Stability Of Structures: Basic Concepts

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Lecture 17: STABILITY OF STRUCTURES: BASIC CONCEPTS 172

TABLE OF CONTENTS

Page

17.1. Introduction 173

17.2. Testing Stability 173

17.2.1. Stability of Static Equilibrium . . . . . . . . . . . . 173

17.2.2. Stability of Dynamic Equilibrium . . . . . . . . . . 174

17.3. Static Stability Loss 175

17.3.1. Buckling Or Snapping? . . . . . . . . . . . . . . 175

17.3.2. Response Diagrams . . . . . . . . . . . . . . . 175

17.3.3. Stability Models . . . . . . . . . . . . . . . . 176

17.3.4. Stability Equations Derivation . . . . . . . . . . . 177

17.4. Exact Versus Linearized Stability Analysis 177

17.4.1. Example 1: The HCR Column: Geometrically Exact Analysis . 178

17.4.2. Example 1: The HCR Column: LPB Analysis . . . . . . 17917.4.3. Example 2: The PCR Column: Geometrically Exact Analysis . 1710

17.4.4. Example 2: The PCR Column - LPB Analysis . . . . . . 1712

17.5. Discrete Stability Analysis As Eigenproblem 1713

17.5.1. Example 3: A Cantilevered Two-Strut Column . . . . . . 1713

17.5.2. Example 4: A Pinned-Pinned Three-Strut Column . . . . . 1715

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173 17.2 TESTING STABILITY

17.1. Introduction

The term stability hasbothinformalandformalmeanings. Asregardstheformer, theAmerican Heritage

Dictionarylists the following three.1. Resistance to sudden change, dislodgment, or overthrow. 2a.

Constancy of character or purpose: tenacity; steadfastness. 2b. Reliability; dependability. Related

verb: to stabilize

. Related adjective: stable

. Opposite terms:stability loss, instability, to destabilize,

unstable.

The formal meaning is found in engineering and sciences, concerning stability ofsystems.* Broadly

speaking, structural stability can be defined as the power to recover equilibrium. It is an essential

requirement for all structures. Jennings provides the following historical sketch:

Masonry structures generally become more stable with increasing dead weight. However

when iron and steel became available in quantity, elastic buckling due to loss of stability of

slender members appeared as a particular hazard.

17.2. Testing Stability

The stability of a mechanical system, and of structures in particular, can be tested (experimentallyor analytically) by observing how it reacts when external disturbances are applied. Here we have to

distinguish between statics and dynamics.

17.2.1. Stability of Static Equilibrium

For simplicity we will assume that the structure under study is elastic, since memory effects such as

plasticity or creep introduce additional complications such as historical dependence, which are beyond

the scope of the course. The applied forces are characterized by a loading parameter, also called a

load factor. Setting =0 means that the structure is unloaded, at which it takes up an equilibrium

configuration C0 = (0)called the undeformed state. Furthermore, assume that this state isstablein

the sense defined below.Asis varied from 0 the structure deforms and assumes equilibrium configurationsC(). These are

assumed to be (I) continuously dependend onand (II) stable for sufficienttly smaller values of.

How is stability tested? Freeze at a specific value, say dwhere dconnotes deformed. The

associated equilibrium configuration is Cd = C(d). Apply aperturbationto Cd, andremoveit. What

sort of perturbation? An action that may disturb the state, for example a tiny load or a small imposed

deflection. More precise restrictions on such perturbations are described later. For now we will

generically calladmissible perturbationsthose that are allowed in the application of the test.

The perturbation triggers subsequent motion of the system. Three possible outcomes are sketched in

Figure 17.1.

S For all admissible perturbations, the structure either returns to the tested configuration Cdorexecutes bounded oscillations about it. The equilibrium atdis calledstable.

* A systemis a functionally related group of components forming or regarded as a collective entity. This definition uses

componentas a generic term that embodies elementorpart,which connote simplicity, as well as subsystem,which

connotes complexity. In this course we shall be concerned aboutmechanical systemsgoverned by Newtonian mechanics,

with focus onstructures.

