[Solid Mechanics and Its Applications] Elementary Continuum Mechanics for Everyone Volume 194 ||...

E. Byskov, Elementary Continuum Mechanics for Everyone,Solid Mechanics and Its Applications 194, DOI: 10.1007/978-94-007-5766-0_18,Ó Springer Science+Business Media Dordrecht 2013

Chapter 18

Elastic Buckling Problemswith Linear PrebucklingBuckling comprises so many important different subtopics18.1 that it is im- Buckling comprises

many important

different subtopics

possible to cover all of them in any detail here. Therefore, we concentrateon some of the problems which are characterized by at least two qualities,namely that they reveal important features of buckling and that they can betreated by analytical means. Among the topics are elastic buckling problemswith linear prebuckling, see Section 18.3, including postbuckling behavior,see Chapter 19, the influence of geometric imperfections, see Chapter 20,in particular Example Ex 20-2, and elastic-plastic stability, which is dealtwith in Chapter 21.

Although bifurcation buckling never occurs in a real structure it is Bifurcation buckling

never occurs in real

structures

customary to consider this phenomenon rather than the more appropri-ate limit load buckling. There are reasons of varying relevance for doingthis. First of all, by nature limit load buckling is nonlinear and thereforeresults in mathematical problems that are more difficult to solve, and be-fore the advent of computers only very simple cases of limit load bucklingcould be handled; so, by tradition, these problems have received less at-tention. Another, better, reason is that many structures experience limitload buckling only because of imperfections, i.e. because of deviations fromthe perfect geometry and perfect load configuration, and that the bifur-cation buckling load in some cases is still a good approximation to thereal load-carrying capacity. Thirdly, for structures that are very imper- Asymptotic

methods may give

good estimates of

the real

load-carrying

capacity

fection sensitive asymptotic methods, which only entail quantities com-puted at the bifurcation load, may provide good estimates of the limit load.These asymptotic methods have been studied since 1945 and are the sub-ject of many books and papers, see e.g. (Koiter 1945), (Budiansky 1966),(Thompson & Hunt 1973), (Budiansky 1974), (Byskov & Hutchinson 1977),

18.1 Among the very important ones are: postbuckling behavior and imperfection sensi-tivity, see e.g. (Koiter 1945), (Thompson & Hunt 1973) and (Budiansky 1974); influence ofplasticity, see e.g. (Hutchinson 1974b); mode interaction, see e.g. (Koiter & Kuiken 1971),(Tvergaard 1973a), and (Byskov & Hutchinson 1977), etc. Time-dependent problems arenot covered.

August 14, 2012 Continuum Mechanics for Everyone Esben Byskov

295

Elastic Buckling

and (Budiansky & Hutchinson 1979). Even in the elastic-plastic case asymp-totic expansions have been developed and applied with success, see in partic-ular (Hutchinson 1974b). Because of their complexity none of these methodswill be introduced in this book, but some indication of their usefulness maybe seen from Example Ex 17-1.5.

In the following we rely on the general equations for the three-dimen-sional continuum derived in Chapter 2, the beam equations from Chapter 7,and the plate equations Chapter 9, which may all be rewritten using theBudiansky-Hutchinson Notation, see Chapter 33. However, in order to makethe exposition self-contained some of the basic equations are also givenbelow.

18.1 Nonlinear PrebucklingHere, we shall be concerned with elastic bifurcation buckling only and limitNonlinear

prebuckling ourselves to the important case of linear prebuckling. First, however, con-sider bifurcation buckling with a nonlinear prebuckling path, see Fig. 18.1.As indicated in Fig. 18.1, the displacements u0 on the prebuckling path as

Postbuckling path u(ux;λ)

uuu0 uc

Prebuckling path u0(ux;λ)

λc

u

λ

λ

Fig. 18.1: Bifurcation buckling with nonlinear prebuckling.

well as the displacements u on the postbuckling path depend nonlinearlyon the scalar load parameter λ

u0 = u0(x;λ) and u = u(x;λ) (18.1)

where also the dependence on the spatial coordinates x is indicated.

However, in many cases a structure responds in a way suggesting that itsLittle bending:

Linear prebuckling

often valid

prebuckling path can be considered linear in the load parameter λ. This isthe case when the prebuckling state entails little or no bending. If bending

Esben Byskov Continuum Mechanics for Everyone August 14, 2012

296

Linear Prebuckling

λc

Prebuckling path λu0(ux)

Postbuckling path u(ux;λ)

u0 uc uu

u

λ

λ

Fig. 18.2: Bifurcation buckling with linear prebuckling.

strains are important in the prebuckling state, linearity of the prebucklingpath is certainly not a good assumption. But, in most cases the engineertries to design a structure, which may experience buckling, in such a waythat the prebuckling state is axial or in-plane and is therefore approximatelylinear. One reason for doing this is that pure compression—or tension—isa very efficient way to carry loads.

18.2 Some PrerequisitesAs before, see Section 33, the generalized strains ε are derived from thegeneralized displacements u according to

Quadratic strainε = l1(u) +12 l2(u) (18.2)

where l1 and l2 denote a linear and a quadratic operator, respectively. Abilinear operator l11 is defined by the following equation

Quadratic

operator l2l2(ua + ub) = l2(ua) + 2 l11(ua,ub) + l2(ub) (18.3)

where ua and ub are two independent displacement fields. Obviously

Bilinear operator

l11l11(ua,ub) = l11(ub,ua) and l11(u,u) = l2(u) (18.4)

The constitutive law is taken to be linearly hyperelastic

Hooke’s “law”σ = H(ε) (18.5)

where σ designates the generalized stresses and H is a linear operator. Weassume that the reciprocal relation

Reciprocal

relationH(εa)εb = H(εb)εa (18.6)


297

Elastic Buckling

holds for all strain fields. Note that this is a point-wise condition because the(18.6) does not entail a dot and therefore an inner product is not implied.This relation is not as restrictive as you may think, which you may realizeif you consider a number of typical elastic structures.

18.3 Linear PrebucklingWe restrict ourselves to cases with linear prebuckling in the sense that theLinear prebuckling

displacement, strain and stress fields on the fundamental path are given by

Linear prebuckling u(x;λ) = λu0(x) , ε(x;λ) = λε0(x) and σ(x;λ) = λσ0(x) (18.7)

where subscript 0 indicates that a quantity is computed on the prebucklingpath, x designates the coordinates, and λ is a scalar load parameter. Thereare other, more general, ways to define linear prebuckling, e.g. assumingthat the bending energy associated with prebuckling is much smaller thanthe axial or membrane energy, but this is the simplest way, which is in mostcases satisfactory. In order to keep the notation less complicated, we do notindicate the dependence on x below.

We impose the additional restriction that

No prebuckling

bendingl11(u0,v) = 0 ∀v (18.8)

In plain language this condition implies that all prebuckling bending effectscan be ignored, as hinted at above. Thus, this restriction is not as limitingas it may seem at a first glance. As consequences of (18.8) we get;

l2(u0) = 0 and ε0 = l1(u0) (18.9)

18.3.1 Principle of Virtual Displacements

In the Principle of Virtual Displacements, see (33.14), the load is given as T ,but here it proves convenient to denote it by λT , where T signifies the loadLoad distribution T

Load intensity λ distribution and the scalar load parameter λ expresses the load intensity.Thus, the principle now is

Principle of virtual

displacementsσ · δε = λT · δu (18.10)

which is valid on the postbuckling as well as on the prebuckling path sinceboth paths are equilibrium states. In (18.10) δε is the variation of strains,δu the variation of displacements, and the dot implies an inner product,i.e. the proper integral over the whole structure or a scalar vector product.

The strain variation δε is defined by

Strain variation δε ε(u+ ǫδu) = ε(u) + ǫδε(u) +O(ǫ2) , |ǫ| ≪ 1 (18.11)

where ǫ often is called the “American epsilon.”“American epsilon” ǫ


298

Linear Prebuckling

Now, (18.2)–(18.4) provide us with

ε(u+ ǫδu) = ε(u) + ǫ{l1(δu) + l11(u, δu)}+O(ǫ2) (18.12)

and comparison with (18.11) furnishes

Strain variation δεδε(u) = l1(δu) + l11(u, δu) (18.13)

which on the prebuckling path simply is

Strain variation in

prebuckling δε0δε0 = l1(δu) (18.14)

because of (18.8).

Above we have exploited the fact that the displacement variation δumay be taken to be the same for all states. Then, on the prebuckling paththe Principle of Virtual Displacements is


displacements in

prebuckling

σ0 · l1(δu) = T · δu (18.15)

and the constitutive relation (18.5), i.e. linear hyperelasticity, provides

Hooke’s “law” in

prebucklingσ0 = H(ε0) (18.16)

18.3.2 Bifurcation Buckling—Classical Critical Load

On the postbuckling path we write the total displacements u as the sumof a contribution λu0 from prebuckling and an increment u, see Fig. 18.2,which are both associated with the same value of the load parameter, asseen from Fig. 18.2

Postbuckling

displacementu = λu0 + u (18.17)

and similarly

Postbuckling

strain and stressε = λε0 + ε and σ = λσ0 + σ (18.18)

As mentioned above, we focus attention on bifurcation buckling which Bifurcation buckling

Classical critical

load = Bifurcation

buckling load

is assumed to take place at the load level λc or λ1. The bifurcation bucklingload is also known as the classical critical load, which indicates that itis associated with bifurcation buckling and not limit load buckling of astructure, be it geometrically perfect or not. Below, we assume that thereis only one buckling mode u1 associated with λc.

From (18.13), (18.14) and (18.17) we get

Postbuckling

strain variationδε = δε0 + l11(u, δu) (18.19)

In postbuckling the principle of virtual displacements (18.10) provides


299

Elastic Buckling

the following expression


displacements in

postbuckling

λσ0 · l1(δu) + σ · l1(δu) + λσ0 · l11(u, δu) + σ · l11(u, δu)

= λT · δu(18.20)

where (18.18) and (18.19) have been utilized.

From (18.2)–(18.4), (18.8)–(18.14) and (18.18) we may get

ε = l1(u) +12 l2(u) (18.21)

With (18.5), (18.16) and (18.18b) we note that

σ = H(ε) (18.22)

When we subtract (18.15) multiplied by λ from (18.20), we may get

σ · l1(δu) + λσ0 · l11(u, δu) + σ · l11(u, δu) = 0 (18.23)

It may immediately be observed that the third term is quadratic in u

while the first two terms are linear. In order to discover bifurcation letλ → λc which at the same time means that any suitable norm ||u|| of ugoes to zero

limλ→λc

||u|| = 0 (18.24)

Let u1 designate u for λ = λc, then (18.21) gives us

ε1 = l1(u1) (18.25)

where ε1 signifies the strain associated with λc and u1.

Further, note that

σ1 = H(ε1) (18.26)

is linear in u1. Then (18.23) providesLinear eigenvalueproblem

Classical criticalload λc

Buckling mode u1

σ1 · l1(δu) + λcσ0 · l11(u1, δu) = 0 (18.27)

This is a Linear Eigenvalue Problem which determines the Classical Crit-ical Load λc, ( often denoted λ1), and its associated Buckling Mode u1. Theshape of the buckling mode is only determined to within a factor, as can beeasily seen from (18.27).

Note that if the buckling mode u1 has been determined, then an expres-sion for the value of the classical critical load λc follows from (18.27) whenu1 is used instead of δu

λc = −σ1 · l1(u1)

σ0 · l2(u1)= −H(l1(u1)) · l1(u1)

σ0 · l2(u1)(18.28)

In many of the following derivations we shall use this formula, which isthe basis for the very useful Rayleigh Quotient, see Section 18.3.5 below.


