Chapter Objectives



To understand the behavior of columns and concept of
critical load and buckling
To determine the axial load needed to buckle a socalled ‘ideal’ column
To determine the ‘effective length’ of a column with
various end-conditions
Copyright © 2011 Pearson Education South Asia Pte Ltd
Buckling of Columns
• The first significant contribution to the theory of the
buckling of columns was made as early as 1744 by
Euler
• His classical approach is still valid, and likely to remain
so, for slender columns possessing a variety of end
restraints.
Ideal column
 It is perfectly straight before loading
 Both ends are pin-supported
 Loads are applied throughout the centroid of the cross section
2
COLUMNS HAVING VARIOUS END-CONDITIONS
3
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4
What is a column? APPLICATIONS?
5
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The selection of structural and machine elements are
based on three characteristics:
i. Strength – the ability of the structure to withstand
the applied load
ii. Stiffness – is the resistance to deformation (i.e. the
structure sufficiently stiff not to deform beyond
permissible limits)
iii. stability – ability to support the given load without
experiencing a sudden change in structure
configuration.
BUCKLING is a stability issue and it is a not
strength issue.
In design issue, there are only two primary concerns,
namely;
1. The strength of the structure i.e. the ability to
support a given or specified load without excessive
stress and
2. The ablity of the structure to support a specified
load without undergoing unacceptable deformations.
7
 The issue of structure stability whereby the ability to support a
given load without experiencing a sudden change in
configuration will be discussed.
 Columns is a structure solely design to support a compressive
load, slender columns with certain height said to be critical to
buckle under certain load i.e. critical load, Pcr.
8
What Is Buckling?
A slender structure that is subjected to a critical load, Pcr (compression),
suddenly experiences a large deformation and it may lose its ability to carry
the load (i.e. as a consequence, the slender structure carry a smaller axial
stress in the regions).
As shown in the following figures, when a slender rod AB is subjected to an
centric axial compressive load P, at a critical load Pcr, the rod bows out, and
it can be said that the rod has buckled i.e. due to the load that enough to
trigger sudden sideways deflection of slender structure.
It might considered that the column is considered
safe if both the following properties are less than
permissible values of;
The average compressive stress given by, σ = P/A
< σall
Deformation, δ = PL/(AE) < δall
9
BUCKLING THEORY
Consider a simplified model consisting of two rigid rods AC and BC connected at C by a pin
and a torsional spring of constant K (Figure 10.3).
If the two rods and forces P and P’ are perfectly aligned, the system will remain in the
position of equilibrium as long as it is not moved by slightly pushed at C slightly to the right.
There are two possibilities could occurs; 1. the system return to its original equilibrium
position or 2. move further away. For the first case, the system is said to be stable while in
the second, it is unstable (Figure 10.4).
10
To determine whether the two-rod system is stable or unstable, consider the Free Body
Diagram for rod AC (Figure 10.5).
If the moment of the second couple M is larger than the moment of the first couple (produced by P
and P’), the system tends to return to its original equilibrium position; THE SYSTEM IS STABLE.
If the moment of the first couple is larger than the moment of the second couple, the system tends to
move away from its original equilibrium position; THE SYSTEM IS UNSTABLE. The load when
the two couples balance each other is called the critical load, Pcr , which is given as
Since sin ∆θ ~ ∆θ, therefore at the onset of buckling
11
The system is stable when P < Pcr and unstable for P > Pcr. Assume that a load P > Pcr has been
applied to the two rods of Fig. 10.3 and the system has been disturbed. Since P > Pcr , the system
will move further away from the vertical and, after some oscillations, will settle into a new
equilibrium position [Figure 5 (Fig. 10.6a)]. Considering the equilibrium of the free body AC (Fig.
10.6b), an equation similar to Eq. (10.1) but involving the finite angle θ, is
Figure 5
12
The value of θ corresponding to the equilibrium position in Fig. 10.6 is obtained by solving Eq.
(10.3) by trial and error. But for any positive value of θ, sin θ < θ. Thus, Eq. (10.3) yields a value of
θ different from zero only when the left-hand member of the equation is larger than one. Recalling
Eq. (10.2), this is true only if P < Pcr . But, if P > Pcr , the second equilibrium position shown in Fig.
10.6 would not exist, and the only possible equilibrium position would be the one corresponding to
θ = 0. Thus, for P < Pcr , the position where θ = 0 must be stable.
3.1
EULER’S FORMULA FOR PIN-ENDED COLUMNS
By referring column AB (Fig. 10.1), the load P < Pcr the position shown in Fig. 10.1 said to be
stabled. If P > Pcr , the slight disturbance will cause the column to buckle (Fig. 10.2).
