van der heijden a.m.a. (ed.) w. t. koiter's elastic stability of solids and structures (cup,...

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W. T. KOITER’S ELASTIC STABILITY OFSOLIDS AND STRUCTURES

This book deals with the elastic stability of solids and structures, forwhich Warner Koiter was the world’s leading expert of his time. Itbegins with fundamental aspects of stability, relating the basic notionsof dynamic stability to more traditional quasi-static approaches. Thebook is concerned not only with buckling, or linear instability, butmost importantly with nonlinear postbuckling behavior and imperfec-tion sensitivity. After laying out the general theory, Koiter applies thetheory to a number of applications, with a chapter devoted to each.These include a variety of beam, plate, and shell structural problemsand some basic continuum elasticity problems. Koiter’s classic resultson the nonlinear buckling and imperfection sensitivity of cylindricaland spherical shells are included. The treatments of both the funda-mental aspects and the applications are completely self-contained. Thisbook was recorded as a detailed set of notes by Arnold van der Heij-den from W. T. Koiter’s last set of lectures on stability theory at TUDelft.

Arnold M. A. van der Heijden has his own consultancy, HESTOCONConsultancy B.V. He received his master’s and Ph.D. degrees, withhonors, in mechanical engineering and applied mechanics under Pro-fessor Koiter. He has been a technical and scientific staff member in theApplied Mechanics Laboratory at Delft University of Technology, anhonorary research Fellow at Harvard University, a professor at Eind-hoven Technical University, a board member of the Department ofApplied Mechanics of the Royal Dutch Institute for Engineers, andco-editor (with J. F. Besseling) of the Koiter symposium book Trendsin Solid Mechanics. Dr. van der Heijden has worked on staff and con-sulted for many corporations, including Royal Dutch Shell (Pernis andThe Hague), ABB Lummus Global, and as a project leader of ATEXat General Electric Advanced Materials, SABIC, and Essent. Hehas done gas explosion calculations for offshore platforms, includingstructural analysis. He is currently working on improvements in safetymanagement for ProRail.

W. T. Koiter’s Elastic Stability ofSolids and Structures

Edited by

Arnold M. A. van der HeijdenTechnische Universiteit Delft

CAMBRIDGE UNIVERSITY PRESS

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Cambridge University Press

The Edinburgh Building, Cambridge CB2 8RU, UK

First published in print format

ISBN-13 978-0-521-51528-3

ISBN-13 978-0-511-43674-1

© Arnold van der Heijden 2009

2008

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Contents

Preface page vii

1. Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Discrete systems 1

2. Continuous Elastic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1 Thermodynamic background 72.2 Theorems on stability and instability 112.3 The stability limit 182.4 Equilibrium states for loads in the neighborhood of the

buckling load 272.5 The influence of imperfections 412.6 On the determination of the energy functional for an elastic body 47

3. Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3.1 The incompressible bar (the problem of the elastica) 553.2 Bar with variable cross section and variable load distribution 593.3 The elastically supported beam 613.4 Simple two-bar frame 673.5 Simple two-bar frame loaded symmetrically 723.6 Bending and torsion of thin-walled cross sections

under compression 783.7 Infinite plate between flat smooth stamps 843.8 Helical spring with a small pitch 1013.9 Torsion of a shaft 1103.10 Torsion of a shaft with a Cardan (Hooke’s) joint 1193.11 Lateral buckling of a beam loaded in bending 1263.12 Buckling of plates loaded in their plane 1373.13 Post-buckling behavior of plates loaded in their plane 1583.14 The “von Karman-Foppl Equations” 166

v

vi Contents

3.15 Buckling and post-buckling behavior of shells using shallowshell theory 169

3.16 Buckling behavior of a spherical shell under uniform externalpressure using the general theory of shells 182

3.17 Buckling of circular cylindrical shells 2013.18 The influence of more-or-less localized short-wave imperfections

on the buckling of circular cylindrical shells underaxial compression 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’sUniversity of Technology, in the academic year 1978–79. Although these noteswere prepared in close collaboration with Professor Koiter, they are written in theauthor’s style. The author is therefore fully responsible for possible errors.

This course covers the entire field of elastic stability, although recent develop-ments in the field of stiffened plates and shells are not included. Hopefully, theselecture 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 thissystem is represented by a position vector q(q1, q2 . . . qn), where qi (i ∈ 1 . . . n) aren independent coordinates. It is assumed that the system is holonomic, i.e., no rela-tions exist between the derivatives of the coordinates, and scleronomic, i.e., the fac-tor time is not explicitly needed in the description of the system.† Let qi be the gen-eralized velocities. The kinetic energy T is then a homogeneous quadratic functionof the generalized velocities, and hence T can be written as

T = 12

aij (q)qiqj . (1.1.1)

When the system is non-sclerononic, terms linear in the velocities and a term inde-pendent of qi must be added. The coefficient aij (q) is called the inertia matrix. Theforces acting upon the system can be expressed by a generalized force vector Qdefined by

Qiδqi = v.w., (1.1.2)

where the right-hand side stands for the virtual work of all the forces acting uponthe system. In general, this expression is not a total differential. However, for animportant class of problems, it is. Systems for which 1.1.2) is a total differential arecalled conservative systems. In that case we have

Qiδqi = −δP (q) , (1.1.3)

where δP (q) is a total differential and P (q) is called the potential energy. In the fol-lowing, we mainly restrict our attention to conservative systems because for elasticsystems, 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 dqk.

1

2 Stability

the Lagrangian equations for a conservative system are

ddt

∂L∂qi

− ∂L∂qi

= 0, (i ∈ 1, . . . , n) . (1.1.5)

Using the expression (1.1.1), we may rewrite this equation to yield

ddt

(aij qi) − 12

ahk,i qhqk + P,i = 0, (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., dampingforces). It is then advantageous to take these into account separately as follows:

Qiδqi = δP (q) + Q∗i δqi, (1.1.7)

where Q∗ is the vector of non-conservative forces. The equations of motion thenread

ddt

∂L∂qi

− ∂L∂qi

= Q∗i , (i ∈ 1 · · · n) . (1.1.8)

These are n second-order ordinary differential equations.Let us now consider the stability of discrete systems. For a system to be in equi-

librium, the velocities (and hence the kinetic energy) have to vanish. This impliesthat 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 doesnot cause large deviations from this state of equilibrium. A disturbance from thestate of equilibrium implies that the velocities are nonzero or that the position dif-fers from the equilibrium position. We can always choose our coordinate systemsuch that the equilibrium position is given by q = 0. Furthermore, we can alwayschoose the potential energy in such a way that it vanishes in the equilibrium posi-tion. Doing so, we may write

P = 12

P,ij (0) qiqj + · · · . (1.1.10)

To be able to give a more exact definition of stability, we need a measure to denotethe deviation from the state of equilibrium. Remembering that in equilibrium wehave q = q = 0, a number ρ (q, q) is introduced with the following properties:

1) ρ (q, q) ≤ 0 for q �= 0 or q �= 0,

2) ρ (q1, q2, q1 + q2) ≤ ρ (q1, q1) ρ (q2, q2)

(triangle inequality), (1.1.11)

3) ρ (αq, αq) = |α| ρ (q, q) α ∈ R.


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 thetime 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 differentmeasures may be used for t = 0 and t > 0. This freedom is of great importance forapplications. For example, suitable choices for ρ are

ρ =[

n∑i=1

(qi)2 +n∑

i=1

(qi)2

]1/2

,

ρ = max∣∣qi∣∣+ max

∣∣qi∣∣ .

For a conservative system, T + P = constant. This well-known result can easily bederived from Lagrange’s equations for a conservative, holonomic, and scleronomicsystem. Multiplying the equations by qi, we obtain

qi ddt

∂L∂qi

− qi ∂L∂qi

= 0

or

ddt

(qi ∂L

∂qi

)− qi ∂L

∂qi− qi ∂L

∂qi= 0

or

ddt

(qi ∂L

∂qi

)− d

dtL = 0.

Using Euler’s theorem for homogeneous quadratic functions, we readily obtain

ddt

(2T) − ddt

(T − P) = 0,

from which follows

T + P = E, (1.1.12)

where E = T(t = 0) + P(t = 0). This equation enables us to make the followingstatement about stability.

Theorem. The equilibrium is stable provided the potential energy is positive-definite.

To prove this theorem, we introduce the following norms:

‖q‖2 =n∑

i=1

(qi)2

‖q‖2 =n∑

i=1

(qi)2.

4 Stability

Let d(c) denote the minimum of P(q) on the hypersphere ‖ q ‖ = c . P(q) is positive-definite 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 orzero, it follows that P < d (c1) for all t. A similar argument holds for a disturbance‖ q(0) ‖ < c 2. Hence, we may choose an arbitrary (small) disturbance and the dis-placements and velocities will always remain within definable bounds.

The converse of this theorem has not yet been proven in all generality. To seesome of the difficulties that are encountered, we consider the following example(one degree of freedom):

P(q) = e−q−2cos q−2.

For q = 0, all the derivatives vanish. However, in the immediate vicinity of the ori-gin there are always negative values of P. In spite of this, the system is stable forsufficiently small disturbances.

Actual physical systems are never exactly conservative, i.e., there is alwayssome dissipation. The approximation by a conservative system is often a very goodapproximation. In the presence of damping forces, we need the Lagrangian equa-tions with an additional term for the non-conservative forces. Multiplying by qi,

qi ddt

∂L∂qi

− qi ∂L∂qi

= Q∗i qi, (i ∈ 1, . . . , n)

from which follows

ddt

(T + P) = Q∗i qi ≡ −D (q, q) . (1.1.13)

Damping implies that the dissipation function D > 0 for q �= 0. We now make thefollowing assumptions:

1) The damping forces have the property that Q∗i → 0 for ‖q ‖ → 0.

2) D(q, qi) > 0 for ‖q‖ �= 0.

3) P (q) does not possess stationary values for ‖ q ‖ < cexcept at q = 0.

Systems satisfying these conditions are called pseudo-conservative. Notice thatbecause of restriction (1), dry friction forces are excluded.

Theorem. In the presence of (positive) damping forces, a system with an indefinitepotential energy is unstable.


Proof. If P is indefinite, consider a disturbance of the equilibrium configuration withzero velocity and negative potential energy. The initial total energy is thus negativeand, as this configuration cannot be in equilibrium, motion must result, as a resultof which energy is dissipated. The total energy must decrease, so the system cannotstay in the vicinity of the origin, which means that the equilibrium configuration isunstable.

The great advantage of this stability theorem is that it does not involve thekinetic energy, and hence the inertia matrix aij (q). For a conservative or pseudo-conservative 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 thebehavior of the wings of an airplane in an airflow. In this case, the forces do notdepend 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 mustrestrict ourselves to elastic systems, i.e., to systems where there is a potential for theinternal energy. Such a potential does not exist when plasticity occurs.

Let us now have a closer look at the stability problem. As mentioned previ-ously, the stability criterion is fully determined by the potential energy P(q). In theequilibrium position, we have chosen P (0) = 0 and q = 0 so that we may write

P (q) = 12

P,ij (0) qiqj + · · · ≡ 12

cij qiqj + · · · , (1.1.14)

where cij denotes the stiffness matrix in the equilibrium position. It follows thatwhen the stiffness matrix is positive-definite, P (q) is positive-definite and the systemis 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 onedirection), then we must consider higher-order terms in the expansion for P. Thiscase is called a critical case of neutral equilibrium. We shall consider this case inmore detail.

It is convenient to transform the quadratic form (1.1.14) to a sum of quadraticterms. If the form is positive-definite, then the coefficients in the transformed formare all positive. Applying this transform to (1.1.14) and denoting the transformedcoordinates again by qi, we may write

P (q) = 12

n∑i=1

ci(qi)2 + · · · + () (q1)3 + · · · . (1.1.15)

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 dominantterm will be (q1)3. This term can attain negative values, and hence the system will beunstable. A necessary condition for stability is that the coefficient of (q1)3 is equal

6 Stability

x

x = y2

x = 2y2

f x, y) < 0(

y

Figure 1.1.1

to zero. A further necessary condition for stability is that the coefficient of (q1)4 ispositive. However, this condition is insufficient, as will be shown in the followingexample. 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 atthe origin, and there is no cubic term, but the coefficient in the quartic term is posi-tive. Hence, the necessary conditions for stability are satisfied. However, this systemis unstable because in an arbitrarily small neighborhood of the origin, P takes onnegative values.

The reason that the conditions mentioned here are not sufficient is that we haverestricted our investigation to straight lines through the origin (see Fig. 1.1.1). Fol-lowing these straight lines, we always find only positive values in a sufficiently smallneighborhood 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 thereason why the conditions imposed are insufficient, it is easy to find a remedy. Tothis end, we consider a line y = constant in the neighborhood of the origin, and weminimize f (x, y) with respect to x, i.e.,

Miny=y1

f (x, y) = x2 − 3xy21 + 2y4

1. (1.1.17)

This yields 2x − 3y21 = 0, and hence x = 3/2 y2

1. Substitution of this value into f (x, y)yields min f (x, y) = −1/4 y4

1, 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 aimis to investigate this equilibrium state.

For an elastic body, the internal energy per unit mass may be represented byU(s, γ), where s denotes the specific entropy and γ is the deformation tensor. Letxi (i = 1, 2, 3) be the components of the position vector x, which describe the posi-tion of a material point in the “fundamental state” I, which is to be investigated. Letu(x) be the displacement vector from the fundamental state (u is a small but finitedisplacement). The corresponding position in the “adjacent state” II is then x + u.The (additional) deformation tensor is now defined by

γij = 12

(ui,j + uj ,i) + 12

uh,iuh,j . (2.1.1)

The fact that the body has undergone deformations to arrive in the fundamentalstate 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∂s

= ∂2U∂s2

.

The specific heat of the material is now defined by

T∂s∂T

= Cγ, (2.1.3)

where Cγ > 0 for a thermodynamically stable material. As ∂2U/∂s2 is positive (non-zero) 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 Continuous Elastic Systems

F (T, γ) is called the free energy.Writing (2.1.4) as a total differential, we find (for fixed γ)

∂F∂T

δT = ∂U∂s

δs − Tδs − sδT. (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 theequilibrium state is equal to the temperature of its surroundings) by TI. A distur-bance 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 isgiven 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,† vheatwill flow out of the body when its surface temperature is higher than that of thesurrounding 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 = 1T

q (per unit time and per unit area). (2.1.8)

For an arbitrary part of the body, the entropy balance is given by∫V

ρs dV =∫A

h · n dA, (2.1.9)

where ρ is the specific mass. This equation only holds in the absence of irreversibleprocesses in the body. When the state of the body also depends on the deformationrates, irreversible processes will occur, which implies entropy production. In thatcase, the entropy balance reads∫

V

ρs dV =∫A

h · n dA +∫V

ρσ dV, σ ≥ 0, (2.1.10)

where σ denotes the entropy production per unit time and mass. This is the moregeneral formulation of the second law of thermodynamics (Clausius-Duhem). Thefirst law of thermodynamics states that the total amount of heat that flows into abody is transformed into internal energy.

Let PL [u (x (t))] be the potential energy of the external loads and let

K [u (x (t))] = 12

∫V

ρ u · u dV

† This is an early formulation by Clausius (1854).

2.1 Thermodynamic background 9

be the kinetic energy. The total energy balance is then given by

ddt

∫V

ρU (s, γ) dV + K [u (x (t))] + PL [u (x (t))]

= − ∫A

q (x (t)) · n dA, (1st law)

(2.1.11)

where we have a negative sign on the right-hand side of this equation because theheat 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

ddt

∫V

ρ [U (s, γ) − TIs] dV + K [u (x (t))] + PL [u (x (t))]

(2.1.12)

=∫V

(TI

T− 1)

q · n dA − TI

∫V

ρσ dV ≤ 0

(Duhem, 1911).Here we have made use of the relation∫

V

ρs dV = ddt

∫V

ρs dV (2.1.13)

The first term on the right-hand side of (2.1.12) is negative because the heat flux isin the outward direction when T > TI, and the second term is negative because theentropy 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) − TIs = U (s, γ) − Ts + (T − TI) s = F (T, γ) + (T − TI)∂F∂T

, (2.1.14)

we obtain

ddt

∫V

ρ

[F (T, γ) + (TI − T)

∂F∂T

]dV + PL + K

≤ 0. (2.1.15)

Duhem (1911) already discussed the stability of a system on the basis of this equa-tion and came to the conclusion that a system is stable when the form between thebraces 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 thetemperature, whereas PL depends on the displacement field. The problem is to sep-arate the influence of the temperature and the displacement field. A straightforwardexpansion

F (T, γ) = F (TI, γ) +(

∂F∂T

)TI

(T − TI) + 12

(∂2F∂T2

)TI

(T − TI)2 + · · ·


does not solve the problem. Following Ericksen (1965), we may write the Taylorexpansion of the free energy at constant deformation γ in the form

F (TI, γ) = F (T, γ) +(

∂F∂T

)T

(TI − T) + 12

(∂2F∂T2

)∗

(TI − T)2, (2.1.16)

where the first derivative is evaluated at the deformation γ and temperature T, andthe second (starred) derivative at the deformation γ and an intermediate tempera-ture T∗ = T + θ (TI − T), where 0 < θ < 1. Using (2.1.16) we may rewrite the termbetween the square brackets in (2.1.15) as follows:

F (T, γ) + (TI − T)∂F∂T

= F (TI, γ) − 12

(∂2F∂T2

)∗

(TI − T)2

= F (TI, γ) + 12

(cγ

T

)∗

(TI − T)2,

(2.1.17)

where we have used the relation

cγ ≡ T∂s∂T

= T∂2F∂T2

.

The first term on the right-hand side of (2.1.17) depends only on the displacementfield. The second term is positive-definite. The energy balance may now be writtenin the form

ddt

∫V

ρF (TI, γ) dV + PL [u (x (t))] + K [u (x (t))]+12

∫V

ρ(cγ

T

)∗

(TI −T)2 dV

≤0.

(2.1.18)

The last two terms in the left-hand member are positive-definite, and the remainingterms depend only on the displacement field. Our energy balance is not affectedwhen we subtract from the expression between the braces a time-independentquantity, ∫

V

ρF (TI, 0) dV.

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 cur-rent state. The potential energy functional P is now defined by

P [u (x (t))] =∫V

W (γ) dV + PL [u (x (t))] . (2.1.20)

In words: The potential energy is equal to the sum of the increase of the elasticenergy 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 con-stants.† The question of stabilty when TI is not constant is still unsolved (which isimportant, 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 needmeasures to be able to specify expressions like “small disturbance” and “not largedeviations.” In our discussion of the stability of continuous systems, which is gov-erned by the character of the potential energy, we need a measure for the displace-ments. A suitable measure is the L2-norm of the displacement field, defined by

∥∥u (x (t))∥∥2 = 1

M

∫V

ρ u (x (t)) · u (x (t)) dV, (2.2.1)

where M is the total mass of the elastic body. We shall employ the same measurefor the initial disturbance. We shall assume that the potential energy functional isregular in the following sense. On every ball ‖u‖ = c in the function space of kine-matically admissible displacement fields u (x (t)) where the radius c is sufficientlysmall, the energy functional P [u (x (t))] has a proper minimum that is a continuousfunction d(c) in the range of c under consideration. The potential energy functionalis called positive-definite if the function d(c) is a (positive) increasing function in arange 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 equilibriumin the fundamental state is stable if the potential energy functional P [u (x (t))] ispositive-definite.

To show this, we introduce the notation

V [u (x (t)) , u (x (t)) , T (x (t))] = P [u (x (t))](2.2.2)

+K [u (x (t))] + 12

∫V

ρ(cγ

T

)∗

(TI − T)2 dV,

where V is the total energy.‡

† In the literature, one frequently encounters vague and loose statements to the effect that bucklingis “rapid” and that it is therefore “reasonable” to assume that the motion is adiabatic. This wouldimply that elastic stability would be governed by the adiabatic elastic constants rather than by theisothermal 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 smallvalues of the compressive load the straight configuration is stable, and this stability is manifestedby a non-vanishing fundamental frequency. This frequency decreases when the critical Euler loadN1 is approached, and it vanishes for N1. The motion at the critical load is thus infinitely slow, andhence isothermal.

‡ Duhem called it the “ballistic energy.”


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 veloc-ities and temperature variations that are both identically zero. The initial value V0 ofthe total energy is then negative, and by (2.1.18) this energy will decrease until themotion comes to a final stop. Because no equilibrium will be possible in the range0 < c < c1 of ‖u‖, it follows that this norm at some time must approach or exceedthe value c1, no matter how small the initial disturbance has been chosen. It followsthat 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 practicalproblems.

The results obtained so far are not as useful as it may seem because so far ithas proved to be impossible to show that the elastic energy functional is positive-definite, even for an elastic body without external loads. To state this problem moreclearly, we consider the potential energy in the elastic body

P [u (x (t))] = PLu (x (t)) +∫V

W (γ) dV, (2.2.5)

where the elastic potential W (γ) is expanded about its value in the fundamentalstate

W (γ) =(

∂W∂γij

)I

γij + 12

(∂2W

∂γij ∂γk

)I

γij γk + · · · (2.2.6)


This expansion contains linear and quadratic terms in the strains, which in theirturn 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). Ithas been demonstrated that it is impossible to show that W is positive-definite onthis 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 systemis unstable.

Let us now consider the first term in (2.2.6). By virtue of

δW = Sij δγij , (2.2.7)

where Sij is a symmetric stress tensor, we have

Sij = 12

(∂W∂γij

+ ∂W∂γj i

)I

≡(

∂W∂γ(ij )

)I

. (2.2.8)

The potential energy may now be written as

P [u (x (t))] =∫V

[12

Sij (ui,j + uj ,i + uh,iuh,j )

(2.2.9)

+ 12

(∂2W

∂γij ∂γk

)I

γij γk + · · ·]

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

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

P1 [u (x (t))] =∫V

12

Sij (ui,j + uj ,i) dV + PL1 [u (x (t))] = 0. (2.2.11)

Hence (2.2.9) may be written as

P [u (x (t))] =∫V

[12

Sij uh,iuh,j + 12

(∂2W

∂γij ∂γk

)I

γij γk + · · ·]

dV + PL2 [u (x (t))] + · · ·

(2.2.12)

In the remaining part of these lectures, we shall restrict ourselves to dead-weightloads, unless mentioned otherwise. Thus,

Pd.w.L [u (x (t))] = PL1 [u (x (t))] , (2.2.13)


so that the discussion of stability is focused on the integral in (2.2.12). We now definea tensor of elastic moduli

EIijk ≡

(∂2W

∂γ(ij )∂γ(k)

)I

, (2.2.14)

where W is written symmetrically with respect to γij and γk. Notice that this is notthe tensor of elastic moduli that is usually used in the theory of elasticity becausethat tensor is defined by

E0ijk ≡

(∂2W

∂γ(ij )∂γ(k)

)0

, (2.2.15)

where the index 0 indicates that the second derivatives of W must be evaluated inthe undeformed state. Then the tensor of elastic moduli for a homogeneous isotropicmaterial is given by

E0ijk = Gγ

[δikδj + δiδjk + 2ν

1 − 2νδij δk

]. (2.2.16)

This tensor gives a complete description of the elastic material when the elasticpotential is given as

W0 = 12

E0ijkγ

0ij γ

0k, (2.2.17)

i.e., when W0 is a homogeneous quadratic form in the strain components. Noticethat here we have used Cartesian coordinates in the undeformed state. The defor-mation tensor is

γij = 12

(ui,j + uj ,i + uh,iuh,j ) , (2.2.18)

where u now denotes the displacements with respect to the undeformed configura-tion and (),i = ∂ () /∂xi, where xi are Cartesian coordinates. The description of W0

by (2.2.17) is in principle only valid for infinitesimally small strains, and even thenonly the linear terms in the strain tensor are important. The expression for the elas-tic potential may now be generalized by assuming that for finite strains (where thequadratic terms in γij may become important), the elastic potential can still be rep-resented by a quadratic function. The fact that quadratic terms in the strain tensormay become important even when the linearized strain tensor θij ,

θij ≡ 12

(ui,j + uj ,i) , (2.2.19)

is small is immediately clear from the fact that∣∣θij∣∣� 1 does not imply that the

linearized rotation tensor

ωij ≡ 12

(ui,j − uj ,i) (2.2.20)

is small.


The variation of W for a small disturbance from the equilibrium configurationis given by

δW = E0ijkγ

0ij δγ

0k + 1

2E0

ijkδγ0ij δγ

0k (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 possi-ble, at least in principle, to find curvilinear coordinates in the undeformed configu-ration that become Cartesian coordinates in the fundamental state I. In a curvilinearsystem, the variation of W is given by

δW = Eαβλµ

0 γ0αβδγ

0λµ + 1

2Eαβλµ

0 δγ0αβδγ

0λµ, (2.2.22)

where the contravariant components of the tensor of elastic moduli are given by

Eαβλµ = G(

gαλ0 gβµ

0 + gαµ

0 gβλ

0 + 2ν

1 − 2νgαβ

0 gλµ

0

). (2.2.23a)

The difference of the metric tensors in the undeformed configuration and in thefundamental state is

gij0 − 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 that12

Eαβλµ

0 γ0αβγ

0λµ = 1

2EI

ijkγij γk [1 + O (ε)]

and hence

12

EIijkγij γk ≈ 1

2Eijk γij γk > 0, (2.2.24)

where Eijk is the tensor of elastic moduli that is used in the theory of elasticity. Thefact that we have approximated the elastic energy with a relative error of O (ε) doesnot affect the positive-definite character of the potential energy if the first term inthe 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 validbecause 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 func-tion in the strains with the classical tensor of elastic moduli implies that wemay also apply additional approximations used in the theory of elasticity,e.g., beam theory, plate, and shell theories.


Example . Consider a simply supported strut, loaded in compression by a forceN.

NN

x

x

w

u*

ϕ

Figure 2.2.1

In the engineering approach, the strut is usually assumed to be inextensible.We have the following relations:

sin ϕ = dwdx

≡ w′ cos ϕdϕ

dx≡ w′′

cos ϕ =√

1 − w′2 dϕ

dx= κ = w′′

√1 − w′2

u∗ = −∫

0

√1 − w′2 dx.

For an inextensible strut, the potential energy is

P [w (x)] =∫

0

12

EIw′′2

1 − w′2 dx +∫

0

N[√

1 − w′2 − 1]

dx. (2.2.25)

Notice that now the potential energy of the external loads is a nonlinear functionof 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 haveassumed that the strut is inextensible, which means(

1 + u′2)+ w′2 = 1 or u′ =√

1 − w′2 − 1.

In this case, we may say that buckling occurs when the energy supplied by theloads is equal to the strain energy in bending. This may not be generalized.

Returning to the general theory and restricting ourselves to dead-weight loadsand to materials that follow the generalized Hooke’s Law, we may write

P [u (x)] =∫V

[12

Sij uh,iuh,j + 12

Eijkγij γk

]dV, (2.2.26)

where the strain tensor is given by

γij = 12

[ui,j + uj ,i + uh,iuh,j

] ≡ θij + 12

uh,iuh,j . (2.2.27)


The potential energy functional may now be written as (only for dead-weight loads)

P [u (x)] =∫V

[12

Sij uh,iuh,j + 12

Eijkθij θk

]dV

+12

∫V

[Eijkθij um,kum, + Eijkθkuh,iuh,j

]dV + 1

8

∫V

Eijkuh,iuh,j um,kum, dV,

(2.2.28)

where we have arranged the terms so that the integrants contain only terms of sec-ond, third, and fourth degree, respectively, in the displacements. Writing

P [u (x)] = P2 [u (x)] + P3 [u (x)] + P4 [u (x)] (2.2.29)

and using the fact that Eijk = Ekij , we have

P2 [u (x)] = 12

∫V

[Sij uh,iuh,j + Eijkθij θk

]dV

P3 [u (x)] = 12

∫V

Eijkθij um,kum, dV (2.2.30)

P4 [u (x)] = 18

∫Eijkuh,iuh,j um,kum, dV.

A positive-definite energy functional means P2 + P3 + P4 ≥ 0. However, becauseP3 and P4 contain only higher-order terms, it is usually sufficient to consider onlyP2. P2 [u (x)] must be positive-definite for all kinematically admissible displacementfields u (x). When ui and ui,j are sufficiently small, then P2 [u (x)] > 0 is a sufficientcondition 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, andu1 (x) is called the buckling mode.

In the following, we shall argue that this case is only possible in slender con-structions. To see this, we first notice that according to (2.2.19) and (2.2.20) we maywrite

ui,j = θij − ωij (2.2.31)

so that

Sij uh,iuh,j = Sij θhiθhj − 2Sij θhiωhj + Sij ωhiωhj . (2.2.32)

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


construction under consideration must be slender, we write

ωij ,k = 12

(ui,j − uj ,i)′k = 12

(ui,jk − uj ,ik) = 12

(uj ,k − uk,j )′i

−12

(ui,k + uk,i)′j = θjk,i − θik,j .

(2.2.33)

For a sufficiently supported construction,

ωij ,k = O(ω

),

where is a characteristic length of the construction. Large values of one or morecomponents of the rotation tensor are only possible if one or more strains are oforder (ω/)∗ where ∗/ � 1. This condition means that the construction has asecond characteristic length ∗ that is considerably smaller than . This implies thatthe construction is slender (e.g., beams, plates, shells).

A rigorous mathematical proof of this scenario was given by Fritz John. Heshowed that when γij = O(ε), ε � 1, and the dimensions of a body are all of thesame 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 materialfollows the generalized Hooke’s Law, the potential energy is given by

P [u (x)] =∫ [

12

Sij uh,iuh,j + 12

Eijkγij γk

]dV. (2.3.1)

As discussed in Section 2.2, stability is primarily determined by the character of P2,given by

P2 [u (x)] =∫ [

12

Sij uh,iuh,j + 12

Eijkθij θk

]dV. (2.3.2)

If the second variation is positive-definite, the equilibrium is stable, and if thesecond variation is indefinite (or negative-definite), the equilibrium is unstable. Thesecond variation is unable to give a valid decision on the stability of instability in thecritical case that it is semi-definite positive. Let u (x) be a minimizing displacementfield. Then

P2 [u (x) + εζ(x)] ≥ P2 [u (x)] (2.3.3)

for all kinematically admissible displacement fields ζ(x) and for sufficiently smallvalues 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] + ε2P2 [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 followsthat

P11 [u, ζ] = 0. (2.3.6)

This equation is the variational equation for neutral equilibrium. From the func-tional (2.3.2), we now obtain for the bilinear term

P11 [u, ζ] =∫ [

Sij uh,iζh,j + Eijkθij12

(ζk, + ζ,k)]

dV, (2.3.7)

where ζk are the Cartesian components of ζ.Due to the symmetry of Eijk in the indices k, we may write

Eijkθij · 12

(ζk, + ζ,k) = Eijkθij ζk,.


Eijkθij = σk. (2.3.8)

Here σk are the stresses corresponding to the linearized strain tensor in the absenceof prestresses. Thus (2.3.7) may now be rewritten as∫

(Sij uh,j + σhj ) ζh,j dV = 0, (2.3.9)

and using the divergence theorem we obtain∫AP

(Sij uh,i + σhj ) ζhnj dA −∫V

(Sij uh,i + σhj ),j ζh dV = 0 for ∀ζ. (2.3.10)

According to the principal theorem in the calculus of variations, we then must have

(Sij uh,i + σhj ),j = 0 in V,† (2.3.11)

(Sij uh,j + σhj ) nj = 0 on AP, ui = 0, 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

ζ=

[(

xi − x∗i

) (xi − x∗

i

)− R2]2 for(xi − x∗

i

) (xi − x∗

i

) ≤ R2

0 for(xi − x∗

i

) (xi − x∗

i

) ≥ 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.


dy

dx1 + ε x(

)dx1+ε y

()d

y

ψ / 2

ψ / 2

ψ / 2

ψτ

/ 2

IFundamental State

IINo Rotation

IIIFinal State

ω

σy

σ x

Sy

SxSxy

Syx rxy

xy

τyx

rxy

Sx

Sy

Sxy

Syx

Figure 2.3.1

Notice that in the absence of prestresses Sij , these equations reduce to the equationsfrom the classical theory of elasticity. Performing differentiation by parts, we obtainfrom (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 aregiven by {

Sij ,j + Xi = 0 in VSij nj = Pi on Ap

, (2.3.14)

where Xi are the mass forces and Pi are prescribed tractions on Ap . Using theseexpressions, we can rewrite (2.3.12) and (2.3.13) as

− Xiuh,i + Sij uh,ij + σhj ,j = 0 in V (2.3.15)

piuh,i + σhj nj = 0 on Ap, ui = 0 on Au. (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 andHencky (cf. C.B. Biezeno and R. Grammel, Engineering Dynamics, Vol. I). Weshall reproduce here their derivation for the two-dimensional case (to simplify theanalysis).

Consider a rectangular material element with dimensions dx and dy in the fun-damental 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 rotationof 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 forcesSx, Sy, and Sxy acting on the deformed element. To restore equilibrium (to a firstapproximation), we add additional (small) forces σx, σy, and τxy (τxy = τyx). Theseadditional forces do not enable us to satisfy the equilibrium of moments exactly.


To reach this goal, skew-symmetric shear forces rxy (ryx = −rxy) must be added. It isobvious that the final rotation does not disturb the equilibrium of moments.

The equilibrium of moments requires

Sxdy dx12ψ+ Syx dx (1 + εy) dy − Sxy dy (1 + εx) dx

− Sy dx12ψdy − rxy dy dx − rxy dx dy = 0,

from which follows

2rxy = 12ψ(Sx + Sy) + Sxy (εy − εx) . (2.3.17)

The equilibrium of forces yields

(Sx + σx − Sxyω) , x + (Syx + τyx − rxy − Syω) , y + X = 0(2.3.18)

(Sy + σy − Syxω) , y + (Sxy + τxy − rxy − Sxω) , x + Y = 0.

In the fundamental state, we have

Sx,x + Syx,y + X = 0

Sy,y + Sxy,x + Y = 0.(2.3.19)

Substitution of these equations into (2.3.18) and using the relation (2.3.17) yields

σx,x + τyx,y − (Sxyω) ,x − (Syω) ,y −14

[ψ(Sx − Sy)] ,y −12

[Sxy (εy − εx)] ,y = 0

τxy,x + σy,y + (Syxω) ,y + (Sxω) ,x +14

[ψ(Sx − Sy)] ,x +12

[Sxy (εy − εx)] ,x = 0.

(2.3.20)

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)

σx,x + τyx,y −[

12

Sxy (v,x −u,y)]

,x

−[

Sy12

(v,x −u,y)]

,y

(2.3.21)− 1

4[(v,x + u,y) (Sx − Sy)]

,y − 12

[Sxy (v,y − u,x)],y = 0.

The general result from Biezeno and Hencky may be written in the form(σij + 1

2Shiθhj − 1

2Shj θhi + Sihωhj

),i

= 0 (2.3.22)

or rewritten as [σij + Shi (θhj + ωhj ) − 1

2Shiθhj − 1

2Shj θhi

],i

= 0

or (σij + Shiuj ,h − 1

2Shiθhj − 1

2Shj θhi

),i

= 0. (2.3.23)


(Equation 2.3.21) is the two-dimensional form of (2.3.22) for j = 1. Our earlierresult (2.3.11) was

(σij + Shiuj ,h),i = 0. (2.3.24)

The additional terms in (2.3.23) can easily be derived from the variational approachby adding to the energy density the term 1

2 Shiθhj θij . This is small, of order O(ε) com-pared to 1

2 Eijkθij θk, in which we had already admitted a relative error of O(ε).Hence it follows that the equations for neutral equilibrium derived by Biezenoand Hencky and those derived by Trefftz are equivalent within the scope of ourtheory.

We now continue with our general discussion of neutral equilibrium, and weconsider the case that

P2 [u (x)] ≥ 0, P2 [u1 (x)] = 0,

where u1(x) is the buckling mode. The question now arises whether u1 is the onlykinematically admissible displacement field for which P2 [u (x)] vanishes. To investi-gate this condition we might look for additional solutions of the equations of neutralequilibrium, but it proves to be more useful to proceed as follows. Consider the setof orthogonal (kinematically admissible) displacement fields. To define orthogonal-ity, 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 argumentsas in the discussion of P2, we find that the bilinear term must vanish,

T11 [u (x) , v (x)] = 0, (2.3.26)

which defines the orthogonality of u and v.A possible choice for T2 is

T2 [u] = 12

∫V

Gui,j ui,j dV, (2.3.27)

where the integrant is a positive-definite quadratic function of the displacement gra-dients. The factor G has been added to give T2 the dimension of energy. The van-ishing of the bilinear term yields

G∫V

ui,j vi,j dV = 0. (2.3.28)

Another suitable choice is

T2 [u] = 12

∫V

ρuiui dV, (2.3.29)


where ρ is the mass density. This choice is motivated by Rayleigh’s principle for thedetermination of the lowest eigenfrequency of small vibrations about an equilibriumconfiguration, which states

ω2 = MinP2 [u (x)]∫

V

12ρuiui dV

. (2.3.30)

In the critical case of neutral equilibrium, ω vanishes, which means that the motionis infinitely slow. It will turn out that the results are independent of the particularchoice 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, u1]

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

, (2.3.32a)


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] = a2P2 [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, andhence also for displacement fields orthogonal to u1. In other words, it follows thatwhen P2 [u] = 0 for u �= u1, P2 [u] also vanishes for a displacement field orthogonalto u1. We may thus restrict ourselves to displacement fields that are orthogonal tou1. As P2 [αu] → 0 for α → 0, it is more suitable to consider the following minimumproblem:

MinT11[u1,u]=0

P2 [u (x)]T2 [u (x)]

= λ2, (2.3.33)

where λ2 is the minimum of the left-hand member. If λ2 = 0, then we have a secondbuckling mode, and when λ2 > 0, u1 is then the unique buckling mode. Let u2 be thesecond buckling mode; then

P2 [u2 + εη]T2 [u2 + εη]

≥ P2 [u2]T2 [u2]

= λ2 (2.3.34)

and

T11 [u2, u1] = T11 [η, u1] = 0. (2.3.35)

† See following equation (3.3.55).


Rewriting (2.3.34), we find

P2 [u2] + εP11 [u2, η] + ε2P2 [η] ≥ λ2{T2 [u2] + εT11 [u2, η] + ε2T2 [η]

}where P2 [u2] − λ2T2 [u2] = 0, so that

ε{P11 [u2, η] − λ2T2 [u2, η]

}+ ε2 {P2 [η] − λ2T2 [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] − λ2T2 [u2, η] = 0 (2.3.37)

for all displacement fields with T11 [u1, η] = 0. In this condition, the displacementfield η is orthogonal to u1. In our original condition of P11 [u, ζ] = 0, ζ was not sub-mitted to this requirement. We shall now show that (2.3.37) also holds for arbitrarydisplacement fields ζ. To show this, we write ζ= tu1 + η, whereη satisfies the orthog-onality condition. Replacing η in (2.3.37) by ζ, we find

P11 [u2, tu1 + η] − λ2T11 [u2, tu1 + η]

= tP11 [u2, u1] + P11 [u2, η] − λ2tT11 [u2, u1] − λ2T11 [u2, η](2.3.38)

= t{P11 [u2, u1] − λ2T11 [u2, u1]

}+ P11 [u2, η] − λ2T11 [u2, η]

= P11 [u2, η] − λ2T11 [u2, η] = 0

It follows that without a loss of generality we may restrict ourselves to displacementfields that are orthogonal to the ones that are already known.

Suppose we have found m linearly independent solutions. The solution ofP11 [u, ζ] = 0 can then be written as

u (x) = ahuh, 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 anarbitrary displacement field u (x)

u (x) = ahuh (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] = ahT11 [uk, uh] + T11 [uk, u] = ak. (2.3.41)

A substitution of (2.3.40) into the energy functional yields

P [u (x)] = P2 [ahuh + u] + P3 [ahuh + u] + P4 [ahuh + u] + · · ·= P2 [ahuh] + P11 [ahuh, u] + P2 [uh]

(2.3.42)+ P3 [ahuh] + P21 [ahuh, u] + P12 [ahuh, u] + P3 [u]

+ P4 [ahuh] + P31 [ahuh, u] + P22 [ahuh, u] + P13 [ahuh, u] + P4 [u] + · · ·.


When the material follows the generalized Hooke’s Law, the expansion terminatesafter P4 [u]; in other cases, higher-order terms follow. Because uh are bucklingmodes, the first term on the right-hand side vanishes, P11 [ahuh, ζ] = 0, for all dis-placement fields ζ, and hence also for u. Because we have already found m bucklingmodes, P2 [u] must satisfy the relation

P2 [u] ≥ λm+1T2 [u] , λm+1 > 0. (2.3.43)

Let us now first consider the case that u = ≡ 0. We then only have to deal with theterms P3 [ahuh] and P4 [ahuh]. A necessary condition for stability is that P3 [ahuh]vanishes and that P4 [ahuh] ≥ 0. Hence it follows that from the mathematical pointof 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 tothe symmetry of the structure. The conditions mentioned are necessary conditions;to obtain sufficient conditions, we consider small values of u. In fact, the stabilityconditions must be satisfied for small deviations from the fundamental state.

Suppose now that the necessary conditions are satisfied; thus the most importantterms in (2.3.42) are

P2 [u] , P21 [ahuh, u] , P4 [ahuh] ,

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,

MinT11[uk,u]=0

ak const.k∈(1,...,m)

(P2 [u] + P21 [ahuh, u]) . (2.3.44)

This is a meaningful minimum problem because P2 is a positive-definite functionaland P21 is a functional linear in u. Let u = v be the solution to this problem; thus fora variation of this field v + εη we have

P2 [v] + εP11 [v, η] + ε2P2 [η] + P21 [ahuh, v] + εP21 [ahuh, η] ≥ P2 [v] + P21 [ahuh, v]

or

ε (P11 [v, η] + P21 [ahuh, η]) + ε2P2 [η] ≥ 0. (2.3.45)

Because this inequality must hold for all sufficiently small values of ε, it follows that

P11 [v, η] + P21 [ahuh, 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 vwith v∗, from (2.3.46) we find

P11 [v − v∗, η] = 0. (2.3.47)


By arguments similar to those following (2.3.37), we may also replace η by ζ, whereζ is an arbitrary displacement field.† Further, we have

T11 [ahuh, 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 com-pletes our proof that v is unique.

Equation (2.3.46) is quadratic in the buckling modes but linear in v, so the solu-tion can be written as

v (x) = ahakvhk (x) , vhk = vkh. (2.3.50)

Substitution of this expression into (2.3.46) yields

P11 [vhk, ζ] + 12

P111 [uh, uk, ζ] = 0, (2.3.51)


P21 [ahuh; ζ] = 12

ahakP111 [uh, uk, ζ] .

For a known field v = ahakvhk, we may now determine the minimum value of theenergy functional (2.3.42). Only retaining the most important terms, we find

Minak=const.

P11 [u] = P4 [ahuh] + P2 [ahakvhk] + P21 [ahuh, akavk] + · · · . (2.3.52)

Using (2.3.46) with ζ= ahakvhk and making use of the relation

P11 [ahakvhk, aamvm] = 2P2 [ahakvhk] ,

we find

2P2 [ahakvhk] + P21 [ahuh, akavk] = 0 (2.3.53)

so that (2.3.52) may be rewritten to yield

Min(ak=const.)

P [u] = P4 [ahuh] − P2 [ahakvhk] + · · ·(2.3.54)

= P4 [ahuh] + 12

P21 [ahuh, akavk] + · · · .

It follows that the (necessary) condition P4 [ahuh] ≥ 0 is not a sufficient con-dition, because a positive-definite term P2 [ahakvhk] is subtracted. Introducing thenotation

P4 [ahuh] − P2 [ahakvhk] ≡ Aijkaiaj 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:

� Aijkaiaj aka ≥ 0 is a necessary condition for stability, and� Aijkaiaj aka as positive-definite (i.e., zero only for a displacement field u = 0 ) is

a sufficient condition for stability.

We shall now show that the results obtained are independent of the particu-lar choice of the auxiliary functional T2. Let T∗

2 be a positive-definite functional(T∗

2 �= T2), so then we may construct a minimizing displacement field

u∗Min = ahakv∗

hk (x) . (2.3.56)

This field satisfies the same equations as the field v = ahakvhk(x) but is subjected toa different orthogonality condition. The difference of these fields must satisfy thehomogeneous 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 = chku, (2.3.58)

which does not satisfy one of the orthogonality conditions implied by T2 and T∗2 . The

minimum of the potential energy functional is now given by

Min(ak=const.)

P (u) = P4 [ahuh] − P2 [ahakv∗hk] + · · · , (2.3.59)

where P2[ahakv∗

hk

]may be written as

P2 [ah ak (vhk − chklul)] = P2 [ah akvhk]

− P11[ah akvhk, ap aq cpqrur

]+ P2 [ah ak chku] . (2.3.60)

The second term on the right-hand side of (2.3.60) vanishes because P11 [ah uh, ζ] = 0for all fields ζ, and hence also for ζ= v, and the last term vanishes because the argu-ment is a linear combination of buckling modes. Hence it follows that the minimumvalue 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 neighborhoodof the buckling load, we must investigate the behavior of the structure for small butfinite deflections. The fundamental state then depends on the load, so it cannot beused as a reference state. We need a fixed reference state, and it is suitable to choosethe undeformed (stress-free) state as the reference state.


