| Practice Problem 10-3 |
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Given: A W 12 x 22 wide-flange shape made of ASTM A36 (E = 36 Msi) structural steel is 30 ft long and is to be used as a vertical column with the bottom end fixed and the top end pinned. The critical radius of gyration for this column is r = 0.848 in. Req'd: Determine the critical buckling stress, scr [ksi]. |
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Solution: From the Euler formula for buckling, the critical buckling stress for a fixed-pinned element is: scr = (p2E·r2) / (Le)2 Where Le for a fixed-pinned element is Le = 0.7·L = 0.7(30 ft)(12 in/ft) = 252 in. The critical buckling stress is then: Pcr = p2(36·106 psi)(0.848 in)2 / (252 in)2 = 4023 psi = 4.02 ksi |
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