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Column Stability and Design

The document discusses the stability and design of columns. It introduces Euler's formula for calculating the critical buckling load of columns with pin-ended beams. The formula shows that the critical stress is dependent on the column's length, moment of inertia, and slenderness ratio. The analysis is then extended to columns with one fixed end, where the equivalent length replaces the actual length in Euler's formula to calculate the critical load and stress. An example problem applies these concepts to an aluminum column with a fixed end at one end and centric loading at the other.

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Zabid Ullah
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231 views23 pages

Column Stability and Design

The document discusses the stability and design of columns. It introduces Euler's formula for calculating the critical buckling load of columns with pin-ended beams. The formula shows that the critical stress is dependent on the column's length, moment of inertia, and slenderness ratio. The analysis is then extended to columns with one fixed end, where the equivalent length replaces the actual length in Euler's formula to calculate the critical load and stress. An example problem applies these concepts to an aluminum column with a fixed end at one end and centric loading at the other.

Uploaded by

Zabid Ullah
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PPT, PDF, TXT or read online on Scribd
You are on page 1/ 23

CHAPTER MECHANICS OF

SOLIDS
10 Columns

GIK Institute of Engineering Sciences and Technology


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

GIK Institute of Engineering Sciences and Technology 10 - 2


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.

GIK Institute of Engineering Sciences and Technology 10 - 3


MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology 10 - 4


MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology 10 - 5


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

GIK Institute of Engineering Sciences and Technology 10 - 6


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.

GIK Institute of Engineering Sciences and Technology 10 - 7


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

GIK Institute of Engineering Sciences and Technology 10 - 8


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.

GIK Institute of Engineering Sciences and Technology 10 - 9


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

GIK Institute of Engineering Sciences and Technology 10 - 10


MECHANICS OF SOLIDS
Extension of Euler’s Formula

GIK Institute of Engineering Sciences and Technology 10 - 11


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

GIK Institute of Engineering Sciences and Technology 10 - 12


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

GIK Institute of Engineering Sciences and Technology 10 - 13


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

GIK Institute of Engineering Sciences and Technology 10 - 14


MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology 10 - 15


MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology 10 - 16


MECHANICS OF SOLIDS
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.

GIK Institute of Engineering Sciences and Technology 10 - 17


MECHANICS OF SOLIDS
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

GIK Institute of Engineering Sciences and Technology 10 - 18


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

GIK Institute of Engineering Sciences and Technology 10 - 19


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

GIK Institute of Engineering Sciences and Technology 10 - 20


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

GIK Institute of Engineering Sciences and Technology 10 - 21


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

GIK Institute of Engineering Sciences and Technology 10 - 22


MECHANICS OF SOLIDS
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

GIK Institute of Engineering Sciences and Technology 10 - 23

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