0

x

I

+ U

II

+ U + u

Undeformed (stress-free) State

FundamentalState

Adjacent State

x x

Figure 2.4.1

The strain tensor in state II is now given by

12

(Ui + ui),j + 12

(Uj + uj ),i + 1

2(Uh + uh),i (Uh + uh),j

= 12

(Ui,j + Uj ,i) + 12

Uh,i, Uh,j (2.4.1)

+ 12

(ui,j + uj ,i) + 12

uh,iuh,j + 12

Uh,i, uh,j + 12

Uh,j , uh,i ≡ �ij + γij

where

�ij ≡ 12

(Ui,j + Uj ,i) + 12

Uh,iUh,j (2.4.2)

is the strain tensor going from the undeformed state to the fundamental state, and

γij ≡ 12

(ui.j + uj ,i) + 12

uh,iuh,j + 12

Uh,iuh,j + 12

Uh,j uh,i (2.4.3)

describes the strains going from the fundamental state to the adjacent state. Noticethat this tensor also depends on U.

The elastic energy density in state II is given by

12

Eijkl (�ij + γij ) (�k + γk) . (2.4.4)

The increment of elastic energy density going from state I to state II is given by

12

Eijk (�ij + γij ) (�k + γk) − 12

Eijk�ij �k = Eijk�ij γk + 12

Eijkγij γk

= skγk + 12

Eijkγij γk. (2.4.5)

This expression is of the same form as our original energy functional but the straintensor is different.

We consider loads that can be written as a unit load multiplied by a load factorλ. This means that the displacement field in the fundamental state also depends onλ, i.e., U = U (x; λ). The increment of the potential energy going from state I tostate 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 fundamentalstate is an equilibrium configuration, and also that this expression depends on U,i.e., it refers to the fundamental state.


For sufficiently small loads, the potential energy will be positive-definite and theequilibrium configuration will be stable and unique (Kirchhoff’s uniqueness theo-rem). Now let λ ≥ 0 be monotonically increasing; then for a certain value of λ, sayλ = λ1, the potential energy will become semi-definite-positive. (The case λ < 0 is adifferent 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) P′2 [u (x) ; λ1] + · · · (2.4.7)

where ()′ ≡ ∂ () /∂λ.As mentioned previously, the potential energy depends on U (x, λ). In the fol-

lowing, we shall assume that the fundamental state is known and the question iswhether there exist equilibrium configurations in the vicinity of the fundamentalstate. Therefore stationary values of P [u (x) ; λ] are considered for nonzero displace-ment 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 shouldbe. At the critical load, the infinitesimal displacement field is a linear combinationof buckling modes, i.e., u = ahuh. Assuming that for small but finite displacementsthe field can also (approximately) be represented by this expression, we write

u (x) = ahuh (x) + u (x) , T11 [uk, u] = 0, (2.4.9)

which is always possible.The energy functional now becomes

P [u (x) ; λ] = P2 [ahuh; λ1] + P11 [ahuh, u; λ1]

+ P2 [u; λ1] + (λ − λ1) P′2 [ahuh; λ1] + (λ − λ1) P′

11 [ahuh, u; λ1]

+ (λ − λ1) P′2 [u; λ1] + · · · + P3 [ahuh; λ1] + P21 [ahuh, u; λ1] + · · ·

+ (λ − λ1) P′3 [ahuh; λ1] + (λ − λ1) P′

21 [ahuh, u; λ1] + · · ·+ P4 [ahuh; λ1] + · · · . (2.4.10)

By known arguments, the first two terms on the right-hand side of (2.4.10) van-ish. We shall now first consider the terms depending on u. First of all, we havethe positive-definite term P2 [u; λ1], thus the term (λ − λ1) P′

11 [ahuh, u; λ1] is impor-tant because it contains u linearly, and is also linear in ahuh and λ − λ1. The term(λ − λ1) P′

2 [u; λ1] may be neglected because it is quadratic in u and is multiplied byλ − λ1. The next important term depending on u is P21 [ahuh, u; λ1], which is linear inu and quadratic in ahuh. Other terms depending on u may be neglected because theyare either of higher order in u or they contain higher-order terms in ahuh or (λ − λ1).We now determine the minimum with respect to u of the three terms mentioned,

Minw.r.t. u

P2 [u; λ1] + (λ − λ1) P′11 [ahuh, u; λ1] + P21 [ahuh, u; λ1] . (2.4.11)


The solution to this minimum problem is unique. Suppose that u∗ is a second solu-tion. 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, butu and u∗ are orthogonal with respect to these buckling modes, so we must haveu − u∗ = 0. Guided by the structure of (2.4.11), we try a solution of the form

uMin (x) = (λ − λ1) ahv′h (x) + ahakvhk (x) . (2.4.13)

Because uMin (x) is the solution to (2.4.11), we have for a variation of this field, sayuMin + εη,

P2 [uMin + εη; λ1] + (λ − λ1) P′11 [ahuh, uMin + εη; λ1] + P21 [ahuh, uMin + εη; λ1]

≥ P2 [uMin; λ1] + (λ − λ1) P′11 [ahuh, uMin; λ1] , for ∀ε. (2.4.14)

This inequality must be satisfied for all sufficiently small values of ε ∈ R, whichimplies

P11 [uMin, η; λ1] + (λ − λ1) P′11 [ahuh, η; λ1] + P21 [ahuh, η; λ1] = 0. (2.4.15)

Substitution of (2.4.13) into (2.4.15) yields

(λ − λ1) P11 [ahv′h, η; λ1] + P11 [ahakvhk, η; λ1]

(2.4.16)+ (λ − λ1) P′

11 [ahuh, η] + P21 [ahuh, η; λ1] = 0.

Because this equation must hold for all (λ − λ1), we must have

P11 [v′h, η; λ1] + P′

11 [uh, η; λ1] = 0 (2.4.17)

and

P11 [vhk, η; λ1] + 12

P′111 [uh, uk; λ1] = 0, (2.4.18)

where for (2.4.18) we have used the relation

P21 [ahuh, η; λ1] = 12

ahakP111 [uh, uk, η; λ1] .

From these linear equations, v′h 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. Toshow this, we first notice that the minimizing displacement field satisfies the equa-tion

F11 [u∗, ζ] + F1 [ζ] = 0 for ∀ζ. (2.4.20)

Now choose ζ= u∗, then

F1 [u∗] = −F11 [u∗, u∗] .


With

F2 [u] = F2

[12

u + 12

u]

= 12

F2 [u] + 14

F11 [u, u]

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

Minw.r.t .u

P2 [u; λ1] + (λ − λ1) P′11 [ahuh, u; λ1] + P21 [ahuh, u; λ1]

= −P2 [(λ − λ1) ahv′h (x) + ahakvhk (x)]

= − (λ − λ1)2 P2 [ahv′h (x) ; λ1] − (λ − λ1) P11 [ahv′

h (x) , ahakvhk (x) ; λ1]

−P2 [ahakvhk (x) ; λ1] . (2.4.21)

Let us now first consider the case that P3 [ahuh; λ1] �≡ 0. In this case, the termP4 [ahuh; 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), andhere the term P2 [ahakvhk (x) ; λ1] may be neglected because it is of fourth degreein the amplitudes of the buckling modes. Further, the term (λ − λ1) P11

[ahv′

h (x),ah ak vh (x) ; λ1] may be neglected because it is of third degree in the amplitudes ofthe buckling modes and multiplied by (λ − λ1). Finally, we may also neglect the firstterm on the right-hand side of (2.4.21) because it is of order O[(λ − λ1)2 a2], wherea represents the order of magnitude of the amplitudes of the buckling mode. Theonly remaining term in (2.4.10) that still is to be discussed is (λ − λ1)P′

2[ahuh; λ1].This term is of order O[(λ − λ1)a2] and may thus be important. Our final result isnow given by

Min(ah=const)

P(u; λ) = (λ − λ1)P′2[ahuh; λ1]

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

In other words, to a first approximation we might have neglected the terms contain-ing u to obtain the same result.

Let us now consider the case P3 [ahuh; λ1] ≡ 0. Now we must take into accountthe term P4 [ahuh; λ1], and hence also the last term in (2.4.21). Further, the term(λ − λ1) P′

2 [ah uh; λ1] is important, whereas the first two terms on the right-handside of (2.4.21) may be neglected compared to this term (under the assumption thatP′

2 [ahuh; λ1] �≡ 0).In this case, our result is

Min(ah=const.)

P (u; λ) = (λ − λ1) P′2 [ahuh; λ1] + P4 [ahuh; λ1] − P2 [ahakvhk; λ1]

+ O[a5, (λ − λ1) a3, (λ − λ1)2 a2]. (2.4.23)


Equilibrium possible for λ > λ λ >

λ −1

λ1

λ1

λ1λ 1

λλ

uu

No equilibrium possible for ,no linear terms in

Figure 2.4.2

Notice that the last two terms in this expression are the terms that decide on thestability of neutral equilibrium (cf. 2.3.55).

In our discussion, we have restricted ourselves to cases where P′2 [ahuh; λ1] �≡ 0,

i.e., we have assumed that the derivative of P2 [ahuh; λ] with respect to the loadparameter λ exists for λ − λ1 (the fundamental state). This means that we haveassumed 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 treatbranch points. Finally, we notice that when P′

2 [ahuh; λ1] �≡ 0, this term must be neg-ative because for λ − λ1 < 0, the equilibrium in the fundamental state is stable, i.e.,

P2 [ahuh; λ] = P2 [ahuh; λ1] + (λ − λ1) P′2 [ahuh; λ1] + · · · ≥ 0 for λ − λ1 < 0.

The first term on the right-hand side is zero, and hence

P′2 [ahuh, λ1] < 0. (2.4.24)

Let us now consider the simple case that m = 1, i.e., there is only one bucklingmode, say u (x) = au1 (x). Let us further introduce the notation

P3 [u1; λ1] = A3, P′2 [u1; λ1] = A′

2. (2.4.25)

First, consider the case A3 �≡ 0, A′2 < 0. Then we obtain from (2.4.22)

Minak=const.

P [u (x) , λ] = (λ − λ1) A′2a2 + A3a3. (2.4.26)

The condition for equilibrium is obtained by differentiating this expression withrespect to a and equating this result to zero,

2 (λ − λ1) A′2a + 3A3a2 = 0. (2.4.27)


unstable

stable

A3 < 0A3 > 0

λ1

λ

a

Figure 2.4.3

The solutions are

a1 = 0 (the fundamental state)(2.4.28)

a2 = −23

(λ − λ1)A′

2

A3(branched equilibrium state).

The stability condition is that the second derivative of (2.4.26) is non-negative,

2 (λ − λ1) A′2 + 6A3a ≥ 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) A′2 ≥ 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 discussedlater).

For λ sufficiently close to λ1, our approximate solution will be a good approx-imation to the actual solution. However, it is not possible to assess the accuracy ofthe approximation.

F

Figure 2.4.4


a aA3 = 0

(I) (II)

A4 > 0 A3 = 0 A4 < 0

λ1

λ

λ1

λ

Figure 2.4.5

Let us now consider the case A3 = 0. From (2.4.23) we obtain

Minak=const.

P [u, λ] = (λ − λ1) A′2a2 + A4a4, (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) A′2a + 4A4a3 = 0, (2.4.32)

from which

a1 = 0 (fundamental state)

a22 = −1

2(λ − λ1)

A′2

A4.

(2.4.33)

Because for real values of a2 the left-hand side is positive, the right-hand side mustbe positive, i.e.,

λ ≷ λ1 for A4 ≷ 0. (2.4.34)

The stability condition reads

2 (λ − λ1) A′2 + 12A4a2 ≥ 0, (2.4.35)

so the fundamental state is stable for (λ < λ1) and unstable for λ > λ1.Substitution of a2

2 into this condition yields

− 4 (λ − λ1) A′2 ≥ 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 andFigure 2.4.6.


F

rigid rod

Figure 2.4.6

Examples of structures with the behavior of (I) are the Euler column and platesloaded in their plane. An example of a structure with the behavior of (II) is given inFigure 2.4.6.

From Figure 2.4.3 and Figure 2.4.5, we see that branched equilibrium states forloads below the critical load are unstable. Later, we shall prove that this is alwaysthe case.† The converse that branched equilibrium states would always be stable forloads exceeding the critical load is not true (except for the single mode case).

Let us now proceed with the discussion of Minah=constP[u, λ], which is given by(2.4.32) when P3[ahuh; λ1] �≡ 0 and by (2.4.23) when this term is identically equal tozero. The term P′

2[ahuh; λ1] is quadratic in the buckling modes and may be trans-formed to a form containing only quadratic terms, which will be normalized so thatthe coefficients are all equal to −1 (remember P′

2[ahu; λ1] ≤ 0), so that we may write

P′2 [ahuh; λ1] = −aiai. (2.4.37)


P3 [ahuh; λ1] ≡ Aijkaiaj ak,(2.4.38)

P4 [ahuh; λ1] − P2 [ahakvhk; λ1] ≡ Aijklaiaj akal.

We may normalize the load factor λ so that at the critical load, λ = λ1 = 1. DenotingMin P [u; λ] by F [ai; λ], our discussion of stability is reduced to the discussion of thequadratic form

F (ai; λ) = (1 − λ) aiai +{

Aijkaiaj ak

Aijkaiaj aka

(2.4.39)

where the term on top is important when Aijkaiaj ak0, else the quartic term must beused.

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


Let δai be a variation of ai,

F (ai + δai; λ) = (1 − λ) aiai + 2 (1 − λ) aiδai + (1 − λ) δaiδai

+{

Aijk (aiaj ak + 3aj akδai + 3akδaiδaj + δaiδaj δak)Aijkl (aiaj akal + 4aj akalδai + 6akalδaiδaj + 4alδaiδaj δak + δaiδaj δakδal)

= F (ai; λ) + δF (ai; λ) + δ2F (ai; λ) + · · · (2.4.40)

so that

δF (ai; λ) =[

2 (1 − λ) ai +{

3Aijkaj ak

4Aijkaj aka

]δai, (2.4.41)

δ2F (ai; λ) =[

1 − λ +{

3Aijkak

6Aijkaka

]δaiδaj . (2.4.42)

The equilibrium equations are obtained from δF = 0, for all variations δai, so that

2 (1 − λ) ai +{

3Aijkaj ak

4Aijkaj aka

= 0. (2.4.43)

The condition for stability is that δ2F ≥ 0,

(1 − λ) δaiδaj +{

3Aijkakδaiδaj

6Aijkakaδaiδaj≥ 0. (2.4.44)

Let δah = bh be a unit vector, i.e., ‖b‖ = bibi = 1, and let ah = aeh, where e is a unitvector and a is the amplitude of a. The equilibrium equations now read

2 (1 − λ) ei +{

3aAijkej ek

4a2Aijkej eke

= 0, (2.4.45)

and the stability condition is now given by

1 − λ +{

3aAijkeibibj

6a2Aijkekebibj≥ 0 (∀ b | bibi = 1) . (2.4.46)

We are now in a position to prove the statement that the branched equilibrium statesare 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, andhence also for b = e, so that

1 − λ +{

3aAijkeiej ej

6a2Aijkeiej eke

≥ 0. (2.4.47)

Multiplying (2.4.45) by ei, we obtain

2 (1 − λ) +{

3aAijkeiej ek

4a2Aijkeiej eke

= 0, (2.4.48)

from which

3aAijkeiej ek = −2 (1 − λ) , (2.4.49)


or when the cubic term is missing,

a2Aijkeiej eke = −12

(1 − λ) . (2.4.50)

Substitution of these expressions into the corresponding form of the stability condi-tion (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

Aijktitj tk resp. Aijktitj tkt

on the unit sphere ‖t‖ = 1. Introducing a Lagrangean multiplier µ, we look for sta-tionary values of

Aijktitj tj + µ (titi − 1) resp. Aijktitj tkt + µ (titi − 1) , (2.4.51)

i.e., the first variation of these forms must vanish. This yields

3Aijktj tk − 2µti = 0 resp. 4Aijktj tkt − 2µti = 0. (2.4.52)

These equations are of the same form as our equilibrium equations – in otherwords, the equilibrium equations yield stationary directions on the unit sphere‖e‖ = 1. Among these stationary directions, there are minimizing directions. Lett = t∗ be a minimizing direction, and let

Aijkt∗i t∗j t∗k = A3 (t∗) = A∗3 ≤ 0 and Aijkt∗i t∗j t∗kt∗ = 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

Aijkleiej bkbl = Min‖t‖=1

Aijktitj tkt. (2.4.54)

Using this property, we obtain from the stability condition (2.4.46)

1 − λ + 6a2A∗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) + 4a2A∗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 beenproved for a minimizing direction t∗. For other solutions of the equilibrium equa-tions, the question of stability cannot be established in general.

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


1

λ

0

0a

A3

* < 0 (steepest descent)

(smallest rise)

A3

* > 0 (steepest rise)

1

λ

0

0a

A3

* = 0

other solutions

stable

unstable

A4

* > 0

(steepest descent)

1

λ

0

0a

A3

* = 0

A4

* < 0

−

Figure 2.4.7

From the equilibrium equations (2.4.45) with e = t∗ and multiplied by t∗i , wefind

a = −23

1 − λ

A∗3

(A∗3 < 0)

(2.4.58)a2 = −1

21 − λ

A∗4

(A∗3 = 0) .

Our results are shown in Figure 2.4.7.When A∗

3 < 0, the dashed line corresponding to the minimizing solution is a lineof steepest descent because the amplitude a for fixed λ is monotonically increasingfor increasing values of A3 ∈ [A∗

3, 0) ⊂ R−. Similarly, the dashed curve for A∗

4 < 0 isa curve of steepest descent. The solid curve for the minimizing solution when A∗

4 > 0is a curve of smallest rise because for fixed λ, the amplitude a is monotonicallydecreasing 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 moreclearly. To this end, we consider the generalized displacement corresponding to theexternal load. This generalized displacement is such that the product of the load andthe displacement is equal to the increase of the elastic energy in the structure. Forexample, consider an elastic bar (Figure 2.4.3),


εlN

Figure 2.4.8

where ε is the generalized displacement. The behavior of this structure is plottedin Figure 2.4.9.

Here the curve I corresponds to the fundamental state, which is stable up tothe critical load N1 and unstable for loads exceeding N1. The branched solution II isstable.

In the fundamental state, for N = NB the elastic energy is given by the area ofODB, 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 ofOACE, and the energy of the load is given by NB(ε)C, the area of the squareOECNB. 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 tothe 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 theequilibrium 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)

}

NB

A1 A

B

I

II

N

N

δN

1

D ∆εl εlEO

C

Figure 2.4.9


1

0

0

λ

0

0ε

11

λ

0

0ε

I

unstable

I

other solutions

∆ε

ε

= a2 =4

9

1− λ( )

A3*

A3*

A3*

A4*

( )2

2

other solutions

A3* < 0

unstable

A3* = 0, A4

* > 0 A3* = 0, A4

* < 0

I

A4*

other solutions

λ −

Figure 2.4.10

In words: The increment of the generalized displacement is equal to the negative ofthe derivative of the potential energy with respect to the external load, evaluated inthe branched equilibrium state.

In general, denoting the generalized displacement by ε, we may now write

�ε = − ddλ

F(aeq

i ; λ) = −∂F

∂λ

(aeq

i ; λ) = aiai = a2. (2.4.62)

It follows that a2 is a natural measure for the increment of the generalized displace-ment �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 = −12

1 − λ

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 isnot possible to assess the accuracy of the results for finite deflections from thefundamental state.

ii. A structure will (unless prohibited) follow the path of steepest descent or weak-est 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 danger-ous imperfections. It is therefore meaningful to consider these imperfections inparticular.

iii. The fact that the slope of the branched equilibrium path is always smaller thanthat of the fundamental path implies that the stiffness of the structure afterbuckling is always decreased.

2.5 The influence of imperfections

In the foregoing discussions, it was assumed that the geometry of the structure andthe distribution of the loads were known exactly. Further, it was assumed that thematerial was elastic and followed the generalized Hooke’s Law. However, in prac-tice the quantities are not known exactly, e.g., it is impossible to manufacture aperfectly straight column of constant cross section, or a perfect cylindrical shell. Thedeviations from the “idealized structure” in the absence of loads are called initialimperfections. 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 theidealized perfect structure, described by the displacement vector U from the unde-formed state, does not represent a configuration of equilibrium of the actual imper-fect structure. The bifurcation point of the perfect structure can also not be a bifur-cation 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 energyfunctional starts with terms quadratic in the displacements. However, in the pres-ence of imperfections the fundamental state is not an equilibrium configuration, andhence the expansion of the potential energy functional will start with a term linearin the displacements. Furthermore, if we restrict ourselves to small initial imper-fections, this term will also be linear in these imperfections. This implies that ourenergy functional F (ai; λ) is replaced by the functional F∗ (ai; λ), defined by

F∗ (ai; λ) = (1 − λ) aiai +{

Aijkaiaj ak

Aijkaiaj aka

+ µBiai, (2.5.1)

where µ is a measure for the magnitude of the imperfections and B is determined bythe shape of the imperfections. It should be noted that our approach is valid underthe 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 +{

3Aijkaj ak

4Aijkaj aka

+ µBi = 0. (2.5.3)


The stability condition is not affected by the linear term µBiai, and is still given by(2.4.44). Writing ai = aei, |e| = 1, the equilibrium equations become

2 (1 − λ) aei +{

3a2Aijkej ek

4a3Aijkej eke

+ µBi = 0, (2.5.4)

and the stability condition becomes

1 − λ +{

3aAijkekbibj

6a2Aijkekebibj≥ 0, (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 ampli-

tude a., i.e., m − 1 unknowns because ‖e‖ = 1. In general, this problem is extremelycomplicated, but for the most dangerous imperfections, the ones with direc-tions corresponding to the minimizing directions t∗, the problem is simplifiedenormously.

We shall now treat this problem in more detail. First, choose the imperfectionsin 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 [ahuh; λ] 0. The equilibrium equation nowreads

2 (1 − λ) ati + µti + 3a2Aijktj tk = 0. (2.5.6)

Multiplying this equation by ti, summing over i, and using the notation Aijktitj tk =A3(t), we obtain

2 (1 − λ) a + µ + 3a2A3(t) = 0. (2.5.7)

This equation is to be solved under the side condition a ∈ R+. The stability condition

reads

1 − λ + 3aAijkbibj tk ≥ 0 for ∀ b| ‖b‖ = 1. (2.5.8)

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

Min‖a‖=‖b‖=‖c‖=1.

Aijkaibj ck = Aijkt∗i t∗j t∗k = A3 (t∗) = A∗3 < 0, (2.5.9†)

we obtain from (2.5.8)

1 − λ + 3aA∗3 ≥ 0, ⇒ λ < 1. (2.5.10)

The stability limit follows from

1 − λ∗ + 3a∗A∗3 = 0, ⇒ λ∗ < 1. (2.5.11)

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


λ λ

a

00

a

1 1

λ*

A3* < 0

µ < 0

−A3*

µ > 0

a*

00

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 besatisfied 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 sta-bility limit

(1 − λ∗)2 − 3µA∗3. (2.5.14)

The general expression for the amplitude of the buckling modes in the presenceof imperfections in the minimizing directions follows from (2.5.7) with t = t∗,

a1,2 = 13A∗

3

[− (1 − λ) ±

√(1 − λ)2 − 3µA∗

3

]. (2.5.15)

These amplitudes are real for (1 − λ)2> 3µA∗

3, and they are both positive whenµ < 0. When µ > 0, only the negative sign can be used because a ∈ R

+. These resultsare shown in Figure 2.5.1.

One easily shows that for µ < 0, the path is stable for a < a∗ and unstable fora > a∗, and that for µ > 0 the path is stable. Notice that due to the presence ofimperfections, the branch point is not reached. The stability limit is now a limitpoint 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 theminimizing 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 + 3a2A3 (e) + µBiei = 0. (2.5.16)

The stability condition (2.5.5) reads

1 − λ + 3aAijkbibj ek ≥ 0 for ∀ b| ‖b‖ = 1. (2.5.17)


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

(1 − λ)2[

2 − A3 (e)C3 (e)

]= 3µBieiC3 (e) . (2.5.21)

For the right-hand side of (2.5.21), we have the estimate∣∣3µBieiC3 (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 [ahuh; λ] ≡ 0 and Aijkaiaj aka indefinite. The

equilibrium equation now reads

2 (1 − λ) aei + 4Aijkej ekea3 + µBi = 0 (2.5.24)

and the stability condition is given by

1 − λ + ba2Aijkbibj eke ≥ 0 for ∀ b | ‖b‖ = 1. (2.5.25)

Because the fourth degree form is indefinite, we have

Min‖t‖=1

Aijktitj tkt = A4 (t∗) = A∗4 < 0. (2.5.26)

Now we consider imperfections in the direction of the minimizing direction, i.e.,Bi = t∗i . From (2.5.24) and with e = t∗, we now obtain

[2 (1 − λ) a + µ] t∗i + 4Aijkt∗i t∗kt∗a3 = 0. (2.5.27)

Multiplying by t∗i and summing over i, we obtain for the amplitude

2 (1 − λ) a + 4A∗4a3 + µ = 0. (2.5.28)


The stability condition (2.5.25) must hold for all b with ‖b‖ = 1, and hence also forb = t∗, so

1 − λ + 6A∗4a2 ≥ 0. (2.5.29)

The stability limit follows from

1 − λ∗ + 6A∗4a∗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

(1 − λ∗)3/2 = ∓34µ

√−6A∗

4. (2.5.32)

As λ∗ < 1, the left-hand side of (2.5.32) is positive and hence we must choose thenegative sign in (2.5.31) when µ < 0. This implies that (2.5.32) can be rewritten toyield

(1 − λ∗)3/2 = 34

|µ|√

−6A∗4. (2.5.33)

Equation (2.5.33) is a formula for the reduction of the stability limit due to imper-fections in the direction of the minimizing direction, corresponding to the line ofsteepest descent.

We shall now show that this reduction is the largest possible reduction. Con-sider imperfections in an arbitrary direction. Multiplying the equilibrium (equation2.5.24) by ei and summing over i, we obtain

2 (1 − λ) a + 4A4 (e) a3 + µBiei = 0. (2.5.34)

Introducing C4 (e) defined by

Minw.r.t. b, ‖b‖=1

e fixed

Aijkbibj eke = C4 (e) , (2.5.35)

we obtain from the stability condition (2.5.25)

1 − λ + 6a2C4 (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 imperfec-tions, it suffices to restrict ourselves to values of λ for which λ < 1. From (2.5.36), itthen follows that C4 (e) < 0. With the stability limit, we have

1 − λ + 6a2C4 (e) = 0, (2.5.38)


λ λ

a a

1 1

λ*λ*

other imperfections

3* < 0

A4* < 0

other imperfections

A3*

A

= 0

Figure 2.5.2

so that

a1,2 = ±√

− 1 − λ

6C4 (e). (2.5.39)


2 (1 − λ)3/2[

1 − A4 (e)3C4 (e)

]= ±µBiei

√−6C4 (e), (2.5.40)

which determines the reduction of the critical load at the stability limit in the pres-ence of imperfections in arbitrary directions. Let us start the discussion of thisexpression with the right-hand member. Because B and e are unit vectors, we haveBiei ≤ 1. Because C4 (e) ∈ [A∗

4, 0) ⊂ R

− we have, by virtue of (2.5.37),√−6C4 (e) ≤√−6A∗

4, so that Biei√−6C4 (e) ≤ √−6A∗

4. On the left-hand side of (2.5.40), the termbetween 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

(1 − λ)3/2 ≥ ∓34µ

√−6A∗

4. (2.5.41)

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

(1 − λ)3/2 ≥ 34

|µ|√

−6A∗4. (2.5.42)

Comparing this expression with (2.5.33), we find that λ∗ yields the largest possiblereduction 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 theminimizing directions always cause the largest reduction of the critical load.† Theresults obtained so far are shown in Figure 2.5.2.

Notice that for imperfections in the direction of the minimizing direction, thecritical load is obtained as a limit point, whereas for other imperfections the criticalload 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 47

00

1 1

00

λ* λ*

A3* < <0 A3

* = 0, A4* 0

µ µ

Figure 2.5.3

Let us finally consider the influence of the magnitude of the imperfections onthe reduction of the critical load for imperfections in the direction of the minimizingdirection. When P3 [ahuh; λ] �≡ 0, the stability limit is given by (2.5.14), from which

dλ∗

d |µ| = 3A∗3

1 − λ∗ (< 0) , (2.5.43)

i.e., for λ∗ = 1 the tangent to the λ∗ − |µ| curve is vertical. Similarly, for the casethat the cubic terms are identically equal to zero, we obtain from (2.5.33)

dλ∗

d |µ| = −√−6A∗

3

2√

1 − λ∗ , (2.5.44)

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 theelastic energy functional is known. In this section, we shall determine the energyfunctional for a new class of problems.

Our line of thought is depicted in Figure 2.6.1:

U

II*I*

O

O*

II

U

u

u

I

u0

Figure 2.6.1


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 com-ponents of the position vector x. The position of this point in the fundamental stateI (equilibrium configuration) is given by x + U, and the position in an adjacent stateII (not necessarily an equilibrium configuration) by x + U + u.

Let O∗ be the undeformed (stress-free) state for the actual (imperfect) struc-ture, obtained by superposition of a displacement field u0 on the state O, withoutintroducing stresses. The position of a material point is now given by the Cartesiancomponents of the position vector x + u0, where u0 = u0 (x). In the state I∗ (not anequilibrium configuration), the position of the material point is x + u0 + U. In theadjacent state II∗, the position is x + u0 + U + u, and we shall try to determine u sothat 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)

g0∗ij = δij + u0

i,j + u0j ,i + u0

h,iu0h,j (2.6.2)

gIij = δij + Ui,j + Uj ,i + Uh,iUh,j ≡ δij + 2�ij (2.6.3)

gI∗ij = δij + Ui,j + Uj ,i + Uh,iUh,j + u0

i,j + u0j ,i + u0

h,iu0h,j

+ Uh,iu0h,j + Uh,j u0

h,i ≡ δij + u0i,j + u0

j ,i + u0h,iu

0h,j + 2�∗

ij

(2.6.4)

gIIij = δij + Ui,j + Uj ,i + Uh,iUh,j + ui,j + uj ,i + uh,iuh,j

+ Uh,iuh,j + Uh,j uh,i ≡ δij + 2�ij + 2γij + Uh,iuh,j + Uh,j uh,i

(2.6.5)

gII∗ij = δij + Ui,j + Uj ,i + Uh,iUh,j + ui,j + uj ,i + uh,iuh,j

+ u0i,j + u0

j ,i + u0h,iu

0h,j + Uh,iuh,j + Uh,j uh,i (2.6.6)+ Uh,iu0

h,j + Uh,j u0h,i + uh,iu0

h,j + uh,j uh,i

≡ δij + 2�∗ij + 2γ∗

ij + u0i,j + u0

j ,i + u0h,iu

0h,j .

For convenience, we note the relations

�ij = 12

(Ui,j + Uj ,i + Uh,iUh,j )

�∗ij = �ij + 1

2(Uh,iu0

h,j + Uh,j u0h,i)

(2.6.7)γij = 1

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

γ∗ij = γij + 1

2

(uh,iu

0h,j + uh,j u0

h,i

)+ 12

(Uh,iuh,j + Uh,j uh,i).


Further, we introduce the following notation for the linearized strain tensors andthe linearized rotation tensors:

θij = 12

(ui,j + uj ,i) �ij = 12

(Ui,j + Uj ,i)(2.6.8)

ωij = 12

(uj ,i − ui,j ) �ij = 12

(Uj ,i − Ui,j ) .

We shall now derive an approximate expression for the energy functional based onthe following assumptions:

1. The elastic energy density follows the generalized Hooke’s law.2. The imperfections are (infinitesimally) small.3. The displacements u are small (but finite).4. The loads are dead-weight loads.5. 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

WII − WI = 12

Eijkl

[�ij + γij + 1

2(Uh,iuh,j + Uh,j uh,i)

]

×[�k + γk + 1

2(Um,kum, + Um,um,k)

]− 1

2Eijk�ij �k

(2.6.9)= Eijk�k

[γij + 1


]

+ 12

Eijk

[γij + 1


] [γk + 1


].

Because we are dealing with dead-weight loads, the terms linear in u are balancedby the energy of the external loads, so the potential energy is given by

P [u] =∫V

{12

Eijk�kuh,iuh,j + 12

Eijk

[γij + 1


](2.6.10)

×[γk + 1


]}dV.

This expression is fully exact for dead-weight loads and a material that follows thegeneralized Hooke’s Law. To simplify this functional, we need estimates for thevarious 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)12

(Uh,iuh,j + Uh,j uh,i) = O (�ω,�θ) = O (�ω) (2.6.12)


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 thefact that the fundamental state is linear. We now have the following estimates:

γij γk = O(θ2, θω2, ω4)γij12

(Uh,iuh,j + Uh,j uh,i)

= O(θ�ω, �ω3)14

(Uh,iuh,j + Uh,j uh,i)(Um,kum, + Um,um,k) (2.6.14)

= O(�2ω2)12�kluh,iuh,j = O(�ω2).

Because buckling in the elastic range only occurs in slender structures, we haveω � θ.

Now consider the linear terms in (2.6.9),

Eijk�k (ui,j + Uh,iuh,j ) = Eijk�k (δih + Uh,i) uh,j .

Because in the energy functional these terms are balanced by the energy of theexternal 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 ofKirchhoff-Trefftz.† It follows that we may write

Sij = λS(1)ij , (2.6.17)

where S(1)ij 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(�3/2ω2) and may thus be neglected compared to O(�ω2). Terms of O(�ω3)and O(�2ω2) in (2.6.14) may also be neglected. This means that the energy func-tional (2.6.10) can be simplified to yield

P [u] =∫V

[12λS(1)

ij uh,iuh,j + 12

Eijkγij γk

]dV (2.6.18)

with a relative error of O(√

�). Notice that this expression is linear in the load factorλ.

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


Let us now consider the actual (imperfect) structure. The increment in the elas-tic energy density going from state I∗ to state II∗ is

W∗II∗ − W∗

I∗ = 12

E∗ijk(�∗

ij + γ∗ij )(�∗

k + γ∗k) − 1

2E∗

ijk�∗ij �

∗k

= E∗ijk�

∗kγ

∗ij + 1

2E∗

ijkγ∗ij γ

∗k = E∗

ijk

[�k + 1

2

(Um,ku0

m, + Um,ku0m,

)]

×[γij + 1

2

(uh,iu0

h.j + uh,j u0h.i

)+ 12

(Uh,iuh,j + Uh,j uh,i)]

(2.6.19)

+ 12

E∗ijk

[γij + 1

2

(uh,iu0

h.j + uh,j u0h.i

)+ 12

(Uh,iuh,j + Uh,j uh,i)]

×[γk + 1

2

(um,ku0

m, + um,lu0m,k

)+ 12

(Um,kum, + Um,um,k)]

,

where E∗ijk is the tensor of elastic moduli in O∗,

E∗ijk = Eijk + O(γ0E), (2.6.20)

so we may replace E∗ijk by Eijk. For our discussion of the terms with the imperfec-

tions, we shall need the following expressions,

θ0 = 12

∣∣u0i,j + u0

j ,i

∣∣max

and ω0 = 12

∣∣u0j ,i − u0

i,j

∣∣max

. (2.6.21)

We now have the following estimates,

�kl(uh,iu0h,j + uh,j u0

h,i) = O(�ωω0, �θω0, �ωθ0, �θθ0)

γij (u0h,iuh,j + u0

h,j uh,i) = O(θω0ω,ω3ω0) (2.6.22)

γij (Um,ku0m, + Um,u0

m,k) = O(θ�ω0, ω2�ω0).

Further, (2.6.19) contains terms that are quadratic in the imperfections, which canbe 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 needso that θ � �. The leading term in (2.6.22) is of order �ωω0. The only other termthat may be important is the term O (�ωθ0), and all other terms are small comparedto these terms. Using (2.6.20) and (2.6.17) and the fact that we have dead-weightloads, we can simplify the energy functional corresponding to (2.6.19) to yield

P∗ [u] = P [u] +∫V

λS(1)ij u0

h,iuh,j dV, (2.6.23)

where P [u] is the functional for the perfect structure defined in (2.6.18). Notice thatthe 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 functionalof 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 thedead-weight load is then −ρX · u (x), where u denotes the displacement of the point


λ λ

a

00

a

1 1

λ*

00

A3* < 0

λ*

A3* = 0, A4

* < 0

Figure 2.6.2

in question. When the load is applied eccentrically, say, in a point y close to x, thepotential 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, thecorrection of the functional of the external loads is linear in the eccentricity of theloads and linear in the loads. It follows that for both the case of imperfection andthe case of eccentric loads, the additional term in the energy functional is of theform λQ [u].

It is now convenient to slightly modify our function F∗ (ai, λ),

F∗ (ai, λ) = (1 − λ) aiai +{

Aijkaiaj ak

Aijkaiaj aka

+ λµBiai (2.6.25)

so that it explicitly shows the linear dependence on the load factor in the lastterm. This expression is valid for λ ≥ 0, e.g., the equilibrium equations, whenP3 (ahuh, 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 thesecurves increases with decreasing values of λ.

In the case that we only have to deal with geometrical imperfections, we canslightly modify the functional (2.6.25). To this end, the imperfections are repre-sented by

u0 = a0huh + u0 (2.6.28)

and the displacement field is written as

u = ahuh + u. (2.6.29)


We now return to (2.6.23), which fully written out, reads

P∗ [u] =∫V

[λS(1)

ij

(12

uh,iuh,j + u0h,iuh,j

)+ 1

2Eijkγij γk

]dV. (2.6.30)

The second variation is

P2 [u, λ] =∫V

[λS(1)

ij12

uh,iuh,j + 12

Eijkθij θk

]dV. (2.6.31)

Expanding (2.6.31) with respect to λ about λ = 1, and using the fact that P2 [u, 1] =0, we obtain

P2 [u, λ] = (1 − λ)∫V

S(1)ij

12

uh,iuh,j dV + · · · ≡ (1 − λ) P′2 [u, 1] + · · · (2.6.32)

Substitution of (2.6.29) into (2.6.32) yields for the integral∫V

12

S(1)ij

[akalukh,iuh,i + akukh,iuh,j + aluh,j uh,i + uh,iuh,j

]dV. (2.6.33)

The orthogonality of u with respect to the buckling modes is now defined by∫V

12

S(1)ij akukh,iuh,j dV = 0. (2.6.34)

The last term in (2.6.33) may be neglected because it is small compared to the firstterm. The buckling modes are now normalized so that∫

V

12

S(1)ij akalukh,iuh,i dV = akal

∫V

S(1)ij ukh,iuh,j dV = akalδk = akak. (2.6.35)

Using (2.6.28) and (2.6.29), we find for the term with the imperfections in (2.6.30),∫V

λS(1)ij

[a0

kalukh,iuh,j + a0kukh,iuh,j + auh,j u0

h,i + u0h,iu

0h,j

]dV. (2.6.36)

The second term vanishes by virtue of (2.6.34), and the third term vanishes by virtueof the orthogonality condition,∫

V

S(1)ij auhu0

h,i dV = 0. (2.6.37)

The last term in (2.6.36) may be neglected because it is small compared to the firstterm, so we have

λa0kal

∫V

S(1)ij ukh,iuh,j dV = 2λa0

kalδk = 2λa0kak, (2.6.38)

where we have made use of (2.6.35).


In (2.6.25), the influence of the imperfections is accounted for by the termλµBiai, where µ < 0, and is equivalent to (2.6.38). Hence, we may modify (2.6.25)to yield

F∗ (ah, λ) = (1 − λ) aiai +{

Aijkaiaj ak

Aijkaiaj aka

− 2λa0i ai. (2.6.39)

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 Nat 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 thedeflected bar. The following relations then hold:

sin ϕ = dwdx

= w′

cos ϕ · ϕ′ = w′′ (3.1.1)

ϕ′ = w′′

cos ϕ= w′′

√1 − w′2 ,

where ϕ′ denotes the curvature of the bar at the corresponding point. The displace-ment of the moveable end is now given by

− � = −∫

0

cos ϕ dx =∫

0

[1 −

√1 − w′2

]dx. (3.1.2)

Here we have made use of the fact that the bar is incompressible, i.e., inextensional.An explicit condition for incompressibility reads

[(1 + u′)2 + w′2](dx)2 = (dx)2 (3.1.3)or

u′ =√

1 − w′2 − 1. (3.1.4)

N N

x

w(x)

l

−∆l

Figure 3.1.1

55

56 Applications

The potential energy of this system is now given by

P [w; N] =∫

0

12

EIϕ′2 dx + N� =∫

0

[12

EJw′′2

1 − w′2 + N(√

1 − w′2 − 1)]

dx.

(3.1.5)

The expression (3.1.5) was already given in (2.2.25), and it was discussed there thatthe potential energy of the external load is a nonlinear function of the displacementdue to the incompressibility condition imposed.

Using the expansion

11 − w′2 = 1 + w′2 + w′4 + · · ·

(3.1.6a)√1 − w′2 = 1 − 1

2w′2 − 1

8w′4 − · · · ,

we may rewrite (3.1.5) to yield

P [w; N] =∫

0

[12

EIw′′2(1 + w′2 + · · ·) − N(

12

w′2 + 18

w′4 + · · ·)]

dx, (3.1.6b)

and hence we have

P2 [w; N] =∫

0

[12

EIw′′2 − 12

Nw′2]

dx (3.1.7)

P3 [w; N] = 0 (3.1.8)

P4 [w; N] =∫

0

[12

EIw′′2w′2 − 18

Nw′4]

dx. (3.1.9)

The variational equation for neutral equilibrium now follows from (3.1.7),

P11 [w, ζ; N] =∫

0

[EIw′′ζ ′′ − Nw′ζ ′] dx = 0. (3.1.10)

Integration by parts yields

EIw′′ζ ′∣∣0 − (EIw′′′ + Nw′) ζ

∣∣0 +

∫0

(EIw′′′′ + Nw′′) ζdx = 0. (3.1.11)

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 + c2x + c3 cos αx + c4 sin αx, (3.1.15)

where we have introduced the parameter

α2 = NEI

. (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 loadnow follows from the smallest root,

N1 = π2EI2

. (3.1.18)

We shall now investigate whether this equilibrium state is stable. The neces-sary condition for stability P3 [w; N] = 0 is satisfied. The displacement field for smalldeflections 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 mustnow consider the minimum problem, as per (2.3.44),

MinT11[w,w]=0

P2 [w] + P21 [w1, w] (3.1.20)

but the last term vanishes because P3 [w; N] ≡ 0. Conseqently, the quantitythat governs stability in the critical state is P4 [w1; N1] = 0 (see 2.3.54). Takingw1(x) = 1 · sin πx/, we find

P4 [w1; N1] =∫

0

[12

EIπ6

6cos2 πx

sin2 πx

− 1

8N1

π4

4cos4 πx

]dx

(3.1.21)

= 12

N1π4

4

∫0

[14

sin2 2πx

− 14

· 14

(1 + cos2 2πx

)2]

dx = 164

π4

4N1,

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

P2 [w1; N] − P2 [w1; N1] = 12

∫0

(N1 − N)w′21 dx = π4

42(N1 − N ) . (3.1.22)

58 Applications

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

F(a; N) = π4

42(N1 − N) a2 + 1

64π4

4N1a4 = π2N1

42

[(1 − λ) a2 + π2

162a4]

, (3.1.23)

where λ = N/N1. The equilibrium condition follows from

∂F∂a

= 0 = π2N1

22a[

(1 − λ) + π2

82a2]

. (3.1.24)

Besides the fundamental solution a = 0, (3.1.23) has the solution

a = 2√

2π

√

λ − 1.†

(3.1.25)

Let us now consider the displacement −�,

− � = 12

∫0

[1 −

√1 − w′2

]dx ≈

∫0

12

w′2 dx

(3.1.26)

= 12

a2

∫0

π2

2cos2 πx

dx = π2a2

2= 2 (λ − 1) ,

where we have made use of (3.1.25). The effective relative shortening is

− �

= 2 (λ − 1) = 2

(NN1

− 1)

. (3.1.27)

Further, we have

∂

∂N

(−�

)= 2

N1= 22

π2EI= 1

EA22

π2i2, (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

0

0

1

E

N1Et = E

π 2i2

2l2

−∆l

l

Figure 3.1.2

† λ = 1.01 already gives a deflection a ≈ 0.1.

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. Com-parison with the exact solution shows that the results are valid up to deflections ofthe 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 forceN(x) and a load distribution P(x), clamped at one end (see Figure 3.2.1). The bar isconsidered to be incompressible.

Consider an infinitesimal part of the bar in the deflected configuration (see Fig-ure 3.2.2).

Due to the deflection of the bar, the load P (x) dx will move over a distance∫ x0 dx [1 − cos ϕ (x)], and hence the contribution to the potential energy is given by

∫0

−P (x) dx

x∫0

[1 − cos ϕ (ξ)] dξ,

so the total potential energy is given by

P [ϕ (x) ; λ] =∫

0

12

B (x) ϕ′2 (x) dx + λ

∫0

P (x)

x∫0

[cos ϕ (ξ) − 1] dξ

dx

(3.2.1)

+ λN

∫0

[cos ϕ (x) − 1] dx,

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

N (x) = N +∫

x

P (ξ) dξ, (3.2.2)

AP x( ) N

l

+x

Figure 3.2.1

[ ]

Figure 3.2.2

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

60 Applications

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

∫0

−N′ (x)

x∫0

[cos ϕ (ξ) − 1] dξ dx

= −N (x)

x∫0

[cos ϕ (ξ) − 1] dξ

∣∣∣∣∣∣

0

+∫

0

N (x) [cos ϕ (x) − 1] dx

= N ()

∫0

[cos ϕ (x) − 1] dx +∫

0

N (x) [cos ϕ (x) − 1] dx,

so that (3.2.1) can be rewritten to yield

P [ϕ (x) ; λ] =∫

0

{12

B (x) ϕ′2 (x) + λN (x) [cos ϕ (x) − 1]}

dx. (3.2.3)

The second, third, and fourth variations are given by

P2 [ϕ (x) ; λ] =∫

0

[12

B (x) ϕ′2 (x) − 12λN (x) ϕ2 (x)

]dx (3.2.4)

P3 [ϕ (x) ; λ] ≡ 0 (3.2.5)

P4 [ϕ (x) ; λ] =∫

0

124

λN (x) ϕ4 (x) dx. (3.2.6)

Because P3 [ϕ (x) ; λ] ≡ 0 at the critical load factor, and where P2 [ϕ (x) ; λ1] is semi-definite-positive, stability is governed by the sign of P4 [ϕ (x) ; λ1]. The conditionP11 [ϕ (x) , ψ(x) ; λ1] = 0 yields

P11 [ϕ (x) , ψ(x) ; λ] =∫

0

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

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

B (x) ϕ′ (x) ψ(x)∣∣0 −

∫0

{[B (x) ϕ′ (x)]′ + λN (x) ϕ (x)

}ψ(x) dx = 0. (3.2.8)

The first term vanishes when the ends of the bar are hinged or clamped. From thesecond 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 (when A is clamped)

B(0)ϕ′ (0) = 0 (when A is hinged)(3.2.10)

B()ϕ′ () = 0.

