Stability Of Structures: Continuous Models

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18Stability OfStructures:

Continuous Models

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Lecture 18: STABILITY OF STRUCTURES: CONTINUOUS MODELS 182

TABLE OF CONTENTS

Page

18.1. Example 1: The Euler Column 18318.1.1. Critical Load Analysis . . . . . . . . . . . . . . 184

18.1.2. Linearization Limitations . . . . . . . . . . . . . 185

18.1.3. Reformulation as Eigenproblem . . . . . . . . . . . 186

18.2. The Fixed-Pinned Column 186

18.2.1. Critical Load Analysis . . . . . . . . . . . . . . 186

18.3. Elastically Restrained Column 188

18.3.1. Problem Description . . . . . . . . . . . . . . . 188

18.3.2. Critical Load Analysis . . . . . . . . . . . . . . 188

18.3.3. Equivalent Spring Constant . . . . . . . . . . . . . 1810

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183 18.1 EXAMPLE 1: THE EULER COLUMN

This Lecture focuses on buckling analysis ofcontinuous modelsof elastic systems. This process

leads to ordinary or partial differential equations. Those can be solved in closed form only for simple

1D structures, such as prismatic columns treated by linearized prebuckling (LPB).

Despite limitations in terms of obtaining analytical solutions, such problems are important for dis-

playing key intrisic features of continuous models. For example, the associated eigenproblem has

an infinite (but denumerable) set of eigenvalues, meaning an infinite number of critical loads. (Thisshould not be surprising, since continuous models have an infinite number of degrees of freedom.)

Of these, the lowest buckling load is of primary interest to designers.

We illustrate the treatmentof continuous modelsusing viaseveralexamples involving elastic columns

with simple boundary conditions. All of the examples have closed form solutions.

18.1. Example 1: The Euler Column

xx

y yy

v(x)

P P

P

B

M(x) =P v(+ as drawn)

P

X X'

v(x)

X'x

X

constantEI

P

B

X

(a) (b) (c)

elastic

AA A

Lassumed

buckled shape

Figure18.1. The Euler column: (a) problem definition; (b) FBD of whole column assuming a

buckled shapev(x)with its amplitude exaggerated for visibility; (c) FBD at distancexfrom top.

The Euler column is shown in Figure 18.1(a). It is a homogeneous, prismatric, elastic column pinned

(hinged, simply supported) at both ends A and B, and subjected to axial loadP . The elastic modulus

isE. Reference axes are chosen as follows:x is longitudinal, with origin at end A;zis normal to the

plane of thefigure and selected so thatI= Izzis theminimumsecond order of inertia of the columncross section aboutz; finallyy lies in the plane of the figure. For example, if the cross section is a

solid rectangle of widthband thicknesst


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185 18.1 EXAMPLE 1: THE EULER COLUMN

are L = n , in which n= 1, 2, . . .is a positive integer. Square both sides for convenience:2L2 = n2 2, replace2 by P/(E I), solve for Pand tag those loads as critical:

Pcr,n=n2 2E I

L2 , n= 1, 2, . . . (18.7)

As can be observed, there is an infinite number of critical loads. These are the nontrivial solutions

of the characteristic equation (18.5). The lowest one is associated with n= 1:

Pcr,1= Pcr= 2E I

L2 . (18.8)

This is called theEuler critical load.

Mode shape 1(n=1)

P = P =

Mode shape 2(n=2)

Mode shape 3(n=3)

cr cr1 L

E I2

2 P =cr2 L

4 E I2

2 P =cr3 L

9 E I2

2

L

L/2

L/2

L/3

L/3

L/3

Figure18.2. First three buckling mode shapes for the Euler column.

The buckling mode shapes associated with the set of critical loads (18.7) are

vcr,n(x) = Bsinnx

L. (18.9)

Ifn= 1 the mode shape is a half sine wave, similar to that drawn in Figure 18.1(c). Ifn= 2 we geta complete sine wave, ifn= 3, one-and-a-half sine wave, etc. These buckling shapes are drawn forn= 1, 2, 3 in Figure 18.2.

18.1.2. Linearization Limitations

TheLPBassumption does yields thecritical loads,but also bringsabout thefollowing inconsistencies,

some of which have an air of paradox.

The foregoing solution says nothingabout the amplitudeof the buckling modes (18.8) in termsof the data. This is a consequence of the LPB assumptions, which led to a simple linear ODE but

filters thaat information. Determination of thispostbucklingbehavior is quite involved, since it

requires solving a nonlinear ODE in terms of elliptic functions. Such analysis shows that the

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buckled columncontiinues to take up on increasing load, albeit very slowly, as long as it remains

elastic.

