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Column Theory - Column Strength Curve
Example I
A W12x50 is used as a column to support a compressive load of 145 kips.
The length is 20 ft and the ends are pinned. Without regard to LRFD or
ASD investigate the stability of the column
For a W12x50
Minimum r  ry  1.96 in.
Maximum
Pcr 
L L 20(12)
 
 122.4
r ry
1.96
 2 EA  2 (29,000)(14.6)
L r 
2

122.4
2
Pcr  278.9 kips 145kips
 278.9 kips
OK – Column is Stable
Example I
A W12x50 is used as a column to support a compressive load of 145 kips.
The length is 20 ft the bottom is fixed and the top is free. Without regard
to LRFD or ASD investigate the stability of the column
For a W12x50
Minimum r  ry  1.96 in.
L L 20(12)
 
 122.4
r ry
1.96
Maximum
Pcr 
 2 EA
KL r 
2

 2 (29,000)(14.6)
(2.1)(122.4)
Pcr  63.248 kips 145kips
2
 63.248 kips
NG – Column is Unstable
AISC Requirements
CHAPTER E pp 16.1-32
Nominal Compressive Strength
Pn  Fcr Ag
AISC Eqtn E3-1
AISC Requirements
LRFD
Pu  c Pn
Pu  Sum of factoredloads
c  resistancefactorfor compression  0.90
c Pn  design compressive strength
AISC Requirements
ASD
Pn
Pa 
c
Pa  Sum of serviceloads
c  safetyfactorfor compression  1.67
Pn c  allowable compressive strength
AISC Requirements
ASD – Allowable Stress
fa  Fa
f a  computedaxial compressive stress  Pa Ag
Fa  allowable axial compressive stress
Fcr
Fcr


 0.6Fcr
c 1.67
To compute Fcr – ELASTIC BUCKLING
Pe 
 2 EA
KL r 
2
Pe
 2E
Fe 

2
A
KL
r


(AISC Eq. E3- 4)
Recall Assumptions
Assumptions
• Column is perfectly straight
• The load is axial, with no eccentricity
• The column is pinned at both ends
To compute Fcr – ELASTIC BUCKLING
Pe
 2E
Fe 

2
A
KL
r


(AISC Eq. E3- 4)
Fcr  0.877Fe
Accounts for Imperfections
To compute Fcr – INELASTIC BUCKLING
Pe 
 2 Et A
KL r 
2
Fy

Fe

Fcr  0.658



 Fy


Design Strength
Alternatively
Fe 
KL r 
KL

r
KL
E
 4.71
r
Fy
 2E
Fe
E
 4.71
Fy
 2E
2
 2E
Fe
Fe  0.44Fy
Inelastic Buckling

In Summary
Fy


KL
E
e
F 

Fy if
 4.71
 0.658

r
Fy




or Fe  0.44Fy
Fcr  




 0.877Fe otherwise
KL
 200
r
Example
A W14x74 of A992 steel has a length of 20 feet and pinned ends.
Compute the design strength for LRFD and the allowable compressive
strength for ASD
Slenderness Ratio
E
29,000
4.71
 4.71
 113
Fy
50
KL KL (1)(20)(12)
Maximum


 96.77  200
r
ry
2.48
KL
E
 4.71
 Inelastic Buckling
r
Fy
Example
Fe 
 E
2
KL r 
2

 (29,000)
2
96.77
2
Fy
Fcr  0.658
Fe
 0.658
 30.56 ksi
50
30.56
 25 .21 ksi
Pn  Fcr Ag  25.21(21.8)  549.6 kips
Example
Pn  Fcr Ag  25.21(21.8)  549.6 kips
LRFD
c Pn  0.9Fcr Ag  (0.9)549.6  495kips
ASD
Fa  0.6Fcr  (0.6)(25.21)  15.31ksi
Fa Ag  15.31(21.8)  330kips