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

λ1N (x) ϕ1 (x) = − [B (x) ϕ′1 (x])′

, (3.2.11)

so that

P4 [ϕ1 (x) ; λ1] =∫

0

− 124

[B (x) ϕ′1 (x)]′ ϕ3

1 (x) dx

= − 124

B (x) ϕ′1 (x) ϕ3

1(x)∣∣0 +

∫0

124

B (x) ϕ′1 (x) · 3ϕ2

1 (x) ϕ′1 (x) dx.

The first term vanishes when the ends are either clamped or hinged, so

P4 [ϕ1 (x) ; λ1] = 18

∫0

B (x)ϕ′21 (x) ϕ2

1 (x) dx > 0, (3.2.12)

and hence the equilibrium at the buckling mode is stable. Notice that N (x) mayalso be a locally tensile force. In the case that A is an elastic hinge, we must add anadditional term to the potential energy given by 1

2 Cϕ2 (0). The boundary conditionfor 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 contri-bution at x = 0,

124

B (0) ϕ′1 (0) ϕ3

1(0) > 0, (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 compres-sion 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 correspondingenergy is 1

2 cw2 (x). The potential energy of this system is now given by

P [w (x) ; N] =∫

0

[12

Bw′′2

1 − w′2 + 12

cw2 + N(√

1 − w′2 − 1)]

dx, (3.3.1)

and hence

P2 [w; N] =∫

0

(12

Bw′′2 + 12

cw2 − 12

Nw′2)

dx (3.3.2)

P3 [w; N] ≡ 0 (3.3.3)

P4 [w; N] =∫

0

(12

Bw′′2w′2 − 18

Nw′4)

dx. (3.3.4)

The variational equation for neutral equilibrium follows from (3.3.2),

P11 [w, ζ; N] =∫

0

(Bw′′2ζ ′′ + cwζ− Nw′ζ ′) dx = 0. (3.3.5)

From integration by parts, we obtain

Bw′′ζ ′∣∣0 +

∫0

(−Bw′′′ζ ′ + cwζ+ Nw′′ζ) dx − Nw′ζ∣∣0

(3.3.6)

= Bw′′ζ ′∣∣0 − Bw′′′ζ

∣∣0 − Nw′ζ

∣∣0 +

∫0

(Bw′′′′ + Nw′′ + cw) ζdx = 0,

from which follows

Bw′′′′ + Nw′′ + cw = 0, (3.3.7)

with the dynamic boundary conditions

Bw′′() = Bw′′(0) = 0. (3.3.8)


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 sinkπx

. (3.3.10)

Notice that each term in this series satisfies the kinematic boundary conditions. Todetermine 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 beforehandwhether the series obtained is convergent. It is therefore more suitable to return tothe second variation as this only requires the second derivative of this series, which isexpected to be convergent in view of the fact that the dynamic boundary conditionsalso contain a second derivative.

Substitution of (3.3.10) into (3.3.2) yields

P2 [w; N] =∫

0

1

2B

( ∞∑k=1

−k2π2

2ak sin

kπx

)2

+ 12

c

( ∞∑k=1

ak sinkπx

)2

− 12

N

( ∞∑k=1

kπ

ak coskπx

)2 dx =

4

∞∑k=1

[B

k4π4

4+ c − N

k2π2

2

]a2

k.

(3.3.11)

This expression is positive when all its coefficients are positive. The coefficient of ak

becomes zero for

N1 = k2π2

2B + 2

k2π2C. (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

dk= 2

kπ2

B − 22

k3π2C = 0, (3.3.13)

from which

kπ

=(

CB

)1/4

. (3.3.14)

However, here we have assumed that k ∈ R+, but k ∈ N

+, so k cannot generally takeon 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

1

10

0 2 3

N

2 BC k = 3= 2k

k= 1

STABLE

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 thevalue of N is for k = 3 and k = 4. From (3.3.12), we obtain for k = 3

N = 9(3.5)2

√BC + 1

9(3.5)2

√BC = 2.096

√BC,

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 stablein the shaded region but not necessarily at the stability limit, i.e., on the curvesbounding the shaded region. Because P3 [w; N] ≡ 0, stability is governed by the signof P4 [w1; N1].

With N1 given by (3.3.12) and

w1(x) = ak sinkπx

, k ∈ N+, (3.3.16)


P4 [w1; N1] =∫

0

[12

B(−k2π2

2ak sin

kπx

)2 (kπ

ak coskπx

)2

− 18

N1

(ak

kπ

coskπx

)2]

dx

=∫

0

(18

Bk6π6

6a4

k sin2 2kπx

− 18

N1k4π4

4a4

k cos4 kπx

)dx (3.3.17)

= 116

Bk6π6

5a4

k − 364

N1k6π6

3a4

k = 164

k2π2

a4

k

(k4π4

4B − 3C

).

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

C = k4π4

34B. (3.3.18)


0

N

2 BC k = 1k = 2

k = 3stable

unstable1

00 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

P4 [w1; NMin] = − 132

a4k

√CB

C < 0, (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] =∫

0

[−1

2(N − N1) a2

kk2π2

2cos2 kπx

]dx

= 14

(N1 − N) a2k

k2π2

(3.3.20)

= 14

(1 − λ) a2kN1

k2π2

,

and for N1 = Nmin and kπ/ given by (3.3.14), we obtain

P2 [w1; N] = 12

(1 − λ) a2kc. (3.3.21)

The function F(ak; λ) is now given by

F(ak; λ) = 12

c

[(1 − λ) a2

k − 116

√CB

a4k

]= 1

2c[

(1 − λ) a2k − 1

16k2π2

2a4

k

], (3.3.22)

Let us now consider the influence of imperfections on the behavior of this struc-ture. 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 −2λa0

k ak, hence

F∗(ak; λ) = 12

c[

(1 − λ) a2k − 1

16k2π2

2a4

k − 2λa0kak

]. (3.3.23)

66 Applications

The equilibrium condition is

∂F∗

∂ak(ak; λ) = 0 = 1

2c[

2 (1 − λ) ak − 14

k2π2

2a3

k − 2λa0k

](3.3.24)

and the stability condition is

∂2F∗

∂a2k

(ak; λ) = 12

c[

2 (1 − λ) − 34

k2π2

2a2

k

]≥ 0. (3.3.25)

The stability limit is reached for

a2k = 8

3(1 − λ∗)

2

k2π2. (3.3.26)

Substitution into the equilibrium condition yields

43

(1 − λ∗)3/2√

6

kπ− 4

9

√6 (1 − λ∗)3/2

kπ− 2λ∗a0

k = 0

or

(1 − λ∗)3/2 = 38

√6

kπ

λ∗a0k. (3.3.27)

This result shows that for small imperfections the critical load is reduced appre-ciably, e.g., ka0

k/ ≈ 0.01 yields λ∗ ≈ 0.91, i.e., a reduction of about 10% in the criti-cal load. Summarizing, we may say that the elastic foundation increases the bucklingload but it also increases the imperfection sensitivity.

Finally, let us calculate the tangent modulus, which determines the stiffness ofthe structure after buckling. The extra shortening after buckling is

− � =∫

0

[√1 − w′2 − 1

]dx ≈

∫0

12

w′2 dx

= 12

∫0

a2k

k2π2

2cos2 kπx

dx = 1

4k2π2

a2

k = 14

√CB

a2k. (3.3.28)

With (3.3.26), we find

−� = 14

√CB

83

(1 − λ∗)2

k2π2

or

− �

= 2

3

√CB

(1 − λ∗)

√BC

= 23

(1 − λ∗) , (3.3.29)

so that

∂ (−�/)∂N

= − 23Nmin

= 1EtA

. (3.3.30)

Hence,

Et = −32

Nmin

A= −3

√BCA

. (3.3.31)

This result is shown in Figure 3.3.4.

3.4 Simple two-bar frame 67

N

N1

E Et = −3I

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 , rigidlyattached 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 respectto y. At point A, the following kinematic relations then hold:

w′ () = v· () = − sin θ (3.4.1)

w () =∫

0

(√1 − v·2 − 1

)dy (3.4.2)

v () =∫

0

(1 −

√1 − w′2

)dx. (3.4.3)

y

CN

B

l

x

l

v y( )

− θA

– w (x)

Figure 3.4.1

68 Applications

N

Figure 3.4.2

The potential energy is now given by

P [v (y) , w (x) ; N] =∫

0

[12

Bw′′2

1 − w′2 + N(√

1 − w′2 − 1)]

dx +∫

0

12

Bv··2

1 − v·2 dy,

(3.4.4)where B is the bending stiffness of the bars. However, it is inconvenient to formulatethe 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 elasticelement of Figure 3.4.2, where both springs have a stiffness c. The correspondingadditional energy (penalty function) is given by

12

c

w () −

∫0

(1 − v·2 − 1

)dy

2

+ 12

c

w () −

∫0

(1 −

√1 − w′2

)dx

2

. (3.4.5)

Using the appropriate expansions, we find the following expressions for the secondand third variation:

P2 [v, w; N, c] =∫

0

(12

Bw′′2 − 12

Nw′2)

dx +∫

0

12

Bv··2 dy + cw2 () + 12

cv2 () (3.4.6)

P3 [v, w; N, c] = 12

cw ()

∫0

v·2dy − 12

cv ()

∫0

w′2 dy. (3.4.7)

In this case, we have P3 �≡ 0.The condition for neutral equilibrium follows from (3.4.6),

P11 [v, η, w, ζ; N, c] =∫

0

(12

Bw′′ζ′′ − 12

Nw′ζ ′)

dx

+∫

0

12

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



0 = Bw′′ζ ′ ∣∣0 − Bw′′′ζ

∣∣0 − Nw′ζ

∣∣0 +

∫0

(Bw′′′′ + Nw′′) ζdx

(3.4.9)

+ Bv··η· ∣∣0 − Bv···η

∣∣0 +

∫0

Bv····η dy + cw () ζ() + cv () η() ,

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 kine-matically admissible functions, and that they must satisfy the kinematic condition(3.4.1), i.e.,

ζ ′ () = η· () . (3.4.11)

We then obtain

Bw′′ () + Bv·· () = 0 (3.4.12)

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 boundaryconditions

w (0) = v (0) = 0(3.4.14)

w′() = v· () = − sin θ.

These boundary conditions and the differential equations are also homogeneousand linear, so they always have the trivial solution v = w = 0. We now introduce theparameter

N/B = α2. (3.4.15)

Choosing solutions for u and v so that the first two conditions of (3.4.13) and thefirst two conditions of (3.4.14) are satisfied, we obtain

w(x) = A1 sin αx + A2x(3.4.16)

v(x) = A3y + A4y3.

From the remaining conditions, we obtain the following set of equations:

−α2A1 sin α + 6A4 = 0

Bα3A1 cos α − N (αA1 cos α + A2) + c (A1 sin α + A2) = 0(3.4.17)−6BA4 + c

(A3 + A4

3) = 0

αA1 cos α + A2 = A3 + 3A42 = − sin θ.

70 Applications

The second of these conditions can be simplified because the first two terms cancel,so that

A1 sin α + A2

(1 − N

c

)= 0. (3.4.18)

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 = 12

sin θ, A42 = −1

2sin θ. (3.4.20)

From the first of the equations of (3.4.17), we now obtain the bifurcation condition

α22 sin α

sin α − α cos α+ 3 = 0 (3.4.21)

or

tan α = α

1 + 13

(α)2. (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,

P3 [v1, w1; N1, c → ∞] = limc→∞

1

2cw1 ()

∫0

v·21 dy − 1

2cv1 ()

∫0

w′21 dx

. (3.4.23)

From (3.4.16) and (3.4.18), we obtain

limc→∞ cw1 () = A2N = 3N sin θ

(α)2 = 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

limc→∞ cv1 () = 6BA4 = −3B sin θ

()2 . (3.4.25)

We now obtain

P3 [v1, w1; N1, c → ∞]= 32

B sin θ

2

∫

0

(A3 + 3A4y2)2 dy +

∫0

(α1A1 cos α1x + A2)2 dx

= 32

B sin θ

2

[A3 + 2A3A4

3 + 95

A24

5 (3.4.26)

+ 12αA2

1

( + 1

2αsin 2α1

)+ 2A1A2 sin α1 + A2

2

].


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

P3 [v1, w1; N1, c → ∞] = 3N1 sin3 θ

(α1)2

[7

20+ 9

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 neg-

ative values of wmax.

λ

Wmax

N

stable

N

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 twocompressive forces N (see Figure 3.5.1).

The stiff joint is again replaced by an elastic element, and the potential energyis now given by

P [v (y) , w (x) ; N] =∫

0

[12

Bw′′2

1 − w′2 + N(√

1 − w′2 − 1)]

dx

+∫

0

[12

Bv··2

1 − v·2 + N(√

1 − v·2 − 1)]

dy

(3.5.1)

+ 12

c

w () −

∫0

[√1 − v·2 − 1

]dy

2

+ 12

c

v () −

∫0

[1 −

√1 − w′2

]dx

2

.

The second variation is now given by

P2 [v, w; N, c] =∫

0

(12

Bw′′2 − 12

Nw′2)

dx

(3.5.2)

+∫

0

(12

Bv··2 − 12

Nv·2)

dy + 12

cw2 () + 12

cv2 () .

N

N

l

l

Figure 3.5.1

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 thestructure and its loading. We will also need the fourth variation,

P4 [v, w; N, c] =∫

0

(12

Bw′′2w′2 − 18

Nw′4)

dx +∫

0

(12

Bv··2v·2 − 18

v·4)

(3.5.3)

+ 12

c

∫

0

12

v·2 dy

2

+ 12

c

∫

0

12

w′2 dx

2

.

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 bendingmoments 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 sinπx

, v1 = a sinπy

,N1

B= π2

2, (3.5.7)

and we have

cw1 () = cv1 () = 0 (3.5.8)

a =

πsin θ. (3.5.9)

To investigate the stability of this neutral equilibrium state, for small displacementsfrom this configuration we write

w (x) = a sinπx

+ w (x) , v (x) = a sinπy

+ v (y) , (3.5.10)

where w and v are orthogonal to w1 and v1, respectively.We now consider (see 2.4.11)

Minw.r.t. v,w

{P2 [v, w; N1, c → ∞] + (N − N1) P′

11 [v1, w1, v, w; N1, c → ∞]

+ P21 [v1, w1, v, w; N1, c → ∞]}. (3.5.11)

74 Applications

Now

P′11 [v1, w1,v, w·; N1, c → ∞] = −

∫0

w′1w′ dx −

∫0

v·1v· dy = 0 (3.5.12)

due to the orthogonality conditions. From (3.5.2) we obtain

P2 [v, w·; N1, c] =∫

0

(12

Bw′′2 − 12

Nw′2)

dx +∫

0

(12

Bv··2 − 12

Nv·2)

dy

(3.5.13)+ 1

2cw2 () + 1

2cv2 () .

From (3.4.7), we obtain

P3 [v1 + v, w1 + w; N1, c] = 12

c [w1 () + w ()]

∫0

(v·2

1 + 2v·1v· + v2)dy

− 12

c [v1 () + v ()]

∫0

(w′2

1 + 2w′1w′ + w2) dx

= 12

cw1 ()

∫0

v·21 dy − 1

2cv1 ()

∫0

w′21 dx

(3.5.14)

+ cw1 ()

∫0

v·1v· dy − cv1 ()

∫0

w′1w′ dx + · · ·

= 12

cw ()

∫0

v·21 dy − 1

2cv ()

∫0

w′21 dx + · · ·

= P21 [v1, w1, v, w·; N1, c] + · · · .

Let v, w be the minimizing solution. The minimizing solution now must satisfythe 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.,

∫0

(Bw′′ζ ′′ − N1w′ζ ′) dx +∫

0

(Bv··η·· − N1v·η·) dy

+ cw () ζ() + cv () η() (3.5.16)

+ 12

cζ()

∫0

v·21 dy − cη()

∫0

w′21 dx

= 0.



Bw′′ζ ′ ∣∣0 − Bw′′′ζ

∣∣0 − N1w′ζ

∣∣0 +

∫0

(Bw′′′′ + Nw′′) ζdx

+ Bv··η· ∣∣0 − Bv···ζ

∣∣0 − N1v·η

∣∣0 +

∫0

(Bv··· + Nv··) ηdy

(3.5.17)+ cw () ζ() + cv () η()

+1

2c

∫0

v·21 dy

ζ() −

1

2c

∫0

w′21 dx

η() = 0,

which yields the differential equations

Bw′′′′ + N1w′′ = 0, Bv···· + N1v·· = 0. (3.5.18)

For the evaluation of the stock terms, we must again note that η and ζ are kine-matically admissible functions, and that they must satisfy the condition (3.5.6) atx = 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

− Bw′′′ () − N1w′ () + cw () + 12

c

∫0

v·21 dy = 0 (3.5.21)

− Bv··· () − N1v· () + cv () − 12

c

∫0

w′21 dx = 0.

In addition to these dynamic boundary conditions, we have the kinematic boundaryconditions

w (0) = v (0) = 0(3.5.22)

w′ () = v· () .

With the exception of the third and fourth equations of (3.5.21), these differen-tial equations and boundary conditions are the same as those for the determinationof v1 and w1. However, notice that due to the fact that the third and fourth equa-tions of (3.5.21) are inhomogeneous, a solution v = w = 0 is not possible. The inho-mogeneous terms are ± 1

4 a2 π2c

. To satisfy these inhomogeneous conditions, we try asolution of the form

v (y) = A∗2y + A∗∗

2 sinπy (3.5.23)

w (x) = A∗1x + A∗∗

1 sinπx

.

76 Applications

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

∫0

v1v dy +∫

0

w1w dx = 0, (3.5.24)

which yields

(A∗1 + A∗

2)2

π2+ 1

2(A∗∗

1 + A∗∗2 ) = 0. (3.5.25)

The boundary condition (3.5.20), the first two conditions of (3.5.21), and the first twoconditions of (3.5.22) are automatically satisfied by (3.5.23). From the third equationof (3.5.21), we obtain(

−Bπ2

2+ N1

π

)A∗∗

1 + (−N1 + c) A∗1 = −1

4a2 π2c

,

where the first term vanishes due to (3.5.7), so that

A∗1 = −a2π2

42 (1 − N1/c). (3.5.26)

Similarly, we obtain from the fourth equation of (3.5.21),

A∗2 = a2π2

42 (1 − N1/c). (3.5.27)

Because A∗1 + A∗

2 = 0, we obtain from the orthogonality condition (3.5.25),

A∗∗1 + A∗∗

2 = 0. (3.5.28)

From the third condition of (3.5.22), we obtain

A∗1 − A∗

2 = (A∗∗1 − A∗∗

2 )π

, (3.5.29)

from which

A∗∗1 = −A∗∗

2 = − a2π

4 (1 − N1/c). (3.5.30)

According to (2.3.54), stability is now governed by the sign of

P4 [v1, w1; N1, c → ∞] + 12

P21 [v1, w1, v, w; N1, c → ∞] . (3.5.31)

From (3.5.3), we obtain

P4 [v1, w1; N1, c] = − 116

Ba4 π6

6+ 1

16π4

2a4c, (3.5.32)

and from (3.5.14),

P21 [v1, w1, v, w; N1, c] = − π4a4c163 (1 − N1/c)

, (3.5.33)


1

λ

Wmax

Figure 3.5.2

so that

limc→∞ P4 [v1, w1; N1, c] + 1

2P21 [v1, w1, v.w; N1, c] = −π6

16Ba4

6< 0. (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

F ≈ N + δ

N. (3.5.35)

Let the deflection of the horizontal bar be

v (y) = a sinπy

;

then δ is given by

δ ≈∫

0

12

v·2 dy = 12

π2

2a2

∫0

cos2 πx

dx = π2a2

42, (3.5.36)

N

N

F

F

δ

δ

Figure 3.5.3

78 Applications

so that

F = N(

1 + π2a2

42

). (3.5.37)

For the post-buckling behavior of the Euler bar, we had

F = F1

(1 + π2a2

82

), (3.5.38)

where F1 is the critical load. From (3.5.37) and (3.5.38), we find

N = F1

(1 − π2a2

82

). (3.5.39)

This means that the equilibrium is already unstable for values of N smaller than thecritical 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 causebending 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 = −yv′0 (x) − zw′

0 (x) + α′ (x) ψ0 (y, z)

v = v0 (x) − (z − z0) α (x) + 12υ(y2 − z2) v′′

0 (x) + υyzw′′0 (x) − υ�0 (y, z) α′′ (x)

w = w0 (x) + (y − y0) α (x) + υyzv′′0 (x) + 1

2υ(z2 − y2)w′′

0 (x) − υ�0 (y, z) α′′ (x) .

(3.6.1)

Here α (x) is the specific angle of torsion, ψ0 (y, z) and ϕ (y, z) are the real and theimaginary 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

∂y= ∂�0

∂z= ψ0 (y, z) , −∂�0

∂z= ∂�0

∂y= ϕ (y, z) . (3.6.2)

t

x

z

y

* y0 , z0( )

Figure 3.6.1

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 wallthickness of the cross section, L is a characteristic wavelength, and b is a character-istic dimension of the cross section. The center of shear is determined by∫

A

zψ0 (y, z) dA =∫A

yψ0 (y, z) dA = 0. (3.6.3)

Further, the resultant of the normal stresses in the x-direction must vanish, whichyields ∫

A

ψ0 (y, z) dA = 0. (3.6.4)

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) ,∂ ()∂y

,∂ ()∂z

= O (() /b) (3.6.5)

|ψ0| , |ϕ| = O(b2) , |�0| , |ψ0| = O(b3),

where b is the width of the cross section. To simplify the calculations, we shall usethe principal axes of inertia, i.e., ∫

A

yzdA = 0. (3.6.6)

We now obtain the following expression for the potential energy:

P [u] =∫V

12

(σx

∂u∂x

+ τxyγxy + τxzγxz

)dV

= 12

∫ {E (−yv′′

0 − zw′′0 + ψ0α

′′)2 + G[α′ ∂ψ0

∂y− (z − z0) α′ + · · ·

]2

(3.6.7)

+G[α′ ∂ψ0

∂z+ (y − y0) α′ + · · ·

]2}

dV,

where the dots represent terms containing third-order derivatives with respect to x.With the torsional stiffness St defined by

St =∫A

G{[

∂ψ0

∂y− (z − z0)

]2

+[∂ψ0

∂z+ (y − y0)

]2}

dA (3.6.8)

and a warping constant � defined by

� =∫A

ψ20 (y, z) dA (3.6.9)

80 Applications

and using the relations (3.6.3), (3.6.4), and (3.6.6) we may rewrite the potentialenergy expression to yield

P [u] =∫

0

(12

EIzv′′20 + 1

2EIyw′′2

0 + 12

E�α′′2)

dx +∫

0

12

Stα′2 dx. (3.6.10)

Expression (3.6.10) is valid for b2/L2 � 1. In addition to this energy, we have theenergy of the prestresses,∫

V

12

Sij uh,iuh,j dV = −12σ

∫V

(u2

,x + v2,x + w2

,x

)dV. (3.6.11)

The term 12

∫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

−12σ

∫V

{[v′

0 − (z − z0) α′ + · · ·]2 + [w′0 + (y − y0) α′ + · · ·]2

}dV

= −∫

0

(12

Nv′20 + 1

2Nw′2

0

)dx − 1

2σ

∫0

∫A

[(y − y0)2 + (z − z0)2

]dAα′2dx (3.6.12)

−∫

0

(Nz0α′v′

0 − Ny0α′w′

0) dx

with ∫A

[(y − y0)2 + (z − z0)2

]dA = Iz0 + Iy0 = [i2

z0+ i2

y0

]A = i2

0A. (3.6.13)

Finally, we obtain for the total potential energy

P [u, N] =∫

0

(12

EIzv′′20 + 1

2EIyv′′2

0 + 12

E�α′′2 − 12

Nv′20 − 1

2Nw′2

0 − 12

Ni20α

′2

(3.6.14)− Nz0α

′v′0 + Ny0α

′w′0 + 1

2Stα

′2)

dx = P2 [u; N] .

Expression (3.6.14) shows that interaction between bending and torsion only occursin the terms with Nz0α

′v′0 and Ny0α

′w′0. This condition implies that when the shear

center coincides with the center of gravity, there is no interaction. This is the casewhen there are two axes of symmetry. When there is only one axis of symmetry, oneof the coupling terms vanishes, i.e., we then have one axis of buckling mode withoutinteraction and one with interaction.

We shall now restrict our discussion further to the case in which the values ofIy, Iz, �, St, and N are constant. The equations of neutral equilibrium follow from

P11 [v0, w0, α, η, ζ, χ; N] =∫

0

[EIzv′′0 η′′ + EIzw′′

0 0ζ′′ + E�α′′χ′′

− Nv′0η

′ − Nw′0ζ

′ − Ni20α

′χ′ − Nz0 (α′η′ + χ′v′0) (3.6.15)

+ Ny0 (α′ζ ′ + χ′w′0) + Stα

′χ′] dx = 0.


By integration by parts, we obtain

EIz

v′′

0 η′|0 − v′′′0 η|0 +

∫0

v′′′0 ηdx

+ EIy

w′′

0 ζ ′|0 − w′′′0 ζ|0 +

∫0

w′′′′0 ζdx

+ E�

α′′χ′|0 − α′′′χ|0 +

∫0

α′′′′χ dx

− Nv′0η

′|0 +∫

0

Nv′′0 ηdx − Nw′

0ζ|0 +∫

0

Nw′′0 ζdx

(3.6.16)

− Ni20α

′χ|0 +∫

0

Ni20α

′′χ dx − Nz0 (α′η+ χv′0) |0

+ Nz0

∫0

(α′′η+ v′′0χ) dx + Ny0 (α′ζ+ w′

0χ) |0

− Ny0

∫0

(α′′ζ+ w′′0χ) dx + Stα

′χ|0 −∫

0

Stα′′χ dx = 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 = (3.6.18)

η = ζ= χ = 0 at x = 0 and x = ,

and for a simply supported beam,

η = ζ= χ = 0 at x = 0 and x = . (3.6.19)

The differential equations are

v′′′′0 + Nv′′

0 + Nz0α′′ = 0

w′′′′0 + Nw′′

0 − Ny0α′′ = 0 (3.6.20)

α′′′′ + Ni20α

′′ + Nz0v′′0 − Ny0w′′

0 − Stα′′ = 0.

For the clamped-clamped beam, the kinematic boundary conditions are

u (0) = u () = 0, v′0 = w′

0 = α′ = 0 at x = 0 and x = . (3.6.21)

For the simply supported beam, we have the dynamic boundary conditions

EIzv′′0 = EIyw′′

0 = E�α′′ = 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 evenderivatives, we try a solution of the form

v0 = B sinπx

, w0 = C sinπx

, α = A sinπx

.† (3.6.23)

These expressions satisfy the boundary conditions (3.6.21) and (3.6.22). Substitutionof these expressions into (3.6.20) yields

(Ny − N) B −Nz0A = 0(Nz − N) C +Ny0A = 0

−Nz0B+ Ny0C +i20 (Nt − N) A = 0,

(3.6.24)

where we have introduced the notation

π2EIz

2= Ny,

π2EIy

2= Nz,

1

i20

(π2E�

2+ S2

)= Nt, (3.6.25)

where Nt is the lowest buckling load due to torsion. The condition for a non-trivialsolution is the vanishing of the determinant, i.e.,

i20 (Ny − N) (Nz − N) (Nt − N) + y2

0N2 (N − Ny) + z20N2 (N − Nz) = 0. (3.6.26)

When y0 and z0 are zero, the roots of this equation are N = (Ny, Nz, Nt). Now letat least y0 or z0 be nonzero. Then for N = min (Ny, Nz, Nt), the left-hand mem-ber of (3.6.26) is negative. The left-hand member can only become positive forN < min (Ny, Nz, Nt). This condition means that the smallest positive root of thisequation is smaller than Ny, Nz, and Nt, i.e., the actual buckling load is smaller thanthe 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 i20 (Nz − N) (Nt − N) − y2

0N2 = 0, (3.6.27)

where for the first case there is no interaction, but there is interaction in the secondcase. Let us now investigate this second case more closely. Equation (3.6.27) may berewritten in the form

N2

(1 − y2

0

i20

)− (Nz + Nt) N + NzNt = 0. (3.6.28)

This equation is symmetric in Nt and Nz; hence without loss of generality we mayassume

Nz/Nt = η < 1. (3.6.29)

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

, and so on.


Introducing the notation N/Nz = λ, we may rewrite the (3.6.28) to yield

λ2

(1 − y2

0

i20

)−(

1 + 1η

)λ + 1

η= 0. (3.6.30)

The smallest root is given by

λ = 1

2(1 − y2

0/i20

) [(1 + 1/η) −√

(1 − 1/η)2 + 4ηy20/i2

0

]. (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

λ = limη→0

1

2(1 − y2

0/i20

) 1η

[1 + η− (1 − η+ 2ηy2

0/i20

)] = 1, (3.6.33)

as expected. For η → 1, we find

λ = 11 + y0/i0

< 1. (3.6.34)

Because y0/i0 < 1, this means

12

< λ < 1. (3.6.35)

This result shows that under combined torsion and bending, the buckling load maybe considerably smaller than Nz or Nt, especially when y0/i0 → 1. An example ofsuch a structure is the thin-walled circular tube with an open cross section (see Fig-ure 3.6.2).

In this case, we have

St = 23πGRt3, � = π

(23π3 − 4

)R5t, y0/i0 = 2√

5. (3.6.36)

For η = 1, we obtain λ = 0.528, i.e., a reduction of 47% of the buckling load. Weemphasize here once again that the theory is only valid for b2/L2 � 1.

2R

R

t

*

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 infi-nite in width, compressed between flat smooth rigid stamps. The material is suchthat 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 = 12

[λ2

1 + λ22 + λ2

3 − 3], (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 takento be unity. We shall assume that the material is incompressible, i.e.,

λ1λ2λ3 = 1, (3.7.3)

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

λ1λ2 = 1. (3.7.4)

We now consider the following states (see Figure 3.7.1), where σ1h is theforce per unit width of the plate. Let (x, y) be the coordinates of a material pointin the undeformed state. The coordinates of this point in the fundamental state aredenoted by (x, y), where

x = λ1x, y = λ2y. (3.7.5)

In the adjacent state, the coordinates are(x, y), where

x = λ1x + u, y = λ2y + v. (3.7.6)

The square of the length of an infinitely small material element in the undeformedstate is given by

(ds)2 = (dx)2 + (dy)2. (3.7.7)

l

h

x

undeformed state

σ1h

σ1h

λ 2h

I

fundamental state

II

σ1 h

σ1 h

adjacent state

λ1l

Figure 3.7.1

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 + λ2

2 (dy)2, (3.7.8)

and in the adjacent state II, it is given by(ds)2 = (dx

)2 + (dy)2 = [(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∣∣∣∣∣ gxx − g gxy

gxy gyy − g

∣∣∣∣∣ = 0, (3.7.11)

and in the corresponding principal directions we have

g11 = λ 21 , g22 = λ 2

2 , (3.7.12)

where λ1 and λ2 are the extension ratios in those directions. The first invariant ofthe metric tensor is

g11 + g22 = λ 21 + λ 2

2 = gxx + gyy. (3.7.13)

The elastic energy density in the adjacent state II is now given by

WII = 12

(gxx + gyy − 2

)(3.7.14)

= 12

[(λ1 + u,x)2 + v2,x + (λ2 + v,y)2 + u2

,y − 2].

In the fundamental state I, the elastic energy density is

WI = 12

(gxx + gyy − 2

) = 12

(λ2

1 + λ22 − 2

), (3.7.15)

so that the increment of the elastic energy density in passing from the fundamentalstate I to the adjacent state II is

WII − WI = λ1u,x + λ2v,y + 12

(u2

,x + u2,y + v2

,x + u2,y

). (3.7.16)

In addition to this energy expression, we shall need an explicit expression forthe incompressibility condition, which with the aid of the second invariant of themetric tensor can be written as

1 = λ1λ2 = λ1λ2 =√

g11g22 =√

gxxgyy − gxy. (3.7.17)

86 Applications

However, it is slightly simpler to derive the incompressibility condition from

dvdv

=∣∣∣∣∣∂x/∂x ∂x/∂y∂y/∂x ∂y/∂y

∣∣∣∣∣ =∣∣∣∣∣λ1 + u,x u,y

v,x λ2 + v,y

∣∣∣∣∣ = λ1λ2,

from which follows

λ2u,x + λ1v,y + u,xv,y − u,yv,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

P [u] =∫

0

h/2∫−h/2

[λ1u,x + λ2v,y + 1

2

(u2

,x + u2,y + v2

,x + v2,y

)]dx dy

(3.7.19)

+ σ1

∫0

h/2∫−h/2

u,x dx dy,

where for the last term we have rewritten the term σ1h∫

u,x dx as a surface inte-gral. 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), wefind

P [u] =∫

0

h/2∫−h/2

[(λ1 − λ2

2

λ1+ σ1

)u,x + 1

2

(u2

,x + u2,y + v2

,x + v2,y

)(3.7.20)

+ λ2

λ1(u,yv,x − u,xv,y)

]dx dy.

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

λ1= −λ1 + 1

λ31

, (3.7.21)

which is the equilibrium condition in the fundamental state. The discussion of theequilibrium in the fundamental state is now reduced to the discussion of

P [u] =∫

0

h/2∫−h/2

[12

(u2

,x + u2,y + v2

,x + v2,y

) +λ2

λ1(u,yv,x − u,xv,y)

]dx dy (3.7.22)

under the nonlinear side condition (3.7.18). Notice that P [u] is a homogeneousquadratic functional, i.e., P [u] = P2 [u]. A necessary condition for stability is thatP2 [u] ≥ 0 for very small displacements from the fundamental state, which meansthat we may linearize the side condition to yield

λ2u,x + λ1v,y = 0. (3.7.23)


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 condi-tion, we now consider the functional

P [u] +∫

0

h/2∫−h/2

p (x, y) (λ2u,x + λ1v,y) dx dy. (3.7.24)

A necessary condition for stability is now

∫0

h/2∫−h/2

[u,xη,x + u,yη,y + v,xζ,x + v,yζ,y

+ λ2

λ1(u,yζ,x + v,xη,y − u,xζ,y − v,yη,x) (3.7.25)

+ (λ2u,x + λ1v,y) δp + p (x, y) (λ2η,x + λ1ζ,y)] dx dy = 0,

for all kinematically admissible displacements η, ζ. By integration by parts, weobtain

h/2∫−h/2

(u,xη)∣∣0 dy +

∫0

(u,yη)∣∣h/2−h/2 dx −

∫0

h/2∫−h/2

(u,xx + u,yy) ηdx dy

+h/2∫

−h/2

(v,xζ)∣∣0 dy +

∫0

(v,yζ)∣∣h/2−h/2 dx −

∫0

h/2∫−h/2

(v,xx + v,yy) ζdx dy

+ λ2

λ1

h/2∫−h/2

(u,yζ)∣∣0 dy −

∫0

(u,xζ)∣∣h/2−h/2 dx +

∫0

(v,xη)∣∣h/2−h/2 dx −

h/2∫−h/2

(v,yη)∣∣0 dy

+ λ2

h/2∫−h/2

(pη)∣∣0 dy − λ2

∫0

h/2∫−h/2

p,xηdx dy + λ1

∫0

(pζ)∣∣h/2−h/2 dx − λ1

∫0

h/2∫−h/2

p,yζdx dy

+∫

0

h/2∫−h/2

(λ2u,x + λ1v,y)δp dx dy = 0. (3.7.26)

This yields the differential equations

u,xx + u,yy + λ2p,x = 0

v,xx + v,yy + λ1p,y = 0 (3.7.27)

λ2u,x + λ1v,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∫−h/2

(u,x − λ2

λ1v,y + λ2p

)dy = 0 at x = (3.7.31)

v,x + λ2

λ1u,y = 0 at x = 0 and x = (3.7.32)

u,y + λ2

λ1v,x = 0 at y = ±h/2 (3.7.33)

v,y − λ2

λ1u,x + λ1p = 0 at y = ±h/2. (3.7.34)

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, andwe have two boundary conditions on each part of the boundary.

We shall now restrict ourselves to the lowest buckling load, and taking intoaccount the boundary conditions (3.7.28) and (3.7.29), we can try a solution foru (x, y) of the form

u (x, y) = U (y) sinπx

. (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) cosπx

, (3.7.37)

and from the second or third of the equations of (3.7.27), we find

v (x, y) = V (y) cosπx

. (3.7.38)

Substitution of these expressions into (3.7.27) yields

U − π2

2U − λ2

π

P = 0

V − π2

2V + λ1P = 0 (3.7.39)

λ1V + λ2π

U = 0,


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

U (y) = Aeµy, V (y) = Beµy, P (y) = Ceµy, (3.7.40)

which yield (µ2 − π2

2

)A − λ2

π

C = 0

(µ2 − π2

2

)B + λ1µC = 0 (3.7.41)

λ2π

A + λ1µB = 0.

The condition for a nontrivial solution is the vanishing of the determinant of thecoefficients, which yields(

µ2 − π2

2

)(−λ2

1µ2 + λ2

2π2

2

)= 0. (3.7.42)

The eigenvalues are thus

µ1,2 = ±π

, µ3,4 = ±λ

π

, (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 canwrite

U (y) =2∑

i=1

Ai1 sinh µiy +2∑

i=1

Ai2 cosh µiy

V (y) =2∑

i=1

Bi1 cosh µiy +2∑

i=1

Bi2 sinh µiy (3.7.44)

P (y) =2∑

i=1

Ci1 sinh µiy +2∑

i=1

Ci2 cosh µiy,

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 intoa symmetric and an anti-symmetric case.

anti-symmetric deformation symmetric deformation

Figure 3.7.2

90 Applications

In the following, we shall discuss the case of anti-symmetric deformation indetail because calculations show that this type of buckling behavior occurs priorto that of symmetric deformation. The last type of deformation first occurs aftera reduction of the length by about 50%. The calculations involved are entirelysimilar to those of the anti-symmetric case.

In the anti-symmetric case, we have

U (y) = A1 sinhπy

+ A2 sinh λπy

V (y) = B1 coshπy

+ B2 cosh λπy

(3.7.45)

P (y) = C1 sinhπy

+ C2 sinh λπy

.

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 − 1λ2

π

A2. (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

U − λπ

V = 0, V − λ

π

U + λ1P = 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

l(1 + λ2) cosh

πh2

A1 + 2λ cosh λπh2

A2 = 0(3.7.49)

−2λ sinhπh2

A1 −(

λ + 1λ

)sinh λ

πh2

A2 = 0.

Notice that λ = λ2/λ1 is the single significant load parameter. Its value follows fromthe condition for a non-trivial solution of (3.7.49), which leads to

(λ2 + 1

)2tanh λθ = 4λ3 tanh θ, (3.7.50)

where we have introduced the geometric parameter

θ = πh2

. (3.7.51)

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


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

(λ2 − 1)2 = 13λ2θ2[(λ2 + 1)2 − 4] + O(θ3). (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 = 43θ2 + O(εθ2), (3.7.54)

so that

λ2

λ1= 1 + 2

3θ2. (3.7.55)

Using the incompressibility condition λ1λ2 = 1, we find

λ2 = 1 + 13θ2 = 1 + π2h2

122(3.7.56)

λ1 = 1 − 13θ2 = 1 − π2h2

122.

This result for the thin plate (h/ � 1) is in complete agreement with the result forthe Euler column, and as it is derived from a two-dimensional theory, it justifies thelatter 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

λ1 = 1√3.383

� 0.55, (3.7.58)

which means a reduction in length of 45%. In this case, the deformations are concen-trated near y = ±h/2. In the case of symmetric deformation, the bifurcation equa-tion 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-symmetricmode (λ ≤ 3.383) will occur prior to the occurrence of bifurcation in a symmetricmode.

We now continue with the discussion of the anti-symmetric case. From the firstof the equations of (3.7.49), we obtain

A2

A1= − (1 + λ2)

2λ

cosh θ

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

u (x, y) = A1

[sinh

πy

− 1 + λ2

2λ

cosh θ

cosh λθsinh

πy

]sin

πx

v (x, y) = A1

[−λ cosh

πy

+ 1 + λ2

2λ

cosh θ

cosh λθcosh λ

πy

]cos

πx

(3.7.61)

p (x, y) = A11 − λ4

2λλ2

π

cosh θ

cosh λθsinh λ

πy

cosπx

.

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

u1 (x, y) = N−1 sinπx

[−2

cosh λθ

λsinh

πy

+ (λ−2 + 1) cosh θ sinhλπy

]

v1 (x, y) = N−1 cosπx

[2 cosh λθ cosh

πy

− (λ−2 + 1) cosh θ coshλπy

](3.7.64)

p1 (x, y) = (λ2N)−1 π

(λ2 − λ−2) cosh θ sinh

λπy

cosπx

.

Having determined the displacement fields for the critical case of neutral equi-librium, we now proceed with the discussion of its stability. Therefore, we considersmall but finite displacements from the equilibrium configuration,

u (x, y) = au1 (x, y) + u (x, y)(3.7.65)

v (x, y) = av1 (x, y) + v (x, y) ,

where u and v are orthogonal (in some suitable sense) to the buckling mode. Stabil-ity is now governed by the behavior of the functional (3.7.22) under the nonlinearside condition (3.7.18). To make full use of the properties of the functional in thecritical case of neutral equilibrium, we consider the functional (3.7.22). Substituting(3.7.65) into this functional, we find

PII − PI =∫

0

h/2∫−h/2

[12

a2 (u21,x+u2

1,y+v21,x+v2

1,y

)+a (u1,xu,x+u1,yu,y + v1,xv,x+v1,yv,y)

+ 12

(u2

,x + u2,y + v2

,x + v2,y

)+ λa2 (u1,yv1,x − u1,xv1,y) (3.7.66)

+ λa (u1,yv,x + u,yv1,x − u1,xv,y − u,xv1,y) +λ (u,yv,x − u,xv,y)]

dx dy,


where the terms involving a2 vanish because u1 and v1 satisfy the equations for thecritical case of neutral equilibrium. Using (3.7.25) with η = u, ζ= v, and u = u1,v =v1, p = p1, we can rewrite the bilinear terms in (3.7.66) to yield

−∫

0

h/2∫−h/2

ap1 (λ2u,x + λ1v,y) dx dy, (3.7.67)

so that our final result for the functional becomes

PII − PI =∫

0

h/2∫−h/2

[12

(u2

,x + u2,y + v2

,x + v2,y

) + λ (u,yv,x − u,xv,y)

(3.7.68)− ap1 (λ2u,x + λ1v,y)

]dx dy.

Substitution of (3.7.65) into the nonlinear side condition (3.7.18) yields

λ2u,x + λ1v,y + a2 (u1,xv1,y − u1,yv1,x)(3.7.69)

+ a (u1,xv,y + u,xv1,x − u1,yv,x − u,yv1,x) + u,xv,y − u,yv,x = 0.

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

−∫

0

h/2∫−h/2

ap1 (λ2u,x + λ1v,y) dx dy

=∫

0

h/2∫−h/2

[a3p1 (u1,xv1,y − u1,yv1,x) + a2p1 (u1,xv,y + u,xv1,y − u1,yv,x − u,yv1,x)

+ ap1 (u,xv,y − u,yv,x)] dx dy, (3.7.70)

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

∫0

sin2 πx

cosπx

dx = 0,

∫0

cos3 πx

dx = 0.

The functional (3.7.68) can now be rewritten in the form

PII − PI =∫

0

h/2∫−h/2

{12

(u2

,x + u2,y + v2

,x + v2,y

)+ λ (u,yv,x − u,xv,y)

+ a2p1 (u1,xv,y + u,xv1,y − u1,yv,x − u,yv1,x) + ap1 (u,xv,y − u,yv,x)

+ p[λ2u,x + λ1v,y + a2 (u1,xv1,y − u1,yv1,x) (3.7.71)

+ a (u1,xv,y + u,xv1,y − u1,yv,x − u,yv1,x) + u,xv,y − u,yv,x]

+µ (u1u + v1v)}

dx dy,

94 Applications

where we have taken into account the nonlinear side condition with the Lagrangianmultiplier p (x, y) and the orthogonality condition

∫0

h/2∫−h/2

(u1u + v1v) dx dy = 0 (3.7.72)

with the multiplier µ.† We now minimize this functional with respect to u, v, p, andµ for a fixed value of the amplitude a. Variation with respect to u yields

∫0

h/2∫−h/2

{u,xδu,x + u,yδu,y + λ (v,xδu,y − v,yδu,x)

+ a2p1 (v1,yδu,x − v1,xδu,y) + ap1 (v,yδu,x − v,xδu,y)(3.7.73)

+ p [λ2δu,x + a (v1,yδu,x − v1,xδu,y) + v,yδu,x − v,xδu,y]

+µu1δu} dx dy = 0.


