Third Edition
CHAPTER
10
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Columns
Lecture Notes:
J. Walt Oler
Texas Tech University
Presented by:
Dr. Mohammed Arafa
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
Third
Edition
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
Design of Columns Under Centric Load
Sample Problem 10.4
Design of Columns Under an Eccentric Load
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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Third
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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 
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L
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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.
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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
M
P



y
2
EI
EI
dx
d2y
P

y0
2
EI
dx
• 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
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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.
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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
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Extension of Euler’s Formula
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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
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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:
ry2

Le, y
ry
Iy
A


1 ab3
12
ab
b2

12
rz
ry 
b
12
2L
b / 12
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

Le, y
ry
0 .7 L
2L

a 12 b / 12
a 0 .7

b
2
a
 0.35
b
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Third
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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  106 psi
138.6 b 2
E = 10.1 x 106 psi
12500 lbs  2 10.1  106 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
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