minimizing solution. The minimizing solution now must satisfy the condition of 3.4.18 P11 [v, w, η, ζ; N1 , c] + P21 [v1 , w1 , η, ζ; N1 , c] = 0,

(3.5.15)

i.e.,

(Bw ζ − N1 w ζ ) dx + 0

(Bv·· η·· − N1 v· η· ) dy

0

+ cw ( ) ζ( ) + cv ( ) η( )   1  + w2 cζ( ) v·2 1 dy − cη( ) 1 dx = 0. 2 0

0

(3.5.16)


75

From integration by parts, we obtain

Bw ζ 0

− Bw ζ 0 − N1 w ζ 0 +

+

Bv·· η· 0

(Bw + Nw ) ζdx

0

− Bv ζ 0 − N1 v· η 0 + ···

(Bv··· + Nv·· ) ηdy

0

+ cw ( ) ζ( ) + cv ( ) η( )     1   1 c w2  +  c v·2 1 dy ζ( ) − 1 dx η( ) = 0, 2 2 0

(3.5.17)

0

which yields the differential equations Bw + N1 w = 0,

Bv···· + N1 v·· = 0.

(3.5.18)

For the evaluation of the stock terms, we must again note that η and ζ are kinematically admissible functions, and that they must satisfy the condition (3.5.6) at x = y = , i.e., η· ( ) = ζ ( ) .

(3.5.19)

Taking into account this condition, we find Bw ( ) + Bv·· ( ) = 0

(3.5.20)

Bw (0) = Bv·· (0) = 0 1 − Bw ( ) − N1 w ( ) + cw ( ) + c v·2 1 dy = 0 2

(3.5.21)

0

1 − Bv··· ( ) − N1 v· ( ) + cv ( ) − c 2

w2 1 dx = 0.

0

In addition to these dynamic boundary conditions, we have the kinematic boundary conditions w (0) = v (0) = 0 w ( ) = v· ( ) .

(3.5.22)

With the exception of the third and fourth equations of (3.5.21), these differential equations and boundary conditions are the same as those for the determination of v1 and w1 . However, notice that due to the fact that the third and fourth equations of (3.5.21) are inhomogeneous, a solution v = w = 0 is not possible. The inho2 mogeneous terms are ± 14 a2 π c . To satisfy these inhomogeneous conditions, we try a solution of the form πy v (y) = A∗2 y + A∗∗ 2 sin (3.5.23) πx ∗ ∗∗ . w (x) = A1 x + A1 sin

76

Applications

These solutions must be orthogonal to the solution v1 , w1 . The orthogonality condition corresponding to (3.5.12) is

v1 v dy +

0

w1 w dx = 0,

(3.5.24)

0

which yields (A∗1 + A∗2 )

1 2 + (A∗∗ + A∗∗ 2 ) = 0. π2 2 1

(3.5.25)

The boundary condition (3.5.20), the first two conditions of (3.5.21), and the first two conditions of (3.5.22) are automatically satisfied by (3.5.23). From the third equation of (3.5.21), we obtain π2 π 1 2 π2 c ∗ −B 2 + N1 A∗∗ , 1 + (−N1 + c ) A1 = − a 4 where the first term vanishes due to (3.5.7), so that A∗1 =

−a2 π2 . 4 2 (1 − N1 /c )

(3.5.26)

Similarly, we obtain from the fourth equation of (3.5.21), A∗2 =

a2 π2 . 4 2 (1 − N1 /c )

(3.5.27)

Because A∗1 + A∗2 = 0, we obtain from the orthogonality condition (3.5.25), ∗∗ A∗∗ 1 + A2 = 0.

(3.5.28)

From the third condition of (3.5.22), we obtain ∗∗ A∗1 − A∗2 = (A∗∗ 1 − A2 )

π ,

(3.5.29)

from which ∗∗ A∗∗ 1 = −A2 = −

a2 π . 4 (1 − N1 /c )

(3.5.30)

According to (2.3.54), stability is now governed by the sign of 1 P4 [v1 , w1 ; N1 , c → ∞] + P21 [v1 , w1 , v, w; N1 , c → ∞] . 2

(3.5.31)

From (3.5.3), we obtain P4 [v1 , w1 ; N1 , c] = −

π6 1 1 π4 4 Ba4 6 + a c, 16 16 2

(3.5.32)

and from (3.5.14), P21 [v1 , w1 , v, w; N1 , c] = −

π4 a4 c , 16 3 (1 − N1 /c )

(3.5.33)


77

λ 1

Wmax Figure 3.5.2

so that 1 π6 Ba4 lim P4 [v1 , w1 ; N1 , c] + P21 [v1 , w1 , v.w; N1 , c] = − < 0. c→∞ 2 16 6

(3.5.34)

The equilibrium state is thus unstable. The behavior is shown in Figure 3.5.2. At first sight, this result may seem surprising as both bars behave like the Euler bar. However, the result may be explained as shown in Figure 3.5.3. Replace the forces N by two statically equivalent forces F (F > N). Let δ be the horizontal and vertical components of the corner point where the loads are applied; then δ F ≈ N + N. (3.5.35) Let the deflection of the horizontal bar be v (y) = a sin

πy ;

then δ is given by δ≈

1 ·2 1 π2 2 v dy = a 2 2 2

0

cos2 0

δ N N

δ

F

F

Figure 3.5.3

π2 a2 πx dx = , 4 2

(3.5.36)

78

Applications

so that

π2 a2 . F =N 1+ 4 2

(3.5.37)

For the post-buckling behavior of the Euler bar, we had π2 a2 F = F1 1 + , 8 2 where F1 is the critical load. From (3.5.37) and (3.5.38), we find π2 a2 . N = F1 1 − 8 2

(3.5.38)

(3.5.39)

This means that the equilibrium is already unstable for values of N smaller than the critical load F1 . 3.6 Bending and torsion of thin-walled cross sections under compression We consider a beam with a thin-walled cross section, loaded by forces that cause bending and torsion, as shown in Figure 3.6.1. Let (y0 , z0 ) be the center of shear. The displacements at a point (x, y, z) are given by u = −yv0 (x) − zw0 (x) + α (x) ψ0 (y, z) 1 v = v0 (x) − (z − z0 ) α (x) + υ y2 − z2 v0 (x) + υyzw0 (x) − υ0 (y, z) α (x) 2 1 w = w0 (x) + (y − y0 ) α (x) + υyzv0 (x) + υ z2 − y2 w0 (x) − υ0 (y, z) α (x) . 2 (3.6.1) Here α (x) is the specific angle of torsion, ψ0 (y, z) and ϕ (y, z) are the real and the imaginary parts of an analytic function F (y + iz) = ψ0 (y, z) + iϕ (y, z), and 0 (y, z) and 0 (y, z) are defined by ∂0 ∂0 = = ψ0 (y, z) , ∂y ∂z

−

∂0 ∂0 = = ϕ (y, z) . ∂z ∂y

z t

*( y0 , z0 ) x

y

Figure 3.6.1

(3.6.2)

3.6 Bending and torsion of thin-walled cross sections under compression

79

The approximations (3.6.1) are valid for t2 /b2 1 and b2 /L2 1, where t is the wall thickness of the cross section, L is a characteristic wavelength, and b is a characteristic dimension of the cross section. The center of shear is determined by zψ0 (y, z) dA = yψ0 (y, z) dA = 0. (3.6.3) A

A

Further, the resultant of the normal stresses in the x-direction must vanish, which yields ψ0 (y, z) dA = 0. (3.6.4) A

The expressions (3.6.1) further guarantee the vanishing of the stresses σy , σz, and σyz. In our calculations, we shall make use of the estimates α (x) = O(α/L), α (x) = O (α/L2 ), and so on ∂ () ∂ () |y − y0 | , |z − z0 | = O (b) , , = O (() /b) ∂y ∂z 2 |ψ0 | , |ϕ| = O b , |0 | , |ψ0 | = O(b3 ),

(3.6.5)

where b is the width of the cross section. To simplify the calculations, we shall use the principal axes of inertia, i.e., yz dA = 0. (3.6.6) A

We now obtain the following expression for the potential energy: 1 ∂u P [u] = σx + τxy γxy + τxzγxz dV 2 ∂x V 2 1 ∂ψ0 2 − (z − z0 ) α + · · · = (3.6.7) E (−yv0 − zw0 + ψ0 α ) + G α 2 ∂y 2 . ∂ψ0 + (y − y0 ) α + · · · +G α dV, ∂z where the dots represent terms containing third-order derivatives with respect to x. With the torsional stiffness St defined by 2 2 . ∂ψ0 ∂ψ0 St = − (z − z0 ) + + (y − y0 ) G dA (3.6.8) ∂y ∂z A

and a warping constant defined by

=

ψ20 (y, z) dA A

(3.6.9)

80

Applications

and using the relations (3.6.3), (3.6.4), and (3.6.6) we may rewrite the potential energy expression to yield P [u] =

1 1 2 1 1 2 2 2 dx + EIzv0 + EIy w0 + Eα St α dx. 2 2 2 2

0

(3.6.10)

0

Expression (3.6.10) is valid for b /L 1. In addition to this energy, we have the energy of the prestresses, 2 1 1 u,x + v2,x + w2,x dV. (3.6.11) Sij uh,i uh,j dV = − σ 2 2 2

2

V

V

The term σu2,x dV may be neglected compared to the term 12 Eu2,x dV in (3.6.7). Hence, the additional energy is given by / 0 1 2 2 − σ [v0 − (z − z0 ) α + · · ·] + [w0 + (y − y0 ) α + · · ·] dV 2 1 2

V

=−

1 2 1 1 2 Nv0 + Nw0 dx − σ (y − y0 )2 + (z − z0 )2 dAα2 dx (3.6.12) 2 2 2 0 A

0

−

(Nz0 α v0 − Ny0 α w0 ) dx

0

with

(y − y0 )2 + (z − z0 )2 dA = Iz0 + Iy0 = i 2z0 + i2y0 A = i20 A.

(3.6.13)

A

Finally, we obtain for the total potential energy P [u, N] = 0

1 1 1 1 2 1 1 2 2 2 2 2 EIzv2 0 + EI y v0 + Eα − Nv0 − Nw0 − Ni 0 α 2 2 2 2 2 2 (3.6.14) 1 2 − Nz0 α v0 + Ny0 α w0 + St α dx = P2 [u; N] . 2

Expression (3.6.14) shows that interaction between bending and torsion only occurs in the terms with Nz0 α v0 and Ny0 α w0 . This condition implies that when the shear center coincides with the center of gravity, there is no interaction. This is the case when there are two axes of symmetry. When there is only one axis of symmetry, one of the coupling terms vanishes, i.e., we then have one axis of buckling mode without interaction and one with interaction. We shall now restrict our discussion further to the case in which the values of Iy , Iz, , St , and N are constant. The equations of neutral equilibrium follow from P11 [v0 , w0 , α, η, ζ, χ; N] = 0

[EIzv0 η + EIzw0 0 ζ + Eα χ − Nv0 η − Nw0 ζ − Ni 20 α χ − Nz0 (α η + χ v0 ) (3.6.15) + Ny0 (α ζ + χ w0 ) + St α χ ] dx = 0.


81

By integration by parts, we obtain        v w EIz v0 η | 0 − v 0 η|0 + 0 ηdx + EI y w0 ζ |0 − w0 ζ|0 + 0 ζdx 0



+ E α χ | 0 − α χ| 0 +

0

 α χ dx

0

− Nv0 η | 0

Nv0

+ 0

− Ni 20 α χ| 0 +

ηdx −

Nw0 ζ| 0

+

Nw0 ζdx

0

(3.6.16)

Ni 20 α χ dx − Nz0 (α η + χv0 ) | 0

0

+ Nz0

(α η + v0 χ) dx + Ny0 (α ζ + w0 χ) | 0

0

− Ny0

(α ζ +

w0 χ) dx

+ St α

χ| 0

−

0

St α χ dx = 0.

0

We now consider a beam that is supported at x = 0 and x = so that v0 (0) = w0 (0) = α (0) = v0 ( ) = w0 ( ) = α ( ) = 0.

(3.6.17)

For a clamped-clamped beam, we have η = ζ = χ = 0

at

x = 0 and x =

η = ζ= χ = 0

at

x = 0 and x = ,

(3.6.18)

and for a simply supported beam, η = ζ = χ = 0 at

x = 0 and x = .

(3.6.19)

The differential equations are v 0 + Nv0 + Nz0 α = 0 w 0 + Nw0 − Ny0 α = 0

α

+

Ni 20 α

+

Nz0 v0

−

(3.6.20)

Ny0 w0

− St α = 0.

For the clamped-clamped beam, the kinematic boundary conditions are u (0) = u ( ) = 0,

v0 = w0 = α = 0

at x = 0 and x = .

(3.6.21)

For the simply supported beam, we have the dynamic boundary conditions EIzv0 = EIy w0 = Eα = 0

at x = 0 and x = ,

(3.6.22)

i.e., the vanishing of the bending moments Mz and My and of the bimoment.† †

The bimoment is not a real moment; its dimension is FL2 .

82

Applications

We shall now consider the case of a simply supported beam in more detail. Because the differential equations and the boundary conditions contain only even derivatives, we try a solution of the form v0 = B sin

πx ,

w0 = C sin

πx ,

α = A sin

πx † .

(3.6.23)

These expressions satisfy the boundary conditions (3.6.21) and (3.6.22). Substitution of these expressions into (3.6.20) yields (Ny − N) B −Nz0 B+

−Nz0 A =0 +Ny0 A =0 (Nz − N) C Ny0 C +i 20 (Nt − N) A = 0,

(3.6.24)

where we have introduced the notation π2 EIz = Ny , 2

π2 EIy = Nz, 2

1 i 20

π2 E + S2 2

= Nt ,

(3.6.25)

where Nt is the lowest buckling load due to torsion. The condition for a non-trivial solution is the vanishing of the determinant, i.e., i 20 (Ny − N) (Nz − N) (Nt − N) + y20 N2 (N − Ny ) + z20 N2 (N − Nz) = 0. (3.6.26) When y0 and z0 are zero, the roots of this equation are N = (Ny , Nz, Nt ). Now let at least y0 or z0 be nonzero. Then for N = min (Ny , Nz, Nt ), the left-hand member of (3.6.26) is negative. The left-hand member can only become positive for N < min (Ny , Nz, Nt ). This condition means that the smallest positive root of this equation is smaller than Ny , Nz , and Nt , i.e., the actual buckling load is smaller than the buckling load in bending and smaller than the buckling load in torsion. Let us now consider the special case that the y-axis is an axis of symmetry, i.e., z0 = 0. Then we have N = Ny

or

i 20 (Nz − N) (Nt − N) − y20 N2 = 0,

(3.6.27)

where for the first case there is no interaction, but there is interaction in the second case. Let us now investigate this second case more closely. Equation (3.6.27) may be rewritten in the form % & y20 2 N 1 − 2 − (Nz + Nt ) N + NzNt = 0. (3.6.28) i0 This equation is symmetric in Nt and Nz; hence without loss of generality we may assume Nz/Nt = η < 1. †

(3.6.29)

We are only interested in the lowest buckling mode k = 1. In general, one writes v0 = 1∞ kπx k=1 Bk sin , and so on.


Introducing the notation N/Nz = λ, we may rewrite the (3.6.28) to yield % & y20 1 1 2 λ + = 0. λ 1− 2 − 1+ η η i0

83

(3.6.30)

The smallest root is given by 1 2 2 2 (1 + 1/η) − (1 − 1/η) + 4ηy0 /i 0 . λ= 2 1 − y20 /i 20

(3.6.31)

Let us now consider the two limiting cases η→ 0

and

η → 1.

(3.6.32)

For η → 0, we may write 1 1 1 + η − 1 − η + 2ηy20 /i 20 = 1, λ = lim η→0 2 1 − y2 /i 2 η 0 0

(3.6.33)

as expected. For η → 1, we find λ=

1 < 1. 1 + y0 /i 0

(3.6.34)

1 < λ < 1. 2

(3.6.35)

Because y0 /i 0 < 1, this means

This result shows that under combined torsion and bending, the buckling load may be considerably smaller than Nz or Nt , especially when y0 /i 0 → 1. An example of such a structure is the thin-walled circular tube with an open cross section (see Figure 3.6.2). In this case, we have 2 3 2 2 3 St = πGRt , = π π − 4 R5 t, y0 /i 0 = √ . (3.6.36) 3 3 5 For η = 1, we obtain λ = 0.528, i.e., a reduction of 47% of the buckling load. We emphasize here once again that the theory is only valid for b2 /L2 1. t R

*

2R

Figure 3.6.2

84

Applications

3.7 Infinite plate between flat smooth stamps We consider a homogeneous isotropic elastic plate of thickness h, length , and infinite in width, compressed between flat smooth rigid stamps. The material is such that it can have finite deformations in the elastic range (a neo-Hookean material), and its elastic energy per unit volume of the undeformed body is given by W=

1 2 λ1 + λ22 + λ23 − 3 , 2

(3.7.1)

where λ1 , λ2 , and λ3 are the extension ratios in the fundamental state I, i.e., λ1 = 1 + ε1 ,

λ2 = 1 + ε2 ,

λ3 = 1 + ε3 ,

(3.7.2)

where ε1 (i = 1, 2, 3) are the principal strains. In (3.7.1), the elastic constant is taken to be unity. We shall assume that the material is incompressible, i.e., λ1 λ2 λ3 = 1,

(3.7.3)

and we consider the case of plane strain, i.e., λ3 = 1, so that λ1 λ2 = 1.

(3.7.4)

We now consider the following states (see Figure 3.7.1), where σ1 h is the force per unit width of the plate. Let (x, y) be the coordinates of a material point in the undeformed state. The coordinates of this point in the fundamental state are denoted by (x, y), where y = λ2 y. In the adjacent state, the coordinates are x, y , where x = λ1 x,

x = λ1 x + u,

(3.7.5)

y = λ2 y + v.

(3.7.6)

The square of the length of an infinitely small material element in the undeformed state is given by (ds)2 = (dx)2 + (dy)2 .

(3.7.7)

σ1 h

h

σ1 h

λ 2h

l

λ1l

II

I

x σ1 h undeformed state

fundamental state Figure 3.7.1

σ1 h adjacent state

3.7 Infinite plate between flat smooth stamps

85

In the fundamental state I, it is given by (ds)2 = (dx)2 + (dy)2 = λ21 (dx)2 + λ22 (dy)2 ,

(3.7.8)

and in the adjacent state II, it is given by 2 2 2 ds = dx + dy = [(λ1 + u,x )2 ] (dx)2 + [(λ2 + v,x )2 u2,y ] (dy)2 + 2 [(λ1 + u,x ) u,y + (λ2 + v,y ) v,y ] dx dy.

(3.7.9)

The components of the metric tensor in the adjacent state are thus gxx = (λ1 + u,x )2 + v2,x gyy = (λ2 + v,y )2 + u2,y

(3.7.10)

gxy = (λ1 + u,x ) u,y + (λ2 + v,y ) v,x . The principal values of the metric tensor follow from g − g gxy xx = 0, gxy gyy − g

(3.7.11)

and in the corresponding principal directions we have g11 = λ12 ,

g22 = λ22 ,

(3.7.12)

where λ1 and λ2 are the extension ratios in those directions. The first invariant of the metric tensor is g11 + g22 = λ12 + λ22 = gxx + gyy .

(3.7.13)

The elastic energy density in the adjacent state II is now given by 1 gxx + gyy − 2 2 1 = [(λ1 + u,x )2 + v2,x + (λ2 + v,y )2 + u2,y − 2]. 2

WII =

(3.7.14)

In the fundamental state I, the elastic energy density is WI =

1 2 1 λ + λ22 − 2 , g + gyy − 2 = 2 xx 2 1

(3.7.15)

so that the increment of the elastic energy density in passing from the fundamental state I to the adjacent state II is WII − WI = λ1 u,x + λ2 v,y +

1 2 u,x + u2,y + v2,x + u2,y . 2

(3.7.16)

In addition to this energy expression, we shall need an explicit expression for the incompressibility condition, which with the aid of the second invariant of the metric tensor can be written as 1 = λ1 λ2 = λ1 λ2 = g11 g22 = gxx gyy − gxy . (3.7.17)

86

Applications

However, it is slightly simpler to derive the incompressibility condition from dv ∂x/∂x ∂x/∂y λ1 + u,x u,y = = λ1 λ2 , = λ2 + v,y dv ∂y/∂x ∂y/∂y v,x from which follows λ2 u,x + λ1 v,y + u,x v,y − u,y v,x = 0.

(3.7.18)

This incompressibility condition is a nonlinear side condition in our problem. The total potential energy per unit width of the plate is now given by h/2 1 2 2 2 2 P [u] = λ1 u,x + λ2 v,y + u + u,y + v,x + v,y dx dy 2 ,x 0 −h/2

(3.7.19)

h/2 + σ1

u,x dx dy, 0 −h/2

where for the last term we have rewritten the term σ1 h u,x dx as a surface integral. The stability condition is now that P [u] > 0 under the nonlinear side condition (3.7.18). Using this side condition to eliminate v,y , in the linear terms in (3.7.19), we find h/2 P [u] = 0 −h/2

λ22 1 2 λ1 − u,x + u2,y + v2,x + v2,y + σ1 u,x + λ1 2

λ2 + (u,y v,x − u,x v,y ) dx dy. λ1

(3.7.20)

The stability condition is still P [u] > 0 under the side condition (3.7.18). A necessary condition that P [u] > 0 is the vanishing of the linear term in P [u], i.e., σ1 = −λ1 +

λ22 1 = −λ1 + 3 , λ1 λ1

(3.7.21)

which is the equilibrium condition in the fundamental state. The discussion of the equilibrium in the fundamental state is now reduced to the discussion of h/2 P [u] = 0 −h/2

λ2 1 2 2 2 2 u + u,y + v,x + v,y + (u,y v,x − u,x v,y ) dx dy (3.7.22) 2 ,x λ1

under the nonlinear side condition (3.7.18). Notice that P [u] is a homogeneous quadratic functional, i.e., P [u] = P2 [u]. A necessary condition for stability is that P2 [u] ≥ 0 for very small displacements from the fundamental state, which means that we may linearize the side condition to yield λ2 u,x + λ1 v,y = 0.

(3.7.23)


87

A sufficient condition for stability is that P2 [u] > 0 under the linear side condition (3.7.23). Introducing a Lagrangian multiplier p (x, y) to take into account the side condition, we now consider the functional h/2 P [u] +

p (x, y) (λ2 u,x + λ1 v,y ) dx dy.

(3.7.24)

0 −h/2

A necessary condition for stability is now h/2 [u,x η,x + u,y η,y + v,x ζ,x + v,y ζ,y 0 −h/2

λ2 (3.7.25) (u,y ζ,x + v,x η,y − u,x ζ,y − v,y η,x ) λ1 + (λ2 u,x + λ1 v,y ) δp + p (x, y) (λ2 η,x + λ1 ζ,y )] dx dy = 0,

+

for all kinematically admissible displacements η, ζ. By integration by parts, we obtain h/2

(u,x η)0 dy +

−h/2

h/2 (u,y η)−h/2 dx −

+

(v,x ζ)0 dy +

−h/2



+

λ2   λ1

(u,xx + u,yy ) ηdx dy 0 −h/2

0

h/2

h/2

h/2 (v,y ζ)−h/2 dx −

(u,y ζ)0 dy −

−h/2

h/2 + λ2 −h/2

h/2 (u,x ζ)−h/2 dx +

0

(pη) dy − λ2

(v,xx + v,yy ) ζdx dy 0 −h/2

0

h/2

h/2

h/2 (v,x η)−h/2 dx −

p ,x ηdx dy + λ1

0

0 −h/2

  (v,y η)0 dy

−h/2

0

h/2

h/2

h/2 (pζ)−h/2 dx − λ1

0

h/2 p ,y ζdx dy 0 −h/2

h/2 +

(λ2 u,x + λ1 v,y )δp dx dy = 0.

(3.7.26)

0 −h/2

This yields the differential equations u,xx + u,yy + λ2 p ,x = 0 v,xx + v,yy + λ1 p ,y = 0

(3.7.27)

λ2 u,x + λ1 v,y = 0. At x = 0 and x = , we have the kinematic boundary conditions u,y (0) = u,y ( ) = 0.

(3.7.28)

88

Applications

and at x = 0, u (0) = 0,

(3.7.29)

which implies η,y (0) = η,y ( ) = 0,

η(0) = 0.

(3.7.30)

Taking into account these conditions, we find for the dynamic boundary conditions h/2 λ2 u,x − v,y + λ2 p dy = 0 at x = λ1

(3.7.31)

−h/2

v,x +

λ2 u,y = 0 λ1

(3.7.32)

λ2 v,x = 0 at y = ±h/2 λ1

(3.7.33)

λ2 u,x + λ1 p = 0 at y = ±h/2. λ1

(3.7.34)

u,y +

v,y −

at x = 0 and x =

The condition (3.7.32) may be simplified to v,x = 0 at x = 0 and x =

(3.7.35)

due to (3.7.28). The system (3.7.27) may easily be shown to be of fourth order, and we have two boundary conditions on each part of the boundary. We shall now restrict ourselves to the lowest buckling load, and taking into account the boundary conditions (3.7.28) and (3.7.29), we can try a solution for u (x, y) of the form u (x, y) = U (y) sin

πx .

(3.7.36)

From the first of the equations of (3.7.27), it then follows that p (x, y) is of the form p (x, y) = P (y) cos

πx ,

(3.7.37)

and from the second or third of the equations of (3.7.27), we find v (x, y) = V (y) cos

πx .

(3.7.38)

Substitution of these expressions into (3.7.27) yields π2 π U − λ2 P = 0 2 π2 V¨ − 2 V + λ1 P˙ = 0 π λ1 V˙ + λ2 U = 0,

U¨ −

(3.7.39)


89

where ()· = d ( ) /dy. This is a set of three homogeneous linear ordinary differential equations, which can be solved by the substitutions U (y) = Aeµy ,

V (y) = Beµy ,

P (y) = Ceµy ,

(3.7.40)

which yield π2 2 µ − 2 A π2 2 µ − 2 B π λ2 A + λ1 µB

π − λ2 C = 0 + λ1 µC = 0

(3.7.41)

= 0.

The condition for a nontrivial solution is the vanishing of the determinant of the coefficients, which yields 2 π2 2 2 2 2π µ − 2 −λ1 µ + λ2 2 = 0. (3.7.42) The eigenvalues are thus π µ1,2 = ± ,

π µ3,4 = ±λ ,

(3.7.43)

where λ = λ2 /λ1 . Because we have two pairs of roots with opposite signs, we can also express the solutions as hyperbolic functions instead of exponential functions. Hence, we can write U (y) =

2

Ai1 sinh µi y +

i=1

V (y) =

2

2

Ai2 cosh µi y

i=1

Bi1 cosh µi y +

i=1

P (y) =

2

2

Bi2 sinh µi y

(3.7.44)

i=1

Ci1 sinh µi y +

i=1

2

Ci2 cosh µi y,

i=1

where µ1 = π/ and µ2 = λπ/ . The solutions with the subscript 1 yield anti-symmetric deformations, and those with the subscript 2 symmetric deformations. Hence, we may split the problem into a symmetric and an anti-symmetric case.

anti-symmetric deformation Figure 3.7.2

symmetric deformation

90

Applications

In the following, we shall discuss the case of anti-symmetric deformation in detail because calculations show that this type of buckling behavior occurs prior to that of symmetric deformation. The last type of deformation first occurs after a reduction of the length by about 50%. The calculations involved are entirely similar to those of the anti-symmetric case. In the anti-symmetric case, we have πy πy + A2 sinh λ πy πy V (y) = B1 cosh + B2 cosh λ πy πy + C2 sinh λ . P (y) = C1 sinh U (y) = A1 sinh

(3.7.45)

For µ = π/ , we find from (3.7.41) C1 = 0,

B1 = −λA1 ,

(3.7.46)

and for µ = λπ/ , we find B2 = −A2 ,

C2 =

λ2 − 1 π A2 . λ2

(3.7.47)

The boundary conditions (3.7.33) and (3.7.34), with the aid of (3.7.36) to (3.7.38), can be rewritten to yield π U˙ − λ V = 0,

π V˙ − λ U + λ1 P = 0,

(3.7.48)

for y = ±h/2. Substitution of (3.7.45) into (3.7.48) and using (3.7.46) and (3.7.47) yields πh πh l 1 + λ2 cosh A1 + 2λ cosh λ A2 = 0 2 2 πh πh 1 A1 − λ + sinh λ A2 = 0. −2λ sinh 2 λ 2

(3.7.49)

Notice that λ = λ2 /λ1 is the single significant load parameter. Its value follows from the condition for a non-trivial solution of (3.7.49), which leads to 2 2 λ + 1 tanh λθ = 4λ3 tanh θ,

(3.7.50)

where we have introduced the geometric parameter θ=

πh . 2

(3.7.51)

Apart from the solution λ = 1, which corresponds to the fundamental state only, (3.7.50) has one solution, λ > 1. Let us consider two limiting cases θ → 0 and θ → ∞.


91

For θ 1, we can use series expansions for the hyperbolic functions. Then (3.7.50) can be rewritten to yield (λ2 − 1)2 =

1 2 2 2 λ θ [(λ + 1)2 − 4] + O(θ3 ). 3

(3.7.52)

As for θ = 0, λ2 = 1 is a root of this equation; we try a solution of the form λ2 = 1 + ε,

0 < ε 1,

(3.7.53)

which yields ε=

4 2 θ + O(εθ2 ), 3

(3.7.54)

so that λ2 2 = 1 + θ2 . λ1 3

(3.7.55)

Using the incompressibility condition λ1 λ2 = 1, we find 1 π2 h2 λ2 = 1 + θ 2 = 1 + 3 12 2 1 π2 h2 . λ1 = 1 − θ 2 = 1 − 3 12 2

(3.7.56)

This result for the thin plate (h/ 1) is in complete agreement with the result for the Euler column, and as it is derived from a two-dimensional theory, it justifies the latter result. For θ → ∞, (3.7.50) reduces to (λ2 + 1)2 = 4λ3 ,

(3.7.57)

from which follows λ = 3.383 (and of course, λ = 1). Hence, it follows that 1 λ1 = √ 0.55, 3.383

(3.7.58)

which means a reduction in length of 45%. In this case, the deformations are concentrated near y = ±h/2. In the case of symmetric deformation, the bifurcation equation becomes 4λ3 tanh λθ = (λ2 + 1)2 tanh θ,

(3.7.59)

which has a root λ ≥ 3.383. Hence, it follows that bifurcation in an anti-symmetric mode (λ ≤ 3.383) will occur prior to the occurrence of bifurcation in a symmetric mode. We now continue with the discussion of the anti-symmetric case. From the first of the equations of (3.7.49), we obtain A2 (1 + λ2 ) cosh θ =− . A1 2λ cosh λθ

(3.7.60)

92

Applications

Using (3.7.36) to (3.7.38), (3.7.45) to (3.7.47), and (3.7.60), we find πy 1 + λ2 cosh θ πx πy sin u (x, y) = A1 sinh − sinh 2λ cosh λθ 2 πy 1 + λ cosh θ πy πx v (x, y) = A1 −λ cosh + cosh λ cos 2λ cosh λθ πy πx 1 − λ4 π cosh θ sinh λ cos . p (x, y) = A1 2λλ2 cosh λθ

(3.7.61)

We now choose the constant A1 so that v (0, 0) = 1, which yields A1 =

−2λ−1 cosh λθ . 2 cosh λθ − (λ−2 + 1) cosh θ

(3.7.62)

Further, we introduce the shorthand 2 cosh λθ − λ−2 + 1 cosh θ ≡ N.