−∫

0

h/2∫−h/2

[u,xx+ u,yy + a2 (p1v1,y),x + a2 (p1v1,y)

,x − a2 (p1v1,x),y

+ a (p1v,y),x − a (p1v,x),y + λ2p,x

+ a (pv1,x),x − a (pv1,x),y + (pv,y),x − (pv,x),y + µu1] δu dx dy (3.7.74)

+∫

0

(u,y + λv,x − a2p1v1,x − ap1v,x − apv1,x − pv,x) δu∣∣∣h/2−h/2 dx

+h/2∫

−h/2

(u,x − λv,y + a2p1v1,y + ap1v,y + λ2p + apv1,y + pv,y) δu∣∣0 dy.

Remembering the geometric boundary conditions u = 0 at x = 0 and u,y = 0 at x =, we obtain

u,xx + u,yy + a2(p1,xv1,y − p1,yv1,x) + a (p1,xv,y − p1,yv,y)(3.7.75)

+ λ2p,x + a(p,xv1,y − p,yv1,y) + p,xv,y − p,yv,x + µu1 = 0,

and the dynamic boundary conditions

u,y + λv,x − p1(a2v1,x + av,x) − p (av1,x + v,x) = 0 for y = ±h/2(3.7.76)

h/2∫−h/2

[u,x − λv,y + (a2p1 + ap)v1,y + (ap1 + p) v,y + λ2p] dy = 0 for x = .

† Notice that µ is independent of x and y.


Variation with respect to v yields

∫0

h/2∫−h/2

[v,x δv,x + v,yδv,y + λ (u,yδv,x − u,xδv,y)

+ a2p1 (u1,xδv,y − u1,yδv,x) + ap1 (u,xδv,y − u,yδv,x)(3.7.77)

+ p{λ1δv,y + a (u1,xδv,y − u1,yδv,x) + u,xδv,y − u,yδv,x

}+ µv1δv] dx dy = 0.


−∫

0

h/2∫−h/2

[v,xx + v,yy + a2 (p1u1,x),y − a2 (p1u1,y),x

+ a (p1u,x),y − a (p1u,y),x + λ1p,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)

+∫

0

(v,y − λu,x + a2p1u1,x + ap1u,x + λ1p + apu1,x + pu,x) δv |h/2−h/2 dx

+h/2∫

−h/2

(v,x + λu,y − a2p1u1,y − ap1u,y − apu1,y − p u,y) δv |0dy = 0,

from which follows

v,xx + v,yy + a2(p1,yu1,x − p1,xu1,y) + a(p1,yu,x − p1,xu,y)(3.7.79)

+ λ1p,y + a(p,yu,x − p,xu1,y) + p,yu,x − p,xu,y + µv1 = 0,


v,y − λu,x + (a2p1 + ap)u1,x + (ap1 + p)u,x + λ1p = 0 for y = ±h/2 (3.7.80)

v,x + λu,y − (a2p1 + ap)u1,y − (ap1 + p)u,y = 0 for x = 0 and x = .

(3.7.81)Variation with respect to p and µ yields the nonlinear side condition and the orthog-onality condition, respectively.

We now restrict ourselves to vanishingly small values of the amplitude a. Hence,we may write

u = u 0 + au1 + a2u2 + · · ·v = v0 + av1 + a2v2 + · · ·

(3.7.82)p = p 0 + ap1 + a2p2 + · · ·µ = µ0 + aµ1 + a2µ2 + · · · .

96 Applications

Introducing these expansions into the differential equations and the correspondingboundary conditions, we find to zeroth order

u 0,xx + u 0

,yy + λ2p 0,x + p 0

,xv0,y − p 0

,yv0,x + µ0u1 = 0

v0,xx + v0

,yy + λ1p 0,y + p 0

,yu 0,x − p 0

,xu 0,y + µ0v1 = 0

(3.7.83)λ2u 0

,x + λ1v0,y + u 0

,xv0,y − u 0

,yv0,x = 0

∫0

h/2∫−h/2

(u1u 0 + v1v0)dx dy = 0,


u0,y + λv0

,x − p0v0,x = 0

v0,y − λu0

,x + p0u0,x + λ1p0 = 0

}for y = ±h/2, (3.7.84)

h/2∫−h/2

(u0

,x − λv0,y + p0v0

,y + λ2p0) dy = 0 for x = (3.7.85)

v0,x + λu0

,y − p0u0,y = 0 for x = 0 and x = , (3.7.86)

and the geometric boundary conditions

u0 = 0 for x = 0 and u0,y = 0 for x = . (3.7.87)

Because the equations and the boundary conditions are homogeneous, the uniquesolution is

u0 = v0 = p0 = µ0 = 0. (3.7.88)

For term linear in a, we obtain

u1,xx + u1

,yy + λ2p1,x + µ1u1 = 0

v1,xx + v1

,yy + λ1p1,y + µ1v1 = 0

(3.7.89)λ2u1

,x + λ1v1,y = 0

∫0

h/2∫−h/2

(u1u1 + v1v1) dx dy = 0,


u1,y + λv1

,x = 0

v1,y − λu1

,x + λ1p1 = 0

}for y = ±h/2 (3.7.90)

h/2∫−h/2

(u1

,x − λv1,y + λ2p1) dy = 0 for x = (3.7.91)


v1,x + λu1

,y = 0 for x = 0 and x = (3.7.92)

and the geometric conditions

u1 = 0 for x = 0 and u1,y = 0 for x = . (3.7.93)

Again, this is a set of homogeneous equations and boundary conditions, so theunique solution is given by

u1 = v1 = p1 = µ1 = 0. (3.7.94)

For the terms quadratic in the amplitude a, we obtain

u2,xx + u2

,yy + λ2p2,x + µ2u1 = p1,yv1,x − p1,xv1,y

v2,xx + v2

,yy + λ1p2,y + µ2v1 = p1,xv1,y − p1,yv1,x

(3.7.95)λ2u2

,x + λ1v2,y = u1,yv1,x − u1,xv1,y

∫0

h/2∫−h/2

(u1u2 + v1v2) dx dy = 0,


u2,y + λv2

,x = p1v1,x

v2,y − λu2

,x + λ1p2 = p1u1,x

}for y = ±h/2 (3.7.96)

h/2∫−h/2

(u2

,x − λv2,y + λ2p2) dy = −

h/2∫−h/2

p1v1,y dy for x = (3.7.97)

v2,x + λu2

,y = p1u1,y for x = 0 and x = , (3.7.98)


u2 = 0 at x = 0 and u2,y = 0 at x = . (3.7.99)

The present equations and boundary conditions are the conditions for the sta-tionary value of the functional

(PII − PI)∗ = a4

∫0

h/2∫−h/2

[12

(u22

,x + u22

,y + v22

,x + v22

,y

)

+ λ(u2

,yv2,x − u2

,xu2,y

)(3.7.100)

+ p1(u1,xv2

,y + u2,xv1,y − u1,yv2

,x − u2,yv1,x

) ]dx dy

under the linear side condition

λ2u2,x + λ1v2

,y + u1,xv1,y − u1,yv1,x = 0 (3.7.101)

98 Applications

and the orthogonality condition

∫0

h/2∫−h/2

(u1u2 + v1v2) dx dy = 0. (3.7.102)

This is the variational problem of a quadratic functional with the linear side condi-tion, and has a second variation that is positive-definite under the requirement oforthogonality 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)∗ = a4A4, (3.7.103)

where A4 is defined by the value of the integral in (3.7.100) evaluated for u2, v2 assolutions of the equations and boundary conditions (3.7.95) to (3.7.99), a necessarycondition for stability is A4 ≥ 0, and a sufficient condition is A4 > 0 in the case ofneutral equilibrium at λ = λcr.

To avoid lengthy calculations, in the following we shall restrict ourselves to thecase of a slender plate (i.e., θ � 1). In that case, from (3.7.55) we have

λcr = 1 + 23θ2 + O

(θ4) = 1 + π2h2

62+ O

(h4

4

). (3.7.104)

Because πy/ is of order θ, we can write the buckling mode (3.7.64) in the form

u1 =[πy

+ O

(θ3)] sin

πx

v1 =[

1 − π2y2

22+ O

(θ4)] cos

πx

(3.7.105)

p1 = π

[2πy2

+ O(θ3)] cos

πx

,

where we must note that the derivative with respect to y of an error term θn is oforder θn−1−1.

We can now evaluate the expressions

p1,xv1,y − p1,yv1,x = π3

3

[1 + O

(θ2)] sin

2πx

p1,yu1,x − p1,xu1,y = 2π4y4

[1 + O

(θ2)] sin

2πx

u1,xv1,y − u1,yv1,x = π2

22

[1 + O

(θ2)] (1 − cos

2πx

)

−p1v1,x = π3y3

[1 + O

(θ2)] sin

2πx

(3.7.106)

p1u1,x = π4y2

4

[1 + O

(θ2)] (1 + cos

2πx

).


p1v1,y = −π4y2

4

[1 + O

(θ2)] (1 + cos

2πx

)

−p1u1,y = −π3y3

[1 + O

(θ2)] sin

2πx

.

The equations (3.7.95) now become

u2,xx + u2

,yy + λ2p2,x = −µ2 πy

sin

πx

− π3

3sin

2πx

v2,xx + v2

,yy + λ1p2,y = −µ2

(1 − π2y2

22

)cos

πx

− 2π4y4

(3.7.107)

λ2u2,x + λ1v2

,y = − π2

22

(1 − cos

2πx2

)∫

0

h/2∫−h/2

[πy

sinπx

u2 +(

1 − π2y2

22

)cos

πx

v2]

dx dy = 0,


u2,y + λv2

,x = ∓π3h23

sin2πx

v2,y − λu2

,x + λ1p2 = π4h2

44

(1 + cos

2πx

)

for y = ±h2

(3.7.108)

h/2∫−h/2

(u2

,x − λv2,y + λ2p2) dy =π4h3

64for x = (3.7.109)

v2,x + λu2

,y = 0 for x = 0 and x = , (3.7.110)


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 leadto terms in u2 and v2 proportional to sin πx/ and cos πx/, respectively, which donot satisfy the orthogonality condition, and hence we require

µ2 = 0. (3.7.112)

We now try a solution in the form

u2 =(

π

8− π3 y2

43

)sin

2πx

− π2

42x + C1

π3 h2

3sin

2πx

+ C2π4 h2

4x

v2 =(

π2 y42

− π4 y3

24

)cos

2πx

− π2

42y + C3

π4 h2y4

cos2πx

+ C4π4 h2

4y (3.7.113)

p2 = −π4 y2

4cos

2πx

− π4 y2

4+ C5

π4 h2

4cos

2πx

+ C6π4 h2

4,

100 Applications

where all numbers C1 − C6 are of order of magnitude unity. We have made use ofthe fact that λ1 and λ2 are of order unity with an error of order π2h2/42, but in someplaces we had to allow for the more accurate relation λ2/λ1 = λ = 1 + π2h2/62.

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(π2

42+ 3

8π4h2

4

)cos

2πx

− π2

42+ C3

π4h2

4cos

2πx

+ C4π4h2

4

− λ

[(π2

42− π4h2

84

)cos

2πx

− π2

42+ C1

2π4h2

4cos

2πx

+ C2π4h2

4

]

+ λ1

(−π4h2

44cos

2πx

− π4h2

44+ C5

π4h2

4cos

2πx

+ C6π4h2

4

)

=(

π2

42+ 3

8π4h2

4− π2

42+ π4h2

84− π4h2

244− π4h2

44+ · · ·

)cos

2πx

− π2

42+ π2

42+ π4h2

244− π4h2

44+ C4

π4h2

4− C2

π4h2

4+ C6

π4h2

4+ · · · (3.7.114)

+(

C3π4h2

4− 2C1

π4h2

4+ C5

π4h2

4

)cos

2πx

=(

524

+ C3 + C5 − 2C1

)π4h2

4cos

2πx

+(

− 524

+ C4 − C2 + C6

)π4h2

4

= − π4h2

44

(1 + cos

2πx

)+(

124

+ C4 − C2 + C6

)π4h2

4

+(

1124

+ C3 + C5 − 2C1

)π4h2

4cos

2πx

.

It follows that

− C2 + C4 + C6 + 124

= 0 (3.7.115)

− 2C1 + C3 + C5 + 1124

= 0. (3.7.116)

For the left-hand member of (3.7.109), we obtain

2{(

πh16

− π3h3

963

)2π

− π2h

82+ C1

π3h3

23

2π

+ C2

π4h3

24

− λ

[π2h82

+ π4h3

164− π2h

82+ C3

π4h3

24+ C4

π4h3

23

]

+ λ2

[−π4h3

244− π4h3

244+ C5

π4h3

24+ C6

π4h3

24

]}(3.7.117)

= −π4h3

34+ π4h3

4(2C1 + C2 − C3 − C4 + C5 + C6)

= π4h3

64+(

2C1 + C2 − C3 − C4 + C5 + C6 − 12

)π4h3

4.

3.8 Helical spring with a small pitch 101

Hence, it follows that

2C1 + C2 − C3 − C4 + C5 + C6 − 12

= 0. (3.7.118)

When the constants C1, C2, and C5 are chosen, the additional constants C3, C4, andC6 can be calculated. For the evaluation of (3.7.103), we only need the dominatingpart of u2 and v2, i.e.,

u2 =(

π

8− π3y2

43

)sin

2πx

− π2

42x

(3.7.119)

v2 =(

π2

42y + π4y3

24

)cos

2πx

− π2

42y.

Substitution of these expressions into (3.7.100) yields

Min (PII − PI)∗ = π6h2a4

1926h[1 + O

(h2/2)] . (3.7.120)

which shows that the critical state of neutral equilibrium is stable.A comparison with Euler column theory requires the evaluation of

P2 [u1, λ = 1] = a2

∫0

h/2∫−h/2

[12

(u2

1,x + u21,y + v2

1,x + v21,y

)+ (u1,yv1,x − u1,xv1,y)]

dx dy.

(3.7.121)Using the expressions (3.7.105), we find

P2 [u1, λ = 1] = a2[

12

π2h

+ 148

π4h3

3+ O

(h5

5

)− 1

2π2h

+ 116

π4h3

3

](3.7.122)

= 112

π4h2a2

4h

[1 + O

(h2

2

)].

The ratio of the expressions (3.7.121) and (3.7.122) is π2a2/162 with a relative errorof order h2/2. This is indeed in full agreement with the result of the Euler columnin 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) andGrammel (1924). Both ignored the shear elasticity of the spring. Shortly thereafter,Biezeno and Koch took into account the effect of shear but they overestimated theeffect, implying as a consequence that any spring, however short, would be liable tobuckling, which is not confirmed by experiment. Later, Haringx (1942) took intoaccount the shear effect correctly. In the first part of this section we shall follow hisapproach.†

† Cf. J. A. Haringx, On highly compressible helical springs and rubber rods, and their application forvibration-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 deforma-tion of one single coil under a certain load will differ little from that of an unclosedcircular ring lying in a flat plane. Consequently, we may imagine the compressedhelical spring as being replaced by a number of similar rings a connected by per-fectly rigid elements b, as represented in Figure 3.8.1.

The influence of the axial force having been discounted as the successive coilsare imagined as being flat after compression of the spring, we now must see whatdistortions occur as a result of the bending moments and transverse forces to betransmitted by these flat coils (see Figure 3.8.2).

In the following, we shall assume that the principal axes of inertia of the crosssection 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 thering and perpendicular to it, respectively, and let St be the torsional stiffness of thecross 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

2πR3u, D = Sb1

πR3v, M = Sb2 · St

πR (Sb2 + St)ϕ. (3.8.1)

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

� = 2πR3

StNn ≡ N

EA, (3.8.2)

N

N

MM

D D

Figure 3.8.2


where the “stretching stiffness” EA is defined by

EA = St

2πR3n. (3.8.3)

Further, for the total displacement due to D we have

vtot = nπR3D

Sb1

≡ DβEA

, β = 2Sb1

St. (3.8.4)

The factor β becomes small for flat cross sections. If t(y) is the thickness of the crosssection, we then have

β = 2

∫ 112 Et3(y) dy∫ 13 Gt3(y) dy

= 1 + υ, (3.8.5)

so that in general, β ≥ 1 + υ. Finally, the total angle of rotation is given by

ϕtot = nπR (Sb2 + St)

Sb2 · St≡ M

B, (3.8.6)

where B is the “bending stiffness,”

B = Sb2 · St

Sb2 + St

1πR

n≡ α EAR2 (3.8.7)

and

α = 2Sb2

Sb2 + St. (3.8.8)

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 displacementsof the transverse shear forces D. (This is not the case for ordinary beams.)

d

d

w

dv

ψ

ds

ψ

Z

N

X

M

D

II

ψ

undeformed fundamental state

adjacent state

dx d x

x + du

Figure 3.8.3

104 Applications

We now have the following relations,

ds = (dx + du) cos ψ+ dw sin ψ(3.8.10)

dv = − (dx + du) sin ψ+ dw cos ψ.

In the fundamental state, we have

N = −λEA (which defines λ) (3.8.11)

x = (1 − λ) x (3.8.12)

so that with ()′ = d (/dx), we have

dsdx

= (1 − λ + u′) cos ψ+ w′ sin ψ

(3.8.13)dvdx

= − (1 − λ + u′) sin ψ+ w′ cos ψ.

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 = dvdx

= − (1 − λ + u′) sin ψ+ w′ cos ψ. (3.8.15)

Further, the curvature of the rod is given by

κ = dψ

dx= ψ′, (3.8.16)

so we can write

N = EAε, D = βEAγ, M = αEAR2κ. (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.(3.8.19)

Z = N sin ψ+ D cos ψ = const. ≡ ζEA

as conditions for the equilibrium of forces, and

M′dx + Dds − Ndv = 0 (3.8.20)


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(3.8.21)

D = Z cos ψ− X sin ψ = (λ sin ψ+ ζcos ψ) EA.

Substituting these expressions into (3.8.20) and using the relations (3.8.13) to(3.8.17), we find

αR2ψ′′ + (λ sin ψ+ ζcos ψ) [(1 − λ + u′) cos ψ+ w′ sin ψ](3.8.22)

− (−λ cos ψ +ζsin ψ) [− (1 − λ + u′) sin ψ+ w′ cos ψ] = 0,

or, upon simplification,

αR2ψ′′ + 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 differentialequation 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(3.8.24)

= (1 − λ cos ψ+ ζsin ψ) cos ψ− 1β

(λ sin ψ+ ζcos ψ) sin ψ

w′ = γ cos ψ+ (1 + ε) sin ψ(3.8.25)

= 1β

(λ sin ψ+ ζcos ψ) cos ψ+ (1 − λ cos ψ+ ζsin ψ) sin ψ.

Substitution of these expressions into (3.8.23) and rearranging terms yields

αR2ψ′′ + λ sin ψ− λ2(

1 − 1β

)sin ψcos ψ

(3.8.26)+ ζ

[cos ψ−

(1 − 1

β

)λ cos 2ψ+ ζ

(1 − 1

β

)sin ψcos ψ

]= 0.

When there are no transverse shear forces in the spring (e.g., for a spring hingedon 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, whichyields

αR2ψ′′ + λψ− λ2(

1 − 1β

)ψ+ ζ

[1 −

(1 − 1

β

)λ

]= 0, (3.8.27)

where we have neglected the term in ζ2. As mentioned previously, when the springis hinged at both ends, ζ= 0. Nonzero values of ζ occur, e.g., in the cases shown inFigure 3.8.4.

106 Applications

Figure 3.8.4

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

∫0

w′ dx = 0, (3.8.28)

which with (3.8.25) yields

∫0

sin ψ− λ

(1 − 1

β

)sin ψ cos ψ+ ζ

(sin2 ψ+ 1

βcos2 ψ

)dx = 0, (3.8.29)

or, linearized,

∫0

{ψ

[1 − λ

(1 − 1

β

)]+ ζ

β

}dx = 0. (3.8.30)

Let us now first consider the case that the spring is hinged at both ends. Thelowest buckling mode is then given by

ψ = cosπx

(3.8.31)

and the bifurcation condition becomes(1 − 1

β

)λ2 − λ + π2αR2

2= 0, (3.8.32)

from which we obtain

λ1 = β

2 (β − 1)−[

β2

4 (β − 1)2 − π2αβ

β − 1R2

2

]1/2

. (3.8.33)

It follows that for sufficiently large values of R/ (short spring), there is no criticalload, i.e., no buckling can occur. We shall discuss this result in more detail later inthis section.

When the spring is clamped at both ends, ψ = 0 for x = 0 and x = , and fora symmetric buckling mode, there are no transverse shear forces, i.e., ζ= 0. Then(3.8.30) is satisfied by

ψ = sin2πx

(3.8.34)


and the bifurcation condition becomes(1 − 1

β

)λ2 − λ + 4π2αR2

2= 0, (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 incre-

ment of the potential energy in passing from the undeformed state to the adjacentstate is given by

PII − PI = 12

EA

∫0

{αR2ψ′2 + [(1 − λ + u′) cos ψ+ w′ sin ψ− 1]2

+β [− (1 − λ + u′) sin ψ+ w′ cos ψ]2}dx (3.8.36)

− 12

EA

∫0

λ2 dx + λEA

∫0

(−λ + u′) dx.

The equations for the equilibrium of forces are obtained by variations of this func-tional with respect to u′ and w′. Variation with respect to u′ yields

∫0

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

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

and because there are no restrictions on δu′, it follows that

[(1 − λ + u′) cos ψ+ w′ sin ψ− 1] cos ψ(3.8.38)

+β [− (1 − λ + u′) sin ψ+ w′ cos ψ] sin ψ+ λ = 0.

Variation with respect to w′ yields

∫0

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

(3.8.39)+β [− (1 − λ + u′) sin ψ+ w′ cos ψ] cos ψ+ c

}δw′dx = 0,

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(3.8.40)

+β [− (1 − λ + u′) sin ψ+ w′ cos ψ] cos ψ+ c = 0.

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 ψand (3.8.40) by cos ψand subtracting the resulting equa-tions, 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

PII − PI = 12

EA

∫0

[αR2ψ′2 + (λ cos ψ+ c sin ψ)

2

(3.8.43)+ 1

β(λ sin ψ− c cos ψ)2 + 2λu′

]dx,

where we have omitted the constant terms in (3.8.36) because they are unimpor-tant 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 ψcos ψ

+ λ sin2 ψ− c sin ψcos ψ = 0,

from which

u′ = −1 + cos ψ+ λ

(1 − 1

β

)sin2 ψ− c

(1 − 1

β

)sin ψcos ψ. (3.8.44)

The functional (3.8.43) can now be rewritten to yield

PII − PI = 12

EA

∫0

{αR2ψ′2 − 2λ (1 − cos ψ)

(3.8.45)+ λ2

[1 +

(1 − 1

β

)sin2 ψ

]+ c2

[sin2 ψ+ 1

βcos2 ψ

]}dx.

The second variation is now given by

P2 [ψ; λ] = 12

EA

∫0

[αR2ψ′2 − λψ2 + λ2

(1 − 1

β

)ψ2 + c2

β

]dx, (3.8.46)

where we have made use of the fact that c � 1. When there are no kinematic con-ditions, c = 0. In this case, the equation for neutral equilibrium is given by

αR2ψ′′ + λ

[1 − λ

(1 − 1

β

)]ψ = 0, (3.8.47)

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 thefourth degree terms, i.e., by

P4 [ψ ; λ] = 12

EA

∫0

[112

λψ4 − 13λ2(

1 − 1β

)ψ4 + c2

(1 − 1

β

)ψ2]

dx. (3.8.48)


For c = 0, this becomes

P4 [ψ ; λ] = 12

EA

∫0

112

[λ − 4λ2

(1 − 1

β

)]ψ4 dx. (3.8.49)

For a hinged-hinged spring, we have already discussed the buckling mode and thecritical load, cf. (3.8.31) to (3.8.33). The critical load λ1 given in (3.8.33) is real for

π2R2

2≤ β

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 signof

f (λ1) = λ1 − 4λ21

(1 − 1

β

). (3.8.51)

Using (3.8.32), we obtain

f (λ1) = λ1 − 4(

λ1 − π2αR2

2

)= 4π2 αR2

2− 3λ1. (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) = β

β − 1− 3β

2 (β − 1)= 3β

2 (1 − β)< 0, (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 = 4π2R2

32α. (3.8.55)

Substitution into the bifurcation condition (3.8.32) yields

β − 1β

(4π2R2

32α

)2

− 4π2R2

32α + π2 αR2

2= 0,

or

π2R2

2α = 3β

16 (β − 1)<

β

4 (β − 1), (3.8.56)

i.e., the equilibrium is unstable for

β

4 (β − 1)≤ λ1 ≤ 3β

4 (β − 1). (3.8.57)

We can now represent our results in the graph in Figure 3.8.5.

110 Applications

β / 2 β − 1( )

β / 4 β − 1( )

2πβ − 1

βα l / R

stable for the buckling mode

ψ = sinπ x

l

STABLE

UNSTABLE

UNSTABLE

λ1

1

Figure 3.8.5

For a spring with a circular cross section, no buckling will occur for

R< 2π

√β − 1

βα = 2π

√1 + 2ν

2 + ν. (3.8.58)

It is still an open question whether the unstable domain on the boundary for valuesof λ1 such that β/4 (β − 1) < λ1 < β/2 (β − 1) does actually exist, or that it stemsfrom 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 (seeFigure 3.9.1).

Let, x, y, z be axes fixed in space, and let y1, z1 be (co-rotating) principle axes ofinertia, which are rotated over an angle kx with respect to the fixed axes y, z. Herek is the torsion angle per unit length.

z1 z

y1

ykx

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 com-ponents 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, andthe 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 = −yv′0 (x) − zw′

0 (x) , (3.9.2)

where ()′ = d () /dx. To obtain an approximate uni-axial state of stress, we assumethe following displacement field,

v (x, y, z) = v0 (x) + 12νv′′

0(y2 − z2) + νw′′0yz

(3.9.3)w (x, y, z) = w0 (x) + νv′′yz + 1

2νw′′

0(z2 − y2).

This field satisfies∂v∂y

= ∂w∂z

= −ν∂u∂x

(3.9.4)ψyz = ∂v

∂z+ ∂w

∂y= 0.

However, there are non-vanishing angles of shear,

ψxy = ∂u∂y

+ ∂v∂x

= 12νv′′

0

(y2 − z2)+ νw′′

0yz. (3.9.5)

Let be the wavelength of the deformation pattern, and let b be a characteristiclength in the cross section. The contribution of ψxy in the energy function is then oforder b4v2

0/6, whereas the contributions due to ∂u/∂x are of order b2v2

0/4. 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

∫0

dx∫ ∫

A

12

E (yv′′0 + zw′′

0)2 dy dz. (3.9.6)

It is now convenient to employ the principle axes of inertia. Using the relations

y = y1 cos kx − z1 sin kx(3.9.7)

z = y1 sin kx + z1 cos kx,

we obtain for the energy density

∫0

[12

EIz1 (v′′0 cos kx + w′′

0 sin kx)2 + 12

EIy1 (w′′0 cos kx − v′′

0 sin kx)2]

dx dy. (3.9.8)

112 Applications

Because we are dealing with buckling, we cannot neglect the stresses in the funda-mental state. Recalling our general expression for the second variation in the caseof dead-weight loads,

P2 [u] =∫ [

12

Sij uh,iuh,j + 12

Eijklθij θk

]dV, (3.9.9)

where we have already evaluated the second term (3.9.8), we are still in need of anexplicit form of the first term. Because for torsion we have only two nonzero stresscomponents, we find

12

Sij uh,iuh,j = τxy (u,xu,y + v,xv,y + w,xw,y) + τxz (u,xu,z + v,xv,z + w,xw,z)

= τxy

{(−yv′′

0 − zw′′0) (−v′

0) +[

v′0 + 1

2νv′′′

0 (y2 − z2) + νw′′′0 yz]

× (νv′′0y + νw′′

0z) +[

w′0 + νv′′′

0 yz + 12νw′′′

0 (z2 − y2)]

(νv′′0z − νw′′

0y)}

(3.9.10)

+ τxz

{(−yv′′

0 − zw′′0) (−w′

0) +[

v′0 + 1

2νv′′′

0 (y2 − z2) + νw′′′0 yz]

× (−νv′′0z + νw′′

0 y) +[

w′0 + νv′′

0yz + 12νw′′′

0 (z2 − y2)]

(νv′′0 y + νw′′

0z)}

.

We shall now argue that the terms with third derivatives may be neglected. Considerthe term ∫ ∫

A

τxy × 12νv′′′

0 (y2 − z2) × νv′′′0 y dx dy = O

(τv2

0b5

5

). (3.9.11)

In our discussion of ψxy, we have already neglected terms of O(Ev2

0b6/6), and have

taken into account terms of O(Ev2

0b4/4). Hence, it follows that we can neglect the

contribution of (3.9.11) in the energy and, similarly, the other terms involving thirdderivatives with respect to x.

The expression (3.9.10) can now be rewritten to yield

12

Sij uh,iuh,j = v′0v′′

0 [yτxy + νyτxy − νzτxz] + w′0w′′

0 [−νyτxy + zτxz + νzτxz](3.9.12)

+ v′0w′′

0 [zτxy + νzτxy + νyτxz] + v′′0w′

0 [νzτxy + yτxz + νyτxz] .

Integrating this expression over the cross-sectional area, we must note that∫ ∫A

yτxy dy dz =∫ ∫

y∂F∂z

dy dz =∮

edge

yFnz ds = Fedge

∮ynz ds = 0, (3.9.13)

where Fedge is the constant value of F at the edge. Similarly, we have∫ ∫A

zτxz dy dz = 0. (3.9.14)

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


with v vanish because∫ ∫A

zτxy dy dz = −12

W∫ ∫

A

yτxz dy dz = 12

W. (3.9.15)

Further, we recall that the torque W is given by

W =∫ ∫

A

(yτxz − zτxy) dy dz, (3.9.16)

so the torsion energy per unit length is given by

12

W (v′′0w′

0 − v′0 w′′

0) . (3.9.17)

The second variation now becomes

P2 [u] =∫

0

[12

EIz1 (v′′0 cos kx + w′′

0 sin kx)2

(3.9.18)+ 1

2EIz1 (w′′

0 cos kx − v′′0 sin kx)2 + 1

2W (v0w′

0 − v′0w0)

]dx.

For a shaft with a circular cross section EIy1 = EIz1 = B, the functional reduces to

P2 [u] =∫

0

[12

B(v′′2

0 + w′′20

)+ 12

W (v0w′0 − v′

0w′′0)]

dx. (3.9.19)

Our further discussion will be restricted to this functional.The condition for neutral equilibrium is

P11 [u, ζ] =∫

0

[12

B(v′′2

0 + w′′20

)(3.9.20)

+ 12

W (v′′0 ζ ′ + w′

0 η′′ − v′0 ζ ′′ − w′′

0 η′)]

dx = 0.

By integration by parts, we obtain(Bv′′

0 + 12

Ww′0

)η′∣∣

0 +(

Bw′′0 − 1

2Wv′

0

)ζ ′∣∣

0

− (Bv′′′0 + Ww′′

0) η′∣∣0 − (Bw′′′

0 − Wv′′0) ζ∣∣0 (3.9.21)

+∫

0

[(Bv′′′0 + Ww′′′

0 ) η+ (Bv′′′′0 − Wv′′′

0 ) ζ] dx = 0,


Bv′′′′0 + Ww′′′

0 = 0, Bw′′′0 − Wv′′′

0 = 0. (3.9.22)

Let us now consider the case that the shaft is supported at its ends. It is nowconvenient 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 conditionsare then

v0 = w0 = 0 for x = ±. (3.9.23)

Modifying the boundaries in (3.9.21) correspondingly, we obtain the dynamicboundary conditions

Bv′′0 + 1

2Ww′

0 = 0, Bw′′0 − 1

2Wv′

0 = 0 for x = ±. (3.9.24)

To solve this problem, we set

v0 = Ceiµx, w0 = Deiµx, (3.9.25)

and substitution into the differential equations yields

µ4BC − iµ3WD = 0(3.9.26)

µ4BD + iµ3WC = 0.

The condition for a non-trivial solution is now

µ8 − µ6 W2

B2= 0, (3.9.27)

i.e.,

µ1, . . . , µ6 = 0, µ7 = WB

, µ8 = −WB

. (3.9.28)

Defining

WB

= µ,

we can write the solution in the form

v0 = C1 cos µx + C2 sin µx + C3x2 + C4x + C5(3.9.29)

w0 = C1 sin µx − C2 cos µx + C6x2 + C7x + C8

because the differential equations are satisfied identically by quadratic polynomials.Due to symmetry, we can split the solution into even and odd functions. Choosingan even function for v0, w0 must be odd due to (3.9.22). The problem with v0 oddand w0 even is readily found from the previous one by interchanging v0 and w0. Theadvantage of splitting up the problem is that we must now only satisfy four boundaryconditions. From (3.9.23), we obtain

v0 (x = ) = C1 cos µ + C32 + C5 = 0

(3.9.30)w0 (x = ) = C1 sin µ + C7

2 = 0

and from (3.9.24),

−12µ2 cos µ BC1 + 2C3B + 1

2WC7 = 0

(3.9.31)12µ2 sin µ BC1 − C3B = 0.


Because C5 only occurs in the first of the equations of (3.9.31), the condition for anon-trivial solution is ∣∣∣∣∣∣∣∣∣

sin µ 0

−12µ2 cos µ 2

12µ

−12µ2 sin µ −µ 0

∣∣∣∣∣∣∣∣∣= 0, (3.9.32)

from which follows

tan µ = −13µ. (3.9.33)

The smallest non-negative root of this transcendental equation is

µ = 1.566π

2, (3.9.34)

which yields

W = 1.566πB2

, (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) = −y1v′0() − zw′

0(), (3.9.36)

which implies that the shear stresses τxy cause a bending moment∫ ∫A

τxyu dy dx = Mz (3.9.37)

and, similarly for the shear stresses,∫ ∫A

τxzu dy dz = −My. (3.9.38)

This result means that the load on the end faces is not directed along the unde-formed axis. Bending moments My and Mz must be added. Evaluating the integrals

z, z1

y, y1

τ xz

τ xy

Figure 3.9.2

116 Applications

in (3.9.37) and (3.9.38), we find

My =∫ ∫

A

τxz(y1v1

0 () + zw′0 ()

)dy1 dz1 = 1

2Wv′

0 ()

(3.9.39)Mz = −

∫ ∫A

τxz (y1v′0 () + zw′

0 ()) dy1 dz1 = 12

Ww′0 () .

These moments are called by Ziegler† semi-tangential moments, i.e., the momentsare 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. Theloading is then non-conservative. Greenhill’s boundary conditions are obtainedfrom our previous problem by superimposing loads

�Mz = −12

Ww′0, �My = −1

2Wv′

0 (3.9.40)

for x = ±. The boundary conditions now become

Bw′′0 − Wv′

0 = 0, Bv′′0 + Ww′

0 = 0, x = ±. (3.9.41)

The differential equations are still given by (3.9.22), and the general solution is


w0 = C1 sin µx − C2 cos µx + C6x2 + C7x + C8,

where µ = W/B. By arguments similar to those used in the previous problem, wemay restrict ourselves to the underlined terms. The kinematic boundary conditionsv0 = w0 = 0 at x = require

C1 cos µ + C32 + C5 = 0

(3.9.43)C1 sin µ + C7 = 0,

and the dynamic boundary conditions become

−2C3W = 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

W = 2πB2

. (3.9.46)

The correctness of this result, obtained from the equations for neutral equilibriumfor a conservative system, is in this case verified by the solution of the dynamicproblem.

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


W

W

l

x

Figure 3.9.3

To show that this equivalent is not always justified, we consider the problemshown in Figure 3.9.3.

The boundary conditions in this case are

v0 = w0 = 0, v′0 = w′

0 = 0, at x = 0(3.9.47)

Bv′′0 + Ww′

0 = 0, Bw′′0 − Wv′

0 = 0

Bv′′′0 + Ww′′

0 = 0, Bw′′′0 − Wv′′

0 = 0

}at x = .

Integrating the differential equations (3.9.22) twice and making use of the boundaryconditions at x = , we find

Bv′′0 + Ww′

0 = 0, Bw′′0 − Wv′

0 = 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(3.9.50)

C2µ = 0, C1µ = 0.

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 whichwe consider a shaft without mass and a concentrated mass m at its end x = (seeFigure 3.9.4).

Because the shaft has no mass, the differential equations are unaltered. Theboundary conditions in this case are

v0 = w0 = v′0 = w′

0 = 0 at x = 0(3.9.51)

Bv′′0 + Ww′

0 = 0, Bw′′0 − Wv′

0 = 0

Bw′′′0 + Ww′′

0 = mv0, Bw′′′0 − Wv′′

0 = mw0

}at x = ,

xl

mW

W

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′′ = mB

V at x = . (3.9.55)

Introducing V = U(x)eiωt into the differential equation, we find

U = D1eiµx + D2x2 + D3x + D4. (3.9.56)

From the kinematic boundary conditions, we obtain

D1 + D4 = 0, iµD1 + D3 = 0. (3.9.57)

Hence, we can write

U = D1(eiµx − 1 − iµx

)+ D2x2. (3.9.58)

From the dynamic boundary conditions, we obtain

−µ2D1 + 2 (1 − iµ) D2 = 0(3.9.59)

−mω2

B

(eiµ − 1 − iµ

)D1 +

(−mω2

B2 + 2iµ

)D2 = 0.

The condition for a non-trivial solution is∣∣∣∣∣∣−µ2 2 (1 − iµ)

−mω2

B

(eiµ − 1 − iµ

) −mω2

B2 + 2iµ

∣∣∣∣∣∣ = 0, (3.9.60)

which yields

mω2

B[µ22 + 2(1 − iµ)(eiµ − 1 − iµ)] − 2iµ3 = 0, (3.9.61)

so that the square of the frequency is given by

ω2 = 2iµ3B/mµ22 + 2 (1 − iµ) (eiµ − 1 − iµ)

. (3.9.62)

It follows that ω2 is complex unless the denominator is purely imaginary, which isnot the case, as we shall show.

The fact that ω2 is complex implies that there is one root with a negative imag-inary part, which implies that V, and hence v0 and w0, increase exponentially withtime. 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 denomi-nator can now be written as

µ22 + 2 (1 − iµ)[

1 + iµ − 12µ22 − 1

6iµ33 + 1

24µ44 + O(µ55) − iµ − 1

]

= µ22[

1 − 1 + iµ − 13

iµ − 13µ22 + 1

12µ22 + O(µ33)

]

= µ33[

23

i − 14µ + O(µ22)

],

so that

ω2 = 3Bm3

[1 − 3

8iµ + O(µ22)

]. (3.9.63)

Although at first sight this result seems alarming, it is not important for actualconstructions because it is virtually impossible to apply a torque the way it isassumed 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 inthe 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 Fig-ure 3.10.1).

Because large rotations may occur, we shall first derive an expression for therotation vector for finite rotations.

Let n be a unit vector along the axis of rotation, and let α be the angle of rota-tion. 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(3.10.2)

= r cos α + (r · n) n (1 − cos α) + (n × r) sin α.

x = −l x = l

z

y

x

Cardan’s joint

W

Figure 3.10.1

120 Applications

e1

θ

e2

e3

n

α

r

r ′α

Figure 3.10.2

Without loss of generality, we may now restrict ourselves to values of α suchthat −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 yielda rotation vector that is the sum of the two corresponding rotation vectors. We nowconsider the rotation of the triad (e1, e2, e3) to obtain the rotation matrix. Let usdenote the rotated triad by

(e′

1, e′2, e′

3

). 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),

e′1 = e1 cos α + ω1

2 sin α/2ω

2 sin α/2(1 − cos α) + ω × e1

2 sin α/2sin α (3.10.4)

with

|ω|2 = ω2 = ω21 + ω2

2 + ω23 = 4 sin2 α/2

cos α = 1 − 2 sin2 α/2 = 1 − 12ω2

ω × e1 = (ω1e1 + ω2e2 + ω3e3) × e1= −ω2e3 + ω3e2

e2

1

e3

′e1

e

′e2

′e3

Figure 3.10.3


and

cos α/2 =√

12

(1 + cos α) =√

1 − 14ω2.

We can then rewrite this expression to yield

e′1 =

[1 − 1

2

(ω2

2 + ω23

)]e1 +

[12ω1ω2 + ω3

(1 − 1

4ω2)1/2

]e2

(3.10.5)

+[

12ω1ω3 − ω2

(1 − 1

4ω2)1/2

]e3.

By a cyclic interchanging of the subscript, we obtain

e′2 =

[12ω2ω1 − ω3

(1 − 1

4ω2)1/2

]e1 +

[1 − 1

2

(ω2

3 + ω21

)]e2

(3.10.6)

+[

12ω2ω3 + ω1

(1 − 1

4ω2)1/2

]e3.

e′3 =

[12ω3ω1 + ω2

(1 − 1

4ω2)1/2

]e1 +

[12ω3ω2 − ω1

(1 − 1

4ω2)1/2

]e2

(3.10.7)+[

1 − 12

(ω2

1 + ω22

)]e3.

We can now write our results in the form

e′i = Rij ej , (3.10.8)

where the rotation matrix Rij is given by

Rij = δij

(1 − 1

2ω2)

+ 12ωiωj + εijkωk

(1 − 1

4ω2)1/2

, (3.10.9)

where εijk is the alternating tensor. Up to now, all the expressions were exact. Forinfinitesimal 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 forthe second variation of the elastic energy. Taking into account quadratic terms, weobtain

Rij = δij

(1 − 1

2ω2)

+ 12ωiωj + εijkωk + O(ω3). (3.10.11)

Before we begin with the discussion of the buckling problem, let us make afew remarks. The Cardan Joint was invented by the physician and mathematicianG. Cardan in Italy in the 16th century, but in most English-speaking countries it isreferred to as Hooke’s joint. The construction is sketched in Figure 3.10.4.

122 Applications

mθ

A ′

x = −l

x = l

W

B

B ′

A

e1

e2

e3

Figure 3.10.4

The bars AA′ and BB′ are rigid and are rigidly connected perpendicular to eachother. 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 , respec-tively. These vectors are expressed in terms of the triad (e1, e2, e3), which is fixed inspace,

= − sin θ e2 + cos θ e3 (3.10.12)m = cos θ e2 + sin θ e3.

Let(e′

1, e′2, e′

3

)be connected to the deformable shaft; thus

′ = − sin θ e′2 + cos θ e′

3. (3.10.13)

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 e′2 + cos θ e′

3) · [cos (θ + ϕ) e2 + sin (θ + ϕ) e3] = 0, (3.10.15)

which yields

− sin θ cos (θ + ϕ)[

1 − 12

(ω2

3 + ω21

)]− sin θ sin (θ + ϕ)

[ω1

(1 − 1

4ω2)1/2

+ 12ω2ω3

]

+ cos θ cos (θ + ϕ)

[−ω1

(1 − 1

4ω2)1/2

+ 12ω3ω2

](3.10.16)


+ cos θ sin (θ + ϕ)[

1 − 12

(ω2

1 + ω22

)] = 0,

or

− cos ϕω1

(1 − 1

4ω2)1/2

+ 12ω2ω3 cos (2θ + ϕ)

(3.10.17)+ sin ϕ

(1 − 1

2ω2

1

)+ 1

2ω2

3 sin θ cos (θ − ϕ) − 12ω2

2 cos θ sin (θ + ϕ) = 0.

For known ω, the angle ϕ can be determined from this (exact) equation. Becausewe only need an approximate solution that is correct up to and including quadraticterms, we can simplify this equation to yield

− ω1 + 12ω 2ω 3 cos 2θ + ϕ + 1

2ω2

3 sin θ cos θ − 12ω2

2 cos θ sin θ = 0, (3.10.18)

from which follows

ϕ = ω1 − 12ω2ω3 cos 2θ − 1

4

(ω2

3 − ω22

)sin 2θ. (3.10.19)

The second variation is now obtained by adding to the functional (3.9.20) theterms due to the rotations at x = ±,

P2 [u] =∫

−

[12

B(v′′2

0 + w′′20

)− 12

W (v′0w′′

0 − v′′0w′

0)]

dx

+ 12

W[ω2ω3 cos 2θ + 1

2

(ω2

3 − ω22

)sin 2θ

]x=

(3.10.20)

− 12

W[ω2ω3 cos 2θ − 1

2

(ω2

3 − ω22

)sin 2θ

]x=−

.

With

ω2 = −w′0, ω3 = v′

0, (3.10.21)

we can rewrite this functional in the form

P2 [u] =∫

−

[12

B(v′′2

0 + w′′20

)− 12

W (v′0w′′

0 − v′′0w′

0)]

dx

+ 12

W[−v′

0w′0 cos 2θ + 1

2

(v′2

0 − w′20

)sin 2θ

]x=

(3.10.22)

+ 12

W[

v′0w′

0 cos 2θ + 12

(v′2

0 − w′20

)sin 2θ

]x=−

.

