Chapter 11 Buckling of Columns

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Chapter

11

BUCKLING OF COLUMNS:11.1 Introduction

11.2 Stabilit o! Structure"

11.# Buc$lin% o! &in'(nded Colu)n"

11.* Colu)n" +ith Other (nd Condition"

11., Critical Stre"": Cla""i!ication o! Colu)n"

11.- (ccentricall Loaded Colu)n" and the Secant For)ula

11.1 Introduction:

• In discussing the analysis and design of various structures in the previous chapters, we

had two primary concerns:

(1) The strength of the structure, i.e. its ability to support a specified load without

experiencing excessive stresses.

() The ability of the structure to support a specified load without undergoing

unacceptable deformations.

! "hapter 1# deal with an introduction to deflection due to loads determined by various

methods.

• In this chapter, another type of failure mode called buc$ilng of columns is discussed.

%ifferent types of buc$ling will also be discussed. &e shall be concerned with stability of

the structure, i.e. with its ability to support a given load without experiencing a sudden

change in its configuration.

• 'ur discussion will relate mainly to column, i.e. to the analysis and design of vertical

prismatic members supporting axial loads. If a beam element is under a compressive load

and its length if the orders of magnitude are larger than either of its other dimensions

such a beam is called a columns. %ue to its sie its axial displacement is going to be very

small compared to its lateral deflection called buckling .

! uite often the buc$ling of column can lead to sudden and dramatic failure. *nd as a

result, special attention must be given to design of column so that they can safely supportthe loads. In loo$ing at columns under this type of loading we are only going to loo$ at

four different types of supports: (a) fixed-free; (b) pinned-pinned ; (c) fixed-pinned ; and

(d) fixed-fixed .



2

! 'ne problem that does not involve relative motion, but

which is closely related to structural dynamics, is

column buc$ling. This similarity is due to the type of

mathematics involved in the analysis of buc$ling of columns.

! The buc$ling occurs primarily due to instability of a

column under axial loading conditions. +uc$ling can

also occur due to secondary instability, which is another

type of instability. *n example of this is the buc$ling a soda straw, where the straw $in$s

at a point along its length due to bending load. *nother example is twist buc$ling of thin

cylinders.

11.2 Stabilit o! Structure":

%ue to imperfections no column is really straight. *t some critical compressive load it will

buc$le. To determine the maximum compressive load (Buc$lin% Load) we assume that

buc$ling has occurred as shown in igure below:

-I/I+0I -T2'%:

! The lateral deflection that occurs is called buc$ling.

! The maximum axial load a column can support when it is on the verge of buc$ling

is called the critical load, P cr .

! 3pring develops restoring force, F =k , while applied load P .

! The border between stability and instability occurs when,

Pδ =kLδ ∨( P−kL ) δ =0 , for any small displacement δ .



Free-body diagram of the buckled segment

#

! This condition is referred to as neutral e4uilibrium. rom the foregoing expression,

we can define the critical load as,

Pcr=kL(a)

! 5hysically, Pcr represents the load for which the system is on the verge of

buc$ling.

! or P< Pcr , the system is in stable e4uilibrium, and if

P> Pcr the system is in

unstable e4uilibrium.

11.# Buc$lin% o! &in'(nded Colu)n":

/oo$ closely at the +% of the left hand end of the beam as in next igure:

-4uating moments at the cut end:

∑ M =0= Pν+ M ( x )=0∴ M ( x )=− Pν

+ut since the deflection of a beam is related with its bending moment distribution,

then:

EI d

2ν

d x2=− Pν⇒ EI

d2

ν

d x2+ Pν=0

which simplifies to:

d2

ν

d x2+( P

EI )ν=0

, where P/ EI is a constant. This expression is in the form of a second order differential

equation.

+y 3etting,



*

0ewrite the second order differential equation in the following type:

d2

ν

d x2+ p

2ν=0

where: p

2

=

P

EI

The general solution of the preceding e4uation is,

ν= A sin px+B cos px

&here A and B are constants, which can be determined using the column6s

$inematic boundary conditions. The constants * and + are evaluated from the end

conditions:

ν (0 )=0∧ν ( L )=0

If B=0 , 7o bending moment exists, so the only logical solution is for:

sin ( p / L )=0 and the only way that this can happen is if :The first re4uirement

yields B=0 , and the second leads to:

A sin pL=0⇒ pL=nπ

The foregoing is satisfied if either, A=0 , or sin pL=0 and p= P/ EI . It

is fulfilled if:

pL=√ P

EI L=nπ (n=1,2,… … .)

rom which,

P=π

2n

2 EI

L2

(n=1,2,… … … … .. )(11.4 )

The values of n define the buc$ling mode shapes.

(uler For)ula:

The value for n=1 , yields:

Pcr=π

2 EI

L2 (11.5)

, which is also called Euler Buckling Load , Pcr , critical or maximum axial load on the

column 8ust before it begins to buc$le.

E : is 9oung6s modulus of elasticity.

I : is the least second moment of area for the column6s cross sectional area



,

L : unsupported length of the column, whose ends are pinned

The preceding result is $nown as -uler6 s ormula the corresponding load is called

the -uler buc$ling /oad

Note: 3ince

Pcr is

proportional

to I , the

column will

buc$le in the

direction corresponding to the minimum value of I , as shown in next igure:

*s defined, I = A r2

, where * is the cross;sectional area and r represents the

radius of gyration about the axis of bending. This gives,

Pcr=π

2 EA

( L/r )2(11.6)

(L/r) is called the slenderness ratio, and r is the radius of

gyration. For Pcr to be minimum, r should be small.