A. Jennings,Structures: From Theory to Practice, Taylor and Francis, London, 2004, Chapter 7.

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S: Stable*

Equilibriumconfiguration Subsequentmotion

Apply anadmissible

perturbation

Motion oscillates aboutequilibrium configuration

or decays toward it

Motion is either unbounded, oroscillates about, or decays toward,another equilibrium configuration

U: Unstable

Transition

between stableand unstable

N: Neutrally stable

* Strictly speaking, Srequires stabilityfor allpossible admissible perturbations

Figure17.1. Stability test outcomes.

U If for at least one admissible perturbation the structure moves to (decays to, or oscillates about)another configuration, ortakes offin an unbounded motion, the equilibrium isunstable.

N The transition from S to N occurs at a value = crcalled the critical load factor. The

configurationCcr = C(cr)at the critical load factor is said to be in neutralequilibrium. The

quantitative determination of this transition is a key objective of stability analysis.

The foregoing classification leaves several gaps and details unanswered.

First, speaking aboutmovingorreturningintroduces time into the picture. Indeed the concept of

stability is necessarilydynamicin nature There is abefore: prior to applying the perturbation, and an

after: what happens upon removing it. Many practical methods to assess critical loads, however, factor

out the time dimension as long as certain conditions are verified. Those are know asstatic criteria.

Second, the concept ofperturbationassmall imposed changeis imprecise. How small is atiny

loador aslight deflection? The idea is made more mathematically precise later when we introduce

linearized stability, also called stability in the small. This is a natural consequence of assuming

infinitesimal configuration changes.

17.2.2. Stability of Dynamic Equilibrium

Stabilityofmotionisa more general topic that includes thestaticcase as a particular one. (Aspreviously

noted, the concept of stability is essentially dynamic in nature.)

Suppose that a mechanical system is moving in apredictablemanner. For example, a bridge oscillates

under wind, an airplane is flying a predefined trajectory under automatic pilot, a satellite orbits the

Earth, the Earth orbits the Sun. What is the sensitivity of such a motion to changes of parameters suchas initial conditions? If the system includes stochastic or chaotic elements, like turbulence, the analysis

will require probabilistic methods.

For example, Bazant and Cedolin in Stability of Structures: Elastic, Inelastic, Fracture and Damage Theories ,Dover, 2003,

comment on p. 144: Failure of structures is a dynamical process, and so it is obviouly more realistic to approach buckling

and instability from a dynamical point of view.

E.g., conservative loading: applied loads derive from a potential.

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175 17.3 STATIC STABILITY LOSS

To make such problems mathematically tractable it is common to restrict the kind of motions in such a

way that abounded reference motioncan be readily defined. For example a bounded periodic motion

of a oscillating structure. Departures under parametric changes are studied. Transition to unbounded

or unpredictable motion is taken as a sign of instability.

An important application of this concept are vibrations of structures that interact with an external or

internalfluid flow: bridges, buildings, airplanes,fluid pipes. The steady speed of the flow may betaken as parameter. At a certain airflow or liquid-flow speed, increasing oscillations may be triggered:

this is called flutter. Or a non-oscillatory unbounded motion happens: this is called divergence. A

famous example offlutter in a civil structure was the collapse of the newly opened Tacoma-Narrows

suspension bridge near Seattle in 1940 under a moderate wind speed of about 40 mph.

Modeling and analysis of dynamic instability is covered in other courses, primarily at the graduate

level because it requires fancier math tools and heavier use of complex calculus. In this course we

consider only static stability. Moreover, the class of problems, methods and examples will be severely

constrained so as tofit into three lectures.

17.3. Static Stability Loss

As just noted, we restrict attention to stability ofstatic equilibrium. Two assumptions are introduced:

Linear Elasticity. The structural material is, and remains, linearly elastic. Displacements and rota-

tions, however, are not necessarily small.

Conservative Loading. The applied loads areconservative, that is, derivable from a potential. For ex-

ample, gravity and hydrostatic loads are conservative. On the other hand, aerodynamic and propulsion

loads (wind gusts on a bridge, rocket thrust, etc) are often nonconservative.