The condition for neutral equilibrium is

P11 [u, ζ] =(

Bv′′0 + 1

2Ww′′

0

)η′ ∣∣− +

(Bw′′

0 − 12

Wv′0

)ς′ ∣∣−

− (Bv′′′0 + Ww′′

0) η∣∣− − (Bw′′′

0 − Wv′′0) ς

∣∣−

+∫

−

[(Bv′′′′0 + Ww′′′

0 ) η+ (Bw′′′′0 − Wv′′′

0 ) ς] dx

124 Applications

+ 12

W [−w′0 cos 2θ + v′

0 sin 2θ] η′ ∣∣x= (3.10.23)

+ 12

W [−v′0 cos 2θ − w′

0 sin 2θ] ς′ ∣∣x=

+ 12

W [w′0 cos 2θ + v′

0 sin 2θ] η′ ∣∣x=−

+ 12

W [v′0 cos 2θ − w′

0 sin 2θ] ς′ ∣∣x=− = 0,

where η and ζ are arbitrary kinematically admissible displacement fields. Assumingthat x = ±, we have the kinematic conditions

v0 (±) = w0 (±) = 0. (3.10.24)

We arrive at the following dynamic boundary conditions at x = ,[Bv′′

0 + 12

Wv′0 sin 2θ + 1

2Ww′

0 (1 − cos 2θ)]

x=

= 0(3.10.25)[

Bw′′0 − 1

2Wv′

0 (1 + cos 2θ) − 12

Ww′0 sin 2θ

]x=

= 0,

and at x = −, we have[−Bv′′

0 + 12

Wv′0 sin 2θ − 1

2Ww′

0 (1 − cos 2θ)]

x=−

= 0[−Bw′′

0 + 12

Wv′0 (1 + cos 2θ) − 1

2Ww′

0 sin 2θ

]x=−

= 0.†

The differential equations are still given by

Bv′′′′0 + Ww′′′′

0 = 0, Bw′′′′0 − Wv′′′

0 = 0. (3.10.26)

As already discussed in our previous examples, the general solution to these equa-tions is


w0 = C1 sin µx − C2 cos µx + C6x2 + C7x + C8

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 mustonly consider the underlined terms in (3.10.27). Introduction of these expressionsinto the kinematic boundary conditions yields

C1 cos µ + C32 + C5 = 0

(3.10.28)C1 sin µ + C7 = 0.

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


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(3.10.29)

µ [(1 − cos 2θ) sin µ + sin 2θ cos µ] C1

+ 2 (1 + cos 2θ) C3 + sin 2θ C7 = 0.

Notice that due to symmetry and anti-symmetry, the boundary conditions atx = − are satisfied automatically. Because the constant C5 only appears in the firstof 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 kinematiccondition. 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 solu-

tion λ = π/2, which is half the value of Greenhill’s result. For θ = π/2, we haveλ = π (Greenhill’s value). Evaluating θ for given values of λ, we obtain the graphshown in Figure 3.10.5.

λ2

π

2

1

0

0 θ02

π1 2

θ2

π

arctan1

1+ υ

stable

v0

w0 symmetric

symmetric

This part of the curvecannot be reached

Figure 3.10.5

126 Applications

The case where w0 is symmetric and v0 is anti-symmetric is easily obtained by rotat-ing 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

θ − θ0 = W

St, (3.10.33)

which yields

θ0 = θ − W

St= θ − λ

BSt

. (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 thisline may be tangent to the curve. The part of the curve above this line then cannot bereached, which implies that the critical torque is a discontinuous function of θ. Fromthe practical point of view, these results are not as important because the criticalvalue 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 inthe 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 bendingin the y-direction will occur besides the bending in the z-direction. The correspond-ing additional displacements in passing from the fundamental state to the adjacentstate are (approximately) given by

u = −yv′0 (x) + ψ0 (y, z) α′ (x)

v = v0 (x) − α (x) (z − z0) + v∗ (x, y, z) (3.11.1)

w = α (x) (y − y0) + w∗ (x, y, z) ,

y0, z0( )

z

y

M*

Figure 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, andmay for our purpose be neglected. In (3.11.1), ψ0 (y, z) is the warping function withrespect to the shear center and α (x) is the torsion angle per unit length. We knowfrom our general theory that for dead-weight loads, the second variation is given by

P2 [u] =∫

V

[12

Sij uh,iuh,j + 12

Eijklθij θkl

]dV. (3.11.2)

The second term in (3.11.2) is readily obtained in the form∫V

12

Eijklθij θkl dV =∫

0

[12

EIzv′′20 + 1

2Stα

′2 + 12

E�α′′2]

dx, (3.11.3)

where � is the warping constant.For the evaluation of the energy in the fundamental state, we notice that

S11 = σx = MzIy

(3.11.4)

is the only non-vanishing stress component. The contribution to the second variationis ∫

V

12

S11(u2

1,1 + u22,1 + u2

3,1

)dV =

∫V

12σx(u2

,x + v2,x + w2

,x

)dV

=∫

12σx

{(−yv0 + ψ0α

′′)2 +[

v′0 − α′ (z − z0) + ∂v∗

∂x

]2

(3.11.5)

+[α′ (y − y0) + ∂w∗

∂x

]2}

dV.

First, notice that the first term between the brackets can be neglected because in it isσxu2

,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

12

Sij uh,iuh,j dV

=∫

dx∫∫

Mz2Iy

[{v′

0 − α′ (z − z0)}2 + α′2 (y − y0)2

]dy dz (3.11.6)

=∫

dx∫∫

Mz2Iy

[(v′

0 + α′z0)2 + α′2y20 − 2α′ (v′

0 + α′z0) z − 2α′2yy0 + α′2z2 + α′2y2]

dy dz.

The first two terms between the brackets vanish after integration because the originof our (y, z) axes is the center of gravity, and the fourth term vanishes after integra-tion because our axes are principle axes of inertia. Carrying out the integration, weobtain ∫

V

12

Sij uh,iuh,j dV

(3.11.7)

=∫

dx[−Mα′ (v′

0 + α′z0) + 12

Mα′2

Iy

∫∫z(y2 + z2) dy dz

].

128 Applications

Introducing the shorthand

z0 − 12Iy

∫∫z(y2 + z2) dy dz = c, (3.11.8)

for the second variation we finally obtain

P2 [u] =∫ [

12

EIzv′′20 + 1

2Stα

′2 + 12

E�α′′2 − Mα′v′0 − cMα′2

]dx. (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 con-

tributions of the shear stresses (according to Saint-Venant’s theory for the bendingof beams) to the elastic energy are∫

12

Sij uh,iuh,j dV =∫

[S12 (u1,1u1,2 + u2,1u2,2 + u3,1u3,2)

+ S13 (u1,1u1,3 + u2,1u2,3 + u3,1u3,3)] dV

=∫ {

τxy

[(−yv′′

0 + ψ0α′′)(

−v′0 + α′ ∂ψ0

∂y

)+ α′ (y − y0) α

]

+ τxz

[(−yv′′

0 + ψ0α′′) α′ ∂ψ0

∂z+ {v′

0 − α′ (z − z0)}

(−α)]}

dV

(3.11.10)

We now notice that∣∣ψ0 (y, z)

∣∣ = O(b2), where b is the height of the beam, so that

∣∣∣∣ψ0∂ψ0

∂yα′α′′

∣∣∣∣ = O(

b3α2

3

)(3.11.11)

and that

∣∣αα′(y − y0)∣∣ = O

(bα2

). (3.11.12)

Further, we have the estimates

∣∣yv′′0v′

0

∣∣ = O

(bv2

0

3

)= O

(b3α2

3

)∣∣∣∣α′yv′′

0∂ψ0

∂y

∣∣∣∣ = O(

b2v0α

3

)= O

(b3α2

3

)(3.11.13)

∣∣α′′ψ0v′0

∣∣ = O(

b2v0α

3

)= O

(b3α2

3

),

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 terms in the second line of (3.11.10), and our final resultbecomes∫

12

Sij uh,iuh,j dV

=∫

dx∫∫

[(yτxy + zτxz) αα′ − αα′τxyy0 + τxzα (v′0 + α′z0)] dy dz (3.11.14)

=∫

dx∫∫

αα′ (yτxy + zτxz) dy dz +∫

M′α (v′0 + 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

=∫∫ {

12

∂

∂y[(y2 + z2)τxy] + 1

2∂

∂z[(y2 + z2)τxz)

}dy dz

(3.11.16)− 1

2

∫∫(y2 + z2)

(∂τxy

∂y+ ∂τxz

∂z

)dy dz

= 12

∮edge

(y2 + z2) (τxyny + τxznz) ds + 12

∫∫(y2 + z2)

∂σx

∂xdy dz.

The line integral vanishes because τxy and τxz vanish at the edge, and using (3.11.4)our final result is∫∫

(yτxy + zτxz) dy dz = 12

M′

Iy

∫∫z(y2 + z2) dy dz

(3.11.17)= M′ (z0 − c) .

The second variation for a beam loaded by a transverse shear force (dead-weight load), applied in the center of shear-by-shear stresses according to Saint-Venant’s theory, is now given by

P2 [u] =∫

0

[12

EIzv′′20 + 1

2Stα

′2 + 12

E�α′′2 − Mα′v′0

− cMα′2 − M′α (v′0 + z0α

′) + M′αα′ (z0 − c)]

dx (3.11.18)

=∫

0

[12

EIzv′′20 + 1

2Stα

′2 + 12

E�α′′2 − (Mα)′ v′0 − c (Mα)′

α′]

dx.

This expression is also (approximately) valid for distributed loads, provided thatthey are distributed conforming to the shear stress distribution of Saint-Venant’stheory. When the transverse shear force is not applied in the center of shear, wemust add the contribution of this force to the energy (see Figure 3.11.2).

130 Applications

y0, z 0( )d

z F

y

*

Figure 3.11.2

When the force is applied in a point (y0, z0 + d), the contribution is

− 12

Fdα2. (3.11.19)

Because this contribution is negative, we have a destabilizing effect. When the forceis 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 appliedin the center of the shear. The necessary condition for neutral equilibrium is

P11 [u, ζ] =∫

0

[EIzv′′

0η′′ + Stα

′ϕ′ + E�α′′ϕ′′

(3.11.20)− (Mα)′

η′ − (Mϕ)′ v′0 − c (Mα)′

ϕ′ − c (Mϕ)′α′] dx = 0,

where ηand ϕ are kinematically admissible fields. By integration by parts, we obtain

EIzv′′0η

′∣∣0 − [EIzv′′′

0 + (Mα)′]η

∣∣∣0+ E�α′′ϕ′ ∣∣

0

+ [Stα′ − E�α′′′ − Mv′

0 − cMα′ − c (Mα)′]ϕ∣∣0 (3.11.21)

+∫

0

{[EIzv′′′′

0 + (Mα)′′]η+ [−Stα

′′ + E�α′′′′ + Mv′′0 + c (Mα)′′ + cMα′′]ϕ}dx = 0,

FF

FF

destabilizing

stabilizing

Figure 3.11.3



EIzv′′′′0 + (Mα)′′ = 0

(3.11.22)E�α′′′′ − Stα

′′ + Mv′′0 + c (Mα)′′ + cMα′′ = 0.

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

EIzv′′0 = 0, E�α′′ = 0. (3.11.24)

ii) Clamped end. Here we have only the kinematic conditions

v0 = 0, α = 0, v′0 = 0, α′ = 0. (3.11.25)

iii) Free end. In this case, we have only dynamic boundary conditions

EIzv′′0 = 0, EIzv′′′

0 + (Mα)′ = 0,(3.11.26)

E�α′′ = 0, Stα′ − E�α′′′ − Mv′

0 − cMα′ − c (Mα)′ = 0.

For narrow cross sections, |ψ0| = O (ht) (see Figure 3.11.4), and the warping con-stant � is of order |�| = O(h3t3), which means the warping is not important, so thatthe elementary theory (� = 0) may be used. This is also true for a T profile (seeFigure 3.11.5).

However, for other profiles we have different conditions (see Figure 3.11.6).

ht

Figure 3.11.4

*

center of shear

Figure 3.11.5

132 Applications

shear center

e.g.

Γ = ( ) = ( )O b t S O b tt5 3, .and

* * *

Figure 3.11.6

Γ = ( ) = ( )O b S O bt6 , .4

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 beamis reduced by one. When � = 0, the underlined terms in (3.11.23) to (3.11.26) mustbe omitted. In the following examples, we shall assume that � = 0.

Consider a simply supported beam loaded by a constant bending couple M. Thedifferential equations then are

EIzv′′′′0 + Mα′′ = 0, −Stα

′′ + Mv′′0 + 2Mcα′′ = 0. (3.11.27)

By choosing

α = A sinkπx

v0 = B sinkπx

, (3.11.28)

the boundary conditions (3.11.23) and (3.11.24) are satisfied identically. Substitutioninto the differential equations yield

k4π4

4EIzB − k2π2

2MA = 0

(3.11.29)

−k2π2

2MB + k2π2

2(St − 2Mc) A = 0.


z

t

h y

St =13

=1

12, ,IG y Iyt3h t3ht 3h =

112

1 + vMcr =

6�

Et3h

2 ( )π

1 + vv

� �cr =

hEt2

2 ( ) 5

8t2Eh

( = 0.28)σππ

≅

Figure 3.11.8. Example of slender rectangular section.

The condition for a non-trivial solution is

M2 = k2π2

3EIzSt

(1 − 2Mc

St

). (3.11.30)

When c = 0, we have

M = ±kπ

√EIzSt. (3.11.31)

In general, we have 2Mc/St � 1, and then

M ≈ ±kπ

√EIzSt

[1 ∓ kπc

√EIz

St

]. (3.11.32)

For example, if t/h = 0.1, h/ = 0.1, then σcr ≈ 0.002E, which for most construc-tion materials is still within the elastic range.

Mcr = π

24

Eht3√2 (1 + υ)

[1 − πh

5

√12

(1 + υ)

]

σcr = πt2

h2

h

E√2 (1 + υ)

[1 − πh

5

√12

(1 + υ)

]

≈ 5π

8E

t2

h2

h

for υ = 0.28 and h/ � 1.

Notice that this is (approximately) the same result as for the rectangular crosssection.

St =Eht

3

24 (1 )EIz0

St=

12

(1

z0 =215

h c =15

h,

EIz0 =148

Eht3 EIy =136

Eth3,

,+ v

)+ v

z

t

hy

Figure 3.11.9. Example of slender triangular section.

134 Applications

l

Fx

z

Figure 3.11.10

Let us now consider a beam clamped at one end and loaded by a trans-verse shear force (applied through the shear center) at the other end (see Fig-ure 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,

(3.11.33)F ( − x) v′′

0 − Stα′′ = 0.


v0 = 0, α = 0, v′0 = 0 at x = 0 (3.11.34)

EIzv′′0 = 0, EIzv′′′

0 + [F ( − x) α]′ = 0,(3.11.35)

Stα′ = 0 at x =

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

EIzv′′0 + F ( − x) α = 0. (3.11.37)

Eliminating v′′0 from (3.11.37) and the second of the equations (3.11.33), we obtain

an equation in α,

α′′ + F2( − x)2

EIzStα = 0 (3.11.38)

to be solved under the conditions

α = 0 at x = 0, α′ = 0 at x = . (3.11.39)

Let us now introduce a new variable η

η = − x

, (3.11.40)


the equation becomes then

d2α

dη2+ F24

EIzStη2α = 0, (3.11.41)

ord2α

dη2+ λη2α = 0, (3.11.42)

where λ is defined by

λ = F 22

EIzSt. (3.11.43)

The equation must be solved under the conditions

α = 0 at η = 1,dα

dη= 0 at η = 0. (3.11.44)

The solutions to (3.11.42) are Bessel functions, but for our purpose it is more con-venient to try a solution of the form

α =∞∑

n=0

Anηn. (3.11.45)

Substitution into the differential equation yields∞∑

n=0

An (n − 1) ηn−2 + λ

∞∑n=0

Anηn+2 = 0. (3.11.46)

The coefficient of ηk is

Ak+2 (k+ 2) (k+ 1) + λAk−2 = 0, (3.11.47)

so that

Ak+2 = − λ

(k+ 1) (k+ 2)Ak−2. (3.11.48)

The solution then becomes

α = A0

(1 − λ

3.4η4 + λ

3.4λ

7.8η8 − λ

3.4λ

7.8λ

11.12η12 + · · ·

). (3.11.49)

This solution satisfies the boundary condition at η = 0. The value of λ follows fromα = 0 for η = 1, i.e., the series between the brackets must vanish for η = 1, whichyields

λ = 16.104, (3.11.50)

so the critical load is

Fcr = 4.013√

EIzSt

2. (3.11.51)

For a beam with a rectangular cross section (t, h) the critical stress becomes

σcr = 4.013E√

2 (1 + υ)

t2

h2

h. (3.11.52)

As we have seen, the problem in this case can be reduced to a single second-orderdifferential 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 beintegrated twice to yield

EIzv′′0 + Mα = const. (3.11.53)

When the bending moment at x = 0 or x = vanishes, the constant is equal to zero.Solving v′′

0 from this equation (for simplicity, we shall assume that the constant isequal to zero), we find

v′′0 = − M

EIzα, (3.11.54)

and substitution into the second of the equations of (3.11.33) yields

α′′ + M2

EIzStα = 0. (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 ∫

0

[12

Stα′2 − M2

2EIzα2]

dx, (3.11.56)

as follows from the variation of the functional, which yields

Stα′δα∣∣0 −∫

0

(Stα

′′ + M2

EIzα

)δα dx = 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

α = ξ − 12ξ2, ξ = x/, (3.11.59)

which satisfies both boundary conditions. Introducing this approximation intothe functional (3.11.56) and setting the result equal to zero, we find when M =F ( − x) ,∫ 1

0

[12

St

2(1 − ξ)2 − F22

2EIz(1 − ξ)2

(ξ − 1

2ξ2)2] dξ = 0, (3.11.60)

which yields

F24

EIzSt= 105

6= 17.5, (3.11.61)

so that

Fcr = 4.18√

EIzSt

2, (3.11.62)

which gives an error of about 4% compared to the exact value.

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. Withoutgoing into details, we note that a plate theory can be derived under the followingassumptions:

i) Normals to the undeformed middle surface remain normal to the deformedmiddle 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 thanthe bending stiffness, which means that after buckling, the displacements perpendic-ular to the middle plane are considerably larger than the displacement in the middleplane.

We now recall that stability in the fundamental state is governed by the func-tional

P[u] =∫V

(12

Sij uh,iuh,j + 12

Eijklγij γk

)dV, (3.12.1)

where γ is the nonlinear strain tensor. For plates, the non-vanishing components ofthe strain tensor are

γ11(z) = u,x + 12

w2,x − zw,xx

γ22(z) = v,y + 12

w2,y − zw,yy (3.12.2)

2γ12(z) = u,y + v,x + w,xw,y − 2zw,xy,

where u(x, y), v(x, y) are the displacements in the mid-plane, w(x, y) is the dis-placement 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 energydensity can now be written as

12

Eijklγij γk = E2 (1 − υ2)

[γ2

11(z) + γ222(z) + 2υγ11(z)γ22(z) + 2(1 − υ)γ2

12(z)].

(3.12.3)Integrating (3.12.3) over the thickness of the plate, we obtain the energy per unitarea,

V = Eh2 (1 − υ2)

{(u,x + 1

2w2

,x

)2

+(

v,y + 12

w2,y

)2

+ 2υ

(u,x + 1

2w2

,x

)(v,y + 1

2w2

,y

)+ 1

2(1 − υ) (u,y + v,x + w,xw,y)2 (3.12.4)

+ h2

12

[w2

,xx + w2,yy + 2υw,xxw,yy + 2(1 − υ)w2

,xy

]},

where the last term represents the bending energy.

138 Applications

Assuming that in the fundamental state the only non-vanishing stress compo-nents are

S11 = Sx, S22 = Sy, S12 = Sxy,

the corresponding energy per unit area of the middle plane is[12

Sxw2,x + 1

2Syw2

,y + Sxyw,xw,y

]h. (3.12.5)

The energy functional for a plate loaded in its plane now becomes

P[u] = Eh12 (1 − υ2)

∫∫ {(u,x + 1

2w2

,x

)2

+(

v,y + 12

w2,y

)2

+ 2υ

(u,x + 1

2w2

,x

)(v,y + 1

2w2

,y

)+ 1

2(1 − υ) (u,y + v,x + w,xw,y)2

(3.12.6)

+ h2

12

[w2

,xx + w2,yy + 2υw,xxw,yy + 2(1 − υ)w2

,xy

] }dx dy

+ 12

h∫∫ [

Sxw2,x + Syw2

,y + 2Sxyw,xw,y]

dx dy.

We now assume that the fundamental state is linear, i.e., that the stresses are pro-portional to the loads. This assumption is valid when the rotations are of the sameorder of magnitude as the strains, which are small. This is the case for a property-supported plate (no rotations, as in a rigid body). Under these assumptions, we canconsider the x-, y-coordinates are the same as the coordinates in the undeformedstate.

Stability is primarily determined by the second variation, which is given by

P2[u] = Eh2 (1 − υ2)

∫∫ {u2

·x + v2,y + 2υu,xv,y + 2(1 − υ) (u,y + v,x)2

+ h2

12

[w2

,xx + w2,yy + 2υw,xxw,yy + 2(1 − υ)w2

,xy

](3.12.7)

+ 1 − υ2

E

(Sxw2

,x + Syw2,y + 2Sxyw,xw,y

)}dx dy.

We may now draw some important conclusions from the structure of this secondvariation:

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 linetogether, which only depend on w, become semi-positive-definite, and that thefirst 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 > 0for ∀a, b, c, |υ| < 1, and a, b, c not vanishing simultaneously.


ii) Because in (3.12.6) the third and fourth line together are semi-positive-definiteand the first two lines together are positive-definite, it follows that in the criticalstate 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 holdfor curved plates and shells.

Let us now derive the conditions for neutral equilibrium, which follow from

P11[u, ζ] = 0 = Eh2

12 (1 − υ2)

∫∫ {w,xxζ,xx + w,yyζ,yy

+υw,xxζ,yy + υw,yyζ,xx + 2 (1 − υ) w,xyζ,xy (3.12.9)

+ 12(1 − υ2)Eh2

[Sxw,xζ,x + Syw,yζ,y + Sxy (w,xζ,y + w,yζ,x)]}

dx dy.

Using the divergence theorem once, with υ as the unit normal vector to the edge,we obtain positive in the outward direction

Eh3

12 (1 − υ2)

∮edge

[w,xx ζ,xυx + w,yyζ,yυy + υw,xxζ,yυy

+ υw,yyζ,xυx + (1 − υ)w,xyζ,xυy + (1 − υ)w,xyζ,yυx] ds

+ Eh3

12 (1 − υ2)

∫∫ {− (w,xx + υw,yy)

,x ζ,x − (w,yy + υw,xx),y ζ,y (3.12.10)

−(1 − υ)w,xyyζ,x − (1 − υ)w,xyxζ,y

+ 12(1 − υ2

)Eh2

[Sxw,xζ,x + Syw,yζ,y + Sxy (w,xζ,y + w,yζ,x)]

}dx dy = 0.

Repeated application of the divergence theorem yields

Eh3

12 (1 − υ2)

∮edge

{[(w,xx + υw,yy) υx + (1 − υ)w,xyυy]

}ζ,x

+ [(w,yy + υw,xx) υy + (1 − υ)w,xyυx] ζ,y

−{[

(w,xx + υw,yy),x + (1 − υ)w,xyy

]υx

+[(w,yy + υw,xx)

,y + (1 − υ)w,xxy

]υy

}ζ (3.12.11)

+ 12(1 − υ2

)Eh2

[(Sxw,x + Sxyw,y) υx + (Syw,y + Sxyw,x) υy] ζ}

dx dy

+ Eh3

12 (1 − υ2)

∫∫ {��w − 12(1 − υ2)

Eh[Sxw,xx + Syw,yy + 2Sxyw,xy

+ (Sx,x + Sxy,y) w,x + (Sy,y + Sxy,x) w,y]}

ζdx dy = 0,

where � is the Laplacian operator. The differential equation for neutral equilibriumis now given by

��w − 12(1 − υ2

)Eh2

(Sxw,xx + Syw,yy + 2Sxyw,xy − Xw,x − Yw,y) = 0, (3.12.12)

140 Applications

where X and Y are the mass forces per unit area. Here we have used the equilibriumequations

Sx,x + Sxy,y + X = 0(3.12.13)

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

Using the relations

ζ,x = ζ,stx + ζ,υυx, ζ,y = ζ,sty + ζ,υυy, (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

∮edge

{[(w,xx + υw,yy) υ2

x + 2(1 − υ)w,xyυxυy + (w,yy + υw,xx) υ2y

]ζ,υ

+ [(w,xx + υw,yy) υxtx + (w,yy + υw,xx) υyty

+ (1 − υ)w,xy (υytx + υxty)] 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

− 12(1 − υ2

)Eh

[(Sxw,x + Sxyw,y) υx + (Syw,y + Sxyw,x) υy]}

ζ

}ds = 0.

Let us now consider a part of the edge curve (see Figure 3.12.1).It then follows that

ty = υx, tx = −υy. (3.12.16)

Further, we shall make use of the relation∫ s2

s1

F(s)ζ,sds = F(s)ζ∣∣∣∣s2

s1

−∫ s2

s1

F,sζds, (3.12.17)

which holds when F(s) is a differentiable function. This implies that the edge curvehas no corners in s1 < s < s2. In the following, we shall assume that the edge curveis a smooth curve. From (3.12.15), we now obtain[

(w,xx + υw,yy) υ2x + 2(1 − υ)w,xyυxυy + (w,yy + υw,xx) υ2

y

]ζ,v∣∣edge

= 0 (3.12.18)

y

x

t

Figure 3.12.1


{(1 − υ)

[(w,yy − w,xx) υxυy + (υ2

x − υ2y

)w,xy

],s

+ (w,xx + w,yy),x υx + (w,xx + w,yy)

,y υy (3.12.19)

− 12(1 − υ2

)Eh

[(Sxw,x + Sxyw,y) υx + (Syw,y + Sxyw,x) υy]}

ζ

∣∣∣∣edge

= 0

(1 − υ)[(w,yy − w,xx) υxυy + (υ2

x − υ2y

)w,xy

]ζ∣∣s2

s1= 0. (3.12.20)

First, we notice that the left-hand side of (3.12.20) vanishes identically along a closedcurve. 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 theedge curve adjacent to the free part is properly supported.

In the following discussion, we shall assume that (3.12.20) is satisfied. Let usnow consider the various boundary conditions.

i) The free edge. Because there are no restrictions imposed on ζand ζ,υ, their coef-ficients must vanish, i.e.,

(w,xx + υw,yy) υ2x + 2(1 − υ)w,xyυxυy + (w,yy + υw,xx) υ2

y = 0, (3.12.21)

which means that the bending moment along the edge vanishes, and

(1 − υ)[(w,yy − w,xx) υxυy + (υ2

x − υ2y

)w,xy

],s

+ (w,xx − w,yy),x υx + (w,xx + w,yy)

,y υy (3.12.22)

− 12(1 − υ2)Eh

[(Sxw,x + Sxyw,y) υx + (Syw,y + Sxyw,x) υy] = 0,

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,xyυxυy + (w,yy + υw,xx)υ2

y = 0 (3.12.23)

in addition to the kinematic condition w = 0.iii) The clamped edge. Here we have the kinematic conditions

w = ∂w∂υ

= 0 at the edge (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

σσ

x = a

x

Figure 3.12.2

In this case, the only non-vanishing stress component in the fundamental stateis Sx = −σ. Further, there are no mass forces, so the differential equation for neutralequilibrium is given by

Eh3

12(1 − υ2)��w + hσw,xx = 0. (3.12.25)

In this case, we have

υx = 1 υy = 0 at x = aυx = −1 υy = 0 at x = 0υy = 1 υx = 0 at y = bυy = −1 υx = 0 at 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)

Similarly,

w,yy = 0 w = 0 at y = 0, y = b. (3.12.28)

Because both the differential equation and the boundary conditions containonly even derivatives, we can attempt a solution of the form

w(x, y) = Cmn sinmπx

asin

mπyb

, (3.12.29)

which satisfies all the boundary conditions.† Substitution into the differential equa-tion yields

Eh3

12(1 − υ2)

(m2π2

a2+ n2π2

b2

)2

− hσm2π2

a2= 0. (3.12.30)

† Here m, n are integers.


For given values of m and n, (3.12.20) is satisfied for

σ = π2Eh2

12(1 − υ2)b2

(m2 b

a+ n2 a

b

)2 1m2

. (3.12.31)

The critical load is obtained for the minimum value of the right-hand term, whichfor fixed m has a minimum value for n = 1, so that we must minimize the expression

σ = π2Eh2

12(1 − υ2)b2

(m2 b2

a2+ 2 + a2

b2m2

)(3.12.32)

with respect to m. To this end, we assume that m is continuous. Differentiating theexpression between the brackets with respect to m2 and setting the result equal tozero, we find

m = ab. (3.12.33)

This is the exact result when a/b is an integer. In other cases, m is the integer that isclosest to a/b. Hence, it follows that

σcr ≥ π2Eh2

3(1 − υ2)b2(3.12.34)

where the equality sign holds when a/b is an integer. For sufficiently large values ofa/b, (3.12.34) is a good approximation,† which means that for a sufficiently long platewe can always approximate a/b by an integer. The critical load for other boundaryconditions is often expressed as

σcr = π2Eh2

3(1 − υ2)k, (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

σcr = π2Eh2

12(1 − υ2)a2

(1 + 2

a2

b2+ a4

b4

). (3.12.36)

In the limiting case a/b → 0, we have

σcr = π2Eh2

12(1 − υ2)a2, (3.12.37)

which also follows from the result for the Euler bar when we take into account thatthere 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. Wehave 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 4.095when 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 sinmπx

asin

nπyb

(3.12.38)

can be differentiated term-wise, which is only admissible when the differentiatedseries is uniformly convergent for 0 ≤ x ≤ a, 0 ≤ y ≤ b. We shall now show that fora plate with simply supported edges, this series can be differentiated term-wise. Todo this, we apply a Fourier transform to (3.12.25), which yields∫ a

0

∫ b

0

[Eh3

12(1 − υ2)��w + hσw,xx

]sin

mπxa

sinnπy

bdx dy = 0. (3.12.39)

Let us first consider the term∫ a

0w,xxxx sin

mπxa

dx = w,xxx sinmπx

a

∣∣∣a0−∫ a

0w,xxx

mπ

acos

mπxa

dx(3.12.40)

= −w,xxmπ

acos

mπxa

∣∣∣a0−∫ a

0w,xx

m2π2

a2sin

mπxa

dx.

Here the stock term vanishes because w,xx = 0 at x = a and x = 0. This stock termdoes not vanish in the case of a clamped edge. Continuing our partial integration,we obtain∫ a

0w,xxxx sin

mπxa

dx = −w,xm2π2

a2sin

mπxa

∣∣∣∣a

0

+∫ a

0w,x

m3π3

a3cos

mπxa

dx (3.12.41)

= wm3π3

a3cos

mπxa

∣∣∣∣a

0+ m4π4

a4

∫ a

0w sin

mπxa

dx.

Here the stock term vanishes for both simply supported edges and clamped edgesbecause then w = 0 at x = 0 and x = a. Similarly, we obtain for simply supportededges ∫ a

0w,xx sin

mπxa

dx = −m2π2

a2

∫ a

0w sin

mπxa

dx(3.12.42)∫ b

0w,yyyy sin

nπya

dy = n4π4

b4

∫ b

0w sin

nπya

dy

∫ a

0

∫ b

0w,xxw,yy sin

mπxa

sinnπy

adx dy

= m2n2π4

a2b2

∫ a

0

∫ b

0w sin

mπxa

sinnπy

bdx dy.

Our final result for the Fourier transform of the differential equation can now bewritten as [

Eh3

12(1 − υ2)

(m2π2

a2+ n2π2

b2

)2

− hσm2π2

a2

]Wmn = 0 (3.12.43)


where

Wmn =∫ a

0

∫ b

0w(x, y) sin

mπxa

sinmπy

bdx dy (3.12.44)

is the Fourier transform of w(x, y). The equation (3.12.43) is valid for all integervalues of m and n, and because Wmn �≡ 0 the expression between the brackets mustvanish, which leads to our earlier result (3.12.31).

Introducing the series (3.12.38) into (3.12.44), we find the following relationbetween Cmn and the Fourier transform of w(x, y):

Wmn = 14

Cmnab. (3.12.45)

To obtain this result, we have interchanged integration and summation, which isadmissible because the series (3.12.38) is a uniformly convergent series. We see thatonly 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

Wm(y) =∫ a

0w(x, y) sin

mπ

adx, (3.12.46)

we find

Eh3

12(1 − υ2)

(d4Wm

dy4− 2

m2π2

a2

d2Wm

dy2+ m4π4

a4Wm

)− hσ

m2π2

a2Wm = 0, (3.12.47)

where we have used the relation∫ a

0

dnwdyn

sinmπx

adx = dnWm

dyn. (3.12.48)

As discussed previously for a short plate (a/b � 1), the wave number is m = 1.Introducing the notation

y = aπ

η, d () /dη = ()·, W1 = W, (3.12.49)

y

σ

x

σ

free edge

s⋅ss⋅s

s⋅s

Figure 3.12.3

146 Applications

we may rewrite (3.12.47) in the form

π2Eh2

12(1 − υ2)a2(W···· − 2W·· + W) − σW = 0. (3.12.50)

Further, it is convenient to write

σ = λπ2Eh2

12(1 − υ2)a2(3.12.51)

so that the differential equation becomes

W···· − 2W·· + W(1 − λ) = 0. (3.12.52)


w,yy + υw,xx = 0, w,yyy + (2 − υ)w,xyy = 0 at y = 0(3.12.53)

w = 0, w,yy = 0 at y = b,

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 equa-tion for the Fourier transform of w(x, y). Writing

w(x, y) = Y(y) sinπxa

, (3.12.55)


W(y) = a2

Y(y). (3.12.56)

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)

W = 0, W·· − υW = 0 at η = πba.

The boundary conditions at x = 0, x = a are satisfied by our choice (3.12.55). Tosolve (3.12.52), we set

W = Ceµη. (3.12.58)

Introducing this expression into the differential equation, we obtain the character-istic equation

µ4 − 2µ2 + 1 − λ = 0, (3.12.59)

from which

µ = ±(1 ±√

λ)1/2

. (3.12.60)

These roots are real when 0 ≤ λ ≤ 1. Because we want solutions that decay withincreasing distance from y = 0, we must only consider the solution

W = C1e−√

1+√λη + C2e−

√1−√

λη. (3.12.61)


Substitution into the boundary conditions at η = 0 yields(1 +

√λ − υ

)C1 + (1 −

√λ − υ

)C2 = 0

(3.12.62)C1(1 +

√λ)1/2(1 − υ −

√λ)+ C2

(1 −

√λ)1/2(1 − υ +

√λ) = 0.

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

d√

λ= 2(1 −

√λ)1/2(1 − υ +

√λ) − 1

2(1 −

√λ)−1/2(1 − υ +

√λ)2

(3.12.64)+ 2(1 +

√λ)1/2(1 − υ −

√λ) − 1

2(1 +

√λ)−1/2(1 − υ −

√λ)2.

At√

λ = 1, df /d√

λ = −∞, and for√

λ = 0 we have df /d√

λ = (1 − υ)(3 + υ), sothat 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)


ε(2 − υ − ε2)2 − (2 + ε2)1/2(ε2 − υ)2,

from which

ε =√

2υ2

(2 − υ)2

[1 + O(ε2)

], (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 = 7781

= 0.9506. (3.12.68)

f λ,υ( )

υ = 0

υ > 0

λ1

Figure 3.12.4

148 Applications

x = aσ

y

σ

− b / 2

b / 2

x

Figure 3.12.5

To a second-order approximation, ε is given by

ε =√

2(2 − υ)2

υ(8 − υ)

[1 + 2υ3(8 − υ)

(2 − υ)4

]1/2

, (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 thatis simply supported at y = 0.

Let us now consider a rectangular plate with clamped edges at y = ±b/2, andsimply supported edges at x = 0 and x = a, loaded in compression by forces σh perunit 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 con-sider the case that the plate is sufficiently long, and that 2 is the wavelength ofthe deformation pattern in the buckled state. The differential equation for neutralequilibrium is

Eh3

12(1 − υ2)��w + σhw,xx = 0, (3.12.71)

and the boundary conditions are

w = 0, w,xx = 0 at x = 0, x = m(≤ a)(3.12.72)

w = 0, w,y = 0 at y = ± b/2.

Because the lowest bifurcation load will occur when m = 1, we try a solution of theform

w(x, y) = W(y) sinπx

. (3.12.73)

Substitution into the differential equation yields

Eh3

12(1 − υ2)

(d4Wdy4

− 2π2

2

d2Wdy2

+ π4

4W)

− σhπ2

2W = 0. (3.12.74)


It is now convenient to write

σ = λπ2Eh2

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

d4Wdy4

− 2π2

2

d2Wdy2

+(

π4

4− 4π4λ

2b2

)W = 0. (3.12.76)

The boundary conditions at x = 0 and x = are satisfied automatically by our choicefor w(x, y), and the conditions at y = ± b/2 read

W = 0,dWdy

= 0 at y = ± b/2. (3.12.77)

To solve (3.12.76), we set

W = Ceπy

µ. (3.12.78)

Introduction of this expression into the differential equation yields the characteristicequation

µ4 − 2µ2 + 1 − 4λ2

b2= 0, (3.12.79)

from which

µ = ±√

1 ± 2

b

√λ. (3.12.80)

As the plate with two clamped edges has a larger stiffness than the simply supportedplate, λ > 1. This means that for 2

√λ/b > 1, there will be two imaginary roots.

Hence, we write

µ1,2 = ±α α =(

1 + 2

b

√λ

)1/2

(3.12.81)

µ3,4 = ± iβ β =(

2

b

√λ − 1

)1/2

.

Because the construction and its loading are symmetric with respect to the x-axis, we can split the solution into a part that is symmetric with respect to y = 0and an anti-symmetric part. Guided by the result for the simply supported plate, weexpect the lowest bifurcation load for the symmetric solution; thus we consider thesolution

W(y) = C1 cosh απy/ + C2 cos βπy/. (3.12.82)

Substitution of the boundary conditions (3.12.77) yields

C1 cosh απb2

+ C2 cos βπb2

= 0(3.12.83)

αC1 sinh απb2

− βC2 sin βπb2

= 0.

150 Applications

The condition for a non-trivial solution is

−β cosh απb2

sin βπb2

− α sinh απb2

cos βπb2

= 0,

or rewritten,

β tan βπb2

+ α tanh απb2

= 0. (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 tran-scendental equation (3.12.83), we now try an approximate solution to our problem.Introducing a dimensionless coordinate η defined by

y = ηb, −12

≤ η ≤ 12, (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)

∫∫ [(�w)2 − 2 (1 − υ)

(w,xxw,yy − w2

,xy

)− 12(1 − υ2)h2

σ

Ew2

,x

]dx dy.

(3.12.88)

The second variation is semi-positive definite for the exact solution. For the exactbuckling load and an assumed (approximate) displacement, the field P2[u] > 0, sothat the application of Rayleigh’s method (also called Rayleigh-Ritz method) wherewe put P2[u] = 0 for an assumed displacement field yields an upper bound for thecritical load.

For the evaluation of (3.12.88), it will be convenient to make use of the followingproperty: ∫∫

S

[w,xxw,yy − w2

,xy

]dx dy = 0 (3.12.89)

for a domain bounded by straight lines (polygonal edge curve) with w = const. atthe boundary ∂S, and ∂w/∂υ = 0 in the corner points at the boundary where υ is theunit 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.


the divergence theorem to (3.12.89), which yields

∫∫S

(w,xxw,yy − w2

,xy

)dx dy = 0

=∫∂S

(w,xxw,yυy − w,xyw,yυx) ds −∫∫

S

(w,xxyw,y − w,xyxw,y)dx dy

=∫∂S

−w,y (w,xxtx + w,xyty) ds (3.12.90)

=∫∂S

−w,y (w,xxdx + w,xydy) = −∫∂S

w,yd (w,x)

= −∫∂S

(w,υυy + w,sty) d (w,υυx + w,stx) .

So far, our result is fully general. When w is constant along ∂S, we have w,s = 0along ∂S, and along a straight line υx and υy are constant, so that for a polygonaledge curve where w = const., we can write

∫∫S

(w,xxw,yy − w2,xy)dx dy = −1

2

n∑i=1

(υxυyw2

,υ

)∣∣∣∣∣si+1

si

(3.12.91)

for a polygonal edge curve with n corner points, where si denotes the ith corner pointand 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 cornerpoints on ∂S, where ∂S is a polygonal curve.

For simply supported and clamped straight edges, we have w = 0, ∂w/∂υ = 0 inthe corner points, and thus the result (3.12.89) applies,† so that for these plates wecan write

P2 [u1] = Eh3

24 (1 − υ2)

∫∫ [(�w)2 − 12

(1 − υ2

)h2

σ

Ew2

,x

]dx dy. (3.12.92)

Using the results

∫0

12 b∫

− 12 b

(�w)2 dx dy =∫

0

12∫

− 12

b[−π2

2

(1 − 8η2 + 16η4)+ 1

b2

(−16 + 108η2)]2

sin2 πx

dxdη

(3.12.93)= 128

b3

(45

+ 64105

C + 128315

C2)

,

† Notice that the limiting case that n → ∞ for a simply supported plate bounded by a regular poly-gonal 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 = π2b2

162, (3.12.94)

and

∫0

12 b∫

− 12 b

w2,x dx dy =

∫0

12∫

12

bπ2

2cos2 πx

(1 − 8η2 + 16η4)2 dx dη

= 64315

16Cb

, (3.12.95)

we obtain from the condition P2[u1] = 0,

σ = Eh2

12(1 − υ2)b2

128(

45

+ 64105

C + 128315

C2)

64315

· 16C. (3.12.96)

Minimizing with respect to C, we obtain

Cmin =√

6332

= 1.403 (3.12.97)

and from (3.12.94),

(/b)min = 0.663. (3.12.98)

The critical load is now

σcr = π2Eh2

3(1 − υ2)b21.7454, (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 (seeFigure 3.12.6).

From (3.12.12), the governing differential equation is now

Eh3

12(1 − υ2)��w − 2τhw,xy = 0. (3.12.100)

τ

τ

x = aτ

τ

y = b

y

x

Figure 3.12.6


y

x

Figure 3.12.7

Notice that the differential equation now contains even and odd derivatives withrespect 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 = 5.35π2Eh2

12(1 − υ2)b2. (3.12.102)

However, the numerical factor is not very accurate. Later, we shall give an upperbound that is slightly smaller.

The exact solution of this problem requires rather tedious calculations, andtherefore we shall try to construct an approximate solution. We try a solution ofthe form

w(x, y) = f sinπ

(x − my) sin

πyb

, (3.12.103)

which satisfies the boundary condition w = 0 at y = 0, y = b. The deflection vanishesat the nodal lines x − my = k (k = 0, ±1, ±2, . . . ).

y

x = l x = 2l

x

Figure 3.12.8

† Cf. R.V. Southwell, Proc. Roy. Soc., London, Series A, 105, 582.

154 Applications

The energy functional is

∫0

b∫0

{Eh3

12(1 − υ2)

[(�w)2 − 2(1 − υ)

(w,xxw,yy − w2

,xy

)]+ τhw,xw,y

}dx dy,

(3.12.104)where the second term vanishes because of (3.12.87). Applying the Rayleigh-Ritzmethod, we obtain

τ = π2Eh2

24(1 − υ2)b2

[1m

(m2 + 1

)2 b2

2+ 1

m2

b2+ 6m + 2

m

], (3.12.105)

where m and b/ are still unknown parameters. Minimizing with respect to (b/2),we obtain

1m

(m2 + 1)2 − 1m

4

b4= 0, (3.12.106)

which yields

2

b2= m2 + 1. (3.12.107)


τ = π2Eh2

24(1 − υ2)b2

(8m + 4

m

). (3.12.108)

Minimizing with respect to m, we obtain

m = 12

√2, (3.12.109)

so that the (upper bound for the) critical load is given by

τcr = π2Eh3

3(1 − υ2)b2

√2, (3.12.110)

which means a factor 1.414 instead of 1.3337 (Southwell). This result is a rathercrude approximation, and is due to the fact that the boundary condition w,yy = 0 isnot satisfied at y = 0, y = b.

Let us now analyze the consequences of our assumption for the displacementfield shown in Figure 3.12.9.

y

w = 0

w = 0

xA

Figure 3.12.9


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) thatis slightly more general than (3.12.102):

w(x, y) = sinπ

(x − ϕ(y))W(y), W(0) = W(b) = 0. (3.12.111)

We then obtain

w,y = −π

W(y)ϕ′(y) cos

π

[x − ϕ(y)] + W′(y) sin

π

[x − ϕ(y)] (3.12.112)

w,yy =[

W(y) − π2

2W(y)ϕ′2(y)

]sin [x − ϕ(y)]

(3.12.113)− [2W′(y)ϕ′(y) + W(y)ϕ′′(y)] cos

π

[x − ϕ(y)]

w,yy(x, 0) = W′′(0) sin[x − ϕ(0)] − 2π

W′(0)ϕ′(0) cos

π

[x − ϕ(0)]. (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 condi-tions are satisfied by choosing

w(x, y) = f sinπ

(x − ϕ(y)) sin

πyb

, (3.12.116)

where

ϕ(y) = mbπ

(1 − cos

πyb

). (3.12.117)

With this choice, the result is

τcr = 1.352π2Eh2

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) =∞∑

k=1

Wk(x) sinkπy

b, (3.12.119)

Figure 3.12.10

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) =∞∑

k=1

(ak cos

πx

+ bk sinπx

)sin

kπyb

. (3.12.121)

Substitution of this expression into (3.12.100) yields

Eh3

12(1 − υ2)

∞∑k=1

(π2

2+ k2π2

b2

)2 (ak cos

πx

+ bk sinπx

)sin

πy

(3.12.122)

− 2τh∞∑

k=1

kπ2

b

(−ak sin

πx

+ bk cosπx

)cos

kπyb

= 0,

where we have assumed that the interchanging of summation and differentiation isadmissible.