(3.7.63)

The expressions (3.7.61) can now be rewritten in the form cosh λθ πy πx λπy −2 sinh + (λ−2 + 1) cosh θ sinh λ πy πx λπy 2 cosh λθ cosh v1 (x, y) = N−1 cos − (λ−2 + 1) cosh θ cosh πx π λπy p 1 (x, y) = (λ2 N)−1 (λ2 − λ−2 ) cosh θ sinh cos . u1 (x, y) = N−1 sin

(3.7.64)

Having determined the displacement fields for the critical case of neutral equilibrium, we now proceed with the discussion of its stability. Therefore, we consider small but finite displacements from the equilibrium configuration, u (x, y) = au1 (x, y) + u (x, y) v (x, y) = av1 (x, y) + v (x, y) ,

(3.7.65)

where u and v are orthogonal (in some suitable sense) to the buckling mode. Stability is now governed by the behavior of the functional (3.7.22) under the nonlinear side condition (3.7.18). To make full use of the properties of the functional in the critical case of neutral equilibrium, we consider the functional (3.7.22). Substituting (3.7.65) into this functional, we find h/2 PII − PI = 0 −h/2

1 2 2 a u1,x +u21,y +v21,x +v21,y +a (u1,x u,x +u1,y u,y + v1,x v,x +v1,y v,y ) 2

+

1 2 u,x + u2,y + v2,x + v2,y + λa2 (u1,y v1,x − u1,x v1,y ) 2

(3.7.66) + λa (u1,y v,x + u,y v1,x − u1,x v,y − u,x v1,y ) +λ (u,y v,x − u,x v,y ) dx dy,


93

where the terms involving a2 vanish because u1 and v1 satisfy the equations for the critical case of neutral equilibrium. Using (3.7.25) with η = u, ζ = v, and u = u1 ,v = v1 , p = p 1 , we can rewrite the bilinear terms in (3.7.66) to yield h/2 ap 1 (λ2 u,x + λ1 v,y ) dx dy,

−

(3.7.67)

0 −h/2

so that our final result for the functional becomes h/2 PII − PI = 0 −h/2

1 2 u,x + u2,y + v2,x + v2,y + λ (u,y v,x − u,x v,y ) 2

− ap 1 (λ2 u,x + λ1 v,y ) dx dy.

(3.7.68)

Substitution of (3.7.65) into the nonlinear side condition (3.7.18) yields λ2 u,x + λ1 v,y + a2 (u1,x v1,y − u1,y v1,x ) + a (u1,x v,y + u,x v1,x − u1,y v,x − u,y v1,x ) + u,x v,y − u,y v,x = 0.

(3.7.69)

We now use this nonlinear condition to rewrite the term involving p1 in (3.7.68), which yields h/2 −

ap 1 (λ2 u,x + λ1 v,y ) dx dy 0 −h/2

h/2 =

3 a p 1 (u1,x v1,y − u1,y v1,x ) + a2 p 1 (u1,x v,y + u,x v1,y − u1,y v,x − u,y v1,x )

0 −h/2

+ ap 1 (u,x v,y − u,y v,x )] dx dy,

(3.7.70)

where the first term on the right-hand side vanishes because

πx πx cos dx = 0, sin

2

0

cos3

πx dx = 0.

0

The functional (3.7.68) can now be rewritten in the form h/2 ! PII − PI = 0 −h/2

1 2 u,x + u2,y + v2,x + v2,y + λ (u,y v,x − u,x v,y ) 2

+ a2 p 1 (u1,x v,y + u,x v1,y − u1,y v,x − u,y v1,x ) + ap 1 (u,x v,y − u,y v,x ) + p[λ2 u,x + λ1 v,y + a2 (u1,x v1,y − u1,y v1,x ) + a (u1,x v,y + u,x v1,y − u1,y v,x − u,y v1,x ) + u,x v,y − u,y v,x ] " + µ (u1 u + v1 v) dx dy,

(3.7.71)

94

Applications

where we have taken into account the nonlinear side condition with the Lagrangian multiplier p (x, y) and the orthogonality condition h/2 (u1 u + v1 v) dx dy = 0

(3.7.72)

0 −h/2

with the multiplier µ.† We now minimize this functional with respect to u, v, p, and µ for a fixed value of the amplitude a. Variation with respect to u yields h/2 {u,x δu,x + u,y δu,y + λ (v,x δu,y − v,y δu,x ) 0 −h/2

+ a2 p 1 (v1,y δu,x − v1,x δu,y ) + ap 1 (v,y δu,x − v,x δu,y ) + p [λ2 δu,x + a (v1,y δu,x − v1,x δu,y ) + v,y δu,x − v,x δu,y ]

(3.7.73)

+ µu1 δu} dx dy = 0. By integration by parts, we obtain h/2 [u,xx + u,yy + a2 (p 1 v1,y ),x + a2 (p 1 v1,y ),x − a2 (p 1 v1,x ),y

− 0 −h/2

+ a (p 1 v,y ),x − a (p 1 v,x ),y + λ2 p ,x + a (pv1,x ),x − a (pv1,x ),y + (pv,y ),x − (pv,x ),y + µu1 ] δu dx dy +

(3.7.74)

h/2 (u,y + λv,x − a2 p 1 v1,x − ap 1 v,x − apv1,x − pv,x ) δu −h/2 dx

0

h/2 +

(u,x − λv,y + a2 p 1 v1,y + ap 1 v,y + λ2 p + apv1,y + pv,y ) δu 0 dy.

−h/2

Remembering the geometric boundary conditions u = 0 at x = 0 and u,y = 0 at x = , we obtain u,xx + u,yy + a2 (p 1,x v1,y − p 1,y v1,x ) + a (p 1,x v,y − p 1,y v,y ) + λ2 p ,x + a(p ,x v1,y − p ,y v1,y ) + p ,x v,y − p ,y v,x + µu1 = 0,

(3.7.75)

and the dynamic boundary conditions u,y + λv,x − p 1 (a2 v1,x + av,x ) − p (av1,x + v,x ) = 0 for

y = ±h/2

h/2 [u,x − λv,y + (a2 p 1 + ap)v1,y + (ap 1 + p) v,y + λ2 p] dy = 0 −h/2

†

Notice that µ is independent of x and y.

for

(3.7.76) x = .


95

Variation with respect to v yields h/2 [v,x δv,x + v,y δv,y + λ (u,y δv,x − u,x δv,y ) 0 −h/2

+ a2 p 1 (u1,x δv,y − u1,y δv,x ) + ap 1 (u,x δv,y − u,y δv,x ) + p λ1 δv,y + a (u1,x δv,y − u1,y δv,x ) + u,x δv,y − u,y δv,x

(3.7.77)

+ µv1 δv] dx dy = 0. Integration by parts yields h/2 −

[v,xx + v,yy + a2 (p 1 u1,x ),y − a2 (p 1 u1,y ),x 0 −h/2

+ a (p 1 u,x ),y − a (p 1 u,y ),x + λ1 p ,y + a (p u1,x ),y − a (pu,y ),x + (p u,x ),y − (p u,y ),x + µv1 ] δv dx dy +

(3.7.78) h/2

(v,y − λu,x + a2 p 1 u1,x + ap 1 u,x + λ1 p + apu1,x + pu,x ) δv |−h/2 dx 0

h/2 (v,x + λu,y − a2 p 1 u1,y − ap 1 u,y − apu1,y − p u,y ) δv | 0 dy = 0,

+ −h/2

from which follows v,xx + v,yy + a2 (p 1,y u1,x − p 1,x u1,y ) + a(p 1,y u,x − p 1,x u,y ) + λ1 p ,y + a(p ,y u,x − p ,x u1,y ) + p ,y u,x − p ,x u,y + µv1 = 0,

(3.7.79)

with the dynamic boundary conditions v,y − λu,x + (a2 p 1 + ap)u1,x + (ap 1 + p)u,x + λ1 p = 0

for

v,x + λu,y − (a2 p 1 + ap)u1,y − (ap 1 + p)u,y = 0 for

x=0

y = ±h/2

(3.7.80)

x = . (3.7.81) Variation with respect to p and µ yields the nonlinear side condition and the orthogonality condition, respectively. We now restrict ourselves to vanishingly small values of the amplitude a. Hence, we may write and

u = u 0 + au1 + a2 u2 + · · · v = v 0 + av1 + a2 v2 + · · · p = p 0 + ap 1 + a2 p 2 + · · · µ = µ0 + aµ1 + a2 µ2 + · · · .

(3.7.82)

96

Applications

Introducing these expansions into the differential equations and the corresponding boundary conditions, we find to zeroth order 0 0 u,xx + u,yy + λ2 p ,x0 + p ,x0 v,y0 − p ,y0 v,x0 + µ0 u1 = 0 0 0 v,xx + v,yy + λ1 p ,y0 + p ,y0 u,x0 − p ,x0 u,y0 + µ0 v1 = 0

λ2 u,x0 + λ1 v,y0 + u,x0 v,y0 − u,y0 v,x0 = 0 h/2

(3.7.83)

u1 u 0 + v1 v 0 dx dy = 0,

0 −h/2

with the dynamic boundary conditions

.

u0,y + λv0,x − p 0 v0,x = 0

for

v0,y − λu0,x + p 0 u0,x + λ1 p 0 = 0 h/2

0 u,x − λv0,y + p 0 v0,y + λ2 p 0 dy = 0

y = ±h/2,

(3.7.84)

x=

(3.7.85)

for

−h/2

v0,x + λu0,y − p 0 u0,y = 0

for

x = 0 and

x = ,

(3.7.86)

and

u0,y = 0 for

x = .

(3.7.87)

and the geometric boundary conditions u0 = 0 for

x=0

Because the equations and the boundary conditions are homogeneous, the unique solution is u0 = v0 = p 0 = µ0 = 0.

(3.7.88)

For term linear in a, we obtain u1,xx + u1,yy + λ2 p 1,x + µ1 u1 = 0 v1,xx + v1,yy + λ1 p 1,y + µ1 v1 = 0

(3.7.89)

λ2 u1,x + λ1 v1,y = 0 h/2

u1 u1 + v1 v1 dx dy = 0,

0 −h/2

with the dynamic boundary conditions u1,y + λv1,x = 0 v1,y − λu1,x + λ1 p 1 = 0 h/2 −h/2

. for

y = ±h/2

1 u,x − λv1,y + λ2 p 1 dy = 0 for

x=

(3.7.90)

(3.7.91)


v1,x + λu1,y = 0

x = 0 and

for

x=

97

(3.7.92)

and the geometric conditions u1 = 0 for

x = 0 and

u1,y = 0

x = .

for

(3.7.93)

Again, this is a set of homogeneous equations and boundary conditions, so the unique solution is given by u1 = v1 = p 1 = µ1 = 0.

(3.7.94)

For the terms quadratic in the amplitude a, we obtain u2,xx + u2,yy + λ2 p 2,x + µ2 u1 = p 1,y v1,x − p 1,x v1,y v2,xx + v2,yy + λ1 p 2,y + µ2 v1 = p 1,x v1,y − p 1,y v1,x λ2 u2,x + λ1 v2,y = u1,y v1,x − u1,x v1,y h/2

(3.7.95)

u1 u2 + v1 v2 dx dy = 0,

0 −h/2


.

u2,y + λv2,x = p 1 v1,x v2,y − λu2,x + λ1 p 2 = p 1 u1,x h/2

2 u,x − λv2,y + λ2 p 2 dy = −

−h/2

for

y = ±h/2

(3.7.96)

h/2 p 1 v1,y dy

x=

for

(3.7.97)

−h/2

v2,x + λu2,y = p 1 u1,y

for

x = 0 and

x = ,

(3.7.98)

x = .

(3.7.99)

and the geometric conditions u2 = 0 at

x = 0 and

u2,y = 0

at

The present equations and boundary conditions are the conditions for the stationary value of the functional ∗

h/2

(PII − PI ) = a

4 0 −h/2

1 22 2 2 2 u,x + u2,y + v2,x + v2,y 2

+ λ u2,y v2,x − u2,x u2,y + p 1 u1,x v2,y + u2,x v1,y − u1,y v2,x − u2,y v1,x dx dy

(3.7.100)

under the linear side condition λ2 u2,x + λ1 v2,y + u1,x v1,y − u1,y v1,x = 0

(3.7.101)

98

Applications

and the orthogonality condition h/2

u1 u2 + v1 v2 dx dy = 0.

(3.7.102)

0 −h/2

This is the variational problem of a quadratic functional with the linear side condition, and has a second variation that is positive-definite under the requirement of orthogonality of the field u2 , v2 with respect to the buckling mode u1 . The solution of the variational problem thus represents a minimum of the functional (PII − PI )∗ . Writing Min (PII − PI )∗ = a4 A4 ,

(3.7.103)

where A4 is defined by the value of the integral in (3.7.100) evaluated for u2 , v2 as solutions of the equations and boundary conditions (3.7.95) to (3.7.99), a necessary condition for stability is A4 ≥ 0, and a sufficient condition is A4 > 0 in the case of neutral equilibrium at λ = λcr . To avoid lengthy calculations, in the following we shall restrict ourselves to the case of a slender plate (i.e., θ 1). In that case, from (3.7.55) we have 4 π2 h2 2 h λcr = 1 + θ2 + O θ4 = 1 + . + O 3 6 2 4

(3.7.104)

Because πy/ is of order θ, we can write the buckling mode (3.7.64) in the form πy πx + O θ3 sin u1 = 4 πx π2 y2 (3.7.105) cos + O θ v1 = 1 − 2 2 πx π 2πy + O θ3 cos , p1 = 2 where we must note that the derivative with respect to y of an error term θn is of order θn−1 −1 . We can now evaluate the expressions 2 2πx π3 1 + O θ sin 3 2 2πx 2π4 y p 1,y u1,x − p 1,x u1,y = 4 1 + O θ sin 2 2πx π2 1 − cos u1,x v1,y − u1,y v1,x = 2 1 + O θ 2 2 2πx π3 y −p 1 v1,x = 3 1 + O θ sin 2 2πx π4 y2 . 1 + cos p 1 u1,x = 4 1 + O θ

p 1,x v1,y − p 1,y v1,x =

(3.7.106)


99

2 2πx π4 y2 1 + O θ 1 + cos 4 3 2πx π y = − 3 1 + O θ2 sin .

p 1 v1,y = − −p 1 u1,y

The equations (3.7.95) now become πx π3 πy 2πx sin − 3 sin 2 2 π πx y 2π4 y cos v2,xx + v2,yy + λ1 p 2,y = −µ2 1 − − 2 2 4 2 2πx π λ2 u2,x + λ1 v2,y = − 2 1 − cos 2 2 h/2 πy πx 2 π2 y2 πx 2 cos u + 1− v dx dy = 0, sin 2 2 u2,xx + u2,yy + λ2 p 2,x = −µ2

(3.7.107)

0 −h/2


 π3 h 2πx    + = ∓ 3 sin  2 2πx  π4 h2   1 + cos v2,y − λu2,x + λ1 p 2 =  4 4 u2,y

λv2,x

h/2 −h/2

2 π4 h3 u,x − λv2,y + λ2 p 2 dy = 4 6

v2,x + λu2,y = 0 for

for

y=±

for

x=

x = 0 and

h 2

x = ,

(3.7.108)

(3.7.109)

(3.7.110)

and the geometric conditions u2 = 0 at

x = 0 and

u2,y = 0

at

x = .

(3.7.111)

The value µ2 in (3.7.107) is still unknown. The terms in (3.7.107) involving µ2 lead to terms in u2 and v2 proportional to sin πx/ and cos πx/ , respectively, which do not satisfy the orthogonality condition, and hence we require µ2 = 0. We now try a solution in the form 2πx π3 y2 π2 π3 h2 π4 h2 π 2πx 2 sin − − + C u = x + C sin x 1 2 8 4 3 4 2 3 4 2 2πx π4 h2 y π4 h2 π y π4 y3 2πx π2 cos v2 = − y + C cos y − + C 3 4 4 2 2 4 4 2 4 4 π4 h2 π4 h2 π4 y2 2πx π4 y2 2πx − + C6 4 , p2 = − cos + C5 4 cos 4 4

(3.7.112)

(3.7.113)

100

Applications

where all numbers C1 − C6 are of order of magnitude unity. We have made use of the fact that λ1 and λ2 are of order unity with an error of order π2 h2 /4 2 , but in some places we had to allow for the more accurate relation λ2 /λ1 = λ = 1 + π2 h2 /6 2 . It is easily verified that the expressions (3.7.113) satisfy the equations (3.7.107) and the first boundary conditions (3.7.109) with relative errors of O h2 /L2 . For the left-hand member of the second of the boundary conditions (3.7.108), we obtain 2 π 2πx π2 π4 h2 π4 h2 3 π4 h2 2πx cos − + C + + C cos 3 4 4 2 8 4 4 2 4 4 2 2πx π2 2π4 h2 π4 h2 π π4 h2 2πx cos − + C −λ − + C cos 1 2 4 2 8 4 4 2 4 4 4 2 π h π4 h2 π4 h2 2πx π4 h2 2πx − + C + λ1 − 4 cos + C cos 5 6 4 4 4 4 4 2 4 2 2 4 2 4 2 4 2 π 3π h π π h π h π h 2πx = + − 2+ − − + · · · cos 4 2 8 4 4 8 4 24 4 4 4 2 2 4 2 4 2 4 2 4 2 4 2 π h π h π h π π π h π h − 2+ 2+ − + C4 4 − C2 4 + C6 4 + · · · (3.7.114) 4 4 4 4 24 4 2πx π4 h2 π4 h2 π4 h2 cos + C3 4 − 2C1 4 + C5 4 4 2 4 2 π h π h 5 5 2πx + C3 + C5 − 2C1 + − + C4 − C2 + C6 = cos 4 24 24 4 4 2 4 2 2πx π h 1 π h 1 + cos + + C4 − C2 + C6 =− 4 4 24 4 4 2 π h 11 2πx + C3 + C5 − 2C1 . + cos 24 4 It follows that − C2 + C4 + C6 +

1 =0 24

(3.7.115)

− 2C1 + C3 + C5 +

11 = 0. 24

(3.7.116)

For the left-hand member of (3.7.109), we obtain ! πh π3 h3 2π π2 h π3 h3 2π π4 h3 2 − − 2 + C1 3 + C2 4 3 16 96 8 2 2 2 4 3 2 4 3 4 3 π h π h π h π h π h −λ + − 2 + C3 4 + C4 3 8 2 16 4 8 2 2 4 3 4 3 4 3 4 3 " π h π h π h π h + λ2 − − + C5 4 + C6 4 4 4 24 24 2 2 π4 h3 π4 h3 = − 4 + 4 (2C1 + C2 − C3 − C4 + C5 + C6 ) 3 π4 h3 1 π4 h3 = + 2C1 + C2 − C3 − C4 + C5 + C6 − . 6 4 2 4

(3.7.117)

3.8 Helical spring with a small pitch

101

Hence, it follows that 2C1 + C2 − C3 − C4 + C5 + C6 −

1 = 0. 2

(3.7.118)

When the constants C1 , C2 , and C5 are chosen, the additional constants C3 , C4 , and C6 can be calculated. For the evaluation of (3.7.103), we only need the dominating part of u2 and v2 , i.e., 2πx π3 y2 π2 π sin − − u2 = x 8 4 3 4 2 (3.7.119) 2 2πx π π4 y3 π2 cos v2 = y + y. − 4 2 2 4 4 2 Substitution of these expressions into (3.7.100) yields Min (PII − PI )∗ =

π6 h2 a4 h 1 + O h2 / 2 . 192 6

(3.7.120)

which shows that the critical state of neutral equilibrium is stable. A comparison with Euler column theory requires the evaluation of h/2 P2 [u1 , λ = 1] = a

2 0 −h/2

1 2 u1,x + u21,y + v21,x + v21,y + (u1,y v1,x − u1,x v1,y ) dx dy. 2 (3.7.121)

Using the expressions (3.7.105), we find 2 5 1π h 1 π2 h 1 π4 h3 1 π4 h3 h P2 [u1 , λ = 1] = a2 − + + + O 2 48 3 5 2 16 3 h2 1 π4 h2 a2 . h 1 + O 2 = 4 12

(3.7.122)

The ratio of the expressions (3.7.121) and (3.7.122) is π2 a2 /16 2 with a relative error of order h2 / 2 . This is indeed in full agreement with the result of the Euler column in the initial post-buckling range, cf. (3.1.23). 3.8 Helical spring with a small pitch The elastic stability of helical springs was first dealt with by Hurlbrink (1910) and Grammel (1924). Both ignored the shear elasticity of the spring. Shortly thereafter, Biezeno and Koch took into account the effect of shear but they overestimated the effect, implying as a consequence that any spring, however short, would be liable to buckling, which is not confirmed by experiment. Later, Haringx (1942) took into account the shear effect correctly. In the first part of this section we shall follow his approach.† †

Cf. J. A. Haringx, On highly compressible helical springs and rubber rods, and their application for vibration-free mountings (Philips Research Laboratories, 1950).

102

Applications

a

b

Figure 3.8.1

If the helical spring has a sufficiently small pitch when compressed, the deformation of one single coil under a certain load will differ little from that of an unclosed circular ring lying in a flat plane. Consequently, we may imagine the compressed helical spring as being replaced by a number of similar rings a connected by perfectly rigid elements b, as represented in Figure 3.8.1. The influence of the axial force having been discounted as the successive coils are imagined as being flat after compression of the spring, we now must see what distortions occur as a result of the bending moments and transverse forces to be transmitted by these flat coils (see Figure 3.8.2). In the following, we shall assume that the principal axes of inertia of the cross section of the ring are in the plane perpendicular to the plane of the ring. Let Sb1 and Sb2 be the bending stiffness of the cross section of the ring in the plane of the ring and perpendicular to it, respectively, and let St be the torsional stiffness of the cross section. Let u, v be the displacements due to N and D, respectively, and let ϕ be the rotation due to M. Elementary calculations show N=

St u, 2πR3

D=

Sb1 v, πR3

M=

Sb2 · St ϕ. πR (Sb2 + St )

(3.8.1)

Haringx now smears the stiffness, i.e., he replaces the spring with n coils by a continuous rod. Let be the length of the spring; then the elongation of the spring is =

2πR3 N , Nn ≡ St EA N D

M

M

D

N

Figure 3.8.2

(3.8.2)


103

where the “stretching stiffness” EA is defined by EA =

St . 2πR3 n

(3.8.3)

Further, for the total displacement due to D we have vtot = n

πR3 D D , ≡ Sb1 βEA

β=2

Sb1 . St

(3.8.4)

The factor β becomes small for flat cross sections. If t(y) is the thickness of the cross section, we then have 1 3 Et (y) dy β = 2 12 = 1 + υ, (3.8.5) 1 Gt3 (y) dy 3 so that in general, β ≥ 1 + υ. Finally, the total angle of rotation is given by πR (Sb2 + St ) M ≡ , Sb2 · St B

(3.8.6)

Sb2 · St 1 ≡ α EA R2 Sb2 + St πR n

(3.8.7)

2Sb2 . Sb2 + St

(3.8.8)

ϕtot = n where B is the “bending stiffness,” B= and

α=

For a circular cross section, the parameters α and β are given by α=

2 (1 + υ) , 2+υ

β = 2 (1 + υ) .

(3.8.9)

We now consider the deformation of an infinitesimally small element of the rod (see Figure 3.8.3). Notice that in our model, the shear deformation is defined by the displacements of the transverse shear forces D. (This is not the case for ordinary beams.)

Z N

d x + du ψ

X ψ

M D

ψ

dw

II dx undeformed

dv

ds

dx fundamental state

adjacent state

Figure 3.8.3

104

Applications

We now have the following relations, ds = (dx + du) cos ψ + dw sin ψ dv = − (dx + du) sin ψ + dw cos ψ.

(3.8.10)

In the fundamental state, we have N = −λEA (which defines λ)

(3.8.11)

x = (1 − λ) x

(3.8.12)

ds = (1 − λ + u ) cos ψ + w sin ψ dx dv = − (1 − λ + u ) sin ψ + w cos ψ. dx

(3.8.13)

so that with () = d (/dx), we have

The elongation per unit length is thus ε = (1 − λ + u ) cos ψ + w sin ψ − 1,

(3.8.14)

and the angle of shear is γ=

dv = − (1 − λ + u ) sin ψ + w cos ψ. dx

(3.8.15)

Further, the curvature of the rod is given by κ=

dψ = ψ , dx

(3.8.16)

so we can write N = EAε,

D = βEAγ,

M = αEAR2 κ.

(3.8.17)

In the fundamental state, X = N = −λEA.

(3.8.18)

In the adjacent state, we have X = N cos ψ − D sin ψ = const. Z = N sin ψ + D cos ψ = const. ≡ ζEA

(3.8.19)

as conditions for the equilibrium of forces, and M dx + Dds − Ndv = 0

(3.8.20)


105

for the equilibrium of moments. Solving N and D from (3.8.19), our results are N = X cos ψ + Z sin ψ = (−λ cos ψ + ζsin ψ) EA D = Z cos ψ − X sin ψ = (λ sin ψ + ζcos ψ) EA.

(3.8.21)

Substituting these expressions into (3.8.20) and using the relations (3.8.13) to (3.8.17), we find αR2 ψ + (λ sin ψ + ζcos ψ) [(1 − λ + u ) cos ψ + w sin ψ] − (−λ cos ψ + ζsin ψ) [− (1 − λ + u ) sin ψ + w cos ψ] = 0,

(3.8.22)

or, upon simplification, αR2 ψ + λw + ζ(1 − λ + u ) = 0.

(3.8.23)

We shall now try to express w and u as functions of ψ, to obtain a differential equation in one variable. To this end, we solve 1 − λ + u and w from (3.8.14) and (3.8.15) and make use of the relations (3.8.17) and (3.8.21). Our result is 1 − λ + u = (1 + ε) cos ψ − γ sin ψ = (1 − λ cos ψ + ζsin ψ) cos ψ −

1 (λ sin ψ + ζcos ψ) sin ψ β

w = γ cos ψ + (1 + ε) sin ψ 1 = (λ sin ψ + ζcos ψ) cos ψ + (1 − λ cos ψ + ζsin ψ) sin ψ. β

(3.8.24)

(3.8.25)

Substitution of these expressions into (3.8.23) and rearranging terms yields 1 2 2 αR ψ + λ sin ψ − λ 1 − sin ψ cos ψ β (3.8.26) 1 1 λ cos 2ψ + ζ 1 − sin ψcos ψ = 0. + ζ cos ψ − 1 − β β When there are no transverse shear forces in the spring (e.g., for a spring hinged on both ends), ζ = 0 and, in other cases, ζ 1 in the case of small deflections ψ, ζ = O (ψ). For the determination of the buckling mode, (3.8.26) can be linearized, which yields 1 1 αR2 ψ + λψ − λ2 1 − ψ+ ζ 1 − 1 − λ = 0, (3.8.27) β β where we have neglected the term in ζ2 . As mentioned previously, when the spring is hinged at both ends, ζ = 0. Nonzero values of ζ occur, e.g., in the cases shown in Figure 3.8.4.

106

Applications

Figure 3.8.4

If ζ is nonzero, the value of ζ is determined from the condition

w dx = 0,

(3.8.28)

0

which with (3.8.25) yields

1 1 2 2 sin ψ cos ψ + ζ sin ψ + cos ψ dx = 0, sin ψ − λ 1 − β β

(3.8.29)

0

or, linearized, " ! 1 ζ ψ 1−λ 1− + dx = 0. β β

(3.8.30)

0

Let us now first consider the case that the spring is hinged at both ends. The lowest buckling mode is then given by πx ψ = cos (3.8.31) and the bifurcation condition becomes 1 π2 αR2 1− λ2 − λ + = 0, (3.8.32) β 2 from which we obtain

1/2 β2 β π2 αβ R2 − λ1 = − . 2 (β − 1) β − 1 2 4 (β − 1)2

(3.8.33)

It follows that for sufficiently large values of R/ (short spring), there is no critical load, i.e., no buckling can occur. We shall discuss this result in more detail later in this section. When the spring is clamped at both ends, ψ = 0 for x = 0 and x = , and for a symmetric buckling mode, there are no transverse shear forces, i.e., ζ = 0. Then (3.8.30) is satisfied by ψ = sin

2πx

(3.8.34)


107

and the bifurcation condition becomes 4π2 αR2 1 = 0, 1− λ2 − λ + β 2

(3.8.35)

i.e., the last term is four times the corresponding term for the hinged-hinged spring. We shall now treat the problem with the general theory of stability. The increment of the potential energy in passing from the undeformed state to the adjacent state is given by 1 PII − PI = EA 2

{αR2 ψ2 + [(1 − λ + u ) cos ψ + w sin ψ − 1]

2

0

+ β [− (1 − λ + u ) sin ψ + w cos ψ] } dx 1 2 − EA λ dx + λEA (−λ + u ) dx. 2 2

0

(3.8.36)

0

The equations for the equilibrium of forces are obtained by variations of this functional with respect to u and w . Variation with respect to u yields 0

[(1 − λ + u ) cos ψ + w sin ψ − 1] cos ψ

− β [− (1 − λ + u ) sin ψ + w cos ψ] sin ψ + λ δu dx = 0,

(3.8.37)

and because there are no restrictions on δu , it follows that [(1 − λ + u ) cos ψ + w sin ψ − 1] cos ψ + β [− (1 − λ + u ) sin ψ + w cos ψ] sin ψ + λ = 0.

(3.8.38)

Variation with respect to w yields 0

[(1 − λ + u ) cos ψ + w sin ψ − 1] sin ψ +β [− (1 − λ + u ) sin ψ + w cos ψ] cos ψ + c δw dx = 0,

(3.8.39)

where we have introduced the Lagrangian multiplier c because δw is not arbitrary, as it must satisfy the kinematic condition 0 δw dx = 0. The second equilibrium equation now reads [(1 − λ + u ) cos ψ + w sin ψ − 1] sin ψ + β [− (1 − λ + u ) sin ψ + w cos ψ] cos ψ + c = 0.

