MEC112

Machine Construction
Lecture 2

Columns Buckling
Ahmed Saleh
Faculty of Engineering
Galala University
 Learning objectives
• Describe the behaviour of columns in terms of stability.

• Develop Euler’s formula for columns, using effective lengths

to account for different end conditions.

• Develop the Secant formula for analysis of eccentrically

loaded columns.

• Discuss Centric & Eccentric Load Design.

2
Introduction
• Stress Analysis studies the internal effects of stress & strain in a solid
body that is subjected to an external loading.
→ Stress is associated with the strength of the material from which the
body is made.
→ Strain is a measure of the deformation of the body (related to stiffness).
→ It also includes the study of the body’s stability when it is subjected to
compressive loading (e.g. columns).

• The stability of a structure indicates its ability to

support a given load without experiencing a
sudden change in configuration.
→ The present discussion is focused on columns, i.e., the
analysis and design of vertical prismatic members
supporting axially compressive loads.

Laboratory test showing

a buckled column.
3
Stability of Structures
• In the design of columns, cross-sectional area is
selected such that:
P
→ Allowable stress is not exceeded =   all
A
PL
→ Deformation falls within specifications =   spec
AE

• It may happen that as the load is applied, the

column buckles, instead of remaining straight. Pin-ended axially Buckled pin-ended
loaded column. column.

• Consider a model with two rigid rods connected at

C by a pin and a torsional spring of constant K.
After a small perturbation:
K ( 2 ) = restoring moment
L L
P sin  = P  = destabilizing moment
2 2

L
• Column is stable (tends to P   K ( 2 )
2
Model column. Free-body diagram
return to aligned orientation) if 4K
of model column.
P  Pcr =
L 4
Stability of Structures
• Assume that a load P > Pcr is applied to the two
rods of the model column. After a perturbation,
the system settles to a new equilibrium
configuration at a finite deflection angle (θ):

L
P sin  = K ( 2 )
2
PL P 
= = Model column in buckled position,
4 K Pcr sin  with free-body diagram of rod AC.

→ Since for any +ve value of θ, sin θ < θ, the assumed configuration is only possible if
P > Pcr.

→ If P < Pcr, the only possible equilibrium position would be at θ = 0, i.e., for P < Pcr,
the position where θ = 0 must be stable.

• The above observation applies to structures & mechanical systems in general and is
used in the next section for the stability of elastic columns.

5
Euler’s Formula for Pin-Ended Columns
• Consider an axially loaded column. After a
small perturbation, the system reaches an
equilibrium configuration such that:
d2y M P
2
= = − y
dx EI EI
d2y P
+ y=0
dx 2 EI
• Solution with assumed configuration can only
be obtained if:
Free-body diagrams of (a) buckled
π 2 EI column and (b) portion AQ.
P  Pcr = 2
L
P
 =   cr =
(
π 2 E Ar 2)=
π2 E
(L r)
2
A L2 A

6
Euler’s Formula for Pin-Ended Columns
• The value of stress corresponding to the
critical load (Pcr):

 2 EI
P  Pcr =
L2
P P
=   cr = cr
A A

 cr =
(
π 2 E Ar 2 )
L2 A
π2 E Plot of critical stress (σcr) versus
= = critical stress L∕r for structural steel.
(L r)
2

L If σcr is larger than the yield

= slenderness ratio
r strength σY, this value is of
no interest, since the column
will yield in compression &
→ The above analysis is limited to perfectly aligned cease to be elastic before it
centric loads. has a chance to buckle.

7
Example 1:
• Consider a 2 m long pin-ended wood column with a square cross section
and E = 13 GPa, σall = 12 MPa. 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 load of (a) 100 kN & (b) 200 kN.

Solution:
(a) For the 100 kN load & using the factor of safety: Pcr = 2.5 ×100 = 250 kN
𝑃𝑐𝑟 𝐿2 (250×103 )×2
→ Using Euler’s formula & solving for I: 𝐼 = = = 𝟕. 𝟕𝟗𝟒 × 𝟏𝟎−𝟔 𝐦4
𝜋2 𝐸 𝜋2 ×(13×109 )

𝑎4
→ For a square of side a: 𝐼= = 7.794 × 10−6 → a = 98.3 mm ≈ 100 mm
12
𝑃 100×1000
→ Check the value of the normal stress in the column: 𝜎=𝐴= = 𝟏𝟎 𝐌𝐏𝐚
0.12

→ Since σ is smaller than the allowable stress (σall), a 100 × 100 mm cross section is
acceptable.