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

Eh2π2

12(1 − υ2)b2

∞∑k=1

(µ2 + k2)2ak sinkπy

b− 2τµ

∞∑k=1

kbk coskπy

b= 0

(3.12.123)Eh2π2

12(1 − υ2)b2

∞∑k=1

(µ2 + k2)2bk sinkπy

b+ 2τµ

∞∑k=1

kak coskπy

b= 0.

Introducing the notation

τ = λπ2Eh2

3(1 − υ2)b2, (3.12.124)

we can rewrite these equations in the form

∞∑k=1

(µ2 + k2)2ak sinkπy

b− 8µλ

∞∑k=1

kbk coskπy

b= 0

(3.12.125)∞∑k=1

(µ2 + k2)2bk sinkπy

b+ 8µλ

∞∑k=1

kak coskπy

b= 0.

Assuming that the left-hand members can be written as a sine series, we multiplyboth sides of these equations by sin j πy/b and integrate (term-wise) from y = 0 toy = b. With

b∫0

sinj πyb

sinkπy

bdy = 1

2bδjk (3.12.126)


b∫0

sinj πyb

coskπy

bdy = j

[1 − (−1)j +k]( j 2 − k2) π

b (= 0 for j = k) , (3.12.127)

we obtain

(µ2 + j 2)2aj − 32π

λµ

∞∑k=1

kj2 ( j 2 − k2)

[1 − (−1)k+j ]bk = 0

(3.12.128)

(µ2 + j 2)bj + 32π

λµ

∞∑k=1

kj2 ( j 2 − k2)

[1 − (−1)k+j ]ak = 0.

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 thematrix of coefficients vanishes, i.e., with

32π

λµ = ρ (3.12.129)

a1 b2 a3 b4 · · ·∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(µ2 + 1)2 2ρ

30

4ρ

15· · ·

2ρ

3(µ2 + 1)2 6ρ

50 · · ·

0 −6ρ

5(µ2 + 9)2 12ρ

7· · ·

4ρ

150

12ρ

7(µ2 + 1)2 · · ·

......

......

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

. (3.12.130)

Taking into account two terms, we obtain

(µ2 + 1)2(µ2 + 4)2 − 4ρ2

9= 0, (3.12.131)

which yields

(µ2 + 1)(µ2 + 4) = 2ρ

3= 64µλ

3π,

so that

λ = 3π

64

(µ3 + 5µ + 4

µ

). (3.12.132)

Minimizing with respect to µ yields

µ4 + 53µ2 = 4

3= 0,

so that

µ2 = −5 + √73

b≈ 0.59 (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)

and with four terms,

λcr = 1.336, (3.12.136)

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

P [u] =∫∫ {

Eh3

12(1 − υ2)

[(�w)2 − 2(1 − υ)

(w,xxw,yy − w2

,xy

)]

+ Eh2(1 − υ2)

[(u,x + 1

2w2

,x

)2

+(

v,y + 12

w2,y

)2

(3.13.1)+ 2υ

(u,x + 1

2w2

,x

)(v,y + 1

2w2

,y

)

+ 12

(1 − υ)(u,y + v,x + w,xw,y)2]

− 12σhw2

,x

}dx dy.

The second variation is given by

P[u] =∫∫ {

Eh3

12(1 − υ2)[(�w)2 − 2(1 − υ)

(w,xxw,yy − w2

,xy

)X]

(3.13.2)+ Eh

2(1 − υ2)

[u2

,x + v2,y + 2υu,xv,y + 1

2(1 − υ)(u,y + v,x)2

]− 1

2σhw2

,x

}dx dy,

and the third and fourth variations are given by

P3[u] = Eh2(1 − υ2)

∫∫ [u,xw2

,x + v,vw2,y + υ(u,xw2

,y + W2,x)

(3.13.3)+ (1 − υ)(u,y + v,x)w,xw,y

]dx dy

and

P4[u] = Eh8(1 − υ2)

∫∫ (w2

,x + w2·y)2

dx dy, (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 byminimizing the functional

P2[u] + a(λ − 1)P′11[u1, u] + a2P21[u1, u], (3.13.6)

where λ = σ/σcr (see (2.4.11). The term

P′11[u1, u] = − Eh2

4(1 − υ2)

∫∫w1,xw,x dx dy (3.13.7)

vanishes due to the orthogonality condition

P21 [u1, u] = Eh2(1 − υ2)

∫∫[(u,x + υv,y) w2

1,x + (v,y + υu,x) w21,y

(3.13.8)+ (1 − υ) (u,y + v,x) w1,xw1,y] dx dy,

and P2 [u] follows from (3.13.2) by replacing u, v, w by u, v, w. We now must min-imize (3.13.6) with respect to u, i.e.,

Minw.r.t.u

P2 [u] + a2P21 [u1, 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 thefirst line in (3.13.2), which is positive-definite, vanishes, which yields

w ≡ 0. (3.13.10)

P2 [u] is now given by

P2 [u] = Eh2 (1 − υ2)

∫∫ [u2

,x + v2,y + 2υu,xv,y + 1

2(1 − υ) (u,y + v,x)2

]dx dy.

(3.13.11)The necessary condition for (3.13.9) to be a minimum is that the variation of thisfunctional with respect to u vanishes, which yields

Eh1 − υ2

∫0

12 b∫

− 12 b

{u,x ξ,x + v,yη,y + υu,xη,y + υv,yξ,x + 1

2(1 − υ) (u,y + v,x) (ξ,y + η,x)

(3.13.12)

+ a2

2

[(ξ,x + υη,y) w2

1,x + (η,y + υξ,x)w21,y + (1 − υ) (ξ,y + η,x) w1,xw1,y

]}dx dy = 0.


b/2∫−b/2

[u,x + υv,y + a2

2

(w2

1,x + υw21,y

)]ξ

∣∣∣∣∣∣∣

0

dy +∫

0

[v,y + υu,x + a2

2

(w2

1,y + υw21,x

)]η

∣∣∣∣∣∣b/2

−b/2

dx

+ (1 − υ)2

∫0

(u,y + v,x + a2w1,xw1,y

)ξ

∣∣∣∣∣∣b/2

−b/2

dx + (1 − υ)2

b/2∫−b/2

(u,y + v,x + a2w1,xw1,y

)η

∣∣∣∣∣∣∣

0

dy

160 Applications

−∫

0

b/2∫−b/2

{u,xx + 1

2(1 − υ) u,yy + 1

2(1 + υ) v,xy

(3.13.13)

+ a2

2

[(w2

1,x + w21,y

),x

+ (1 − υ) (w1,xw1,yy − w1,yw1,xy)]}

ξ

+{

v,yy + 12

(1 − υ) v,xx + 12

(1 + υ) u,xy

+ a2

2

[(w2

1,x + w21,y

),y

+ (1 − υ)(w1,yw1,xx − w1,xw1,xy

)]η

}dx dy = 0,


u,xx + 12

(1 − υ) u,yy + 12

(1 + υ)v,xy

= −a2

2

[(w2

1,x + w21,y

),x

+ (1 − υ) (w1,xw1,yy − w1,yw1,xy)]

(3.13.14)v,yy + 1

2(1 − υ)v,xx + 1

2(1 + υ)u,xy

= −a2

2

[(w2

1,x + w21,y

),y

+ (1 − υ) (w1,yw1,xx − w1,xw1,xy)].

We shall now discuss various boundary conditions. First, notice that for thedetermination of the buckling load we must only deal with w, whereas for the post-buckling only in-plane quantities (u, v) must be dealt with. We shall now adaptthis theory to the problem of a plate with many fields, loaded in compressionby forces σh per unit length. The plate is simply supported at x = ± /2, and aty = ±((2k+ 1)/2)b (k = 0, 1, 2, . . .) in such a way that in-plane displacements arefree (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)

σ σ

y

x

l

b

Figure 3.13.1


b σ

σ

l

y

x

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)

w1 = cosπx

cosπy

≡ cos λx cos µy. (3.13.16)

The equations for u and v now become

u,xx + 12

(1 − υ)u,yy + 12

(1 + υ)v,xy

= −a2

4λ[(λ2 − υµ2) sin 2λx + (λ2 + µ2) sin 2λx cos 2µy],

(3.13.17)v,yy + 1

2(1 − υ)v,xx + 1

2(1 + υ)u,xy

= −a2

4µ[(µ2 − υλ2) sin 2µy + (λ2 + µ2) sin 2µy cos 2λx].

By inspection, we notice that particular solutions of (3.13.17) are of the form

upart = A sin 2λx + B sin 2λx cos 2µy(3.13.18)

vpart = C sin 2µy + D sin 2µy cos 2λx.


4λ2A ≡ a2

4λ(λ2 − υµ2)

[4λ2 + 2(1 − υ)µ2]B + 2(1 + υ)λµD ≡ a2

4λ(λ2 + µ2)

(3.13.19)

2(1 + υ)λµB + [4µ2 + 2(1 − υ)λ2]D ≡ a2

4µ(λ2 + µ2)

4µ2C = a2

4µ(µ2 − υλ2),

from which

A = a2

16λ2 − υµ2

λ, C = a2

16µ2 − υλ2

µ,

(3.13.20)

B = a2

16λ, D = a2

16µ.

162 Applications

The particular solutions are now given by

upart = a2

16λ[(λ2 − υµ2) sin 2λx + λ2 sin 2λx cos 2µy]

(3.13.21)

vpart = a2

16µ[(µ2 − υλ2) sin 2µy + µ2 cos 2λx sin 2µy].

To satisfy the boundary conditions, we must add solutions of the homogeneousequations. With the condition (3.13.15), the boundary conditions become

u = const.u,y + v,x = 0

}at x = ± /2

(3.13.22)v = const.

u,y + v,x = 0

}at y = ± /2,

which means that the shear strains vanish along the edge of the plate. Writing the fullsolution as the sum of the solution of the homogeneous equations and the particularsolution

u = uh + upart, v = vh + vpart, (3.13.23)

we find that uh and vh must satisfy

uh,xx + 12

(1 − υ)uh,yy + 12

(1 + υ)vh,xy = 0(3.13.24)

vh,yy + 12

(1 − υ)vh,xx + 12

(1 + υ)uh,xy = 0

uh = const.uh,y + vh,x = 0

}at x = ± /2

(3.13.25)vh = const.

uh,y + vh,x = 0

}at y = ± /2.

It is easy to verify that

uh = −a2ε1x and vh = −a2ε2y (3.13.26)

satisfy both the equations and the boundary conditions. The unknown strains −a2ε1,−a2ε2 follow from the conditions

b/2∫−b/2

(u,x + υv,y + a2

2w2

1,x

)dy = 0 at x = ± /2

(3.13.27)/2∫

−/2

(v,y + υu,x + a2

2w2

1,y

)dx = 0 at y = ± b/2,


which yields

ε1 + υε2 = 18

(λ2 + υµ2)(3.13.28)

υε1 + ε2 = 18

(υλ2 + µ2),

from which we obtain the expressions for the strains ε1, ε2,

ε1 = 18λ2, ε2 = 1

8µ2. (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

P [au1 + u] =∫∫ {

Eh3a2

12(1 − υ2)

[(�w1)2 − 2(1 − υ)

(w1,xxw1,yy − w2

1,xy

)]

+ Eh2(1 − υ2)

[(u,x + 1

2a2w2

1,x

)2

+(

v,y + 12

a2w21,y

)2

(3.13.30)

+ 2υ

(u,x + a2

2w2

1,x

)(v,y + a2

2w2

1,y

)]

+ 12

(1 − υ)(u,x + v,y + a2w1,xw1,y

)2 − 12σha2w2

1,x

}dx dy.

Making use of the fact that∫∫Eh3

12(1 − υ2)

[(�w1)2 − 2(1 − υ)

(w1,xxw1,yy − w2

1,xy

)]dx dy

= 12σcrh

∫∫w2

1,xdx dy

and the fact that u and v are the solutions of the minimum problem, we have

Eha2

12(1 − υ2)

∫∫ [u,xw2

1,x + v,yw21,y + υ

(u,xw2

1,y + v,yw21,x

)+ (1 − υ) (u,x + v,y) w1,xw1,y

]dx dy (3.13.31)

= − Eh1 − υ2

∫∫ [u2

.x + v2,y + 2υu,xv,y + 1

2(1 − υ) (u,x + v,y)2

]dx dy.

Then, we may write

P [au1 + u] = Eh8 (1 − υ2)

/2∫−/2

b/2∫−b/2

(w2

1,x + w21,y

)2dx dy

− Eh2(1 − υ2)

/2∫−/2

b/2∫−b/2

[u2

,x + v2,y + 2υu,xv,y + 1

2(1 − υ) (u,y + v,x)2

]dx dy

+ 12

(σcr − σ)h

/2∫−/2

b/2∫−b/2

w21,xdx dy = F(a). (3.13.32)

164 Applications

Evaluating these integrals, we find

F(a) = a2

8(σcr + σ)hbλ2 + a4

256Ehb(λ4 + µ4). (3.13.33)

The equilibrium equation follows from

∂F∂a

= 0 = 14

ahb

[λ2(σcr − σ) + E

16a2(λ4 + µ4)

], (3.13.34)

which yields

a2 = 16(σ − σcr)λ2

E(λ4 + µ4)= 16π2

2

1π4

4+ π4

b4

σ − σcr

E. (3.13.35)

For the case = b, we have

a2 = 8b2

π2

σ − σcr

E= 8h2

3(1 − υ2)

(σ

σcr− 1)

, (3.13.36)

where we have made use of (3.12.34). This means that for σ = 2σcr, we have

a = h

√8

3(1 − υ2)≈ 1.7h ( for υ = .25), (3.13.37)

i.e., when the load is twice the critical load, the deflection is still only 1.7 times theplate thickness.

The displacement of the load σ after buckling is

2σ

E− u

(x =

2

)=

2

( σ

E+ a2ε1

)≡

2εtot (3.13.38)

so that

εtot = σ

E+ 2

σ − σcr

Eλ4

λ4 + µ4, (3.13.39)

and for = b, we have λ = µ, and hence

εtot = σ

12

E− σcr

E≡ σ

Et− σcr

E. (3.13.40)

The so-called tangent modulus Et is half the modulus E. This behavior is shown inFigure 3.13.3.

σEt

σ cr

E

ε tot

asymptotic result

actual behavior

Figure 3.13.3


σ σ bl

Figure 3.13.4

However, it should be noted that our solution is only valid in a small neighbor-hood of the critical load. Comparison of our result with exact numerical calculationsshow that for sufficiently long plates, the range of validity is moderately large. How-ever, for short plates the range of validity is extremely small. Let us consider thiscase in some detail.

From (3.12.37), the critical load is then given by

σcr = π2Eh2

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 = 16σ − σcr

λ2E= 16

σ − σcr

E2

h2. (3.13.42)

The total strain εtot is now given by

εtot = σ

E+ a2

8h2

2= σ

13

E− 2σcr

E, (3.13.43)

so the tangent modulus now is given by Et = 1/3E. This result has an extremelysmall range of validity because almost immediately after buckling, the plate behaveslike a buckled bar.

This result is valid for the lowest bifurcation load, with only one wave.

σ

Et

σ cr

E

ε tot

asymptotic result

actual behavior

Figure 3.13.5

166 Applications

From (3.12.36), the exact formula for the critical load for a plate is

σcr = Eh2

12(1 − υ2)λ2(

1 + 2λ2

µ2+ λ4

µ4

). (3.13.44)

For fixed λ and, say, λ/µ = 10−2 for one wave, we have

σ(1)cr ≈ Eh2

12(1 − υ2)λ2 (1 + 2.10−4). (3.13.45)

When there are two waves, we have λ/µ = 2.10−2 and

σ(2)cr ≈ Eh2

12(1 − υ2)λ2 (1 + 8.10−4), (3.13.46)

which is very close to the lowest bifurcation load. This result shows that the lowestbifurcation point is a cluster point.

Let us finally make some remarks regarding the results obtained. We haveassumed that the edges at x = ± remain straight. This assumption holds exactlyfor sufficiently long plates because the nodal lines are straight lines. However, forshort plates this must be accomplished with an edge beam, without constraining thestrain long the edge. This condition can be done between two smooth stamps in atesting machine.

3.14 The “von Karman-Foppl Equations”

The post-buckling behavior of plates can also be analyzed using the so-called “vonKarman-Foppl equations.” For the derivation of these equations, we consider thedeviation from the undeformed state. The strains in the mid-plane are

γxx = u,x + 12

w2,x, γyy = v,y + 1

2w2

,y,

(3.14.1)2γxy = u,x + v,y + w,xw,y.

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

Eh2(1 − υ2)

∫∫ {(u,x + 1

2w2

,x

)2

+(

v,y + 12

w2,y

)2

+ 2υ

(u,x + 1

2w2

,x

) (v,y + 1

2w2

,y

)

+ 12

(1 − υ) (u,y + v,x + w,xw,y)2 (3.14.4)

+ h2

12

[w2

,xx + w2,yy + 2υw,xxw,yy + 2(1 − υ)w2

,xy

]}dx dy.

3.14 The “von Karman-Foppl Equations” 167

This expression is in full agreement with (3.12.4) because the fundamental state islinear. Carrying out the variations with respect to u and v, we obtain a stationaryvalue when∫∫ {

Eh1 − υ2

[u,x + 1

2w2

,x + υ

(v,y + 1

2w2

,y

)]δu,x

+ Eh1 − υ2

[υ

(u,x + 1

2w2

,x

)+ v,y + 1

2w2

,y

]δv,y (3.14.5)

+ Eh2(1 + υ)

(u,y + v,x + w,xw,y) (δu,y + δv,x)}

dx dy = 0.

Introducing the relations

Eh1 − υ2

[u,x + 1

2w2

,x + υ

(v,y + 1

2w2

,y

)]= Nx

Eh1 − υ2

[υ

(u,x + 1

2w2

,x

)+(

v,y + 12

w2,y

)]= Ny (3.14.6)

Eh2(1 + υ)

(u,y + v,x + w,xw,y) = Nxy,

we can rewrite this expression to yield∫∫[Nxδu,x + Nyδv,y + Nxy (δu,y + δv,x)]dx dy = 0. (3.14.7)

Applying the divergence theorem, we obtain∫∂S

[(Nxυx + Nxyυy) δu + (Nyυy + Nxyυx) δv]ds

(3.14.8)−∫∫

S

[(Nx,x + Nxy,y) δu + (Ny,y + Nxy,x) δv]dx dy = 0,

which yields the in-plane equilibrium equations

Nx,x + Nxy,y = 0, Ny,y + Nxy,x = 0, (3.14.9)

and the in-plane boundary conditions, when either u, v are zero or free,

(Nxυx + Nxyυy) δu∣∣∂S = 0

(3.14.10)(Nyυy + Nxyυx) δv

∣∣∂S = 0.

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 withrespect to x and y, respectively, we obtain

Nx = F∗,yy + f (y), Ny = F∗

,xx + g(x),

or defining

f (y) = f ∗,yy, g(x) = g∗

,xx,

Nx = F∗,yy + f ∗

,yy, 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,

F,xx = F∗,xx + g∗

,xx = Nx, 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∫∫ {

Nxδ

(12

w2,x

)+ Nyδ

(12

w2,y

)+ Nxyδ(w,xw,y)

+ Eh3

12(1 − υ2)[(w,xx + υw,yy) δw,xx + (w,yy + υw,xx) δw,yy (3.14.12)

+ 2(1 − υ) w,xyδw,xy]}

dx dy = 0.

Equation (3.14.12) is fully identical with (3.12.9) when S is replaced by N, so wemay use all the results derived from that expression. In particular, we mention thedifferential equation

Eh3

12(1 − υ2)��w − Nxw,xx − Nyw,yy − 2Nxyw,xy = 0. (3.14.13)

Using (3.14.11), we can rewrite (3.14.13) in the form

Eh3

12(1 − υ2)��w − F,yyw,xx + 2F,xyw,xy − F,xxw,yy = 0. (3.14.14)

The equation for the stress function is obtained from the condition that the displace-ment 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:

u,x + 12

w2,y = 1

Eh(F,yy − υF,xx)

v,y + 12

w2,y = 1

Eh(F,xx − υF,yy) (3.14.15)

u,y + v,x + w,xw,y = −2(1 − υ)Eh

F,xy.

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 differentiatingthe first equation with respect to y twice, and adding the second equation differen-tiated twice with respect to x, and subtracting the last equation differentiated withrespect to x and y. The result is

1Eh

��F = w2,xy − w,xxw,yy. (3.14.16)

This equation and (3.14.14), without the bending term, were first derived by Foppl.Later, von Karman derived the full equations.

Notice that in this approach there are only two unknowns, Airy’s stress functionand the displacement w, whereas in the general theory we use the three displace-ments 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 small-est radius of curvature, shallow shell theory may be used. Shallow shell theory canbe derived as follows.

Consider a surface Z(x, y), where x and y are Cartesian coordinates in a tangentplane to the surface (see Figure 3.15.1), chosen so that the projection of the lineswith 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). Inthe 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

(3.15.1)= (1 + Z2

,x

)(dx)2 + 2Z,xZ,y dx dy + (1 + Z2

,y

)dy2.

For the description of the surface in the deformed state, we shall make use of the factthat the in-plane displacements u, v are considerably smaller than the displacement

z y

x

w v

u

Figure 3.15.1

170 Applications

w, perpendicular to the mid-plane, i.e.,

|u|, |v| � |w|, (3.15.2)

and that

w2,x, w2

,y � 1. (3.15.3)

Under these conditions, a point (x, y, z) in the undeformed state moves to

(x + u − Z,xw, y + v − Z,yw, z + w)

in the deformed state. The square of the length of a line element in the deformedstate is now

(ds)2 = (dx + du − Z,xxw dx − Z,xyw dy − Z,x dw)2

(3.15.4)+ (dy + dv − Z,yyw dy − Z,xyw dx − Z,y dw)2 + (dz + dw)2,

where Z,xy = 0 because x and y are the principal directions of curvature,

(ds)2 − (ds)2 = 2(dx)2[

u,x − Z,xxw + 12

w2,x

+ 12

(u,x − Z,xxw + Z,xw,x)2 + 12

(v,x − Z,yw,x)2]

+ 2dx dy [u,y + v,x + w,xw,y

+ (u,x − Z,xxw − Z,xw,x)(u,y − Z,xw,y) (3.15.5)

+ (v,y − Z,yyw − Z,yw,y)(v,x − Z,yw,x)]

+ 2(dy)2[

v,y − Z,yyw + 12

w2,y

+ 12

(v,y − Z,yyw + Z,yw,y)2 + 12

(u,y − Z,xw,y)2]

.

Neglecting higher order terms, except for w2,x, w2

,y and w,x, w,y, we obtain

(ds)2 − (ds)2 = 2(

u,x − Z,xxw + 12

w2,x

)(dx)2

+ 2(u,y + v,x + w,xw,y) dx dy(3.15.6)

+ 2(

u,y − Z,yyw + 12

w2,y

)(dy)2

≡ 2γxx(dx)2 + 4γxy dx dy + 2γyy(dx)2.

Hence, the strains are given by

γxx = u,x − wR1

+ 12

w2,x

2γxy = u,y + v,x + w,xw,y (3.15.7)

γyy = v,y − wR2

+ 12

w2,y


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 neighbor-hood of the tangent point. Under the assumption that the loads are dead-weightloads, which cause stresses σx, σy, and τxy in the fundamental state, the energy isgiven by

P[u] = Eh2(1 − υ 2)

∫∫ {(u,x − w

R1+ 1

2w2

,x

) 2

+(

v,y − wR2

+ 12

w2,y

)2

+ 2υ

(u,x− w

R1+ 1

2w2

,x

)(v,y− w

R2+ 1

2w2

,y

)+ 1

2(1−υ)(u,y+v,x+w,xw,y)2

(3.15.10)

+ h2

12

[w2

,xx + w2,yy + 2υw,xxw,yy + 2(1 − υ)w2

,xy

]

+ 1 − υ 2

E

(σxw2

,x + σyw2,y + 2τxyw,xw,y

)}dx dy,

where the radii of the curvature can be assumed to be constants. Later we shall showthat 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

P2[u] = Eh2(1 − υ 2)

∫∫ {(u,x − w

R1

) 2

+(

v,y − wR2

)2

+ 2υ

(u,x − w

R1

)(v,y − w

R2

)+ 1

2(1 − υ)(u,y + v,x)2

(3.15.11)

+ h2

12

[w2

,xx + w2,yy + 2υw,xxw,yy + 2(1 − υ)w2

,xy

]+ 1 − υ 2

E

(σxw2

,x + σyw2,y + 2τxyw,xw,y

)}dx dy

P3[u] = Eh2(1 − υ 2)

∫∫ {[u,x − w

R1+ υ

(v,y − w

R2

)]w2

,x

(3.15.12)+[

v,y − wR2

+ υ

(u,x − w

R1

)]w2

,y + (1 − υ)(u,y + v,x)w,xw,y

}dx dy.

The variational equation for neutral equilibrium is P11[u, ζ] = 0, where ζ= (ξ, η, ζ) isa kinematically admissible displacement field,

P11[u, ζ] = Eh1 − υ 2

∫∫ [(u,x − w

R1

) (ξ,x − ζ

R1

)+(

v,y − wR2

)(η,y − ζ

R2

)

+υ

(u,x − w

R1

)(η,y − ζ

R2

)+ υ

(v,y − w

R2

)(ξ,x − ζ

R1

)

172 Applications

+ 12

(1 − υ)(u,y + v,x)(ξ,y + η,x) (3.15.13)

+ h2

12

{w,xxζ,xx + w,yyζ,yy + υw,xxζ,yy + υw,yyζ,xx + 2(1 − υ)w,xyζ,xy

}+ 1 − υ 2

E(σxw,xζ,x + σyw,yζ,y + τxyw,xζ,y + τxyw,yζ,x)

]dx dy.

Repeated application of the divergence theorem yields

Eh1 − υ 2

∮∂S

[[{(u,x − w

R1

)+ υ

(v,y − w

R2

)}υx + 1

2(1 − υ)(u,y + v,x)υy

]ξ

+[{

v,y − wR2

+ υ

(u,x − w

R1

)}υy + 1

2(1 − υ)(u,y + v,x)υx

]η

+ h2

12

[{(w,xx + υw,yy) υx + (1 − υ)w,xyυy

}ζ,x

+ {(w,yy + υw,xx)υy + (1 − υ)w,xyυx}ζ,y]

+{

1 − υ 2

E[σxw,xυx + σyw,yυy + τxy(w,xυ,y + w,yυx)]

(3.15.14)

− h2

12[(w,xx + w,yy),xυx + (w,xx + w,yy),yυy]

}ζ

]ds

− Eh1 − υ 2

∫∫s

{u,xx + 1

2(1 − υ)u,yy + 1

2(1 + υ)v,xy −

(1

R1+ υ

R2

)w·x

}ξ

+{

v,yy + 12

(1 − υ)v,xx + 12

(1 + υ)u,xy −(

1R2

+ υ

R1

)w,y

}η

{−(

1R1

+ υ

R2

)u,x −

(1

R2+ υ

R1

)v,y +

(1

R21

+ 2υ

R1R2+ 1

R22

)w

+ h2

12��w− 1 − υ 2

E

{(σxw,x),x+(σyw,y),y+(τxyw,x),y+(τxyw,y),x

}}ζ

]dx dy=0,

where υ is the unit normal vector at the edge. This yields the differential equations

u,xx + 12

(1 − υ)u,yy + 12

(1 + υ)v,xy −(

1R1

+ υ

R2

)w,x = 0

(3.15.15)v,yy + 1

2(1 − υ)v,xx + 1

2(1 + υ)u,xy −

(1

R2+ υ

R1

)w,y = 0

h2

12��w −

(1

R1+ υ

R2

)u,x −

(1

R2+ υ

R1

)v,y +

(1

R21

+ 2υ

R1R2+ 1

R22

)w

(3.15.16)

− 1 − υ 2

E(σxw,xx + σyw,yy + 2τxyw,xy) = 0,

where we have used the fact that σx, σy, and τxy (approximately) satisfy the equa-tions

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,[{(u,x − w

R1

)+ υ

(v,y − w

R2

)}υx + 1

2(1 − υ)(u,y + v,x)υy

]ξ

∣∣∣∣∂S

= 0(3.15.18)[{(

v,y − wR2

)+ υ

(u,x − w

R1

)}υy + 1

2(1 − υ)(u,y + v,x)υx

]η

∣∣∣∣∂S

= 0.

For the reduction of the remaining terms, we make use of the relations

ζ,x = −υyζ,s + υxζ,v, ζ,y = υxζ,s + υyζ,v, (3.15.19)

and (3.12.17) to obtain[(w,xx + υw,yy)υ2

x + 2(1 − υ)w,xyυxυy + (w,yy + υw,xx)υ2y

]ζ,v∣∣∂S

= 0 (3.15.20){(1 − υ)

[(w,yy − w,xx)υxυy + w,xy

(υ2

x − υ2y

)],s

+ (w,xx + w,yy),xυx + (w,xx + w,yy),yυy (3.15.21)

− 12(1 − υ 2)Eh3

[(σxw,x + τxyw,y)υx + (σyw,y + τxyw,x)υy]}

ζ

∣∣∣∣∂S

= 0.

Notice that only the in-plane boundary conditions are affected by the curvature ofthe surface.

We shall now apply this theory to a spherical shell under a uniform externalpressure p per unit area of the middle surface.† The stresses in the fundamentalstate are given by

σx = σy = −σ = −pR2h

, τxy = 0. (3.15.22)

The equations now read

u,xx + 12

(1 − υ)u,yy + 12

(1 + υ)v,xy − 1 + υ

Rw,x = 0

(3.15.23)v,yy + 1

2(1 − υ)v,xx + 1

2(1 + υ)u,xy − 1 + υ

Rw,y = 0

Eh2

12(1 − υ 2)��w − E

(1 − υ)R

(u,x + v,y − 2

wR

)+ σ(w,xx + w,yy) = 0. (3.15.24)

We try a solution of the form

u = A sin px/R cos qy/R

v = Bcos px/R sin qy/R (3.15.25)

w = C cos px/R cos qy/R, (p, q ∈ R).


−[

p2 + 12

(1 − υ)q2]

A − 12

(1 + υ)pqB + (1 + υ)pC = 0(3.15.26)

− 12

(1 + υ)pqA −[

12

(1 − υ)p2 + q2]

B + (1 + υ)qC = 0.

† See the remark below (3.15.10).

174 Applications

We can now solve A/C and B/C from these equations, which yield

AC

= (1 + υ)p

p2 + q2,

BC

= (1 + υ)q

p2 + 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 aresatisfied.

Substitution of the displacement field (3.15.25) into the third equilibrium equa-tion and using (3.15.27), we obtain

σ

E= 1

p2 + q2+ h2

4c2R2(p2 + q2), c =

√3(1 − υ 2). (3.15.28)

We still must minimize this expression with respect to p and q, but because theexpression only contains the combination p2 + q2, we can minimize with respect tothis parameter, which yields

p2 + q2 = 2cRh

. (3.15.29)

This means that our assumptions are satisfied as R/h � 1. The critical load is nowgiven by

σcr = EhcR

. (3.15.30)

Notice that the critical load is proportional to h/R, whereas for plates the criticalload was proportional to h2/b2.

Because p and q must only satisfy the condition (3.15.29), there is an infinitenumber of buckling modes. In the p − q plane, (3.15.29) represents a circle withradius p0 = √2cR/h (see Figure 3.15.2).

Because we have a continuous spectrum, the displacement field is given by inte-grals instead of a series. However, for our purpose we may consider some discretevalues, pi, qi. Because for the evaluation of P3[u1] we must deal with cubic terms,

2cR/ h

(

) 1/2

p

q

Figure 3.15.2


we can consider the discrete pairs

(p1, q1), (p2, q2), (p3, q3), (3.15.31)

which in the evaluation of P3 [u1] lead to integrals of the form∫ (sincos

p1xR

sincos

p2xR

sin p3xR

)dx, (3.15.32)

which can be expressed in integrals of the form∫sincos

(p1 ± p2 ± p3)xR

dx. (3.15.33)

For large values of x, these integrals vanish† except when the argument of the cosineis equal to zero, i.e.,

p1 ± p2 ± p3 = 0, (3.15.34)

and similarly for the terms in y,

q1 ± q2 ± q3 = 0. (3.15.35)

This means that the points (p1, q1), (p2, q2), (p3, q3) divide the circle in three equalparts. We shall now consider the special case of (see Figure 3.15.2)

(p1, q1) = (p0, 0), (p2, q2) =(

−12

p0,12

p0

√3)

,

(3.15.36)(p3, q3) =

(−1

2p0,−1

2p0

√3)

.

Let us introduce the notation

p = 12

p0, q = 12

√3 p0; (3.15.37)

then our displacement field is given by

u = (1 + υ)h2

2cR

(2pa0 sin 2p

xR

+ pa1 sin pxR

cos qyR

)

v = (1 + υ)h2

2cR

(qa1 cos p

xR

sin qyR

)(3.15.38)

w = h(

a0 cos 2pxR

+ a1 cos pxR

cos qyR

),

where we have used (3.15.27).For the evaluation of P3[u1], we shall need the following expressions,

u,x − wR

= hR

[υa0 cos 2p

xR

− 14

(3 − υ)a1 cos pxR

cos qyR

]

v,y − wR

= hR

[−a0 cos 2p

xR

− 14

(1 − 3υ)a1 cos pxR

cos qyR

]

† Let L � x � R, e.g., x = O(R3/4h1/4

), so that our shallow shell assumptions still hold approxi-

mately.

176 Applications

u,x − wR

+ υ(

v,y − wR

)= h

R

[−3

4a1(1 − υ 2) cos p

xR

cos qyR

]

v,y − wR

+ υ(

u,x − wR

)= h

R

[−(1 − υ 2)a0 cos 2p

xR

− 14

(1 − υ 2)a1 cos pxR

cos qyR

]

u,y + v,x = (1 + υ)hR

[−1

2

√3 a1 sin p

xR

sin qyR

](3.15.39)

w,x = 2phR

(−a0 sin 2p

xR

− 12

a1 sin pxR

cos qyR

)

w,y = 2phR

(−1

2

√3 a1 cos p

xR

sin qyR

).


( ) = 14R2

R∫−R

R∫−R

( )dx dy, (3.15.40)

which is the average value of (), we obtain

[u,x − w

R+ υ

(v,y − w

R

)]w2

,x +[v,y − w

R+ υ

(u,x − w

R

)]w2

,y

+ (1 − υ)(u,y + v,x)w,xw,y

= p2 h3

R3

[−3

8(1 − υ 2)a0a2

1 − 38

(1 − υ 2)a0a21 − 3

8(1 − υ 2)a0a2

1

](3.15.41)

= − 916

(1 − υ 2)ch2

R2a0a2

1,

so the average value of P2[u1] is

P3[u1] = − 932

Ech3

R2a0a2

1. (3.15.42)

For the evaluation of P2[u; λ], we need

w2·x + w2·y = 4p2 h2

R2

(12

a20 + 1

4a2

1

), (3.15.43)

so that

P2[u1; λ] = (1 − λ)Eh3

R2

(12

a20 + 1

4a2

1

). (3.15.44)

In a sufficiently small neighborhood of the critical load, the average energy P[u, λ]is given by

F(ai, λ) = P[u, λ] = Eh3

R2

[12

(1 − λ)(

a20 + 1

2a2

1

)− 9c

32a0a2

1

]. (3.15.45)


The equilibrium equations are given by

∂F∂a0

= Eh3

R2

[(1 − λ)a0 − 9c

32a2

1

]= 0

(3.15.46)∂F∂a1

= Eh3

R2

[12

(1 − λ)a1 − 9c16

a0a1

]= 0.

From the second of these equations, we obtain the non-trivial solution

a0 = 89c

(1 − λ), (3.15.47)

and from the first equation,

a1 = ±169c

(1 − λ) = ±2a0. (3.15.48)

It is sufficient to consider only the positive solution because the negative sign isobtained by shifting the origin of our coordinate system.

As

∂2F

∂a20

∂2F

∂a21

−(

∂2F∂a0∂a1

)2

= −(

Eh3

R2

)2

(1 − λ)2 < 0, (3.15.49)

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 = 815

(1 − λ), a1 = 1615

(1 − λ), (3.15.50)

so that for λ = 1/2, a0 = 4/15, a1 = 8/15, w = 0.8h. This means that for deflectionsof the order of the plate thickness, the load is reduced to about 50% of the criticalload.

Assuming that our average energy expression holds on the whole surface of thesphere, the total energy is given by

4πR2F(ai, λ). (3.15.51)

The work of the uniform pressure p is −p�V, where �V is the volume incrementpassing from the fundamental path to the branched path. It follows that

4πR2 ∂F∂p

= −�V, (3.15.52)

or with

p = 2λEh2

cR2(3.15.53)

2πR4cEh2

∂F∂λ

= −�V. (3.15.54)

Using (3.15.45), we obtain

�V = −64π

27chR2(1 − λ)2. (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 = −4πR2wcr = −σcr

E(1 − υ)R · 4πR = −4π

c(1 − υ)hR2, (3.15.56)

so that

�Vtot

�Vcr= 16

27(1 − υ)(1 − λ)2 + λ. (3.15.57)

This relation is illustrated in Figure 3.15.3. Notice that there is no stable equilibriumstate in the vicinity of the bifurcation point, and that the load drops very rapidly afterbifurcation, which means that the shell is very imperfection-sensitive. We will comeback to this later.

Let us now first have a closer look at the buckling mode,

w = ha0

(cos 2p

xR

+ 2 cos pxR

cos p√

3yR

). (3.15.58)


2pxR

= α, 2pyR

= β, (2.15.59)

we can rewrite this expression as

w = ha0f (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, wenotice that

f max = 3 for α = 4j π, β = 4kπ√3

,

(3.15.61)α = (2j + 2)π, β = (2k+ 2)

π√3.


Necessary conditions for a minimum value of f (α, β) are

∂ f∂ α

= − sin α − sin α/2 cos√

3 β/2 = 0(3.15.62)

∂ f∂ β

= −√

3 cos α/2 sin√

3β/2 = 0.

From the second equation, we find

cos α/2 = 0, α = (2k+ 1)π,

sin√

3 β/2 = 0, β = 2j π√3

,(3.15.63)

and from the first equation,

sin α/2 = 0, α = 2kπ,(3.15.64)

2 cos α/2 + cos√

3 β/2 = 0.

Both equations are satisfied when

α = (2k+ 1)π, β = 2j + 1√3

π (3.15.65)

β = 2jπ√3, α =

±2π

3+ 4kπ ( j odd)

±4π

3+ 4kπ ( j even)

(3.15.66)

α = 2kπ, β = 2j π√3

. (3.15.67)

We now obtain the following values for f (α, β),

f(

(2k+ 1)π,2j + 1√

3π

)= 1

f(

±2π

3+ 4kπ,

2j π√3

)=

+12

j even

−32

j odd(3.15.68)

f(

±4π

3+ 4kπ,

2j π√3

)= −3

2

f(

2kπ,2j π√

3

)={

−1 j + k odd3 j + k even.

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 modeis artificially fixed, which may be done by fitting a solid sphere with a radius slightly

180 Applications

−3π −2π −π 0 π 2π 3π

3

3

3 / 2

3 / 2

3 3

3

3

3

3

3 / 2 3 / 2

/ 2

α

β

3 / 2

Figure 3.15.4

smaller than that of the shell inside the shell.† The experimental results are in agree-ment with our buckling pattern.

As we have seen previously, the load drops rapidly after bifurcation, whichimplies that the shell is very imperfection-sensitive. To get quantitative informa-tion about the imperfection sensitivity, we assume imperfections corresponding tothe buckling mode, which according to the general theory are the worst type ofimperfections.

Let the imperfections be given by

w0 = h[a0 cos 2p

xR

+ a1 cos pxR

cos√

3 pyR

], (3.15.69)

where a1 = 2a0. In the presence of imperfections, the function F(ai; λ) (see 3.15.45)must be replaced by (see 2.6.39)

F∗(ai; λ) = Eh3

R2

[12

(1 − λ)(

a20 + 1

2a2

1

)− 9c

32a0a2

1 − λ

(a0a0 + 1

2a0a1

)]. (3.15.70)

With a1 = 2a0 and a0 and a1 given by (3.15.47) and (3.15.48), we can rewrite thisexpression in the form

F∗(ai; λ) = Eh3

2R2

[3(1 − λ)a2

0 − 9c4

a31 − 6λa0a0

]. (3.15.71)

† Cf. R. L. Carlson, R. L. Sendelbeck, and N. J. Hoff, Experimental studies of the buckling ofcomplete spherical shells. Experimental Mechanics, 7, no. 7, 281–288 (1967).

3.15 Buckling behavior of a spherical shell under uniform external pressure 181

The equilibrium equation then reads

∂F∗

∂a0= Eh3

2R2

[6(1 − λ)a0 − 27c

4a2

0 − 6λa0

]= 0. (3.15.72)

The minimum value of a0 is obtained when the discriminant of this quadratic equa-tion vanishes, which yields

(1 − λ∗)2 = 9c2

λ∗a0. (3.15.73)

The corresponding amplitude is given by

a∗0 = 4(1 − λ∗)

9c. (3.15.74)

For λ∗ = 0.5, we have a0 = 1/15 (for υ = 0.28), so the maximum imperfection ampli-tude is 3a0 = 1/5, i.e., imperfections of about 20% of the shell thickness reducethe critical load by 50%. This result shows that the shell is extremely imperfection-sensitive.

The present approach is due to Hutchinson.† Let us now reconsider theapproach. The starting point was shallow shell theory, so the results are only valid ina sufficiently small neighborhood of the tangent point, i.e., x, y � R. However, forour evaluation of the average value of the energy we have carried out the integra-tion over the domain −R ≤ x, y ≤ R. Furthermore, we have ignored the fact thatthe hexagonal buckling pattern cannot be extended over the complete sphericalshell, which implies a relaxation of the geometric conditions, which should result indecreased stability in the post-buckling range. This means that Hutchinson’s analy-sis yields a lower bound for the post-buckling load. The difficulties with the domainof validity for x and y can be overcome by using a displacement field (u∗, v∗, w∗),defined by

(u∗, v∗, w∗) = (u, v, w) exp[−1

2µ2(x2 + y2)/R2

], (3.15.75)

where 1 � µ2 � p2 + q2. Under this restriction, the exponential factor can be con-sidered to be a constant factor when differentiation is carried out, which means thatthis 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 imper-fections.‡ In the next section, we shall treat the problem of the spherical shell againusing the general theory of shells, and verify the validity of results obtained in thissection.

† 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 usingthe general theory of shells

In this section, we shall again consider the behavior of a spherical shell under uni-form external pressure, but now we shall employ the general theory of shells. Forthe 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 basevectors are aα = r,α = ∂r/∂xα. The reciprocal base is defined by aα · aβ = δ

βα. 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 = 1

2εαβaα × aβ whereεαβ is the contravariant alternating tensor. The second fundamental tensor is speci-fied by bαβ = n · r,αβ.

A point in shell space is identified by its distance z to the middle surface and bythe surface coordinates of its projection on to the middle surface. The coordinate z isorthonormal to the middle surface. The shell faces z = ± 1

2 h, where h is the constantshell thickness, are surfaces parallel to the mid-surface.

The covariant components of the spatial metric tensor gij are specified by

gαβ = aαβ − 2zbαβ + z2cαβ

(3.16.1)g13 = g23 = 0, g33 = 1,

where cαβ = bκαbκβ 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√ga

= 1 − 2zH + z2K, (3.16.2)

where H = 12 bα

α is the mean curvature in a point (x1, x2) of the middle surface andK = b1

1b22 − b1

2b21 is the Gaussian curvature.

The edge of the shell is assumed to be a ruled surface formed by normals to themiddle surface along an edge curve on this surface. Let υ be the unit vector in thetangent plane, normal to the edge curve and positive outward. The positive senseon 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 dis-placement field,

u(xκ) = uαaα + wn. (3.16.3)

The strain tensor in the mid-plane is given by

γαβ = θαβ + 12

aκλ(θκα − ωκα)(θλβ − ωλβ) + 12ϕαϕβ, (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).


where θαβ is the linearized strain tensor,

θαβ = 12

(uα|β + uβ|α) − bαβw (3.16.5)

ωαβ = 12

(uβ|α − uα|β) (3.16.6)

is a rotation in the tangent plane to the shell, and

ϕα = w,α + bκαuκ (3.16.7)

is a rotation of the normal vector n. The linearized tensor of changes of curvature is

ραβ = 12

(ϕα|β + ϕβ|α) − 12

(bκ

αωκβ + bκβωκα

). (3.16.8)

After this survey of expressions from shell theory, we return to our stability prob-lem.

According to the general theory of elastic stability (2.6.18), the energy func-tional for a three-dimensional body is given by

P[u] =∫V

(12

Sij uh,iuh,j + 12

Eijkγij γk

)dV, (3.16.9)

where u denotes the three-dimensional increment of the displacement field, passingfrom the fundamental state to the adjacent state, and γij is the strain tensor

γij = 12

(ui,j + uj ,i + uh,iuh,j ). (3.16.10)

The second variation of this energy is given by

P2[u] =∫V

(12

Sij uh,iuh,j + 12

Eijkθij θk

)dV, (3.16.11)

where θij is the linearized strain tensor,

θij = 12

(ui,j + uj ,i). (3.16.12)

In shell theory, the state of stress is assumed to be approximately plain andparallel to the mid-surface. This results in a decoupling of the membrane energyand the bending energy. The energy functional can then be written as

P[u] =∫A

[12

Sij uh,iuh,j + h2

Eαβλµ

(γαβγλµ + h2

12ραβρλµ

)]dA, (3.16.13)

where the integration is to be carried out over the mid-surface of the shell. HereEαβλµ is the tensor of elastic moduli corresponding to a plate state of stress

Eαβλµ = G[

aαλaβµ + aαµaβλ + 2υ

1 − υaαβaλµ

], (3.16.14)

184 Applications

where G is the shear modulus,

G = E2(1 + υ)

. (3.16.15)

The second term in (3.16.13) can now be rewritten to yield

12

hEαβλµ

(γαβγλµ + h2

12ραβρλµ

)(3.16.16)

= Eh2(1 − υ 2)

{(1 − υ)γαβγαβ + υ (γ κ

κ )2 + h2

12

[(1 − υ)ραβραβ + υ(ρκ

κ)2]} .