300

Linear Prebuckling

18.3.3 Higher Bifurcation Loads

Repeating the above analysis, we can get a linear eigenvalue problem for Higher bifurcation

loadseach higher buckling mode uJ associated with the bifurcation load λJ , whereJ = 2, 3, ...

Higher bifurcation

loadsσJ · l1(δu) + λJσ0 · l11(uJ , δu) = 0 (18.29)

Since all buckling modes are kinematically admissible in the strictestsense, i.e. they satisfy the homogeneous kinematic boundary conditions, wemay take δu = uK in (18.29)

σJ · l1(uK) + λJσ0 · l11(uJ ,uK) = 0 (18.30)

Similarly, we may write a relation analogous to (18.29), but with λKand uK instead of λJ and uJ and introduce δu = uJ to get

σK · l1(uJ ) + λKσ0 · l11(uK ,uJ) = 0 (18.31)

Relations similar to (18.25) hold for εJ and εK . Further, the reciprocalrelation (18.6) gives us

σJ · εK = σK · εJ (18.32)

When we subtract (18.31) from (18.30) we get

(λJ − λK)σ0 · l11(uJ ,uK) = 0 (18.33)

where the symmetry of l11 has been exploited.

Equation (18.33) shows that each pair of buckling modes uJ and uK

with λJ 6= λK are orthogonal in the so-called energy sense

Orthogonality of

buckling modesσ0 · l11(uJ ,uK) = 0 , λJ 6= λK (18.34)

For a pair of buckling modes uM and uN that are associated with λM =λN we have the freedom to choose them to be orthogonal in the senseof (18.34). For convenience, consistency, and continuity with the case ofseparate buckling loads we shall take all buckling modes to be orthogonalin the following fashion:18.2

Orthogonality of

buckling modesσ0 · l11(uJ ,uK) = 0 , J 6= K (18.35)

In most cases we wish to normalize the buckling modes in a particularway, sometimes in a geometrical way such that the maximum displacementis equal to a plate thickness or a beam depth, but under other circumstances

18.2 It would seem unreasonable if two buckling modes with coincident buckling were tobe taken to be orthogonal in some other sense than (18.34), while two buckling modesassociated with buckling loads that are only an iota apart indeed are orthogonal accordingto (18.34).


301

Linear Prebuckling

in the following energy sense such that

σ0 · l2(uJ) (18.36)

takes a certain value for all uJ .

The first example below is concerned with buckling of the Euler Column,which was treated in Example Ex 7-3, but here is given as an example ofinterpretation of the above formulas, which are in the Budiansky-HutchinsonNotation.

Ex 18-1 The Euler ColumnLet us reconsider the Euler Column analyzed in Example Ex 7-3 inThe Euler Columnlight of the present theory and notation.

w

λP

xL

Fig. Ex. 18-1.1: The Euler Column.

Before we proceed it is necessary to investigate whether an assumptionof linear prebuckling is valid. If we consider a column made of a com-mon structural material such as steel, wood, concrete, or even somekind of plastic, it seems obvious that when an axial load is applied asshown in Fig. Ex. 18-1.1 the displacements are purely axial and verysmall until buckling occurs.18.3 Provided that this is true, we may as-sume that prebuckling is linear. Also, the fact that prebuckling clearlydoes not involve bending in this case supports the postulate of linearprebuckling.

Ex 18-1.1 Generalized Quantities and Interpretations of theBudiansky-Hutchinson Notation

First we need the generalized quantities, i.e. the displacements, strainsand stresses according to the kinematically moderately nonlinear Ber-noulli-Euler beam theory.18.4

The generalized displacements are the axial displacement componentu and the transverse displacement component w, and thus

18.3 You may try the experiment with a plastic ruler shown in Fig. 16.2. Probably youwill realize that the axial deformation is extremely small before bifurcation, i.e. buckling,takes place.18.4 In Example Ex 18-2 we choose another beam theory, namely the Timoshenko beamtheory, which also accounts for shear strains.


302

The Euler Column

Generalizeddisplacements

u =

[u1

u2

]≡[u

w

](Ex. 18-1.1)

The generalized strains are the axial strain ε and the bending strain κ

Generalizedstrains

ε =

[ε1

ε2

]≡[ε

κ

]=

[u′ + 1

2(w′)

2

w′′

](Ex. 18-1.2)

where prime indicates differentiation with respect to the axial coordi-nate x. Furthermore, see (7.12) and (7.14), or Section 7.7 where thefollowing interpretations are also given.

Now, (Ex. 18-1.2) in connection with (18.3) shows that we may inter-pret l1 and l2 as

Linear andquadraticoperators

l1(u) =

[u′

w′′

]and l2(u) =

[(w′)

2

0

](Ex. 18-1.3)

Similarly, the interpretation of l11 is

Bilinear operatorl11(ua,ub) =

[w′

aw′

b

0

](Ex. 18-1.4)

where subscripts a and b indicate two different displacement fields.

According to this theory, the generalized stresses are the axial force Nand the bending moment M

Generalizedstresses

σ =

[σ1

σ2

]≡[N

M

](Ex. 18-1.5)

Finally, the constitutive operator H is

Hooke’s “law”H =

[EA 0

0 EI

](Ex. 18-1.6)

where EA designates the axial and EI the bending stiffness, respec-tively. Later we restrict ourselves to the case where the beam propertiesare constant along the length.

When we have established the boundary conditions and with the aboveformulas in hand, we may easily write the variational equations for theprebuckling state and for buckling.

Ex 18-1.2 Boundary Conditions

As usual, we need the boundary conditions for the structural problem.Here they are

Boundaryconditions

u(0) = 0 , w(0) = 0 , M(0) = 0 ⇒ w′′(0) = 0

N(L) = −P , w(L) = 0 , M(L) = 0 ⇒ w′′(L) = 0(Ex. 18-1.7)

where we have written the static boundary conditions on M in termsof the second derivative of the displacement.


303

Linear Prebuckling

Below, the indication of prebuckling and buckling by subscripts 0 and

1, respectively, is changed such that the numbers are placed on top of

the quantities, e.g.0

N instead of N0.18.5 The reason for doing this may

not be obvious in this example, but in Example Ex 18-3 the originalnotation would result in a cluttering of subscripts designating prebuck-ling and buckling in combination with subscripts indicating coordinateaxes.

Ex 18-1.3 Prebuckling

The variational statement (18.15) becomes

Prebuckling

∫ L

0

(0

Nδu′ +0

Mδw′′

)dx = (−P )δu(L) (Ex. 18-1.8)

which provides the solution18.6

Prebucklingstresses

0

N = −P and0

M = 0 (Ex. 18-1.9)

and

Prebucklingdisplacements

0u = − P

EAx and

0w = 0 (Ex. 18-1.10)

Ex 18-1.4 Buckling

Here, (18.27) becomes

Buckling

0 =

∫ L

0

(1

Nδu′ +1

Mδw′′

)dx

+λc

∫ L

0

(0

N1w′δw′

)dx

(Ex. 18-1.11)

or, because of (Ex. 18-1.9)

0 =

∫ L

0

(1

Nδu′ +1

Mδw′′

)dx

−λcP

∫ L

0

(1w′δw′

)dx

(Ex. 18-1.12)

Integration by parts provides

0 =

[1

Nδu

]L

0

+

[1

Mδw′

]L

0

−[(

1

M ′ + λcPw′)δw

]L

0

−∫ L

0

1

N ′δudx

+

∫ L

0

(1

M ′′ + λcP1w′′

)δwdx

(Ex. 18-1.13)

The axial component1u of the buckling mode u1 is seen to vanish. ThenBuckling axial

displacement1u = 0 its transverse component

1w may be determined by the relatively simple

18.5 You might ask why don’t I always use numbers on top to indicate prebuckling andbuckling (and postbuckling, if relevant). The reason is that I find the use of subscriptsmore aesthetically pleasing.18.6 Since the prebuckling state is (assumed to be) linear only one solution is possible.


304

The Euler Column

differential equation in1w below, which is a linear eigenvalue problem

in1w, where it is assumed that the bending stiffness is independent of

the axial coordinateBucklingeigenvalueproblem

1

M ′′ + λcP1w′′ = 0 ⇒ EI

1wiv + λcP

1w′′ = 0 (Ex. 18-1.14)

with boundary conditions

Boundaryconditions

1w(0) = 0 ,

1w(L) = 0 ,

1w′′(0) = 0 ,

1w′′(L) = 0 (Ex. 18-1.15)

Together, the equations (Ex. 18-1.14) and (Ex. 18-1.15) constitute alinear eigenvalue problem with the solution18.7

Displacements ofthe solution

uJ (x) = 0

wJ (x) = ξJ sin(Jπx

L

)}J = 1, 2, 3, . . . (Ex. 18-1.16)

where the amplitudes ξJ are undetermined, as usual, and

Buckling loadsλJ = J2 π2EI

PL2, J = 1, 2, 3, . . . (Ex. 18-1.17)

The smallest of these values, λ1, is the Classical Critical Load, alsocalled the Euler Load, λc

Classical criticalload = Eulerbuckling load

λc =π2EI

PL2(Ex. 18-1.18)

with the associated buckling mode1w

Buckling mode1w = ξ1 sin(πx/L) (Ex. 18-1.19)

The expressions (Ex. 18-1.16)–(Ex. 18-1.19) probably constitute themost famous of all solutions to elastic buckling problems.

It is a characteristic feature of elastic structures with infinitely manydegrees of freedom that bifurcation is possible at infinitely many distinctvalues of the load parameter. The reason for emphasizing the words elasticand distinct is that structures which experience bifurcation in the plasticregime may bifurcate over intervals of the load, see e.g. Chapter 21.

In other structures, such as shells and long plates the bifurcation loadsare usually not as well separated as is the case for the Euler Column. With-out going into any details, it may be worth mentioning that this can lead toMode Interaction, see Chapter 20, with the consequence that the structurebecomes strongly imperfection sensitive. Example Ex 20-2 provides someinsight into this phenomenon.

18.7 You know this from your calculus course, but if you have forgotten the answer youcan just insert the postulated solution into the differential equation and the boundaryconditions to verify it.


305

Linear Prebuckling

Ex 18-2 A Pinned-Pinned Column Analyzedby Timoshenko TheoryAs an example of the influence of the shear flexibility on the bucklingload of a column we consider the same column that was analyzed inA Timoshenko

Column Examples Ex 7-3 and Ex 18-1, but perform the analysis using Timo-shenko beam theory and refer to Fig. Ex. 18-1.1. We shall employ thesame procedure as in Example Ex 18-1, but shorten it somewhat.


Here, the generalized displacements are the axial displacement compo-nent u, the transverse displacement component w, and the rotation ω


u =

u1

u2

u3

≡

u

w

ω

(Ex. 18-2.1)

and the generalized strains are the axial strain ε, the shear strain ϕand the bending strain κ, i.e.

Generalizedstrains

ε =

ε1

ε2

ε3

≡

ε

ϕ

κ

=

u′ + 1

2(w′)

2

w′ − ω

ω′

(Ex. 18-2.2)

The interpretation of l1 and l2 is


l1(u) =

u′

w′ − ω

ω′

and l2(u) =

(w′)

2

0

0

(Ex. 18-2.3)

where prime, as before, indicates differentiation with respect to theaxial coordinate x. Similarly, the interpretation of l11 is

Bilinear operator l11(ua,ub) =

w′

aw′

b

0

0

(Ex. 18-2.4)

where subscripts a and b indicate two different displacement fields.

The generalized stresses are the axial force N , the shear force V andthe bending moment M

Generalizedstresses

σ =

σ1

σ2

σ3

≡

NVM

(Ex. 18-2.5)

Finally, the constitutive operator H is

Hooke’s “law” H =

EA 0 00 GAe 00 0 EI

(Ex. 18-2.6)

where we in this example assume that the beam properties are constantalong the length.


