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 .
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! '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 .
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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,
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*
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
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,
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:
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! 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*
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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:
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(a)ple 11.#:
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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:
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( 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:
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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.*:
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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""
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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.,:
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