1. Compression Members

2. COLUMN STABILITY
A. Flexural Buckling
Elastic Buckling
Inelastic Buckling
Yielding
B. Local Buckling – Section E7 pp
and B4 pp
C. Lateral Torsional Buckling

3. AISC Requirements CHAPTER E pp 16.1-32 Nominal Compressive Strength
AISC Eqtn E3-1

4. AISC Requirements
LRFD

5. In Summary

6. In Summary - Definition of 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
Torsional Buckling
2-axis of symmetry
Flexural Torsional Buckling
1 axis of symmetry
Flexural Torsional Buckling
No axis of symmetry
AISC Eqtn
E4-4
AISC Eqtn
E4-5
AISC Eqtn
E4-6

7. Assumption : Strength Governed by Flexural Buckling
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)

8. 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

9. 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
Required Design Strength
Enter Column Tables with KL=(1)(26)=26 ft OK

10. 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
Required Allowable Strength
Enter Column Tables with KL=(1)(26)=26 ft OK

11. 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
Enter Column Tables with KL=(1)(24)=24 ft
No Footnote: No need to check for local buckling

12. IF COLUMNS NOT APPLICABLE
Assume a value for Fcr
Determine required area
LRFD
ASD

13. IF COLUMNS NOT APPLICABLE
3 Select a shape that satisfies area requirement
4 Compute Fcr for the trial shape
Revise if necessary
If available strength too close to required value try next tabulated value
Else repeat 1-4 using Fcr of trial shape
6 Check local stability and revise if necessary

14. 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
Try
Required Area

15. 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 OK
Slenderness OK
Euler’s Stress
Elastic Buckling
Slenderness Limit
ELASTIC BUCKLING

16. 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
Design Strength NG

17. 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

18. 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 OK
Slenderness OK
Euler’s Stress
Elastic Buckling
Slenderness Limit
ELASTIC BUCKLING

19. 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
Design Strength NG
This is very close, try next larger size

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 18x130
Available Area
Slenderness OK
Euler’s Stress
Elastic Buckling
Slenderness Limit
ELASTIC BUCKLING

21. 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
Design Strength OK

22. More on Effective Length Factor

23. Effective Length Factor-Alingnment Charts
Use alignment charts (Structural Stability Research Council SSRC)
AISC Commentary Figure C-C2.3 nad C-C2.4 p
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

24. Example
Joint A
Joint B
Joint C
Sway Uninhibited
Pinned End

25. Example AISC Commentary Figure C-C2.3 nad C-C2.4 p 16-.1-241 COLUMN AB
COLUMN BC

26. More on Effective Length
Violated

27. Alingnment Charts & Inelastic Behavior
Stiffness Reduction Factor
SRF: Table 4-21 AISC Manual pp 4-317

28. Example
Compute Stiffness Reduction Factor per LRFD for an axial compressive stress of 25 ksi and Fy=50 ksi