306

A Timoshenko Column


The boundary conditions are

Boundaryconditions

u(0) = 0 , w(0) = 0 , M(0) = 0 ⇒ ω′(0) = 0

N(L) = −P , w(L) = 0 , M(L) = 0 ⇒ ω′(L) = 0(Ex. 18-2.7)

As in Example Ex 18-2 prebuckling and buckling are denoted by 0 and

1, respectively, e.g.0

N .


In view of our experience from Example Ex 18-2 we postulate

Prebucklingstresses.Prebucklingdisplacements

0

N = −P ,0

V = 0 and0

M = 0

0u = − P

EA,

0ω = 0 and

0w = 0

(Ex. 18-2.8)

which are seen to fulfill (7.48) in prebuckling.

Ex 18-2.4 Buckling

Here, (18.27) becomes

0 =

∫ L

0

(1

Nδu′ +1

V (δw′ − δω) +1

Mδω′

)dx

+λc

∫ L

0

0

N1w′δw′dx

(Ex. 18-2.9)

or, because of (Ex. 18-2.8a)

Buckling problem

0 =

∫ L

0

(1

Nδu′ +1

V(δ

1w′ − δ

1ω)+

1

Mδω′

)dx

−λcP

∫ L

0

1w′δw′dx

(Ex. 18-2.10)

Here, only one term entails δu implying that

1

N ≡ 0 (Ex. 18-2.11)

which we shall exploit in the following. Now, integration by partsprovides

0 = +

[(1

V − λcP1w′′

)δw

]L

0

+

[1

Mδω

]L

0

−∫ L

0

(1

M ′ +1

V

)δω dx

−∫ L

0

(1

V ′ − λcP1w′′

)δw dx

(Ex. 18-2.12)

The components1w and

1ω of the buckling mode may be determined by


307

Linear Prebuckling

the relatively simple differential equations below

Buckling problem

1

V ′ − λcP1w′′ = 0

1

M ′ +1

V = 0

(Ex. 18-2.13)

With the strain-displacement relations and the constitutive relationsinserted this becomes

Buckling problemGAe(

1w′′ − 1

ω′)− λcP1w′′ = 0

EI1ω′′ +GAe(

1w′ − 1

ω) = 0(Ex. 18-2.14)

The boundary conditions are

Bucklingboundaryconditions

1w(0) = 01w(L) = 01

M(0) = 0 ⇒ 1ω′(0) = 0

1

M(L) = 0 ⇒ 1ω′(L) = 0

(Ex. 18-2.15)

Together, the equations (Ex. 18-2.14) and (Ex. 18-2.15) constitute alinear eigenvalue problem. In the present context we shall only be con-cerned with the one associated with the lowest eigenvalue, namely λc.

When we utilize that0

N is constant and equal to −P we may find thesolution

Solution tobuckling problem

1w(x) = ξ1 sin

(πx

L

)

1ω(x) = ξ1

1

1 + P /(GAe)

( πL

)cos(πx

L

)

= ξ1λc

(πL

)cos(πx

L

)

λc =1

1 + P /(GAe)=

(1 + π2 EI

GAeL2

)−1

(Ex. 18-2.16)

When we recognize that λ = 1 corresponds to the buckling load ac-cording to the Bernoulli-Euler theory we may see that shear flexibilitylowers the value. If, for example, the length of a column with a solidcross-section is about 5 times the depth of a square cross-section, theload-carrying capacity according to the Timoshenko theory is about7.5% lower than the prediction from using Bernoulli-Euler theory.18.8

For a sandwich column the values computed by these theories lie evenfurther.

The second type of examples of interpretation and application of the abovegeneral formulas is concerned with plate buckling, which is somewhat more

18.8 For such a stubby column plasticity may very well enter making the above elasticanalysis problematic. On the other hand, sandwich columns with a soft core, which isthe common design, may very well buckle without any trace of plasticity.


308

Buckling of an Elastic Plate

complicated than column buckling if for no other reason then because thereare two spatial coordinates instead of one. It may be worth mentioning that,while the Euler Column is postbuckling neutral, i.e. after buckling the loadstays the same independently of the total compression,18.9 plates exhibitstable postbuckling behavior in that the plate is able to support greaterloads after buckling has occurred. One consequence is that after bucklingthe plate still possesses stiffness towards in-plane compression, while thecolumn is infinitely flexible with respect to additional axial compression.

Ex 18-3 Buckling of an Elastic PlateAs mentioned in Chapter 9 plates may carry transverse as well as Plate bucklingin-plane loads, and here we concentrate on the case of in-plane load

λP2

λP2

λP1λP1

Fig. Ex. 18-3.1: Simply supported plate with lubri-cated grips.

on elastic plates. If the load is predominantly transverse the platemay fail because the stresses become too large, but if the load is in-plane the plate may buckle at fairly low stresses in much the same wayas the Euler Column. To be specific, consider the simply supportedplate shown in Fig. Ex. 18-3.1. In the transverse direction the plate issimply supported, i.e. the transverse displacement component w andthe bending momentMnn vanish on the boundary. In its own plane thesides of the plate remain straight, but are subjected to loads throughinfinitely rigid lubricated grips, see Fig. Ex. 18-3.1, and thus the in-plane shear stress Nnt and the displacement derivative un,n vanish onthe boundary. The boundary conditions are given in Fig. Ex. 18-3.2,while the load is shown in Fig. Ex. 18-3.1.

18.9 This is not completely true, but a consequence of the beam theory used here. Onthe other hand, the rotations of the column have to be large before a more exact theorygives answers that differ noticeably from the ones provided by the present theory, seeExamples Ex 8-2 and Ex 7-2.


309

Linear Prebuckling

It may be worth mentioning that the static boundary conditions, whichare quite obvious, also follow from the variational statement, i.e. theprinciple of virtual work for the plate. If we so choose, we may de-rive the static boundary conditions separately for the prebuckling andbuckling problems from their associated variational statements.

N12 = 0u1 = 0w = 0

M11 = 0

a0

b

0

N12 = 0u1,2 = 0w = 0M11 = 0

x1

x2

︷︸︸︷N21 = 0 u2 = 0w = 0 M22 = 0

N21 = 0 u2,1 = 0w = 0 M22 = 0︸︷︷︸

Fig. Ex. 18-3.2: Plate geometry, coordinate systemand boundary conditions.

As far as possible, we follow the same procedure below as in Exam-ple Ex 18-1 and begin by realizing that, for similar reasons as for theLinear prebuckling

reasonable Euler Column, an assumption of linear prebuckling is valid here, too.

We assume that the plate is isotropic and homogeneous, i.e. that itsproperties are independent of direction and coordinate and may there-fore exploit the expressions derived in Section 9.2. The plate thicknessis t, ν designates Poisson’s Ratio, and E denotes Young’s Modulus. Theplate is rectangular with sides a and b, as shown in Fig. Ex. 18-3.2.


As in Example Ex 18-1 we need the generalized displacements, strainsand stresses, here according to the von Karman plate theory developedin Chapter 9, particularly in Sections 9.2 and 9.2.5.

Here, the generalized displacements are the in-plane displacement com-ponents uα and the transverse displacement component w, and thus


u =

u1

u2

u3

≡

u1

u2

w

(Ex. 18-3.1)

It proves convenient to choose the generalized strains as εj , j = 1, . . . , 6,see (Ex. 18-3.2) below, instead of the the membrane strains εαβ to-


310


gether with the bending strains καβ . The main reason for not makingthe latter choice is that it would have been at the expense of having afourth order constitutive tensor Dαβγδ, resulting in more complicatedequations than the ones below.

Generalizedstrains

ε =

ε1

ε2

ε3

ε4

ε5

ε6

≡

ε11

ε22

2ε12

κ11

κ22

2κ12

=

u1,1

u2,2

u1,2 + u2,1

w,11

w,22

2w,12

+ 12

(w,1)2

(w,2)2

2w,1w,2

0

0

0

(Ex. 18-3.2)

Then, the interpretation of the operators l1, l2 and l11 is


l1(u) =

u1,1

u2,2

u1,2 + u2,1

w,11

w,22

2w,12

and l2(u) =

(w,1)2

(w,2)2

2w,1w,2

0

0

0

(Ex. 18-3.3)

and

Bilinear operatorl11(ua,ub) =

wa,1w

b,1

wa,2w

b,2

wa,1w

b,2 + wa

,2wb,1

0

0

0

(Ex. 18-3.4)

where subscripts a and b and superscripts a and b indicate two differentdisplacement fields.

The generalized stresses are the membrane stresses Nαβ and the bend-ing and torsional moments Mαβ arranged according to

Generalizedstresses

σ =

σ1

σ2

σ3

σ4

σ5

σ6

≡

N11

N22

N12

M11

M22

M12

(Ex. 18-3.5)

which means that the constitutive operator H is

Hooke’s “law”H =

DM[DH

] [0]

[0]

DB[DH

]

(Ex. 18-3.6)


311

Linear Prebuckling

where[DH

], DM and DB are given by (9.45), (9.46) and (9.48), re-

spectively

[DH

]≡

1 ν 0ν 1 0

0 0 12(1− ν)

(Ex. 18-3.7)

DM ≡ Et

1− ν2(Ex. 18-3.8)

DB ≡ Et3

12(1− ν2)(Ex. 18-3.9)

and the matrix [0] is

[0] ≡

0 0 0

0 0 0

0 0 0

(Ex. 18-3.10)

In order to formulate variational equations for the prebuckling andbuckling states we need the boundary conditions, which are given be-low.


As usual, we need the boundary conditions for the structural problem.They are already noted in Fig. Ex. 18-3.2, but are recapitulated belowfor completeness

Boundaryconditions

w = 0 , M11 = 0 , u1 = 0 , N12 = 0 for x1 = 0

w = 0 , M11 = 0 , u1,2 = 0 , N12 = 0 for x1 = a

w = 0 , M22 = 0 , u2 = 0 , N12 = 0 for x2 = 0

w = 0 , M22 = 0 , u2,1 = 0 , N12 = 0 for x2 = b

(Ex. 18-3.11)

In addition to these boundary conditions we must fulfill the integralconditions

Loads = integralof stresses on

boundary

λP1 =

∫ b

0

N11dx2 for x1 = 0 and x1 = a

λP2 =

∫ a

0

N22dx1 for x2 = 0 and x2 = b

(Ex. 18-3.12)

From (Ex. 18-3.11) we conclude

w = 0 , w,2 = 0 , w,22 = 0 at (x1 = 0) ∨ (x1 = a)

w = 0 , w,1 = 0 , w,11 = 0 at (x2 = 0) ∨ (x2 = b)(Ex. 18-3.13)

and, in view of the definition (9.1b) of the bending strain and theconstitutive relation given by e.g. (9.61), we may therefore rewrite(Ex. 18-3.11)

Boundaryconditions

w = 0 , w,11 = 0 , u1 = 0 , N12 = 0 for x1 = 0

w = 0 , w,11 = 0 , u1,2 = 0 , N12 = 0 for x1 = a

w = 0 , w,22 = 0 , u2 = 0 , N12 = 0 for x2 = 0

w = 0 , w,22 = 0 , u2,1 = 0 , N12 = 0 for x2 = b

(Ex. 18-3.14)


312


Ex 18-3.2.1 Boundary Conditions and the Airy Stress Func-tion

If we wish to employ the Airy Stress Function Φ, see Section 9.2.5, the The Airy StressFunction Φboundary conditions on the zeroth and first derivatives of the in-plane

displacements deserve special care because the stress function is associ-ated with the second derivatives. In the following we derive boundaryconditions at x1 = a and mention that the equivalent conditions atthe other edges follow by symmetry. The first boundary condition,however, is straightforward and follows directly from the fact that thein-plane shear stress vanishes on the boundary