The main reason for the second restriction is that loss of stability under nonconservative loads is

inherently dynamic in nature, and thus lies beyond the scope of the course.

17.3.1. Buckling Or Snapping?

Under the foregoing restrictions, two types of instabilities may occur:

Bifurcation. Structural engineers use the more familiar name bucklingfor this one. The structure

reaches abifurcation point, at which two or more equilibrium paths intersect. What happens after the

bifurcation point is traversed is called post-bucklingbehavior.

Snapping. Structural engineers use the termsnap-throughorsnap bucklingfor this one. The structure

reaches a limit pointat which the load reaches a maximum. What happens after the limit point is

traversed is calledpost-snappingbehavior.

Bifurcation pointsandlimit pointsare instances ofcritical points. The importance of critical points in

static stability analysis stems from the following property:

Transition from stability to instability can only occur at critical points

Reaching a critical point may lead to immediate destruction (collapse) of the structure. This depends

on its post-buckling or post-snapping behavior, and nature of the material. For some scenarios the

knowledge of such behavior is important since immediate collapse may lead to loss of life. On the

other hand, there are some configurations where the structure keeps resisting significant or even

increasing loads after traversing a critical point. Such load-sustainingdesigns are obviously

preferable from a safety standpoint.

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Equilibrium paths

Equilibriumpath

Load or loadparameter

Load or loadparameter

Representativedeflection

Reference state

Limit point(snap)

Reference state

Representativedeflection

Initial linear response

Bifurcationpoint (buckling)

B

R R

LTerms in redarethose in commonuse by structuralengineers

(b)(a)

Figure 17.2. Graphical representation of static equilibrium paths and their critical points: (a)

a response path with no critical points; (b) multiple response paths showing occurrence of two

critical points types: bifurcation point (B) and limit point (L).

17.3.2. Response Diagrams

To illustrate the occurrence of static instability as well as critical points we will often display load-

deflection response diagrams. This is a plot of equilibrium configurations taken by a structure as a

load, or load parameter, is gradually and continuously varied. The load (or load parameter) is plotted

along theverticalaxis whilea judiciouslychosenrepresentativedeflectionis plotted along thehorizontal

axes. A common convention is to take zero deflection at zero load. This defines the reference state,

labeled as point R in such plots.

A continuous set of equilibrium configurations forms an equilibrium path. Such paths are illustrated

in Figure 17.2. The plot in Figure 17.2(a) shows a response path with no critical points. On the other

hand, the plot in Figure 17.2(b) depicts the occurrence of two critical points: one bifurcation and one

limit point. Those points are labeled as B and L, respectively.

17.3.3. Stability Models

Stability models of actual structures fall into two categories:

Continuous. Such models have an infinite number of degrees of freedom (DOF). They lead to ordinary

or partial differential equations (ODEs or PDEs) in space, from which stability equations may be

derived by perturbation techniques. Obtaining nontrivial solutions of such equations generally leads

to trascendental eigenproblems, even if the underlying model is linear.

Discrete. These models have afinite number of DOF in space. Where do these come from? Often they

emerge as discrete approximations to the underlying continuum models. Two common discretization

techniques are:(1) Lumped parameter models, in which theflexibility of the structure is localized at a finite number

of places. One common model of this type for columns: joint-hinged rigid struts supported by

extensional or torsional springs at the joints.

(2) Finite element modelsthat include the so-called geometric stiffnesseffects.

Students should be familiar with this visualization technique if they have done tension or torsion mechanical tests.

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177 17.4 EXACT VERSUS LINEARIZED STABILITY ANALYSIS

Stabilityequationsfordiscretemodels maybe constructed usingvariousdevices. Forlumped parameter

models one may resort to either perturbed equilibrium equations built via FBDs, or to energy methods.

For FEM models only energy methods are practical. All techniques eventually lead to matrix stability

equations that take the form of an algebraic eigenproblem.

Both continuous and lumped-parameter discrete models for column problems are presented later in

this Lecture, and in the following two Lectures.

