CHAPTER
10
MECHANICS OF
MATERIALS
Columns
Dr. Atta ur Rehman Shah
(atta.shah@hitecuni.edu.pk)
Website: https://sites.google.com/view/atta85
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Columns
Stability of Structures
Euler’s Formula for Pin-Ended Beams
Extension of Euler’s Formula
Sample Problem 10.1
Eccentric Loading; The Secant Formula
Sample Problem 10.2
10 - 2
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Stability of Structures
• In the design of columns, cross-sectional area is
selected such that
- allowable stress is not exceeded

P
  all
A
- deformation falls within specifications

PL
  spec
AE
• After these design calculations, may discover
that the column is unstable under loading and
that it suddenly becomes sharply curved or
buckles.
10 - 3
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Stability of Structures
• Consider model with two rods and torsional
spring. After a small perturbation,
K 2   restoring moment
L
L
P sin   P   destabiliz ing moment
2
2
• Column is stable (tends to return to aligned
orientation) if
L
P   K 2 
2
P  Pcr 
4K
L
10 - 4
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Stability of Structures
• Assume that a load P is applied. After a
perturbation, the system settles to a new
equilibrium configuration at a finite
deflection angle.
L
P sin   K 2 
2
PL
P



4 K Pcr sin 
• Noting that sin <  , the assumed
configuration is only possible if P > Pcr.
10 - 5
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Euler’s Formula for Pin-Ended Beams
• Consider an axially loaded beam.
After a small perturbation, the system
reaches an equilibrium configuration
such that
d2y
dx

2
d2y
dx

2
M
P

y
EI
EI
P
y0
EI
• Solution with assumed configuration
can only be obtained if
P  Pcr 
 2 EI
L2
 
P
 2 E Ar 2
 2E
    cr 

2
A
L A
L r 2
10 - 6
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Euler’s Formula for Pin-Ended Beams
• The value of stress corresponding to
the critical load,
 2 EI
P  Pcr 

 cr 

L2
P
P
  cr  cr
A
A
 
 2 E Ar 2
L2 A
 2E
L r 
2
 critical stress
L
 slenderness ratio
r
• Preceding analysis is limited to
centric loadings.
10 - 7
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Extension of Euler’s Formula
• 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,
Pcr 
 cr 
 2 EI
L2e
 2E
Le r 2
Le  2 L  equivalent length
10 - 8
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Extension of Euler’s Formula
10 - 9
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 10.1
An aluminum column of length L and
rectangular cross-section has a fixed end at B
and 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.
a) Determine the ratio a/b of the two sides of
the cross-section corresponding to the most
efficient design against buckling.
L = 20 in.
E = 10.1 x 106 psi
b) Design the most efficient cross-section for
the column.
P = 5 kips
FS = 2.5
10 - 10
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 10.1
SOLUTION:
The most efficient design occurs when the
resistance to buckling is equal in both planes of
symmetry. This occurs when the slenderness
ratios are equal.
• Buckling in xy Plane:
1 ba 3
2
I
a
2
z
rz   12

A
ab
12
Le, z
rz

rz 
a
12
0.7 L
a 12
• Most efficient design:
Le, z
• Buckling in xz Plane:
3
1
b2
2 I y 12 ab
ry 


A
Le, y
ry

ab
2L
b / 12
12
rz
ry 
b
12

Le, y
ry
0.7 L
2L

a 12 b / 12
a 0.7

b
2
a
 0.35
b
10 - 11
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 10.1
• Design:
Le
2L
220 in  138 .6



ry b 12 b 12
b
Pcr  FS P  2.55 kips   12.5 kips
 cr 
 cr 
Pcr 12500 lbs

0.35b b
A
 2E

2
Le r 


 2 10.1  10 6 psi
138 .6 b 2
E = 10.1 x 106 psi
12500 lbs  2 10.1  10 6 psi

0.35b b
138 .6 b 2
P = 5 kips
b  1.620 in.
L = 20 in.
FS = 2.5


a  0.35b  0.567 in.
a/b = 0.35
10 - 12
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Eccentric Loading; The Secant Formula
• Eccentric loading is equivalent to a centric
load and a couple.
• Bending occurs for any nonzero eccentricity.
Question of buckling becomes whether the
resulting deflection is excessive.
• The deflection become infinite when P = Pcr
d2y
dx