In-plane shearstress onboundary

N12 = 0 ⇒ Φ,12 = 0 on x1 = a (Ex. 18-3.15)

In order to arrive at the next boundary condition we compute18.10

N21,2 = −Φ,122 = −Φ,221 on x1 = a (Ex. 18-3.16)

and

0 = ε12,2 = 12(u1,22 + u2,12 + w,12w,2 + w,1w,22)

= 12(0 + u2,12 + 0 + 0) on x1 = a

(Ex. 18-3.17)

Further, compute

ε22,1 = u2,21 + w,21w,2 = 0 + 0 on x1 = a (Ex. 18-3.18)

Also,

0 = ε22,1 =1

E(1− ν2)(N22,1 + νN11,1)

⇒ 0 = Φ,111 + νΦ,221 = Φ,111 + νΦ,122

= Φ,111 + 0

⇒ Φ,111 = 0 on x1 = a

(Ex. 18-3.19)

The remaining boundary condition on Φ at x1 = a comes from thecondition (Ex. 18-3.12a)

λP1 =

∫ b

0

N11dx2 =

∫ b

0

Φ,22dx2

⇒ λP1 = Φ,2(a, b)− Φ,2(a, 0)

(Ex. 18-3.20)

To summarize, the boundary conditions on Φ are

Boundaryconditions on Φ

Φ,12 = 0 , Φ,111 = 0 at x1 = 0 and at x1 = a

Φ,12 = 0 , Φ,222 = 0 at x2 = 0 and at x2 = b

Φ,2(a, b)−Φ,2(a, 0) = λP1

Φ,1(a, b)−Φ,1(0, b) = λP2

(Ex. 18-3.21)

18.10 If the following derivations seem less than self-evident to you, you are not the firstone. On the other hand, due to the nature of Φ, it must be clear that displacementboundary conditions must be converted into conditions on stress and after that intostrains and, finally, into derivatives of Φ.


313

Linear Prebuckling


Recall (Ex. 18-3.11) and let ∆α indicate the homogeneous in-planedisplacements at the edges x1 = a and x2 = b, respectively. Then,δu1(a, x2) = δ∆1 and δu2(x1, b) = δ∆2, and the variational statement(18.15) becomes

Principle of virtualdisplacements in

prebuckling

∫ a

0

∫ b

0

(0

Nαβ12(δuα,β + δuβ,α) +

0

Mαβδw,αβ

)dx1dx2

= − P1δ∆1 − P2δ∆2

(Ex. 18-3.22)

or∫ a

0

∫ b

0

(0

Nαβδuα,β +0

Mαβδw,αβ

)dx1dx2

= − P1δ∆1 − P2δ∆2

(Ex. 18-3.23)

because of the symmetry of Nαβ and therefore also of0

Nαβ .

Rewriting of the left-hand side of (Ex. 18-3.23) followed by applicationof the Divergence Theorem yields

∫ a

0

∫ b

0

(0

Nαβδuα,β +0

Mαβδw,αβ

)dx1dx2

=

∫ a

0

∫ b

0

((0

Nαβδuα

)

,β

−0

Nαβ,βδuα

)dx1dx2

+

∫ a

0

∫ b

0

((0

Mαβδw,α

)

,β

−(

0

Mαβ,βδw

)

,α

+0

Mαβ,βαδw

)dx1dx2

=

∮

Γ

0

NαβnαδuαdΓ−∫ a

0

∫ b

0

0

Nαβ,βδuαdx1dx2

+

∮

Γ

0

Mαβnβδw,αdΓ−∮

Γ

0

Mαβ,βnαδwdΓ

+

∫ a

0

∫ b

0

0

Mαβ,βαδwdx1dx2

(Ex. 18-3.24)

The first integral vanishes on x1 = 0 and on x2 = 0, and the remainingcontributions are

∫ b

0

0

N11δu1dx2

∣∣∣∣x1=a

+

∫ 0

a

0

N22δu2(−dx1)

∣∣∣∣x2=b

= δ∆1

∫ b

0

0

N11dx2

∣∣∣∣x1=a

+ δ∆2

∫ a

0

0

N22dx1

∣∣∣∣x2=b

(Ex. 18-3.25)

which must balance the contributions from the applied loads, i.e. theright-hand side of (Ex. 18-3.23). This, however, does not provide anyuseful statements regarding the boundary conditions on the membranestresses Nαβ , only on the resultants.


314


Since the variations δuα of the in-plane displacements uα are arbitraryin the field the second term provides the field equations for in-planeequilibrium

Prebucklingin-plane staticfield equation

0

Nαβ,β = 0 (Ex. 18-3.26)

At the faces δw,n is arbitrary, while δw,t = 0, and thus the third term

yields the boundary conditions on the edge moments0

Mnn Prebucklingboundarycondition onmoment

0

Mnn = 0 , x1 = (0, a) and x2 = (0, b) (Ex. 18-3.27)

The fourth term vanishes because δw = 0 on all sides and thus itfurnishes no information.

There is no transverse load on the plate and, therefore, the fifth termresults in a statement of the transverse equilibrium

Prebucklingtransverse staticfield equation

0

Mαβ,βα = 0 (Ex. 18-3.28)

Of course, we could have obtained these equilibrium equations fromthe general expressions of Section 9.1.6 by linearization and after aslight change of notation, but I judge it instructive to furnish a specificexample.

A possible solution to (Ex. 18-3.25)–(Ex. 18-3.28) is18.11

Prebucklingsolution

0

N11 = − P1

b,

0

N22 = − P2

a,

0

N12 = 00

Mαβ = 0 ,0w = 0

(Ex. 18-3.29)

which may be verified by insertion into the field equations and theboundary conditions when the constitutive relations are observed.

From (Ex. 18-3.29) it is obvious that the prebuckling state is in-plane,as expected.

Ex 18-3.3.1 The Airy Stress Function

A formulation in terms of the Airy Stress Function Φ has its advan-tages, and below we give the relevant equations for the prebucklingstate. After linearization, the equations derived in Sections Ex 18-3.2.1 and 9.2.5 may provide the expressions below. The field equationsbecome, see (9.72) and (9.73)

Prebuckling staticfield equations

∇40

Φ = 0 and DB∇4 0w = 0 (Ex. 18-3.30)

and the in-plane boundary conditions follow from (Ex. 18-3.21)

Prebucklingin-plane staticboundaryconditions

0

Φ,12 = 0 ,0

Φ,111 = 0 at (x1 = 0) ∨ (x1 = a)0

Φ,12 = 0 ,0

Φ,222 = 0 at (x2 = 0) ∨ (x2 = b)0

Φ,2(a, b)−0

Φ,2(a, 0) = P1

0

Φ,1(a, b)−0

Φ,1(0, b) = P2

(Ex. 18-3.31)

18.11 Usually, I like to derive solutions rather than rely on guesswork, which is then provedcorrect, but in the present case it seems justified to me.


315

Linear Prebuckling

The boundary conditions on the transverse displacement component0w are, see (Ex. 18-3.11)

Prebucklingboundary

conditions ontransverse

displacement

0w = 0 ,

0w,11 = 0 for x1 = 0

0w = 0 ,

0w,11 = 0 for x1 = a

0w = 0 ,

0w,22 = 0 for x2 = 0

0w = 0 ,

0w,22 = 0 for x2 = b

(Ex. 18-3.32)

Apart from inconsequential linear contributions to0

Φ the solution in

terms of0

Φ and0w is

Prebucklingsolution

0

Φ = − P1

2b(x2)

2 − P2

2a(x1)

2 and0w = 0 (Ex. 18-3.33)

which, of course, agrees with the solution (Ex. 18-3.29).

Ex 18-3.4 Buckling

As was the case for the prebuckling state we formulate and solve thegoverning equations both as variational statements and in terms of wand Φ. Here, the eigenvalue problem(18.27) becomes

Principle of virtualdisplacements in

bucklingEigenvalue

problem

0 =

∫ a

0

∫ b

0

(1

Nαβδuα,β +1

Mαβδw,αβ

)dx1dx2

+ λc

∫ a

0

∫ b

0

0

Nαβ1w,αδw,βdx1dx2

(Ex. 18-3.34)

where the symmetry of0

Nαβ has been exploited.

The first integral may be handled in the same way as the integral onthe right-hand side of (Ex. 18-3.23). Therefore, we concentrate on thesecond integral. By use of the Divergence Theorem we get

λc

∫ a

0

∫ b

0

0

Nαβ1w,αδw,βdx1dx2

= λc

∫ a

0

∫ b

0

((0

Nαβ1w,αδw

)

,β

−(

0

Nαβ1w,α

)

,β

δw

)dx1dx2

= λc

∮

Γ

0

Nαβ1w,αnβδwdΓ− λc

∫ a

0

∫ b

0

(0

Nαβ1w,α

)

,β

δwdx1dx2

(Ex. 18-3.35)

Here, the first integral vanishes because δw = 0 on the boundary. Thesecond integral and the second term of the first integral in (Ex. 18-3.34) must balance each other since these are the only field terms thatentail δw

0 =

∫ a

0

∫ b

0

1

Mαβ,βαδwdx1dx2

−λc

∫ a

0

∫ b

0

(0

Nαβ1w,α

)

,β

δwdx1dx2 = 0

(Ex. 18-3.36)

Since δw is arbitrary over the interior of the plate this may provide the


316


differential equation, the eigenvalue problem Buckling staticfield equation.Eigenvalueproblem

1

Mαβ,βα − λc

(0

Nαβ1w,α

)

,β

= 0 (Ex. 18-3.37)

When we utilize the ∇-notation we may rewrite the first term, seealso (Ex. 18-3.30) and (9.73), and by use of the prebuckling solution(Ex. 18-3.29) the second term is simplified with the result

Buckling problemDB∇4 1w + λc

(P1

b

1w,11 +

P2

a

1w,22

)= 0 (Ex. 18-3.38)

It may be verified that the in-plane displacements and forces associatedwith buckling all vanish

In-plane bucklingdisplacements orforces vanish

1uα = 0 and

1

Nαβ = 0 (Ex. 18-3.39)

Thus, we are left with the task of solving (Ex. 18-3.38) with its asso-ciated boundary conditions

Bucklingboundaryconditions ontransversedisplacement

1w = 0 ,

1w,11 = 0 for x1 = 0

1w = 0 ,

1w,11 = 0 for x1 = a

1w = 0 ,

1w,22 = 0 for x2 = 0

1w = 0 ,

1w,22 = 0 for x2 = b

(Ex. 18-3.40)

which follow directly from (Ex. 18-3.14) or may be derived from thevariational statement (Ex. 18-3.34) of the buckling problem. Since

the ∇-term as well as the1w,αβ-terms involve differentiation of only

even order, a solution in terms of trigonometric functions presents anobvious possibility. The boundary conditions preclude cosines, andtherefore we investigate whether

Buckling solution?1w = ξ sin

(mπ

x1

a

)sin(nπ

x2

b

)(Ex. 18-3.41)

which satisfies all boundary conditions also satisfies the differentialequation (Ex. 18-3.38). In (Ex. 18-3.40) ξ is an amplitude, m and nare positive integers. After rearrangement of terms (Ex. 18-3.40) and(Ex. 18-3.38) yield

0 =

(DB

((mπ

a

)4+ 2

(mπ

a

)2 (nπ

b

)2+(nπ

b

)4)

− λc

(P1

b

(mπ

a

)2+P2

a

(nπ

b

)2))

×ξ sin(mπ

x1

a

)sin(nπ

x2

b

)

(Ex. 18-3.42)

Nontrivial solutions, i.e. solutions for ξ 6= 0, require that the coefficientof the sines vanish with the result that

Bifurcation loadsλc(m,n) = DB

((mπ

a

)2+(nπ

b

)2)2

P1

b

(mπ

a

)2+P2

a

(nπ

b

)2 (Ex. 18-3.43)


317

Linear Prebuckling

All these values of λc denote bifurcation loads, but in general we areonly interested in the lowest one, which—maybe surprisingly—is notalways the value for (m,n) = (1, 1), as we shall see below.

