Mukavemet II
Strength of
Materials II
Mukavemet II
Strength of
Materials II
Stabilite (Burkulma)
Stability (Buckling)
Teaching Slides
Chapter 10:
Stability (Buckling)
Chapter Outline
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Stability of Structures
Euler’s Formula for Pin-Ended Columns
Extension of Euler’s Formula to Columns with
Other End Conditions
Eccentric Loading; the Secant Formula
Design of Columns under a Centric Load
Design of Columns under an Eccentric Load
Stability (Buckling)
FIGURE Buckling demonstrations.
Stability (Buckling)
Stability (Buckling)
In Fig. a, the ball is said to be in stable equilibrium because, if it is
slightly displaced to one side and then released, it will move back
toward the equilibrium position at the bottom of the “valley.’’
Stability (Buckling)
If the ball is on a perfectly flat, level surface, as in Fig. b,
it is said to be in a neutral-equilibrium configuration. If
slightly displaced to either side, it has no tendency to
move either away from or toward the original position,
since it is in equilibrium in the displaced position as well
as the original position.
Stability (Buckling)
Stability (Buckling)
THE IDEAL PIN-ENDED COLUMN; EULER BUCKLING LOAD
To investigate the stability of real columns, we begin by
considering the ideal pin-ended column, as illustrated in Fig.10:
We make the following simplifying assumptions:
 The column is initially perfectly straight, and it is made of
linearly elastic material.
 The column is free to rotate, at its ends, about frictionless pins;
that is, it is restrained like a simply supported beam. Each pin
passes through the centroid of the cross section.
 The column is symmetric about the xy plane, and any lateral
deflection of the column takes place in the xy plane.
 The column is loaded by an axial compressive force P applied
by the pins.
Stability (Buckling)
However, if a load P=Pcr is applied to the column,
the straight configuration becomes a neutralequilibrium configuration, and neighboring
configurations, like the buckled shape in Fig.
10.6b, also satisfy equilibrium requirements.
Therefore, to determine the value of the critical
load, Pcr, and the shape of the buckled column, we
will determine the value of the load P such that the
(slightly) bent shape of the column in Fig. 10.6b is
an equilibrium configuration.
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Stability (Buckling)
Equilibrium of the Buckled Column
First, using the FBD of Fig. 10.6c, we get
Ax=P (from ΣFx=0 ).
Ay=0 [from ΣMB=0], and By=0.
Therefore, on the FBD in Fig. 10.6d, we show
only a vertical force P acting on the pin at A, and
we show no horizontal force V(x) at section x.
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The sign convention adopted for the
moment M(x) in Fig. 10.6d is a
positive bending moment. From Fig.
10.6d we get
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Differential Equation of Equilibrium, and End Conditions:
Substituting M(x) into the moment-curvature equation,
we obtain elastic curve equation as below:
EIv" ( x)  M ( x)   Pv( x)
or
d 2 v M ( x)
P

 v
2
dx
EI
EI
or
EIv" ( x)  Pv( x)  0
This is the differential equation that governs the
deflected shape of a pin-ended column.
It is a homogenous, linear, second-order, ordinary
differential equation.
The boundary conditions for the pin-ended member are
v(0)  0
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and
v ( L)  0
Solution of the Differential Equation:
The term v(x) means that we cannot simply integrate twice to get
the solution. When EI is constant, there is a simple solution to this
equation.
It is an ordinary differential equation with constant coefficients. For
the uniform column, therefore, it becomes
EI v" P v  0
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P
 
EI
2
v" v  0
2
The general solution to this homogeneous equation is
v( x)  C1 sin x  C2 cosx
We seek a value of and constants of integration C1
and C2 such that the two boundary conditions of this
homogeneous equation are satisfied.
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Thus,
v(0)  0
 C2  0
v( L)  0
 C1 sin L  0
Obviously, if the constants C1 and C2 equal to zero, the
deflection v(x) is zero everywhere.
We just have the original straight configuration.
Stability (Buckling)
For an alternative equilibrium configuration, since C1 can
not be zero, sin(λ 1L) must be zero. So, the characteristic
equation is satisfied with C1≠0.
sin n L   0
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n
n 
,
L
n  1, 2,....
Combining the equations above gives the following
formula for the possible buckling loads:
P
2
 
EI
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Pn  n
 EI
2
2
2
L
The deflection curve that corresponds to each load Pn
is obtained by combining the following Equations.
v( x)  C1 sin x  C2 cosx
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and
sin n L  0
we get
 nx 
vx   C sin