2
 Py  Pe
EI
  P  
  1
ymax  e sec

  2 Pcr  
Pcr 
 2 EI
L2e
• Maximum stress
 max 

P   ymax  e c 
1

A 

r2
P  ec  1 P Le 

1  sec
A  r 2  2 EA r 
10 - 13
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Eccentric Loading; The Secant Formula
 max   Y 
P  ec  1 P Le 

1  2 sec
A r
 2 EA r 
10 - 14
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 10.2
The uniform column consists of an 8-ft section
of structural tubing having the cross-section
shown.
a) Using Euler’s formula and a factor of safety
of two, determine the allowable centric load
for the column and the corresponding
normal stress.
E  29  10 6 psi.
b) Assuming that the allowable load, found in
part a, is applied at a point 0.75 in. from the
geometric axis of the column, determine the
horizontal deflection of the top of the
column and the maximum normal stress in
the column.
10 - 15
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 10.2
SOLUTION:
• Maximum allowable centric load:
- Effective length,
Le  28 ft   16 ft  192 in.
- Critical load,
Pcr 
 2 EI

2
Le


 2 29  106 psi 8.0 in 4
192 in 2

 62.1 kips
- Allowable load,
P
62.1 kips
Pall  cr 
FS
2
Pall  31.1 kips
Pall 31.1 kips

A
3.54 in 2
  8.79 ksi

10 - 16
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 10.2
• Eccentric load:
- End deflection,
  P  
  1
ym  e sec

2
P
cr

 

    
 0.075 in sec
  1
2
2
 
 
ym  0.939 in.
- Maximum normal stress,
P  ec   P 

 m  1  2 sec

A
r
 2 Pcr 
31.1 kips  0.75 in 2 in    

1
sec

2 
2
2
2


3.54 in 
1.50 in 
 m  22.0 ksi
10 - 17
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Problem 10.17
A column of 22-ft effective length is made by welding two 9 x
0.5-in. plates to a W8 x 35 as shown. Determine the allowable
centric load if a factor of safety of 2.3 is required. Use E = 29 x
106 psi.
10 - 18
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Problem 10.21
The uniform brass bar AB has a rectangular cross section and is
supported by pins and brackets as shown. Each end of the bar can
rotate freely about a horizontal axis through the pin, but rotation
about a vertical axis is prevented by the brackets. (a) Determine
the ratio b/d for which the factor of safety is the same about the
horizontal and vertical axes. (b) Determine the factor of safety if
P = 1.8 kips , L = 7 ft, d = 1.5 in., and E = 29 x 106 psi .
10 - 19
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Problem 10.23
A 1-in.-square aluminum strut is
maintained in the position shown by a
pin support at A and by sets of rollers at
B and C that prevent rotation of the strut
in the plane of the figure. Knowing that
LAB = 3 ft, LBC = 4 ft , and LCD = 1 ft,
determine the allowable load P using a
factor of safety with respect to buckling
of 3.2. Consider only buckling in the
plane of the figure and use E = 10.4 x
106 psi.
10 - 20
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Problem 10.24
Column ABC has a uniform
rectangular cross section with
b = 12 mm and d = 22 mm. The
column is braced in the xz
plane at its midpoint C and
carries a centric load P of
magnitude 3.8 kN. Knowing
that a factor of safety of 3.2 is
required, determine the largest
allowable length L. Use E =
200 GPa.
10 - 21
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Problem 10.36
A brass pipe having the cross
section shown has an axial
load P applied 5 mm from its
geometric axis. Using E =
120 GPa, determine (a) the
load P for which the
horizontal deflection at the
midpoint C is 5 mm, (b) the
corresponding
maximum
stress in the column.
10 - 22
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Problem 10.39
The line of action of the axial
load P is parallel to the
geometric axis of the column and
applied at a point located on the
x-axis at a distance e = 12 mm
from the geometric axis of the
W310 x 60 rolled-steel column
BC. Assuming that L = 7.0 m and
using E = 200 GPa, determine
(a) the load P for which the
horizontal deflection of the
midpoint C of the column is 15
mm, (b) the corresponding
maximum stress in the column.
10 - 23