8
Example 1:
• Consider a 2 m long pin-ended wood column with a square cross section
and E = 13 GPa, σall = 12 MPa. 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 load of (a) 100 kN & (b) 200 kN.

Solution:
(b) For the 200 kN load & using the factor of safety: Pcr = 2.5 ×200 = 500 kN
𝑃𝑐𝑟 𝐿2 (500×103 )×2
→ Using Euler’s formula & solving for I: 𝐼 = 𝜋2 𝐸
= 𝜋2 ×(13×109 )
= 𝟏𝟓. 𝟓𝟗 × 𝟏𝟎−𝟔 𝐦4
𝑎4
→ For a square of side a: 𝐼= = 15.59 × 10−6 → a = 116.95 mm
12
𝑃 200×1000
→ Check the value of the normal stress: 𝜎=𝐴= = 𝟏𝟒. 𝟔𝟐𝐌𝐏𝐚
0.116952

→ Since σ is larger than σall, the dimension obtained is not acceptable, and the cross
section must be designed on the basis of its resistance to compression.
𝑃 200×1000
→ 𝐴 = 𝑎2 = 𝜎 = = 16.67 × 10−3 m2 → a = 129.1 mm
𝑎𝑙𝑙 12×106

→ A 130 × 130 mm cross section is acceptable. 9

Euler’s Formula for Columns with Other End
Conditions
• A column with one fixed and one free end, will
behave as the upper-half of a pin-connected
column.
• The critical loading is calculated from Euler’s
formula:

π 2 EI
Pcr = 2
Le
Effective length of a fixed-free
 E2
column of length L is equivalent to
 cr = a pin-ended column of length 2L.
( Le r )
2

Le = 2 L = equivalent length

10
Euler’s Formula for Columns with Other End
Conditions
• Effective length of column for various end conditions.

11
Example 2:
• An aluminum column of length L & a rectangular cross-section
has a fixed end at B & supports a centric load at A. Two smooth
and rounded fixed plates restrain end A from moving in one of
the vertical planes of symmetry but allow it to move in the
other plane. Knowing that L = 0.5 m, E = 70 GPa, P = 20 kN & a
factor of safety (FS) of 2.5 is required: (a) Determine the ratio
a/b of the two sides of the cross section corresponding to the
most efficient design against buckling. (b) Design the most
efficient cross section for the column.
Solution:
→ The most efficient design is when the critical stresses corresponding to the two possible buckling
modes are equal. This occurs when the effective slenderness ratios (Le/r) are equal.
(a) Buckling in xy Plane: Buckling in xz Plane: Most efficient design
1 3 1 ab3 Le, z Le, y
2 I z 12 ba a2 a Iy b2 b =
rz = = = rz = ry2 = = 12 = ry = rz ry
A ab 12 12 A ab 12 12
0 .7 L 2L a
Le, z 0 .7 L Le, y
=
2L = = 0.35
= a 12 b / 12 b
rz a 12 ry b / 12
a 0 .7
=
b 2 12
Example 2:
(b) Design: The data obtained in (a) can be used to size the cross section.

Le 2L 2 ( 0.5 m ) 3.464 L = 0.5 m

= = =
ry b 12 b 12 b E = 70 GPa
P = 20 kN
Pcr = ( FS) P = ( 2.5)( 20 kN ) = 50 kN
FS = 2.5
Pcr 50000 N a/b = 0.35
 cr = =
A ( 0.35b ) b

 cr =
 2E
=
(
 2 70  10 9 Pa )
( Le r )2 ( 3.464 b)2
2
50000 N  70  10 Pa
=
( 9
) b = 39.7 mm
( 0.35b) b ( 3.464 b)2 a = 0.35b = 13.89 mm