(3.8.40)

Multiplying (3.8.38) by cos ψand (3.8.40) by sin ψand adding the resulting equations, we obtain (1 − λ + u ) cos ψ + w sin ψ − 1 + λ cos ψ + c sin ψ = 0.

(3.8.41)

108

Applications

Multiplying (3.8.38) by sin ψ and (3.8.40) by cos ψ and subtracting the resulting equations, we find β [(1 − λ + u ) sin ψ − w cos ψ] + λ sin ψ − c cos ψ = 0.

(3.8.42)

Using (3.8.41) and (3.8.42), we can rewrite the functional (3.8.36) to yield 1 PII − PI = EA 2

αR2 ψ2 + (λ cos ψ + c sin ψ)

2

0

1 + (λ sin ψ − c cos ψ)2 + 2λu dx, β

(3.8.43)

where we have omitted the constant terms in (3.8.36) because they are unimportant for our further discussion. We now still need an expression for u . Multiplying (3.8.41) by β cos ψ and (3.8.42) by sin ψ and adding the resulting equations, we find β (1 − λ + u ) − β cos ψ + βλ cos2 ψ + βc sin ψ cos ψ + λ sin2 ψ − c sin ψ cos ψ = 0, from which

1 1 sin2 ψ − c 1 − sin ψcos ψ. u = −1 + cos ψ + λ 1 − β β

(3.8.44)

The functional (3.8.43) can now be rewritten to yield ! 1 αR2 ψ2 − 2λ (1 − cos ψ) PII − PI = EA 2 0 " 1 1 2 2 2 2 2 sin ψ + c sin ψ + cos ψ dx. +λ 1+ 1− β β

(3.8.45)

The second variation is now given by 1 P2 [ψ; λ] = EA 2

1 c2 αR2 ψ2 − λψ2 + λ2 1 − ψ2 + dx, β β

(3.8.46)

0

where we have made use of the fact that c 1. When there are no kinematic conditions, c = 0. In this case, the equation for neutral equilibrium is given by 1 2 αR ψ + λ 1 − λ 1 − ψ = 0, (3.8.47) β in full agreement with (3.8.27) with ζ = 0. There are no cubic terms, so P21 [au, u] = 0. The stability in the critical case of neutral equilibrium is thus governed by the fourth degree terms, i.e., by 1 P4 [ψ ; λ] = EA 2

0

1 1 1 1 2 4 4 2 λψ − λ 1 − ψ +c 1− ψ2 dx. 12 3 β β

(3.8.48)


109

For c = 0, this becomes 1 P4 [ψ ; λ] = EA 2

1 1 2 λ − 4λ 1 − ψ4 dx. 12 β

(3.8.49)

0

For a hinged-hinged spring, we have already discussed the buckling mode and the critical load, cf. (3.8.31) to (3.8.33). The critical load λ1 given in (3.8.33) is real for π2 R2 β . ≤ 2 4α (β − 1)

(3.8.50)

The sign of P4 [ψ; λ] is determined by the coefficient of ψ4 in (3.8.49), i.e., by the sign of 1 2 f (λ1 ) = λ1 − 4λ1 1 − . (3.8.51) β Using (3.8.32), we obtain

π2 αR2 f (λ1 ) = λ1 − 4 λ1 − 2

= 4π2

αR2 − 3λ1 . 2

(3.8.52)

Let us now first consider the critical case where the equality sign holds in (3.8.50), i.e., λ1 =

β . 2 (β − 1)

(3.8.53)

In this case, f (λ1 ) =

3β 3β β − = < 0, β − 1 2 (β − 1) 2 (1 − β)

(3.8.54)

so the equilibrium is unstable in that point. Let us now investigate what happens in the points where f (λ1 ) = 0, i.e., for λ1 =

4π2 R2 α. 3 2

(3.8.55)

Substitution into the bifurcation condition (3.8.32) yields β−1 β

2 2 4π2 R2 4π2 R2 2 αR α − α + π = 0, 3 2 3 2 2

or β π2 R2 3β < , α= 2 16 (β − 1) 4 (β − 1)

(3.8.56)

i.e., the equilibrium is unstable for β 3β ≤ λ1 ≤ . 4 (β − 1) 4 (β − 1) We can now represent our results in the graph in Figure 3.8.5.

(3.8.57)

110

Applications

λ1 1

stable for the buckling mode

β / 2 ( β − 1)

ψ = sin

UNSTABLE

πx l

UNSTABLE β / 4 ( β − 1) STABLE

2π

β −1 α β

l/R Figure 3.8.5

For a spring with a circular cross section, no buckling will occur for β−1 1 + 2ν < 2π α = 2π . R β 2+ν

(3.8.58)

It is still an open question whether the unstable domain on the boundary for values of λ1 such that β/4 (β − 1) < λ1 < β/2 (β − 1) does actually exist, or that it stems from the approximation that the spring is built up from flat circular rings. 3.9 Torsion of a shaft We consider a shaft of arbitrary (constant) cross section loaded by a torque W (see Figure 3.9.1). Let, x, y, z be axes fixed in space, and let y1 , z1 be (co-rotating) principle axes of inertia, which are rotated over an angle kx with respect to the fixed axes y, z. Here k is the torsion angle per unit length.

z

z1

y1 kx y

0

Figure 3.9.1

3.9 Torsion of a shaft

111

In the fundamental state, we have pure torsion, and the only non-vanishing components of the stress tensor are τxy =

∂F , ∂z

τxz = −

∂F , ∂y

(3.9.1)

where F is the stress function of torsion. In the adjacent state, bending will occur, and the point 0 will have displacements v0 (x) , w0 (x) along the y and z axes, respectively. Assuming that Bernoulli’s hypothesis holds, the displacement u of a point (x, y, z) is given by u = −yv0 (x) − zw0 (x) ,

(3.9.2)

where () = d () /dx. To obtain an approximate uni-axial state of stress, we assume the following displacement field, 1 v (x, y, z) = v0 (x) + νv0 (y2 − z2 ) + νw0 yz 2 1 w (x, y, z) = w0 (x) + νv yz + νw0 (z2 − y2 ). 2

(3.9.3)

∂w ∂u ∂v = = −ν ∂y ∂z ∂x ∂v ∂w ψyz = + = 0. ∂z ∂y

(3.9.4)

This field satisfies

However, there are non-vanishing angles of shear, ψxy =

1 ∂u ∂v + = νv0 y2 − z2 + νw0 yz. ∂y ∂x 2

(3.9.5)

Let be the wavelength of the deformation pattern, and let b be a characteristic length in the cross section. The contribution of ψxy in the energy function is then of order b4 v20 / 6 , whereas the contributions due to ∂u/∂x are of order b2 v20 / 4 . Under the assumption b2 / 2 1, we can now neglect the contribution of ψxy in the energy, which results in a relative error of O(b2 / 2 ). The energy density due to bending is given by

dx

1 2 E (yv0 + zw0 ) dy dz. 2

(3.9.6)

A

0

It is now convenient to employ the principle axes of inertia. Using the relations y = y1 cos kx − z1 sin kx z = y1 sin kx + z1 cos kx,

(3.9.7)

we obtain for the energy density 0

1 1 2 2 EIz1 (v0 cos kx + w0 sin kx) + EIy1 (w0 cos kx − v0 sin kx) dx dy. (3.9.8) 2 2

112

Applications

Because we are dealing with buckling, we cannot neglect the stresses in the fundamental state. Recalling our general expression for the second variation in the case of dead-weight loads, 1 1 (3.9.9) P2 [u] = Sij uh,i uh,j + Eijkl θij θk dV, 2 2 where we have already evaluated the second term (3.9.8), we are still in need of an explicit form of the first term. Because for torsion we have only two nonzero stress components, we find 1 Sij uh,i uh,j = τxy (u,x u,y + v,x v,y + w,x w,y ) + τxz (u,x u,z + v,x v,z + w,x w,z) 2 ! 1 2 2 = τxy (−yv0 − zw0 ) (−v0 ) + v0 + νv0 (y − z ) + νw0 yz 2 " 1 2 2 (3.9.10) × (νv0 y + νw0 z) + w0 + νv0 yz + νw0 (z − y ) (νv0 z − νw0 y) 2 ! 1 + τxz (−yv0 − zw0 ) (−w0 ) + v0 + νv (y2 − z2 ) + νw 0 yz 2 0 " 1 2 2 × (−νv0 z + νw0 y) + w0 + νv0 yz + νw0 (z − y ) (νv0 y + νw0 z) . 2 We shall now argue that the terms with third derivatives may be neglected. Consider the term % & 2 5 τv b 1 2 0 τxy × νv0 (y − z2 ) × νv . (3.9.11) 0 y dx dy = O 2 5 A

In our discussion of ψxy , we have already neglected terms of O Ev20 b6 / 6 , and have 2 4 4 taken into account terms of O Ev0 b / . Hence, it follows that we can neglect the contribution of (3.9.11) in the energy and, similarly, the other terms involving third derivatives with respect to x. The expression (3.9.10) can now be rewritten to yield 1 Sij uh,i uh,j = v0 v0 [yτxy + νyτxy − νzτxz] + w0 w0 [−νyτxy + zτxz + νzτxz] 2 (3.9.12) + v0 w0 [zτxy + νzτxy + νyτxz] + v0 w0 [νzτxy + yτxz + νyτxz] . Integrating this expression over the cross-sectional area, we must note that 4 4 ∂F dy dz = yτxy dy dz = y yFnz ds = Fedge ynz ds = 0, (3.9.13) ∂z A

edge

where Fedge is the constant value of F at the edge. Similarly, we have zτxz dy dz = 0.

(3.9.14)

A

This implies that after integration, the first and the second term between brackets in (3.9.12) vanish and in the third and the fourth term between brackets the terms


with v vanish because

1 zτxy dy dz = − W 2

113

yτxz dy dz =

A

1 W. 2

(3.9.15)

A

Further, we recall that the torque W is given by ˙ W= (yτxz − zτxy ) dy dz,

(3.9.16)

A

so the torsion energy per unit length is given by 1 W (v0 w0 − v0 w0 ) . 2

(3.9.17)

The second variation now becomes P2 [u] =

1 2 EIz1 (v0 cos kx + w0 sin kx) 2

0

1 1 2 + EIz1 (w0 cos kx − v0 sin kx) + W (v0 w0 − v0 w0 ) dx. 2 2

(3.9.18)

For a shaft with a circular cross section EIy1 = EIz1 = B, the functional reduces to 1 1 2 2 B v0 + w0 + W (v0 w0 − v0 w0 ) dx. 2 2

P2 [u] =

(3.9.19)

0

Our further discussion will be restricted to this functional. The condition for neutral equilibrium is P11 [u, ζ] = 0

1 2 B v0 + w2 0 2

1 + W (v0 ζ + w0 η − v0 ζ − w0 η ) dx = 0. 2

By integration by parts, we obtain 1 1 Bv0 + Ww0 η 0 + Bw0 − Wv0 ζ 0 2 2 − (Bv + Ww ) η − (Bw − Wv ) ζ 0

+

0

0

0

0

0

(3.9.20)

(3.9.21)

[(Bv 0 + Ww0 ) η + (Bv0 − Wv0 ) ζ] dx = 0,

0

from which we obtain Bv 0 + Ww0 = 0,

Bw 0 − Wv0 = 0.

(3.9.22)

Let us now consider the case that the shaft is supported at its ends. It is now convenient to choose the origin of our coordinate system in the middle of the shaft,

114

Applications

and let − ≤ x ≤ , i.e., we consider a shaft of length 2 . The geometric conditions are then v0 = w0 = 0 for

x = ± .

(3.9.23)

Modifying the boundaries in (3.9.21) correspondingly, we obtain the dynamic boundary conditions 1 Bv0 + Ww0 = 0, 2

1 Bw0 − Wv0 = 0 for 2

x = ± .

(3.9.24)

To solve this problem, we set v0 = Ceiµx ,

w0 = Deiµx ,

(3.9.25)

and substitution into the differential equations yields µ4 BC − iµ3 WD = 0

(3.9.26)

µ4 BD + iµ3 WC = 0. The condition for a non-trivial solution is now µ8 − µ6

W2 = 0, B2

(3.9.27)

i.e., µ1 , . . . , µ6 = 0,

µ7 =

W , B

µ8 = −

W . B

(3.9.28)

Defining W = µ, B we can write the solution in the form v0 = C1 cos µx + C2 sin µx + C3 x2 + C4 x + C5 w0 = C1 sin µx − C2 cos µx + C6 x2 + C7 x + C8

(3.9.29)

because the differential equations are satisfied identically by quadratic polynomials. Due to symmetry, we can split the solution into even and odd functions. Choosing an even function for v0 , w0 must be odd due to (3.9.22). The problem with v0 odd and w0 even is readily found from the previous one by interchanging v0 and w0 . The advantage of splitting up the problem is that we must now only satisfy four boundary conditions. From (3.9.23), we obtain v0 (x = ) = C1 cos µ + C3 2 + C5 = 0 w0 (x = ) = C1 sin µ + C7 2 = 0

(3.9.30)

and from (3.9.24), 1 1 − µ2 cos µ BC1 + 2C3 B + WC7 = 0 2 2 1 2 µ sin µ BC1 − C3 B = 0. 2

(3.9.31)


115

Because C5 only occurs in the first of the equations of (3.9.31), the condition for a non-trivial solution is sin µ 0 1 2 1 − µ cos µ µ 2 (3.9.32) 2 2 = 0, 1 2 − µ sin µ −µ 0 2 from which follows 1 tan µ = − µ . 3

(3.9.33)

The smallest non-negative root of this transcendental equation is π µ = 1.566 , 2

(3.9.34)

which yields W = 1.566π

B , 2

(3.9.35)

which is the critical value of the torque W. Let us now examine the boundary conditions more closely. We consider the end cross section at x = (see Figure 3.9.2). The axes y, z are fixed in space, and y1 , z1 are co-rotating axes. The displacement u( , y1 , z1 ) is given by u( , y1 , z,1 ) = −y1 v0 ( ) − zw0 ( ), which implies that the shear stresses τxy cause a bending moment τxy u dy dx = Mz

(3.9.36)

(3.9.37)

A

and, similarly for the shear stresses, τxzu dy dz = −My .

(3.9.38)

A

This result means that the load on the end faces is not directed along the undeformed axis. Bending moments My and Mz must be added. Evaluating the integrals z, z1 τ xz τ xy y, y1

Figure 3.9.2

116

Applications

in (3.9.37) and (3.9.38), we find 1 My = τxz y1 v10 ( ) + zw0 ( ) dy1 dz1 = Wv0 ( ) 2 A 1 τxz (y1 v0 ( ) + zw0 ( )) dy1 dz1 = Ww0 ( ) . Mz = − 2

(3.9.39)

A

These moments are called by Ziegler† semi-tangential moments, i.e., the moments are obtained by multiplying the torque by half the angle of rotation. In the problem treated previously, the loading was conservative. Greenhill‡ treated a similar problem, but in his case the bending moments were absent. The loading is then non-conservative. Greenhill’s boundary conditions are obtained from our previous problem by superimposing loads 1 1 Mz = − Ww0 , My = − Wv0 2 2 for x = ± . The boundary conditions now become Bw0 − Wv0 = 0,

Bv0 + Ww0 = 0,

x = ± .

(3.9.40)

(3.9.41)

The differential equations are still given by (3.9.22), and the general solution is v0 = C1 cos µx + C2 sin µx + C3 x2 + C4 x + C5 w0 = C1 sin µx − C2 cos µx + C6 x2 + C7 x + C8 ,

(3.9.42)

where µ = W/B. By arguments similar to those used in the previous problem, we may restrict ourselves to the underlined terms. The kinematic boundary conditions v0 = w0 = 0 at x = require C1 cos µ + C3 2 + C5 = 0 C1 sin µ + C7 = 0,

(3.9.43)

and the dynamic boundary conditions become −2C3 W = 0

(3.9.44)

2BC3 + WC7 = 0, which yields C3 = C7 = 0,

C5 = −C1 cos µ ,

C1 sin µ = 0,

(3.9.45)

and hence, µ = π, so that the critical torque is given by B . (3.9.46) 2 The correctness of this result, obtained from the equations for neutral equilibrium for a conservative system, is in this case verified by the solution of the dynamic problem. W = 2π

† ‡

Cf. H. Ziegler, Principles of Structural Stability (Blaisdell Publ. Comp.), p. 124. Cf. A. G. Greenhill, Proc. Inst. Mech. Engrs., 182 (1883).


117 W W

x l Figure 3.9.3

To show that this equivalent is not always justified, we consider the problem shown in Figure 3.9.3. The boundary conditions in this case are v0 = w0 = 0, Bv0 + Ww0 Bv 0 + Ww0

v0 = w0 = 0,

at x = 0 . =0

Bw0 − Wv0 Bw 0 − Wv0

= 0, = 0,

(3.9.47) at x = .

=0

Integrating the differential equations (3.9.22) twice and making use of the boundary conditions at x = , we find Bv0 + Ww0 = 0,

Bw0 − Wv0 = 0.

(3.9.48)

The general solution to these equations reads v0 = C1 cos µx + C2 sin µx + C3

(3.9.49)

w0 = C1 sin µx − C2 cos µx + C4 . From the kinematic conditions at x = 0, we find C1 + C3 = 0,

−C2 + C4 = 0

C2 µ = 0,

C1 µ = 0.

(3.9.50)

These results mean that for µ = 0, there is no neutral equilibrium (Ziegler, 1950). The dynamic problem yields increasing amplitudes for non-vanishing W (flutter). We shall analyze this phenomenon in the following (simpler) problem, in which we consider a shaft without mass and a concentrated mass m at its end x = (see Figure 3.9.4). Because the shaft has no mass, the differential equations are unaltered. The boundary conditions in this case are v0 = w0 = v0 = w0 = 0 at x = 0 Bv0 + Ww0 = 0,

Bw0 − Wv0 = 0

.

(3.9.51) at x = ,

v0 , Bw ¨0 Bw 0 + Ww0 = m¨ 0 − Wv0 = mw

W m x

W l Figure 3.9.4

118

Applications

where () = ∂ () /∂x and ()· = ∂ () /∂t. Introducing the complex function V(x, t) = v0 (x, t) + iw0 (x, t) ,

(3.9.52)

we may combine the two differential equations to yield V − iµV = 0,

(3.9.53)

where µ = W/B. The kinematic boundary conditions now become V = V = 0

at x = 0,

(3.9.54)

and the dynamic boundary conditions now read V − iµV = 0,

V − iµV =

m V¨ at x = . B

(3.9.55)

Introducing V = U(x)eiωt into the differential equation, we find U = D1 eiµx + D2 x2 + D3 x + D4 .

(3.9.56)

From the kinematic boundary conditions, we obtain D1 + D4 = 0,

iµD1 + D3 = 0.

(3.9.57)

U = D1 eiµx − 1 − iµx + D2 x2 .

(3.9.58)

Hence, we can write

From the dynamic boundary conditions, we obtain −µ2 D1 + 2 (1 − iµ ) D2 = 0 −

mω2 2 mω2 iµ e − 1 − iµ D1 + − + 2iµ D2 = 0. B B

The condition for a non-trivial solution is −µ2 2 (1 − iµ ) = 0, mω2 mω2 2 − eiµ − 1 − iµ − + 2iµ B B

(3.9.59)

(3.9.60)

which yields mω2 2 2 [µ + 2(1 − iµ )(eiµ − 1 − iµ )] − 2iµ3 = 0, B

(3.9.61)

so that the square of the frequency is given by ω2 =

2iµ3 B/m . µ2 2 + 2 (1 − iµ ) (eiµ − 1 − iµ )

(3.9.62)

It follows that ω2 is complex unless the denominator is purely imaginary, which is not the case, as we shall show. The fact that ω2 is complex implies that there is one root with a negative imaginary part, which implies that V, and hence v0 and w0 , increase exponentially with time. This fact means that for µ = 0, i.e., when there is a torque, the system is always

3.10 Torsion of a shaft with a Cardan (Hooke’s) joint

119

unstable. To show that ω2 is complex, we consider small values of ω . The denominator can now be written as 1 1 1 µ2 2 + 2 (1 − iµ ) 1 + iµ − µ2 2 − iµ3 3 + µ4 4 + O(µ5 5 ) − iµ − 1 2 6 24 1 1 1 = µ2 2 1 − 1 + iµ − iµ − µ2 2 + µ2 2 + O(µ3 3 ) 3 3 12 2 1 i − µ + O(µ2 2 ) , = µ3 3 3 4 so that 3 3B 2 2 1 − iµ + O(µ ) . ω = m 3 8 2

(3.9.63)

Although at first sight this result seems alarming, it is not important for actual constructions because it is virtually impossible to apply a torque the way it is assumed in this problem. To apply a torque, one usually makes use of the universal (Cardan) joint, and then the system is conservative. We shall treat this problem in the next section. 3.10 Torsion of a shaft with a Cardan (Hooke’s) joint We consider a shaft loaded in torsion with a Cardan (Hooke’s) joint (see Figure 3.10.1). Because large rotations may occur, we shall first derive an expression for the rotation vector for finite rotations. Let n be a unit vector along the axis of rotation, and let α be the angle of rotation. The position vector of a point is denoted by r (see Figure 3.10.2). The position of the point after rotation is denoted by r , and is given by r = r cos θ e1 + r sin θ cos α e2 + r sin θ sin α e3 ,

(3.10.1)

which can be rewritten to yield r = (r · n) n + {r − (r · n) n} cos α + (n × r) sin α = r cos α + (r · n) n (1 − cos α) + (n × r) sin α.

z y x = −l Cardan’s joint

x

Figure 3.10.1

W x=l

(3.10.2)

120

Applications

e2

r

θ

r′

α

α n

e1

e3 Figure 3.10.2

Without loss of generality, we may now restrict ourselves to values of α such that −π ≤ α ≤ π. We now introduce a rotation vector ω defined by ω = n · 2 sin α/2

(3.10.3)

with components ωi (i ∈ {1, 2, 3}) . Notice that two subsequent rotations do not yield a rotation vector that is the sum of the two corresponding rotation vectors. We now consider the rotation of the triad (e1 , e2 , e3 ) to obtain the rotation matrix. Let us denote the rotated triad by e1 , e2 , e3 . The triad (e1 , e2 , e3 ) is fixed to the body under consideration (see Figure 3.10.3). With r = e1 and n = ω/(2 sin α/2), we obtain from (3.10.2), e1 = e1 cos α +

ω ω1 ω × e1 sin α (1 − cos α) + 2 sin α/2 2 sin α/2 2 sin α/2

with |ω|2 = ω2 = ω21 + ω22 + ω23 = 4 sin2 α/2 1 cos α = 1 − 2 sin2 α/2 = 1 − ω2 2 ω × e1 = (ω1 e1 + ω2 e2 + ω3 e3 ) × e1 = − ω2 e3 + ω3 e2 e2 e′2 e1′

e1

e3

e′3 Figure 3.10.3

(3.10.4)


and

) cos α/2 =

1 (1 + cos α) = 2

)

121

1 1 − ω2 . 4

We can then rewrite this expression to yield 1 2 1 1 2 1/2 2 ω + ω3 e1 + e1 = 1 − e2 ω1 ω2 + ω3 1 − ω 2 2 2 4 1 2 1/2 1 ω1 ω3 − ω2 1 − ω + e3 . 2 4 By a cyclic interchanging of the subscript, we obtain 1 1 2 1/2 1 2 ω2 ω1 − ω3 1 − ω ω3 + ω21 e2 e2 = e1 + 1 − 2 4 2 1/2 1 1 ω2 ω3 + ω1 1 − ω2 + e3 . 2 4

(3.10.5)

(3.10.6)

e3

1 1 1 2 1/2 1 2 1/2 = e1 + e2 ω3 ω1 + ω2 1 − ω ω3 ω2 − ω1 1 − ω 2 4 2 4 (3.10.7) 1 2 2 ω + ω2 e3 . + 1− 2 1

We can now write our results in the form ei = Rij ej , where the rotation matrix Rij is given by 1/2 1 1 1 Rij = δij 1 − ω2 + ωi ωj + εijkωk 1 − ω2 , 2 2 4

(3.10.8)

(3.10.9)

where εijk is the alternating tensor. Up to now, all the expressions were exact. For infinitesimal rotations, the rotation matrix can be linearized, Rij = δij + εijkωk + O(ω2 ).

(3.10.10)

For our purpose, this expression is insufficient because we need quadratic terms for the second variation of the elastic energy. Taking into account quadratic terms, we obtain 1 2 1 Rij = δij 1 − ω + ωi ωj + εijkωk + O(ω3 ). (3.10.11) 2 2 Before we begin with the discussion of the buckling problem, let us make a few remarks. The Cardan Joint was invented by the physician and mathematician G. Cardan in Italy in the 16th century, but in most English-speaking countries it is referred to as Hooke’s joint. The construction is sketched in Figure 3.10.4.

122

Applications

e3

B′ A′ A

m θ e2

e1 B x = −l

x=l

W Figure 3.10.4

The bars AA and BB are rigid and are rigidly connected perpendicular to each other. The semi-rings are also rigid. The bearings in A, A , B, B are frictionless. Along the bars AA and BB , we introduce the unit vectors m and , respectively. These vectors are expressed in terms of the triad (e1 , e2 , e3 ), which is fixed in space, = − sin θ e2 + cos θ e3 m = cos θ e2 + sin θ e3 .

(3.10.12)

= − sin θ e2 + cos θ e3 .

(3.10.13)

Let e1 , e2 , e3 be connected to the deformable shaft; thus

Let ϕ be the torsion angle at x = − , so m = cos (θ + ϕ) e2 + sin (θ + ϕ) e3 .

(3.10.14)

Because and m are perpendicular to each other, we have (− sin θ e2 + cos θ e3 ) · [cos (θ + ϕ) e2 + sin (θ + ϕ) e3 ] = 0, which yields

(3.10.15)

1 2 1 2 1/2 1 2 ω + ω1 − sin θ sin (θ + ϕ) ω1 1 − ω − sin θ cos (θ + ϕ) 1 − + ω2 ω3 2 3 4 2 1/2 1 1 + ω3 ω2 + cos θ cos (θ + ϕ) −ω1 1 − ω2 (3.10.16) 4 2


123

1 2 ω1 + ω22 = 0, + cos θ sin (θ + ϕ) 1 − 2 or

1/2 1 1 − cos ϕ ω1 1 − ω2 + ω2 ω3 cos (2θ + ϕ) 4 2 (3.10.17) 1 1 1 + sin ϕ 1 − ω21 + ω23 sin θ cos (θ − ϕ) − ω22 cos θ sin (θ + ϕ) = 0. 2 2 2

For known ω, the angle ϕ can be determined from this (exact) equation. Because we only need an approximate solution that is correct up to and including quadratic terms, we can simplify this equation to yield 1 1 1 − ω1 + ω 2 ω 3 cos 2θ + ϕ + ω23 sin θ cos θ − ω22 cos θ sin θ = 0, 2 2 2

(3.10.18)

from which follows 1 1 2 ϕ = ω1 − ω2 ω3 cos 2θ − ω3 − ω22 sin 2θ. 2 4

(3.10.19)

The second variation is now obtained by adding to the functional (3.9.20) the terms due to the rotations at x = ± , 1 1 2 2 P2 [u] = B v0 + w0 − W (v0 w0 − v0 w0 ) dx 2 − 2 1 1 2 2 + W ω2 ω3 cos 2θ + ω − ω2 sin 2θ (3.10.20) 2 2 3 x= 1 1 2 ω3 − ω22 sin 2θ − W ω2 ω3 cos 2θ − . 2 2 x=− With ω2 = −w0 ,

ω3 = v0 ,

we can rewrite this functional in the form 1 1 2 dx B v0 + w2 W − w − v w P2 [u] = (v ) 0 0 0 0 0 2 − 2 1 1 2 + W −v0 w0 cos 2θ + v0 − w2 sin 2θ 0 2 2 x= 1 1 2 2 v − w0 sin 2θ + W v0 w0 cos 2θ + . 2 2 0 x=− The condition for neutral equilibrium is 1 1 P11 [u, ζ] = Bv0 + Ww0 η − + Bw0 − Wv0 ς − 2 2 − (Bv0 + Ww0 ) η − − (Bw0 − Wv0 ) ς − + [(Bv 0 + Ww0 ) η + (Bw0 − Wv0 ) ς] dx −

(3.10.21)

(3.10.22)

124

Applications

1 + W [−w0 cos 2θ + v0 sin 2θ] η x= 2 1 + W [−v0 cos 2θ − w0 sin 2θ] ς x= 2 1 + W [w0 cos 2θ + v0 sin 2θ] η x=− 2 1 + W [v0 cos 2θ − w0 sin 2θ] ς x=− = 0, 2

(3.10.23)

where η and ζ are arbitrary kinematically admissible displacement fields. Assuming that x = ± , we have the kinematic conditions v0 (± ) = w0 (± ) = 0. We arrive at the following dynamic boundary conditions at x = , 1 1 Bv0 + Wv0 sin 2θ + Ww0 (1 − cos 2θ) =0 2 2 x= 1 1 = 0, Bw0 − Wv0 (1 + cos 2θ) − Ww0 sin 2θ 2 2 x=

(3.10.24)

(3.10.25)

and at x = − , we have 1 1 −Bv0 + Wv0 sin 2θ − Ww0 (1 − cos 2θ) =0 2 2 x=− 1 1 −Bw0 + Wv0 (1 + cos 2θ) − Ww0 sin 2θ = 0.† 2 2 x=− The differential equations are still given by Bv 0 + Ww0 = 0,

Bw 0 − Wv0 = 0.

(3.10.26)

As already discussed in our previous examples, the general solution to these equations is v0 = C1 cos µx + C2 sin µx + C3 x2 + C4 x + C5 w0 = C1 sin µx − C2 cos µx + C6 x2 + C7 x + C8

(3.10.27)

where µ = W/B. We shall now first discuss the case that v0 is an even function in x. Then w0 is an odd function in x, as follows from the differential equation. This means that we must only consider the underlined terms in (3.10.27). Introduction of these expressions into the kinematic boundary conditions yields C1 cos µ + C3 2 + C5 = 0 C1 sin µ + C7 = 0. †

(3.10.28)

These boundary conditions can also be derived directly by requiring that the moments along and m vanish.