Differentiating (18.7) three times gives a nonzero transverse shear load Vy (x)= E I Bv cr1=E I( 3/L3)Bcos(x/L). Evaluation at x=0 or x= Lgives a nonzero transverse shearforce whereas Figure 18.1(b) shows zero horizontal reactions there. A similar inconsistency

arises if one differentiates four times. This gives a nonzero lateral load p(x)but no such loadexists. Thisparadoxindicates that the sine-wave result must be corrected once the buckling

mode amplitude becomesfinite.

The analysis does not take into account imperfections such an initially crooked column, or axialload eccentricity. Such an analysis is beyond the level of this course, but one basic result is used

in the justification of theSouthwell plotexperimental procedure described in Lecture 19.

18.1.3. Reformulation as Eigenproblem

Lecture 17 emphasized that a discrete stability problem handled through the LPB assumptions may

be presented as a matrix eigenproblem. This is also the case for the linearized continuous problem,

but it requires a reformulation. For the Euler column, the two end boundary conditionsv(0) = 0 andv(L) = 0 may be written conjointly as

1 0

cos L sin L

A

B

=

0

0

. (18.10)

This is a matrix eigenproblem in = P/(E I), but note that (unlike the discrete case) this variabledoes not appear linearly. For a nontrivial solution, meaning that Aand B are not both zero, the

determinant of the matrix on the left side must vanish. This leads to the condition

sin L= 0 (18.11)which agrees with that previously found from the characteristic equation (18.5). This reformulation

becomes useful in more complicated cases.

18.2. The Fixed-Pinned Column

As second example we consider the configuration shown in Figure 18.3(a). The buckled column is

drawn in Figure 18.3(b) in a deflected position. The most notable difference with respect tp the Euler

column of Figure 18.1(a) is the presence of additional reaction forces: the lateral reactions RAand

RB , and thefixed end momentMB . These are positive as shown in Figure 18.3(b).

18.2.1. Critical Load Analysis

Equilibrium ofyforces in Figure 18.3(b) gives RA= RB . Equilibrium of moments taken withrespect to either A or B yields MB= RA/L. Next consider the FBD of the portion AX shown inFigure 18.3(c). Equilibrium with respect to Xgives the non-homogeneous ODE

E Iv (x) + Pv(x) = RAx=MBx

L. (18.12)

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187 18.2 THE FIXED-PINNED COLUMN

xx

y yy

v(x)

P P

PB

B

M(x)(+ as drawn)

P

X X'

v(x)

X'

X

constantEI

P

B

X

(a) (b) (c)

elastic

A

A A

Lassumed

buckled shape

RA

AR = RB

M

x

RA

RA

Figure18.3. Columnfixed at one end and simply supported (hinged, pinned) at the other.

The LHS is the same as in (18.1), but the RHS is no longer zero. Dividing through byE Iand setting

2 = P/(E I)gives after some manipulations

v(x) + 2 v(x) = 2MBx

P L. (18.13)

The general solution is the sum of the homogeneous solutionvH(x) = Asin x +Bcos xand theparticular solutionv P (x) = MBx/(P L):

v(x)

=Asin x

+Bcos x

+

MBx

P L

. (18.14)

The three kinemtic BCs arevA= v(0) = 0,vB (0) = v(L) = 0 andv B= v(L) = 0. These providethree equations:

B + MBL

= 0, A sin L +Bcos L + 0, A l MBP L

= 0. (18.15)

Solving these equations simultaneously one obtains the characteristic equation

tan L= L . (18.16)

The samllest roots of this transcendental equation, to 4 places, is L=

4.493. The corresponding

critical load is

Pcr=20.19E I

L2 . (18.17)

Since 20.19 2.05 2 this critical load is approximately twice that of the Euler column. Thusfixingone end has substantially increased the critical load.

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x

x

y,vy

v(x)

PP P

P

L

RA

RA

RA

M = kB

B

B

AM(x)=P v+R xx

P

P

RA

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

k

AA

CB

Belastic beam

beam-columnX

Figure18.4. Column with elastic restraint: a torsional spring, at end B.

18.3. Elastically Restrained Column

The title configuration is of interest because it covers the three boundary condition cases tested in the

Column Buckling Lab.as special cases.

