# Compression Members.

## Presentation on theme: "Compression Members."— Presentation transcript:

Compression Members

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

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

AISC Requirements LRFD

In Summary

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

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)

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 Required Design Strength Enter Column Tables with KL=(1)(26)=26 ft 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 Required Allowable Strength Enter Column Tables with KL=(1)(26)=26 ft 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 Enter Column Tables with KL=(1)(24)=24 ft No Footnote: No need to check for local buckling

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

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

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

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

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

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

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

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

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

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

More on Effective Length Factor

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

Example Joint A Joint B Joint C Sway Uninhibited Pinned End

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

More on Effective Length
Violated

Alingnment Charts & Inelastic Behavior
Stiffness Reduction Factor 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