Since a column is like a beam placed in a vertical position and subjected to an axial load, P and
denote by x the distance from end A to a point Q denote by y the deflection (Fig. 10.7a). The
positive x axis is vertical and directed downward, and the y axis is horizontal and directed to the
right. Considering the equilibrium of the free body AQ (Fig. 10.7b),
Considering the equilibrium of the free body AQ (Fig. 10.7b), the bending moment M at Q is M =
-Py, substitute M in the beam defelction theory, gives;
13
This equation is a linear, homogeneous differential equation of the second
order with constant coefficients. Lets’
which is the same as the differential equation for simple harmonic motion,
except the independent variable is now the distance x instead of the time t.
The general solution of Eq. (10.7) is;
and is easily checked by calculating d2y/dx2 and substituting for y and d2y/dx2
Eq. (10.7). Recalling the boundary conditions that must be satisfied at ends A
and B of the column (Fig. 10.7a), make x =0, y = 0 in Eq. (10.8), and find
that B = 0. Substituting x = L, y = 0, obtain
This equation is satisfied if either A = 0 or sin pL = 0. If the first of these conditions is satisfied,
Eq. (10.8) reduces to y = 0 and the column is straight (Fig. 10.1). For the second condition to be
satisfied, pL = nπ, or substituting for p from (10.6) and solving for P,
The smallest value of P defined by Eq. (10.10) is that corresponding to
n = 1. Thus,
This expression is known as Euler’s formula, substituting this expression for P
into Eq. (10.6), the value for p into Eq. (10.8), and recalling that B = 0,
which is the equation of the elastic curve after the column has buckled (Fig. 10.2). Note that the
maximum deflection ym = A is indeterminate.
This is because the differential Eq. (10.5) is a linearized approximation of the governing
differential equation for the elastic curve. If P < Pcr , the condition sin pL = 0 cannot be satisfied,
and the solution of Eq. (10.12) does not exist. Then we must have A = 0, and the only possible
configuration for the column is a straight one. Thus, for P < Pcr the straight configuration of Fig.
10.1 is stable.
14
In a column with a circular or square cross section, the moment of inertia I is the same about any
centroidal axis, and the column is as likely to buckle in one plane as another (except for the
restraints that can be imposed by the end connections). For other cross-sectional shapes, the
critical load should be found by making I = Imin in Eq. (10.11a). If it occurs, buckling will take
place in a plane perpendicular to the corresponding principal axis of inertia. The stress
corresponding to the critical load, Pcr is the critical stress σcr . Recalling Eq. (10.11a) and setting I
= Ar2, where A is the cross-sectional area and r its radius of gyration gives
The quantity L/r is the slenderness ratio of the column. The minimum value of the radius of
gyration, rmin should be used to obtain the slenderness ratio, L/r and the critical stress, σcr in a
column.
Equation (10.13a) shows that the critical stress is proportional to the modulus of elasticity of the
material and inversely proportional to the square of the slenderness ratio of the column. The plot
of σcr versus L/r is shown in Fig. 10.8 for structural steel, assuming E = 200 GPa and σy = 250
MPa (i.e. no factor of safety taken into account). Also, if σcr obtained from Eq. (10.13a) or from
the curve of Fig. 10.8 is larger than the yield strength σy, this value is of no interest, since the
column will yield in compression and cease to be elastic before it has a chance to buckle.
15
Example 1
• A 7m long steel tube having the
cross section shown is to be
used as a pin-ended column.
Determine the maximum
allowable axial load the column
can support so that it does not
buckle or yield. Take the yield
stress of 250MPa.
16
Solution
17
18
3.2
EULER’S FORMULA FOR COLUMNS WITH OTHER END CONDITIONS
3.2.1
One free end A and one fixed end B (Free-Fixed Connections)
A column with one free end A supporting a load P and one fixed end B (Fig. 10.10a) behaves as
the upper half of a pin-connected column (Fig. 10.10b). The critical load, Pcr for the column can be
obtained from Euler’s formula Eq. (10.11a). By using a column length equal to twice the actual
length L i.e. the effective length Le = 2L and substitute in Euler’s formula (10.11a), gives
19
where, Le/r known as column effective slenderness ratio
3.2.2
ONE FIXED END A AND ONE FIXED END B (FIXED-FIXED CONNECTIONS)
The effective length Le = 0.5L in Euler’s formula (10.11a), gives
20
3.2.3 ONE FIXED END B AND ONE PIN-CONNECTED END A (FIXED-PIN
CONNECTIONS)
where the effective length is Le = 0.699L ~ 0.7L
21
22
23
24
25
Buckling Problems
Example 1: A 7m long steel tube having the cross section shown is to be used as a pin-ended
column. Determine the maximum allowable axial load the column can support so that it does not
buckle or yield. Take the yield stress of 250MPa.
26
Example 2: A 2 m long pin ended column with a square cross section is to be made of wood.
Assuming E=13 GPa and allowable stress of 12 MPa (σall=12 MPa) and using a factor of safety of
2.5 to calculate Euler’s critical load for buckling. Determine the size of the cross section if the
column is to safely support a 100 kN load.
PCR  FS 100  2.5 100  250kN