Notice that the elastic energy density is expressed in terms of invariants of the straintensor and the tensor of changes of curvature.

Let Nαβ be the membrane stresses in the fundamental state, due to dead-weightloads. The energy functional can then be written as

P[u] =∫A

{Eh

2(1 − υ 2)

[(1 − υ)γ αβγαβ + υ(γ κ

κ )2] + h2

12

[(1 − υ)ραβραβ + υ(ρκ

κ)2]

+ 12

Nαβ[aκλ (θκα − ωκα) (θλβ − ωλβ) + ϕαϕβ

]}dA. (3.16.17)

In the case of uniform pressure p, the fundamental state is nonlinear, because thepotential energy p�V is proportional to the volume V.

Let us now consider this functional more closely. Let V0 be the volume in theundeformed state. The volume in the deformed state is then

V =∫V0

16

εijk εpqr yi,p yj ,q yk,r dV0, (3.16.18)

where yi are the coordinates in the deformed state of a point with coordinates xi inthe undeformed state. Let u be the displacement field, then

yi = xi + ui. (3.16.19)

Expression (3.16.18) can then be written as

V =∫V0

16

εijk εpqr (δip + ui,p)(δj q + uj ,q)(δkr + uk,r) dV0

= V0 +∫V0

12

εijk εpqrδipδj quk,r dV0 (3.16.20)

+∫V

12

εijk εpqrδipuj ,quk,r dV0 +∫V0

16

εijk εpqrui,puj ,quk,r dV0.

For the term linear in u, we obtain∫V0

12

εijk εpqr δip δj quk,r dV0 =∫V0

12

(δj qδkr − δj rδkq) δj quk,r dV0

=∫V0

δkruk,r dV0 =∫

uk,k dV0 =∫A

uk nk dA =∫A

w dA,

(3.16.21)


where we have used the divergence theorem on a closed surface so that there areno stock terms. The reduction of the term quadratic in u is considerably moredifficult.∫

V0

12

εijk εpqrδipuj ,quk,r dV0 =∫V0

12

(δj qδkr − δj rδkq) uj ,quk,r dV0

=∫V0

12

(uj ,j uk,k − uj ,kuk,j ) dV0

=∫A

(12

uj ,j uknk

)dA −

∫V0

12

uj ,jkuk dV0 (3.16.22)

−∫A

12

uj ,kuknj dA +∫�0

12

uj ,kj uk d�0

=∫A

12

(uj ,j nk − uj ,knj ) uk dA.

Because the integrand is an invariant, it can readily be rewritten in general coordi-nates, which yields ∫

A

12

(uj∥∥j uknk − uj

∥∥kuknj)

dA, (3.16.23)

where a subscript preceded by a double vertical bar denotes three-dimensionalcovariant differentiation with respect to the undeformed metric. In our coor-dinate 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

−∫A

12

(uj∥∥j w − w

∥∥kuk ) dA = −∫A

12

(uα ‖αw + w ‖3w − w ‖αuα − w ‖3w

)dA

(3.16.25)

= −∫A

12

(uα ‖αw − w ‖αuα

)dA.

Now

ui ‖α = ui,α + �i

kαuk, (3.16.26)

where �ikα is a Christoffel symbol of the second kind, so that

uα ‖α = uα,α + �α

βαuβ + �α3αw

(3.16.27)w ‖α = u3 ‖α = w,α + �3

βαuβ + �33αw.

186 Applications

The Christoffel symbols can be expressed as

�α3α = gαk�k3α = gαβ�β3α + gα3�33α

= gαβ 12

(gβ 3,α + gβα,3 − g3α,β) = −gαβbαβ = −bαα, (3.16.28)

where we have made use of (3.16.1),

�3βα = g3k�kβα = g33�3βα = 1

2(g3β,α + g3α,β − gβα,3) = bαβ (3.16.29)

�33α = g3k�k3α = g33�33α = 1

2(g33,α + g3α,3 − g3α,3) = 0. (3.16.30)

Here, �ijk are Christoffel symbols of the first kind.Using the relation

uα∣∣α = uα

,α + �αβαuβ , (3.16.31)

which is the covariant derivative of uα with respect to the surface coordinates, wecan now rewrite (3.16.25) to yield

−∫A

12

[(uα∣∣α − bα

αw)

w − (w,α + bαβuβ)

uα]

dA = −∫A

12

(wθαα − ϕαuα) dA.

(3.16.32)The energy of the uniform pressure load is now

− p∫A

wA − p∫ (

12

wθαα − 1

2ϕαuα

)dA − p

∫V0

16

εijk εpqr ui,p uj ,q uk,r dV0. (3.16.33)

Because p = O(N/R), where N denotes the maximum of the membrane forces andR the minimum radius of curvature, the second term is of order O (Nwθα

α/R) =O(Nw2/R2), whereas the first term is of order O(Nw/R). The potential energy ofthe membrane stresses in the fundamental state is O(Nw2/L2), so that for L/R � 1the 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 followsthat for L/R � 1 (shallow shell theory), only the first (linear) term must be takeninto account, which means that to a first approximation, uniform external pressuremay be considered as a dead-weight load. This implies that the energy functional isgiven by (3.16.17).

We shall now apply these results to a spherical shell under uniform externalpressure. However, we shall also take into account the second term in (3.16.33) andshow that this term may indeed be neglected. The basic property of a spherical shellis that the first and second fundamental tensors are proportional to each other,

bαβ = 1R

aαβ, bαβ = 1

Raα

β, bαβ = 1R

aαβ, (3.16.34)

where R is the radius of the middle surface. The sign convention in (3.16.34) impliesthat the positive direction of n points inward. The linearized strain tensor (3.16.5)


can now be written as

θαβ = 12

(uα|β + uβ|α) − 1R

aαβw. (3.16.35)

The rotation of n (3.16.7) now reads

ϕα = w,α + 1R

aα, (3.16.36)

and the linearized tensor of changes of curvature (3.16.8) is now given by

ραβ = w∣∣αβ + 1

2R(uα|β + uβ|α ) = w

∣∣αβ + aαβ

wR2

+ 1R

θαβ. (3.16.37)

In the following, we shall make use of the property that the tangential displace-ment field can always be expressed in terms of the two invariants

uα = ϕ,α + εαλ ψ∣∣λ .† (3.16.38)

To prove this property, we notice that

uα |α = ϕ∣∣αα + ε.αλψ

∣∣.λα = �ϕ + εαλψ|λα = �ϕ, (3.16.39)

where the term εαλψ|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 uniquesolution‡ for sufficiently smooth displacement fields. To get an equation for ψ, wemultiply (3.16.38) by εαµ to obtain

uαεαµ = εαµϕ,α + εαµεαλψ

∣∣λ = εαµϕ,α + ψ|µ ,

so that

ψ,µ = ε.αµ(uα − ϕ,α). (3.16.40)

This equation is integrable when the right-hand side of

ψ∣∣µυ = ε.αµ (uα|µ − ϕ |αυ ) (3.16.41)

is symmetric in µ and υ, so that the left-hand term must also be symmetric, whichmeans

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

θαβ = ϕ∣∣αβ + 1

2εαλψ

∣∣.λβ + 12εβλψ

∣∣.λα − 1R

aαβw (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 + 1R

ϕ,α + 1R

εαλψ∣∣λ (3.16.44)

ραβ = w∣∣αβ + 1

Rϕ∣∣αβ + 1

2Rεαλψ

∣∣.λβ + 12R

ε·βλψ∣∣.λα (3.16.45)

We now return to our energy functional (3.16.17). The fundamental state of aspherical shell under uniform external pressure p is a membrane state of stress, andthe contravariant tensor of stress resultants is

Nαβ = σhaαβ = 12

pRaαβ. (3.16.46)

It will be convenient to introduce a dimensionless load parameter λ, defined by

σ = EhcR

λ, c =√

3(1 − υ 2). (3.16.47)

Also taking into account the second term of (3.16.33), the second variation of ourenergy functional can now be written as

P2[u; λ] =∫A

{Eh

2(1 − υ 2)

[(1 − υ)θαβθαβ + υ(θκ

κ)2 + h2

12

{(1 − υ)ραβρ

αβ + υ(θκκ)2}]

− Eh2

2cRλ[(θαβ − ωαβ)(θαβ − ωαβ) + ϕαϕ

α]+ Eh2

cR2λ(wθα

α − uαϕα

)}dA.

(3.16.48)

Let us now consider the term∫A

θαβθαβ dA =

∫ (ϕ∣∣αβ + 1

2εαλψ

∣∣.λβ + 12εβλψ

∣∣.λα− 1R

aαβw)(

ϕ∣∣αβ + 1

2ε.αλ w

∣∣λβ(3.16.49)

+ 12ε.

βλψ∣∣λα− 1

Raαβw

)dA.

We shall now show that the coupling terms between ψ and ϕ and between ψ and wvanish. The term ∫

A

aαβwε.αλψ∣∣λβ dA =

∫A

wεβλψ∣∣λβ dA = 0 (3.16.50)

because ψ∣∣λβ is symmetric. Thus,∫

ϕ∣∣αβε

α·λψ∣∣λβ dA =

∫ϕ∣∣αβε

αλψ∣∣·βλ

dA

=∫ (

ϕ∣∣αβε

αλψ∣∣β )∣∣∣∣

λ

dA −∫

ϕ∣∣αβλ εαλψ

∣∣β dA.

The first of these integrals vanishes after the application of the divergence theoremon the closed surface. Changing the order of covariant differentiation in the remain-ing integral, we can write

−∫A

ϕ∣∣αβλε

αλψ∣∣β dA = −

∫A

ϕ∣∣αλβ εαλψ

∣∣β dA −∫A

Rκ·αβλϕ,κε

αλψ∣∣β dA.


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

·αε.βλ. (3.16.51)

For a sphere K = R−2, we can write

−∫A

Rκ·αβλϕ,κε

αλψ∣∣β dA = − 1

R2

∫A

εκ·αεβλε

αλϕ,κψ|α dA = − 1R2

∫A

εκαϕ |κ ψ|α dA

= − 1R2

∫A

(εκαϕ |κ ψ)∣∣α dA + 1

R2

∫A

εκαϕ |καψdA = 0.

The first of these integrals vanishes after the application of the divergence theoremon 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∫

A

θαβθαβ dA =

∫A

[(ϕ∣∣αβ − 1

Raαβw

) (ϕ∣∣αβ − 1

Raαβw

)

+ 14

(εαλψ

∣∣λ·β + εβλψ∣∣λ·α ) (εα

·κψ∣∣κβ + εβ

·κψ|κα )] dA(3.16.52)

=∫A

(ϕ∣∣αβϕ∣∣αβ − 2

Rw�ϕ + 2

R2w2)

dA

+12

∫A

(ψ∣∣λ·β ψ

∣∣∣·βλ + εαλεβ·κψ∣∣λ·β ψ|κα

)dA.

Let us now consider various terms separately,

∫A

ϕ∣∣αβ ϕ

∣∣αβ dA =∫A

(ϕ∣∣αβϕ

∣∣β

)∣∣∣∣∣∣α

dA −∫A

ϕ∣∣αβ··α ϕ

∣∣β dA.

Applying the divergence theorem on the first term, this term vanishes for a closedsurface. Changing the order of differentiation, in the second term we obtain∫

A

ϕ∣∣αβ ϕ

∣∣αβ dA =∫A

ϕ∣∣αβ·α ϕ∣∣β dA −

∫A

1R2

εκαεβ·αϕ,κϕ,β dA

= −∫A

(�ϕ)∣∣βϕ ∣∣β dA − 1

R2

∫A

ϕ |κ ϕ,κ dA

= −∫ (

(�ϕ)∣∣β ϕ)∣∣

βdA +

∫ϕ��ϕ dA − 1

R2

∫(ϕ |κϕ )

∣∣κ

dA

+ 1R2

∫ϕ�ϕ dA.

190 Applications

Applying the divergence theorem to the first and the third terms, these terms vanishbecause we are dealing with a closed surface. Our final result can now be written as∫

A

ϕ∣∣αβ ϕ

∣∣αβ dA =∫A

(ϕ��ϕ + 1

R2ϕ�ϕ

)dA, (3.16.53)

or in the equivalent form∫A

ϕ∣∣αβ ϕ

∣∣αβ dA =∫A

{(�ϕ)2 + 1

R2ϕ�ϕ

}dA. (3.16.54)

Similarly,∫A

ψ∣∣λ·β ψ

∣∣∣·βλ dA =∫A

(ψ∣∣λ·β ψ

∣∣β )∣∣λ

dA −∫A

ψ∣∣λ·βλψ

∣∣β dA

= −∫A

ψ∣∣λ·λβ ψ

∣∣β dA −∫A

1R2

εκλεβλψ,κψ∣∣β dA

= −∫A

(�ψ)∣∣βψ∣∣β dA − 1

R2

∫A

ψ∣∣λ ψ∣∣λ dA

(3.16.55)= −

∫A

((�ψ) ψ

∣∣β )∣∣β

dA +∫A

(�ψ)2 dA

− 1R2

∫A

(ψ∣∣λ ψ)∣∣

λdA + 1

R2

∫A

ψ�ψdA

=∫A

[(�ψ)2 + 1

R2ψ�ψ

]dA ≡

∫A

(ψ��ψ+ 1

R2ψ�ψ

)dA.

Finally,∫A

εαλεβ·κψ∣∣λ·β ψ|κα dA =

∫A

εαλεβκψ∣∣λ·β ψ

∣∣ακ dA

=∫A

(εαλε

βκψ∣∣λ·βψ∣∣α)∣∣κ dA −

∫A

εαλεβκψ∣∣λβκψ|α dA

= −12εαλε

βκ(ψ∣∣λ·βκ − ψ

∣∣λ·κβ )ψ|α dA.

Interchanging the order of covariant differentiation in the second term, we obtain∫A

εαλεβ·κψ∣∣λ·β ψ|κα dA = 1

2R2

∫A

εαλεβκεµλεκβψ,µψ|α dA

= − 12R2

∫A

2δµαψ∣∣αψ,µ dA = − 1

R2

∫A

ψ|µψ|µ dA (3.16.56)

= − 1R2

∫A

(ψ|µψ)∣∣µ

dA + 1R2

∫A

ψ�ψdA = 1R2

∫A

ψ�ψdA.


Using these results, we may rewrite (3.16.52) to yield

∫A

θαβθαβ dA =

∫A

[(�ϕ)2 + 1

R2ϕ�ϕ − 2

Rw�ϕ + 2

R2w2

+ 12

(�ψ)2 + 12R2

ψ�ψ+ 12R2

ψ�ψ

]dA (3.16.57)

=∫A

[(�ϕ)2 + 1

R2ϕ�ϕ − 2

Rw�ϕ + 2

R2w2 + 1

2(�ψ)2 + 1

R2ψ�ψ

]dA.

We now return to (3.16.48) and consider the term

∫A

(θκκ)2 dA =

∫A

(�ϕ − 2

Rw)2

dA, (3.16.58)

so that the membrane energy is given by

P(m)2 [u, λ] = Eh

2 (1 − υ 2)

∫A

[(�ϕ)2 + 1 − υ

R2ϕ�ϕ − 2 (1 + υ)

Rw�ϕ

(3.16.59)

+ 2 (1 + υ)R2

w2 + 12

(1 − υ) �ψ

(�ψ+ 2

R2ψ

)]dA.

We shall now proceed with the evaluation of the bending terms in (3.16.48).First, consider∫

A

ραβραβ dA =

∫A

[w∣∣αβ + 1

Rϕ∣∣αβ + 1

2R

(εαλψ

∣∣λ·β + εβλψ∣∣λα

)](3.16.60)

×[

w∣∣αβ + 1

Rϕ∣∣αβ + 1

2R

(εα·κψ∣∣κβ + εβ

·κψ|κα )] dA.

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 toshow that the coupling terms between ψ and w vanish. To show this, consider theterm∫

A

w∣∣αβε

α·κψ∣∣κβ dA =

∫A

w∣∣αβε

ακ ψ∣∣β·κ dA =

∫A

(εακψ

∣∣β·κw,α

)∣∣β

dA −∫A

εακψ

∣∣∣·βκβw,α dA.

Applying the divergence theorem to the first integral, this term vanishes on a closedsurface. Changing the order of covariant differentiation in the second term, weobtain∫A

w∣∣αβε

α·κψ∣∣κβ dA = −

∫A

εακψ

∣∣∣·ββ·κw,α dA − 1R2

∫εακελβεκβψ,λw,α dA

= −∫A

(εακ�ψw,α

)∣∣κ

dA +∫A

εακ�ψw |ακ dA − 1R2

∫A

εακψ,κw,α dA.

192 Applications

Applying the divergence theorem to the first of these integrals, this term vanishesbecause we are dealing with a closed surface. The second integral vanishes becausew |ακ is symmetric in α and κ. Hence, we are left only with the third term, which canbe rewritten as

− 1R2

∫A

(εακψw,α)∣∣κ

dA + 1R2

∫A

εακψw |ακ dA = 0,

where the first term vanishes on a closed surface and the second term vanishes dueto symmetry in α and κ of w |ακ . Our result is thus∫

A

w∣∣αβ εα

·κψ∣∣κβ dA = 0. (3.16.61)

This means that (3.16.60) can be rewritten in the form∫A

ραβραβ dA =

∫A

[(w∣∣αβ + 1

Rϕ∣∣αβ

)(w∣∣αβ + 1

Rϕ∣∣αβ

)

+ 14R2

(εαλψ

∣∣λ·β + εβλψ∣∣λ·α) (εα

·κψ∣∣κβ + εβ

κψ|κα)]

dA(3.16.62)

=∫A

[w∣∣αβ w

∣∣αβ + 2R

w∣∣αβ ϕ∣∣αβ+ 1

R2ϕ∣∣αβ ϕ∣∣αβ

+ 12R2

(ψ∣∣λ·β ψ

∣∣∣·βλ + εαλεβ·κψ∣∣λ·β ψ|κα

)]dA,

where the terms involving ψ have been evaluated in (3.16.55) and (3.16.56). Theterm containing only ϕ is given in (3.16.53) or (3.16.54), which are obviously alsovalid when ϕ is replaced by w.

Completely analogously, we obtain∫A

w∣∣αβw

∣∣αβ dA =∫A

�ϕ(�w + w

R2

)dA =

∫A

�w(�ϕ + ϕ

R2

)dA,(3.16.63)

so that (3.16.62) can be rewritten to yield∫A

ραβραβ dA =

∫A

[�w

(�w + w

R2

)+ 1

R�w

(�ϕ + ϕ

R2

)

+ 1R

�ϕ(�w + w

R2

)+ 1

R2�ϕ(�ϕ + ϕ

R2

)(3.16.64)

+ 12R2

�ψ(�ψ+ ψ

R2

)+ 1

2R4ψ�ψ

]dA.

We now return to (3.16.48) and consider the term

∫A

(ρκκ)2 dA =

∫A

(�w + 1

R�ϕ

)2

dA, (3.16.65)


so that the bending energy is given by

P(b)2 [u, λ] = Eh2

2 (1 − υ 2)h2

12

∫A

[(1 − υ)

2R2�ψ

(�ψ+ 2ψ

R2

)(3.16.66)

+(

�w + 1R

�ϕ

)2

+ (1 − υ)(

�w + 1R

�ϕ

)( wR2

+ ϕ

R3

)]dA.

Next, we must determine the contribution P2 [u, λ] of the stresses in the funda-mental state. From (3.16.48), consider the integral∫

A

[(θαβ − ωαβ)

(θαβ − ωαβ

)+ ϕαϕα]

dA. (3.16.67)

From (3.16.5) and (3.16.6), we obtain

θαβ − ωαβ = uα|β − bαβ w = uα|β − 1R

aαβw, (3.16.68)

and using (3.16.38), we can rewrite this expression in the form

θαβ − ωαβ = ϕ∣∣αβ + εαλψ

∣∣λ·β − 1R

aαβw. (3.16.69)

The integral (3.16.67) can now be written as∫A

[(ϕ∣∣αβ + εαλψ

∣∣λ·β − 1R

aαβw)(

ϕ∣∣αβ + εα

·κψ∣∣κβ − 1

Raαβw

)(3.16.70)

+(

w,α + 1R

ϕ,α + 1R

εαλψ∣∣λ)(w

∣∣∣∣α + 1R

ϕ

∣∣∣∣ α+ 1R

εα·κψ|κ

)]dA,

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 andw vanish, so we can rewrite this integral in the form∫

A

[(ϕ∣∣αβ − 1

Raαβw

)(ϕ∣∣αβ − 1

Raαβw

)+(

w |α + 1R

ϕ |α)(

w |α + 1R

ϕ |α)

(3.16.71)+ ψ

∣∣κ·βψ∣∣·βκ + 1R2

ψ∣∣λ ψ|λ

]dA,

Consider∫A

(ϕ∣∣αβ ϕ

∣∣αβ − 2wR

ϕ∣∣αα + 2

R2w2)

dA =∫A

[(�ϕ)

(�ϕ+ 1

R2ϕ

)− 2w

R�ϕ+ 2

R2w2]

dA,

where we have made use of (3.16.54). For the second term in (3.16.71), we obtain∫A

(w |αw |α+ 2

Rw |αϕ |α+ 1

R2ϕ |α ϕ |α

)dA

=∫A

[(w |α w)

∣∣α

− w�w + 2R

(ϕw |α )∣∣α

− 2R

w�ϕ + 1R2

(ϕ |α ϕ)∣∣α

− 1R2

ϕ�ϕ

]dA

194 Applications

= −∫A

[w�w + 2

Rw�ϕ + 1

R2ϕ�ϕ

]dA.

For the last two terms in (3.16.71), we obtain∫A

(ψ∣∣κ·β ψ

∣∣·βκ + 1

R2ψ∣∣λψ|λ

)dA =

∫A

[(�ψ)

(�ψ+ 1

R2ψ

)− 1

R2ψ�ψ

]dA

=∫A

(�ψ)2 dA,

where we have made use of (3.16.55). The contribution of the stresses in the funda-mental state to the energy functional is thus given by

P(s)2 [u; λ] = Eh2

2cRλ

∫A

[(�ϕ)2 + 1

R2ϕ�ϕ − 2w

R�ϕ + 2

R2w2

− w�w − 2R

w�ϕ − 1R2

ϕ�ϕ + (�ψ)2]

dA (3.16.72)

= −Eh2

2cRλ2∫A

[(�ϕ)2 − 4w

R�ϕ − w�w + 2

R2w2 + (�ψ)2

]dA.

Finally, we shall consider the last term in (3.16.48),∫A

[wθα

α − uαϕα

]dA =

∫A

[w(

�ϕ − 2R

w)

−(

w

∣∣∣∣α + 1R

ϕ

∣∣∣∣ α + 1R

εα·λψ∣∣∣∣λ)

(ϕ,α + εακψ|κ )]

dA(3.16.73)

=∫A

[w�ϕ − 2

Rw2 −

(w |α + 1

Rϕ |α)

ϕ |α − 1R

ψ|κψ|κ]

dA

=∫A

[w�ϕ − 2

Rw2 + w�ϕ + 1

Rϕ�ϕ + 1

Rψ�ψ

]dA.

The second variation of the potential energy is now given by

P2 [u; λ] = Eh2 (1 − υ 2)

∫A

[(�ϕ)2 + 1 − υ

R2ϕ�ϕ − 2 (1 + υ)

Rw�ϕ

+ 2 (1 + υ)R2

w2 + 12

(1 − υ) �ψ

(�ψ+ 2

R2ϕ

)

+ h2

12

{1 − υ

2R2�ψ

(�ψ+ 2ψ

R2

)+(

�w + 1R

�ϕ

)2

(3.16.74)

+ (1 − υ)(

�w + 1R

�ϕ

)( wR2

+ ϕ

R3

)}

− (1−υ2) σ

E

{(�ϕ)2+ 2

R2ϕ�ϕ−w�w − 2

R2w2 + �ψ

(�ψ+ 2

Rψ

)}]dA.


To be in the elastic range, we must have σ/E <<< 1(σ/E = O

(10−3

))for most

engineering materials, which means that the underlined terms in (3.16.74) may beneglected, implying a relative error of O (σ/E) compared to unity. Further, we mayneglect the term with ψ in the third line compared to that in the second line, whichintroduces a relative error of O

(h2/R2

), which is smaller than the error in shell the-

ory. Similarly, we may neglect the term with (�ϕ)2 in the third line compared tothe corresponding term in the first line, again with a relative error of O

(h2/R2

). To

show that the term 2R−1h2�w�ϕ in the third line may also be neglected, we employthe inequality

2hR

∫A

h�w�ϕ dA ≤ hR

∫A

{(h�w)2 + (�ϕ)2

}dA, (3.16.75)

which shows that this term may be neglected in comparison with the sum of theterms h2 (�w)2 in the third line and (�ϕ)2 in the first line, introducing a relativeerror of O (h/R), which is consistent with the error inherent to the shell equations.However, we shall accept this error. To discuss the term h2w�w/R2 in the fourthline, we consider the inequality

2∫A

h2

R2w�w dA ≤ h

R

∫A

{w2

R2+ h2 (�w)2

}dA, (3.16.76)

which shows that this term may be neglected compared to the sum of the terms withw2/R2 in the second line and h2 (�w)2 in the third line, again with a relative error ofO (h/R). The terms

(h2/R2

)w�ϕ/R and

(h2/R2

)ϕ�ϕ/R2 in the fourth line may be

neglected compared to the terms w�ϕ/R and ϕ�ϕ/R2 in the first line with a relativeerror of O

(h2/R2

). Using the relation

∫A

ϕ�w dA =∫A

w�ϕ dA, ∀ϕ, w|w, ϕ ∈ C2, (3.16.77)

it follows that the term h2ϕ�w/R3 in the fourth line may also be neglected with arelative error of O

(h2/R2

). This means that the second variation (3.16.74) can be

simplified to yield

P2 [u; λ] = Eh2 (1 − υ 2)

∫A

[(�ϕ)2 + 1 − υ

R2ϕ�ϕ − 2 (1 + υ)

Rw�ϕ

+ 2 (1 + υ)R2

w2 + 12

(1 − υ) �ψ

(�ψ+ 2

R2ψ

)

+ h2

12(�w)2 + (1 − υ 2) σ

Ew�w

]dA[

1 + O(

hR

)]. (3.16.78)

A necessary condition for stability is that P2 [u; λ] is positive-definite. Because ψ

appears uncoupled from ϕ and w, we shall first consider the contribution of the

196 Applications

terms with ψ, i.e., we consider∫A

�ψ

(�ψ+ 2

R2ψ

)dA. (3.16.79)

Any continuously differentiable function on a spherical surface can be expanded ina series of spherical surface harmonics. Hence, we can write

ψ =∞∑

n = 0

CnSn (xα) , (3.16.80)

where Sn (xα) is a spherical surface harmonic of integral degree n, characterized bythe differential equation

�Sn (xα) = −n (n + 1)R2

Sn (xα) . (3.16.81)

Sn (xα) can be represented by

Sn (xα) = a0Pn (cos θ) +n∑

m= 1

(am cos mβ + bm sin mβ)Pmn (cos θ) , (3.16.82)

where θ and β are surface coordinates (see Figure 3.16.1).Pn (cos θ) is a Legendre polynomial of degree n, and Pm

n (cos θ) is a Legendrefunction of the first kind, of degree n and order m. Further, S0 (xα) = constant.

Let n1 and n2 be two distinct integral numbers, then spherical surface harmonicsof degree n1 and n2, respectively, satisfy the orthogonality relation∫

A

Sn1 (xα) Sn2 (xα) dA = 0. (3.16.83)

Using this relation and (3.16.82), we can rewrite (3.16.79) to yield

1R4

∫A

∞∑n = 0

{[n (n + 1) − 1]2 − 1

}C2

nS2n dA. (3.16.84)

z

y

x

Rθ

β

Figure 3.16.1


This expression vanishes for n = 0 and n = 1 and is positive for n > 1. Because n = 1implies a rigid body motion, we are only interested in values of n > 1. This meansthat the terms with ψ are positive-definite, so a minimum of P2 [u; λ] can only beobtained when the terms with ψ vanish, i.e.,

�ψ = 0 or �ψ+ 2R2

ψ = 0. (3.16.85)

Later, we shall see that only the first of these equations holds. From (3.16.6), therotation in the tangent plane is

� = 12εαβωαβ = 1

2εαβ(ϕ∣∣βα + εβλψ

∣∣λα − ϕ

∣∣αβ − εαλψ

∣∣λβ

) = −12ψ∣∣λλ = −1

2�ψ,

(3.16.86)so that the first of the equations of (3.16.86) means that the rotation � around thenormal vanishes identically in all buckling modes.

We may now restrict our discussion to the functional

P2 [u, ω; λ] = Eh2 (1 − υ 2)

∫A

[(�ϕ)2 + 1 − υ

R2ϕ�ϕ − 2 (1 + υ)

Rw�ϕ

+2 (1 + υ)R2

w2 +h2

12(�w)2 + (1 − υ 2) σ

Ew�w

]dA. (3.16.87)

A necessary condition for a minimum value of this functional is that its variationwith respect to ϕ and w vanishes. For the reduction of the various terms, we needthe following results:∫

A

�ϕ�δϕ dA =∫A

�ϕδϕ∣∣αα dA = −

∫A

�ϕ |α δϕ |α dA

= −∫A

�ϕ |α δϕ |α dA =∫A

�ϕ∣∣αα δϕ dA =

∫A

��ϕδϕ dA (3.16.88)

delta∫A

w�ϕ dA =∫A

�ϕδw dA +∫A

wδϕ∣∣αα dA =

∫A

�ϕδw dA +∫A

�wδϕ dA.

Our result can now be written as∫A

[2��ϕ δϕ + 2 (1 − υ)

R2�ϕδϕ − 2 (1 + υ)

R�ϕδw

−2 (1 + υ)R

�wδϕ + 2 (1 + υ)R2

wδw + 2h2

12��wδw (3.16.89)

+2(1 − υ 2) σ

E�wδw

]dA = 0, ∀δϕ, δw,

which yields the following equations of neutral equilibrium,

��ϕ + 1 − υ

R2�ϕ − 1 + υ

R�w = 0

(3.16.90)h2

12��w + (1 − υ 2) σ

E�w + (1 + υ)

R2w − (1 + υ)

R�ϕ = 0.

198 Applications

Expand ϕ and w in a series of spherical surface harmonics,

ϕ = R∞∑

n = 0

DnSn, w =∞∑

n = 0

CnSn. (3.16.91)

Assuming that these series can be differentiated term-wise four times and using therelation (3.16.81), we obtain from (3.16.90) the following equations:{

[n (n + 1)]2 − (1 − υ) n (n + 1)}

Dn + (1 − υ) n (n + 1) Cn = 0(3.16.92)

(1 − υ) n (n + 1) Dn +{

h2

12R2[n (n + 1)]2 − (1 − υ 2) σ

En (n + 1)

}Cn = 0.

The condition for a non-trivial solution is(1 − υ 2) [n (n + 1) − 2] + [n (n + 1) − (1 − υ)]

(3.16.93)

×[

h2

12R2n2 (n + 1)2 − (1 − υ 2) σ

En (n + 1)

]= 0,

which yields

σ

E= n (n + 1) − 2

n (n + 1) [n (n + 1) − (1 − υ)]+ h2

12R2

n (n + 1)1 − υ 2

. (3.16.94)

For values of n of O(1), the first term on the right-hand side is of order unity, whichmeans σ/E = O (1). Because we must stay in the elastic range, we require σ/E � 1,which is only possible for n � 1. This means that (3.16.94) can be simplified to yield

σ

E= 1

n (n + 1)+ h2

12R2

n (n + 1)1 − υ2

. (3.16.95)

It follows that because n � 1, the wavelength L of the deformation pattern is small,i.e., L/R � 1, which means that we can restrict ourselves to shallow shell theory.The functional (3.16.78) can now further be simplified to yield

P2 [u; λ] = Eh2 (1 − υ 2)

∫A

[(�ϕ)2 − 2 (1 + υ)

wR

�ϕ + 12

(1 − υ) (�ψ) 2

(3.16.96)

+h2

12(�w)2 + (1 − υ 2) σ

Ew�w

]dA[

1 + O(

hR

+ L2

R2

)].

Omitting the error term, we rewrite this functional in the form

P2 [u; λ] = Eh2 (1 − υ 2)

∫A

{[�ϕ − (1 + υ)

wR

]2+ 1

2(1 − υ) (�ψ)2

(3.16.97)

− (1 + υ2) w2

R2+ h2

12(�w)2 + (1 − υ 2) σ

Ew�w

}dA.

Notice that for the present simplified functional, we did not need any a prioriassumptions with respect to the wavelength of the deformation pattern. The con-dition that L/R � 1 follows from the condition that σ/E � 1 to stay in the elasticrange.


In our present functional, we only have quadratic terms with positive constants,except for the load term. For the functional to become semi-positive-definite, theterm with ψmust vanish, which leads to the first of the equations of (3.16.85), andthe term with ϕ must also vanish, which yields the equation

�ϕ = (1 + υ)wR

. (3.16.98)

This admits of a single-valued and non-singular solution for the potential ϕ if theintegral of the normal deflection w over the entire surface of the shell vanishes forall buckling modes. To show this, apply the operator

∫A () dA to both sides of this

equation. The left-hand member then vanishes (divergence theorem), so the right-hand side must also vanish. This condition is satisfied because w =∑∞

n CnSn (xα)and

∫A SndA = 0 (n ≥ 1). Apart from irrelevant arbitrary constants, the solution to

(3.16.98) is

ϕ (xα) = −∞∑n

1 + υ

n (n + 1)RCnSn (xα) . (3.16.99)

Let us now return to the functional (3.16.97). The remaining terms contain onlyw, so that a stationary value of P2 [u; λ] is obtained by putting the variation withrespect to w equal to zero. Using (3.16.91), the condition for a non-trivial solutionyields again the expression for σ/E given in (3.16.96). To obtain the critical valueof σ/E, we minimize this expression with respect to n (n + 1), which yields (for acontinuous variable n∗)

n∗ (n∗ + 1) = 2cRh

, (3.16.100)

so that

σcr = EhcR

. (3.16.101)

Notice that this is in complete agreement with our earlier result of (3.15.30).The positive root n∗ of (3.16.100) is in general non-integral. However, the near-

est integers do not differ more from n∗ than O(n∗−1

), which means a relative error

of O (h/R), and which is negligible within the framework of shell theory. The crit-ical load factor λ1 is therefore always equal to unity within the limits of accu-racy of shell theory. The associated buckling mode is any spherical surface har-monic of integral degree [n∗] or [n∗] + 1 associated with the integer that yields thesmallest value of the right-hand member of (3.16.95). The corresponding bucklingmode is

wn = ah Sn (xα) , (3.16.102)

where A is the amplitude factor and Sn (xα) is given in (3.16.82). This representationshows that at the same value of the load factor λ1, there are 2n + 1 independentbuckling modes.

200 Applications

In the special case that R/h is such that (3.16.100) yields two integer values forn∗, the right-hand side of (3.16.95) is the same for both values. In that case, thecorresponding buckling mode is given by

wn = ah Sn (xα) + a∗h Sn−1 (xα) , (3.16.103)

so that in this case, we have 4n independent buckling modes. Further, we have

ψ = 0, ∀n. (3.16.104)

A necessary condition for stability at the critical bifurcation load is that thethird variation vanishes identically for any linear combination of buckling modes.Returning to (3.16.17) and remembering that γαβ = θαβ + 1

2 w,αw,β, we see that theonly contribution to P3 [u] originates from the membrane strains,

P3[u] = Eh2(1 − υ 2)

∫A

[(1 − υ)θαβw

∣∣αw∣∣β + υθα

αw,κw∣∣κ] dA

= Eh2(1 − υ 2)

∫A

{(1 − υ)

[12

(uα|β + uβ|α) − aαβ

wR

]w∣∣αw∣∣β (3.16.105)

+ υ(uα∣∣α

− 2wR

)w,κw∣∣κ} dA.

For the evaluation of this functional, we recall that

ϕ = − (1 − υ)Rn(n + 1)

w, ψ = 0, uα = ϕ,α = − (1 − υ)Rn(n + 1)

w,α, (3.16.106)

so the functional can be rewritten to yield

P3[ahuh] = Eh2(1 − υ 2)

∫A

{(1 − υ)

[− (1 − υ)R

n(n + 1)w∣∣αβ

− aαβ

wR

]w∣∣αw∣∣β

+υ

[− (1 − υ)R

n(n + 1)�w − 2

wR

]w,κw

∣∣κ} dA

(3.16.107)

= Eh2(1 − υ 2)

∫A

[− (1 − υ 2)R

n(n + 1)w∣∣αβ

w∣∣αw∣∣β − (1 − υ)

wR

w∣∣βw∣∣β

− υ(1 − υ)wR

w,κw∣∣κ] dA,

where we have used (3.16.81).Let us now consider the various terms∫

A

w∣∣αβ

w∣∣αw∣∣β dA =

∫A

(w∣∣αw∣∣α)∣∣βw∣∣β dA −

∫A

w∣∣αw∣∣αβw∣∣β dA

=∫A

[(w∣∣αw∣∣αw∣∣β) ∣∣

β− w

∣∣αw∣∣αw∣∣ββ

]dA −

∫A

w∣∣αβ

w∣∣αw∣∣β dA.

3.17 Buckling of circular cylindrical shells 201

so that (because the first term vanishes on a closed surface)

∫w∣∣αβ

w∣∣αw∣∣β = −1

2

∫A

w∣∣αw∣∣α�w dA = n (n + 1)

2R2

∫A

w∣∣αw∣∣αw dA

(3.16.108)

= n (n + 1)4R2

∫A

[(w2w

∣∣α) ∣∣α

− w2�w]

dA = n2 (n + 1)4R4

∫A

w3 dA

∫A

w∣∣β w∣∣β w dA = n (n + 1)

2R2

∫A

w3 dA. (3.16.109)

The final result for P3 [ahuh] is now

P3 [ahuh] = −38

EhR3

n (n + 1)∫A

w3 dA = −34

cER2

∫A

w3 dA, (3.16.110)

where we have made use of (3.16.100). For the evaluation of this term and a furtherdiscussion of the problem we refer to Koiter’s paper.†

The result is that P3 [ahuh] vanishes if the degree of the buckling mode is odd,but it does not vanish when the degree of the buckling mode is even. This meansthat in the latter case, the critical bifurcation point is unstable. In the case of an evendegree of the buckling mode, stability is governed by the sign of A4; cf. (3.4.38) and(3.4.53). It turns out that A4 is negative for both even and odd degrees of bucklingmodes, so the critical bifurcation point is unstable.

3.17 Buckling of circular cylindrical shells

In the following discussion, we shall assume that the fundamental state is a mem-brane state of stress. (This assumption is usually violated in experiments.) Thesecond variation of the energy functional is then given by

P2 [u] =∫A

{Eh

2 (1 − υ 2)

[(1 − υ) θαβθ

αβ + υ (θκκ)2 + h2

12

{(1 − υ) ραβρ

αβ + υ (ρκκ)2}]

+ 12

Nαβ [(θκ·α − ωκ

·α) (θκβ − ωκβ) + ϕαϕβ] (3.17.1)

+ 12

P (wθαα − uαϕα)

}dA.

It is convenient to employ Cartesian coordinates (x, y) on the surface, and let n bethe unit vector normal to this surface and positive in the outward direction. Let Rbe the radius of the shell, and then we can introduce dimensionless coordinates so

† W.T. Koiter, The nonlinear buckling problem of a complete spherical shell under uniform externalpressure. Proc. Kon Ned. Ak Wit., Series B, 72, 40–123 (1969).

202 Applications

that

∂

∂x( ) ≡ 1

R( )′

,∂

∂y( ) ≡ 1

R( ) . (3.17.2)

In the following, we shall use the Koiter-Sanders strain-displacement relations forcylindrical shells.†

θxx = 1R

u′ θyy = 1R

(v· + w) θxy = 12R

(u· + v′)

ωxy = 12

(v′ − u·) ϕx = 1R

w′ ϕy = 1R

(w· − v)

ρxx = 1R2

w′′ ρyy = 1R2

(w· − v)·ρxy = 1

R2

(w′· − 3

4v′ + 1

4u·). (3.17.3)

Because we are dealing with Cartesian coordinates, the functional (3.17.1) can read-ily be rewritten to yield

P2[u] =∫A

Eh2 (1 − υ 2)

{θ2

xx + θ2yy + 2υθxxθyy + 2 (1 − υ) θ2

xy

+ h2

12

[ρ2

xx + ρ2yy + 2υρxxρyy + 2 (1 − υ) ρ2

xy

]}dA

(3.17.4)

+∫A

{12

Nxx

[θ2

xx + (θyx − ωyx)2 + ϕ2x

]+ 1

2Nyy

[θ2

yy + (θxy − ωxy)2 + ϕ2y

]

+12

P [w (θxx + θyy) − uϕx − vϕy]}

dA.

The term with Nxy is absent because there are no shear forces. The terms Nxxθ2xx and

Nyyθ2yy can be neglected compared to the terms Eh

(θ2

xx + θ2yy + 2υθxxθyy

).

Koiter‡ has shown that within the context of the classical theory of shells, thechanges of curvature can be altered by adding terms of the order of the strain inthe middle surface divided by the minimum radius of curvature without altering theaccuracy of the theory. We now consider the following modified energy functional,

P∗2 [u] = P2 [u] +

∫A

Eh3

24 (1 − υ 2)

[−2 (1 − υ)

Rθxxρyy + 2

Rθyy (ρxx + ρyy)

(3.17.5)+ 1

R2θ 2

yy + 2 (1 − υ)R

θxyρxy − 3 (1 − υ)2R2

θ 2xy

]dA.

Koiter has shown that an upper bound for the change in strain energy is∣∣P∗2 [u] − P2[u]

∣∣ < 0.467hR

P2[u][

1 + O(

hR

)]. (3.17.6)

† Cf. W. T. Koiter, Summary of equations for modified simplistic possible accurate linear theory ofthin circular cylindrical shells. Rep. Lab. For Appl. Mech. Delft, 442.

‡ W. T. Koiter, A consistent first approximation in the general theory of thin elastic shells. Proc.IUTAM Symp. On Theory of Thin Elastic Shells (Delft, Aug. 1959; North-Holland Publ. Co., 1960),pp. 12–33.


Substitution of the relations (3.17.3) into this modified energy functional yields

P∗2 [u] =

∫A

Eh2 (1 − υ2) R2

{u′2 + (v· + w)2 + 2υu′(v· + w) + 1

2(1 − υ) (u· + v′)2

+ h2

12R2

[w′′ 2 + w··2 + 2υw′′w·· + 2 (1 − υ) w′·2 + 2w (w′′ + w··) + w2]

+ 2(1 − υ)h2

12R2[u′v· − u·v′ + u·w′· − u′w·· + v·w′′ − v′w′·]

}dA (3.17.7)

+∫A

{Nxx

2R2

(v′2 + w′2)+ Nyy

2R2

{u·2 + (w· − v)2

]

+ P2R

(u′w − uw′ + v·w − vw· + v2 + w2)}dA

This modified functional has the advantage that the first term with h2/R2 containsonly the displacement w, whereas the second term with h2/R2 contains only expres-sions of the divergence type, which do not alter the differential equation. Consider,e.g.,

R−1∫A

(u′v· − u·v′) dA =∫∂S

(u′vvy − u·vvx) ds

−∫A

(u′· − u·′)v dA = −∫∂S

u·vvx ds,

because at x = 0 and x = , vy = 0. Alternatively, we can also write

R−1∫

(u′v· − u·v′) dA =∫∂S

(uv·vx − uv′vy) ds

−∫A

(v·′ − v′·) u dA =∫∂S

uv·vx ds,

i.e., these terms contribute to the boundary conditions.Let us now introduce the following load parameters,

λx = (1 − υ2) Nxx

Eh, λy = (1 − υ2) Nyy

Eh, Nyy = −PR; (3.17.8)

then the last integral in (3.17.7) can be rewritten in the formEh

2 (1 − υ 2) R2

∫A

[λx(v′ 2 + w′ 2)

(3.17.9)+ λy

(u·2 + w·2 − w·v − u′w + uw′ − v·w − w2)] dA.

A necessary condition for neutral equilibrium is that

P11[u, ζ] = 0 = Eh2 (1 − υ 2) R2

∫A

{2u′ξ′ + 2 (v· + w) (η· + ζ)

+ 2υu′ (η· + ζ) + 2υξ′ (v· + w) + (1 − υ) (u· + v′) (ξ· + η′)

+ h2

12R2[2w′′ζ ′′ + 2w··ζ ·· + 2υw′′ζ ·· + 2υζ ′′w··

204 Applications

+ 4 (1 − υ) w′·ζ ′· + 2w (ζ ′′ + ζ ··) + 2ζ(w′′ + w··) + 2wζ](3.17.10)

+ 2 (1 − υ)12

h2

R2[u′η· + ξ′v· − u·η′ − ξ·v′

+ u·ζ ′· + ξ·w′· − u′ζ ·· − ξ′w·· + v·ζ ′′ + η·w′′ − v′ζ ′· − η′w′·]

+ 2λx (v′η′ + w′ζ ′) + λy (2u·ξ· + 2w·ζ· − w·η− ζ ·v

− u′ζ− ξ′w + uζ ′ + ξw′ − v·ζ− η·w − 2wζ)}

dA

for all kinematically admissible displacement fields (ξ, η, ζ).Let us first consider all terms with ξ,

R2

2π∫0

∫0

{[u′ + υ (v· + w) + (1 − υ)

12h2

R2{v· − w··} − 1

2λyw

]ξ′

+[

12

(1 − υ) (u· + v′) + (1 − υ)12

h2

R2(w′· − v′) + λyu·

]ξ· (3.17.11)

12λyw′ξ

}dθ dx/R = 0.