→ Estimate the critical stress (σcr) and compare it to the tensile yield strength (σY)
for aluminum alloys.
→ The calculated critical Euler buckling stress can never be taken to exceed the yield
strength of the material.
13
Eccentric Loading & the Secant Formula
• Eccentric loading is equivalent to a centric load & a couple.
• Bending occurs for any nonzero eccentricity. Question of
buckling becomes whether the resulting deflection is excessive.
• The deflection becomes infinite when P = Pcr
d 2 y −P Pe
= y − Column with an eccentric
dx 2 EI EI
load modelled as a column
  P L  with an equivalent centric
ymax = e sec   − 1 force-couple load.
  EI 2  

• Maximum stress (which occurs where the

bending moment is maximum).
P  ( ymax + e ) c 
 max = 1 + 
A r2 
P  ec  1 P Le  
=  1 + sec   
A  r 2  2 EA r  
Secant formula plots for buckling
This Secant Eq. can be used with any end conditions, as in eccentrically loaded columns.
long as the appropriate value is used for the critical load.
14
Example 3:
• The uniform column AB consists of 2.4 m section of
structural tubing with the cross section shown. (a) Using
Euler’s formula and a factor of safety of 2, determine the
allowable centric load for the column & the corresponding
normal stress. (b) Assuming that the allowable load found
in part (a) is applied at a point 18 mm from the geometric
axis of the column, determine the horizontal deflection of
the top of the column & the maximum normal stress in the
column.

Solution:
(a) Maximum allowable centric load:
E = 200 GPa
→ Effective length: Le = 2 × 2.4 = 4.8 m
𝜋2 𝐸𝐼 𝜋2 ×(200×109 )×(3.33×10−6 )
→ Critical load: 𝑃𝑐𝑟 = = = 285.3 kN
𝐿2𝑒 4.82

𝑃𝑐𝑟 285.3
→ Allowable centric load: 𝑃𝑎𝑙𝑙 = = = 𝟏𝟒𝟐. 𝟕 kN
𝐹𝑆 2
𝑃𝑎𝑙𝑙 142.7×103
→ Allowable normal stress: 𝜎𝑎𝑙𝑙 = = = 𝟔𝟐. 𝟓 MPa
𝐴 2284×10−6
15
Example 3:
(b) Eccentric load:
→ Horizontal deflection of the top of the column:
  P  
ym = e sec   − 1
  2 Pcr  
    
= (18 mm ) sec   − 1
  2 2 

ym = 22.5 mm
P  ec   P 
→ Maximum normal stress:  m = 1 + 2 sec  
A  r  2 Pcr  

142.7  103 N  ( 0.018 m )( 0.05 m )   

= 1 + sec   
2284  10 −6 m 2  ( 0.038 m ) 2  2 2 

 m = 150.2 MPa

16
Design Considerations
• The preceding sections determined the critical load of a column using
Euler’s formula and investigated the deformations & stresses in
eccentrically loaded columns using the secant formula.
→ In each case, all stresses remained below the proportional limit & the column was
initially a straight, homogeneous prism. Real columns fall short of such an
idealization.

• To account for the differences between idealized columns & real

columns, design normally is based on empirical formulas developed
from laboratory tests (and adopted by professional organizations).
• Columns design is typically based on the following categorization:
→ Long columns with central loading.
→ Intermediate-length columns with central loading.
→ Columns with eccentric loading.
→ Struts or short columns with eccentric loading.
17
Recommended Values for End-Condition Constant
• As mentioned before for pin-ended columns,
the critical load is given by Euler’s formula:

• Analogous to the use of effective length, other

end conditions can be accounted for with the
addition of an end-condition constant C.

→ In practice it is difficult to fix the column ends so that the factor C = 2 or C = 4 would
apply. Therefore, some designers never use a value of C greater than unity.
→ More conservative values of C are often used, as shown in the table below.

18
Design of Long Columns with Central Loading
• Using I = Ak2, where A is the area and k is the radius of gyration,
Euler’s formula can be expressed as:

→ l/k is the slenderness ratio, used to classify columns according to length categories.
→ Pcr/A is the critical unit load, the load per unit area necessary to place the column
in a condition of unstable equilibrium.

