Chapter 9 Buckling of Columns
9.0 INTRODUCTION TO BCUKLING (SI&4th:649-652)
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; (2) the ability of the structure to support a
specified load without undergoing unacceptable deformations. In this chapter, we 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. Our 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. Due to its size its
axial displacement is going to be very small compared to its lateral deflection called buckling.
Quite often the buckling of column can lead to sudden and dramatic failure. And as a result,
special attention must be given to design of column so that they can safely support the loads.
In looking at columns under this type of loading we are only going to look at three different
types of supports: pin-ended, doubly built-in and cantilever.
9.1 SLENDER PIN-ENDED COLUMN (SI 649-657; 4th:652-657; 3rd Ed p.653661)
Due to imperfections no column is really straight. At some critical compressive load it will
buckle. To determine the maximum compressive load (Buckling Load) we assume that
buckling has occurred as shown in Fig. 9.1,
y,v
P
P
x
L
Fig. 9.1 Deflection column due to applied compressive load P
Look closely at the FBD of the left hand end of the beam as in Fig. 9.2:
M(x)
y,v
V(x)
P
P
v
x
Fig. 9.2 FBD of section of length x of deflected column
Equating moments at the cut end:
∑ M = 0 = Pv + M (x ) = 0
∴ M ( x ) = − Pv
(9.1)
But since the deflection of a beam is related with its bending moment distribution, then:
d 2v
EI 2 = − Pv
(9.2)
dx
d 2v  P 
which simplifies to:
+  v = 0
(9.3)
dx 2  EI 
where P/EI is a constant. This expression is in the form of a second order differential
equation of the following type:
Lecture Notes of Mechanics of Solids, Chapter 9
1
d 2v
+ α 2v = 0
(9.4)
2
dx
P
α2 =
where:
(9.5)
EI
The solution of this equation is:
v = A cos(αx ) + B sin(αx )
(9.6)
where A and B are constants, which can be determined using the column’s kinematic
boundary conditions.
Kinematic Boundary Conditions
at x = 0, v = 0: 0 = A + 0, giving that A = 0
at x = L , v = 0, then: 0 = B sin(αL )
If B = 0, No bending moment exists, so the only logical solution is for: sin(αL ) = 0 and the
only way that this can happen is if :
α L = nπ ,
(9.7)
where n = 1,2,3,L . But since: α 2 =
P  nπ 
= 
EI  L 
2
(9.8)
then we get that buckling load as:
π 2 EI
P = n2 2
L
The values of 'n' define the buckling mode shapes, as in Fig. 9.3:
P1
First mode of buckling P1 =
π 2 EI
(9.9)
P1
L2
P2
P2
Second mode of buckling P2 =
4π 2 EI
L2
P3
P3
Third mode of buckling
P3 =
9π 2 EI
L2
Fig. 9.3 First three modes of buckling loads
Critical Buckling Load
However, since P1 < P2 < P3, the column buckles at P1 and never gets to P2 or P3 unless
bracing is placed at the points where v = 0 to prevent buckling at lower loads.
The critical load for a pin ended column is therefore:
π 2 EI
PCrit = 2 = PE
(9.10)
L
which is also called Euler Buckling Load,
PCrit
Critical or maximum axial load on the column just before it begins to buckle
E
Young’s modulus of elasticity
I
least second moment of area for the column’s cross sectional area