Ex 18-3.4.1 Specific Plate

As an example, let a = 2b, P1 = P and P2 = 0 and introduce

m ≡(mn

)(Ex. 18-3.44)

and get

λc(m, n) =π2DB

4P b

(m2 + 4

)2n2

m2(Ex. 18-3.45)

Clearly, the larger n the larger the value of λc(m, n) for a fixed value ofm. Therefore, we minimize λc(m, n) with respect to m, which requires

0 = 4m3(m2 + 4

)− 2m

(m2 + 4

)2

⇒ 0 = (m− 2)(m+ 2)(m2 + 4)m ⇒ m = 2

⇒ m = 2n

(Ex. 18-3.46)

Since the lowest possible value of n is 1 we get the minimum of λc(m,n)for18.12

m = 2 and n = 1 (Ex. 18-3.47)

with the associated value

Buckling load λc = λc(2, 1) =4π2DB

P b(Ex. 18-3.48)

and buckling mode

Buckling mode1w = ξ sin

(πx1

b

)sin(πx2

b

)(Ex. 18-3.49)

Fig. Ex. 18-3.3: Buckling mode of plate.α ≡ a/b = 2/1.

Thus, the plate does not “want” to buckle in what may otherwise beconsidered the simplest mode, namely with one half-wave in both di-rections. The fact that the plate has two half-waves in the x1-directionsThe buckling mode

is not the one thatappears to be thesimplest possible

shows that the buckling load is unaltered if a = b because at x1 = b thesame conditions as the boundary conditions at x1 = 2b are satisfiedby the solution (Ex. 18-3.49). By similar reasoning we may furtherconclude that as long as the length of the plate is an integer multipleof its width the buckling load is given by (Ex. 18-3.48).

18.12 Note that both n and m are positive integers, as required.


318


Ex 18-3.4.2 Long Plates

In view of the results for the plate that is twice as long as it is wideit may be interesting to investigate the buckling load of longer andshorter plates. In order to do so, let us normalize the buckling loadson the buckling load of the square plate with side length b, namelyλc of (Ex. 18-3.48). We shall only consider load in the x1-directionand may therefore utilize (Ex. 18-3.43) with n = 1 because any largervalue of n would increase λc. When we normalize in the way indicatedabove, the only parameters left are the ratio α ≡ a/b of the lengthsof the two faces and the number of axial half-waves m with the resultthat the normalized buckling load λc becomes

Normalizedbuckling load oflong plate

λc =

(α2 +m2

2αm

)2, α ≡ a

b(Ex. 18-3.50)

From Fig. Ex. 18-3.50 it is clear that that the buckling load increasesdramatically when a/b is lower than, say 2/3. On the other hand, when

m = 5m = 4m = 3m = 2m = 1

α

λc

543210

2.5

2

1.5

1

0.5

0

Fig. Ex. 18-3.4: Normalized buckling load of longplates.The parameter α ≡ a/b.

a/b is increased from the value 1, then the buckling load becomes moreand more independent of a/b.

Ex 18-3.4.3 The Airy Stress Function

The w-Φ-notation is also applicable to the buckling problem. In line w-Φ-notationwith the procedure from Section 18.3.2 we may write

Φ = λ0

Φ+ Φ and w = λ0w + w (Ex. 18-3.51)


319

Linear Prebuckling

From (9.75) we may then get

0 =DB∇4(λ0w + w)

−eαγeβδ

((λ

0

Φ+ Φ

)

,δγ

(λ

0w + w

))

,β

(Ex. 18-3.52)

When we exploit our knowledge of0

Φ and0w (Ex. 18-3.52) becomes

DB∇4w − eαγeβδ

((λ

0

Φ+ Φ

)

,δγ

w

)

,β

= 0 (Ex. 18-3.53)

and because Φ as well as w is small

DB∇4w − λeαγeβδ

(0

Φ,δγw

)

,β

= 0 (Ex. 18-3.54)

When we insert (Ex. 18-3.50) in (9.72) we may get

∇4

(λ

0

Φ+ Φ

)

=Et

((λ

0w + w

)2,12

−(λ

0w + w

),11

(λ

0w + w

),22

) (Ex. 18-3.55)

Because of the smallness of Φ and w and our knowledge that0w vanishes

∇4Φ = 0 (Ex. 18-3.56)

For buckling, the in-plane boundary conditions are

Buckling in-planeboundaryconditions

1

Φ,12 = 0 ,1

Φ,111 = 0 at x1 = 0 and at x1 = a1

Φ,12 = 0 ,1

Φ,222 = 0 at x2 = 0 and at x2 = b1

Φ,2(a, b)−1

Φ,2(a, 0) = 01

Φ,1(a, b)−1

Φ,1(0, b) = 0

(Ex. 18-3.57)

We may readily verify that the solution found above satisfies the equa-tions given in the w-Φ-notation.

18.3.4 Expansion Theorem

From calculus it is known that any function w can be expanded in terms ofExpansion theorem

eigenfunctions, here the buckling modes uj

Number of modes

Mw = ξjuj , sum over j = 1, . . . ,M (18.37)

where the number of nodes M may be infinite.

The amplitudes ξJ may be determined from (18.37) in conjunction withthe orthogonality condition (18.35). Simply write

σ0 · l11(w,uJ ) = σ0 · l11(ξjuj ,uJ) (18.38)

and exploit the fact that the buckling modes are—or may be taken to be—


320

The Rayleigh Quotient

mutually orthogonal in the sense of (18.35) to get

Participation ξJof uJ in w

ξJ =σ0 · l11(w,uJ)

σ0 · l2(uJ )(18.39)

where ξJ is the participation of uJ in w.

18.3.5 Numerical and Approximate Solutions, the Ray-leigh Quotient

In many cases it is either impossible or not worth the effort to try to deter- Approximate valuesof buckling load

Rayleigh Quotient

and Rayleigh-Ritz

Procedure

mine the exact value of λc and the exact shape of u1. More often than not,we shall therefore resort to numerical or approximate solutions. The Ray-leigh Quotient is a very strong tool in this connection, in particular whenused in the Rayleigh-Ritz Procedure which we establish below.

The Rayleigh Quotient Λ[φ], which is a functional of the kinematicallyadmissible displacement field φ(x), is here defined by

Rayleigh Quotient

Λ[φ]Λ[φ] ≡ −H(l1(φ)) · l1(φ)

σ0 · l2(φ)(18.40)

A comparison between (18.40) and (18.27) with δu = u1 shows that

Λ[u1] = λcΛ[u1] = λc (18.41)

If we are able to guess a function φ that is close to u1 in shape it maytherefore seem reasonable that the Rayleigh Quotient provides us with agood estimate of λc. It is our intention to show that this is true.

18.3.6 Stationarity of the Rayleigh Quotient

Let us investigate Λ[φ] in the neighborhood of u1. In order to do this write

φ = u1 + ǫw , |ǫ| ≪ 1 (18.42)

and insert it into the Rayleigh Quotient to get

Λ[φ] = −H(l1(u1 + ǫw)) · l1(u1 + ǫw)

σ0 · l2(u1 + ǫw)(18.43)

which may be differentiated with respect to ǫ to furnish

∂Λ

∂ǫ=(2σ0 · l11(u1 + ǫw,w)H(l1(u1 + ǫw)) · l1(u1 + ǫw)

− 2σ0 · l2(u1 + ǫw)H(l1(u1 + ǫw)) · l1(w))

×(σ0 · l2(u1 + ǫw)

)−2

(18.44)

For ǫ = 0 the value of this is

∂Λ

∂ǫ

∣∣∣∣ǫ=0

=2σ0 · l11(u1,w)σ1 · ε1 − 2σ0 · l2(u1)σ1 · l1(w)

(σ0 · l2(u1)

)2 (18.45)


321

Linear Prebuckling

From (18.27) with δu = u1 we get

σ1 · ε1 = −λcσ0 · l2(u1) (18.46)

where (18.25) has been introduced.

Now, (18.45) yieldsThe Rayleigh

Quotient Λ[φ] is

stationary for

φ = u1

∂Λ

∂ǫ

∣∣∣∣ǫ=0

= −2σ1 · l1(w) + λcσ0 · l11(u1,w)

σ0 · l2(u1)= 0 (18.47)

The numerator is the eigenvalue problem (18.27) for u1 and λc wherew has been substituted for δu. Therefore, the right-hand side of (18.47)vanishes, and the Rayleigh Quotient Λ[φ] is therefore stationary for φ = u1.

18.3.7 Minimum Property of the Rayleigh Quotient?

The stationarity of the Rayleigh Quotient is, in itself, an important property,Is Λ(φ) minimum

for φ = u1? but if we were able to show that the Rayleigh Quotient attains a minimumfor φ = u1 this tool becomes much stronger because, given two differentapproximations φa and φb, then we could immediately tell which is thebetter. In order to prove this we could compute ∂2Λ/∂ǫ2 for ǫ = 0 and seeif this is always positive. A rather lengthy computation would provide uswith

∂2Λ

∂ǫ2

∣∣∣∣ǫ=0

= 2λc

Λ[w]

H(l1(w)) · l1(w)

σ1 · ε1(Λ[w]− λ1) (18.48)

which is not very helpful in itself. We shall therefore employ a more directapproach. Omitting an unessential factor (recall that the buckling modeis determined to within an arbitrary factor), any kinematically admissiblefield φ can be written as

Number of modes

Mφ = u1 + ajuj , sum over j = 2, 3, . . . ,M (18.49)

whereM , as mentioned above, may be infinite, aj are coefficients, which areindependent of the spatial coordinates, and u1,u2, . . . ,uM are the bucklingmodes associated with λ1(= λc), λ2, . . . , λM ordered such that

λj+1 ≥ λj ≥ λj−1 ≥ . . . ≥ λ1 > 0 (18.50)

If all eigenvalues are negative the ordering must be changed accordingly,and Λ[φ] will prove to take its maximum value for φ = u1. Cases with aProblems when

some eigenvalues

are negative and

others are positive

number of negative eigenvalues, while the rest are positive, are not coveredby the following, and we shall assume that all eigenvalues are of the samesign, in particular positive, as indicated in (18.50). Actually, cases withboth positive and negative eigenvalues need not be contrived since the load


322

The Rayleigh Quotient

on a structure may very well induce compression in parts of the structurefor λ > 0 and tension in other parts.

λP λP

λPλP

Fig. 18.5: Frame with positive and negative buckling loads.

As an example of this, consider the frame in Fig. 18.5 where it must beclear that it may buckle for some negative value of λ as well as for a positiveone because for λ < 0 the columns which were in tension for λ > 0 nowexperience compression while the horizontal member then is in tension.18.13

The numerator on the right-hand side of (18.40) becomes

Sum from index

= 2, not 1H(l1(φ)) · l1(φ) = H(l1(u1 + ajuj)) · l1(u1 + akuk) (18.51)

where a repeated lower-case index indicates summation from 2 to M . Thisnotation is employed throughout the rest of this proof.