250 103 N 2m 
P l2
6
4
1
a4
3

7
.
794

10
m
PCR  2  I  CR2  I 
I

a

a

 7.794 10 6 m 4 
 2 13 109 Pa
l
 E
12
12
P
100kN
 
 10 MPa 
   all  10  12
A 0.100m 2
a  98.3mm  100mm
 2 EI

2

Example 3: The A-36 steel W200 x 46 member shown in Fig. 13–8 is to be used as a pinconnected column. Determine the largest axial load it can support before it either begins to buckle
or the steel yields.
27
Assignment (Buckling)
Q1. The structural member shown is to be used as a pin-connected column. Determine the largest
axial load it can support before it either begins to buckle or the steel yields. Section properties are
given as follows:
Q2. A 3m column with the following cross section is constructed of material with E=13GPa and is
simply supported at its two ends.
a. Determine Euler buckling load
b. Determine stress associated with the buckling load
28
EULER BUCKLING
29
30
31
32
33
34
35
36
37
38
39
40
CRITICAL LOAD
•
•
Long slender members subjected to an axial
compressive force are called columns, and the
lateral deflection that occurs is called buckling.
The maximum axial load that a column can support
when it is on the verge of buckling is called the critical
load.
41
Copyright © 2011 Pearson Education South Asia Pte Ltd
CRITICAL LOAD (cont)
•
From the free-body diagram:
•
For small θ, tan ≈ θ,
2P tan   F  k    k L / 2
2 P  k  L / 2 
Pcr  kL / 4 (independent of  )
P  kL / 4 (stable equilibrium)
P  kL / 4 (unstable equilibrium)
•
Note:
This loading (Pcr = kL/4) represents a case of the mechanism being in neutral
equilibrium. Since Pcr is independent of θ, any slight disturbance given to the
mechanism will not cause it to move further out of equilibrium, nor will it be
restored to its original position. Instead, the bars will remain in the deflected
position.
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42
IDEAL COLUMN
•
Ideal column
 It is perfectly straight before loading
 Both ends are pin-supported
 Loads are applied throughout the centroid of the
cross section
•
Behavior
– When P < Pcr, the column remains straight.
– When P = Pcr,
d 2v
EI 2  M   Pv
dx
d 2v  P 
  v  0
2
dx  EI 
 P 
 P 