−R2

2π∫0

∫0

[u′′ + υ (v·′ + w′) + (1 − υ)

12h2

R2(v·′ − w··′) − 1

2λyw′

+ 12

(1 − υ) (u·· + v′·) + 1 − υ

12h2

R2(w′·· − v′·) + λy

(u·· − 1

2w′)]

dθ dx/R(3.17.12)

+ R2

2π∫0

[u′ + υ (v· + w) + (1 − υ)

12h2

R2(v· − w··) − 1

2λyw

]ξ

∣∣∣∣∣∣

0

dθ

+ R2

∫0

[12

(1 − υ) (u· + v′) + (1 − υ)12

h2

R2w′· − v′ + λyu

]ξ

∣∣∣∣2π

0

dxR

= 0,

which yields the equilibrium equation

u′′ + 12

(1 − υ) u·· + 12

(1 + υ) v·′ + υw′ + λy (u·· − w′) = 0. (3.17.13)

In the following, we shall consider the case of a simply supported edge at x = 0and x = , i.e., we have the kinematic conditions

v = w = 0 at x = 0, x = . (3.17.14)

We then obtain from (3.17.12) the dynamic boundary conditions

u′ + υ (v· + w) + 1 − υ

12h2

R2(v· − w··) − 1

2λyw = 0 at x = 0, x = (3.17.15)

which with (3.17.14) reduce to

u′ = 0 at x = 0, x = . (3.17.16)


The second line integral in (3.17.12) vanishes identically because all the functionsmust be single-valued at θ = 0, and θ = 2π.

Let us now consider the terms with η,

R2

2π∫0

∫0

{[v· + w + υu′ + (1 − υ)

12h2

R2(u′ + w′′) − 1

2λyw

]η·

×[+1

2(1 − υ) (u· + v′) + (1 − υ)

12h2

R2(−u· − w′·) + λxv′

]η′ (3.17.17)

− 12λyw·η

}dθ

dxR

= 0.


−R2

2π∫0

∫0

[v·· + w· + υu′· + (1 − υ)

12h2

R2(u′· + w′′·) − 1

2λyw·

+ 12

(1 − υ) (u·′ + v′′) − (1 − υ)12

h2

R2(u·′ + w′·′) + λxv′′ + 1

2λyw·

]ηdθ

dxR

(3.17.18)

+ R2

∫0

[v· + w + υu′ + (1 − υ)

12h2

R2(u′ + w′′) − 1

2λyw

]η

∣∣∣∣∣∣2π

0

dxR

+ R2

2π∫0

[12

(1 − υ) (u· + v′) − (1 − υ)12

h2

R2(u· + w′·) + λxv′

]η

∣∣∣∣

0dθ = 0,

which yields the equilibrium equation

v·· + 12

(1 − υ) v′′ + 12

(1 + υ) u′· + w· + λxv′′ = 0, (3.17.19)

and in the case of a simply supported edge, both line integrals vanish.Finally, we consider the terms with ζ,

2π∫0

∫0

{[v· + w + υu′ + h2

12R2(w′′ + w·· + w) + 1

2λy (−u′ − v· − 2w)

]ζ

+[λxw′ + 1

2λyu]

ζ ′ + λy

(w· − 1

2v)

ζ ·

+ h2

12R2[w′′ + υw·· + w + (1 − υ) v·] ζ ′′ (3.17.20)

+ h2

12R2[w·· + υw′′ − (1 − υ) u′ + w] ζ ··

+ h2

12R2[2 (1 − υ) w′· + (1 − υ) u· − (1 − υ) v′] ζ ′·

}dθ

dxR

= 0.

206 Applications

By integration by parts once, we obtain

2π∫0

∫0

{[v· + w + υu′ + h2

12R2(w′′ + w·· + w) − λxw′′ − λy (u′ + w·· + w)

]ζ

− h2

12R2[w′′′ + υw··′ + w′ + (1 − υ) v·′] ζ ′

− h2

12R2[w··· + υw′′· − (1 − υ) u·′ + w· + 2 (1 − υ) w′·′ + (1 − υ) u·′

− (1 − υ) v′′] ζ ·}

dθdxR

(3.17.21)

+2π∫

0

(λxw′ + 1

2λyu)

ζ

∣∣∣∣∣∣

0

dθ +∫

0

λy

(w· − 1

2v)

ζ

∣∣∣∣∣∣2π

0

dxR

+ h2

12R2

2π∫0

{[w′′ + υw·· + w + (1 − υ) v·] ζ ′

+ [2 (1 − υ) w′· + (1 − υ) u· − (1 − υ) v′] ζ ′} |0 dθ

+ h2

12R2

∫0

(w·· + υw′′ − (1 − υ) u′ + w) ζ ·

∣∣∣∣∣∣2π

0

dxR

= 0,

where the second and the fourth line integrals vanish because the displacements andtheir derivatives are periodic in the y-direction. For simply supported edges, the firstline integral and the second term in the third line integral vanish. Integrating byparts once again, we obtain

2π∫0

∫0

{v· + w + υu′ + h2

12R2[w′′ + w·· + w + w′′′′ + υw··′′ + w′′ + (1 − υ) v·′′

+ w···· + υw′′·· − (1 − υ) u′·· + w·· + 2 (1 − υ) w′′·· + (1 − υ) u′·· − (1 − υ) v′′·]

− λxw′′ − λy (u′ + w·· + w)}

ζ dθdxR

+ h2

12R2

2π∫0

[w′′ + υw·· + w + (1 − υ) v·] ζ ′∣∣0 dθ (3.17.22)

− h2

12R2

2π∫0

[w′′′ + υw··′ + w′ + (1 − υ) v·′] ζ∣∣0 dθ

− h2

12R2

∫0

[w··· + υw′′ − (1 − υ) u′· + w· + 2 (1 − υ) w′′·

+ (1 − υ) u′· − (1 − υ) v′′] ζ∣∣2π

0

dxR

= 0.


Here, the last line integral vanishes because the displacements and their derivativesare periodic in the y-direction. For simply supported edges, the second of the lineintegrals vanishes. We now obtain the third equilibrium equation,

v· + w + υu′ + h2

12R2[w′′′′ + 2w′′·· + w···· + 2 (w′′ + w··) + w]

(3.17.23)− λxw′′ − λy (u′ + w·· + w) = 0,

and the dynamic boundary conditions

w′′ + υw·· + w + (1 − υ) v· = 0 at x = 0, x = , (3.17.24)

which with (3.17.14) can be simplified to yield

w′′ = 0 at x = 0, x = . (3.17.25)

Notice that (3.17.23) can be written as

v· + υu′ + w + h2

12R2(� + 1)2 w − λxw′′ − λy (u′ + w·· + w) = 0,

where

� = R2 (∂ 2/∂x2 + ∂ 2/∂y2) . (3.17.26)

The equations (3.17.13), (3.17.15), and (3.17.26) are known as the Koiter-MorleyEquations.

We now try to find a solution of these equations of the form

u = Acos (px/R) sin (qy/R)

v = B sin (px/R) cos (qy/R) (3.17.27)

w = C sin (px/R) sin (qy/R) ,

where p = kπR/, k ∈ N+, and q ∈ N

+. These expressions satisfy all the bound-

ary conditions of (3.17.14), (3.17.16), and (3.17.24). Substitution of these expres-

sions into the Koiter-Morley equations yields three homogeneous equations for

the unknown constants A, B, and C. The condition for a non-trivial solution is

then∣∣∣∣∣∣∣∣∣∣

p2 + 12

(1 − υ) q2 + λyq2 12

(1 + υ) pq −υ p + λyp12

(1 + υ) pq12

(1 − υ) p2 + λxp2 −q

−υ p + λyp −q 1 + h2

2R2

(p2 + q2 − 1

)2 + λxp2 + λy(q2 − 1

)

∣∣∣∣∣∣∣∣∣∣= 0.

(3.17.28)After some algebra, we find the following result:

12

(1 − υ)[(

1 − υ2) p4 + h2

12R2

(p2 + q2)2 (p2 + q2 − 1

)2]

+ p2

{12

(1 − υ)(p2 + q2)2 −υ2p2 +

[p2 + 1

2(1 − υ) q2

] [1 + h2

12R2

(p2 + q2 − 1

)2]}λx

208 Applications

+{

12

(1 − υ) q2[(

p2 + q2)2 − 3p2 − q2]

− 12

(1 − υ) (1 − 2υ) p4 (3.17.29)

+ h2

12R2

(p2 + q2 − 1

)2 [12

(1 − υ) p2q2 + q4

]}λy

+ [q2 (q2 − 1)− p2] [1

2(1 − υ) p2 + q2

]λ2

y + · · · = 0.

Because λx, λy � 1, the underlined terms may be neglected compared to the prin-cipal terms. Further, we may neglect the higher order terms in λx and λy, e.g., thecoefficient of λ2

y is of the same order of magnitude as the coefficient of λy, so that theterm with λ2

y may be neglected compared to the term with λy. The resulting equationnow becomes

(1 − υ2)p4 + h2

12R2

(p2 + q2)2 (p2 + q2 − 1

)2 + λxp2[(

p2 + q2)2 + q2]

(3.17.30)+ λyq2

[(p2 + q2)2 − (3p2 + q2)] = 0.

We shall now discus some special cases:

i) q = 0. In this case, the term with λy vanishes, i.e., there is no influence of theexternal pressure P. The resulting equation now reads

(1 − υ2) p4 + h2

12R2p4 (p2 − 1

)2 + λxp6 = 0, (3.17.31)

from which (for p �= 0 )

λx = −1 − υ2

p2− h2

12R2

(p2 − 1

)2p2

. (3.17.32)

To stay in the elastic range, we must have large values of p, which means thatwe can neglect unity with respect to p2, so that the expression for λx can besimplified to yield

λx = −1 − υ2

p2− h2

12R2p2. (3.17.33)

Notice that λx is negative, i.e., the cylinder is loaded in compression. The crit-ical value for λx follows from (3.17.33) by minimizing the right-hand side withrespect to p2, which yields

(p2)

min = 2cRh

, (λx)min = − (1 − υ2) hcR

. (3.17.34)

Notice the(p2)

min is indeed a large number, as R/h � 1.ii) q = 1. In this case, the equation becomes

(1 − υ2) p4 + h2

12R2p4 (p2 + 1

)2 + λxp2[(

p2 + 1)2 + 1

](3.17.35)

+ λy(p4 − p2) = 0.


Because λy � 1, the term p4 in its coefficient may be neglected compared to theterm

(1 − υ2

)p4. We must now distinguish various possibilities:

a) p � 1: In this case, we recover the results for q = 0.b) p � 1: The equation can now be reduced to(

1 − υ2) p2 + 2λx − λy = 0, (3.17.36)

which yields (with k = 1)

2λx − λy = − (1 − υ2)p2 = − (1 − υ2) π2R2

2. (3.17.37)

This is the case of Euler buckling of the cylinder, which can easily be seen bysetting λy = 0. In this case, using (3.17.8) we have

F = −2πRNxx = π2E2

πR3h, (3.17.38)

where πR3h is the moment of inertia of the cylinder. When λx = 0, λy is pos-itive, namely,

λy = (1 − υ2) π2R2

2. (3.17.39)

This is a case of buckling under internal uniform pressure P.Using (3.17.8), we can write

F = −PπR2 = π2E2

πR3h, (3.17.40)

which is again the Euler load.c) p = O(1) : In this case, λx is of order unity, which is not possible in the elastic

range. Hence, values of p = O (1) must be excluded.

Before proceeding with the general discussion of (3.17.30), we notice that thisis the equation of a straight line in the λx, λy plane. Let pi, qi (i ∈ N

+) denoteminimizing values of p and q, respectively. The corresponding values (λx)cr , (λy)cr

then lie on a straight line i, which is the boundary between a stable and an unstableregion. Let Ds

i be the stable region. Let Ds = ∩i=1...nDsi ; then Ds is convex (i.e., if

x, y ∈ Ds, then µx + (1 − µ) y ∈ Ds, 0 ≤ µ ≤ 1 ), if the fundamental state is linear.(The origin (λx, λy) = (0, 0) ∈ Ds.)

To prove this property we proceed as follows: Let S(k)ij be the stress due to a

unit load in the loading case k. The total stress under n loads can then be written

Pπ R2Pπ R2 −P

Figure 3.17.1

210 Applications

as Sij =∑nk=1 λkS(k)

ij (linearity). Substitution of this expression into the functionalP2 [u] yields

P2 [u] =∫V

[n∑

k=1

λkS(k)ij uh,iuh,j + 1

2Eijklθij θk

]dV. (3.17.41)

The necessary condition for stability is that

P2 [u] ≥ 0 ∀u that are kinemafically admissible, (3.17.42)

where the equality sign holds on ∂Ds. Let λ(1)k and λ

(2)k be two load factors ∈ Ds +

∂Ds; then

µP2

[u; λ(1)

k

]+ (1 − µ) P2

[u; λ(2)

k

]≥ 0, 0 ≤ µ ≤ 1, (3.17.43)

and hence

∫V

{n∑

k=1

[µλ

(1)k + (1 − µ) λ

(2)k

]S(k)

ij + 12

Eijklθij θk

}dV ≥ 0 (3.17.44)

so that µλ(1)k + (1 − µ) λ

(2)k ∈ Ds + ∂Ds, q.e.d.

Let us now return to the buckling equation (3.17.30) and let us first consider thecase that there is no axial loading, λx = 0. In this case, λy is given by

λy = −(1 − υ2

)p4 + h2

12R2

(p2 + q2)2 (p2 + q2 − 1

)2q2[(p2 + q2)2 − (3p2 + q2)

] . (3.17.45)

The critical value of λy is obtained by minimizing (3.17.45) with respect to p2 andq2. We shall not carry out these calculations, but we notice that λy will attain itsminimum value for the smallest value of p, i.e., p2 = π2R2/2.

Let us now consider the special case of a very long cylinder, R/ → 0. In thiscase, λy is given by

λy = − h2

12R2

(q2 − 1

). (3.17.46)

The critical value is attained for the smallest possible value of q. However, q = 0cannot be taken into account because then the limit process is not valid. The valueq = 1, the Euler buckling, also can not be taken, so that the minimum value for q isq = 2,

λycr = − h2

4R2. (3.17.47)

This is a long cylinder (or a ring) under uniform external pressure (see Fig-ure 3.17.2).


P P = −E

4 1− υ2( )h3

R3

Figure 3.17.2

Let us now consider the case that λy = 0, i.e., we have only axial loading. Wethen have

− λx =(1 − υ2

)p2

(p2 + q2)2 + q2+ h2

12R2

(p2 + q2

)2 (p2 + q2 − 1

)2p2[(p2 + q2)2 + q2

] . (3.17.48)

Let us now consider the case q = O (1) but q �= 1. Then, to stay in the elastic range,we must have either p2 � 1 or p2 � 1.

Let us first consider the case p2 � 1. In this case, (3.17.48) can be simplified toyield

− λx =(1 − υ2

)p2

q2 (q2 + 1)+ h2

12R2

q2(q2 − 1

)2p2 (q2 + 1)

. (3.17.49)

Minimizing this expression with respect to p2, we obtain

1 − υ2

q2 (q2 + 1)− h2

12R2

q2(q2 − 1

)p4 (q2 + 1)

= 0,

from which

p2 = h2cR

q2 (q2 − 1). (3.17.50)

Because p = kπR/ � 1, we must have

= kπR

√2cR

h1

q√

q2 − 1� 1, (3.17.51)

i.e., a sufficiently long cylinder. The corresponding critical load is then

(−λx)cr = (1 − υ2) hcR

q2 − 1q2 + 1

. (3.17.52)

For q = 2, we have

(−λx)cr = 0.6(1 − υ2) h

cR. (3.17.53)

212 Applications

Let us next consider the case p2 � 1, q = O (1), q �= 1. In this case, (3.17.48)can be simplified to yield

− λx = 1 − υ2

p2+ h2

12R2p2. (3.17.54)

Minimizing this expression with respect to p2, we obtain

(p2)min = 2cRh

, (−λx)cr = (1 − υ2)h

cR. (3.17.55)

Notice that this critical value is always larger than the value given in (3.17.48), whichmeans that for q = O (1) the critical load is given by (3.17.52), and this result holdsfor long cylinders.

Let us now consider the case q � 1, i.e., a short wavelength (shallow shell the-ory). Then (3.17.48) can be simplified to yield

− λx = (1 − υ2)p2

(p2 + q2)+ h2

12R2

(p2 + q2)2

p2. (3.17.56)

Minimizing this expression with respect to p2/(p2 + q2)2, we find[(p2 + q2)2

p2

]min

= 2cRh

, (3.17.57)

so that

(−λx)cr = (1 − υ2)h

cR, or (−σx)cr = −Eh

cR. (3.17.58)

This result holds for cylinders of moderate length. As we are only interested in pos-itive values of p and q, (3.17.57) can be rewritten as

p2 + q2 =√

2cR/h p. (3.17.59)

For p, q ∈ R+, this is the equation of a circle with radius 1

2

√2cR/h in the p, q plane

(see Figure 3.17.3).However, because q ∈ N

+, we have only discrete values, and to a given value ofq there corresponds two values of p, say, pq1 and pq2 . This means that for the same

q

pq1pq2 p

0p

p0

= 2cR / h

Figure 3.17.3


buckling load, we have buckling modes of the form

(cossin

)pq1

pq2

x/R

(

cossin

)(q y/R) . (3.17.60)

Let us now consider a particular combination of buckling modes represented by

w = h{b0 cos(p0x/R) +

∑q

[cq1 cos(pq1 x/R) + cq2 cos(pq2 x/R)] cos(qy/R)

(3.17.61)+ cm cos(mx/R) cos(my/R)

},

where the last term must only be taken into account when m = 12 p0 = √cR/2h. The

corresponding in-plane displacements are then

u = h{− υ

p0b0 sin(p0x/R) −

∑q

pq1

(υp2

q1− q2

)(p2

q1+ q2

)2 cq1 sin(pq1 x/R) cos(qy/R)

−∑

q

pq2

(υp2

q2− q2

)(p2

q2+ q2

)2 cq2 sin(pq2 x/R) cos(qy/R) (3.17.62)

+ 1 − υ

4mcm sin(mx/R) cos(my/R)

}

v = h

{−∑

q

q[(2 + υ) p2

q1+ q2

](p2

q1+ q2

)2 cq1 cos(pq1 x/R) sin(qy/R)

−∑

q

q[(2 + υ) p2

q2+ q2

](p2

q2+ q2

)2 cq2 cos(pq2 x/R) sin(qy/R) (3.17.63)

− 3 + υ

4mcm cos(mx/R) sin(my/R)

}.

A necessary condition for stability at the critical load is that the third variationvanishes. Because we are dealing with the case q � 1 or p � 1, we can use (3.15.12),which is valid for shallow shell theory. However, we must note that in the presentcase we have chosen W positive in the outward direction, so that in the notationadopted in this section, we have

P3 [u] = Eh3

12 (1 − υ2) R3

∫A

{[u′ + υ (v· + w)] w′2

(3.17.64)+ (υu′ + v′ + w) w·2 + (1 − υ) (u· + v′) w′w·} dA.

For the evaluation of this integral, one must deal with integrands of the form(cossin

)(q1y/R)

(cossin

)(q2y/R)

(cossin

)(q3y/R)

(cossin

)(p1x/R)

(3.17.65)

×(

cossin

)(p2x/R)

(cossin

)(p3y/R) ,

214 Applications

which can be reduced to(cossin

)(p1 ± p2 ± p3)

xR

,

(cossin

)(q1 ± q2 ± q3)

yR

. (3.17.66)

For a sufficiently long cylinder, the only non-vanishing terms (after integration) arethe cosine terms with an argument of zero; i.e., we must only consider the combina-tions

p1 ± p2 ± p3 = 0, q1 ± q2 ± q3 = 0, (3.17.67)

under the side condition that (pi, qi) (i = 1, 2, 3) lies on the semicircle (3.17.59). Theonly combination that satisfies these conditions is

p1,2 = 12

p0 ± 12

√p2

0 − 4q2, 0/q ≤ 12

p0(3.17.68)

p3 = p0, q3 = 0.

The result of a fairly lengthy but elementary calculation is

P3 [u] = 3π

4Eh4

R2

(∑q

q2b0cq1 cq2 + 12

mb0c2m

)�= 0, (3.17.69)

which means that the equilibrium at the critical load is unstable.Let us now consider the post-buckling behavior. Because P3 [u] �≡ 0, the post-

buckling behavior is determined by the functional

P2 [ahuh; λ] + P3 [ahuh] , (3.17.70)

and in both terms, shallow shell theory is used. For the evaluation of the first term,we notice that at the critical load,∫

A

12σxcr

hR2

w′2 dA +∫A

12

hEαβλµ

(θαβθλµ + h2

12ραβρλµ

)dA = 0, (3.17.71)

so that

P2 [ahuh; λ] = 12

(1 − λ)Eh2

cR3

∫A

w′2 dA, (3.17.72)

where λ = −σxcR/Eh. Using (3.17.61) and (3.17.69), we obtain

P2 [ahuh; λ] + P3 [ahuh] = π

4cEh4

R2

{(1 − λ)

[2p2

0 b20

(3.17.73)

+∑

q

(p2

q1c2

q1+ p2

q2c2

q2

)+ m2c2m

]+ 3c

[∑q

q2b0cq1 cq2 + 12

m2b0c2m

]}

The equations of equilibrium are obtained by taking the variations with respect tob0, cq1 , cq2 , and cm, respectively, which yields

4 (1 − λ) p20b0 + 3c

(∑q

q2cq1 cq2 + 12

m2c2m

)= 0


2 (1 − λ) p2q1

cq1 + 3cq2b0cq2 = 0(3.17.74)

2 (1 − λ) p2q2

cq2 + 3cq2b0cq1 = 0

2 (1 − λ) m2cm + 3cm2b0cm = 0.

The second and third of these equations contain only cq1 and cq2 , and have a non-trivial solution when

4 (1 − λ)2 p2q1

p2q2

− 9c2q4b20 = 0. (3.17.75)

Bearing in mind that pq1 and pq2 lie on the semicircle and satisfy the relationpq1 pq2 = q2 (see Figure 3.17.4), we can simplify (3.17.75) to yield

4 (1 − λ)2 − 9cb20 = 0, (3.17.76)

from which

3cb0 = ±2 (1 − λ) . (3.17.77)

The last of the equations (3.17.74) has the solution

3cb0 = −2 (1 − λ) ∨ cm = 0. (3.17.78)

When the first of these solutions is chosen, compatible with (3.17.77), cm remainsundetermined. This result is due to the fact that we have not taken into account thehigher order terms P4 [ahuh]. With this choice for b0, we find from the second andthird of the equations of (3.17.75),

cq2

cq1

= − 2 (1 − λ) p2q1

3cq2 [−2 (1 − λ) /3c]= pq1

pq2

(317.79)

so that pq1 cq1 = pq2 cq2 .The first of the equations of (3.17.74) can be written as

∑q

q2cq1 cq2 + 12

m2c2m = 8 (1 − λ)2

9c2p2

0, (3.17.80)

q

q

pq1pq2

p

Figure 3.17.4

216 Applications

or, using (3.17.79) and the relations pq1 pq2 = q2 and p20 = 2cR/h,

∑q

p2q1

c2q1

+ 12

m2c2m = 16 (1 − λ)2

9cRh

. (3.17.81)

Now b0, the amplitude of the axisymmetric deflection is small, namely,

b0 = −2 (1 − λ)3c

≈ −0.4 (1 − λ) for c = 5/3, (3.17.82)

i.e., 40% of the shell thickness, when λ = 0 . cm is the amplitude of the non-symmetric buckling mode, and for cq1 = 0 is given by

c2m = 32 (1 − λ)2

9cRh

1m2

, (3.17.83)

which with m = 12 p0 = √cR/2h can be rewritten in the form

cm = ±8 (1 − λ)3c

, (3.17.84)

i.e., |cm| = 4b0, twice the shell thickness, much larger than the amplitude of theaxisymmetric buckling mode.

Although the amplitudes of the buckling modes are indeterminate, the shorten-ing of the shell is determined. From (2.4.61), the general formula for the additionalshortening is given by

�ε = −∂P [ueq; N]∂N

. (3.17.85)

The compression force N is given by

N = 2πRhσ = 2π

cEh2λ, (3.17.86)

and the elastic energy is given by (3.17.73). Using this relation, we find

�ε = − ∂P∂N

= −∂P∂λ

c2πEh2

(3.17.87)

= 18

h2

R2

{2p2

0b20 +∑

q

(p2

q1c2

q1+ p2

q2c2

q2

)+ m2c2m

},

which with (3.17.79) and (3.17.81) can be rewritten to yield

�ε = 23

h

cR(1 − λ)2 = 2

3εcr (1 − λ)2

, (3.17.88)

so that

�ε

εcr= 2

3(1 − λ)2

. (3.17.89)

Notice that this formula is valid for all post-buckling paths. This relation is shown inthe load-generalized displacement diagram (see Figure 3.17.5).


1

1

unstable (3.17.90)

unstable (general theory)

λ =σ

σ cr

2 / 3 ε / εcr

Figure 3.17.5

Because

∂2P

∂b20

= π

cEh4

R2p2

0 (1 − λ) < 0 for λ > 1, (3.17.90)

it follows that for λ > 1 the equilibrium path is unstable. Because there are no sta-ble equilibrium configurations in the vicinity of the bifurcation point, the shell willcollapse explosively.

Because the post-buckling behavior of the perfect cylinder is unstable, the cylin-der is an imperfection-sensitive structure. To get quantitative results for the imper-fection sensitivity, we consider first the case of an axisymmetric imperfection in thedirection of the buckling mode. Let this imperfection be given by

w0 = κh cos p0x/R. (3.17.91)

According to the general theory, the term 12σhw2

,x in the energy functional must bereplaced by

σh(

12

w2,x + w0xw,x

)(3.17.92)

so that in the presence of the axisymmetric imperfection, the energy functional isgiven by

P2[u, u0; λ] + P3[u] = P2[u; λ] + P3[u](3.17.93)

− π

cEh4

R2p2

0b0κλ.

The extra term contains only the amplitude b0, so that only the first of the equilib-rium equations (3.17.76) is affected. This equation must be replaced by

4 (1 − λ) p20b0 + 3c

(∑q

q2cq1 cq2 + 12

m2c2m

)− 4λp2

0κ = 0. (3.17.94)

Because prior to buckling cq1 , cq2 , and cm are zero, we readily obtain

b0 = λ

1 − λκ. (3.17.95)

218 Applications

0

0 .2 .4 .6

λ

1

κ

Figure 3.17.6

The values of b0 for a non-trivial solution for cq1 , cq2 of the other equilibriumequations follow from (3.17.76), so we must have

b20 = 4 (1 − λ)2

9c= λ2

(1 − λ)2 κ 2, (3.17.96)

from which follows

(1 − λ)2 = 32

cλ |κ| ≈ 52λ |κ| for c = 5

2. (3.17.97)

In this case, cq1 and cq2 are nonzero, which means that branching from the axisym-metric mode to a non-axisymmetric buckling mode will occur. It follows that smallimperfections reduce the buckling load considerably, e.g., a reduction of the buck-ling load by 50% is obtained when κ = 1/5, say, for an amplitude of the imperfectionof 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 thebuckling mode,

w0 = κh [cos(p0x/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-symmetricmode is four times that of the symmetric mode. In this case, the energy functional isgiven by

P2[u, u0; λ] + P3[u] = P2[u; λ] + P3[u](3.17.99)

− π

cEh4

R2

(p2

0b0 + 2m2cm)

uλ.

The equilibrium equations now read

4 (1 − λ) p20b0 + 3c

(∑q

q2cq1 cq2 + 12

m2c2m

)− 4λp2

0κ = 0

2 (1 − λ) p2q1

cq1 + 3cq2b0cq2 = 0(3.17.100)

2 (1 − λ) p2q2

cq2 + 3cq2b0cq1 = 0

2 (1 − λ) m2cm + 3cm2b0cm − 8λm2κ = 0.


For λ = 0, these equations yield the null solution. For λ < λcr, cq1 and cq2 are equalto zero, and the first and last of these equations are simplified to read

4 (1 − λ) b0 + 38

cc2m − 4λκ = 0

(3.17.101)2 (1 − λ) cm + 3cb0cm − 8λκ = 0

and have a solution for cm = 4b0. The resulting equation for b0 then reads as

b20 + 2

3c(1 − λ) b0 − 2

3cλκ = 0, (3.17.102)

which yields

b0 = −1 − λ

3c±√

(1 − λ)2

9c2+ 2

3cλκ. (3.17.103)

We must choose the positive solution because the result of the negative solutiondoes not refer to the undeformed state.

The expression under the square-root sign is always positive when κ > 0, andthen the imperfection has a stabilizing effect. However, for κ < 0 the expressionmay 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 axisymmetricimperfection. In this case, we do not have a branch point but a limit point. Thebehavior is shown in Figure 3.17.7. Notice that an imperfection of 5% of the shellthickness (κ = 0.05) yields a reduction of the critical load by 50%.

λ

b0

stable

unstable

1

b0 = −2

3c1 − λ( )

Figure 3.17.7

220 Applications

3π −− 2π −π 0 π 2π 3π

2π

π

−π

−2π

−5

−5

−5

−5

−5

−5

−5

−5 −5

−5−5

−5

3 3

3 3

3

3

3 3

33

1 1 1 12

2

2

2

2

0

0 0

0

00

0

β

α

3

Figure 3.17.8

To get an impression of the pattern of the imperfection, we consider thefunction

f (α, β) = cos 2α + 4 cos α cos β. (3.17.105)

For {α = (2j + 1) π

β = (2k+ 1) π

}and

{α = 2j πβ = 2kπ

}, (3.17.106)

we have f (α, β) = 5, i.e., a maximum inward deflection. The maximum outwarddeflection is f (α, β) = −3 and is obtained for{

α = (2j + 1) π

β = 2kπ

}and

{α = 2j π

β = (2k+ 1) π

}. (3.17.107)

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 Lock-heed.† These experiments were carried out with a mandrel with a radius slightlysmaller than that of the cylinder, inside the cylinder. The tests were filmed with ahigh- speed camera (8,000 images per second) to record the dynamic behavior dur-ing buckling.

† B. O. Almroh, A. M. C. Holmes, and D. O. Brush, An experimental study of the buckling ofcylinders 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

µ2/ m2 = 0.1

mx / R

w0

5

4

3

2

1

0 2 4 6 8 10

Figure 3.18.1

3.18 The influence of more-or-less localized short-wave imperfections on thebuckling of circular cylindrical shells under axial compression†

So far, we have only considered overall imperfections. In practice, one also oftenencounters localized imperfections (dimples), which are usually inward deflections.To investigate the influence of these local imperfections, we consider an imperfec-tion of the form

w0 = κk[cos(p0x/R) + 4 cos(mx/R) cos(my/R)] e− 12 µ2(x2+y2) /R2

p0 = 2m,

(3.18.1)i.e., an imperfection in the direction of the buckling mode that is rapidly decayingwith increasing values of x and y (even for small values of µ2/m2). To get a feelingfor the behavior of this imperfection, we have drawn the graph for µ2/m2 = 0.1 (seeFigure 3.18.1).

To solve this problem, we shall employ the Rayleigh-Ritz method, and weassume deflections of the form

w = hb0 [cos( p0x/R) + 4 cos(mx/R) cos(my/R)] e− 12 µ2(x2+y2) /R2

u = hb0

[− υ

2msin( p0x/R) + 1 − υ

msin(mx/R) cos(my/R)

]e− 1

2 µ2(x2+y2) /R2(3.18.2)

v = −hb03 + υ

mcos(mx/R) sin(my/R) e− 1

2 µ2(x2+y2) /R2.

† Cf. Koiter’s paper under the same title, Rep. Lab. For Appl. Mech. Delft 534, I. N. Vekua Anniver-sary volume (Moscow, 1978), pp. 242–244.

222 Applications

Let us now consider the term/2∫

−/2

2π∫0

−12σhw2

·x dx dy, (3.18.3)

which can be evaluated to yield

−12λ

Eh2

cRh2b2

0

/2∫−/2

2π∫0

[−p0

Rsin(p0x/R) − 4m

Rsin(mx/R) cos(my/R)

− µ2xR2

{cos(p0x/R) + 4 cos(mx/R) cos(my/R)

}]2

e−µ2(x2+y2) /R2dx dy

= −12λ

Eh4

cR3b2

0

/2∫−/2

2π∫0

{[p0 sin(p0x/R) + 4msin(mx/R) cos(my/R)]2

(3.18.4)

+2µ2xR

[p0 sin(p0x/R) + 4msin(mx/R) cos(my/R)]

× [cos(p0x/R) + 4 cos(mx/R) cos(my/R)]

+µ4x2

R2[cos(p0x/R) + 4 cos(mx/R) cos(my/R)]2

}e−µ2(x2+y2) /R2

dx dy.

Let us now first consider integration in the x-direction. First, we notice that the firstterm in the integrand leads to integrals in the x-direction of the form

/2∫−/2

cos (nx/R) e−µ2x2/R2dx. (3.18.5)

The second term leads to integrals of the form

/2∫−/2

xR

sin (nx/R) e−µ2x2/R2dx, (3.18.6)

and the third term to/2∫

−/2

x2

R2cos (nx/R) e−µ2x2/R2

dx. (3.18.7)

Because the integrants are rapidly decaying functions, we can replace the bound-aries of the integrals by −∞ to ∞, so that we must deal with the following integrals,

∞∫−∞

cos(nx/R) e−µ2x2/R2dx,

∞∫−∞

xR

sin (nx/R) e−µ2x2/R2dx,

(3.18.8)∞∫−∞

x2

R2cos (nx/R) e−µ2x2/R2

dx.


A ′

A

− R

iy

CB ′

x0

R

B

ia

2

Figure 3.18.2

Let us consider the first of these integrals. It is convenient to consider the inte-gral ∮

C

e−z2dz (3.18.9)

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 ∣∣∣e−z2∣∣∣ = ∣∣∣e−(x2−y2)e−2ixy

∣∣∣ = e−R2ey2

< e−R2ea2/4,

and this expression tends uniformly to zero as R tends to infinity. Thus, the portionsof the whole integral that arise from the vertical sides tend to zero, and if we carryout the passage to the limit R → ∞ and note that dz = d(x + 1

2 ia) = dx, on A′B′ wecan express the result of Cauchy’s theorem as follows,

∞∫−∞

e−(x+ 12 ia)2

dx −∞∫

−∞e−x2

dx = 0.

The second integral is easily evaluated by noticing that

∞∫

−∞e−x2

dx

2

=∞∫

−∞

∞∫−∞

e−(x2+y2) dx dy =∞∫

−∞

2π∫0

e−r2r dr dθ = π,

so that we have

ea2/4

∞∫−∞

e−x2(cos ax − i sin ax) dx = √

π,

and because sin ax is an odd function, its contribution in the integral vanishes so that

∞∫−∞

cos ax e−x2dx = √

π e−a2/4. (3.18.10)

224 Applications

The result for the first of the integrals in (3.18.8) is then

∞∫−∞

cos(nx/R) e−µ2x2/R dx = Rµ

√π e

− n2

4µ2 . (3.18.11)

Noting that n is either zero or a multiple of m (m � 1), it follows that only the casen = 0 must be taken into account.

Let us now consider the other integrals in (3.18.8). Differentiating (3.18.11) withrespect to n, we obtain

∞∫−∞

xR

sin(nx/R) e−µ2x2/R2dx = R

√π

2µ3ne

− n2

4µ2 , (3.18.12)

which is zero for n = 0 and is exponentially small for large values of n, and may thusbe neglected. Differentiating (3.18.12) with respect to n, we obtain

∞∫−∞

x2

R2cos(nx/R) e−µ2x2/R2

dx = R√

π

2µ3

(1 − n

2µ2

)e− n2

4µ2 . (3.18.13)

For large values of n this integral vanishes, and for n = 0 the contribution isR

√π/2µ3. Summarizing our results, we find that the contribution of the first term in

(3.18.4) is of order m2R√

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 contri-

bution of the third term compared with the first term, making a relative error in theenergy of order µ2/m2.

Let us now evaluate the first term using the fact that the integrand is also expo-nentially decaying in the y-direction, so that the boundaries of integration 0 to 2π

can be replaced by −∞ to ∞,

∞∫−∞

∞∫−∞

[2msin(2mx/R) + 4msin(mx/R) cos(my/R)]2e−µ2(x2+y2) /R2dx dy

= 4m2

∞∫−∞

∞∫−∞

[sin2(2mx/R) + 4 sin(2mx/R) sin(mx/R) cos(my/R)

+ 4 sin2(mx/R) cos2(my/R)]e−µ2(x2+y2) /R2dx dy

= 4m2

∞∫−∞

∞∫−∞

{32

− cos(2mx/R) − 12

cos(4mx/R) − cos(2my/R) (3.18.14)

+ 2 [cos(mx/R) − cos(3mx/R)] cos(my/R)

+ cos(2mx/R) cos(2my/R)}

e−µ2(x2+y2) /R2dx dy

= 4m2[

32

R2

µ2π + exponentially small terms

].


The corresponding term for an imperfection of the form (3.17.98) is 4m2πR, whichmeans 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

F∗∗ = 3π

2Eh3

µ2

[(1 − λ) b2

0 + 23

cb30 − 2λκb0

]. (3.18.15)

The equilibrium equation now reads

2 (1 − λ) b0 + 2cb20 − 2λκ = 0, (3.18.16)

from which

b0 = −1 − λ

2c±√

(1 − λ)2

4c2+ λκ

c. (3.18.17)

For positive values of κ, the term under the square-root sign is always positive, whichmeans that an imperfection in the outward direction is stabilizing. However, fornegative 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 isreplaced by a factor 4; i.e., a local imperfection is equally harmful as an imperfectionof 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 secondapproximation with an error of order µ4/m4. His improved approximation yieldsthe following equation for the critical load:[

1 + (d − ε)µ2

m2− λ∗

]2

= −4λ∗cκ[

1 + (e − ε)µ2

m2

], (3.18.19)

where

ε = 38, e = 7 + 5υ − 6υ2

12 (1 − υ2),

(3.18.20)d = 1

9b(1 − υ2){(3 − υ) [υ2 + 2 (1 − υ)2 + 2 (3 + υ)2] + 80

(1 − υ2)},

For υ = 0.3, we have

ε = 0.375, e = 0.729, d = 1.540, (3.18.21)

and the equation for the critical load then reads(1 + 1.165

µ2

m2− λ∗

)2

= −4λ∗cκ(

1 + 0.354µ2

m2

). (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 cylin-drical shells with more-or-less localized shortwave imperfections. W.T.H.D., Rep. 88.

226 Applications

λ *

µ2/ m2 = 0

µ2/m2 = 0.1

cκ

1.2

1.0

0.8

0.6

0.4

0.2

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

1.116

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 yieldsan upper bound for the critical load. However, we have also neglected terms inpowers 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 dueto damage, and these imperfections are in the inward direction, and are thereforestrongly destabilizing. It is further interesting to note that even the carefully manu-factured cylindrical shells used for the Lockheed tests† had local imperfections.

† Cf. footnote at the end of Section 3.17.

Selected Publications of W. T. Koiter onElastic Stability Theory

Over de stabiliteit van het elastisch evenwicht (On the stability of elastic equilibrium). Proef-schrift T. H. Delft (1945).

On the stability of elastic equilibrium (Washington, 1967), 6 + 202 pp. (NASA TechnicalTranslation F-10, 833, Clearinghouse US Dept. of Commerce/Nat. Bur. of Standards N67-25033).

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., ReportTR-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). North-Holland 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-ment Postcritique des Plaques Utilisees en Construction Metallique, Liege (1963).

“Elastic stability and post-buckling behaviour.” Proc. Symp. Nonlinear Problems, Universityof 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 axialcompression. 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 Selected Publications of W. T. Koiter on Elastic Stability Theory

“The nonlinear buckling problem of a complete spherical shell under uniform external pres-sure.” 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, HarvardUniversity, June 1974, Springer-Verlag (1976), 133–148.

“A basic open problem in the theory of elastic stability.” Joint IUTAM/IMU Symposium onApplications of Methods of Functional Analysis to Problems of Mechanics. Resumes desconferences, September 1 to 6, 1075, Marseille. Part XXIX (Marseille, 1975).

“Buckling of a flexible shaft under torque loads transmitted by cardan joints.” Ingenieur-Archiv, 49, 369–373 (1980).

Index

Adjacent state, 8Airy’s stress function, 167Almroth, B.O., 220Alternating tensor, 121

Beam bending, 126Biezeno, C.B., 20, 101Bifurcation condition, 70, 106Bifurcation point, 41, 166, 178, 201Bimoment, 81Branched, 33, 177Branch point, 32, 40, 219Brush,D.O., 220Buckling mode, 17, 22

Cardan,G., 119Carlson,R.F., 180Cauchy’s theorem, 223Center of shear, 78, 126, 131Characteristic equation, 146Characteristic length, 18, 111Chistoffel symbol, 185Clamped, 59, 81, 106, 131, 141Clausius-Duhem, 8Conservative systems, 1, 116Critical load, 11, 29, 35, 47Curvilinear, 15

Damping forces, 2, 4Dead weight loads, 13, 18, 41, 112, 171Deflection, 5, 27Deformation tensor, 7Destabilizing effect, 130Displacement field, 9, 11Divergence theorem, 19, 139, 167, 172, 185Duhem,P., 9, 11Dynamic boundary conditions, 62, 73, 88, 94, 114,

204

Eccentric loads, 51Effective relative shortening, 58

Elastic hinge, 61Elastic potential, 12Energy functional, 10, 12, 29, 47, 138, 158,

183Entropy, 7Entropy production, 8Equilibrium, 2Equilibrium equations, 20, 36, 41, 140, 167Euler column, 35, 91, 101

Foppl,A., 188Fourier series, 63Fourier transform, 144Frame, 67Free energy, 8Fritz,John, 18Fundamental path, 41, 177Fundamental state, 7, 20

Gaussian curvature, 182, 189Generalized displacement, 38, 216Grammel,R., 20, 101Greenhill,A.G., 116, 225Gristchak,V.Z., 181, 225

Haringx,J.A., 101Heat flux, 8Helical spring, 101Hencky, 20Hoff,N.J., 180Holmes,A.M.C., 220Holonomic, 1Hooke’s law, 16, 25, 40Hurlbrink, 101Hutchinson,J.W., 181Hypersphere, 4

Imperfection, 41Incompressible, 55, 84Indefinite, 4, 11, 44Inextensible, 16

229

230 Index

Irreversible, 8Isothermal, 10

Karman,von T., 166Kinematic boundary conditions, 63, 81, 116Kinematically admissible, 11, 56Kinetic energy, 1Kinetic potential, 1Kirchhoff,G., 29, 50Kirchhoff-Trefftz stress tensor, 50Koch, 101Koiter,W.T., 35, 37, 42, 59, 72, 181, 188, 201, 202,

207, 221Koiter-Morley equations, 207

Lagrangean multiplier, 87, 107Lagrangian equations, 2Laplacian operator, 139Legendre function, 196Legendre polynomial, 196Lekkerkerker,J.G., 61Limit point, 43, 219Lower bound, 44, 181

Mean curvature, 182Measures, 3Metric tensor, 15, 48, 85, 182Minimizing direction, 37Minimizing displacement field, 18Minimum problem, 23, 57, 163Multilinear forms, 37

Necessary condition for stability, 5, 25, 57Neo-Hookean material, 84Neutral equilibrium, 5, 17, 56, 80, 117, 130, 171,

197Non-conservative forces, 2

Plate, 84, 137Positive-definite, 3, 11Post-buckling, 78, 101, 158, 169Potential energy, 1, 4Principle axes of inertia, 110, 127

Radius of gyration, 58Rayleigh’s principle, 23Rayleigh-Ritz method, 150, 221

Riemann-christoffel tensor, 189Rotation, 14, 121, 138, 166, 183Rotation tensor, 49Rotation vector, 119

Saint Venant,B. de, 128Scleronomic, 1Second law of thermodynamics, 8Second variation, 17Sendelbeck,R.L., 180Shallow shell theory, 169Shear modulus, 17, 184Shell, 15, 41, 139, 169Side condition, 25, 42, 68, 214Simmonds,J.G., 182Southwell,R.V., 153Specific mass, 8Stability, 1Stability criterion, 3Stability limit, 18Stabilizing effect, 130, 219Stationary, 2Steepest descent, 38Strain, 13Strain tensor, 14Stress tensor, 13Strut, 11, 15Sufficient condition for stability, 17Surface coordinates, 182

Tangent modulus, 66, 164Tensor of elastic moduli, 14Torsion, 78, 102, 110Torsional stiffness, 79, 102Total differential, 1Total energy, 5Trefftz,E., 20

Unstable, 12Upper bound, 150, 202, 226

Van der Neut,A., 187

Warping constant, 79, 127Weakest ascent, 40

Ziegler,H., 116

van der heijden a.m.a. (ed.) w. t. koiter's elastic stability of solids and structures (cup,...

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