17.3.4. Stability Equations Derivation

The equations that determine critical points are called characteristic equationsin the applied mathe-

matics literature.* In structural engineering the names stability equationsand buckling equationsare

common. For structural stability analysis, two methods are favored for deriving those equations.

Equilibrium Method. The equilibrium equations of the structure are established in a perturbed

equilibrium configuration. This is usually done with Free Body Diagrams (FBD). The resulting

equations are examined for the occurrence of nontrivial perturbed equilibrium configurations. Such

perturbed configurations are obtained by disturbing parametrized equilibrium positions by admissible

buckling modes. If those equilibrium equations are linearized for small perturbations, one obtainsan algebraic eigenproblem. The eigenvalues give values of critical loads while eigenvectors yield the

buckling mode shapes.

Energy Method. The total potential energy of the system is established in terms of the degrees of

freedom. The Jacobian matrix of the potential energy function, taken with respect to those degrees of

freedom, is established and tested for positive definiteness as the load parameter (or set of parameters)

is varied. Loss of such property occurs at critical points. These may be in turn categorized into

bifurcation and limit points according to a subsequent eigenvector analysis.

The energy method is more general for structures subject to conservative loading. It has two major

practical advantages: (1) merges naturally with the FEM formulation and so it can be efficiently

implemented in general-purpose codes, and (2) requires no a prioriassumptions as regards admissible

perturbations. But since energy methods are not covered at the undergraduate level, only equilibrium

methods will be presented here. Such techniques are necessarily restricted to simple 1D problems

amenable to FBDs, but that approach is sufficient for an introduction to stability analysis.

17.4. Exact Versus Linearized Stability Analysis

As noted above, we will use only the equilibrium methodto set up stability equations. Its key feature

is that FBDs must take the perturbed configuration into account. For certain simple problems it is

possible to establish the equations using the exact geometryof the deflected structure. Such analyses

will be called geometrically exact. A couple of examples follow to illustrate exact versus linearized

results.

* Often this term is restricted to the determinantal form of the stability eigenproblem. This is the equation whose roots give

the eigenvalues, which can be interpreted physically as critical loads, or as critical load parameters.

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AA'

A

B B

Brigid

k cr

P = P

Equilibriumpath of straight

(untilted) column

Equilibrium pathof tilted column

Bifurcationpoint

ref

P = Prefv =L sin

P

L

L

(a) (b) (c)

Unstable

Stable

A

; ; ;

M = k

Figure 17.3. Geometrically exact analysis of the hinged cantilevered rigid (HRC) column: (a) untilted

column, (b) tilted column, (c) equilibrium paths intersecting at bifurcation point.

17.4.1. Example 1: The HCR Column: Geometrically Exact Analysis

Consider the configuration depicted in Figure 17.3(a). A rigidstrut of lengthLstabilized by a torsional

spring of stiffnessk>0 isaxially loadedbya verticaldead load P = Pre f, inwhich Pref isa reference

load and a dimensionless load parameter. The load remains vertical as the columntilts. (Observe that

khas the physical dimension of force length, i.e. of a moment.) This configuration will be called

a hinged cantilevered rigid column, or HCR column for brevity. The definition P = k/Lrenders

dimensionless, which is convenient for result presentation. As state parameter we pick the tilt angle

as most appropriate for hand analysis.

For sufficiently small Pthecolumnremains verticalas in Figure17.3(a), with = 0. The only possible

buckled shape is the tilted column shown in Figure 17.3(b). This Figure depicts the FBD required to

analyze equilibrium of the tilted column. Notice that is notassumed small. Taking moments with

respect to the hinge B as sketched we obtain the following equilibrium equation in terms ofand:

k = Pre f d = Pre fL sin k Pre f

Lsin = 0. (17.1)

The equation on the right hand side has the two solutions

= 0 for any, =k

Pre fL

sin . (17.2)

These pertain to the untilted ( = 0) and tilted ( = 0) equilibrium paths, respectively. Sincelim(/ sin ) 1 as 0, the two paths intersect when

cr =k

PrefL, or Pcr =

k

L. (17.3)

The two paths are plotted in Figure 17.3(c). The intersection (17.3) characterizes a bifurcation point B.

