Compression Members
COLUMN STABILITY
A. Flexural Buckling
• Elastic Buckling
• Inelastic Buckling
• Yielding
B. Local Buckling – Section E7 pp 16.1-39
and B4 pp 16.1-14
C. Lateral Torsional Buckling
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 factored loads
c  resistance factor for compressio n  0.90
c Pn  design compressiv e strength
In Summary
Fy


KL
E
e
F 

Fy if
 4.71
 0.658


r
Fy




or Fe  0.44 Fy
Fcr  




 0.877 Fe otherwise
KL
 200
r
In Summary - Definition of Fe
Fe:
Elastic Buckling Stress corresponding to the controlling mode of
failure (flexural, torsional or flexural torsional)
Theory of Elastic Stability (Timoshenko & Gere 1961)
Flexural Buckling
 2E
Fe 
KL / r 2
Torsional Buckling Flexural Torsional
2-axis of symmetry Buckling
1 axis of symmetry
AISC Eqtn
E4-4
AISC Eqtn
E4-5
Flexural Torsional
Buckling
No axis of symmetry
AISC Eqtn
E4-6
Column Design Tables
Assumption : Strength Governed by Flexural Buckling
Check Local Buckling
Column Design Tables
Design strength of selected shapes for effective length KL
Table 4-1 to 4-2, (pp 4-10 to 4-316)
Critical Stress for Slenderness KL/r
table 4.22 pp (4-318 to 4-322)
Design of Members in Compression
• Selection of an economical shape: Find lightest shape
• Usually category is defined beforehand, e.g. W, WT etc
• Usually overall nominal dimensions defined in advance
because of architectural and other requirements.
USE OF COLUMN LOAD TABLES
IF NOT APPLICABLE - TRIAL AND ERROR
EXAMPLE I – COLUMN LOAD TABLES
A compression member is subjected to service loads pf 165 dead and 535
kips live. The member is 26 feet long and pinned at each end
LRFD
Calculate factored load
Pu  1.2 D  1.6 L  1.2(165)  1.6(535)  1,054 kips
Required Design Strength
c Pn  1,054 kips
Enter Column Tables with KL=(1)(26)=26 ft
W14 X 145 design strength :1,230 kips  Pu
OK
EXAMPLE I – COLUMN LOAD TABLES
A compression member is subjected to service loads pf 165 dead and 535
kips live. The member is 26 feet long and pinned at each end
ASD
Calculate factored load
Pa  D  L  (165)  (535)  700 kips
Required Allowable Strength
Pn
 700 kips
c
Enter Column Tables with KL=(1)(26)=26 ft
W14 X 132 design strength : 702 kips  Pa
OK
EXAMPLE Ii – COLUMN LOAD TABLES
Select the lightest W-shape that can resist a service dead load of 62.5
kips and a service live load of 125 kips. The effective length is 24 feet.
Use ASTM A992 steel
LRFD
Calculate factored load and required strength
Pu  1.2 D  1.6L  1.2(62.5)  1.6(125)  275 kips  c Pn
Enter Column Tables with KL=(1)(24)=24 ft
W 8 :No W8 with φc Pn  275 kips
W10 :W10 X 54, c Pn  282 kips
W12 :W10 X 58, c Pn  293 kips
W 14 :W 14 X 61, c Pn  293 kips
No Footnote: No need to
check for local buckling
IF COLUMNS NOT APPLICABLE
1. Assume a value for Fcr
Fcr  Fy
2. Determine required area
LRFD
Pu
c Fcr Ag  Pu  Ag 
c Fcr
ASD
Pa
Pa
0.6 Fcr 
 Ag 
Ag
0.6Fcr
IF COLUMNS NOT APPLICABLE
3
Select a shape that satisfies area requirement
4
Compute Fcr for the trial shape
5
Revise if necessary
•
•
6
If available strength too close to required value try next tabulated value
Else repeat 1-4 using Fcr of trial shape
Check local stability and revise if necessary
Example
Select a W18 shape of A992 steel that can resist a service dead load of 100
kips amd a service live load of 300 kips. Effective length KL=26 ft
Calculate factored load and required strength
Pu  1.2D  1.6L  1.2(100)  1.6(300)  600 kips
Try
Fcr 
Required Area
2
Fy  33 ksi
3
Pu
600
Ag 

 20.2 in 2
c Fcr 0.933
Example
Select a W18 shape of A992 steel that can resist a service dead load of 100
kips amd a service live load of 300 kips. Effective length KL=26 ft
Try W 18x71
Available Area
Slenderness
Euler’s Stress
Elastic Buckling
Slenderness Limit
Ag  20.8 in 2  20.2
OK
KL 26 12

 183.5  200
rmin
1.70
Fe 
 2E
KL / r 
2

 2 (29,000)
183.5
2
OK
 8.5 ksi
E
29,000
4.71
 4.71
 113
Fy
50
ELASTIC BUCKLING
KL

 183.5
rmin
Example
Select a W18 shape of A992 steel that can resist a service dead load of 100
kips amd a service live load of 300 kips. Effective length KL=26 ft
Critical Stress
Fcr  0.877 Fe  0.8778.5  7.455 ksi
Design Strength
c Pn  c Fcr Ag  0.9(7.455)( 20.8)  140 kips
 600 kips
NG
Example
Select a W18 shape of A992 steel that can resist a service dead load of 100
kips amd a service live load of 300 kips. Effective length KL=26 ft
Assume NEW
Critical Stress
Required Area
 33  7.455 
Fcr  20 ksi  

