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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 0.933 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 0.8778.5 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 0.920 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 0.87721.27 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 0.87721.42 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 0.65850 Fe 50 Fe 35.61 ksi Fcr (elastic) 0.877 Fe 0.87735.61 31.23 a Fcr (inelastic) Fcr ( elastic) 27.28 0.890 31.23 Fy