Download presentation

Presentation is loading. Please wait.

Published byJeffrey Foulks Modified over 2 years ago

1
Compression Members

2
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

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 F e Elastic Buckling Stress corresponding to the controlling mode of failure (flexural, torsional or flexural torsional) Fe:Fe: Theory of Elastic Stability (Timoshenko & Gere 1961) Flexural BucklingTorsional 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
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 1.Assume a value for F cr 2.Determine required area LRFD ASD

13
IF COLUMNS NOT APPLICABLE 3Select a shape that satisfies area requirement 4Compute F cr for the trial shape 5Revise if necessary If available strength too close to required value try next tabulated value Else repeat 1-4 using F cr of trial shape 6Check 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 Slenderness OK Available Area Eulers 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 NG Design Strength

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 Required Area Assume NEW Critical Stress

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 Slenderness OK Available Area Eulers 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 Slenderness OK Available Area Eulers 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 OK Design Strength

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 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

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

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

26
More on Effective Length Violated

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

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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google