L
unsupported length of the column, whose ends are pinned
Lecture Notes of Mechanics of Solids, Chapter 9
2
9.2 BUILT-IN COLUMN (SI&4th: 658-668; 3rd Ed p.662-672)
The critical load for other columns can be expressed in terms of the critical buckling load for
a pin- ended column PE. A built-in column looks like Fig. 9.4:
P
L
A
B
Zero Bending Moment
L/4
L/2
P
P
L/4
P
LE
Fig. 9.4 Built-in column at both ends showing the effective pin-ended length
From symmetry conditions, at the points of inflection
d 2v
= 0 = M (x )
dx 2
which occurs at 1/4L points. Thus the middle half of the column can be taken out and treated
as a pin-ended column of length LE = L/2 as shown in Fig. 9.4. The critical load for this half
length is then :
π 2 EI 4π 2 EI
PCrit = 2 =
= 4 PE
(9.11)
LE
L2
9.3 CANTILEVER COLUMN
L=LE/2
P
A
B
P
LE
Fig. 9.5 Cantilever column and its effective length
This is similar to previous case. However, this span is equivalent to 1/2 of the Euler span LE,
as illustrated in Fig. 9.5, thus:
π 2 EI π 2 EI PE
(9.12)
PCrit = 2 =
=
4
4 L2
LE
Note: Since PCrit is proportional to I, the column will buckle in the direction corresponding to
the minimum value of I, as shown in Fig. 9.6:
Lecture Notes of Mechanics of Solids, Chapter 9
3
Buckling Direction
Cross-section
y
P
y
z
P
z
A
h
x
b
Iy>Iz
Fig. 9.6 Column cross section showing the direction of buckling (here: I z =
bh 3
hb 3
< Iy =
)
12
12
9.4 CRITICAL COLUMN STRESS
A column can either fail due to the material yielding, or because the column buckles, it is of
interest to the engineer to determine when this point of transition occurs.
Consider the Euler buckling equation 9.10
π 2 EI
PE = 2
L
Because of the large deflection caused by buckling, the least second moment of area term I
can be expressed as follows:
I = Ar 2
(9.13)
where: A is the cross sectional area and r is called radius of gyration of the cross sectional
area, i.e. r = I / A . Note that the smallest radius of gyration of the column, i.e. the least
second moment of area I should be taken in order to find the critical stress.
Dividing the buckling equation by A, gives:
P
π2 E
(9.14)
σE = E =
A (L / r )2
where: σE is the compressive stress in the column and must not exceed the yield stress σY of the
material, i.e. σE<σY, L / r is called the slenderness ratio, it is a measure of the column's flexibility.
If this equation is plotted for steel it gives:
σx
σY
240MPa
σ Crit =
89
π2 E
( L / r )2
L/r
Fig. 9.7 Critical stress vs slenderness ratio for steel
For a column not to fail by either yielding or buckling, its stress must remain underneath this
diagram in Fig. 9.7.
Example 9.1 A 2m long pin ended column of square cross section. Assuming E=12.5GPa,
σallow=12MPa for compression parallel to the grain, and using a factor of safety of 2.5 in
computing Euler’s critical load for buckling, determining the size of the cross section if the
column is to safely support (a) a P = 100kN load and (b) a P = 200kN load.
Lecture Notes of Mechanics of Solids, Chapter 9
4
a
A
P
B
Section a-a
y
I
P
s
z
a
s
Part (a)
Second moment of area
Iz = Iy =
1 3 s4
ss =
12
12
Buckling criterion
F fail 