125

From the dynamic boundary conditions at x = , we obtain µ2 [(1 + cos 2θ) cos µ + sin 2θ sin µ ] C1 −2 (2 + µ sin 2θ) C3 − µ (1 − cos 2θ) C7 = 0 µ [(1 − cos 2θ) sin µ + sin 2θ cos µ ] C1

(3.10.29)

+ 2 (1 + cos 2θ) C3 + sin 2θ C7 = 0. Notice that due to symmetry and anti-symmetry, the boundary conditions at x = − are satisfied automatically. Because the constant C5 only appears in the first of the kinematic conditions that does not contain C7 , we can first calculate C1 , C3 , and C7 from the remaining equations and then determine C5 from the first kinematic condition. The condition for a non-trivial solution of these three equations is sin µ [−2(2 + µ sin 2θ) sin θ + 2µ (1 − cos2 2θ)] + {µ2 [(1 + cos 2θ) cos µ + sin 2θ sin µ ]2 (1 + cos 2θ)

(3.10.30)

+ 2(2 + µ sin 2θ)µ[(1 − cos 2θ) sin µ + sin 2θ cos µ ]} = 0, which can be reduced to λ sin λ tan2 θ + [(λ2 − 1) sin λ + λ cos λ] tan θ + λ2 cos λ = 0

(3.10.31)

where λ = µ . First, notice that there are no solutions for λ 1. For θ = 0, we have the solution λ = π/2, which is half the value of Greenhill’s result. For θ = π/2, we have λ = π (Greenhill’s value). Evaluating θ for given values of λ, we obtain the graph shown in Figure 3.10.5. λ

2 π

stable

2

v0 symmetric w0 symmetric This part of the curve cannot be reached 1

arctan 0 0

θ0

2 π

1

1 1+υ 2 θ Figure 3.10.5

2 π

126

Applications

The case where w0 is symmetric and v0 is anti-symmetric is easily obtained by rotating the axes, which yield v0 → w0 , w0 → −v0 for θ → θ + π/2. Instead of (3.10.31), we obtain λ sin λcotan2 θ − [(λ2 − 1) sin λ + λ cos λ] cotan θ + λ2 cos λ = 0,

(3.10.32)

which has roots that are shifted over a distance π/2 compared to the roots of (3.10.31). The corresponding curve is drawn as a dashed curve in Figure 3.10.5. The torsion angle follows from W θ − θ0 = , (3.10.33) St which yields θ0 = θ −

W B =θ−λ . St St

(3.10.34)

This result is a straight line in the λ − θ graph. For a circular shaft, we have θ0 = θ − λ (1 + υ) ,

(3.10.35)

and its tangent in the λ − θ plane is arctan (1/1 + υ) < π/4, which implies that this line may be tangent to the curve. The part of the curve above this line then cannot be reached, which implies that the critical torque is a discontinuous function of θ. From the practical point of view, these results are not as important because the critical value of the torque is very large. 3.11 Lateral buckling of a beam loaded in bending We consider a beam with a slender cross section loaded by a bending moment M in the direction of a principle axis of inertia (see Figure 3.11.1). The center of shear is (y0 , z0 ). When M is sufficiently large, torsion and bending in the y-direction will occur besides the bending in the z-direction. The corresponding additional displacements in passing from the fundamental state to the adjacent state are (approximately) given by u = −yv0 (x) + ψ0 (y, z) α (x) v = v0 (x) − α (x) (z − z0 ) + v∗ (x, y, z) ∗

w = α (x) (y − y0 ) + w (x, y, z) , z

*

( y0, z0 ) M y

Figure 3.11.1

(3.11.1)

3.11 Lateral buckling of a beam loaded in bending

127

where the starred functions are added to obtain a uni-axial state of stress. However, the values of these functions are small compared to those of the other terms, and may for our purpose be neglected. In (3.11.1), ψ0 (y, z) is the warping function with respect to the shear center and α (x) is the torsion angle per unit length. We know from our general theory that for dead-weight loads, the second variation is given by 1 1 (3.11.2) P2 [u] = Sij uh,i uh,j + Eijkl θij θkl dV. 2 V 2 The second term in (3.11.2) is readily obtained in the form 1 1 1 2 1 2 dx, (3.11.3) + α + Eijkl θij θkl dV = EIzv2 S Eα t 0 2 2 2 2 0 V

where is the warping constant. For the evaluation of the energy in the fundamental state, we notice that Mz S11 = σx = (3.11.4) Iy is the only non-vanishing stress component. The contribution to the second variation is 2 1 1 2 2 2 S11 u1,1 + u2,1 + u3,1 dV = σx u,x + v2,x + w2,x dV 2 2 V V 1 ∂v∗ 2 2 σx (−yv0 + ψ0 α ) + v0 − α (z − z0 ) + (3.11.5) = 2 ∂x . ∗ 2 ∂w + α (y − y0 ) + dV. ∂x First, notice that the first term between the brackets can be neglected because in it is σx u2,x , which is always small compared to the term Eu2,x , which is taken into account in (3.11.3). Second, the terms with an asterisk can be neglected as already discussed, so that we obtain V

1 Sij uh,i uh,j 2 = dx = dx

dV 2 Mz v0 − α (z − z0 ) + α2 (y − y0 )2 dy dz (3.11.6) 2Iy Mz 2 (v0 + α z0 ) + α2 y20 − 2α (v0 + α z0 ) z − 2α2 yy0 + α2 z2 + α2 y2 dy dz. 2Iy

The first two terms between the brackets vanish after integration because the origin of our (y, z) axes is the center of gravity, and the fourth term vanishes after integration because our axes are principle axes of inertia. Carrying out the integration, we obtain 1 Sij uh,i uh,j dV 2 V (3.11.7) 1 Mα2 2 2 = dx −Mα (v0 + α z0 ) + z(y + z ) dy dz . 2 Iy

128

Applications

Introducing the shorthand 1 z0 − 2Iy

z(y2 + z2 ) dy dz = c,

(3.11.8)

for the second variation we finally obtain 1 1 2 1 2 2 + α + − Mα v − cMα P2 [u] = dx. EIzv2 S Eα t 0 0 2 2 2

(3.11.9† )

Notice that when the y-axis is an axis of symmetry, c = 0. When the beam is loaded by a shear force through the center of shear, the contributions of the shear stresses (according to Saint-Venant’s theory for the bending of beams) to the elastic energy are 1 Sij uh,i uh,j dV = [S12 (u1,1 u1,2 + u2,1 u2,2 + u3,1 u3,2 ) 2 + S13 (u1,1 u1,3 + u2,1 u2,3 + u3,1 u3,3 )] dV ! ∂ψ0 = τxy (−yv0 + ψ0 α ) −v0 + α + α (y − y0 ) α ∂y " ∂ψ0 + τxz (−yv0 + ψ0 α ) α + v0 − α (z − z0 ) (−α) dV ∂z (3.11.10) We now notice that ψ0 (y, z) = O(b2 ), where b is the height of the beam, so that 3 2 ∂ψ0 bα ψ0 αα =O ∂y 3

(3.11.11)

2 αα (y − y0 ) = O bα .

(3.11.12)

and that

Further, we have the estimates % & 3 2 2 bv bα 0 yv v = O =O 0 0 3 3 2 3 2 ∂ψ0 = O b v0 α = O b α α yv 0 ∂y 3 3 2 3 2 α ψ0 v = O b v0 α = O b α , 0 3 3

(3.11.13)

so that by only taking into account the term αα (y − y0 ) in the first line of (3.11.10), we make a relative error of order b2 / 2 , which is admissible for b2 / 2 1. A similar †

In the literature, the term with c is often missing. However, this is not always admissible.


argument holds for becomes 1 Sij uh,i uh,j 2 = dx = dx

129

terms in the second line of (3.11.10), and our final result

dV [(yτxy + zτxz) αα − αα τxy y0 + τxzα (v0 + α z0 )] dy dz (3.11.14) αα (yτxy + zτxz) dy dz + M α (v0 + α z0 ) dx.

Here we have made use of the fact that τxy dy dz = 0, τxz dy dz = D = M ,

(3.11.15)

where D (x) is the transverse shear force in a cross section. The first term in (3.11.14) can now be rewritten by noticing that (yτxy + zτxz)dy dz " ! 1 ∂ 1 ∂ [(y2 + z2 )τxy ] + [(y2 + z2 )τxz) dy dz = 2 ∂y 2 ∂z (3.11.16) ∂τxy ∂τxz 1 + dy dz − (y2 + z2 ) 2 ∂y ∂z 4 1 ∂σx 1 = dy dz. (y2 + z2 ) (τxy ny + τxznz) ds + (y2 + z2 ) 2 2 ∂x edge

The line integral vanishes because τxy and τxz vanish at the edge, and using (3.11.4) our final result is 1 M z y2 + z2 dy dz (yτxy + zτxz) dy dz = 2 Iy (3.11.17) = M (z0 − c) . The second variation for a beam loaded by a transverse shear force (deadweight load), applied in the center of shear-by-shear stresses according to SaintVenant’s theory, is now given by 1 1 2 1 2 P2 [u] = EIzv2 0 + St α + Eα − Mα v0 2 2 2 0 (3.11.18) − cMα2 − M α (v0 + z0 α ) + M αα (z0 − c) dx = 0

1 1 2 1 2 2 EIzv0 + St α + Eα − (Mα) v0 − c (Mα) α dx. 2 2 2

This expression is also (approximately) valid for distributed loads, provided that they are distributed conforming to the shear stress distribution of Saint-Venant’s theory. When the transverse shear force is not applied in the center of shear, we must add the contribution of this force to the energy (see Figure 3.11.2).

130

Applications z

F

d

*

( y 0, z 0 ) y

Figure 3.11.2

When the force is applied in a point (y0 , z0 + d), the contribution is 1 − Fd α2 . (3.11.19) 2 Because this contribution is negative, we have a destabilizing effect. When the force is applied in (y0 , z0 − d), this will be a stabilizing effect (see Figure 3.11.3). For our further discussion, we shall now assume that the shear force is applied in the center of the shear. The necessary condition for neutral equilibrium is P11 [u, ζ] = EIzv0 η + St α ϕ + Eα ϕ 0 (3.11.20) − (Mα) η − (Mϕ) v0 − c (Mα) ϕ − c (Mϕ) α dx = 0, where η and ϕ are kinematically admissible fields. By integration by parts, we obtain EIzv η − EIzv + (Mα) η + Eα ϕ 0

0

0

0

0

+ St α − Eα − Mv0 − cMα − c (Mα) ϕ0 (3.11.21) EIzv + 0 + (Mα) η + −St α + Eα + Mv0 + c (Mα) + cMα ϕ dx = 0, 0

F

F

destabilizing

F stabilizing Figure 3.11.3

F


131

from which we obtain EIzv 0 + (Mα) = 0

Eα − St α + Mv0 + c (Mα) + cMα = 0.

(3.11.22)

We now consider the following boundary conditions: i) Simply supported end. In this case, the kinematic conditions are v0 = 0,

α = 0.

(3.11.23)

It then follows that the dynamic boundary conditions are EIzv0 = 0,

Eα = 0.

(3.11.24)

ii) Clamped end. Here we have only the kinematic conditions v0 = 0,

v0 = 0,

α = 0,

α = 0.

(3.11.25)

iii) Free end. In this case, we have only dynamic boundary conditions EIzv0 = 0,

EIzv 0 + (Mα) = 0,

Eα = 0,

St α − Eα − Mv0 − cMα − c (Mα) = 0.

(3.11.26)

For narrow cross sections, |ψ0 | = O (ht) (see Figure 3.11.4), and the warping constant is of order || = O(h3 t3 ), which means the warping is not important, so that the elementary theory ( = 0) may be used. This is also true for a T profile (see Figure 3.11.5). However, for other profiles we have different conditions (see Figure 3.11.6).

t

h

Figure 3.11.4

center of shear *

Figure 3.11.5

132

Applications

* shear center

*

e.g.

*

( )

(

b 3t .

(

b4 .

Γ = O b 5t , and St = O

)

Figure 3.11.6

( )

Γ = O b6 ,

St = O

)

Figure 3.11.7

For a massive cross section, we have Figure 3.11.7. For = 0, the order of the system of differential equations (3.11.22) is reduced by two, and therefore the number of boundary conditions at each end of the beam is reduced by one. When = 0, the underlined terms in (3.11.23) to (3.11.26) must be omitted. In the following examples, we shall assume that = 0. Consider a simply supported beam loaded by a constant bending couple M. The differential equations then are EIzv 0 + Mα = 0,

−St α + Mv0 + 2Mcα = 0.

(3.11.27)

By choosing α = A sin

kπx kπx v0 = B sin ,

(3.11.28)

the boundary conditions (3.11.23) and (3.11.24) are satisfied identically. Substitution into the differential equations yield k4 π4 k2 π2 EI B − MA = 0 z 4 2 k2 π2 k2 π2 − 2 MB + 2 (St − 2Mc) A = 0.

(3.11.29)


z

St =

h

1 Gt3h, 3

Iy =

Mcr =

y

σcr = t

2 (1 + v)

Iy =

1 3 th 12

Et3 h

π 6

Et2

π h

1 3 th , 12

133

2 (1 + v) ≅

5π t2 E 8 h

(v = 0.28)

Figure 3.11.8. Example of slender rectangular section.

The condition for a non-trivial solution is M2 =

2Mc k2 π2 1 − . EI S z t 3 St

(3.11.30)

kπ EIzSt .

(3.11.31)

When c = 0, we have M=±

In general, we have 2Mc/St 1, and then kπ kπc M≈± EIzSt 1 ∓

EIz . St

(3.11.32)

For example, if t/h = 0.1, h/ = 0.1, then σcr ≈ 0.002E, which for most construction materials is still within the elastic range. ) Eht3 π πh 1 Mcr = 1− (1 + υ) 24 2 (1 + υ) 5 2 ) E t2 h πh 1 σcr = π 2 1− (1 + υ) h 2 (1 + υ) 5 2 ≈

5π t2 h E 8 h2

for

υ = 0.28

and

h/ 1.

Notice that this is (approximately) the same result as for the rectangular cross section. z

h

2 1 h, c = h 15 5 1 1 Eht 3 , EIy = Eth3 EIz0 = 48 36 z0 =

t

y

3

Eht , St = 24 (1 + v)

EIz0 1 = (1 + v) St 2

Figure 3.11.9. Example of slender triangular section.

134

Applications z

x F l Figure 3.11.10

Let us now consider a beam clamped at one end and loaded by a transverse shear force (applied through the shear center) at the other end (see Figure 3.11.10). We further assume that the cross section is such that = 0, c = 0. In this case, the differential equations are EIzv 0 + [F ( − x) α] = 0,

F ( − x) v0 − St α = 0.

(3.11.33)

The boundary conditions are v0 = 0,

α = 0,

EIzv0 = 0, St α = 0

v0 = 0

at x = 0

EIzv 0 + [F ( − x) α] = 0,

at x =

(3.11.34)

(3.11.35)

Integrating the first equation and taking into account the second condition of (3.11.35), we obtain EIzv 0 + [F ( − x) α] = 0.

(3.11.36)

Integrating once again and taking into account the first and the third conditions of (3.11.35), we obtain EIzv0 + F ( − x) α = 0.

(3.11.37)

Eliminating v0 from (3.11.37) and the second of the equations (3.11.33), we obtain an equation in α, α +

F 2 ( − x)2 α=0 EIzSt

(3.11.38)

to be solved under the conditions α=0

at x = 0,

α = 0 at x = .

Let us now introduce a new variable η −x , η=

(3.11.39)

(3.11.40)


135

the equation becomes then d 2α F 2 4 2 + η α = 0, 2 dη EIzSt

(3.11.41)

d 2α + λη2 α = 0, dη2

(3.11.42)

or

where λ is defined by F 2 2 . (3.11.43) EIzSt The equation must be solved under the conditions dα = 0 at η = 0. (3.11.44) α = 0 at η = 1, dη The solutions to (3.11.42) are Bessel functions, but for our purpose it is more convenient to try a solution of the form λ=

α=

∞

An ηn .

(3.11.45)

n=0

Substitution into the differential equation yields ∞

An (n − 1) ηn−2 + λ

n=0

∞

An ηn+2 = 0.

(3.11.46)

n=0

The coefficient of ηk is Ak+2 (k + 2) (k + 1) + λAk−2 = 0,

(3.11.47)

so that Ak+2 = −

λ Ak−2 . (k + 1) (k + 2)

(3.11.48)

The solution then becomes λ 4 λ 12 λ λ 8 λ λ η + η − η + ··· . (3.11.49) α = A0 1 − 3.4 3.4 7.8 3.4 7.8 11.12 This solution satisfies the boundary condition at η = 0. The value of λ follows from α = 0 for η = 1, i.e., the series between the brackets must vanish for η = 1, which yields λ = 16.104, so the critical load is

(3.11.50)

√

EIzSt . (3.11.51) 2 For a beam with a rectangular cross section (t, h) the critical stress becomes Fcr = 4.013

σcr = 4.013

t2 h . 2 (1 + υ) h2 E

(3.11.52)

As we have seen, the problem in this case can be reduced to a single second-order differential equation. This reduction is always possible when we have a beam with

136

Applications

at least one free edge. In this case, the first of the equations of (3.11.33) can be integrated twice to yield EIzv0 + Mα = const.

(3.11.53)

When the bending moment at x = 0 or x = vanishes, the constant is equal to zero. Solving v0 from this equation (for simplicity, we shall assume that the constant is equal to zero), we find v0 = −

M α, EIz

(3.11.54)

and substitution into the second of the equations of (3.11.33) yields α +

M2 α = 0. EIzSt

(3.11.55)

This equation also holds for a distributed load. An approach to get an approximate solution follows from the fact that (3.11.55) minimizes the functional 1 2 M2 2 (3.11.56) St α − α dx, 2 2EIz 0 as follows from the variation of the functional, which yields M2 St α + St α δα0 − α δα dx = 0. EIz 0

(3.11.57)

Let the boundary conditions be α=0

at x = 0,

α = 0

at x = ,

(3.11.58)

in agreement with (3.11.57). We now assume 1 α = ξ − ξ2 , ξ = x/ , (3.11.59) 2 which satisfies both boundary conditions. Introducing this approximation into the functional (3.11.56) and setting the result equal to zero, we find when M = F ( − x) , 1 1 St 1 2 2 F 2 2 2 2 dξ = 0, (3.11.60) (1 − ξ) − (1 − ξ) ξ − ξ 2 2 2EIz 2 0 which yields 105 F 2 4 = = 17.5, EIzSt 6 so that

(3.11.61)

√

EIzSt , 2 which gives an error of about 4% compared to the exact value. Fcr = 4.18

(3.11.62)

3.12 Buckling of plates loaded in their plane

137

3.12 Buckling of plates loaded in their plane We consider a homogeneous, isotropic elastic flat plate, loaded in its plane. Without going into details, we note that a plate theory can be derived under the following assumptions: i) Normals to the undeformed middle surface remain normal to the deformed middle surface. ii) Changes in length of these normals may be neglected. iii) The state of stress is approximately plane and parallel to the middle surface. Further, we note that the stiffness of the plate in its plane is considerably larger than the bending stiffness, which means that after buckling, the displacements perpendicular to the middle plane are considerably larger than the displacement in the middle plane. We now recall that stability in the fundamental state is governed by the functional 1 1 P[u] = (3.12.1) Sij uh,i uh,j + Eijkl γij γk dV, 2 2 V

where γ is the nonlinear strain tensor. For plates, the non-vanishing components of the strain tensor are 1 γ11 (z) = u,x + w2,x − zw,xx 2 1 γ22 (z) = v,y + w2,y − zw,yy 2 2γ12 (z) = u,y + v,x + w,x w,y − 2zw,xy ,

(3.12.2)

where u(x, y), v(x, y) are the displacements in the mid-plane, w(x, y) is the displacement perpendicular to the mid-plane, and z is the distance to the mid-plane (−h/2 ≤ z ≤ h/2). The contribution of the second term in (3.12.1) to the energy density can now be written as 2 1 E 2 2 Eijkl γij γk = γ (z) + γ22 (z) + 2υγ11 (z)γ22 (z) + 2(1 − υ)γ12 (z) . 2 2 (1 − υ2 ) 11 (3.12.3) Integrating (3.12.3) over the thickness of the plate, we obtain the energy per unit area, Eh 1 2 2 1 2 2 V= u,x + w,x + v,y + w,y 2 (1 − υ2 ) 2 2 1 1 1 v,y + w2,y + (1 − υ) (u,y + v,x + w,x w,y )2 (3.12.4) + 2υ u,x + w2,x 2 2 2 . h2 2 2 2 w + w,yy + 2υw,xx w,yy + 2(1 − υ)w,xy , + 12 ,xx where the last term represents the bending energy.

138

Applications

Assuming that in the fundamental state the only non-vanishing stress components are S11 = Sx ,

S22 = Sy ,

S12 = Sxy ,

the corresponding energy per unit area of the middle plane is 1 1 2 2 Sx w,x + Sy w,y + Sxy w,x w,y h. 2 2 The energy functional for a plate loaded in its plane now becomes Eh 1 2 2 1 2 2 u,x + w,x + v,y + w,y P[u] = 12 (1 − υ2 ) 2 2 1 1 1 v,y + w2,y + (1 − υ) (u,y + v,x + w,x w,y )2 + 2υ u,x + w2,x 2 2 2 . h2 2 2 2 w + w,yy + 2υw,xx w,yy + 2(1 − υ)w,xy dx dy + 12 ,xx 1 Sx w2,x + Sy w2,y + 2Sxy w,x w,y dx dy. + h 2

(3.12.5)

(3.12.6)

We now assume that the fundamental state is linear, i.e., that the stresses are proportional to the loads. This assumption is valid when the rotations are of the same order of magnitude as the strains, which are small. This is the case for a propertysupported plate (no rotations, as in a rigid body). Under these assumptions, we can consider the x-, y-coordinates are the same as the coordinates in the undeformed state. Stability is primarily determined by the second variation, which is given by ! Eh P2 [u] = u2·x + v2,y + 2υu,x v,y + 2(1 − υ) (u,y + v,x )2 2 (1 − υ2 ) h2 2 w,xx + w2,yy + 2υw,xx w,yy + 2(1 − υ)w2,xy (3.12.7) + 12 " 1 − υ2 Sx w2,x + Sy w2,y + 2Sxy w,x w,y dx dy. + E We may now draw some important conclusions from the structure of this second variation: i) The first line in the integrand is positive-definite.† At neutral equilibrium, P2 [u] is semi-positive definite, which implies that the second and the third line together, which only depend on w, become semi-positive-definite, and that the first line must vanish at neutral equilibrium. This means that u1 (x, y) = v1 (x, y) ≡ 0

(3.12.8)

at neutral equilibrium. †

To show this we recall that a2 + b2 + 2υab + 2(1 − υ)c2 = (a + υb)2 + (1 − υ)b2 + 2(1 − υ)c2 > 0 for ∀a, b, c, |υ| < 1, and a, b, c not vanishing simultaneously.


139

ii) Because in (3.12.6) the third and fourth line together are semi-positive-definite and the first two lines together are positive-definite, it follows that in the critical state of neutral equilibrium P[u] > 0, so that for flat plates loaded in their plane, the critical state of neutral equilibrium is stable. This conclusion does not hold for curved plates and shells. Let us now derive the conditions for neutral equilibrium, which follow from ! Eh2 w,xx ζ,xx + w,yy ζ,yy P11 [u, ζ] = 0 = 12 (1 − υ2 ) + υw,xx ζ,yy + υw,yy ζ,xx + 2 (1 − υ) w,xy ζ,xy (3.12.9) " 2 12(1 − υ ) + [Sx w,x ζ,x + Sy w,y ζ,y + Sxy (w,x ζ,y + w,y ζ,x )] dx dy. Eh2 Using the divergence theorem once, with υ as the unit normal vector to the edge, we obtain positive in the outward direction 4 Eh3 [w,xx ζ,x υx + w,yy ζ,y υy + υw,xx ζ,y υy 12 (1 − υ2 ) edge

+ υw,yy ζ,x υx + (1 − υ)w,xy ζ,x υy + (1 − υ)w,xy ζ,y υx ] ds Eh3 − (w,xx + υw,yy ),x ζ,x − (w,yy + υw,xx ),y ζ,y (3.12.10) + 12 (1 − υ2 ) −(1 − υ)w,xyy ζ,x − (1 − υ)w,xyx ζ,y . 12 1 − υ2 + [Sx w,x ζ,x + Sy w,y ζ,y + Sxy (w,x ζ,y + w,y ζ,x )] dx dy = 0. Eh2 Repeated application of the divergence theorem yields 4 Eh3 [(w,xx + υw,yy ) υx + (1 − υ)w,xy υy ] ζ,x 2 12 (1 − υ ) edge

+ [(w,yy + υw,xx ) υy + (1 − υ)w,xy υx ] ζ,y / − (w,xx + υw,yy ),x + (1 − υ)w,xyy υx 0 (3.12.11) + (w,yy + υw,xx ),y + (1 − υ)w,xxy υy ζ 12 1 − υ2 + w + S w + w + S w υ υ ζ dx dy (S ) ) [(S ] x ,x xy ,y x y ,y xy ,x y Eh2 ! 12(1 − υ2 ) Eh3 w − + [Sx w,xx + Sy w,yy + 2Sxy w,xy 12 (1 − υ2 ) Eh " + (Sx,x + Sxy,y ) w,x + (Sy,y + Sxy,x ) w,y ] ζ dx dy = 0, where is the Laplacian operator. The differential equation for neutral equilibrium is now given by 12 1 − υ2 w − (Sx w,xx + Sy w,yy + 2Sxy w,xy − Xw,x − Yw,y ) = 0, (3.12.12) Eh2

140

Applications

where X and Y are the mass forces per unit area. Here we have used the equilibrium equations Sx,x + Sxy,y + X = 0

(3.12.13)

Sxy,y + Sy,y + Y = 0. Using the relations ζ,x = ζ,s tx + ζ,υ υx ,

ζ,y = ζ,s ty + ζ,υ υy ,

(3.12.14)

where t is the unit tangent vector to the edge, we can rewrite the line integral in (3.12.11) to yield 4 (w,xx + υw,yy ) υ2x + 2(1 − υ)w,xy υx υy + (w,yy + υw,xx ) υ2y ζ,υ edge

+ [(w,xx + υw,yy ) υx tx + (w,yy + υw,xx ) υy ty + (1 − υ)w,xy (υy tx + υx ty )] ζ,s (3.12.15) / − [(w,xx + υw,yy ),x + (1 − υ)w,xyy ]υx + [(w,yy + υw,xx ),y + (1 − υ)w,xxy ] υy . 0 12 1 − υ2 − [(Sx w,x + Sxy w,y ) υx + (Sy w,y + Sxy w,x ) υy ] ζ ds = 0. Eh Let us now consider a part of the edge curve (see Figure 3.12.1). It then follows that ty = υx ,

tx = −υy .

Further, we shall make use of the relation s2 s2 F (s)ζ,s ds = F (s)ζ − s1

s1

(3.12.16)

s2

F,s ζ ds,

(3.12.17)

s1

which holds when F (s) is a differentiable function. This implies that the edge curve has no corners in s1 < s < s2 . In the following, we shall assume that the edge curve is a smooth curve. From (3.12.15), we now obtain (w,xx + υw,yy ) υ2x + 2(1 − υ)w,xy υx υy + (w,yy + υw,xx ) υ2y ζ,v edge = 0 (3.12.18) y

t

x Figure 3.12.1


141

! (1 − υ) (w,yy − w,xx ) υx υy + υ2x − υ2y w,xy ,s + (w,xx + w,yy ),x υx + (w,xx + w,yy ),y υy (3.12.19) " 12 1 − υ 2 − =0 [(Sx w,x + Sxy w,y ) υx + (Sy w,y + Sxy w,x ) υy ] ζ Eh edge s2 (1 − υ) (w,yy − w,xx ) υx υy + υ2x − υ2y w,xy ζ s = 0. 1

(3.12.20)

First, we notice that the left-hand side of (3.12.20) vanishes identically along a closed curve. Further, it vanishes along parts of the edge that are supported such that w = 0, and it also vanishes along a free edge provided that the edge in the parts of the edge curve adjacent to the free part is properly supported. In the following discussion, we shall assume that (3.12.20) is satisfied. Let us now consider the various boundary conditions. i) The free edge. Because there are no restrictions imposed on ζ and ζ,υ , their coefficients must vanish, i.e., (w,xx + υw,yy ) υ2x + 2(1 − υ)w,xy υx υy + (w,yy + υw,xx ) υ2y = 0,

(3.12.21)

which means that the bending moment along the edge vanishes, and (1 − υ) (w,yy − w,xx ) υx υy + υ2x − υ2y w,xy ,s + (w,xx − w,yy ),x υx + (w,xx + w,yy ),y υy −

(3.12.22)

12(1 − υ ) [(Sx w,x + Sxy w,y ) υx + (Sy w,y + Sxy w,x ) υy ] = 0, Eh 2

which means that the reduced transverse shear force vanishes at the edge. ii) The simply supported edge. Here ζ ≡ 0 along the edge, so (3.12.19) is satisfied, so that we have (w,xx + υw,yy )υ2x + 2(1 − υ)w,xy υx υy + (w,yy + υw,xx )υ2y = 0

(3.12.23)

in addition to the kinematic condition w = 0. iii) The clamped edge. Here we have the kinematic conditions w=

∂w = 0 at the edge ∂υ

(3.12.24)

so that ζ = ζ,υ = 0 along the edge, which implies that the conditions (3.12.18) and (3.12.19) are satisfied automatically. Let us now apply our result to a square plate, simply supported at its edges, loaded in compression by forces σ per unit length at its edges at x = 0 and x = a (see Figure 3.12.2).

142

Applications y y=b σ

σ x x=a Figure 3.12.2

In this case, the only non-vanishing stress component in the fundamental state is Sx = −σ. Further, there are no mass forces, so the differential equation for neutral equilibrium is given by Eh3 w + hσw,xx = 0. 12(1 − υ2 )

(3.12.25)

In this case, we have υx υx υy υy

=1 = −1 =1 = −1

υy υy υx υx

=0 =0 =0 =0

at at at at

x=a x=0 y=b y=0

so that the boundary conditions x = a are w,xx + υw,yy = 0,

w = 0.

(3.12.26)

Because w ≡ 0 at x = a, w,yy = 0, so that we can rewrite the boundary conditions as w,xx = 0 w = 0 at x = a,

x = 0.

(3.12.27)

w,yy = 0 w = 0

y = b.

(3.12.28)

Similarly, at y = 0,

Because both the differential equation and the boundary conditions contain only even derivatives, we can attempt a solution of the form w(x, y) = Cmn sin

mπy mπx sin , a b

(3.12.29)

which satisfies all the boundary conditions.† Substitution into the differential equation yields 2 2 2 Eh3 mπ n2 π2 m2 π2 + − hσ 2 = 0. (3.12.30) 2 2 2 12(1 − υ ) a b a †

Here m, n are integers.