v  C1 sin 
x   C2 cos
x 
 EI 
 EI 
43
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44
IDEAL COLUMN (cont)
•
Since v = 0 at x = 0, then C2 = 0
•
 P 
L   0
Since v = 0 at x = L, then C1 sin 
 EI 
•
•
•
 P 
L   0
Therefore, sin 
 EI 
 P 
L   n
Which is satisfied if 
 EI 
Or
n 2 2 EI
P
where n  1,2,3,...
2
L
45
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IDEAL COLUMN (cont)
•
 2 EI
Smallest value at P is when n = 1, thus Pcr  2
L
 2E
Corresponding stress is  cr 
2
•
Where r = √ (I/A) is called ‘radius of gyration’
•
(L/r) is called the ‘slenderness ratio’.
•
The critical-stress curves are
hyperbolic, valid only for σcr
is below yield stress
•
L / r
46
Copyright © 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 1
The A-36 steel W200 x 46 member shown in Fig. 13–8 is to be
used as a pin-connected column. Determine the largest axial
load it can support before it either begins to buckle or the steel
yields.
47
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48
EXAMPLE 1 (cont)
Solutions
• From Appendix B,
A  5890 mm 2 , I x  45.5 106 mm 4 , I y  15.3 106 mm 4
• By inspection, buckling will occur about the y–y axis.
Pcr 
 2 EI
L2

 2 200 106 15.3 10 4 1 / 10004
42
 1887.6 kN
• When fully loaded, the average compressive stress in the column is
 cr 
Pcr 1887.6 1000

 320.5 N/mm 2
A
5890
• Since this stress exceeds the yield stress,
250 
P
 P  1472.5 kN  1.47 MN (Ans)
5890
49
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COLUMNS HAVING VARIOUS END-CONDITIONS
•
Consider the moment-deflection equation for the
cantilevered column, which is fixed at the base.
d 2v
EI 2  P  v 
dx
V  2 v  2 where 2 
P
EI
•
The solution is v  C1 sin x  C2 cosx  
•
Since v = 0 at x = 0, so that C2  
•
Also, v'  C1 cosx  C2 sin x
•
Since v’ = 0 at x = 0, so that C1  0
50
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COLUMNS HAVING VARIOUS END-CONDITIONS
(cont)
v   1 cosx 
•
Hence
•
Since v = δ at x = L, thus
cosL   0 or L 
•
n
2
The smallest critical load occurs when n = 1, thus
Pcr 
 2 EI
2
4L
or Pcr 
 2 EI
 KL 
2
 with k  2 
K is called the ‘effective-length factor’
Le  KL
51
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COLUMNS HAVING VARIOUS END-CONDITIONS
(cont)
•
K for various end conditions:
52
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EXAMPLE 2
A W150 x 24 steel column is 8 m long and is fixed at its ends as
shown in Fig. 13–11a. Its load-carrying capacity is increased by
bracing it about the y–y (weak) axis using struts that are
assumed to be pin connected to its mid-height. Determine the
load it can support so that the column does not buckle nor the
material exceed the yield stress. Take Est = 200 GPa and
σY = 410 MPa.
53
Copyright © 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 2 (cont)
Solutions
• Effective length for buckling about the x–x and y–y axis is
KLx  0.58  4 m  4000 mm
KLy  0.78 / 2  2.8 m  2800 mm
• From the table in Appendix B,
I x  13.410 6  mm 4
I y  1.83106  mm 4
• Applying Eq. 13–11,
 Pcr  x 
 Pcr  y 
 EI x
2
 KL  x
2
 2 EI y
 KL  y
2

 2  200  13.4 106  

 2  200  1.83 106  
4000
2
2800
2
 1653.2 kN (1)
 460.8 kN (2)
54
Copyright © 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 2 (cont)
Solutions
• By comparison, buckling will occur about the y–y axis.
• The average compressive stress in the column is
 cr 
Pcr