The diagram shows four branches emanating from B. Three are stable (full lines) and one is unstable

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(dashed line). Note that the applied load may rise beyond the critical Pcr = cr Pre f = k/Lby moving

to a tilted configuration. It is not difficult to show that the maximum load occurs if 180, for

which P ; this is a consequence of the assumption that the column is rigid (and that it may fully

rotate by that amount about the hinge).

17.4.2. Example 1: The HCR Column: LPB Analysis

The geometrically exactanalysis that leads to (17.2)has theadvantageof providinga complete solution.

In particular, it shows what happens after the bifurcation point B is traversed. For this configuration

the structure maintains load-bearing capabilities while tilted, which is the hallmark of a safe design.

But for more complicated cases this approach becomes impractical as it involves solving systems of

nonlinear algebraic or differential equations. Even for the Euler column presented in Lecture 18, a

geometrically exact analysis leads to elliptic functions.

Often the engineer is interested only in the critical load. This is especially true in preliminary design

scenarios, when the main objective is to assess safety factors against buckling. If so, it is more practical

to work with a linearizedversion of the problem. Technically the full technical name is linearized

prebuckling(LPB) analysis. This approach relies on the following assumptions: Deformations prior to buckling are neglected. This permits the analysis to be carried out in the

referenceconfiguration geometry.

Perturbations of the equilibrium configuration produce only infinitesimaldisplacements and ro-

tations.

The structure remains linearly elastic up to buckling.

Both structure and loading do not exhibit any imperfections.

The critical state is a bifurcation point.

We apply these rules to theHCRcolumnofFigure17.3(a). The equilibriumequation (17.1) is linearized

by assuming an infinitesimal tilt angle


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; ;

A

B

P = Pref

(a)

;

;

;

kA'

A

B

P = Prefv =L sin

u =L (1 cos)

P

(b)

A

A

;

;

;

rigid

L

L

CC'

Figure 17.4. Geometrically exact analysis of a propped cantilevered rigid (PRC) column with

extensional spring remaining horizontal: (a) untilted column, (b) tilted column.

A'A

B

P = Prefv =L sin

P

(a)A

F = k vA A

Lcos

Equilibriumpath of straight

(untilted) column

Equilibrium pathof tilted column

Bifurcationpoint

(b)

UnstableStable

+9090

cr = k L/Pref

Figure 17.5. Geometrically exact analysis of a propped cantilevered rigid (PRC) column with

extensionalspring remaininghorizontal: (a)FBDfor tiltedequilibrium, (b)equilibriumpaths intersecting

at bifurcation point.

17.4.3. Example 2: The PCR Column: Geometrically Exact Analysis

Consider next theconfiguration pictured in Figure17.4(a); this differs from theHRCcolumnin the type

of stabilizing spring. A rigid strut of lengthLis hinged at B and supports a vertical load P = Pre fat

end A. The load remains vertical as the column tilts. The column is propped by an extensional spring

of stiffness kattached to A. This configuration will be called apropped cantilevered rigid column; or

PCR column for short. The only DOF is the tilt angle.

For the geometrically exact analysis is it important to know what happens to the spring as the column

tilts. One possible assumption is that it remains horizontal, as depicted in Figure 17.4(b). If so the

FBD in the tilted configuration will be as shown in Figure 17.5(a). Taking moments aboutByields

Pre f sin = k L sin cos . (17.5)

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; ;

A

B

P = Pref(a)

;

;

kA'

A

B

P = Prefv =L sin

u =L (1 cos)

P

(b)

A

A

rigid

L

L

C C

L

;

;

Figure17.6. Geometrically exact analysis of a propped cantilevered rigid (PRC) column with wall-attached

extensional spring: (a) untilted column, (b) tilted column.