The orthogonality condition (18.34) which can also be written as

Buckling modes

are orthogonalσJ · εK = 0 , J 6= K (18.52)

and simplifies (18.51) significantly

H(l1(φ)) · l1(φ) = σ1 · ε1 +M∑

J=2

a2JσJ · εJ (18.53)

Utilize (18.29) with δu = u1,u2, . . . ,M to get

H(l1(φ)) · l1(φ) = −λ1σ0 · l2(u1)−M∑

J=2

a2JλJσ0 · l2(uJ) (18.54)

Normalize all the eigenfunctions such that

σ0 · l2(uJ) = q , J = 1, 2, . . . ,M (18.55)

18.13 For what it is worth, to my experience, these cases seldom present severe problemsas regards the usefulness of the Rayleigh Quotient.


323

Linear Prebuckling

Then, if q is some positive, arbitrary, but fixed number, (18.54) becomes

H(l1(φ)) · l1(φ) = −(λ1 +

M∑

J=2

a2JλJ

)q (18.56)

The denominator of (18.40) is

σ0 · l2(φ) = σ0 · l11(u1 + ajuj,u1 + akuk) (18.57)

The orthogonality condition and the normalization make it possible towrite (18.57) as

σ0 · l2(φ) = q

(1 +

M∑

J=2

a2J

)(18.58)

When we rewrite (18.56)

H(l1(φ)) · l1(φ) = −λ1(1 +

M∑

J=2

a2JλJλ1

)q (18.59)

we may get

Λ[ap] = −−λ1

(1 +

M∑

J=2

a2JλJλ1

)q

(1 +

M∑

J=2

a2J

)q

(18.60)

Recall that

λJ ≥ λ1 (18.61)

This expression, in conjunction with (18.60), shows thatWhen all λj > 0:

Λ[φ] is minimum

for φ = u1

Λ[φ] ≥ λ1 (18.62)

Thus, we have proved that the Rayleigh Quotient— provided that allbuckling loads are positive—is minimum for the buckling mode u1 associ-ated with the lowest buckling load λ1, or λc.

18.3.8 The Rayleigh-Ritz Procedure

Instead of working with one assumed displacement field φ(x) we may writeThe Rayleigh-Ritz

Procedure works

with more than one

trial function

it as a sum of a number, say N , assumed functions, or trial functions,φj(x), j = 1, 2, . . . , N , that are all kinematically admissible

φ(x) = vjφj(x) , sum over j = 1, 2, . . . , N (18.63)

where we return to the usual habit of letting a repeated lower-case indexindicate summation from 1 to N , vj are coefficients, which do not depend on


324

The Rayleigh-Ritz Procedure

the spatial coordinates, and where all the Trial Functions φj(x) are chosen Trial functions

φj(x)once and for all. Below, we omit the indication that the fields φj and σ0depend on the spatial coordinate x. The expression (18.40) for the RayleighQuotient now is

Λ[vp] = −H(l1(vjφj)) · l1(vkφk)

σ0 · l2(vmφm)(18.64)

or, because vj are constants

Λ[vp] = −vjvkH(l1(φj)) · l1(φk)

vmvnσ0 · l11(φm,φn)(18.65)

Note that, in general, φm and φn are not orthogonal.

For simplicity introduce the “stiffness matrix”Kjk

“Stiffness matrix”

KjkKjk = H(l1(φj)) · l1(φk) (18.66)

That Kjk is closely connected with the stiffness of the structure may beseen from reading Part V.

Further, define the so-called Geometric (Stiffness) Matrix 18.14 KGmn

“Geometric

(stiffness) matrix”

KGmn

KGmn = σ0 · l11(φm,φn) (18.67)

which expresses the (de)stabilizing effect of the in-plane or membrane loads—stabilizing when the load is predominantly dominated by tension.

Then,

Λ[vp] = − KjkvjvkKG

mnvmvn(18.68)

We know that the best approximation to λ1 we can obtain from (18.68)is the lowest and therefore we make Λ stationary with respect to all vp, i.e.we require that

Demand that

Λ(vp) is

stationary

∂Λ

∂vp= 0 (18.69)

This gives us

KpkvkKGmnvmvn −KjkvjvkK

Gpnvn = 0 (18.70)

18.14 Here, we have utilized a common Finite Element Method nomenclature, see Part V,according to which Kjk is termed the Stiffness Matrix, and KG

mn is the Geometric (Stiff-

ness) Matrix. It is immediately observed that both matrices Kjk and KGmn are symmetric

in their indices.


325

Linear Prebuckling

With a change of dummy indices

(KGmnvmvnKpj −KmnvmvnK

Gpj)vj = 0 (18.71)

Utilize (18.68) to get a matrix eigenvalueproblemBuckling ∼

matrix eigenvalue

problem

(Kpj +Λ[vm]KG

pj

)vj = 0 (18.72)

which provides us with N eigenvalues λJ and their associated eigenvectors

v(J)j . The lowest of these eigenvalues is Λ1 which is our best approximationto λ1 becauseLowest eigenvalue

Λ1 ' λ1

with eigenvector

v(1)j

Λ1 ≥ λ1 (18.73)

and

ΛJ ≥ Λ1 , J = 2, 3, ..., N (18.74)

In numerical applications it is usually such that Λ1 determined from(18.72) is not the best estimate we can get. The value is improved if wecompute Λ[vp] from (18.68) on the basis of the values of vp found from(18.72).

The Rayleigh-Ritz procedure can be extended to cover the higher orderbuckling modes and their associated buckling loads. We do not, however,intend to include this here.

18.3.9 Another Finite Element Notation

Occasionally, we shall employ a notation that is more in accord with anotherFinite element

notation of the most common ones in Finite Element literature. According to thatnotation a column matrix with elements vj is denoted {v}, a row matrixwith elements Bj by [B], and a two-dimensional matrix with elements Kjk

is given by [K]. Since this notation is less explicit than the one used above itLess information,

but no unnecessary

details

gives less information but, on the other hand, the reader is not bothered byinformation that sometimes is so detailed that it obscures the main messageof an equation. Further, in finite element literature, there is a standard setof matrix names, which we introduce below through rewriting some of theabove formulas.

Equation (18.63), which gives the displacement interpolation, becomesDisplacement

interpolation

matrix [N ]

{u} = [N ]{v} (18.75)

while the linear strain contribution {e} is given byStrain distribution

matrix [B]{e} = [B]{v} (18.76)

where the strain distribution matrix [B] is derived from [N ] by proper(and obvious) differentiations according to the linear part of the strain-displacement relation, i.e. from l1(u). Note that [N ] and {v} correspond toφj and vj , respectively.


326

Finite Element Notation

The linear constitutive operator H is replaced by a constitutive matrix[D] such that the linear part {s} of the stress vector {σ} is

Constitutive

matrix [D]{s} = [D]{e} = [D][B]{v} (18.77)

The geometric matrix [G] is given by the nonlinear term of the strain-displacement relation, i.e. by l2(u), and is therefore derived from [N ] byproper differentiations. Thus, we can write the total strain matrix {ε}

Geometric matrix

[G]{ε} = [B]{v}+ 1

2

∑

k

{jk}({v}T [G]T [G]{v}

)k

(18.78)

where the meaning of {jk} follows from Examples Ex 18-4.1 and Ex 18-4.2. In the second term of the right-hand side of (18.78) we sum over Second term of

(18.78) explained

later

the nonlinear strain components, whose number usually is smaller than thetotal number of strain components. The above mentioned examples oughtto clarify the way this term may be interpreted for theoretical purposes.

Ex 18-4 Interpretation of∑

k

({jk}{v}T [G]T [G]{v}

)k

Ex 18-4.1 Bernoulli-Euler Beam

Consider a Bernoulli-Euler beam and recall (7.12) and (7.14) whichmay be collected to

Generalizedstrains {ε}{ε} =

[ε

κ

]=

[u′

w′′

]+ 1

2

[(w′)2

0

](Ex. 18-4.1)

Utilizing the finite element notation this may be written

Generalizedstrains {ε}{ε} = [N ]{v} + 1

2

[1

0

]{v}T [G]T [G]{v} (Ex. 18-4.2)

Ex 18-4.2 von Karman Plate

For a von Karman plate there are 6 generalized strains, namely 3 mem-brane strains and 3 bending strains, but only the membrane strainscontain nonlinear terms. Thus, the situation becomes more compli-cated. Recall (9.1) or, more convenient for our present purpose, (Ex. 18-3.2) which, with a slight change of notation is

Generalizedstrains {ε}{ε} =

ε1

ε2

ε3

ε4

ε5

ε6

≡

ε11

ε22

2ε12

κ11

κ22

2κ12

=

u1,1

u2,2

u1,2 + u2,1

w,11

w,22

2w,12

+ 12

(w,1)2

(w,2)2

2w,1w,2

0

0

0

(Ex. 18-4.3)

Define [Gα] , α = [1, 2] by

w,1 = [G1]{v} and w,2 = [G2]{v} (Ex. 18-4.4)


327

Linear Prebuckling

then we may write (Ex. 18-4.3) in the following form

{ε} = [N ]{v}+ 12{j1}{v}T [G1]

T [G1]{v}

+ 12{j2}{v}T [G2]

T [G2]{v}

+ 12{j3}{v}T [G1]

T [G2]{v}

+ 12{j3}{v}T [G2]

T [G1]{v}

(Ex. 18-4.5)

where

{j1}T ≡ [ 1 , 0 , 0 , 0 , 0 , 0 ]

{j2}T ≡ [ 0 , 1 , 0 , 0 , 0 , 0 ]

{j3}T ≡ [ 0 , 0 , 1 , 0 , 0 , 0 ]

(Ex. 18-4.6)

Based on these examples interpretation of∑

k{jk}{v}T [G]T [G]{v} for othertypes of structures ought to be fairly easy.

Now, the stress vector {σ} is

Stress vector {σ} {σ} = [D]{ε} = [D]([B]{v}+ 1

2

∑

k

({jk}{v}T [G]T [G]{v}

)k

)(18.79)

Again, the nonlinear term must be handled the way described above.

With the above formulas in hand, we can compute the (linear) stiffnessmatrix [K]

Linear stiffness

matrix [K][K] = [B]T · ([D][B]) (18.80)

The geometric (stiffness) matrix [KG] isGeometric

(stiffness) matrix

[KG]

[KG] = σ0 ·([G]T [G]

)(18.81)

where the dot implies integration over the structure, and in (18.80) and(18.81) we have taken the liberty of mixing the Budiansky-Hutchinson No-tation with the Finite Element Notation.

Then, the Rayleigh Quotient (18.68) becomes

The Rayleigh

Quotient Λ[{v}] Λ[{v}] = − {v}T [K]{v}{v}T [KG]{v}

(18.82)

and the eigenvalue problem (18.72) is

Matrix eigenvalue

problem

([K] +Λm[KG]

){v} = {0} (18.83)

Practical Implementation of∑

k

({jk}{v}T [G]T [G]{v}

)k

As often is the case interpretations the above kind are useful in theoreticalDon’t use {jk} in

implementations context, but rather inappropriate in connection with practical implementa-tion in a computer program, see also page 446 and page 449, because wedon’t want to multiply by zero if it can be avoided.


328

Buckling of Roorda’s Frame by the Rayleigh-Ritz Procedure 329

18.3.10 A Word of Caution

In applications you should avoid establishing Λ as a function of vj and thendifferentiate with respect to vp. It is always easier and more efficient tosolve the eigenvalue problem (18.72) or (18.83).

18.3.11 Examples of Application of the Rayleigh Quo-tient and the Rayleigh-Ritz Procedure

In most cases it is not possible to find an exact value of the classical crit-ical load λc, and in general we must therefore resort to some numerical orapproximate method.18.15 By far the most commonly used method is toexploit the minimum property of the Rayleigh Quotient,18.16 in particularin connection with the Rayleigh-Ritz Procedure. However, in order to deter-mine the validity of the results obtained in this way, we study examples forwhich exact solutions are known.