A
460.8 103 
3060
 150.6 N/mm2  150.6 MPa
• Since this stress is less than the yield stress,
buckling will occur before the material yields.
• Thus, Pcr  461 kN (Ans)
• From Eq. 13–12 it can be seen that buckling will always
occur about the column axis having the largest
slenderness ratio, since it will give a small critical stress
55
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EULER “VALIDITY LIMIT”
• Euler theory is unsafe for small L/r ratios. It is useful,
therefore, to determine the limiting value of L/r below
which the Euler theory should not be applied; this is
termed the validity limit.
r
r
r
Hearn EJ. Mechanics of materials 2, 3rd edition, Butterworth-Heinemann, UK, 1997.
56
EULER “VALIDITY LIMIT”
• The validity limit is taken to be the point where the Euler
σcr equals the yield or crushing stress σY, i.e. the point
where the strut load
P  Y A
• Now the Euler load can be written in the form
Pcr 
 2 EI
 KL 
2
=
 2 EAr 2
 KL 
2
• Therefore in the limiting condition
P  Pcr   Y A=
 2 EAr 2
 KL 
2
L
 2E
 =
r
K 2 Y
• The value of this expression will vary with the type of
end condition; as an example, low carbon steel struts
with pinned ends give L/r80.
57
Hearn EJ. Mechanics of materials 2, 3rd edition, Butterworth-Heinemann, UK, 1997.
RANKINE-GORDON FORMULA
• Rankine-Gordon formula is a combination of the Euler
and crushing loads for a strut
1
1
1


PR Pcr PC
• For very short struts Pcr is very large; 1/Pcr can therefore
be neglected and PR = PC.
• For very long struts Pcr is very small and 1/Pcr is very
large so that 1/PC can be neglected. Thus PR = Pcr.
• This formula is therefore valid for extreme values of L/r.
It is also found to be fairly accurate for the intermediate
values in the range under consideration.
58
Hearn EJ. Mechanics of materials 2, 3rd edition, Butterworth-Heinemann, UK, 1997.
RANKINE-GORDON FORMULA
• Thus, re-writing the formula in terms of stresses,
1
1
1


 R A  cr A  Y A
 cr Y
Y
R 

 cr   Y [1  ( Y /  cr )]
• For example, for a strut with both ends pinned
 2E
 cr 
2
L / r
• Therefore
R 
Y
2

Y  L  
1  2   
  E  r  

Y
2

 L 
1  a   
 r  

59
Hearn EJ. Mechanics of materials 2, 3rd edition, Butterworth-Heinemann, UK, 1997.
RANKINE-GORDON FORMULA
• Therefore Rankine load
PR 
Y A
1  a ( L r ) 2 
• Typical values of a for use in the Rankine formula are
given in Table below.
60
Hearn EJ. Mechanics of materials 2, 3rd edition, Butterworth-Heinemann, UK, 1997.
SECANT FORMULA
•
For design of a column subjected to eccentric load,
consider the moment-curvature equation
EIV ' '  M   Pe  v 
or v' '2v' '  2e where 2 
•
P
EI
The solution is v  C1 sin x  C2 cosx  e
Boundary conditions:
x = 0, v = 0;
x = L, v = 0
61
Copyright © 2011 Pearson Education South Asia Pte Ltd
SECANT FORMULA (cont)
•
Since v = 0 at x = 0, so C2 = e
•
Since v = 0 at x = L, so C1  e[1  cos( L)] / sin( L)
•
From trigonometric properties
•
 L 
 L 
 L 
1  cos   L   2sin 2 
and
sin

L

2sin
cos







 2 
 2 
 2 
 L 
So, C1  e tan  2 
 
•
 





v

e
tan
sin

x

cos

x

1
Hence,
 2

  

•
And vmax x  L 2  e sec   L 2   1
•
Note: A nonlinear relationship
occurs between the load P and
the defection v. As a result, the
principle of superposition does
not apply here.
Copyright © 2011 Pearson Education South Asia Pte Ltd
62
SECANT FORMULA (cont)
•
Maximum moment occurs at the column’s midpoint, i.e.
M  P  e  vmax 
•
 L 
or M  Pe sec 

2


Hence the maximum stress is
 max 
P Mc P Pe  c
 L 

 
sec 

A I
A
I
2


Or,  max
 L  P  
P  ec
 1  2 sec 

  

A  r
 2r  EA   
63
Copyright © 2011 Pearson Education South Asia Pte Ltd
SECANT FORMULA (cont)
Design curves:
• The above equation for maximum stress σmax is
transcendental, and cannot be solved explicitly.
• Graphs to aid designer are available.
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