 , we make the required critical load as
Using given Factor of Safety FS=2.5  FS =
Fallow 

PCrit ≥ FS × P = 2.5 × 100kN = 250 × 10 3 N
Based on Euler’s formula, Eq. (9.10), we have
250 × 10 3 L2
π 2 EI
PCrit = 2 ≥ 250 × 10 3 N
∴I ≥
L
π2 E
250 × 10 3 L2
250 × 10 3 × 2 2
4
×
12
=
× 12 = 0.0993m = 99.3mm
π2 E
π 2 × 12.5 × 10 9
P
P
∴ A = s2 ≥
Stress criterion
σ = =≤ σ allow
A
σ allow
or:
s B1 ≥ 4
100 × 10 3
= 0.0913m = 91.3mm
σ allow
12 × 10 6
Comparing the results from these two criteria, we have s ≥ max{s B 2 , sσ 2 } = 99.3mm . In this
case, the design is taken against the buckling criterion. Finally, one may select a round-up
amount, e.g. s = 100mm, as the design of the size of cross section.
i.e. s σ1 ≥
P
=
Part (b)
Buckling criterion
PCrit ≥ FS × P = 2.5 × 200kN = 500 × 10 3 N
Step 2: Euler’s formula PCrit =
π 2 EI
≥ 500 × 10 3 N
2
L
∴I ≥
500 × 10 3 L2
π2 E
500 × 10 3 L2
500 × 10 3 × 2 2
4
or:
sB2 ≥
× 12 =
× 12 = 0.1181m = 118.1mm
π2 E
π 2 × 12.5 × 10 9
P
P
σ = =≤ σ allow
∴ A = s2 ≥
Stress criterion
A
σ allow
4
200 × 10 3
i.e. s σ 2 ≥
=
= 0.1291m = 129.1mm
σ allow
12 × 10 6
Comparing the results from these two criteria, we have s ≥ max{s B 2 , sσ 2 } = 129.1mm . In this
case, the design is taken against the stress criterion. One may select s = 130mm as the design
of the size of cross section.
P
Lecture Notes of Mechanics of Solids, Chapter 9
5
Example 9.2 Determine the largest load P which may be applied to the structure as shown.
Assume that E=200GPa, allowable vertical deflection at point A δallow=0.5mm and allowable
compressive and tensile stress σallow=50MPa.
Cross section for AB & AC
B
Pin A
FAB
30º
A
3m
30º
A
FAC
P
h=50mm
C
y
z
8m
P
b=100mm
Step 1: Determine the members’ internal forces
+ ↑ ∑ Fy = 0 = FAB sin 30° − P
∴ FAB = 2 P
(+ tensile force)
+ → ∑ Fx = 0 = FAB cos 30° + FAC ∴ FAC = − 3P (- compressive force)
Step 2: Buckling criterion FAB is in tension, we do not considered its buckling. But bar AC is
a strut and we need to check for buckling. I about y and z is computed respectively


bh 3 0.1 × 0.53
hb 3 0.5 × 0.13
−6 4 
−6 4 
Iz =


=
=
×
=
=
=
×
1
.
04267
10
m
<
I
41
.
667
10
m 
y

 
12
12
12
12

 

∴ PCrit ,AC =
π 2 E AC I AC
L2AC
=
(
)(
π 2 × 200 × 10 9 × 1.04267 × 10 −6
8
2
) = 32.128kN
But FAC = PCrit ,AC = − 3PB , ∴ PB = PCrit / 3 = 18.55kN
Step 3: Strength criterion Consider tensile and compressive stresses in AB and AC respectively.
F
50 × 10 6
2P
σ AB = AB =
≤ σ allow = 50 × 10 6
P=
= 125kN
AAB 0.05 × 0.1
400
FAC
3P
50 × 10 6
6
σ AC =
=
≤ σ allow = 50 × 10
P=
= 144.3kN
AAC 0.05 × 0.1
400
From stress criterion, the maximum allowable load should be the smallest one i.e. Pσ=125kN
Step 4: Stiffness criterion Consider vertical deflection at point A using Castigliano’s method.
F 2L
F2 L
F2 L
Total strain energy due to axial forces: U = ∑ i i = AB AB + AC AC
2 E AB AAB 2 E AC AAC
i 2 E i Ai
Fi 2 Li
∂ 
The displacement can be then computed as: ∆ P =
∑
∂P  i 2 Ei Ai
Member
Fi (N)
AB
AC
2P
- 3P
(
∂Fi ∂P
2
- 3
) (
Li(m)
Ai (m2)
6
8
0.05
0.05
)

∂F  L
 = ∑ (Fi ) i  i


 ∂P  Ei Ai
 i
Fi (Li Ei Ai )(∂Fi ∂P )
2.4×10-9P
2.4×10-9P



Thus we have: ∆P = 2.4 × 10 −9 P + 2.4 × 10 −9 P = 4.8 × 10 −9 P ≤ δ allow
0.0005
∴ Pδ =
= 104.17kN
4.8 × 10 −9
Step 5: Determine the maximum allowable load P from the above three criteria
Clearly, for the safety reason, we should pick the lowest level as the allowable load
P = min{PB , Pσ , Pδ } = 18.55kN
Lecture Notes of Mechanics of Solids, Chapter 9
6