It may be worth mentioning that today the Finite Element Method is For complicated

geometries, use the

Finite Element

Method

usually applied when the geometry of the structure in question is compli-cated. For structures of the kind investigated below, the Rayleigh-Ritz Pro-cedure, with assumed displacement fields that are continuous everywhere,suffices. On the other hand, the study of Roorda’s Frame in Example Ex 18-5 entails a simple application of the Finite Element Method, and in all ex-amples the notation of that method is employed because of its clarity andease.

Ex 18-5 Roorda’s Frame—Application of theRayleigh-Ritz ProcedureAs an example of application of the Rayleigh Quotient in connection Rayleigh-Ritz

Procedure appliedto Roorda’s frame

with the Rayleigh-Ritz Procedure let us consider the famous Roorda’s

Frame, see Fig. Ex. 18-5.1, whose elastic imperfection sensitivity wasdetermined experimentally by Roorda (1965). Later its postbucklingbehavior and imperfection sensitivity were analyzed by Koiter (1966),and its elastic-plastic postbuckling and imperfection sensitivity were in-vestigated by Byskov (1989). In the present analysis—as in Koiter’s—we shall assume that the members of the frame are inextensible with Inextensibility ⇒

linear prebucklingthe result that the prebuckling path entails no displacements but onlyaxial forces

Linear prebucklingN0AB = −P = −EI

L2and N0

BC = 0 (Ex. 18-5.1)

Thus, the prebuckling state is linear in the strictest sense. As a refer-ence, we shall occasionally cite Koiter’s results (Koiter 1966), but wedo not include his analysis here—beautiful as it is.

18.15 By an approximate method I mean one that furnishes formulas, while a numericalmethod results in numbers.18.16 Recall that minimum is guarantied when all buckling loads are positive, otherwiseRayleigh’s Quotient is only stationary.


Linear Prebuckling

A

B C

λP

L

ξ

L

Fig. Ex. 18-5.1: Roorda’s Frame.

The Rayleigh Quotient (18.40) is

Rayleigh quotientΛ[φ]

Λ[φ] = −H(l1(φ)) · l1(φ)

σ0 · l2(φ)(Ex. 18-5.2)

In the present case

H(l1(φ)) · l1(φ) =

∫ B

A

EI(w′′

1 )2dx+

∫ C

B

EI(w′′

1 )2dx(Ex. 18-5.3)

and

σ0 · l2(φ) =

∫ B

A

N0AB(w′

1)2dx+

∫ C

B

N0BC (w′

1)2dx

=

∫ B

A

N0AB(w′

1)2dx

(Ex. 18-5.4)

The denominator in the Rayleigh Quotient does not contain contribu-tions from the axial displacement component u1, only from the trans-verse component w1. If we allowed the frame to be extensible in buck-ling, then the numerator would entail a term that is quadratic in u1

and, since the Rayleigh Quotient attains a minimum for the exact so-Only transversedisplacements in

bucklinglution, it is immediately seen that u1 must vanish and that the onlydisplacement component of the buckling mode is w1. Therefore, wemay write the Rayleigh Quotient as

Rayleigh quotientΛ[w]

Λ[w] = −

∫ B

A

EI(w′′)2dx+

∫ C

B

EI(w′′)2dx

∫ B

A

N0AB(w′)2dx

(Ex. 18-5.5)

where w signifies the approximation to the transverse component ofthe buckling mode u1.

18.17

18.17 Don’t let the minus sign bother you, recall that N0AB < 0.


330


Ex 18-5.1 One-Parameter Solution

In AB and in BC the simplest admissible displacement fields, which One-ParameterSolutionalso satisfy all static boundary conditions,18.18 namely that the bending

moments vanish at the supports, are given byTrial functionwhich satisfy thestatic boundaryconditions

wAB = L(− 1

2ζ3 + 1

2ζ)ξ

wBC = L(− 1

2ζ3 + 3

2ζ2 − ζ

)ξ

(Ex. 18-5.6)

where

ζ ≡ x

L(Ex. 18-5.7)

With (Ex. 18-5.6) in hand we could utilize (Ex. 18-5.5) directly, butbecause we later increase the number of fields we use the finite elementnotation given in Section 18.3.9.

By comparing (Ex. 18-5.6) with (18.75) we may make the followingidentifications

{v(1)} = {ξ} (Ex. 18-5.8)

where subscript (1) indicates that this is our first approximation, and

Displacementinterpolationmatrices [N ]

[NAB ] =[L(− 1

2ζ3 + 1

2ζ)]

[NBC ] =[L(− 1

2ζ3 + 3

2ζ2 − ζ

)] (Ex. 18-5.9)

Compare (18.78) with the expression for the axial strain ε, see (7.12)

ε ≡ u′ + 12

(w′)2

(Ex. 18-5.10)

and get

Geometric matrix[G]

[GAB ] =1

L

d[NAB ]

dζ= [− 3

2ζ2 + 1

2]

[GBC ] =1

L

d[NBC ]

dζ= [− 3

2ζ2 + 3ζ − 1]

(Ex. 18-5.11)

Since the only non-vanishing buckling strain is the bending strain κ Only bending strainin bucklingwhich is defined by, see (7.14)

κ ≡ w′′ (Ex. 18-5.12)

we find

Strain distributionmatrix [B]

[BAB ] =1

L2

d2[NAB ]

dζ2=

1

L[−3ζ]

[BBC ] =1

L2

d2[NBC ]

dζ2=

1

L[−3ζ + 3]

(Ex. 18-5.13)

18.18 It is not necessary that the trial functions satisfy any static conditions, but, some-times, with a given number of trial functions we get a better solution if the static boundaryconditions are fulfilled, but see Example Ex 32-4, in particular the comments page 571and Example Ex 32-1, page 551 where a similar situation is discussed. In the presentcase, it is easy to find such functions, but for other structures it may not be.


Linear Prebuckling

The constitutive matrix [D] isConstitutivematrix [D]

[D] = [EI ] (Ex. 18-5.14)

Now, it is easily found that

Contributions tostiffness matrix

[KAB ] =

[+3

EI

L

]=EI

L[+3]

[KBC ] =

[+3

EI

L

]=EI

L[+3]

(Ex. 18-5.15)

and that

Contributions togeometric

(stiffness) matrix

[KGAB ] =

[−P L

5

]=EI

L

[− 1

5

]

[KGBC ] = [0]

(Ex. 18-5.16)

For the whole structureEigenvalue

problem([K] + λ[KG]) {v(1)} = {0} (Ex. 18-5.17)

withStiffness matrix

[K]and geometric

(stiffness) matrix[KG]

[K] =EI

L[+6]

[KG] =

[−P L

5

]=EI

L[− 1

5]

(Ex. 18-5.18)

The solution is

λ(1) ≫ λex λ(1) = 30 (Ex. 18-5.19)

while the exact value—to 14 significant digits—is given by

λex λex = 13.885 942 905 965 (Ex. 18-5.20)

The value of λ(1), given by (Ex. 18-5.19), is clearly not satisfactory. Thereason is that the assumption for [NAB ] given by (Ex. 18-5.9a) does notaccount for the change in displacement field caused by the influence ofthe compressive axial force in AB. Therefore, the expression for [KAB ],see (Ex. 18-5.15a), is inaccurate, while the expression for [KBC ] givenby (Ex. 18-5.15b) is exact.

The presence of the beam BC has increased the buckling load byaround 40% over the buckling load of the Euler Column which is givenby the value λEuler

c = π2.

Ex 18-5.2 Two-Parameter Solution

Since the displacement field utilized above is exact in BC, we focusTwo-ParameterSolution attention on AB. There are, of course, many possible ways to improve

the displacement field, e.g. dividing AB into a number of finite ele-ments. We shall, however, pursue a different course in that we stilllet AB be an undivided part of the frame, but increase the number ofdisplacement functions in that column. This may be done in many dif-ferent ways. However, because of the inextensibility of the beams the


332


additional term(s) in [NAB ] must fulfill the condition that the bucklingmode entails vanishing displacements at A and at B. Furthermore, wechoose the additional term(s) in [NAB ] such that they provide no ad-ditional rotation at B. This may seem like an arbitrary choice, but thereason for doing this is that the buckling mode amplitude is then eas-ily identified as the displacement field parameter associated with thefirst term in [NAB ]. Especially in the computation of the postbucklingcoefficient a, see Example Ex 19-1 this proves very convenient.

Finally, if we can satisfy the static boundary condition that the bend-ing moment MA at A vanishes, we may expect a good approximationto the column displacements. It is easily verified that the followingassumption conforms to our requirements Displacement

interpolationmatrix [N ]

[NAB ] = L[− 12ζ3 + 1

2ζ ; +ζ4 − 3

2ζ3 + 1

2ζ] (Ex. 18-5.21)

with [NBC ] unchanged. Differentiations yieldGeometric matrix[G]

[GAB ] = [N ′

AB ] = [− 32ζ2 + 1

2; +4ζ3 − 9

2ζ2 + 1

2] (Ex. 18-5.22)

and furtherStrain distributionmatrix [B]

[BAB ] =1

L[−3ζ ; +12ζ2 − 9ζ] (Ex. 18-5.23)

Compute the stiffness matrix [KAB ]

[KAB ] =

∫ B

A

EI [BAB ]T [BAB ]dx

=EI

L2L

∫ 1

0

[BAB ]T [BAB ]dζ

=EI

L

∫ 1

0

[−3ζ

+12ζ2 − 9ζ

][−3ζ ; +12ζ2 − 9ζ

]dζ

(Ex. 18-5.24)

with the result

Stiffness matrix[K]

[KAB ] =EI

L

[+3 0

0 + 95

](Ex. 18-5.25)

It may be worthwhile mentioning that the expression for [KAB ] doesnot entail off-diagonal terms, which means that the two elements in[NAB ] are mutually orthogonal. This is desirable in any approximationbecause it results in numerically stable systems of equations and lesswork.

In a similar way, we compute the geometric (stiffness) matrix [KGAB ]

[KGAB ] =

∫ B

A

N0AB [GAB ]T [GAB ]dx

= −PL∫ 1

0

[GAB ]T [GAB ]dζ

= −PL∫ 1

0

[− 3

2ζ2 + 1

2

+4ζ3 − 92ζ2 + 1

2

]

[− 3

2ζ2 + 1

2; +4ζ3 − 9

2ζ2 + 1

2

]dζ

(Ex. 18-5.26)


Linear Prebuckling

The result isContribution to

geometric(stiffness) matrix

[KGAB ]

[KGAB ] = −EI

L

[+ 1

5+ 1

10

+ 110

+ 335

](Ex. 18-5.27)

As before, the first element in [K] is twice that of [KAB ] because thebeam BC also contributes stiffness against rotation of B. Then, theeigenvalue problem is still given in the form of (Ex. 18-5.17), whichresults in the equation

+1

140λ2 − 153

175λ+

54

5= 0 (Ex. 18-5.28)

with the solution

λ(2) ≈ λex

λ(2) = min(+ 306

5± 6

5

√1551

)= 13.94071520

cf. λex = 13.8859735(Ex. 18-5.29)

where, as indicated, only the smallest solution is of any interest in thepresent connection. The relative error on the approximate solution isonly 0.4%, which is sufficiently accurate for almost any purpose. Theeigenvector, i.e. the approximation to the buckling mode, associatedwith λ(2) is {v}(2), with

buckling mode {vAB}(2) =[

1

(9 +√1551)/21

]=

[1

2.303 939 873

](Ex. 18-5.30)

where we have normalized the buckling mode such that ξ = 1 corre-sponds to a rotation of one radian of point B. For BC

{vBC}(2) = [1] (Ex. 18-5.31)

It is, of course, possible to continue the above process, and for the sakeof completeness I give the expression for [NAB ]

More terms in[NAB ]

[NAB ] = [− 12ζ3 + 1

2ζ ; ζ4 − 3

2ζ3 + 1

2ζ ; . . . ;

. . . ; ζj+2 − j+1jζj+1 + 1

jζ ; . . . ]

(Ex. 18-5.32)

and the values of the approximations to λc are

λ(1) = 30 , λ(1) = 2.160 458

λ(2) = 13.940 715 20 , λ(2) = 1.003 944

λ(3) = 13.921 293 88 , λ(3) = 1.002 545

λ(4) = 13.886 049 19 , λ(4) = 1.000 007

λ(5) = 13.885 973 47 , λ(5) = 1.000 002

λ(6) = 13.885 973 47 , λ(6) = 1.000 002

(Ex. 18-5.33)

where λ(j) ≡ λ(j)/λex .