143

For given values of m and n, (3.12.20) is satisfied for π2 Eh2 σ= 12(1 − υ2 )b2

2 1 2b 2a m +n . a b m2

(3.12.31)

The critical load is obtained for the minimum value of the right-hand term, which for fixed m has a minimum value for n = 1, so that we must minimize the expression 2 π2 Eh2 a2 2b σ= m (3.12.32) + 2 + 12(1 − υ2 )b2 a2 b2 m2 with respect to m. To this end, we assume that m is continuous. Differentiating the expression between the brackets with respect to m2 and setting the result equal to zero, we find m=

a . b

(3.12.33)

This is the exact result when a/b is an integer. In other cases, m is the integer that is closest to a/b. Hence, it follows that σcr ≥

π2 Eh2 3(1 − υ 2 )b2

(3.12.34)

where the equality sign holds when a/ b is an integer. For sufficiently large values of a/b, (3.12.34) is a good approximation,† which means that for a sufficiently long plate we can always approximate a/b by an integer. The critical load for other boundary conditions is often expressed as σcr =

π2 Eh2 k, 3(1 − υ2 )

(3.12.35)

i.e., it is expressed as a multiple of the critical load for a simply supported load. When a/b < 1, the smallest wave number is m = 1, so we can write the critical load in the form a2 π2 Eh2 a4 σcr = 1+2 2 + 4 . (3.12.36) 12(1 − υ2 )a2 b b In the limiting case a/b → 0, we have σcr =

π2 Eh2 , 12(1 − υ2 )a2

(3.12.37)

which also follows from the result for the Euler bar when we take into account that there is no anti-elastic bending, which accounts for the factor (1 − υ2 )−1 . Let us now consider the solution of our buckling problem in more detail. We have attempted a solution in the form (3.12.29), which means that we have silently †

E.g., a/b = 3.5 yields for the factor between the brackets in (3.12.32), 4.072 for m = 4 and 4.095 when m = 3, which means an error of 2.5% when (3.12.34) is used.

144

Applications

assumed that the infinite series w(x, y) =

∞ ∞ m=1 n=1

Cmn sin

nπy mπx sin a b

(3.12.38)

can be differentiated term-wise, which is only admissible when the differentiated series is uniformly convergent for 0 ≤ x ≤ a, 0 ≤ y ≤ b. We shall now show that for a plate with simply supported edges, this series can be differentiated term-wise. To do this, we apply a Fourier transform to (3.12.25), which yields a b Eh3 mπx nπy w + hσw sin dx dy = 0. (3.12.39) ,xx sin 2) 12(1 − υ a b 0 0 Let us first consider the term a a mπx mπ mπx mπx a dx = w,xxx sin cos dx w,xxxx sin w,xxx − a a 0 a a 0 0 (3.12.40) a mπ m2 π2 mπx mπx a = −w,xx w,xx 2 sin cos dx. − a a 0 a a 0 Here the stock term vanishes because w,xx = 0 at x = a and x = 0. This stock term does not vanish in the case of a clamped edge. Continuing our partial integration, we obtain a a m2 π2 mπx mπx w,xxxx sin dx = −w,x 2 sin a a a 0 0 a m3 π3 mπx dx (3.12.41) + w,x 3 cos a a 0 a m3 π3 mπx m4 π4 a mπx = w 3 cos dx. + w sin a a 0 a4 a 0 Here the stock term vanishes for both simply supported edges and clamped edges because then w = 0 at x = 0 and x = a. Similarly, we obtain for simply supported edges a m2 π2 a mπx mπx dx = − 2 dx w,xx sin w sin a a a 0 0 (3.12.42) b n4 π4 b nπy nπy dy = 4 dy w,yyyy sin w sin a b a 0 0 a

b

nπy mπx sin dx dy a a 0 0 m2 n2 π4 a b mπx nπy = w sin sin dx dy. 2 2 ab a b 0 0 w,xx w,yy sin

Our final result for the Fourier transform of the differential equation can now be written as 2 2 2 Eh3 mπ n2 π2 m2 π2 + 2 − hσ 2 (3.12.43) Wmn = 0 12(1 − υ2 ) a2 b a


where Wmn =

a

b

w(x, y) sin 0

0

mπx mπy sin dx dy a b

145

(3.12.44)

is the Fourier transform of w(x, y). The equation (3.12.43) is valid for all integer values of m and n, and because Wmn ≡ 0 the expression between the brackets must vanish, which leads to our earlier result (3.12.31). Introducing the series (3.12.38) into (3.12.44), we find the following relation between Cmn and the Fourier transform of w(x, y): Wmn =

1 Cmn ab. 4

(3.12.45)

To obtain this result, we have interchanged integration and summation, which is admissible because the series (3.12.38) is a uniformly convergent series. We see that only one term of the series is left, which justifies our earlier approach. Let us now consider the influence of a free edge at y = 0 for the case that a/b 1, when the other edges are simply supported (see Figure 3.12.3). In this case, we shall only apply a Fourier transform in the x direction because w = w = 0 at x = 0 and x = a, but w = 0 at y = 0. With a mπ dx, (3.12.46) Wm(y) = w(x, y) sin a 0 we find Eh3 12(1 − υ2 )

m2 π2 d4 Wm m2 π2 d2 Wm m4 π4 − hσ − 2 + W Wm = 0, m dy4 a2 dy 2 a4 a2

where we have used the relation a n dn Wm d w mπx dx = sin . n a dyn 0 dy

(3.12.47)

(3.12.48)

As discussed previously for a short plate (a/b 1), the wave number is m = 1. Introducing the notation a y = η, d () /dη = ()· , W1 = W, (3.12.49) π y

σ

s⋅s

s⋅s

s⋅s

σ

x free edge Figure 3.12.3

146

Applications

we may rewrite (3.12.47) in the form π2 Eh2 (W···· − 2W·· + W) − σW = 0. 12(1 − υ2 )a2

(3.12.50)

Further, it is convenient to write σ=λ

π2 Eh2 12(1 − υ2 )a2

(3.12.51)

so that the differential equation becomes W···· − 2W·· + W(1 − λ) = 0.

(3.12.52)

The boundary conditions are w,yy + υw,xx = 0, w = 0,

w,yyy + (2 − υ)w,xyy = 0 at y = 0

w,yy = 0 at y = b,

(3.12.53)

and w = 0,

w,xx = 0

at x = 0,

x = a.

(3.12.54)

We have already used these last conditions in the derivation of the differential equation for the Fourier transform of w(x, y). Writing πx w(x, y) = Y(y) sin , (3.12.55) a we obtain from (3.12.47) a W(y) = Y(y). (3.12.56) 2 Because the boundary conditions are homogeneous, we can now express them as W·· − υW = 0,

W··· − (2 − υ)W· = 0

at η = 0

(3.12.57) b W·· − υW = 0 at η = π . a The boundary conditions at x = 0, x = a are satisfied by our choice (3.12.55). To solve (3.12.52), we set W = 0,

W = Ceµη.

(3.12.58)

Introducing this expression into the differential equation, we obtain the characteristic equation

from which

µ4 − 2µ2 + 1 − λ = 0,

(3.12.59)

√ 1/2 µ=± 1± λ .

(3.12.60)

These roots are real when 0 ≤ λ ≤ 1. Because we want solutions that decay with increasing distance from y = 0, we must only consider the solution √ √ √ √ W = C1 e− 1+ λη + C2 e− 1− λη. (3.12.61)


Substitution into the boundary conditions at η = 0 yields √ √ 1 + λ − υ C1 + 1 − λ − υ C2 = 0 √ 1/2 √ √ 1/2 √ 1 − υ − λ + C2 1 − λ 1 − υ + λ = 0. C1 1 + λ

147

(3.12.62)

The condition for a non-trivial solution is now √ √ √ √ f (λ, υ) ≡ (1 − λ)1/2 (1 − υ + λ)2 − (1 + λ)1/2 (1 − υ − λ)2 = 0. (3.12.63) √ For λ = 0, we have f (λ, υ) = 0, and for λ = 1 we have f (λ, υ) = − 2υ2 . To get a better picture of the curve f (λ, υ) = 0, we also determine the derivative √ √ √ √ df 1 √ = 2(1 − λ)1/2 (1 − υ + λ) − (1 − λ)−1/2 (1 − υ + λ)2 2 d λ (3.12.64) √ √ √ √ 1/2 1 + 2(1 + λ) (1 − υ − λ) − (1 + λ)−1/2 (1 − υ − λ)2 . 2 √ √ √ √ At λ = 1, df /d λ = −∞, and for λ = 0 we have df /d λ = (1 − υ)(3 + υ), so that we get the picture shown in Figure 3.12.4. √ It follows that for υ > 0, the root λ will be close to 1, so we set √ λ = 1 − ε2 . (3.12.65) Substitution into (3.12.63) yields ε(2 − υ − ε2 )2 − (2 + ε2 )1/2 (ε2 − υ)2 , from which

√

ε=

2υ2 1 + O(ε2 ) , 2 (2 − υ)

(3.12.66)

so that λcr ≈ 1 −

4υ4 . (2 − υ)4

(3.12.67)

The largest reduction is obtained for υ = 1/2, where ε ≈ 0.157 and λcr =

77 = 0.9506. 81

(3.12.68)

f ( λ,υ )

υ=0 υ>0 1

Figure 3.12.4

λ

148

Applications y b/2

σ

x=a

σ

x

−b/2 Figure 3.12.5

To a second-order approximation, ε is given by √ 1/2 2υ3 (8 − υ) 2(2 − υ)2 1+ ε= , υ(8 − υ) (2 − υ)4

(3.12.69)

which for υ = 1/2 yields ε = 0.1444, and λcr = 0.9585. The “exact” root is ε = 0.1486, so that λcr = 0.9563

(exact),

(3.12.70)

which means a reduction of the critical load of about 5% compared to the plate that is simply supported at y = 0. Let us now consider a rectangular plate with clamped edges at y = ±b/2, and simply supported edges at x = 0 and x = a, loaded in compression by forces σh per unit length at x = 0 and x = a (see Figure 3.12.5). However, the clamped edges are allowed to slide in the x-direction. We consider the case that the plate is sufficiently long, and that 2 is the wavelength of the deformation pattern in the buckled state. The differential equation for neutral equilibrium is Eh3 w + σhw,xx = 0, 12(1 − υ2 )

(3.12.71)

and the boundary conditions are w = 0,

w,xx = 0 at x = 0,

w = 0,

w,y = 0

x = m (≤ a)

at y = ± b/2.

(3.12.72)

Because the lowest bifurcation load will occur when m = 1, we try a solution of the form w(x, y) = W(y) sin

πx .

Substitution into the differential equation yields 4 Eh3 d W 2π2 d2 W π4 π2 − + W − σh W = 0. 12(1 − υ2 ) dy 4 2 dy 2 4 2

(3.12.73)

(3.12.74)


149

It is now convenient to write σ=λ

π2 Eh2 , 3(1 − υ2 )b2

(3.12.75)

i.e., the load is expressed as a multiple of the critical load for a sufficiently long plate, simply supported at its edges. The differential equation can now be written as 4 d4 W 2π2 d2 W π 4π4 λ − 2 + − 2 2 W = 0. (3.12.76) dy 4 dy 2 4 b The boundary conditions at x = 0 and x = are satisfied automatically by our choice for w(x, y), and the conditions at y = ± b/2 read dW = 0 at y = ± b/2. (3.12.77) W = 0, dy To solve (3.12.76), we set πy

W = Ce µ .

(3.12.78)

Introduction of this expression into the differential equation yields the characteristic equation µ4 − 2µ2 + 1 − 4λ from which

) µ=± 1±2

2 = 0, b2

√ λ. b

(3.12.79)

(3.12.80)

As the plate with two clamped edges has a larger stiffness than the simply supported √ plate, λ > 1. This means that for 2 λ/b > 1, there will be two imaginary roots. Hence, we write √ 1/2 µ1,2 = ± α α = 1 + 2 λ b (3.12.81) 1/2 √ λ−1 . µ3,4 = ± iβ β = 2 b Because the construction and its loading are symmetric with respect to the xaxis, we can split the solution into a part that is symmetric with respect to y = 0 and an anti-symmetric part. Guided by the result for the simply supported plate, we expect the lowest bifurcation load for the symmetric solution; thus we consider the solution W(y) = C1 cosh απy/ + C2 cos βπy/ .

(3.12.82)

Substitution of the boundary conditions (3.12.77) yields πb πb + C2 cos β =0 2 2 πb πb αC1 sinh α − βC2 sin β = 0. 2 2

C1 cosh α

(3.12.83)

150

Applications

The condition for a non-trivial solution is −β cosh α

πb πb πb πb sin β − α sinh α cos β = 0, 2 2 2 2

or rewritten, β tan β

πb πb + α tanh α = 0. 2 2

(3.12.84)

This is an equation for λ with b/ as a parameter. Minimizing with respect to b/ yields (numerical calculations) ( /b)minimizing = 0.66,

λcr = 1.7425.†

(3.12.85)

To avoid the rather tedious calculations involved with the solution of the transcendental equation (3.12.83), we now try an approximate solution to our problem. Introducing a dimensionless coordinate η defined by y = ηb,

−

1 1 ≤ η≤ , 2 2

(3.12.86)

we assume the following symmetric polynomial for W, W(η) = 1 − 8η2 + 16η4 ,

(3.12.87)

which satisfies the boundary conditions W = dW/dη = 0

for

η = ± 1/2.

Because u and v are zero at bifurcation, it follows from (3.12.7) that we can write P2 [u1 ] =

Eh3 24(1 − υ2 )

12(1 − υ2 ) σ 2 w dx dy. (w)2 − 2 (1 − υ) w,xx w,yy − w2,xy − h2 E ,x (3.12.88)

The second variation is semi-positive definite for the exact solution. For the exact buckling load and an assumed (approximate) displacement, the field P2 [u] > 0, so that the application of Rayleigh’s method (also called Rayleigh-Ritz method) where we put P2 [u] = 0 for an assumed displacement field yields an upper bound for the critical load. For the evaluation of (3.12.88), it will be convenient to make use of the following property: w,xx w,yy − w2,xy dx dy = 0 (3.12.89) S

for a domain bounded by straight lines (polygonal edge curve) with w = const. at the boundary ∂S, and ∂w/∂υ = 0 in the corner points at the boundary where υ is the unit normal to ∂S, positive in the outward direction. To show this property, we apply †

√ This result is valid for /b ≥ 1/2 λcr = 0.37878.


151

the divergence theorem to (3.12.89), which yields

w,xx w,yy − w2,xy dx dy = 0

S

(w,xx w,y υy − w,xy w,y υx ) ds −

=

(w,xxy w,y − w,xyx w,y )dx dy S

∂S

−w,y (w,xx tx + w,xy ty ) ds

=

(3.12.90)

∂S

−w,y (w,xx dx + w,xy dy) = −

= ∂S

w,y d (w,x ) ∂S

(w,υ υy + w,s ty ) d (w,υ υx + w,s tx ) .

=− ∂S

So far, our result is fully general. When w is constant along ∂S, we have w,s = 0 along ∂S, and along a straight line υx and υy are constant, so that for a polygonal edge curve where w = const., we can write (w,xx w,yy −

w2,xy )dx dy

si+1 n 1 2 υx υy w,υ =− 2 i=1

S

(3.12.91)

si

for a polygonal edge curve with n corner points, where si denotes the ith corner point and sn+1 ≡ s0 . When w,υ = 0, in the corner points we obtain the result (3.12.89). Notice that this result is valid for ∀w w ∈ C3 , w = 0 on ∂S, ∂w/∂υ = 0 in the corner points on ∂S, where ∂S is a polygonal curve. For simply supported and clamped straight edges, we have w = 0, ∂w/∂υ = 0 in the corner points, and thus the result (3.12.89) applies,† so that for these plates we can write 2 12 1 − υ σ Eh3 P2 [u1 ] = w2 dx dy. (3.12.92) (w)2 − 24 (1 − υ2 ) h2 E ,x Using the results 1

1

2 b

2 2

(w) dx dy = 0−1b 2

0−1 2

128 = 3 b †

2 2 πx π2 1 dxdη b − 2 1 − 8η2 + 16η4 + 2 −16 + 108η2 sin b

4 64 128 2 + C+ C , 5 105 315

(3.12.93)

Notice that the limiting case that n → ∞ for a simply supported plate bounded by a regular polygonal curve with n corner points is not a simply supported but a clamped circular plate, since now ∂w/∂υ = 0 along the edge.

152

Applications

where C=

π2 b2 , 16 2

(3.12.94)

and 1

1

2 b

2 w2,x dx dy =

0−1b

b 0

2

=

2 πx π2 1 − 8η2 + 16η4 dx dη cos2 2

1 2

64 16C , 315 b

(3.12.95)

we obtain from the condition P2 [u1 ] = 0, 64 128 2 4 + C + C 128 Eh2 5 105 315 . σ= 2 2 64 12(1 − υ )b · 16C 315 Minimizing with respect to C, we obtain ) Cmin =

63 = 1.403 32

(3.12.96)

(3.12.97)

and from (3.12.94), ( /b)min = 0.663.

(3.12.98)

The critical load is now σcr =

π2 Eh2 1.7454, 3(1 − υ2 )b2

(3.12.99)

which is about 0.2% higher than the exact value. As a last example, we shall treat a rectangular plate, simply supported at its edges and loaded in shear by shear forces τh per unit length along its edges (see Figure 3.12.6). From (3.12.12), the governing differential equation is now Eh3 w − 2τhw,xy = 0. 12(1 − υ2 )

(3.12.100)

y y=b

τ

τ

τ

τ

Figure 3.12.6

x=a

x


153

y

x Figure 3.12.7

Notice that the differential equation now contains even and odd derivatives with respect to x and y. In the following, we shall assume that the plate is infinitely long, so that we do not take into account the boundary conditions on the vertical ends. The boundary conditions at y = 0, y = b read w = 0,

w,yy = 0

at y = 0,

y = b.

(3.12.101)

The exact solution to this problem was given by Southwell,† and his result was τcr = 5.35

π2 Eh2 . 12(1 − υ2 )b2

(3.12.102)

However, the numerical factor is not very accurate. Later, we shall give an upper bound that is slightly smaller. The exact solution of this problem requires rather tedious calculations, and therefore we shall try to construct an approximate solution. We try a solution of the form w(x, y) = f sin

π πy (x − my) sin , b

(3.12.103)

which satisfies the boundary condition w = 0 at y = 0, y = b. The deflection vanishes at the nodal lines x − my = k (k = 0, ±1, ±2, . . . ).

y

x x=l

x = 2l

Figure 3.12.8 †

Cf. R.V. Southwell, Proc. Roy. Soc., London, Series A, 105, 582.

154

Applications

The energy functional is b ! 0 0

" Eh3 2 2 (w) − 2(1 − υ) w,xx w,yy − w,xy + τhw,x w,y dx dy, 12(1 − υ2 )

(3.12.104) where the second term vanishes because of (3.12.87). Applying the Rayleigh-Ritz method, we obtain 2 b2 π2 Eh2 1 2 1 2 2 τ= m +1 , (3.12.105) + + 6m + 24(1 − υ2 )b2 m 2 m b2 m where m and b/ are still unknown parameters. Minimizing with respect to (b/ 2 ), we obtain 1 2 1 4 (m + 1)2 − = 0, m m b4

(3.12.106)

2 = m2 + 1. b2

(3.12.107)

which yields

Substitution into (3.12.104) yields τ=

π2 Eh2 24(1 − υ2 )b2

8m +

4 . m

(3.12.108)

Minimizing with respect to m, we obtain m=

1 2, 2

(3.12.109)

so that the (upper bound for the) critical load is given by τcr =

π2 Eh3 √ 2, 3(1 − υ2 )b2

(3.12.110)

which means a factor 1.414 instead of 1.3337 (Southwell). This result is a rather crude approximation, and is due to the fact that the boundary condition w,yy = 0 is not satisfied at y = 0, y = b. Let us now analyze the consequences of our assumption for the displacement field shown in Figure 3.12.9. y

w=0

A

x w=0 Figure 3.12.9


155

Because w = 0 along the edge and along the nodal line, we must have w,x = w,y = 0 in A but not w,yy = 0. To satisfy this condition, we try a form for w(x, y) that is slightly more general than (3.12.102): π w(x, y) = sin (x − ϕ(y))W(y), W(0) = W(b) = 0. (3.12.111) We then obtain π π π w,y = − W(y)ϕ (y) cos [x − ϕ(y)] + W (y) sin [x − ϕ(y)] w,yy

π2 2 = W(y) − 2 W(y)ϕ (y) sin [x − ϕ(y)] π − [2W (y)ϕ (y) + W(y)ϕ (y)] cos [x − ϕ(y)]

π π w,yy (x, 0) = W (0) sin[x − ϕ(0)] − 2 W (0)ϕ (0) cos [x − ϕ(0)].

(3.12.112)

(3.12.113)

(3.12.114)

Because w,y (x, 0) is free, W (0) = 0, so that the conditions for w,yy (x, 0) = 0 are W (0) = 0,

ϕ (0),

(3.12.115)

which means that the nodal line x = ϕ(y) is perpendicular to the edge. These conditions are satisfied by choosing π πy w(x, y) = f sin (x − ϕ(y)) sin , (3.12.116) b where ϕ(y) =

πy mb 1 − cos . π b

(3.12.117)

With this choice, the result is τcr = 1.352

π2 Eh2 , 3(1 − υ2 )

(3.12.118)

which is already very close to the exact result. Let us now finally consider a more exact approach. We try a solution of the form w(x, y) =

∞

Wk(x) sin

k=1

Figure 3.12.10

kπy , b

(3.12.119)

156

Applications

which satisfies the boundary conditions at y = 0 and y = b term-wise. Furthermore, this function must be periodic in x direction, Wk(x + 2 ) = Wk(x),

(3.12.120)

where is the half wavelength. Hence, we assume w(x, y) =

∞ kπy πx πx + bk sin sin . ak cos b

(3.12.121)

k=1

Substitution of this expression into (3.12.100) yields 2 ∞ 2 Eh3 π πy k2 π2 πx πx + b sin a + cos sin k k 12(1 − υ2 ) 2 b2 k=1

∞ kπy kπ2 πx πx − 2τh −ak sin + bk cos cos = 0, b b

(3.12.122)

k=1

where we have assumed that the interchanging of summation and differentiation is admissible. Because this expression must hold for all x and y, the coefficients of cos πx/ and sin πx/ must vanish, which with b/ = µ yields ∞ ∞ Eh2 π2 kπy kπy 2 2 2 − 2τµ =0 (µ + k ) ak sin kbk cos 12(1 − υ2 )b2 b b k=1 k=1 (3.12.123) ∞ ∞ Eh2 π2 kπy kπy 2 2 2 + 2τµ = 0. (µ + k ) bk sin kak cos 12(1 − υ2 )b2 b b k=1

k=1

Introducing the notation τ=λ

π2 Eh2 , 3(1 − υ2 )b2

(3.12.124)

we can rewrite these equations in the form ∞ k=1 ∞ k=1

∞

(µ2 + k2 )2 ak sin

kπy kπy − 8µλ =0 kbk cos b b

kπy + 8µλ (µ + k ) bk sin b 2

2 2

k=1 ∞ k=1

kπy = 0. kak cos b

(3.12.125)

Assuming that the left-hand members can be written as a sine series, we multiply both sides of these equations by sin j πy/b and integrate (term-wise) from y = 0 to y = b. With b sin 0

kπy 1 j πy sin dy = bδjk b b 2

(3.12.126)


b sin 0

kπy [1 − (−1)j +k] j πy cos dy = j b b b (j 2 − k2 ) π

157

(= 0 for j = k) ,

(3.12.127)

we obtain ∞

(µ2 + j 2 )2 aj −

kj 32 1 − (−1)k+j bk = 0 λµ π 2 ( j 2 − k2 ) k=1

32 (µ + j )bj + λµ π 2

2

∞ k=1

kj 1 − (−1)k+j ak = 0. 2 2 2 (j − k )

(3.12.128)

There are two infinite sets of linear algebraic equations for the unknowns ak, bk (k = 1, 2, . . .). The condition for a non-trivial solution is that the determinant of the matrix of coefficients vanishes, i.e., with 32 λµ = ρ π a1 2 (µ + 1)2 2ρ 3 0 4ρ 15 .. .

b2 2ρ 3 2 (µ + 1)2 6ρ − 5 0 .. .

(3.12.129)

a3

b4

0 6ρ 5 (µ2 + 9)2 12ρ 7 .. .

4ρ 15 0 12ρ 7 2 (µ + 1)2 .. .

···

· · · · · · · · · . · · ·

(3.12.130)

Taking into account two terms, we obtain (µ2 + 1)2 (µ2 + 4)2 −

4ρ2 = 0, 9

(3.12.131)

which yields (µ2 + 1)(µ2 + 4) = so that λ=

64µλ 2ρ = , 3 3π

3π 4 µ3 + 5µ + . 64 µ

(3.12.132)

Minimizing with respect to µ yields 5 4 µ4 + µ2 = = 0, 3 3 so that

√ −5 + 73 µ = ≈ 0.59 b 2

(3.12.133)

158

Applications

and λcr = 1.393

(“exact” λ = 1.336).

(3.12.134)

Taking into account three terms, we find λcr = 1.339,

(3.12.135)

λcr = 1.336,

(3.12.136)

and with four terms,

which shows that this process converges rapidly with the exact result. 3.13 Post-buckling behavior of plates loaded in their plane For a square plate loaded by compressive forces σ per unit length in the x-direction, the energy functional reads ! Eh3 P [u] = (w)2 − 2(1 − υ) w,xx w,yy − w2,xy 2 12(1 − υ ) 1 2 2 1 2 2 Eh u,x + w,x + v,y + w,y + 2(1 − υ2 ) 2 2 (3.13.1) 1 2 1 2 + 2υ u,x + w,x v,y + w,y 2 2 " 1 1 2 2 + (1 − υ)(u,y + v,x + w,x w,y ) − σhw,x dx dy. 2 2 The second variation is given by ! Eh3 [(w)2 − 2(1 − υ) w,xx w,yy − w2,xy X] P[u] = 2 12(1 − υ ) "(3.13.2) 1 Eh 1 2 2 2 2 − dx dy, u (1 − υ)(u σhw + + v + 2υu v + + v ) ,x ,y ,y ,x ,y ,x 2(1 − υ2 ) ,x 2 2 and the third and fourth variations are given by Eh 2 P3 [u] = ) u,x w2,x + v,v w2,y + υ(u,x w2,y + W,x 2 2(1 − υ ) + (1 − υ)(u,y + v,x )w,x w,y dx dy and P4 [u] =

Eh 8(1 − υ2 )

2 2 w,x + w2·y dx dy,

(3.13.3)

(3.13.4)

respectively. The displacement field for small but finite deflections from the fundamental state is written as u = au1 + u,

(3.13.5)

3.13 Post-buckling behavior of plates loaded in their plane

159

where u is orthogonal to the buckling mode. The equations for u are obtained by minimizing the functional P2 [u] + a(λ − 1)P11 [u1 , u] + a2 P21 [u1 , u],

(3.13.6)

where λ = σ/σcr (see (2.4.11). The term [u1 , u] = − P11

Eh2 4(1 − υ2 )

w1,x w,x dx dy

vanishes due to the orthogonality condition Eh P21 [u1 , u] = [(u,x + υv,y ) w21,x + (v,y + υu,x ) w21,y 2(1 − υ2 ) + (1 − υ) (u,y + v,x ) w1,x w1,y ] dx dy,

(3.13.7)

(3.13.8)

and P2 [u] follows from (3.13.2) by replacing u, v, w by u, v, w. We now must minimize (3.13.6) with respect to u, i.e., Min P2 [u] + a2 P21 [u1 , u] .

w.r.t.u

(3.13.9)

Because P21 [u1 , u] does not contain w, we must minimize P2 [u] with respect to w. P2 [u] is positive-definite, and its minimum with respect to w is obtained when the first line in (3.13.2), which is positive-definite, vanishes, which yields w ≡ 0.

(3.13.10)

P2 [u] is now given by

1 2 2 2 u,x + v,y + 2υu,x v,y + (1 − υ) (u,y + v,x ) dx dy. 2 (3.13.11) The necessary condition for (3.13.9) to be a minimum is that the variation of this functional with respect to u vanishes, which yields Eh P2 [u] = 2 (1 − υ2 )

1

Eh 1 − υ2

2 b ! 1 u,x ξ,x + v,y η,y + υu,x η,y + υv,y ξ,x + (1 − υ) (u,y + v,x ) (ξ,y + η,x ) 2 0−1b 2

" (3.13.12) a2 2 2 + (ξ,x + υη,y ) w1,x + (η,y + υξ,x )w1,y + (1 − υ) (ξ,y + η,x ) w1,x w1,y dx dy = 0. 2 Integration by parts yields b/2 b/2 2 2 a a u,x + υv,y + v,y + υu,x + dx w21,x + υw21,y ξ dy + w21,y + υw21,x η 2 2 −b/2 0 −b/2 0 b/2 b/2 (1 − υ) (1 − υ) 2 2 + u,y + v,x + a w1,x w1,y ξ dx + u,y + v,x + a w1,x w1,y η dy 2 2 0 −b/2 −b/2 0

160

Applications

b/2 ! 1 1 − u,xx + (1 − υ) u,yy + (1 + υ) v,xy 2 2 0 −b/2

" a2 2 w1,x + w21,y ,x + (1 − υ) (w1,x w1,yy − w1,y w1,xy ) ξ 2 ! 1 1 + v,yy + (1 − υ) v,xx + (1 + υ) u,xy 2 2 " a2 2 + w1,x + w21,y ,y + (1 − υ) w1,y w1,xx − w1,x w1,xy η dx dy = 0, 2

(3.13.13)

+

from which we obtain 1 1 (1 − υ) u,yy + (1 + υ)v,xy 2 2 a2 2 w1,x + w21,y ,x + (1 − υ) (w1,x w1,yy − w1,y w1,xy ) =− 2 1 1 v,yy + (1 − υ)v,xx + (1 + υ)u,xy 2 2 a2 2 w1,x + w21,y ,y + (1 − υ) (w1,y w1,xx − w1,x w1,xy ) . =− 2 u,xx +

(3.13.14)

We shall now discuss various boundary conditions. First, notice that for the determination of the buckling load we must only deal with w, whereas for the postbuckling only in-plane quantities (u, v) must be dealt with. We shall now adapt this theory to the problem of a plate with many fields, loaded in compression by forces σh per unit length. The plate is simply supported at x = ± /2, and at y = ±((2k + 1)/2)b (k = 0, 1, 2, . . .) in such a way that in-plane displacements are free (see Figure 3.13.1). We shall now consider the field {x, y ||x| ≤ /2, |y| ≤ b/2}. The edges y = ±b/2 remain straight (due to symmetry), so that v = const. at y = ± b/2.