2


Pu
600
Ag 

 33.3 in 2
c Fcr 0.920
Example
Select a W18 shape of A992 steel that can resist a service dead load of 100
kips amd a service live load of 300 kips. Effective length KL=26 ft
Try W 18x119
Available Area
Slenderness
Euler’s Stress
Elastic Buckling
Slenderness Limit
Ag  35.1 in 2  33.3
OK
KL 26 12

 116.0  200
rmin
2.69
Fe 
 2E
KL / r 
2

 2 (29,000)
116.0
2
OK
 21.27 ksi
E
29,000
4.71
 4.71
 113
Fy
50
ELASTIC BUCKLING
KL

 116
rmin
Example
Select a W18 shape of A992 steel that can resist a service dead load of 100
kips amd a service live load of 300 kips. Effective length KL=26 ft
Critical Stress
Fcr  0.877 Fe  0.87721.27  18.65 ksi
Design Strength
c Pn  c Fcr Ag  0.9(18.65)(35.1)  589 kips
This is very close, try next larger size
 600 kips
NG
Example
Select a W18 shape of A992 steel that can resist a service dead load of 100
kips amd a service live load of 300 kips. Effective length KL=26 ft
Try W 18x130
Ag  38.2 in 2
Available Area
Slenderness
Euler’s Stress
Elastic Buckling
Slenderness Limit
KL 26 12

 115.6  200
rmin
2.70
Fe 
 2E
KL / r 
2

 2 (29,000)
115.6
2
OK
 21.42 ksi
E
29,000
4.71
 4.71
 113
Fy
50
ELASTIC BUCKLING
KL

 115.6
rmin
Example
Select a W18 shape of A992 steel that can resist a service dead load of 100
kips amd a service live load of 300 kips. Effective length KL=26 ft
Critical Stress
Fcr  0.877 Fe  0.87721.42  18.79 ksi
Design Strength
c Pn  c Fcr Ag  0.9(18.79)(38.2)  646 kips
 600 kips
OK
More Length
on Effective
Length in
Factor
Effective
of Columns
Frames
I c Lc
Ig Lg
A
Ig Lg
Assumptions
• All columns under
consideration reach buckling
Simultaneously
Ic Lc
B
• All joints are rigid
• Consider members lying in the
plane of buckling
Define:
I

G
I
c
A
g
Lc
Lg
I

G 
I
• All members have constant A
c
Lc
g
Lg
B
• Elastic Behavior
Effective Length Factor-Alingnment Charts
Use alignment charts (Structural Stability Research Council SSRC)
AISC Commentary Figure C-C2.3 nad C-C2.4 p 16-.1-241
Connections to foundations
(a) Hinge
G is infinite - Use G=10
(b) Fixed
G=0 - Use G=1.0
Assumption of Elastic Behavior is violated when
Inelastic Flexural Buckling
E
KL
4.71

Fy rmin
Example
W12x96
W24x55
12’
Lc
g
Lg
Joint A
GA 
B W24x68
W12x120
15’
833 / 12  1070 / 12
 0.94
1350 / 20  1830 / 18
Joint B
1070 / 12  1070 / 15
GB 
 0.95
1350 / 20  1830 / 18
C
20’
c
W24x68
A
W12x120
W24x55
12’
I

G
I
18’
Joint C
Sway Uninhibited
Pinned End
GC  10.0
Example
AISC Commentary Figure C-C2.3 nad C-C2.4 p 16-.1-241
COLUMN AB
COLUMN BC
GA  0.94
Gc  10.0
GB  0.95
GB  0.95
K x  1.3
K x  1.85
olumns in Frames
More on Effective Length
Assumptions
• All columns under
consideration reach buckling
Simultaneously
•
All joints are rigid
•
Consider members lying in the
plane of buckling
•
All members have constant A
•
Elastic Behavior
Violated
Alingnment Charts & Inelastic Behavior
Elastic
Fcr 
Ginelastic

 2E
KL

r
2

 2 Et
KL
t c
g
Inelastic
Fcr 
EI


 EI

Lc
Et
 Gelastic
Lg E
Ginelastic   a  Gelastic
Stiffness
Reduction
Factor
a 
Fcr ( inelastic)
Fcr ( elastic)
Et

E
2
r
SRF: Table 4-21 AISC Manual pp 4-317
Example
Compute Stiffness Reduction Factor per LRFD for an axial compressive stress of
25 ksi and Fy=50 ksi
Pu
 25 ksi
Ag
P
25
F
Fcr (inelastic)  u 
 27.78 ksi  0.658 y
c Ag 0.9
Fe
27.78 ksi  0.65850 Fe 50  Fe  35.61 ksi
Fcr (elastic)  0.877 Fe  0.87735.61  31.23
a 
Fcr (inelastic)
Fcr ( elastic)
27.28

 0.890
31.23

Fy