Question

1) A 20' tall fixed-fixed steel column is fixed the top and the bottom in both directions. The column carries an axial force of 200 kips. The modulus of elasticity is 29,000 ksi and the yield strength is 42 ksi. Design the cross-section by calculating the required moment of inertia of the column (in in4) to prevent yielding and buckling using a factor of safety of 2. [Ans. To check: Imin - 20.124 in4 based on buckling design]

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Answer 1

Given: - Both ends fixed = > : - K = 0.5 P = 200 Kips E = 29,000 Kips F. D.S = 2 l = 20: - 20 times 12 = 240 is rightarrow Using backing formula P_cr = pi^2 E I min/F.D.S times Le^2 = > 200 = pi^2 times 29, 000 = I min/ 2 times (0.5 times 240)^2 = > I min = 20. 124 in^4 Now using the table F - 1 (a), properties of side flange sections - U S C S units the most suitable section is W 12 times 35 which has I of 24.5 in^4 Properties of this c/s (W 12 times 35) rightarrow wt 1 foot rightarrow 35.0 rightarrow Hinge width rightarrow 6.56 rightarrow Area (in^2) rightarrow 10.3 rightarrow Hange thickness rightarrow 0.520 rightarrow Depth (in) rightarrow 12.5 rightarrow Web thick rightarrow 0.300 (in)