(3.13.15)

y

σ

σ

b

x

l

Figure 3.13.1


161

y

σ

b

x σ l Figure 3.13.2

Further, we shall assume that at x = ± /2 the supports are such that u = const. For the rectangular plate simply supported at its edges, the buckling mode is (see (3.12.29) πy πx w1 = cos cos ≡ cos λx cos µy. (3.13.16) The equations for u and v now become 1 1 u,xx + (1 − υ)u,yy + (1 + υ)v,xy 2 2 a2 2 2 = − λ[(λ − υµ ) sin 2λx + (λ2 + µ2 ) sin 2λx cos 2µy], 4 (3.13.17) 1 1 v,yy + (1 − υ)v,xx + (1 + υ)u,xy 2 2 a2 2 2 = − µ[(µ − υλ ) sin 2µy + (λ2 + µ2 ) sin 2µy cos 2λx]. 4 By inspection, we notice that particular solutions of (3.13.17) are of the form upart = A sin 2λx + B sin 2λx cos 2µy vpart = C sin 2µy + D sin 2µy cos 2λx.

(3.13.18)

Substitution into (3.13.17) yields 4λ2 A ≡

a2 λ(λ2 − υµ2 ) 4

[4λ2 + 2(1 − υ)µ2 ]B + 2(1 + υ)λµD ≡

a2 λ(λ2 + µ2 ) 4

a2 2(1 + υ)λµB + [4µ + 2(1 − υ)λ ]D ≡ µ(λ2 + µ2 ) 4 a2 2 2 2 4µ C = µ(µ − υλ ), 4 2

(3.13.19)

2

from which a2 λ2 − υµ2 , 16 λ a2 λ, B= 16 A=

a2 µ2 − υλ2 , 16 µ a2 D= µ. 16

C=

(3.13.20)

162

Applications

The particular solutions are now given by a2 [(λ2 − υµ2 ) sin 2λx + λ2 sin 2λx cos 2µy] 16λ a2 = [(µ2 − υλ2 ) sin 2µy + µ2 cos 2λx sin 2µy]. 16µ

upart = vpart

(3.13.21)

To satisfy the boundary conditions, we must add solutions of the homogeneous equations. With the condition (3.13.15), the boundary conditions become . u = const. at x = ± /2 u,y + v,x = 0 (3.13.22) . v = const. at y = ± /2, u,y + v,x = 0 which means that the shear strains vanish along the edge of the plate. Writing the full solution as the sum of the solution of the homogeneous equations and the particular solution u = uh + upart ,

v = vh + vpart ,

(3.13.23)

we find that uh and vh must satisfy 1 1 uh,xx + (1 − υ)uh,yy + (1 + υ)vh,xy = 0 2 2 1 1 vh,yy + (1 − υ)vh,xx + (1 + υ)uh,xy = 0 2 2 . uh = const. at x = ± /2 uh,y + vh,x = 0 . vh = const. at y = ± /2. uh,y + vh,x = 0

(3.13.24)

(3.13.25)

It is easy to verify that uh = −a2 ε1 x and

vh = −a2 ε2 y

(3.13.26)

satisfy both the equations and the boundary conditions. The unknown strains −a2 ε1 , −a2 ε2 follow from the conditions b/2 a2 2 u,x + υv,y + w1,x dy = 0 at x = ± /2 2

−b/2

/2 a2 2 v,y + υu,x + w1,y dx = 0 2

− /2

(3.13.27) at y = ± b/2,


163

which yields 1 2 (λ + υµ2 ) 8 1 υε1 + ε2 = (υλ2 + µ2 ), 8 ε1 + υε2 =

(3.13.28)

from which we obtain the expressions for the strains ε1 , ε2 , ε1 =

1 2 λ, 8

ε2 =

1 2 µ . 8

(3.13.29)

The energy in the plate is given by (3.13.7), where u = au1 + u, so that with u1 = v1 ≡ 0 and w ≡ 0, we obtain ! Eh3 a2 2 2 P [au1 + u] = − 2(1 − υ) w w − w (w ) 1 1,xx 1,yy 1,xy 12(1 − υ2 ) 2 2 1 1 Eh u,x + a2 w21,x + v,y + a2 w21,y + 2(1 − υ2 ) 2 2 (3.13.30) a2 2 a2 2 v,y + w1,y + 2υ u,x + w1,x 2 2 " 2 1 1 + (1 − υ) u,x + v,y + a2 w1,x w1,y − σha2 w21,x dx dy. 2 2 Making use of the fact that Eh3 (w1 )2 − 2(1 − υ) w1,xx w1,yy − w21,xy dx dy 2 12(1 − υ ) 1 = σcr h w21,x dx dy 2 and the fact that u and v are the solutions of the minimum problem, we have Eha2 u,x w21,x + v,y w21,y + υ u,x w21,y + v,y w21,x 2 12(1 − υ ) + (1 − υ) (u,x + v,y ) w1,x w1,y dx dy (3.13.31) Eh 1 =− u2.x + v2,y + 2υu,x v,y + (1 − υ) (u,x + v,y )2 dx dy. 1 − υ2 2 Then, we may write Eh P [au1 + u] = 8 (1 − υ2 )

/2 b/2

2 2 w1,x + w21,y dx dy

− /2 −b/2

Eh − 2(1 − υ2 )

/2 b/2 u2,x − /2 −b/2

1 + (σcr − σ)h 2

+

v2,y

1 2 + 2υu,x v,y + (1 − υ) (u,y + v,x ) dx dy 2

/2 b/2

− /2 −b/2

w21,x dx dy = F (a).

(3.13.32)

164

Applications

Evaluating these integrals, we find a4 a2 (σcr + σ)hb λ2 + Ehb (λ4 + µ4 ). 8 256 The equilibrium equation follows from ∂F 1 E = 0 = ahb λ2 (σcr − σ) + a2 (λ4 + µ4 ) , ∂a 4 16 which yields F (a) =

(3.13.33)

(3.13.34)

16π2 1 σ − σcr (σ − σcr )λ2 = . (3.13.35) E(λ4 + µ4 ) 2 π4 E π4 + 4 b4 For the case = b, we have 8h2 σ 8b2 σ − σcr 2 a = 2 = −1 , (3.13.36) π E 3(1 − υ2 ) σcr where we have made use of (3.12.34). This means that for σ = 2σcr , we have a2 = 16

8 ≈ 1.7h 3(1 − υ 2 )

a=h

( for υ = .25),

(3.13.37)

i.e., when the load is twice the critical load, the deflection is still only 1.7 times the plate thickness. The displacement of the load σ after buckling is σ σ = (3.13.38) −u x= + a2 ε1 ≡ εtot 2E 2 2 E 2 so that σ − σcr λ4 σ εtot = + 2 , (3.13.39) E E λ4 + µ4 and for = b, we have λ = µ, and hence σcr σ σ σcr − ≡ . (3.13.40) εtot = − 1 E Et E E 2 The so-called tangent modulus Et is half the modulus E. This behavior is shown in Figure 3.13.3. σ Et σ cr

asymptotic result actual behavior

E

ε tot Figure 3.13.3


σ l

σ

165

b

Figure 3.13.4

However, it should be noted that our solution is only valid in a small neighborhood of the critical load. Comparison of our result with exact numerical calculations show that for sufficiently long plates, the range of validity is moderately large. However, for short plates the range of validity is extremely small. Let us consider this case in some detail. From (3.12.37), the critical load is then given by σcr =

π2 Eh2 , 12(1 − υ2 ) 2

(3.13.41)

and for /b → 0, we have µ/λ → 0 so that from (3.13.35) we obtain a2 = 16

σ − σcr σ − σcr 2 . = 16 λ2 E E h2

(3.13.42)

The total strain εtot is now given by εtot =

2σcr a2 h2 σ σ − + , = 1 E 8 2 E E 3

(3.13.43)

so the tangent modulus now is given by Et = 1/3E. This result has an extremely small range of validity because almost immediately after buckling, the plate behaves like a buckled bar. This result is valid for the lowest bifurcation load, with only one wave. σ asymptotic result

Et σ cr

actual behavior E

ε tot Figure 3.13.5

166

Applications

From (3.12.36), the exact formula for the critical load for a plate is Eh2 λ4 λ2 2 λ σcr = + 1 + 2 . 12(1 − υ2 ) µ2 µ4

(3.13.44)

For fixed λ and, say, λ/µ = 10−2 for one wave, we have (1)

σcr ≈

Eh2 λ2 12(1 − υ2 )

(1 + 2.10−4 ).

(3.13.45)

When there are two waves, we have λ/µ = 2.10−2 and (2)

σcr ≈

Eh2 λ2 12(1 − υ2 )

(1 + 8.10−4 ),

(3.13.46)

which is very close to the lowest bifurcation load. This result shows that the lowest bifurcation point is a cluster point. Let us finally make some remarks regarding the results obtained. We have assumed that the edges at x = ± remain straight. This assumption holds exactly for sufficiently long plates because the nodal lines are straight lines. However, for short plates this must be accomplished with an edge beam, without constraining the strain long the edge. This condition can be done between two smooth stamps in a testing machine. ´ an-F ´ ¨ 3.14 The “von Karm oppl Equations” The post-buckling behavior of plates can also be analyzed using the so-called “von ´ an-F ´ ¨ Karm oppl equations.” For the derivation of these equations, we consider the deviation from the undeformed state. The strains in the mid-plane are 1 1 γxx = u,x + w2,x , γyy = v,y + w2,y , 2 2 2γxy = u,x + v,y + w,x w,y .

(3.14.1)

The curvatures are given by w,xx ,

w,yy ,

w,xy .

(3.14.2)

These expressions hold for sufficiently small rotations, i.e., |w,x |2 ,

|w,y |2 1.

(3.14.3)

The elastic energy with respect to the undeformed state is now given by 1 2 2 1 2 2 1 2 1 2 Eh v u w w w w + + v + + 2υ u + + ,x ,y ,x ,y 2(1 − υ2 ) 2 ,x 2 ,y 2 ,x 2 ,y 1 + (1 − υ) (u,y + v,x + w,x w,y )2 2 . h2 2 2 2 w,xx + w,yy + 2υw,xx w,yy + 2(1 − υ)w,xy dx dy. + 12

(3.14.4)

3.14 The “von K´arm´an-Foppl Equations” ¨

167

This expression is in full agreement with (3.12.4) because the fundamental state is linear. Carrying out the variations with respect to u and v, we obtain a stationary value when ! Eh 1 2 1 2 u,x + w,x + υ v,y + w,y δu,x 1 − υ2 2 2 Eh 1 2 1 2 υ u,x + w,x + v,y + w,y δv,y + (3.14.5) 1 − υ2 2 2 " Eh + (u,y + v,x + w,x w,y ) (δu,y + δv,x ) dx dy = 0. 2(1 + υ) Introducing the relations Eh 1 2 1 2 u,x + w,x + υ v,y + w,y = Nx 1 − υ2 2 2 Eh 1 1 υ u,x + w2,x + v,y + w2,y = Ny 1 − υ2 2 2 Eh (u,y + v,x + w,x w,y ) = Nxy , 2(1 + υ) we can rewrite this expression to yield [Nx δu,x + Ny δv,y + Nxy (δu,y + δv,x )]dx dy = 0.

(3.14.6)

(3.14.7)

Applying the divergence theorem, we obtain [(Nx υx + Nxy υy ) δu + (Ny υy + Nxy υx ) δv]ds ∂S

(3.14.8) [(Nx,x + Nxy,y ) δu + (Ny,y + Nxy,x ) δv]dx dy = 0,

− S

which yields the in-plane equilibrium equations Nx,x + Nxy,y = 0,

Ny,y + Nxy,x = 0,

and the in-plane boundary conditions, when either u, v are zero or free, (Nx υx + Nxy υy ) δu∂S = 0 (Ny υy + Nxy υx ) δv∂S = 0.

(3.14.9)

(3.14.10)

The general solution to (3.14.9) is Nx = F,yy ,

Ny = F,xx ,

Nxy = −F,xy ,

(3.14.11)

where F (x, y) is Airy’s stress function. It is easily verified that (3.14.11) is a solution of (3.14.9). To show that (3.14.11) ∗ is the general solution, we write Nxy = −F,xy , then we obtain from the first equation

168

Applications

∗ ∗ Nx,x = F,xyy , and from the second Ny,y = F,xyx . Integrating these expressions with respect to x and y, respectively, we obtain ∗ Nx = F,yy + f (y),

∗ Ny = F,xx + g(x),

or defining ∗ f (y) = f ,yy , ∗ ∗ + f ,yy , Nx = F,yy

g(x) = g∗,xx , ∗ Ny = F,xx + g∗,xx .

We now define a function F (x, y) by F (x, y) = F ∗ (x, y) + f ∗ (y) + g∗ (x). Then ∗ F,xy = F,xy = −Nxy , ∗ + g∗,xx = Nx , F,xx = F,xx

∗ ∗ F,yy = F,yy + f ,yy = Ny ,

which concludes our proof that (3.14.11) is the general solution. We now return to (3.14.4) and take the variation with respect to w to obtain the equilibrium equation in the direction normal to the undeformed mid-plane. Using (3.14.6), we obtain ! 1 2 1 2 Nx δ w,x + Ny δ w,y + Nxy δ(w,x w,y ) 2 2 3 Eh + [(w,xx + υw,yy ) δw,xx + (w,yy + υw,xx ) δw,yy (3.14.12) 12(1 − υ2 ) " + 2(1 − υ) w,xy δw,xy ] dx dy = 0. Equation (3.14.12) is fully identical with (3.12.9) when S is replaced by N, so we may use all the results derived from that expression. In particular, we mention the differential equation Eh3 w − Nx w,xx − Ny w,yy − 2Nxy w,xy = 0. 12(1 − υ2 )

(3.14.13)

Using (3.14.11), we can rewrite (3.14.13) in the form Eh3 w − F,yy w,xx + 2F,xy w,xy − F,xx w,yy = 0. 12(1 − υ2 )

(3.14.14)

The equation for the stress function is obtained from the condition that the displacement field is compatible. To this end, we write the inverse of (3.14.6), making use of (3.14.11), which yields the following expressions: 1 u,x + w2,y = 2 1 v,y + w2,y = 2

1 (F,yy − υF,xx ) Eh 1 (F,xx − υF,yy ) Eh 2(1 − υ) F,xy . u,y + v,x + w,x w,y = − Eh

(3.14.15)

3.15 Buckling and post-buckling behavior of shells using shallow shell theory 169

The displacements u and v are eliminated from these equations by differentiating the first equation with respect to y twice, and adding the second equation differentiated twice with respect to x, and subtracting the last equation differentiated with respect to x and y. The result is 1 F = w2,xy − w,xx w,yy . Eh

(3.14.16)

¨ This equation and (3.14.14), without the bending term, were first derived by Foppl. ´ an ´ derived the full equations. Later, von Karm Notice that in this approach there are only two unknowns, Airy’s stress function and the displacement w, whereas in the general theory we use the three displacements as unknowns. 3.15 Buckling and post-buckling behavior of shells using shallow shell theory In cases where the wavelength of the buckling mode is small compared to the smallest radius of curvature, shallow shell theory may be used. Shallow shell theory can be derived as follows. Consider a surface Z(x, y), where x and y are Cartesian coordinates in a tangent plane to the surface (see Figure 3.15.1), chosen so that the projection of the lines with principal radius of curvature on the tangent plane coincide with the x and y axis, respectively. The z axis is perpendicular to the tangent plane (see Figure 3.15.1). In the undeformed state, the square of a line element on the surface Z(x, y) is given by (ds)2 = (dx)2 + (dy)2 + (dz)2 = 1 + Z2,x (dx)2 + 2Z,x Z,y dx dy + 1 + Z2,y dy2 .

(3.15.1)

For the description of the surface in the deformed state, we shall make use of the fact that the in-plane displacements u, v are considerably smaller than the displacement

z

y

w

v u x

Figure 3.15.1

170

Applications

w, perpendicular to the mid-plane, i.e., |u|, |v| |w|,

(3.15.2)

w2,x , w2,y 1.

(3.15.3)

and that

Under these conditions, a point (x, y, z) in the undeformed state moves to (x + u − Z,x w, y + v − Z,y w, z + w) in the deformed state. The square of the length of a line element in the deformed state is now (d¯s)2 = (dx + du − Z,xx w dx − Z,xy w dy − Z,x dw)2 + (dy + dv − Z,yy w dy − Z,xy w dx − Z,y dw)2 + (dz + dw)2 ,

(3.15.4)

where Z,xy = 0 because x and y are the principal directions of curvature, 1 2 2 2 (d¯s) − (ds) = 2(dx) u,x − Z,xx w + w2,x 2 1 1 2 2 + (u,x − Z,xx w + Z,x w,x ) + (v,x − Z,y w,x ) 2 2 + 2dx dy [u,y + v,x + w,x w,y + (u,x − Z,xx w − Z,x w,x )(u,y − Z,x w,y )

(3.15.5)

+ (v,y − Z,yy w − Z,y w,y )(v,x − Z,y w,x )] 1 + 2(dy)2 v,y − Z,yy w + w2,y 2 1 1 + (v,y − Z,yy w + Z,y w,y )2 + (u,y − Z,x w,y )2 . 2 2 Neglecting higher order terms, except for w2,x , w2,y and w,x , w,y , we obtain 1 (d¯s)2 − (ds)2 = 2 u,x − Z,xx w + w2,x (dx)2 2 + 2(u,y + v,x + w,x w,y ) dx dy 1 + 2 u,y − Z,yy w + w2,y (dy)2 2

(3.15.6)

≡ 2γxx (dx)2 + 4γxy dx dy + 2γyy (dx)2 . Hence, the strains are given by w 1 + w2,x R1 2 = u,y + v,x + w,x w,y w 1 = v,y − + w2,y R2 2

γxx = u,x − 2γxy γyy

(3.15.7)


where R1 = Z,xx ,

R2 = Z,yy ,

(3.15.8)

are the principle radii of curvature. The changes of curvature are given by ρxx = w,xx ,

ρyy = w,yy ,

ρxy = w,xy .

(3.15.9)

Notice that all our expressions are only valid in a sufficiently small neighborhood of the tangent point. Under the assumption that the loads are dead-weight loads, which cause stresses σx , σy , and τxy in the fundamental state, the energy is given by ! Eh w 1 2 2 w 1 2 2 P[u] = u,x − + w,x + v,y − + w,y 2(1 − υ 2 ) R1 2 R2 2 1 w 1 w 1 v,y − + w2,y + (1−υ)(u,y +v,x +w,x w,y )2 + 2υ u,x − + w2,x R1 2 R2 2 2 (3.15.10) h2 2 2 2 w + w,yy + 2υw,xx w,yy + 2(1 − υ)w,xy + 12 ,xx " 1 − υ2 2 2 σx w,x + σy w,y + 2τxy w,x w,y dx dy, + E where the radii of the curvature can be assumed to be constants. Later we shall show that this expression may also be used for a shallow shell under uniform pressure (which is not a dead-weight load). The second and the third variations are given by ! Eh w 2 w 2 P2 [u] = u,x − + v,y − 2(1 − υ 2 ) R1 R2 1 w w v,y − + (1 − υ)(u,y + v,x )2 + 2υ u,x − R1 R2 2 (3.15.11) 2 h 2 2 2 w + w,yy + 2υw,xx w,yy + 2(1 − υ)w,xy + 12 ,xx " 1 − υ2 2 2 σx w,x + σy w,y + 2τxy w,x w,y dx dy + E ! Eh w w w2,x u P3 [u] = − + υ v − ,x ,y 2(1 − υ 2 ) R1 R2 " (3.15.12) w w w2,y + (1 − υ)(u,y + v,x )w,x w,y dx dy. + υ u,x − + v,y − R2 R1 The variational equation for neutral equilibrium is P11 [u, ζ] = 0, where ζ = (ξ, η, ζ) is a kinematically admissible displacement field, Eh w ζ w ζ u,x − ξ,x − + v,y − η,y − P11 [u, ζ] = 1 − υ2 R1 R1 R2 R2 w ζ w ζ η,y − + υ v,y − ξ,x − + υ u,x − R1 R2 R2 R1

172

Applications

1 + (1 − υ)(u,y + v,x )(ξ,y + η,x ) (3.15.13) 2 h2 w,xx ζ,xx + w,yy ζ,yy + υw,xx ζ,yy + υw,yy ζ,xx + 2(1 − υ)w,xy ζ,xy + 12 1 − υ2 (σx w,x ζ,x + σy w,y ζ,y + τxy w,x ζ,y + τxy w,y ζ,x ) dx dy. + E Repeated application of the divergence theorem yields " 4 ! Eh w w 1 u + υ v υ (1 − υ)(u − − + + v )υ ,x ,y x ,y ,x y ξ 1 − υ2 R1 R2 2 ∂S ! " w w 1 + v,y − + υ u,x − υy + (1 − υ)(u,y + v,x )υx η R2 R1 2 h2 + (w,xx + υw,yy ) υx + (1 − υ)w,xy υy ζ,x 12 + (w,yy + υw,xx )υy + (1 − υ)w,xy υx ζ,y ! 1 − υ2 + [σx w,x υx + σy w,y υy + τxy (w,x υ,y + w,y υx )] E (3.15.14) " h2 − [(w,xx + w,yy ),x υx + (w,xx + w,yy ),y υy ] ζ ds 12 " ! Eh 1 1 1 υ u,xx + (1 − υ)u,yy + (1 + υ)v,xy − w·x ξ − + 1 − υ2 2 2 R1 R2 s " ! 1 1 1 υ w,y η + v,yy + (1 − υ)v,xx + (1 + υ)u,xy − + 2 2 R2 R1 % & ! 1 1 υ 1 υ 2υ 1 − u,x − v,y + + + + + 2 w R1 R2 R2 R1 R1 R2 R21 R2 " 2 2 1−υ h + w − (σx w,x ),x +(σy w,y ),y +(τxy w,x ),y +(τxy w,y ),x ζ dx dy = 0, 12 E where υ is the unit normal vector at the edge. This yields the differential equations 1 1 υ 1 w,x = 0 u,xx + (1 − υ)u,yy + (1 + υ)v,xy − + 2 2 R1 R2 (3.15.15) 1 1 1 υ w,y = 0 v,yy + (1 − υ)v,xx + (1 + υ)u,xy − + 2 2 R2 R1 h2 w − 12

υ 1 + R1 R2

u,x −

1 υ + R2 R1

%

v,y +

1 2υ 1 + + 2 2 R1 R2 R1 R2

1 − υ2 − (σx w,xx + σy w,yy + 2τxy w,xy ) = 0, E

& w (3.15.16)

where we have used the fact that σx , σy , and τxy (approximately) satisfy the equations σx,x + τyx,y = 0,

τxy,x + σy,y = 0.

(3.15.17)


From the line integral, we obtain the conditions in the mid-plane, " ! w w 1 + υ v,y − υx + (1 − υ)(u,y + v,x )υy ξ = 0 u,x − R1 R2 2 ! " ∂S w w 1 v,y − + υ u,x − υy + (1 − υ)(u,y + v,x )υx η = 0. R2 R1 2 ∂S

(3.15.18)

For the reduction of the remaining terms, we make use of the relations ζ,x = −υy ζ,s + υx ζ,v ,

ζ,y = υx ζ,s + υy ζ,v ,

and (3.12.17) to obtain (w,xx + υw,yy )υ2x + 2(1 − υ)w,xy υx υy + (w,yy + υw,xx )υ2y ζ,v ∂S = 0 ! (1 − υ) (w,yy − w,xx )υx υy + w,xy υ2x − υ2y

(3.15.19)

(3.15.20)

,s

+ (w,xx + w,yy ),x υx + (w,xx + w,yy ),y υy

" 12(1 − υ 2 ) ζ = 0. − w + τ w )υ + (σ w + τ w )υ [(σ ] x ,x xy ,y x y ,y xy ,x y 3 Eh ∂S

(3.15.21)

Notice that only the in-plane boundary conditions are affected by the curvature of the surface. We shall now apply this theory to a spherical shell under a uniform external pressure p per unit area of the middle surface.† The stresses in the fundamental state are given by σx = σy = −σ = −

pR , 2h

τxy = 0.

(3.15.22)

The equations now read 1 1 1+υ (1 − υ)u,yy + (1 + υ)v,xy − w,x = 0 2 2 R 1 1 1+υ w,y = 0 v,yy + (1 − υ)v,xx + (1 + υ)u,xy − 2 2 R E w Eh2 w − u + σ(w,xx + w,yy ) = 0. + v − 2 ,x ,y 12(1 − υ 2 ) (1 − υ)R R u,xx +

(3.15.23)

(3.15.24)

We try a solution of the form u = A sin px/R cos qy/R v = B cos px/R sin qy/R w = C cos px/R cos qy/R,

(3.15.25) (p, q ∈ R).

Substitution into (3.15.23) yields 1 1 2 2 − p + (1 − υ)q A − (1 + υ)pqB + (1 + υ)pC = 0 2 2 1 1 2 2 − (1 + υ)pqA − (1 − υ)p + q B + (1 + υ)qC = 0. 2 2 †

See the remark below (3.15.10).

(3.15.26)

174

Applications

We can now solve A/C and B/C from these equations, which yield A p , = (1 + υ) 2 C p + q2

B q . = (1 + υ) 2 C p + q2

(3.15.27)

In our derivation of the shallow shell equations, we have use the assumption |w| |u|, |v|, which means A/C, B/C 1, so that for p or q 1, our assumptions are satisfied. Substitution of the displacement field (3.15.25) into the third equilibrium equation and using (3.15.27), we obtain σ 1 h2 2 2 = 2 3(1 − υ 2 ). + (p + q ), c = (3.15.28) E p + q2 4c2 R2 We still must minimize this expression with respect to p and q, but because the expression only contains the combination p 2 + q2 , we can minimize with respect to this parameter, which yields p 2 + q2 =

2cR . h

(3.15.29)

This means that our assumptions are satisfied as R/h 1. The critical load is now given by Eh . cR

σcr =

(3.15.30)

Notice that the critical load is proportional to h/R, whereas for plates the critical load was proportional to h2 / b2 . Because p and q must only satisfy the condition (3.15.29), there is an infinite number of buckling modes. In the p − q plane, (3.15.29) represents a circle with radius p 0 = 2cR/h (see Figure 3.15.2). Because we have a continuous spectrum, the displacement field is given by integrals instead of a series. However, for our purpose we may consider some discrete values, p i , qi . Because for the evaluation of P3 [u1 ] we must deal with cubic terms, q

( 2c

R

p /h

) 1/ 2

Figure 3.15.2


we can consider the discrete pairs (p 1 , q1 ),

(p 2 , q2 ),

(p 3 , q3 ),

which in the evaluation of P3 [u1 ] lead to integrals of the form & % x sin x sin x p1 p2 sin p 3 dx, cos R cos R R which can be expressed in integrals of the form x sin (p 1 ± p 2 ± p 3 ) dx. cos R

(3.15.31)

(3.15.32)

(3.15.33)

For large values of x, these integrals vanish† except when the argument of the cosine is equal to zero, i.e., p 1 ± p 2 ± p 3 = 0,

(3.15.34)

q1 ± q2 ± q3 = 0.

(3.15.35)

and similarly for the terms in y,

This means that the points (p 1 , q1 ), (p 2 , q2 ), (p 3 , q3 ) divide the circle in three equal parts. We shall now consider the special case of (see Figure 3.15.2) 1 1 √ (p 1 , q1 ) = (p 0 , 0), (p 2 , q2 ) = − p 0 , p 0 3 , 2 2 (3.15.36) 1 1 √ (p 3 , q3 ) = − p 0 , − p 0 3 . 2 2 Let us introduce the notation 1√ 1 p0, q = 3 p0; 2 2 then our displacement field is given by h2 y x x u = (1 + υ) 2pa0 sin 2p + pa1 sin p cos q 2cR R R R h2 y x v = (1 + υ) qa1 cos p sin q 2cR R R y x x , w = h a0 cos 2p + a1 cos p cos q R R R p=

(3.15.37)

(3.15.38)

where we have used (3.15.27). For the evaluation of P3 [u1 ], we shall need the following expressions, h 1 y w x x u,x − = υa0 cos 2p − (3 − υ)a1 cos p cos q R R R 4 R R w x x h 1 y −a0 cos 2p − (1 − 3υ)a1 cos p cos q v,y − = R R R 4 R R †

Let L x R, e.g., x = O R3/4 h1/4 , so that our shallow shell assumptions still hold approximately.

176

Applications

h 3 w x y = − a1 (1 − υ 2 ) cos p cos q R R 4 R R h 1 y w x x 2 2 = −(1 − υ )a0 cos 2p − (1 − υ )a1 cos p cos q R R R 4 R R 1√ y h x − (3.15.39) 3 a1 sin p sin q u,y + v,x = (1 + υ) R 2 R R ph x x 1 y w,x = 2 −a0 sin 2p − a1 sin p cos q R R 2 R R ph x 1√ y 3 a1 cos p sin q w,y = 2 − . R 2 R R

w + υ v,y − R w v,y − + υ u,x − R

u,x −

Introducing the notation 1 )= 4R2

(

R R (

)dx dy,

(3.15.40)

−R −R

which is the average value of (), we obtain

u,x −

w w 2 w w 2 + υ v,y − w,x + v,y − + υ u,x − w,y R R R R

+ (1 − υ)(u,y + v,x )w,x w,y 3 3 3 3 2h 2 2 2 2 2 2 = p 3 − (1 − υ )a0 a1 − (1 − υ )a0 a1 − (1 − υ )a0 a1 R 8 8 8 =−

(3.15.41)

9 ch2 (1 − υ 2 ) 2 a0 a21 , 16 R

so the average value of P2 [u1 ] is P3 [u1 ] = −

9 Ech3 a0 a21 . 32 R2

(3.15.42)

For the evaluation of P2 [u; λ], we need w2·x

+

w2·y

h2 = 4p 2 R

2

1 2 1 2 a + a , 2 0 4 1

(3.15.43)

so that P2 [u1 ; λ] = (1 − λ)

Eh3 R2

1 2 1 2 a0 + a1 . 2 4

(3.15.44)

In a sufficiently small neighborhood of the critical load, the average energy P[u, λ] is given by Eh3 F (ai , λ) = P[u, λ] = 2 R

1 9c 1 2 2 2 (1 − λ) a0 + a1 − a0 a1 . 2 2 32

(3.15.45)


The equilibrium equations are given by ∂F Eh3 9c 2 = 2 (1 − λ)a0 − a1 = 0 ∂a0 R 32 Eh3 1 9c ∂F (1 − λ)a1 − a0 a1 = 0. = 2 ∂a1 R 2 16

(3.15.46)

From the second of these equations, we obtain the non-trivial solution a0 =

8 (1 − λ), 9c

(3.15.47)

and from the first equation, a1 = ±

16 (1 − λ) = ±2a0 . 9c

(3.15.48)

It is sufficient to consider only the positive solution because the negative sign is obtained by shifting the origin of our coordinate system. As 2 2 3 2 ∂2 F ∂ 2 F ∂ F Eh − =− (1 − λ)2 < 0, (3.15.49) 2 2 ∂a0 ∂a1 R2 ∂a0 ∂a1 the equilibrium states are unstable for both λ < 1 and λ > 1. Notice that for υ = 0.272 and c = 5/3, we have a0 =

8 (1 − λ), 15

a1 =

16 (1 − λ), 15

(3.15.50)

so that for λ = 1/2, a0 = 4/15, a1 = 8/15, w = 0.8h. This means that for deflections of the order of the plate thickness, the load is reduced to about 50% of the critical load. Assuming that our average energy expression holds on the whole surface of the sphere, the total energy is given by 4πR2 F (ai , λ).

