CHAPTER MECHANICS OF

SOLIDS
10 Columns

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MECHANICS OF SOLIDS
Columns

Stability of Structures
Euler’s Formula for Pin-Ended Beams
Extension of Euler’s Formula
Design of Columns Under Centric Load
Design of Columns Under an Eccentric Load

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MECHANICS OF SOLIDS
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.

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MECHANICS OF SOLIDS
Stability of Structures

• Consider model with two rods and torsional

spring. After a small perturbation,
K 2   restoring moment
L L
P sin   P   destabilizing moment
2 2

• Column is stable (tends to return to aligned

orientation) if
L
P   K 2 
2
4K
P  Pcr 
L

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MECHANICS OF SOLIDS
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 SOLIDS
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
2
   y
dx EI EI

d2y P
2
 y0
dx EI

• Solution with assumed configuration

can only be obtained if
 2 EI
P  Pcr 
L2
P
    cr 
 
 2 E Ar 2

 2E
A 2
L A L r 2

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MECHANICS OF SOLIDS
Euler’s Formula for Pin-Ended Beams
• The value of stress corresponding to
the critical load,
 2 EI
P  Pcr 
L2
P P
   cr  cr
A A

 cr 
 
 2 E Ar 2
L2 A
 2E
  critical stress
L r 
2

L
 slenderness ratio
r

• Preceding analysis is limited to

centric loadings.

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MECHANICS OF SOLIDS
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,
 2 EI
Pcr 
L2e

 2E
 cr 
Le r 2
Le  2 L  equivalent length

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Extension of Euler’s Formula

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MECHANICS OF SOLIDS
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. b) Design the most efficient cross-section for
the column.
E = 10.1 x 106 psi
P = 5 kips
FS = 2.5

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MECHANICS OF SOLIDS
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 ba3 2
2 I a a
rz   12
z  rz 
A ab 12 12
Le, z 0.7 L
 • Most efficient design:
rz a 12 Le, y
Le, z

• Buckling in xz Plane: rz ry
1 3
2 I y 12 ab b2 b 0 .7 L 2L
ry    ry  
A ab 12 12 a 12 b / 12
Le, y 2L a 0 .7 a
   0.35
ry b / 12 b 2 b

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MECHANICS OF SOLIDS
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
Pcr 12500 lbs
 cr  
A 0.35b b

 cr 
 2E


 2 10.1  106 psi 
Le r 2
138.6 b 2
L = 20 in. 
12500 lbs  2 10.1  106 psi


0.35b b 138.6 b 2
E = 10.1 x 106 psi
P = 5 kips b  1.620 in.
a  0.35b  0.567 in.
FS = 2.5
a/b = 0.35

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Design of Columns Under Centric Load
• Previous analyses assumed
stresses below the proportional
limit and initially straight,
homogeneous columns

• Experimental data demonstrate

- for large Le/r, cr follows
Euler’s formula and depends
upon E but not Y.

- for small Le/r, cr is

determined by the yield
strength Y and not E.

- for intermediate Le/r, cr

depends on both Y and E.

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Design of Columns Under Centric Load
• For Le/r > Cc
Structural Steel
 2E  cr
 cr   all 
American Inst. of Steel Construction Le / r 2 FS
FS  1.92

• For Le/r > Cc

 Le / r 2   cr
 cr   Y 1  2 
 all 
 2Cc  FS
3
5 3 L / r 1 L / r 
FS   e   e 
3 8 Cc 8  Cc 

• At Le/r = Cc
2 2 E
 cr  1
2 Y
Cc2 
Y

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MECHANICS OF SOLIDS
Design of Columns Under Centric Load
• Alloy 6061-T6
Aluminum Le/r < 66:
Aluminum Association, Inc.  all  20.2  0.126Le / r  ksi
 139  0.868Le / r  MPa

Le/r > 66:

51000 ksi 351  103 MPa
 all  
Le / r 2
Le / r 2

• Alloy 2014-T6
Le/r < 55:
 all  30.7  0.23Le / r  ksi
 212  1.585Le / r  MPa
Le/r > 66:
54000 ksi 372  103 MPa
 all  
Le / r  2
Le / r 2

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MECHANICS OF SOLIDS
Sample Problem 10.4

SOLUTION:
• With the diameter unknown, the
slenderness ration can not be evaluated.
Must make an assumption on which
slenderness ratio regime to utilize.

• Calculate required diameter for

assumed slenderness ratio regime.

• Evaluate slenderness ratio and verify

initial assumption. Repeat if necessary.
Using the aluminum alloy2014-T6,
determine the smallest diameter rod
which can be used to support the centric
load P = 60 kN if a) L = 750 mm,
b) L = 300 mm

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MECHANICS OF SOLIDS
Sample Problem 10.4
• For L = 750 mm, assume L/r > 55

• Determine cylinder radius:

P 372  103 MPa
 all  
A L r 2
60  103 N 372  103 MPa
 c  18.44 mm
c 2
 0.750 m 
2
 
 c/2 

• Check slenderness ratio assumption:

c  cylinder radius
L L 750mm
r  radius of gyration    81.3  55
r c / 2 18.44 mm 
I c 4 4 c assumption was correct
  
A c 2 2
d  2c  36.9 mm

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MECHANICS OF SOLIDS
Sample Problem 10.4
• For L = 300 mm, assume L/r < 55

• Determine cylinder radius:

P   L 
 all   212  1.585  MPa
A   r 
60  103 N   0.3 m  6
 212  1.585   10 Pa
c 2   c / 2 
c  12.00 mm

• Check slenderness ratio assumption:

L L 300 mm
   50  55
r c / 2 12.00 mm 

assumption was correct

d  2c  24.0 mm

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Design of Columns Under an Eccentric Load
• An eccentric load P can be replaced by a
centric load P and a couple M = Pe.

• Normal stresses can be found from

superposing the stresses due to the centric
load and couple,
   centric   bending
P Mc
 max  
A I

• Allowable stress method:

P Mc
   all
A I

• Interaction method:
P A Mc I
 1
 all centric  all bending

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