A'A

B

P = Prefv =L sin

u =L (1 cos)

P

A

A

C

= angle A'CApositive CW

F = k dd : tiltingspringelongation

A

s

s

Figure17.7. Geometrically exact analysis of a propped cantilevered rigid (PRC) column with wall-attached

extensional spring: (a) FBD for tilted equilibrium, (b) equilibrium paths intersecting at bifurcation point.

60 40 20 20 40

1

1

2

3

4

60 40 20 20 40 60 80

1

0.75

0.5

0.25

0.25

0.5

0.75

1

60 40 20 20 40 60 80

0.2

0.4

0.6

0.8

1

60 40 20 20 40 60 80

1

0.5

0.5

1

1.5

2

= 1/4 = 1/2

= 1 = 5

angle (deg) angle (deg)

angle (deg)

angle (deg)

Figure 17.8. Geometrically exact analysis of a propped cantilevered rigid (PRC) column with wall-attached

extensionalspring: vs./response diagramsforfourvaluesof (definedin Figure17.6(a)),with Pre f = k= L = 1.

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This has two solutions

= 0 for any , =k L

Prefcos . (17.6)

These yield the vertical and tilted equilibrium paths, respectively, plotted in Figure 17.5(b) on the

versus plane. The two paths intersect at = k L/Pre f

, which is a bifurcation point. Consequently

the critical load parameter is cr = kL/Pre fand the critical load is Pcr = cr Pref = k L . Of the 4

branches that emanate from the bifurcation point B, only one (the =0 path for < cris stable.

Once B is reached the tilted column supports only a decreasing load P , which vanishes at = 90.

Consequently this configuration is poor from the standpoint of post-buckling safety. three are unstable.

Another reasonable assumption is that the spring attachment point to the wall, called C in Figure

17.6(a), stays fixed. The distance AC is parametrized with respect to the column length as L , in

whichis dimensionless. If the column tilts, the spring also tilts as pictured in Figure 17.6(b). The

geometrically exact FBD for this case is shown in Figure 17.6(b). Ascan be observed, it is considerably

more involved than for the stay-horizontal case. We quote only thefinal result for the equilibrium path

equations:

= 0 for any, = kLPref

cos + sin + sin

. (17.7)

Figure 17.8 shows response plots on theversusplane for the four cases = 14, = 1

2 = 1 and

=5, drawn with Pre f =1, k =1, and L =1. Although the bifurcation points are in the same

location, the post-buckling response is no longer symmetric for . That deviation from symmetry is

most conspicuous when


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1713 17.5 DISCRETE STABILITY ANALYSIS AS EIGENPROBLEM

vA

vB

(b)(a) P P

B

A

C

rigid

ri

gid

k

3k/2B

A A'

C

L/2

L/2

L

vdeflections+ to the right

B'

Figure17.9. Cantilevered two-strut column. (a): structure; (b): deflected shape defined by two DOFvAand vB ;

17.5. Discrete Stability Analysis As Eigenproblem

When the discrete model has n 2 DOFs, the determination of critical loads by the LPB approach

leads to a matrix algebraic eigenproblem of order n. The neigenvalues provide the critical loads, and

the corresponding eigenvectors give the associated buckling shapes. This is illustrated through several

examples worked out below.

17.5.1. Example 3: A Cantilevered Two-Strut Column

The problem investigated here in detail is the buckling of the two-hinged-cantilever column depicted

in Figure 17.9(a). Rigid links of equal lengthL/2 are connected at the joints A,Band at the bottom

Cby frictionless hinges. The column is propped by two extensional springs attached at A and B withthe stiffnesses shown in the figure.

Data: Length Land spring stiffness k. Required: Set up the stability equations as a matrix eigen-

problem,find critical loads in terms of the data and display buckling shapes as separate diagrams.

Draw an arbitrary, but kinematically admissible, buckling shape as in Figure 17.9(b). This shape is

completely defined by the two lateral deflectionsvAand vBof points A and B, respectively, positive

to the right. Note: The deflections are highly exaggerated in thefigure for visibility; they are actually

infinitesimally small.