(3.15.51)

The work of the uniform pressure p is −pV, where V is the volume increment passing from the fundamental path to the branched path. It follows that 4πR2

∂F = −V, ∂p

(3.15.52)

or with p = 2λ

Eh2 cR2

2πR4 c ∂F = −V. Eh2 ∂λ

(3.15.53)

(3.15.54)

Using (3.15.45), we obtain V = −

64π 2 hR (1 − λ)2 . 27c

(3.15.55)

178

Applications

λ=

p pcr

1

1

∆vtot ∆vcr

Figure 3.15.3

The volume change at the critical load is Vcr = −4πR2 wcr = −

4π σcr (1 − υ)R · 4πR = − (1 − υ)hR2 , E c

(3.15.56)

so that Vtot 16 (1 − λ)2 + λ. = Vcr 27(1 − υ)

(3.15.57)

This relation is illustrated in Figure 3.15.3. Notice that there is no stable equilibrium state in the vicinity of the bifurcation point, and that the load drops very rapidly after bifurcation, which means that the shell is very imperfection-sensitive. We will come back to this later. Let us now first have a closer look at the buckling mode, √ y x x w = ha0 cos 2p + 2 cos p cos p 3 . (3.15.58) R R R Introducing the notation 2p

x = α, R

2p

y = β, R

(2.15.59)

we can rewrite this expression as w = ha0 f (α, β),

f (α, β) = cos α + 2 cos α/2 cos

√

3β/2.

(3.15.60)

To recognize the buckling pattern, we investigate f (α, β) more closely. First, we notice that 4kπ β= √ , 3 π β = (2k + 2) √ . 3

f max = 3 for α = 4j π, α = (2j + 2)π,

(3.15.61)


Necessary conditions for a minimum value of f (α, β) are √ ∂f = − sin α − sin α/2 cos 3 β/2 = 0 ∂α √ √ ∂f = − 3 cos α/2 sin 3β/2 = 0. ∂β

(3.15.62)

From the second equation, we find cos α/2 = 0, sin

α = (2k + 1)π,

√ 2j π 3 β/2 = 0, β = √ , 3

(3.15.63)

and from the first equation, sin α/2 = 0,

α = 2kπ, √ 2 cos α/2 + cos 3 β/2 = 0.

(3.15.64)

Both equations are satisfied when α = (2k + 1)π,

π β = 2j √ , 3

β=

2j + 1 √ π 3

(3.15.65)

 2π   + 4kπ ( j odd) ± 3 α=  4π  ± + 4kπ ( j even) 3

(3.15.66)

2j π β= √ . 3

(3.15.67)

α = 2kπ,

We now obtain the following values for f (α, β), 2j + 1 f (2k + 1)π, √ π =1 3  1   j even + 2π 2j π 2 f ± + 4kπ, √ =  3 3  −3 j odd 2 3 4π 2j π =− f ± + 4kπ, √ 3 2 3 2j π −1 j + k odd = f 2kπ, √ 3 j + k even. 3

(3.15.68)

We may now draw the graph shown in Figure 3.15.4. We have a pattern of regular hexagons. Ordinarily, this pattern is not observed because this buckling mode is unstable. It can be observed when the buckling mode is artificially fixed, which may be done by fitting a solid sphere with a radius slightly

180

Applications β 3

−2π

3

3/ 2

3

3/2 −3π

3/2

3/2

3

−π

π

0

3

2π

3π

α

3 3/2

3/2

3

Figure 3.15.4

smaller than that of the shell inside the shell.† The experimental results are in agreement with our buckling pattern. As we have seen previously, the load drops rapidly after bifurcation, which implies that the shell is very imperfection-sensitive. To get quantitative information about the imperfection sensitivity, we assume imperfections corresponding to the buckling mode, which according to the general theory are the worst type of imperfections. Let the imperfections be given by √ x x y w0 = h a¯ 0 cos 2p + a¯ 1 cos p cos 3 p , (3.15.69) R R R where a¯ 1 = 2a¯ 0 . In the presence of imperfections, the function F (ai ; λ) (see 3.15.45) must be replaced by (see 2.6.39) 9c Eh3 1 1 1 (1 − λ) a20 + a21 − a0 a21 − λ a0 a¯ 0 + a0 a¯ 1 . (3.15.70) F ∗ (ai ; λ) = 2 R 2 2 32 2 With a¯ 1 = 2a¯ 0 and a0 and a1 given by (3.15.47) and (3.15.48), we can rewrite this expression in the form Eh3 9c 3 2 F ∗ (ai ; λ) = 3(1 − λ)a . (3.15.71) − − 6λ a ¯ a a 0 0 0 2R2 4 1 †

Cf. R. L. Carlson, R. L. Sendelbeck, and N. J. Hoff, Experimental studies of the buckling of complete spherical shells. Experimental Mechanics, 7, no. 7, 281–288 (1967).

3.15 Buckling behavior of a spherical shell under uniform external pressure

The equilibrium equation then reads Eh3 27c 2 ∂F ∗ a = − − 6λ a ¯ 6(1 − λ)a 0 0 = 0. ∂a0 2R2 4 0

181

(3.15.72)

The minimum value of a0 is obtained when the discriminant of this quadratic equation vanishes, which yields (1 − λ∗ )2 =

9c ∗ λ a¯ 0 . 2

(3.15.73)

The corresponding amplitude is given by a∗0 =

4(1 − λ∗ ) . 9c

(3.15.74)

For λ∗ = 0.5, we have a¯ 0 = 1/15 (for υ = 0.28), so the maximum imperfection amplitude is 3a¯ 0 = 1/5, i.e., imperfections of about 20% of the shell thickness reduce the critical load by 50%. This result shows that the shell is extremely imperfectionsensitive. The present approach is due to Hutchinson.† Let us now reconsider the approach. The starting point was shallow shell theory, so the results are only valid in a sufficiently small neighborhood of the tangent point, i.e., x, y R. However, for our evaluation of the average value of the energy we have carried out the integration over the domain −R ≤ x, y ≤ R. Furthermore, we have ignored the fact that the hexagonal buckling pattern cannot be extended over the complete spherical shell, which implies a relaxation of the geometric conditions, which should result in decreased stability in the post-buckling range. This means that Hutchinson’s analysis yields a lower bound for the post-buckling load. The difficulties with the domain of validity for x and y can be overcome by using a displacement field (u∗ , v∗ , w∗ ), defined by 1 (3.15.75) (u∗ , v∗ , w∗ ) = (u, v, w) exp − µ2 (x2 + y2 )/R2 , 2 where 1 µ2 p 2 + q2 . Under this restriction, the exponential factor can be considered to be a constant factor when differentiation is carried out, which means that this modified displacement field satisfies the same equations as our original field. Furthermore, (u∗ , v∗ , w∗ ) ∼ (u, v, w) for sufficiently small values of x and y, and (u∗ , v∗ , w∗ ) → 0 for sufficiently large values of x and y. Such a type of displacement field may also be used to describe localized imperfections.‡ In the next section, we shall treat the problem of the spherical shell again using the general theory of shells, and verify the validity of results obtained in this section. † ‡

J. W. Hutchinson, Imperfections sensitivity of externally pressurized spherical shells. Journal Appl. Mech. 34, 49–55 (1967). Cf. W. T. Koiter, The influence of more or less localized short-wave imperfections . . . . Rep. Lab. For Appl. Mech. Delft, 534. V. Z. Gristchak, Asymptotic formula for . . . WTHD rep. No. 88.

182

Applications

3.16 Buckling behavior of a spherical shell under uniform external pressure using the general theory of shells In this section, we shall again consider the behavior of a spherical shell under uniform external pressure, but now we shall employ the general theory of shells. For the derivation of the equations of the theory of shells, we refer to the literature.† However, we shall give a short survey of the most important quantities. Let r(xα ) (α ∈ 1, 2) be the position vector from a fixed origin in space to a generic point on the middle surface of the undeformed shell. The tangential base β vectors are aα = r,α = ∂r/∂xα . The reciprocal base is defined by aα · aβ = δα . The covariant and contravariant metric tensors are given by aαβ = aα · aβ and aαβ = aα · aβ , respectively. The unit normal to the middle surface is n = 12 εαβ aα × aβ where εαβ is the contravariant alternating tensor. The second fundamental tensor is specified by bαβ = n · r,αβ . A point in shell space is identified by its distance z to the middle surface and by the surface coordinates of its projection on to the middle surface. The coordinate z is orthonormal to the middle surface. The shell faces z = ± 12 h, where h is the constant shell thickness, are surfaces parallel to the mid-surface. The covariant components of the spatial metric tensor gij are specified by gαβ = aαβ − 2zbαβ + z2 cαβ g13 = g23 = 0,

g33 = 1,

(3.16.1)

where cαβ = bκα bκβ is the third fundamental tensor. On the mid-surface, we have gαβ = aαβ . Let g be the determinant of gij and a be the determinant of ααβ ; then ) g = 1 − 2z H + z2 K, (3.16.2) a where H = 12 bαα is the mean curvature in a point (x1 , x2 ) of the middle surface and K = b11 b22 − b12 b21 is the Gaussian curvature. The edge of the shell is assumed to be a ruled surface formed by normals to the middle surface along an edge curve on this surface. Let υ be the unit vector in the tangent plane, normal to the edge curve and positive outward. The positive sense on the edge curve is defined by the tangential unit vector t = n × υ. A deformation of the middle surface is described by the two-dimensional displacement field, u(xκ ) = uα aα + wn.

(3.16.3)

The strain tensor in the mid-plane is given by 1 1 γαβ = θαβ + aκλ (θκα − ωκα )(θλβ − ωλβ ) + ϕα ϕβ , 2 2 †

(3.16.4)

Cf. W. T. Koiter and J. G. Simmonds, Foundations of shell theory. Proc. 13th IUTAM Congress, (Springer Verlag, 1972), 150–176 (also WTHD Rep. No. 40).


183

where θαβ is the linearized strain tensor, θαβ =

1 (uα|β + uβ|α ) − bαβ w 2

(3.16.5)

1 (uβ|α − uα|β ) 2

(3.16.6)

ωαβ =

is a rotation in the tangent plane to the shell, and ϕα = w,α + bκα uκ

(3.16.7)

is a rotation of the normal vector n. The linearized tensor of changes of curvature is ραβ =

1 1 κ bα ωκβ + bκβ ωκα . (ϕα|β + ϕβ|α ) − 2 2

(3.16.8)

After this survey of expressions from shell theory, we return to our stability problem. According to the general theory of elastic stability (2.6.18), the energy functional for a three-dimensional body is given by 1 1 P[u] = (3.16.9) Sij uh,i uh,j + Eijk γij γk dV, 2 2 V

where u denotes the three-dimensional increment of the displacement field, passing from the fundamental state to the adjacent state, and γij is the strain tensor γij =

1 (ui,j + uj ,i + uh,i uh,j ). 2

The second variation of this energy is given by 1 1 Sij uh,i uh,j + Eijk θij θk dV, P2 [u] = 2 2

(3.16.10)

(3.16.11)

V

where θij is the linearized strain tensor, θij =

1 (ui,j + uj ,i ). 2

(3.16.12)

In shell theory, the state of stress is assumed to be approximately plain and parallel to the mid-surface. This results in a decoupling of the membrane energy and the bending energy. The energy functional can then be written as 1 h αβλµ h2 P[u] = γαβ γλµ + ραβ ρλµ dA, (3.16.13) Sij uh,i uh,j + E 2 2 12 A

where the integration is to be carried out over the mid-surface of the shell. Here Eαβλµ is the tensor of elastic moduli corresponding to a plate state of stress 2υ αβ λµ αβλµ αλ βµ αµ βλ a a , (3.16.14) =G a a +a a + E 1−υ

184

Applications

where G is the shear modulus, E . (3.16.15) 2(1 + υ) The second term in (3.16.13) can now be rewritten to yield 1 αβλµ h2 γαβ γλµ + ραβ ρλµ (3.16.16) hE 2 12 ! " Eh h2 αβ κ 2 αβ κ 2 = (1 − υ)γ γαβ + υ (γκ ) + (1 − υ)ρ ραβ + υ(ρκ ) . 2(1 − υ 2 ) 12 G=

Notice that the elastic energy density is expressed in terms of invariants of the strain tensor and the tensor of changes of curvature. Let Nαβ be the membrane stresses in the fundamental state, due to dead-weight loads. The energy functional can then be written as ! h2 Eh αβ κ 2 P[u] = (1 − υ)γ (1 − υ)ραβ ραβ + υ(ρκκ )2 + γ + υ(γ ) αβ κ 2(1 − υ 2 ) 12 A " 1 αβ κλ + N a (θκα − ωκα ) (θλβ − ωλβ ) + ϕα ϕβ dA. (3.16.17) 2 In the case of uniform pressure p, the fundamental state is nonlinear, because the potential energy pV is proportional to the volume V. Let us now consider this functional more closely. Let V0 be the volume in the undeformed state. The volume in the deformed state is then 1 V= εijk εpqr yi,p yj ,q yk,r dV0 , (3.16.18) 6 V0

where yi are the coordinates in the deformed state of a point with coordinates xi in the undeformed state. Let u be the displacement field, then yi = xi + ui .

(3.16.19)

Expression (3.16.18) can then be written as 1 εijk εpqr (δip + ui,p )(δj q + uj ,q)(δkr + uk,r ) dV0 V= 6 V0 1 εijk εpqr δip δj quk,r dV0 = V0 + (3.16.20) 2 V0 1 1 + εijk εpqr δip uj ,quk,r dV0 + εijk εpqr ui,p uj ,quk,r dV0 . 2 6 V

V0

For the term linear in u, we obtain 1 1 εijk εpqr δip δj quk,r dV0 = (δj qδkr − δj r δkq) δj quk,r dV0 2 2 V0 V0 = δkr uk,r dV0 = uk,k dV0 = uk nk dA = w dA, V0

A

A

(3.16.21)


185

where we have used the divergence theorem on a closed surface so that there are no stock terms. The reduction of the term quadratic in u is considerably more difficult. 1 1 εijk εpqr δip uj ,quk,r dV0 = (δj qδkr − δj r δkq) uj ,quk,r dV0 2 2 V0

V0

1 (uj ,j uk,k − uj ,kuk,j ) dV0 2

= V0

= A

− =

1 1 uj ,j uknk dA − uj ,jkuk dV0 2 2

(3.16.22)

V0

1 uj ,kuknj dA + 2

A

1 uj ,kj uk d0 2

0

1 (uj ,j nk − uj ,knj ) uk dA. 2

A

Because the integrand is an invariant, it can readily be rewritten in general coordinates, which yields 1 j u j uknk − uj kuknj dA, (3.16.23) 2 A

where a subscript preceded by a double vertical bar denotes three-dimensional covariant differentiation with respect to the undeformed metric. In our coordinate system on the shell surface, n is perpendicular to the coordinates xα , so that n1 = n2 = 0,

n3 = −1,

(3.16.24)

which means that (3.16.23) can be written as k 1 j 1 α u j w − w ku dA = − u α w + w 3 w − w α uα − w 3 w dA − 2 2 A

A

=−

1 α u α w − w α uα dA. 2

(3.16.25)

A

Now ui α = ui,α + ikα uk,

(3.16.26)

where ikα is a Christoffel symbol of the second kind, so that uα α = uα,α + αβα uβ + α3α w w α = u3 α = w,α + 3βα uβ + 33α w.

(3.16.27)

186

Applications

The Christoffel symbols can be expressed as α3α = gαkk3α = gαβ β3α + gα3 33α 1 = gαβ (gβ 3,α + gβα,3 − g3α,β ) = −gαβ bαβ = −bαα , 2

(3.16.28)

where we have made use of (3.16.1), 3βα = g3kkβα = g33 3βα = 33α = g3kk3α = g33 33α =

1 (g3β,α + g3α,β − gβα,3 ) = bαβ 2

(3.16.29)

1 (g33,α + g3α,3 − g3α,3 ) = 0. 2

(3.16.30)

Here, ijk are Christoffel symbols of the first kind. Using the relation uα α = uα,α + αβ α uβ ,

(3.16.31)

which is the covariant derivative of uα with respect to the surface coordinates, we can now rewrite (3.16.25) to yield 1 α 1 − u α − bαα w w − w,α + bαβ uβ uα dA = − (wθαα − ϕα uα ) dA. 2 2 A

A

(3.16.32) The energy of the uniform pressure load is now 1 α 1 1 α − p wA − p wθα − ϕα u dA − p εijk εpqr ui,p uj ,q uk,r dV0 . (3.16.33) 2 2 6 V0

A

Because p = O(N/R), where N denotes the maximum of the membrane forces and R the minimum radius of curvature, the second term is of order O (Nwθαα /R) = O(Nw2 /R2 ), whereas the first term is of order O(Nw/R). The potential energy of the membrane stresses in the fundamental state is O(Nw2 /L2 ), so that for L/R 1 the contribution to the energy of the second term in (3.16.33) may be neglected. The third term, which is even smaller, may then certainly be neglected. It follows that for L/R 1 (shallow shell theory), only the first (linear) term must be taken into account, which means that to a first approximation, uniform external pressure may be considered as a dead-weight load. This implies that the energy functional is given by (3.16.17). We shall now apply these results to a spherical shell under uniform external pressure. However, we shall also take into account the second term in (3.16.33) and show that this term may indeed be neglected. The basic property of a spherical shell is that the first and second fundamental tensors are proportional to each other, bαβ =

1 aαβ , R

bαβ =

1 α a , R β

bαβ =

1 αβ a , R

(3.16.34)

where R is the radius of the middle surface. The sign convention in (3.16.34) implies that the positive direction of n points inward. The linearized strain tensor (3.16.5)


187

can now be written as θαβ =

1 1 (uα|β + uβ|α ) − aαβ w. 2 R

(3.16.35)

The rotation of n (3.16.7) now reads ϕα = w,α +

1 aα , R

(3.16.36)

and the linearized tensor of changes of curvature (3.16.8) is now given by w 1 1 ραβ = w αβ + (uα|β + uβ|α ) = w αβ + aαβ 2 + θαβ . 2R R R

(3.16.37)

In the following, we shall make use of the property that the tangential displacement field can always be expressed in terms of the two invariants uα = ϕ,α + εαλ ψλ .† (3.16.38) To prove this property, we notice that uα |α = ϕ αα + ε.αλ ψ.λα = ϕ + εαλ ψ|λα = ϕ,

(3.16.39)

where the term εαλ ψ|λα vanishes because ψ|λα is symmetric, and is the Laplacian. This is for a known displacement field for an equation with ϕ, which has a unique solution‡ for sufficiently smooth displacement fields. To get an equation for ψ, we multiply (3.16.38) by εαµ to obtain uα εαµ = εαµ ϕ,α + εαµ εαλ ψλ = εαµ ϕ,α + ψ|µ , so that ψ,µ = ε.αµ (uα − ϕ,α ). This equation is integrable when the right-hand side of ψµυ = ε.αµ (uα|µ − ϕ |αυ )

(3.16.40)

(3.16.41)

is symmetric in µ and υ, so that the left-hand term must also be symmetric, which means εµυ ε.αµ (uα|υ − ϕ |αυ ) = uα |α − ϕ = 0.

(3.16.42)

This is exactly the equation for ϕ, and this completes our proof. As we shall show, the advantage of expressing uα in terms of ϕ and ψ is that we obtain a separate equation for ψ in which ϕ and w do not appear. Using (3.16.38), we can rewrite (3.16.35) to (3.16.37) to yield 1 1 1 θαβ = ϕ αβ + εαλ ψ.λβ + εβλ ψ.λα − aαβ w 2 2 R † ‡

(3.16.43)

Cf. A. van der Neut, De elastische stabiliteit vaan den dunwandigen bol (Thesis, Delft; H. J. Paris, Amsterdam, 1932). Apart from an arbitrary constant, which is unimportant as only derivatives are needed.

188

Applications

ϕα = w,α +

1 1 ϕ,α + εαλ ψλ R R

1 1 1 ραβ = w αβ + ϕ αβ + εαλ ψ.λβ + ε·βλ ψ.λα R 2R 2R

(3.16.44)

(3.16.45)

We now return to our energy functional (3.16.17). The fundamental state of a spherical shell under uniform external pressure p is a membrane state of stress, and the contravariant tensor of stress resultants is 1 pRaαβ . (3.16.46) 2 It will be convenient to introduce a dimensionless load parameter λ, defined by Eh σ= λ, c = 3(1 − υ 2 ). (3.16.47) cR Nαβ = σhaαβ =

Also taking into account the second term of (3.16.33), the second variation of our energy functional can now be written as ! Eh h2 αβ κ 2 αβ κ 2 (1 − υ)ραβ ρ + υ(θ κ ) P2 [u; λ] = (1 − υ)θ θαβ + υ(θ κ ) + 2(1 − υ 2 ) 12 A " Eh2 α Eh2 αβ αβ α α λ (θαβ − ωαβ )(θ − ω ) + ϕα ϕ + λ wθα − u ϕα dA. − 2cR cR2 (3.16.48) Let us now consider the term 1 1 1 1 ϕ αβ + εαλ ψ.λβ + εβλ ψ.λα − aαβ w θαβ θαβ dA = ϕ αβ + ε.αλ w λβ 2 2 R 2 A (3.16.49) 1 β λα 1 αβ + ε.λ ψ − a w dA. 2 R We shall now show that the coupling terms between ψ and ϕ and between ψ and w vanish. The term α λβ aαβ wε.λ ψ dA = wεβλ ψλβ dA = 0 (3.16.50) A

A

because ψλβ is symmetric. Thus, λβ ·β ϕ αβ εα·λ ψ dA = ϕ αβ εαλ ψλ dA αλ β = ϕ αβ ε ψ dA − ϕ αβλ εαλ ψβ dA. λ

The first of these integrals vanishes after the application of the divergence theorem on the closed surface. Changing the order of covariant differentiation in the remaining integral, we can write − ϕ αβλ εαλ ψβ dA = − ϕ αλβ εαλ ψβ dA − Rκ·αβλ ϕ,κ εαλ ψβ dA. A

A

A


189

Here the first integral on the right-hand side vanishes because ϕ αλβ is symmetric in α and λ ; Rκ·αβλ is the Riemann-Christoffel tensor of the surface, which is related to the Gaussian curvature by Rκ·αβλ = Kεκ·α ε.βλ .

(3.16.51)

For a sphere K = R−2 , we can write 1 1 κ αλ β κ αλ α − R·αβλ ϕ,κ ε ψ dA = − 2 ε·α εβλ ε ϕ,κ ψ| dA = − 2 εκα ϕ |κ ψ|α dA R R A A A 1 1 α =− 2 (εκα ϕ |κ ψ) dA + 2 εκα ϕ |κα ψdA = 0. R R A

A

The first of these integrals vanishes after the application of the divergence theorem on a closed surface, and the second integral vanishes because ϕ |κα is symmetric in κ and α. We may now rewrite (3.16.49) in the form 1 1 θαβ θαβ dA = ϕ αβ − aαβ w ϕ αβ − aαβ w R R A A λ λ α κβ 1 β κα + εαλ ψ ·β + εβλ ψ ·α ε·κ ψ + ε·κ ψ| dA 4 (3.16.52) αβ 2 2 2 = − wϕ + 2 w dA ϕ αβ ϕ R R A 1 λ ·β ψ ·β ψλ + εαλ εβ·κ ψλ·β ψ|κα dA. + 2 A

Let us now consider various terms separately, αβ αβ ϕ ϕ β dA − ϕ αβ ϕ αβ ϕ dA = ··α ϕ β dA. A

A

α

A

Applying the divergence theorem on the first term, this term vanishes for a closed surface. Changing the order of differentiation, in the second term we obtain 1 κα β β dA − ϕ αβ ϕ αβ dA = ϕ αβ ϕ ε ε·α ϕ,κ ϕ,β dA ·α R2 A A A β 1 = − (ϕ) ϕ β dA − 2 ϕ |κ ϕ,κ dA R A A β 1 =− (ϕ) ϕ β dA + ϕϕ dA − 2 (ϕ |κ ϕ )κ dA R 1 + 2 ϕϕ dA. R

190

Applications

Applying the divergence theorem to the first and the third terms, these terms vanish because we are dealing with a closed surface. Our final result can now be written as 1 ϕϕ + 2 ϕϕ dA, ϕ αβ ϕ αβ dA = (3.16.53) R A

A

or in the equivalent form " ! αβ 1 2 ϕ αβ ϕ dA = (ϕ) + 2 ϕϕ dA. R A

(3.16.54)

A

Similarly, λ β ·β ψ·β ψ λ dA − ψλ·βλ ψβ dA ψλ·β ψλ dA = A

A

A

β 1 κλ dA ε ε ψ ψ βλ ,κ R2 A A β 1 = − (ψ)β ψ dA − 2 ψλ ψλ dA R A A β =− (ψ) ψ β dA + (ψ)2 dA =−

ψλ·λβ ψβ dA −

A

(3.16.55)

A

λ 1 1 − 2 ψ ψ λ dA + 2 ψψdA R R A A 1 1 2 ψψ + 2 ψψ dA. = (ψ) + 2 ψψ dA ≡ R R A

Finally,

εαλ εβ·κ ψλ·β

A

κα

ψ|

εαλ εβκ ψλ·β ψακ dA

dA =

A

A

α εαλ εβκ ψλ·β ψ κ dA −

= A

εαλ εβκ ψλβκ ψ|α dA

A

1 = − εαλ εβκ ψλ·βκ − ψλ·κβ ψ|α dA. 2 Interchanging the order of covariant differentiation in the second term, we obtain 1 εαλ εβ·κ ψλ·β ψ|κα dA = εαλ εβκ εµλ εκβ ψ,µ ψ|α dA 2R2 A A 1 1 µ α = − 2 2δα ψ ψ,µ dA = − 2 ψ|µ ψ|µ dA (3.16.56) 2R R A A 1 1 1 = − 2 (ψ|µ ψ)µ dA + 2 ψψdA = 2 ψψdA. R R R A

A

A


191

Using these results, we may rewrite (3.16.52) to yield 1 2 2 αβ θαβ θ dA = (ϕ)2 + 2 ϕϕ − wϕ + 2 w2 R R R A

A

1 1 1 2 + (ψ) + ψψ + ψψ dA (3.16.57) 2 2R2 2R2 2 1 2 1 1 = (ϕ)2 + 2 ϕϕ − wϕ + 2 w2 + (ψ)2 + 2 ψψ dA. R R R 2 R A

We now return to (3.16.48) and consider the term (θκκ )2 dA = A

2 2 ϕ − w dA, R

(3.16.58)

A

so that the membrane energy is given by 1−υ 2 (1 + υ) Eh (m) wϕ P2 [u, λ] = ϕϕ − (ϕ)2 + 2 (1 − υ 2 ) R2 R A

2 (1 + υ) 2 1 2 + w + (1 − υ) ψ ψ + 2 ψ dA. R2 2 R

(3.16.59)

We shall now proceed with the evaluation of the bending terms in (3.16.48). First, consider λ λ 1 1 αβ w αβ + ϕ αβ + εαλ ψ ·β + εβλ ψ α ραβ ρ dA = R 2R A A (3.16.60) αβ 1 αβ 1 α κβ β κα ε ψ + ε·κ ψ| dA. + ϕ + × w R 2R ·κ We shall show that here again the coupling terms between ψ and w and ψ and ϕ vanish. Because w and ϕ appear in this expression equivalently, it is sufficient to show that the coupling terms between ψ and w vanish. To show this, consider the term ακ β ·β ε ψ·κ w,α β dA − εακ ψκβ w,α dA. w αβ εα·κ ψκβ dA = w αβ εακ ψβ·κ dA = A

A

A

A

Applying the divergence theorem to the first integral, this term vanishes on a closed surface. Changing the order of covariant differentiation in the second term, we obtain 1 ·β w αβ εα·κ ψκβ dA = − εακ ψβ·κ w,α dA − 2 εακ ελβ εκβ ψ,λ w,α dA R A A 1 ακ ε ψw,α κ dA + εακ ψw |ακ dA − 2 εακ ψ,κ w,α dA. =− R A

A

A

192

Applications

Applying the divergence theorem to the first of these integrals, this term vanishes because we are dealing with a closed surface. The second integral vanishes because w |ακ is symmetric in α and κ. Hence, we are left only with the third term, which can be rewritten as 1 1 − 2 (εακ ψw,α )κ dA + 2 εακ ψw |ακ dA = 0, R R A

A

where the first term vanishes on a closed surface and the second term vanishes due to symmetry in α and κ of w |ακ . Our result is thus w αβ εα·κ ψ κβ dA = 0. (3.16.61) A

This means that (3.16.60) can be rewritten in the form 1 1 w αβ + ϕ αβ w αβ + ϕ αβ ραβ ραβ dA = R R A

A

λ λ α κβ 1 β κα dA εαλ ψ ·β + εβλ ψ ·α (ε·κ ψ + εκ ψ| + 4R2 (3.16.62) αβ αβ αβ 2 1 = w αβ w + w αβ ϕ + 2 ϕ αβ ϕ R R A 1 λ ·β β λ κα ψ dA, + ψ + ε ε ψ ψ| αλ ·κ ·β ·β λ 2R2

where the terms involving ψ have been evaluated in (3.16.55) and (3.16.56). The term containing only ϕ is given in (3.16.53) or (3.16.54), which are obviously also valid when ϕ is replaced by w. Completely analogously, we obtain w ϕ w αβ w αβ dA = ϕ w + 2 dA = w ϕ + 2 dA, (3.16.63) R R A

A

A

so that (3.16.62) can be rewritten to yield w 1 ϕ w w + 2 + w ϕ + 2 ραβ ραβ dA = R R R A

A

w 1 ϕ 1 + ϕ w + 2 + 2 ϕ ϕ + 2 R R R R 1 1 ψ + ψ ψ + 2 + ψψ dA. 2R2 R 2R4

(3.16.64)

We now return to (3.16.48) and consider the term (ρκκ )2 dA = A

2 1 w + ϕ dA, R A

(3.16.65)