• Tests show vulnerability to failure near point Q. Euler curve

• Since buckling is sudden and catastrophic, a

conservative approach near Q is desired.
• Point T is usually defined such that Pcr/A = Sy/2,
giving:

Plotting Pcr/A vs l/k, with

→ For (l/k) > (l/k)1, use Euler’s formula. C = 1 gives curve PQR.
→ For (l/k) ≤ (l/k)1, use a parabolic curve between Sy & T.
19
Euler Equations for Specific Cross Sections
• Round Cross Section

Pcr C 2 E
= Solving for d
 d 4 l ( d /4)  2
2

• Rectangular Cross Section

Solving for b or h

20
Intermediate-Length Columns with Central Loading
• For intermediate-length columns, where (l/k) Euler curve

≤ (l/k)1, use a parabolic curve between Sy & T.

General form of parabola:
2
Pcr l
= a − b 
A  k
→ If parabola starts at Sy, then a = Sy
→ If parabola fits tangent to Euler curve at T, then
2
 Sy  1
b= 
 2  CE

• This results in a parabolic formula, also known as J.B. Johnson formula

21
J.B. Johnson Equations for Specific Cross Sections
• Round Cross Section

2
Pcr  Sy  1
= Sy −  Solving for d
( d 4
2
)  2 ( d 4)  CE

• Rectangular Cross Section

2
 
Pcr Sy 1 Solving
= Sy −  
bh ( )
 2 h 12  CE for b or h

22
Design of Columns with Eccentric Loading
• The maximum bending moment at midspan:

• The maximum compressive stress (axial + bending):

P Mc P Mc
c = + = +
A I A Ak 2

• Using Syc as the maximum value of σc & solving for P/A,

the secant column formula is obtained:
Comparison of secant
& Euler Eqs. for steel
with Sy = 280 MPa.

→ ec/k2 is the eccentricity ratio.

→ Design charts of the secant column formula

for various eccentricity ratio can be prepared
for a given material strength.
23
Example 4:
• Specify the diameter of a round column 1.5 m long that is to carry a maximum
load estimated to be 22 kN. Use a design factor nd = 4 and consider the ends as
pinned (rounded). The column material selected has a minimum yield strength of
500 MPa and a modulus of elasticity of 207 GPa.
Solution:
→ The column is to be designed for a critical load of: 𝑃cr = 𝑛𝑑 𝑃 = 4 × 22 = 88 kN
→ Using Euler’s formula for round cross section with C = 1 gives:
14
 64 P l 
2 14
 64 ( 88) 103 (1.5) 2 
d =  3 cr  = 3  = 0.0375 m = 37.5 mm
  CE  ( )
  (1) 207 10 
9

→ Table A–17 (in Shigley’s Mechanical Engineering Design) shows that the preferred
size is 40 mm. The slenderness ratio for this size is:
l
=
l
=
1.5 103 ( )
= 150
k d 4 40 4
→ Estimate (l ∕ k)1 to verify this is an Euler column [for which l ∕ k > (l ∕ k)1]:

( )
 2 2 (1) 207 109
12

12
 l   2 CE 
2

  =  =  = 90.4 < 150

k 1  S y   500 10 ( ) 6
 24
Example 5:
• Repeat Example 4 for a fixed-fixed column with a rectangular cross section with
b = 4h. Round up the results to the next higher whole millimetre.

Solution:
→ From Example 4, Pcr = 88 kN & from the table on “slide 18”, C = 1.2
→ Using Euler’s formula for rectangular cross section & setting b = 4h:
 3( 88)103 (1.5) 2 
14 14
 3P l 
2
h =  2 cr  = 2 9
= 0.0222 m = 22.2 mm
  CE    (1.2 )( 207 ) 10 

→ Rounding up the next higher whole millimetre gives h = 23 mm & b = 92 mm. The
slenderness ratio for this size is:
l 12 l 12 (1.5)
= = = 226
k h 0.023
→ Estimate (l ∕ k)1 to verify this is an Euler column [for which l ∕ k > (l ∕ k)1]:

( )
 2 2 (1.2) 207 109 
12 12
 l   2 CE 
2

  =  =  = 90.40 < 226

k 1  S y   500( )
10 6

25
Example 6:
• Repeat Example 4 with l = 375 mm.

Solution:
→ Using Euler’s formula for round cross section with C = 1 gives:
14
 64 P l 
2 14
 64 ( 88)103 ( 0.375) 2 
d =  3 cr  =  = 0.0187 m = 18.7 mm
  CE    (1) 207 10
3
( )9

l 4l 4 ( 375)
→ From Table A–17, d = 20 mm. The slenderness ratio is: = = = 75
k d 20

→ From Ex. 4, (l ∕ k)1 = 90.4. Since (l ∕ k) < (l ∕ k)1, the column is not an Euler column
and the parabolic equation must be used:

( ) ( )
500 106 ( 0.375) 
12
 Pcr S yl 2 
12
 88 103 2

d = 2 + 2  = 2 + 

 y S  CE    ( 500 ) 10 6
 2
()
1 207 10 9
( )
= 0.0190 m = 19.0 mm

→ From Table A–17, d = 20 mm.

→ The above answer is not substantially different from what was obtained from
Euler’s formula since we were close to point T on Euler curve.
26
Example 7:
• Choose a set of dimensions for a rectangular link that is to carry a maximum
compressive load of 5 N. The material selected has a minimum yield strength of
75 kPa & a modulus of elasticity E = 30 MPa. Use a design factor of 4 & an end
condition constant C = 1 for buckling in the weakest direction, and design for (a) a
length of 15 cm & (b) a length of 8 cm with a minimum thickness of 1.5 cm.
Solution:
12
 l   2 CE 
12
2
 2 2 (1)(30)(106 ) 
(a) The limiting slenderness ratio:   =  =  = 88.9
k 1  S y   75(10)3 
→ By using Pcr = ndP = 4×5 = 20 N, Euler’s formula & the parabolic equation are
solved, using various h values, to generate the table below.
→ The table shows that a cross section of 0.625 by 0.76 cm which is marginally
suitable, gives the least area.
(b) An approach similar to that in part (a)
is used with l = 8 cm. All trials are found to
be in the J. B. Johnson region of l/k values.
A minimum area occurs when the section
is a near square. Thus a cross section of
1.5 × 1.8 cm is found to be suitable & safe. 27
Design of Struts or Short Compression Members
• Strut: a short member loaded in compression.

• If eccentricity exists, maximum stress is at B with

axial compression & bending:

→ Note that it is not a function of length.

→ The above Eq. differs from the secant equation as it
assumes small effect of bending deflection.

• If bending deflection is limited to 1 % of e, the

limiting slenderness ratio for the strut is (refer to
Euler curve):

28
Example 8:
• A workpiece is clamped to a milling machine
table by a bolt tightened to a tension of 2000 N.
The clamp contact is offset from the centroidal
axis of the strut by a distance e = 0.10 cm, as
shown. The strut, or block, is 1 cm square and 4
cm long. Determine the maximum compressive
stress in the block .
Solution:
→ A = bh = 1×1 = 1 cm2, I = bh3/12 = (1×13)/12 = 0.0833 cm4, k2 = I/A = 0.0833 cm2
& l/k = 4/(0.08331/2) = 13.9. The limiting slenderness ratio is given by:
( ) 
1( 30) 106
12
12
l  AE 
  = 0.282   = 0.282  = 48.8
k 2 P   1000 

→ Before having to treat the block using the secant formula, it could be as long as:
l = 48.8k = 48.8×(0.08331/2) = 14.1 cm
→ Thus, the maximum compressive stress is given by:

29
Reading assignment
• Chapter 10 “Columns” – “Beer & Johnston, Mechanics of Materials”
• Chapter 4 “Deflection and Stiffness” – from Section 4-11 to 4-17,
“Shigley’s Mechanical Engineering Design”
• Chapter 4 “Stress, Strain, and Deflection” – Section 4-16, “Norton,
Machine Design: An Integrated Approach”

• Watch the following video on Buckling:

https://www.youtube.com/watch?v=21G7LA2DcGQ

• Watch the following video on the Radius of gyration:

https://www.youtube.com/watch?v=_bbH2BX1bHE

30