Figure 17.10(c1,c2,c3) shows free body diagrams (FBD) of links AB, BC and of the whole column

ABC, respectively. To facilitate visualization applied forces are pictured in black, spring forces in blue,

whereas internal reactions are shown in red.

The two lateral deflections vAandvBare taken as independent degreesof freedom(DOF).Consequently,

two equilibrium equations are required obtained from FBDs are required. Three combinations of two

FBDs are possible: AB and BC, AB and whole column, or BC and whole column. In all FBDs

translational force equilibrium holds identically by construction. The remaining moment equilibrium

conditions become part of the matrix stability equation. Since the three FBDs are linearly dependent

Adapted from E. P. Popov,Engineering Mechanics of Solids

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vA

A

vB

BB

A'

B

B

C

(c1)

(c2)

P

P

P

P

vB

A

B'

B'

F

AA

AF +F

BAF +F

BAF +F

F

B BF = k v , F = 3kv /2BAF +F

P

vA

A

B

A'

B

C

(c3)

vB

A

B'

F

F

P

Applied forcesin black, springforces in blue,reactions in red

FBD oflink AB

FBD of wholecolumn

FBD oflink BC

Restoring springforces + to the left

Figure17.10. Cantilevered two-strut column. (c1,c2,c3): FBDs of link AB, link BC and whole column,

respectively. FBD color convention: applied forces in black, spring foces in blue, reactions in red.

[check thefigure: merging (c1) and (c2) gives (c3)], the results of the analysis should be exactly the

same, although the stability matrix equations will be different. For the ensuing derivation we pick (c1)

and (c3).

For link AB, taking moments with respect to the displaced hinge point B, positive CW, givesMB = P(vA vB ) FA(

12L) = P(vA vB )

12LkvA = 0 (17.10)

For the whole column ABC, taking moments with respect to the hinge C, positive CW, givesMC = PvA FAL FB(

12L) = PvA kLvA

32

k( 12L)vB = 0 (17.11)

Characteristic Equation. Combining (17.10) and (17.11) in a matrix equation gives the stability

equation (also known as the characteristic equation):P 1

2kL P

P kL 34kL

vAvB

=

0

0

, or Av = 0. (17.12)

MatrixAis called the stability matrix or characteristic matrix. System (17.12) has two kinds of

solutions. The trivial solutionv = 0, or vA = vB = 0, corresponds to the undeflected (vertical)

column, which holds for any P . To find the critical loads we must have v = 0, that is, at least one of the

{vA, vB} must be nonzero. ConsequentlyAmust be singular or, equivalenet, have zero determinant.Since the entries ofAdepend on P , this is an algebraic eigenproblem in P .

Critical Loads. The eigenvalues are the solution of the characteristic equation

det(A) = 0, or P2 74kL P + 3

8k2L2 = 0 (17.13)

Solving this quadratic for Pgives the two critical loads:

Pcr1 = 14kL = 0.25 kL , Pcr2 =

32kL = 1.50 kL . (17.14)

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1

1

P = P = 0.25 kLcr cr1 P =1.50 kLcr2

A'

B' 2/3

1A'

B'

Figure17.11. Buckling mode shapes for the cantilevered two-strut column.

The smallest one is the critical load:

Pcr = Pcr1 = 14kL = 0.25 kL . (17.15)

Buckling Mode Shapes. Replacing Pcr1into (17.14) gives the following equation to determine the

eigenvectorv1:

kL

4

3 3

1 1

vA1vB1

=

0

0

v1 =

vA1vB1

= c1

1

1

(17.16)

where c1is an arbitrary nonzero scaling factor.

Replacing Pcr2intoAgives the following equation to determine the eigenvectorv2:

kL

4

2 3

4 6

vA2vB2

=

0

0

v2 =

vA2vB2

= c2

3/2

1

(17.17)

where c2is an arbitrary nonzero scaling factor.

One common nomalization condition for eigenvectors is to make their largest component equal to +1.

This is done for (17.16) and (17.17) by taking c1 =1 or c1 = 1 (either one works), and c2 =2/3,

respectively. The two eigenvectors normalized as per that criterion are

v1 =

1

1

for Pcr1, v2 =

1

2/3

for Pcr2. (17.18)

The buckling shapes defined by (17.18) are plotted in Figure 17.11.