193

so that the bending energy is given by h2 2ψ Eh2 (1 − υ) (b) P2 [u, λ] = ψ ψ + 2 2 (1 − υ 2 ) 12 2R2 R A

(3.16.66) 2 1 w ϕ 1 + 3 + w + ϕ + (1 − υ) w + ϕ dA. R R R2 R Next, we must determine the contribution P2 [u, λ] of the stresses in the fundamental state. From (3.16.48), consider the integral (3.16.67) (θαβ − ωαβ ) θαβ − ωαβ + ϕα ϕα dA. A

From (3.16.5) and (3.16.6), we obtain 1 aαβ w, R and using (3.16.38), we can rewrite this expression in the form θαβ − ωαβ = uα|β − bαβ w = uα|β −

(3.16.68)

1 θαβ − ωαβ = ϕ αβ + εαλ ψλ·β − aαβ w. (3.16.69) R The integral (3.16.67) can now be written as λ αβ 1 1 αβ α κβ ϕ αβ + εαλ ψ ·β − aαβ w ϕ + ε·κ ψ − a w R R A (3.16.70) 1 1 1 1 w α + ϕ α + εα·κ ψ|κ dA, + w,α + ϕ,α + εαλ ψλ R R R R where we have used the relation (3.16.44). By arguments similar to those used for the evaluation of the membrane terms and the bending terms, it follows that the coupling terms between ψ and ϕ and ψ and w vanish, so we can rewrite this integral in the form αβ 1 1 αβ 1 1 α α ϕ αβ − aαβ w w| + ϕ| ϕ − a w + w |α + ϕ | α R R R R A (3.16.71) ·β 1 + ψκ·β ψκ + 2 ψλ ψ|λ dA, R Consider αβ 2w α 2 2 1 2 2 2w ϕ αβ ϕ − ϕ + w dA = ϕ+ 2 w dA, (ϕ) ϕ+ 2 ϕ − R α R2 R R R A

A

where we have made use of (3.16.54). For the second term in (3.16.71), we obtain 2 1 w |α w |α + w |α ϕ |α + 2 ϕ |α ϕ |α dA R R A 1 2 2 1 = (w |α w)α − ww + (ϕw |α )α − wϕ + 2 (ϕ |α ϕ)α − 2 ϕϕ dA R R R R A

194

Applications

ww +

=−

2 1 wϕ + 2 ϕϕ dA. R R

A

For the last two terms in (3.16.71), we obtain κ ·β 1 1 λ 1 ψ ·β ψ κ + 2 ψ ψ|λ dA = (ψ) ψ + 2 ψ − 2 ψψ dA R R R A A = (ψ)2 dA, A

where we have made use of (3.16.55). The contribution of the stresses in the fundamental state to the energy functional is thus given by 1 2w Eh2 2 (s) P2 [u; λ] = λ ϕ + 2 w2 (ϕ)2 + 2 ϕϕ − 2cR R R R A 1 2 (3.16.72) − ww − wϕ − 2 ϕϕ + (ψ)2 dA R R 2 Eh2 2 4w λ ϕ − ww + 2 w2 + (ψ)2 dA. =− (ϕ)2 − 2cR R R A

Finally, we shall consider the last term in (3.16.48), 2 w ϕ − w wθαα − uα ϕα dA = R A A α 1 α 1 α λ κ − w + ϕ + ε·λ ψ (ϕ,α + εακ ψ| ) dA R R 2 2 1 α 1 α κ = wϕ − w − w | + ϕ | ϕ |α − ψ|κ ψ| dA R R R A 2 1 1 wϕ − w2 + wϕ + ϕϕ + ψψ dA. = R R R

(3.16.73)

A

The second variation of the potential energy is now given by 1−υ 2 (1 + υ) Eh P2 [u; λ] = wϕ ϕϕ − (ϕ)2 + 2 (1 − υ 2 ) R2 R A 2 (1 + υ) 2 1 2 + w + ϕ − υ) ψ ψ + (1 R2 2 R2 ! 2 1 h2 1 − υ 2ψ + ψ ψ + 2 + w + ϕ (3.16.74) 12 2R2 R R " 1 ϕ w + + (1 − υ) w + ϕ R R2 R3 . σ 2 2 2 2 − 1−υ2 dA. (ϕ) + 2 ϕϕ−ww − 2 w2 + ψ ψ + ψ E R R R


195

To be in the elastic range, we must have σ/E 1, 2 c R2 0 ∂b0

(3.17.90)

it follows that for λ > 1 the equilibrium path is unstable. Because there are no stable equilibrium configurations in the vicinity of the bifurcation point, the shell will collapse explosively. Because the post-buckling behavior of the perfect cylinder is unstable, the cylinder is an imperfection-sensitive structure. To get quantitative results for the imperfection sensitivity, we consider first the case of an axisymmetric imperfection in the direction of the buckling mode. Let this imperfection be given by w0 = κh cos p 0 x/R.

(3.17.91)

According to the general theory, the term 12 σhw2,x in the energy functional must be replaced by 1 2 (3.17.92) w,x + w0x w,x σh 2 so that in the presence of the axisymmetric imperfection, the energy functional is given by P2 [u, u0 ; λ] + P3 [u] = P2 [u; λ] + P3 [u]

(3.17.93) π Eh4 2 − p b0 κλ. c R2 0 The extra term contains only the amplitude b0 , so that only the first of the equilibrium equations (3.17.76) is affected. This equation must be replaced by % & 1 4 (1 − λ) p 20 b0 + 3c q2 cq1 cq2 + m2 c2m − 4λp 20 κ = 0. (3.17.94) 2 q Because prior to buckling cq1 , cq2 , and cm are zero, we readily obtain λ κ. b0 = 1−λ

(3.17.95)

218

Applications λ 1

0 0

.2

.4

.6

κ

Figure 3.17.6

The values of b0 for a non-trivial solution for cq1 , cq2 of the other equilibrium equations follow from (3.17.76), so we must have b20 =

λ2 4 (1 − λ)2 = κ 2, 9c (1 − λ)2

(3.17.96)

from which follows 5 5 3 cλ |κ| ≈ λ |κ| for c = . (3.17.97) 2 2 2 In this case, cq1 and cq2 are nonzero, which means that branching from the axisymmetric mode to a non-axisymmetric buckling mode will occur. It follows that small imperfections reduce the buckling load considerably, e.g., a reduction of the buckling load by 50% is obtained when κ = 1/5, say, for an amplitude of the imperfection of 20% of the shell thickness. This effect is shown in Figure 3.17.6. Let us now consider a more general imperfection again in the direction of the buckling mode, (1 − λ)2 =

w0 = κh [cos(p 0 x/R) + 4 cos(mx/R) cos(my/R)] ,

(3.17.98)

where we have used the fact that cm = 4b0 ; i.e., the amplitude of the non-symmetric mode is four times that of the symmetric mode. In this case, the energy functional is given by P2 [u, u0 ; λ] + P3 [u] = P2 [u; λ] + P3 [u] π Eh4 2 − p 0 b0 + 2m2 cm uλ. 2 c R

(3.17.99)

The equilibrium equations now read % & 1 2 2 2 2 4 (1 − λ) p 0 b0 + 3c q cq1 cq2 + m cm − 4λp 20 κ = 0 2 q 2 (1 − λ) p 2q1 cq1 + 3cq2 b0 cq2 = 0 2 (1 − λ) p 2q2 cq2 + 3cq2 b0 cq1 = 0 2 (1 − λ) m2 cm + 3cm2 b0 cm − 8λm2 κ = 0.

(3.17.100)

3.17 Buckling of circular cylindrical shells

219

For λ = 0, these equations yield the null solution. For λ < λcr , cq1 and cq2 are equal to zero, and the first and last of these equations are simplified to read 3 4 (1 − λ) b0 + cc2m − 4λκ = 0 8 2 (1 − λ) cm + 3cb0 cm − 8λκ = 0

(3.17.101)

and have a solution for cm = 4b0 . The resulting equation for b0 then reads as b20 +

2 2 (1 − λ) b0 − λκ = 0, 3c 3c

(3.17.102)

which yields

b0 = −

1−λ ± 3c

2 (1 − λ)2 + λκ. 2 9c 3c

(3.17.103)

We must choose the positive solution because the result of the negative solution does not refer to the undeformed state. The expression under the square-root sign is always positive when κ > 0, and then the imperfection has a stabilizing effect. However, for κ < 0 the expression may become negative, and the stability limit is reached when (1 − λ)2 = −6cλκ,

κ < 0.

(3.17.104)

Notice that now (1 − λ)2 is four times larger than in the case of an axisymmetric imperfection. In this case, we do not have a branch point but a limit point. The behavior is shown in Figure 3.17.7. Notice that an imperfection of 5% of the shell thickness (κ = 0.05) yields a reduction of the critical load by 50%. λ

b0 = −

2 (1 − λ ) 3c

1 stable unstable

b0 Figure 3.17.7

220

Applications

α 2π

−5

2 3

1

1 π

3 0

−5 −3π

−π

0 −π

−5

3

0 23 −5

3

3

−5

−5

3

−2π

−5 0 1 2

−2π

−5 1

0

−5 2π

3 π

0

3π

β

3 −5

2

0

0 −5

3

2 −5

3

Figure 3.17.8

To get an impression of the pattern of the imperfection, we consider the function f (α, β) = cos 2α + 4 cos α cos β. For

α = (2j + 1) π β = (2k + 1) π

.

and

α = 2j π β = 2kπ

(3.17.105) . ,

(3.17.106)

we have f (α, β) = 5, i.e., a maximum inward deflection. The maximum outward deflection is f (α, β) = −3 and is obtained for . . α = 2j π α = (2j + 1) π . (3.17.107) and β = (2k + 1) π β = 2kπ Additional information is obtained by considering equipotential linesf (α, β) = const. The results are shown in Figure 3.17.8. These results are in agreement with experimental results obtained at Lockheed.† These experiments were carried out with a mandrel with a radius slightly smaller than that of the cylinder, inside the cylinder. The tests were filmed with a high- speed camera (8,000 images per second) to record the dynamic behavior during buckling. †

B. O. Almroh, A. M. C. Holmes, and D. O. Brush, An experimental study of the buckling of cylinders under axial compression. Rep. 6-90-63-104 of Lockheed Missiles and Space Company, Palo Alto, California (1963).

3.18 The influence of more-or-less localized short-wave imperfections

221

w0 5 µ 2 / m2 = 0.1

4 3 2 1 0

2

4

6

8

10

mx / R

Figure 3.18.1

3.18 The influence of more-or-less localized short-wave imperfections on the buckling of circular cylindrical shells under axial compression† So far, we have only considered overall imperfections. In practice, one also often encounters localized imperfections (dimples), which are usually inward deflections. To investigate the influence of these local imperfections, we consider an imperfection of the form 1 2 2 2 2 w0 = κk[cos(p 0 x/R) + 4 cos(mx/R) cos(my/R)] e− 2 µ (x +y ) /R

p 0 = 2m, (3.18.1) i.e., an imperfection in the direction of the buckling mode that is rapidly decaying with increasing values of x and y (even for small values of µ2 /m2 ). To get a feeling for the behavior of this imperfection, we have drawn the graph for µ2 /m2 = 0.1 (see Figure 3.18.1). To solve this problem, we shall employ the Rayleigh-Ritz method, and we assume deflections of the form 1 2 2 2 2 w = hb0 [cos( p 0 x/R) + 4 cos(mx/R) cos(my/R)] e− 2 µ (x +y ) /R 1 2 υ 1−υ 2 2 2 u = hb0 − sin( p 0 x/R) + sin(mx/R) cos(my/R) e− 2 µ (x +y ) /R 2m m 1 2 3+υ 2 2 2 cos(mx/R) sin(my/R) e− 2 µ (x +y ) /R . v = −hb0 m

†

(3.18.2)

Cf. Koiter’s paper under the same title, Rep. Lab. For Appl. Mech. Delft 534, I. N. Vekua Anniversary volume (Moscow, 1978), pp. 242–244.

222

Applications

Let us now consider the term /2 2π − /2 0

1 − σhw2·x dx dy, 2

(3.18.3)

which can be evaluated to yield 1 Eh2 2 2 − λ h b0 2 cR

/2 2π 4m p0 − sin(p 0 x/R) − sin(mx/R) cos(my/R) R R

− /2 0

2 −µ2 (x2 +y2 ) /R2 µ x e dx dy − 2 cos(p 0 x/R) + 4 cos(mx/R) cos(my/R) R 2

1 Eh4 = − λ 3 b20 2 cR

/2 2π / [p 0 sin(p 0 x/R) + 4m sin(mx/R) cos(my/R)]2

− /2 0

2

µ x +2 [p 0 sin(p 0 x/R) + 4m sin(mx/R) cos(my/R)] R × [cos(p 0 x/R) + 4 cos(mx/R) cos(my/R)] 0 µ4 x2 2 2 2 2 + 2 [cos(p 0 x/R) + 4 cos(mx/R) cos(my/R)]2 e−µ (x +y ) /R dx dy. R

(3.18.4)

Let us now first consider integration in the x-direction. First, we notice that the first term in the integrand leads to integrals in the x-direction of the form /2

cos (nx/R) e−µ

2 2

x /R2

dx.

(3.18.5)

− /2

The second term leads to integrals of the form /2 − /2

x 2 2 2 sin (nx/R) e−µ x /R dx, R

(3.18.6)

x2 2 2 2 cos (nx/R) e−µ x /R dx. R2

(3.18.7)

and the third term to /2 − /2

Because the integrants are rapidly decaying functions, we can replace the boundaries of the integrals by −∞ to ∞, so that we must deal with the following integrals, ∞

−µ2 x2 /R2

cos(nx/R) e −∞ ∞

−∞

∞ dx, −∞

2

x 2 2 2 cos (nx/R) e−µ x /R dx. 2 R

x 2 2 2 sin (nx/R) e−µ x /R dx, R (3.18.8)


223

iy i

A′

a 2

C B′

−R

R 0

A

B

x

Figure 3.18.2

Let us consider the first of these integrals. It is convenient to consider the integral 4

e−z dz 2

(3.18.9)

C

in the complex plane, where the contour C is shown in Figure 3.18.2. By Cauchy’s theorem, the value of the integral is zero. On the vertical sides, we have 2 2 2 2 2 −z −(x2 −y2 ) −2ixy e e = e = e−R ey < e−R ea /4 , and this expression tends uniformly to zero as R tends to infinity. Thus, the portions of the whole integral that arise from the vertical sides tend to zero, and if we carry out the passage to the limit R → ∞ and note that dz = d(x + 12 ia) = dx, on A B we can express the result of Cauchy’s theorem as follows, ∞

2

−(x+ 12 ia)

e

∞ dx −

−∞

e−x dx = 0. 2

−∞

The second integral is easily evaluated by noticing that  

2

∞

−x

e

2

dx =

−∞

∞ ∞

−(x2 +y2 )

e

∞ 2π dx dy =

−∞ −∞

e−r r dr dθ = π, 2

−∞ 0

so that we have a2 /4

∞

e

e−x (cos ax − i sin ax) dx = 2

√

π,

−∞

and because sin ax is an odd function, its contribution in the integral vanishes so that ∞ −∞

cos ax e−x dx = 2

√ −a2 /4 πe .

(3.18.10)

224

Applications

The result for the first of the integrals in (3.18.8) is then ∞ −∞

cos(nx/R) e−µ

2 2

x /R

dx =

R √ − 4µn22 πe . µ

(3.18.11)

Noting that n is either zero or a multiple of m (m 1), it follows that only the case n = 0 must be taken into account. Let us now consider the other integrals in (3.18.8). Differentiating (3.18.11) with respect to n, we obtain ∞ −∞

√ x R π − 4µn22 −µ2 x2 /R2 sin(nx/R) e dx = ne , R 2µ3

(3.18.12)

which is zero for n = 0 and is exponentially small for large values of n, and may thus be neglected. Differentiating (3.18.12) with respect to n, we obtain √ 2 n x2 R π − n −µ2 x2 /R2 1 − e 4µ2 . cos(nx/R) e dx = 2 3 2 R 2µ 2µ

∞ −∞

(3.18.13)

For large values of n this integral vanishes, and for n = 0 the contribution is √ R π/2µ3 . Summarizing our results, we find that the contribution of the first term in √ (3.18.4) is of order m2 R π/µ, whereas the contribution of the third term in (3.18.4) √ is of order (1/4)µ2 R π/µ. For small values of µ2 /m2 , we may neglect the contribution of the third term compared with the first term, making a relative error in the energy of order µ2 /m2 . Let us now evaluate the first term using the fact that the integrand is also exponentially decaying in the y-direction, so that the boundaries of integration 0 to 2π can be replaced by −∞ to ∞, ∞ ∞

2 2 2 2 [2m sin(2mx/R) + 4m sin(mx/R) cos(my/R)]2 e−µ (x +y ) /R dx dy

−∞ −∞

∞ ∞ = 4m

[sin2 (2mx/R) + 4 sin(2mx/R) sin(mx/R) cos(my/R)

2 −∞ −∞

2 2 2 2 + 4 sin2 (mx/R) cos2 (my/R)]e−µ (x +y ) /R dx dy ∞ ∞ ! 1 3 2 − cos(2mx/R) − cos(4mx/R) − cos(2my/R) = 4m 2 2

−∞ −∞

+ 2 [cos(mx/R) − cos(3mx/R)] cos(my/R) " 2 2 2 2 + cos(2mx/R) cos(2my/R) e−µ (x +y ) /R dx dy = 4m

2

3 R2 π + exponentially small terms . 2 µ2

(3.18.14)


225

The corresponding term for an imperfection of the form (3.17.98) is 4m2 π R, which means that we can use the results from the previous section if we replace by R/2µ2 . By similar arguments, we find that for the cubic terms must be replaced by R/3µ2 , so the energy expression becomes 3π Eh3 2 3 ∗∗ 2 F = (3.18.15) (1 − λ) b0 + cb0 − 2λκb0 . 2 µ2 3 The equilibrium equation now reads 2 (1 − λ) b0 + 2cb20 − 2λκ = 0,

(3.18.16)

from which b0 = −

1−λ ± 2c

λκ (1 − λ)2 . + 4c2 c

(3.18.17)

For positive values of κ, the term under the square-root sign is always positive, which means that an imperfection in the outward direction is stabilizing. However, for negative values of κ a limit point will occur, i.e., when (1 − λ∗ )2 = −4λ∗ cκ,

κ < 0.

(3.18.18)

When we compare this expression with (3.17.104), we see that the factor 6 is replaced by a factor 4; i.e., a local imperfection is equally harmful as an imperfection of the corresponding periodic type, with an amplitude reduced by a factor 2/3. To investigate the accuracy of the present first-order approximation, Gristchak,† who worked as a research fellow in our laboratory, calculated a second approximation with an error of order µ4 /m4 . His improved approximation yields the following equation for the critical load: 2 µ2 µ2 1 + (d − ε) 2 − λ∗ = −4λ∗ cκ 1 + (e − ε) 2 , (3.18.19) m m where 7 + 5υ − 6υ2 , 12 (1 − υ2 )

ε=

3 , 8

d=

1 {(3 − υ) [υ2 + 2 (1 − υ)2 + 2 (3 + υ)2 ] + 80 1 − υ2 }, 2 9b (1 − υ )

e=

(3.18.20)

For υ = 0.3, we have ε = 0.375,

e = 0.729,

d = 1.540,

and the equation for the critical load then reads 2 µ2 µ2 ∗ ∗ 1 + 1.165 2 − λ = −4λ cκ 1 + 0.354 2 . m m

(3.18.21)

(3.18.22)

In the graph of Figure 3.18.3, the relation between λ∗ and cκ is given for υ = 0.3. †

V. Z. Gristchak, Asymptotic formula for the buckling stress of axially compressed circular cylindrical shells with more-or-less localized shortwave imperfections. W.T.H.D., Rep. 88.

226

Applications

λ* 1.2 1.116 1.0

0.8

0.6 µ 2/m 2 = 0.1 0.4 µ 2/ m 2 = 0 0.2 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

cκ

Figure 3.18.3

These results show that the first approximation is indeed a good approximation. In our analysis, we have employed the Rayleigh-Ritz method, which in itself yields an upper bound for the critical load. However, we have also neglected terms in powers of µ/m, so it is not certain that the present results yield an upper bound. Let us finally remark that actual shells usually have local imperfections due to damage, and these imperfections are in the inward direction, and are therefore strongly destabilizing. It is further interesting to note that even the carefully manufactured cylindrical shells used for the Lockheed tests† had local imperfections.

†

Cf. footnote at the end of Section 3.17.


Over de stabiliteit van het elastisch evenwicht (On the stability of elastic equilibrium). Proefschrift T. H. Delft (1945). On the stability of elastic equilibrium (Washington, 1967), 6 + 202 pp. (NASA Technical Translation F-10, 833, Clearinghouse US Dept. of Commerce/Nat. Bur. of Standards N6725033). The stability of elastic equilibrium (Palo Alto, Cal., 1970), 13 + 306 pp. (Stanford Univ., Dept. of Aeronaut. and Astronaut. Report AD 704124/Air Force Flight Dynamics Lab., Report TR-70-25). “Buckling and post buckling behaviour of a cylindrical panel under axial compression,” Reports and Transactions National Aeronautical Research Institute, 20 (1956) pp. 71–84 (Nat. Aero. Res. Inst. Report No. S476). “A consistent first approximation in the general theory of thin elastic shells.” Proc. I. U. T. A. M. Symposium on the Theory of Thin Elastic Shells (Deflt, August 1959). NorthHolland Publishing Cy., Amsterdam, 12 (1960). “A sysematic simplification of the general equations in the linear theory of thin shells.” Proc. Kon. Ned. Ak. Wet., B64, 612–619 (1961). Introduction to the post-buckling behaviour of flat plates. Actes du Colloque sur le Comporte´ en Construction Metallique, ´ ` (1963). ment Postcritique des Plaques Utilisees Liege “Elastic stability and post-buckling behaviour.” Proc. Symp. Nonlinear Problems, University of Wisconsin Press, Madison, 257–275 (1963). “The concept of stability of equilibrium for continuous bodies.” Proc. Kon. Ned. Ak. Wet., B66, 173–177 (1963). “The effect of axisymmetric imperfections on the buckling of cylindrical shells under axial compression. Proc. Kon. Ned. Ak. Wet., B66, 265–279 (1963). “Knik van schalen (buckling of shells).” De Ingenieur, 76, 033–041 (1964). “On the instability of equilibrium in the absence of a minimum of the potential energy.” Proc. Kon. Ned. Wet., B68, 107–113 (1965). “The energy criterion of stability for continuous elastic bodies.” Proc. Kon. Ned. Ak. Wet., B68, 178–202 (1965). “Post buckling analysis of a simple two-bar frame.” Recent Progress in Applied Mechanics (The Folke Odqvist Volume), Almqvist & Wiksell, Stockholm (1967) 337–354. “On the nonlinear theory of thin elastic shells” Proc. Kon. Ned. Ak. Wet., B69, 1–54. “General equations of elastic stablity for thin shells.”Proc. Symposium on the Theory of Shells, April 4–6, 1066, Houston, Texas, U. S. A. (Houston, Texas U. S. A., 1967) pp. 187– 227. “A sufficient condition for the stability of shallow shells.” Proc. Kon. Ned. Ak. Wet., B70, 367–375 (1967). 227

228


“The nonlinear buckling problem of a complete spherical shell under uniform external pressure.” Proc. Kon. Ned. Ak. Wet., B72, 40–123 (1969). “Postbuckling theory. Jointly with J. W. Hutchinson.” App. Mech. Reviews, 23, 1353–1366 (1970). “Thermodynamics of elastic stablity.” Proc. 3rd Canadian Congress of Applied Mechanics, Calgary, May 1971, 29–37. “An alternative approach to the interaction of local and overall buckling in stiffened panels. Jointly with M. Pignataro. Proc. IUTAM Symposium on Buckling of Structures, Harvard University, June 1974, Springer-Verlag (1976), 133–148. “A basic open problem in the theory of elastic stability.” Joint IUTAM/IMU Symposium on ´ ´ des Applications of Methods of Functional Analysis to Problems of Mechanics. Resum es ´ conferences, September 1 to 6, 1075, Marseille. Part XXIX (Marseille, 1975). “Buckling of a flexible shaft under torque loads transmitted by cardan joints.” IngenieurArchiv, 49, 369–373 (1980).

Index

Adjacent state, 8 Airy’s stress function, 167 Almroth, B.O., 220 Alternating tensor, 121 Beam bending, 126 Biezeno, C.B., 20, 101 Bifurcation condition, 70, 106 Bifurcation point, 41, 166, 178, 201 Bimoment, 81 Branched, 33, 177 Branch point, 32, 40, 219 Brush,D.O., 220 Buckling mode, 17, 22 Cardan,G., 119 Carlson,R.F., 180 Cauchy’s theorem, 223 Center of shear, 78, 126, 131 Characteristic equation, 146 Characteristic length, 18, 111 Chistoffel symbol, 185 Clamped, 59, 81, 106, 131, 141 Clausius-Duhem, 8 Conservative systems, 1, 116 Critical load, 11, 29, 35, 47 Curvilinear, 15 Damping forces, 2, 4 Dead weight loads, 13, 18, 41, 112, 171 Deflection, 5, 27 Deformation tensor, 7 Destabilizing effect, 130 Displacement field, 9, 11 Divergence theorem, 19, 139, 167, 172, 185 Duhem,P., 9, 11 Dynamic boundary conditions, 62, 73, 88, 94, 114, 204 Eccentric loads, 51 Effective relative shortening, 58

Elastic hinge, 61 Elastic potential, 12 Energy functional, 10, 12, 29, 47, 138, 158, 183 Entropy, 7 Entropy production, 8 Equilibrium, 2 Equilibrium equations, 20, 36, 41, 140, 167 Euler column, 35, 91, 101 ¨ Foppl,A., 188 Fourier series, 63 Fourier transform, 144 Frame, 67 Free energy, 8 Fritz,John, 18 Fundamental path, 41, 177 Fundamental state, 7, 20 Gaussian curvature, 182, 189 Generalized displacement, 38, 216 Grammel,R., 20, 101 Greenhill,A.G., 116, 225 Gristchak,V.Z., 181, 225 Haringx,J.A., 101 Heat flux, 8 Helical spring, 101 Hencky, 20 Hoff,N.J., 180 Holmes,A.M.C., 220 Holonomic, 1 Hooke’s law, 16, 25, 40 Hurlbrink, 101 Hutchinson,J.W., 181 Hypersphere, 4 Imperfection, 41 Incompressible, 55, 84 Indefinite, 4, 11, 44 Inextensible, 16

229

230 Irreversible, 8 Isothermal, 10 Karman,von T., 166 Kinematic boundary conditions, 63, 81, 116 Kinematically admissible, 11, 56 Kinetic energy, 1 Kinetic potential, 1 Kirchhoff,G., 29, 50 Kirchhoff-Trefftz stress tensor, 50 Koch, 101 Koiter,W.T., 35, 37, 42, 59, 72, 181, 188, 201, 202, 207, 221 Koiter-Morley equations, 207 Lagrangean multiplier, 87, 107 Lagrangian equations, 2 Laplacian operator, 139 Legendre function, 196 Legendre polynomial, 196 Lekkerkerker,J.G., 61 Limit point, 43, 219 Lower bound, 44, 181 Mean curvature, 182 Measures, 3 Metric tensor, 15, 48, 85, 182 Minimizing direction, 37 Minimizing displacement field, 18 Minimum problem, 23, 57, 163 Multilinear forms, 37 Necessary condition for stability, 5, 25, 57 Neo-Hookean material, 84 Neutral equilibrium, 5, 17, 56, 80, 117, 130, 171, 197 Non-conservative forces, 2 Plate, 84, 137 Positive-definite, 3, 11 Post-buckling, 78, 101, 158, 169 Potential energy, 1, 4 Principle axes of inertia, 110, 127 Radius of gyration, 58 Rayleigh’s principle, 23 Rayleigh-Ritz method, 150, 221

Index Riemann-christoffel tensor, 189 Rotation, 14, 121, 138, 166, 183 Rotation tensor, 49 Rotation vector, 119 Saint Venant,B. de, 128 Scleronomic, 1 Second law of thermodynamics, 8 Second variation, 17 Sendelbeck,R.L., 180 Shallow shell theory, 169 Shear modulus, 17, 184 Shell, 15, 41, 139, 169 Side condition, 25, 42, 68, 214 Simmonds,J.G., 182 Southwell,R.V., 153 Specific mass, 8 Stability, 1 Stability criterion, 3 Stability limit, 18 Stabilizing effect, 130, 219 Stationary, 2 Steepest descent, 38 Strain, 13 Strain tensor, 14 Stress tensor, 13 Strut, 11, 15 Sufficient condition for stability, 17 Surface coordinates, 182 Tangent modulus, 66, 164 Tensor of elastic moduli, 14 Torsion, 78, 102, 110 Torsional stiffness, 79, 102 Total differential, 1 Total energy, 5 Trefftz,E., 20 Unstable, 12 Upper bound, 150, 202, 226 Van der Neut,A., 187 Warping constant, 79, 127 Weakest ascent, 40 Ziegler,H., 116

W. T. Koiter's Elastic Stability of Solids and Structures

W. T. Koiter's elastic stability of solids and structures

Theory of Elastic Stability

Theory of elastic stability

Nonlinear Theory of Elastic Stability (Mechanics of Elastic Stability)

Stability of Structures: Elastic, Inelastic, Fracture and Damage Theories

Mechanics of Elastic Structures

General Theory of Elastic Stability

Basic Theory of Plates and Elastic Stability

Stability of Stochastic Elastic and Viscoelastic Systems

Elastic waves in solids 2

Wave motion in elastic solids

Wave propagation in elastic solids

The Electronic Structures of Solids

The dynamic stability of elastic systems

Encyclopedia of Computational Mechanics. Solids and Structures

Mathematical models for elastic structures

Inelastic Analysis of Solids and Structures

Mathematical Models for Elastic Structures

Mechanics Of Elastic Structures With Inclined Members

Mathematical Models for Elastic Structures

Mechanics of Elastic Structures with Inclined Members

Stability of Structures: Principles and Applications

Dislocations in Solids: Elastic Theory v. 1 (Dislocations in Solids)

Non-linear Modeling and Analysis of Solids and Structures

Non-linear Modeling and Analysis of Solids and Structures

Complex Inorganic Solids: Structural, Stability, and Magnetic Properties of Alloys

Non-linear Modeling and Analysis of Solids and Structures

On the stability of gaseous stellar structures

Non-linear Modeling and Analysis of Solids and Structures

Theory of Arched Structures. Strength, Stability, Vibration

W. T. Koiter's Elastic Stability of Solids and Structures