18.3.1. Problem Description

Consider the modification of the classical Euler column pictured in Figure 18.4(a). The column is

simply supported (pinned) at both ends A and B, and axially loaded by P . The rotation atBis further

restrained by a torsional spring with stiffness k.

ColumnABhas lengthLand constant flexural rigidityE I, in whichI Izzis the minimum momentof inertia of the cross section that controls buckling. (For a rectangular cross section of width bandthickness t < b,I= Izz= b t3/12.) The column is simply supported (pinned) at both endsAandB,and axially loaded by P . The rotation atBis further restrained by a torsional spring with stiffness k.

Selectxas shown and consider the buckling shape v(x)sketched in Figure A.1(b). The end rotation

at Bis B= v(L), positive CCW. The spring at Bapplies a restoring end moment MB= kB .For convenience in obtaining dimensionless equations we define

k= E IL

(18.18)

where is a numerical coefficient. If= 0 the problem reduces to that of the classical Euler column,whereas if we obtain the case of a column simply supported at Aandfixed (clamped) atB.

18.3.2. Critical Load Analysis

The FBD of the complete column is shown in Figure 18.4(b). Taking moments with respect toBone

finds that

RA= MB /L= kB /L= E IB /L2. (18.19)

Prior to Fall 2010, this used to be Experimental Lab 3. It is now aLab-Homework, meaning that the results are presented

as part of an assigned Homework rather than a formal Lab Report.

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189 18.3 ELASTICALLY RESTRAINED COLUMN

Now cut the column at section Xat distancexas sketched in Figure 18.4(c). Moment equilibrium

with respect toXyields the second order differential equation

E Iv + Pv= RAx= kBx

L= E IBx

L2 , (18.20)

which results from equating the bending moment M(x) = E Iv to Pv +RAx . Dividing throughbyE Iand calling2 = P/E Iproduces the canonical form

v + 2 v= BxL2

. (18.21)

The general solution of (18.21) is the sum of the homogeneous and particular solutions:

v(x) = C1sin x + C2cos x Bx

2L2. (18.22)

Sincev(0)= 0, C2= 0. The rotation is (x)= v(x)= C1 cos x B /(2L2). Evaluatingthis atx= LyieldsB= (L) = C1 cos L B /(

2

L

2

), from which we can solve for the endrotation:

B= C13L2

+ 2L2cos L . (18.23)

Inserting this into the solution (18.22) with C2=0 gives v(x)= C1sin x C1xcos L/(+2L2). Applying now the second boundary condition: v(L) = 0, furnishes the stability equation

v(L) = C1

sin L L + 2L2cos L

= 0. (18.24)

For buckling to occur, C1=0, and the expression in parentheses in (18.24) must vanish. Calling=

L, which is also a dimensionless variable, we obtain the trascendental equation

tan = 2 + (18.25)

For a given 0 we seek the solutioncr >0 of (18.25) closest to zero. If= 0 there is no closedform solution and it is better to proceed numerically, using for example a Newton solver. It is easily

shown that for= [0, ], cr


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The critical load Pcrand the effective lengthLe(tabulated above) are

Pcr=2crE I

L2 =

2E I

L2e, Le=

crL (18.26)

This solution also applies to the buckling of a column AB rigidly connected at B to an elastic

beam BC, as illustrated in Figure 1(d). This is actually one of the configurations to be tested in

Lab 3. All that is needed is to work out the appropriate value of k= MBCB /BCB obtained from amoment-deflection analysis of beamBCsubjected to an applied end momentMBCB = MB .

18.3.3. Equivalent Spring Constant

Suppose the restraining elastic beamBCinFigureA.1has lengthLr= LBC, modulusErandmomentof inertia Iraboutz. Assume a simply support condition at C. Under an end moment MBapplied

as pictured in Figure 18.4(d), the restraining beam deflects by vr= MBxr(L2rx2r)/(6ErIrLr),in which xris the distance from C. The end rotation isB

= (dvr/dx)

|xr

Lr

=MBLr/(3ErIr),

whence the equivalent torsional spring stiffness is k= MB /B is 3ErIr/Lr. Comparing with (1)shows that= 3(L/Lr)(ErIr)/(E I). If the restraining beam is fabricated of the same material andhas the same cross section dimensions of the beam-column, as is the case in Lab 3, Er= E,Ir= I,and= 3L/Lr.

E. M. Popov,Engineering Mechanics of Solids, Prentice Hall, NJ, 1999, case 5 of Table 10, p. 853.

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Stability Of Structures: Continuous Models

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