The effort involved in establishing and solving the eigenvalue problemgrows geometrically with the number of terms and, even with as few asthree terms in [NAB ] it is wise to resort to some analytic manipulationprogram, such as maxima, Mathematica, Maple, MuPAD, or the olderMuMath.


334

Buckling of a Plane Plate by the Rayleigh-Ritz Procedure 335

In view of our success with Roorda’s Frame we may try the Rayleigh-Ritzprocedure on another kind of structure, namely a plate. The geometry issimpler than that of Roorda’s Frame, but, as we have seen in Example Ex 18-3, the buckling mode of the plate depends strongly on the ratio between itstwo sides. Therefore, if we utilized the buckling mode for a square plate toanalyze a long plate we would get very erroneous results because it cannotmodel a buckling mode with many half-waves. From Fig. Ex. 18-3.4 it mustbe clear that such an assumption would predict much too high values ofthe buckling load. For instance, if the plate is twice as long as it is widethe overshoot would be about 50%, see the curve for m = 1. Even if weextended the investigation to many more terms than the ones used belowand saw convergence of the estimates of the buckling load, we would still notget close to the correct result if we did not include terms that allowed formore buckles. As always, engineering insight into the nature of the problem Engineering insight

is importantat hand is pertinent.

Ex 18-6 Plate Buckling—Rayleigh-Ritz Pro-cedureAs our second example of the Rayleigh Quotient and the Rayleigh-Ritz Rayleigh-Ritz

ProcedureProcedurewe return to the problem of buckling of a plane plate, seeExample Ex 18-3, and exploit the information about the prebucklingstate given there, see (Ex. 18-3.29)

0

N11 = − P1

b,

0

N22 = − P2

a,

0

N12 = 00

Mαβ = 0 ,0w = 0

(Ex. 18-6.1)

We shall also rely on the interpretation of the Budiansky-Hutchinson

notation given in Example Ex 18-3.1. In order to keep the analysissimple we limit ourselves to the case of a square plate with equal com-pression in both in-plane directions, i.e.

Square plateEqual compression

b = a and P1 = P2 = P (Ex. 18-6.2)

As in the example that was concerned with Roorda’s Frame, see Ex-ample Ex 18-5, the Rayleigh Quotient (18.40) is

Rayleigh QuotientΛ[φ] = −H(l1(φ)) · l1(φ)

σ0 · l2(φ)(Ex. 18-6.3)

where,

H(l1(φ)) · l1(φ) =

∫

A

DBαβγδ

1

καβ

1

κδγdA

σ0 · l2(φ) =

∫

A

0

Nαβ

1

w,α

1

w,βdA

(Ex. 18-6.3)

and tilde (˜) indicates that the field is an approximate one.

When we introduce the strain-displacement relation (9.1b) the Rayleigh


Linear Prebuckling

Quotient may be written in terms of displacement derivatives1

w,α and1

w,αβ only. The expression then is

Rayleigh Quotient Λ[w] = −

∫

A

DBαβγδ

1

w,αβ

1

w,δγdA∫

A

0

Nαβ

1

w,α

1

w,βdA

(Ex. 18-6.4)

Ex 18-6.1 One-Parameter Solution

Recall that the approximating displacement field does not need to sat-One-Parametersolution isfy the static boundary conditions and choose the simplest nondimen-

sional field possibleAssumed w1 need

not satisfy anystatic conditions

1

w = 16ζ1(1− ζ1)ζ2(1− ζ2) (Ex. 18-6.5)

where

ζα ≡ xα

a, α = (1, 2) (Ex. 18-6.6)

We may cast this in the Finite Element notation and write the dis-placement as

w = [N ]{v} (Ex. 18-6.7)

with the displacement interpolation matrix [N ]18.19Displacementinterpolationmatrix [N ]

[N ] = a[16ζ1(1− ζ1)ζ2(1− ζ2)] (Ex. 18-6.8)

which provides the geometric matrices [G1] and [G2]

Geometricmatrices [G]α

[G1] = [N ],1 = [16ζ2 (−1 + ζ2) (−1 + 2ζ1)]

[G2] = [N ],2 = [16ζ1 (−1 + 2ζ2) (−1 + ζ1)](Ex. 18-6.9)

and the strain distribution matrix [B]

Strain distributionmatrix [B]

[B] =

[N ],11

[N ],22

2[N ],12

=

1

a

32ζ2 (−1 + ζ2)

32ζ1 (−1 + ζ1)

(−32 + 64ζ2) (−1 + 2ζ1)

(Ex. 18-6.10)

where the strains and stresses are generalized in the same fashion asin Example Ex 18-3.1.

Then, the stiffness matrix [K] is

Stiffness matrix[K]

[K] =

∫

A

DB [B]T[DH

][B]dA =

[563245

DB]

(Ex. 18-6.11)

where the expressions for DB and[DH

]may be found in Exam-

ple Ex 18-3.1, and where the geometric (stiffness) matrix becomesGeometric

(stiffness) matrix[KG]

[KG] =

∫

A

0

Nαβ [Gα]T [Gβ ]dA =

[− 256

45aP]

(Ex. 18-6.12)

18.19 The term “Displacement interpolation matrix” may be a little misleading here inthat the displacement is not interpolated between nodes, but, still, the meaning ought tobe clear.


336


Since the eigenvalue problem that determines the buckling load is givenas

Eigenvalueproblem

([K] + λ[KG]

){v1} = {0} (Ex. 18-6.13)

it is easily found that the first approximation λ(1) to λc is

λ(1) off by 11%

λ(1) =5632

256

DB

aP= 22

DB

aP

cf. λex = 2π2 DB

aP= 19.739 209

DB

aP

(Ex. 18-6.14)

Thus, the error on λ(1) is about 11%. This is half the error of the one-parameter solution to the Euler Column in Example Ex 32-5.1, whichat first may be surprising because the assumed displacement fieldsare of the same order in both cases. However, compare (Ex. 18-6.10)with (Ex. 32-5.5b) and realize that while the latter predicts a constantbending moment the former shows that the displacement assumption(Ex. 18-6.5) results in bending and torsional moments that vary overthe plate, although κ11 is constant in the x1-direction (the ζ1-direction)and κ22 is constant in the x2-direction. This means that neither ofthe bending moments M11 and M22 nor the torsional moment M12

vanishes independently of the value of ν. It is thus reasonable thatthe displacement assumption for the plate provides a somewhat betterapproximation to the classical critical load.

Ex 18-6.2 Two-Parameter Solution

An error of 11% is probably more than an engineer is willing to accept, Two-ParameterSolutionso we try to improve the solution by adding one more term to the

displacement approximation Displacementinterpolationmatrix [N ]

[N ] = a[16ζ1(1− ζ1)ζ2(1− ζ2) ;

256 (ζ1)2 (1− ζ1)

2 (ζ2)2 (1− ζ2)

2]

(Ex. 18-6.15)

Now, the elements of [G1] are

Geometric matrix[G]

G1(1) = N(1),1 = 16ζ2 (−1 + ζ2) (−1 + 2ζ1)

G1(2) = N(2),1 = 512 (ζ2)2 ζ1 (−1 + ζ2)

2

× (−1 + 2ζ1) (−1 + ζ1)

(Ex. 18-6.16)

and the elements of [G2] become

Geometric matrix[G]

G2(1) = N(2),1 = 16ζ1 (−1 + 2ζ2) (−1 + ζ1)

G2(2) = N(2),2 = 512 (ζ1)2 ζ2 (−1 + 2ζ2)

× (−1 + ζ2) (−1 + ζ1)2

(Ex. 18-6.17)

Similarly, the elements of the first row of [B] are

First row of straindistributionmatrix [B]

B(1, 1) = N(1),11 =1

a32ζ2 (−1 + ζ2)

B(1, 2) = N(2),11 =1

a512 (ζ2)

2 (−1 + ζ2)2

×(1 + 6 (ζ1)

2 − 6ζ1)

(Ex. 18-6.18)


Linear Prebuckling

The elements of the second row of [B] are

Second row ofstrain distribution

matrix [B]

B(2, 1) = N(1),22 =1

a32ζ1 (−1 + ζ1)

B(2, 2) = N(2),22 =1

a512 (ζ1)

2 (−1 + ζ1)2

×(1− 6ζ2 + 6 (ζ2)

2)

(Ex. 18-6.19)

The elements of the third row of [B] are

Third row ofstrain distribution

matrix [B]

B(3, 1) = 2N(1),12 =1

a(−32 + 64ζ2) (−1 + 2ζ1)

B(3, 2) = 2N(2),12 =1

a2048ζ1ζ2

× (−1 + 2ζ2) (−1 + ζ2)

× (−1 + 2ζ1) (−1 + ζ1)

(Ex. 18-6.20)

With the above expressions for [G1], [G2] and [B] in hand it is a fairlystraightforward, but tedious,18.20 process to compute [K] and [KG],and only the final results for these matrices are given below

Stiffness matrix[K]

[K] = DB

5632

45

8192

225

8192

225

262144

1225

(Ex. 18-6.21)

and

Geometric(stiffness) matrix

[KG][KG] = −P a

256

45

2048

525

2048

525

131072

33075

(Ex. 18-6.22)

The ensuing eigenvalue problem isEigenvalue

problem

([K] + λ[KG]

){v} = {0} (Ex. 18-6.23)

see also (18.83). In this case the easiest way to get the solution is todetermine the roots of the determinant of the coefficient matrix to {v}

∣∣[K] + λ[KG]∣∣ = 0 (Ex. 18-6.24)

which becomes

631494410242480625

− 106367549447441875

(λaP

DB

)+ 54525952

7441875

(λaP

DB

)2= 0(Ex. 18-6.25)

with the solution

λ =DB

aP

4

13

[317−

√63790

317 +√63790

]=DB

aP

[19.825 593

175.251 330

](Ex. 18-6.26)

Here, we are only interested in the smaller of these eigenvalues, whichis our second approximation to λc

18.20 I left this task for maxima to handle.


338


The error on our final approximation, namely

λ(2) ≈ λcλ(2) = 19.825 593DB

aP(Ex. 18-6.27)

is as little as 0.4%, which must be deemed acceptable for all practicalpurposes.

18.3.12 Concluding Comments on the Examples Above

It seems obvious that the results obtained in the preceding examples arevery accurate although the effort involved was moderate. One might objectthat the analytical derivations involved in the plate examples, in particularin Example Ex 18-6.2, are lengthy. However, left to some analytic manipula-tion program, such as maxima, Mathematica, Maple, MuPAD, or even the oldMuMath, the two-parameter solution in Example Ex 18-6.2 only requires aprogram of about 50–100 lines including print-out commands. Furthermore,with only minor modifications the program can be used for more terms withthe comment that the run-times may become prohibitively long.


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