 L 
The function that represents the shape of the deflected column
is called a mode shape, or buckling mode.
The constant C, which determines the direction (sign) and
amplitude of the deflection, is arbitrary, but it must be small.
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The existence of neighboring
equilibrium configurations is
analogous to the fact that the ball
in Fig. 10.3b can be placed at
neighboring locations on the flat,
horizontal surface and still be in
equilibrium.
Stability (Buckling)
The value of P at which buckling will actually occur is
n 2 2 EI
obviously the smallest value given by Eq. Pn 
2
L
(n=1).Thus, the critical load is
The critical load for an ideal column is known as the Euler
buckling load, who was the first to establish a theory of
buckling of columns.
Leonhard Euler (1707–1783) is the famous Swiss mathematician
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Let us examine some important implications of the
Euler buckling-load formula. We can express this in
terms of critical (buckling) stress.
 cr 
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 E
2
Le / r 
2
Le

r
 cr
 2E
 2

 cr
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 2E
 2

 cr
 2E
 2

FIGURE: Critical stress versus slenderness ratio for structural steel columns.
For slenderness ratios less than 100, the critical stress is not meaningful.
Stability (Buckling)
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EXAMPLE
A 2-m-long pin-ended column of square cross
section is to be made of wood. Assuming E=13 GPa,
σall=12 MPa, and using a factor of safety of 2.5 in
computing Euler’s critical load for buckling,
determine the size of the cross section if the column
is to safely support
a) a 100-kN load,
b) a 200-kN load.
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a) For the 100-kN Load.
Using the given factor of safety, we make
in Euler’s formula and solve for I. We have
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Recalling that, for a square of side a, we have
, we write
We check the value of the normal stress in the column:
Since σ is smaller than the allowable stress, a 100x100-mm cross
section is acceptable.
b) For the 200-kN Load.
Solving again buckling equation for I, but making now
Pcr=2.5(200)=500 kN, we have
The value of the normal stress is
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Since this value is larger than the allowable stress, the dimension
obtained is not acceptable, and we must select the cross section
on the basis of its resistance to compression. We write
A 130x130-mm cross section is acceptable
EXTENSION OF EULER’S FORMULA TO
COLUMNS WITH OTHER END CONDITIONS
Euler’s formula was derived in the preceding section for
a column that was pin-connected at both ends.
Now the critical load Pcr will be determined for columns with
different end conditions.
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Column with One Free End
In the case of a column with one free end
A supporting a load P and one fixed end
B (Fig.-a), it is observed that the column
will behave as the upper half of a pinconnected column (Fig.-b).
Stability (Buckling)
The critical load for the column of Fig.-a is
thus the same as for the pin-ended column
of Fig.-b and can be obtained from Euler’s
formula by using a column length equal to
twice the actual length L of the given
column.
Stability (Buckling)
We say that the effective length Le of the column is equal to 2L
and substitute Le=2L in Euler’s formula:
Pcr 
Pcr 
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 2 EI
2 L 
2
 2 EI
L2e

 2 EI
2
4L
Le=2L
The critical stress is found in a similar way from the formula
 cr 
 E
2
Le / r 
2
Le

r
 cr
 E
 2

2
The quantity Le /r is referred to as the effective slenderness ratio
of the column and, in the case considered here, is equal to 2L/r.
Stability (Buckling)
Column with Two Fixed Ends
Consider next a column with
two fixed ends A and B
supporting a load P (Fig. 10.10).
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The symmetry of the supports
and of the loading about a
horizontal axis through the
midpoint C requires that the
shear at C and the horizontal
components of the reactions at A
and B be zero (Fig. 10.11).
Stability (Buckling)
It follows that the restraints
imposed upon the upper half AC
of the column by the support at
A and by the lower half CB are
identical (Fig. 10.12).
Portion AC must thus be symmetric about its
midpoint D, and this point must be a point of
inflection, where the bending moment is
zero. A similar reasoning shows that the
bending moment at the midpoint E of the
lower half of the column must also be zero
(Fig. 10.13a).
Since the bending moment at the ends
of a pin-ended column is zero, it follows
that the portion DE of the column of Fig.
10.13a must behave as a pin ended
column (Fig. 10.13b). We thus conclude
that the effective length of a column with
two fixed ends is Le =L/2.
Pcr 
 EI
2
L / 2 
2

 EI
2
2
L 4
Le=L/2
Pcr 
 2 EI
L2e
EXAMPLE
What is the maximum compressive load that can be
applied to an aluminum-alloy compression member of
length L=4 m if the member is loaded in a manner that
permits free rotation at its ends and if a factor of safety
of 1.5 against buckling failure is to be applied?
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Stability (Buckling)
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Physically, the effective length of a column is the distance
between points of zero moment when the column is deflected in
its fundamental elastic buckling mode. Figure 10.13 illustrates the
effective lengths of columns with several types of end conditions.
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STABILITY OF STRUCTURES
Let us deal with design of a column AB of length L
to support a given load P (Fig. 10.1). The column
exposed to a centric axial load P will be pinconnected at both ends.
Stability (Buckling)
However, it may happen that, as the
load is applied, the column will buckle;
instead of remaining straight, it will
suddenly become sharply curved
(Fig. 10.2). Photo 10.1 shows a
column that has been loaded so that it
is no longer straight; the column has
buckled. Clearly, a column that
buckles under the load it is to support
is not properly designed.
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EULER’S FORMULA FOR PIN-ENDED COLUMNS
Stability (Buckling)
Our approach will be to determine the
conditions under which the
configuration of Fig. 10.2 is possible.
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