From AISC Eq E32 or Table 4 22 pg 4 323 obtain \u03c6 c F cr 298 ksi based on KLr 75

From aisc eq e32 or table 4 22 pg 4 323 obtain φ c f

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From AISC Eq. E3.2 or Table 4-22 (pg. 4-323), obtain φ c F cr = 29.8 ksi, based on KL/r = 75. 3.Compute required ?𝑔=𝑃?𝜑𝑐?𝑐?=27429.8= 9.2 sq. in. 4. Select section based on buckling in weak direction, ( KL/r ) y . From Column Load Tables in Part 4 of AISC manual, select a W 10×33, the lightest section that has the required area. 5. Check the W 10 × 33 section: A =9.71 sq in., r x / r y =2.16, r y =1.94, KL/r y = 99, φ c F cr = 22 ksi (Note: ( KL/r x ) = ( KL/r y )( r x / r y )= 99/2.16 =45.8 (not govern).) φ c P n = φ c F cr A g = 22(9.71) = 213.62 kips (< P u = 274) NG Note that Table 4-22 provides φ c F cr for given values of KL/r . Other possible sections are: Section Area (sq.in.) KL/r y φ c F cr (ksi) φ c P n (kips) W 8 × 40 11.7 94.1 23.60 276 (lightest section) W 10 × 45 13.3 95.5 22.75 307 W 12 × 45 13.1 98.5 22.15 290
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88 P P P P 15 ft 15 ft 15 ft 15 ft Potential buckled shape Example : Select the lightest Grade 50 W-section for a pin-end column of a braced structure main column length 30 feet 100 kips dead and 200 kips live compressive loads weak direction support at mid-height Solution : 1. Factored load: P u =1.2 D +1.6 L =1.2(100) + 1.6(200) = 440 kips 2. For hinged ends, K =1 and effective length are ( KL ) x = 30 feet and ( KL ) y = 15 feet 3. AISC Column Load Tables are computed assuming ( KL/r ) y controls. Enter these tables with the effective length ( KL ) y = 15 ft and P u = 440 kips. 4. Starting with the shallow W10 sections and working towards deeper sections Section φ c P n (kips) r x /r y A g (in 2 ) W10 × 49 449 1.71 14.4 W12 × 53 478 2.11 15.6 5. Try 2 12 53: 15.6 in ; 2.48 in; 2.11 (> 2.0) ; 5.23 in x g y x y r W A r r r Since ( KL ) x = 2( KL ) y , if r x /r y ≥ 2, weak axis (i.e. bending about y -y axis) controls and tabular loads give the correct answer. That is, 30(12) 15(12) 68.8 72.6 5.23 2.48 x y KL KL r r Thus W 12 × 53 is acceptable. ( φ c P n = 478 kips (as tabulated) Try 2 10 49: 14.4 in ; 2.56 in; 1.71 (<2.0) ; 4.34 in x g y x y r W A r r r with strong axis bending controls. That is, 30(12) 15(12) 82.95 70.32 4.34 2.56 x y KL KL r r . (a) Using Table 4-22, c 82.95, 27.2 ksi cr KL F r ; φ c P n = φ c F cr A g = 27.2(14.4) = 392 kips (b) Using Column Table, compute equivalent ( KL ) y corresponding to ( KL ) x = 30 ft by setting (??) ? ? ? = (??) ?,𝑒??𝑖?𝑎𝑙𝑒𝑛? ? ? Equivalent ( KL ) y = (KL) x /( r x / r y )= 30/1.71 = 17.54ft For ( KL ) y = 17.54 ft, W10×49's design strength is 393 kips (< P u = 440) (NG) 5. Check section W 12 × 53: ( KL/r ) y = 15(12)/2.48 = 72.5, φ c F cr = 30.65 ksi. φ c P n = φ c F cr A g = 30.65(15.6) = 478.1 kips (> P u = 440) Use section W 12 × 53. (( KL ) y,equivalent = 30/2.11 ≤ 15ft, strong axis ok.)
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89 More on Effective Length Factor K Basic Cases: L -unsupported length, KL -effective length In order to take into account of the different boundary conditions, the unsupported length L is replaced by the effective length KL in the calculation of buckling stresses and the slenderness ratio L/r is replaced by the effective slenderness ratio KL/r .
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  • Spring '14
  • GregoryG.Deierlein
  • Column, Buckling, KL, AISC, effective length

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