(a)ple 11.1

Gi/en: * m long pin ended column of s4uare cross section. *ssuming E <1.=>5a,

σ all=12 MPa for compression parallel to the grain, and using a factor of safety of .= in

computing -uler6s critical load for buc$ling.

0eter)ine: the sie of the cross section if the column is to safely support (a) a

P=100k N load and (b) a P=200k N load.

Solution:



! E"ecti#e lengths of columns for #arious end conditions$ (a) %&ed-free; (b) 'inned-'inned; (c) %&ed-'inn

-

11.* Colu)n" +ith Other (nd Condition":

he end conditions$ (a) %&ed-free; (b) 'inned-'inned; (c) %&ed-

'inned; (d) %&ed-%&ed as shon in Figure .!* a"ect the Euler+s

e,uation (.) as*



Pcr=C π

2 EI

L2 (a)

* here is a constant de'endent u'on the end conditions. he #alues

of for the common cases shon in Figure .!a* b* c* and are /.0* *

0./!* and !* res'ecti#ely.

1ote that the critical load* Pcr * increases as the degree of

constraint increases. E,uation (a) can be made to resemble the fundamental case if

L2/C * is re'laced by Le

2

* in hich Le * is called the

efective length. hus the Euler+s formula can be ritten in the

form$

Pcr=π

2 EI

Le

2 (11.5' )

2s shon in the Figure .!* the efective length is the distance

beteen the inection points on the elastic cur#es* or points of

zero moment .

his length is often e&'ressed in terms of an e"ecti#e length

factor K $ Le= KL

.

If e set I = A r2

* the foregoing e,uation becomes$

Pcr= π

2 EA

( Le/r )2(11.6' )

Le /r is called the efective slenderness ratio of the column.

(a)ple 11.2:



(a)ple 11.#:



3



14

5C6U5L (FF(C6I7( L(NG68S 9 AISC :

11., Critical Stre"": Cla""i!ication o! Colu)n":

C;I6IC5L S6;(SS: LONG COLUMNS:

σ cr= Pcr

A = π

2

E

( Le /r )2 (11.9 )

• The corresponding portion "% of the curve shown in igure 11.? is labeled

as (uler<" Cur/e.

• The critical value of the slenderness ratio that fixes the lower limit of this

curve is found by e4uatingσ cr to the proportional limit

σ p of the specific

material:



11

( Le

r )c

=π √ E

σ p(11.10)

• or example, in the case of structural steel with,

¿210GPa∧σ p=250 MPa , the foregoing results in ( ( Le /r )=90

).

S8O;6 COLUMNS O; S6;U6S (3teel: L/r<30 ): for this case, failure

occurs by yielding or by crushing, without buc$ling, at stresses exceeding the

proportional limit of the material. 3o the maximum stress:

σ max= P

A (11.11)

• The horiontal line AB in igure 11.?, is represent the strength limit of short

column. It is e4ual to the yield stress or ultimate stress in compression.

IN6(;M(0I56( COLUMNS:

• ost structural columns lie in a region between short and long columns,

represented by part BC in ig. 11.?.

• 3uch intermediate @ length columns do not fail by direct compression or by

elastic instability.

• 2ence, the critical stress may be expressed by Engesser’s tangent modulus

formula:

σ cr= Pt

A=

π 2 Et

( Le /r )2(11.12)

• Johnson’s formula: has the form:



12

σ cr=σ −1

E ( σ

2 π

Le

r )2

(11.13 a)

*lternatively, in terms of the critical load :

Pcr=σ cr A=σ A [1−σ ( Le/r )2

4 π 2 E ](11.13!)

(a)ple 11.*:



1#

11.- (ccentricall Loaded Colu)n" and the Secant For)ula:

EI d

2ν

d x2+ P ( ν+e )=0

d2

ν

d x2+ p

2

ν=− p

2

e

, where, p= P/ EI , as before. The general solution of the foregoing is:

ν= A sin px+Bcos px−e

2ere the first two terms represent the homogeneous solution and −e the

particular solution.

To obtain constants A and B, the boundary condition:

ν ( L/2 )=ν (− L/2 )=0 , are applied, with the following results:

A=0B= e

cos [√ P / EI ( L/2 ) ]

The e4uation of the deflection curve is therefore:

ν=e [ cos √ P /( EI ) x

cos√ P L2/(4 EI )

−1](11.17) -xpressed in the term of the critical load Pcr=π

2

EI / L2

, the midspan ( x=0 )

deflection is:

νmax=e ["ec ( π 2 √ P

Pcr )−1](11.18)

*s P approaches P cr , the maximum deflection, vmax increases without bound.

Mai)u) "tre""



1*

σ max= P

A+

Mc

I =

P

A+

M max c

A r2 (11.19)

∵ M max=− P ( νm#x−e ) , together with e4uation (11.1?) into e4uation (11.1A), we obtain:

σ max= P

A [1+ ec

r2 "ec( π

2 √ P

Pcr )](11.20a)

*lternatively,

σ max= P

A [1+ ec

r2 "ec( L

2r √ P

AE )](11.20!)

, which is $nown as the secant formula, applies only when the maximum stress remainswithin the elastic limit of the material.

5llo+able "tre""

The buc$ling load determined from e4uation (11.#) is converted to limit or yield load

P by P= P B

σ max=σ . +y so doing, the secant formula becomes:

σ = P

A

[1+

ec

r2

"ec ( L

2 r

√

P

AE )](11.21)

*fter finding the limiting load, we can determine the allowable axial load Pall from :

Pall= P

n "

(11.22)

S8O;6 COLUMNS O; S6;U6S

σ max= P

A

(1+

ec

r

2

)(11.23)

(a)ple 11.,:



1,



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Chapter 11 Buckling of Columns

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