17.5.2. Example 4: A Pinned-Pinned Three-Strut Column

This is a slight varint of one treated in Jennings (op. cit.). Figure 17.12(a) shows a column built with

three rigid struts of equal length, pinned at the joints, and with two extensional spring of stiffness k

as lateral supports at the internal joints B and C. Determine the critical loads of this configuration by

linearization.

Figure 17.12(b) depeicts how the structure might displace after buckling. Since the structs are rigid

and cannot change length, joints A, B and C will necesasrily moved down while the springs tilt;

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vB

vC

(c)(a)

B

C

D

rigid

rigid

rigid

k

k

vdeflections+ to the right

L/3

L/3

L/3

A

(b)

B'

C'

D

A'

B'

C'

D

A

P>Pcr PP

F = k vB B

F = k vC C

B C

(1/3) (F +2F ) =

(1/3) k (v +2v )B C

B C

(1/3) (2F +F ) =

(1/3) k (2v +v )B C

Figure 17.12. Stability analysis of three-strut column propped by extensional springs: (a) referenceconfiguration; (b) admissible buckled shape showing realistic geometry change (struts do not change length);

(c) linearized configuration: B and C displace by small horizontal movements whereas A stays fixed.

such geometry is correct for geometrically exact analysis. The linearized version is shown in Figure

17.9(c); here B and C move horizontally byvBand vD, respectively, and A does not move vertically.

The horizontal reactions RAand RDshown in Figure 17.9(c) are obtained in terms ofFBand FDby

statics on taking moments with respect to D and A, respectively.

The two FBDs used to form the stability equations are shown in Figure 17.13. We take moments with

respect to Cin the left FBD and with respect to Bin the right FBD:

MC = PvC FB

L

3+RA

2L

3= 0,

MB = PvB FC

L

3+RD

2L

3= 0. (17.19)

vB

vC

B'

C'

A

P

P P

F = k vC C

F = k vB B

vB

vC

B'

C'

D

PF = k vB B

F = k vC C

B C

R = (1/3) (F +2F )

= (1/3) k (v +2v )B C

B C

R = (1/3) (2F +F )

= (1/3) k (2v +v )B CA

D

Figure17.13. Free Body Diagrams for the 3-strut column of Figure 17.12.

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P = kL/9cr1 P = kL/3cr2

1 1

11

Buckling mode #1

(antisymmetric)

Buckling mode #2

(symmetric)

B

C

D

A

B

C

D

A

Figure17.14. Critical loads and buckling mode shapes for the 3-strut column of Figure 17.12.

Substituting RA = (2FB + FC)/3, RD = (FB + 2FC)/3, FB = kvBand FC = kvC, and casting

(17.19) in matrix form we obtain the stability equations

k L

9

2 1

1 2

vBvC

= P

vBvc

, (17.20)

This is a 2 2 matrix eigenproblem with Pas the eigenvariable. The characteristic equation is

C(P) = det

k L

9

2 1

1 2

P

1 0

0 1

=

k2L2

27

4 k P L

9+ P2. (17.21)

This is a quadratic polynomial in P . Setting C(P) = 0 gives the two critical loads: Pcr1 = kL/9 and

Pcr2 = kL/3, as roots. The corresponding eigenvectors, normalized to 1 as largest entry, are

v1 =

1

1

, v2 =

1

1

. (17.22)

These are plotted in Figure 17.14. Note that the mode shape corresponding to the lowest critical load,

Pcr1 = kL/9, is antisymmetric. This is in sharp contrast to the Euler column buckling treated in

Lecture 18.

An alternative eigenvalue problem is obtained on premultiplying by the inverse of the LHS matrix in

(17.20). This gives3P

k L

2 1

1 2

vBvc

=

vBvC

, (17.23)

an eigenproblem which has the same solutions. This is the form derived by Jennings (except for scale

factors).

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Stability Of Structures: Basic Concepts