P where The aluminum W 310 x 202 (wide flange) column is subjected to an eccentric axial load P as shown. The maximum compressive stress in the column is given by the so-called secant formula: L o max 0 [1+% sec (√)) ở = P/A = average stress A = 25 800 mm² = cross-sectional area of the column e = 85 mm = eccentricity of the load c = 170 mm = half depth of the column r = 142 mm = radius of gyration of the cross section L = 7100 mm = length of the column E = 71 x 10⁹ Pa = modulus of elasticity Determine the maximum load P that the column can carry if the maximum stress is not to exceed 120 × 10^6 Pa. Terminate the computation when Σa < 0.0005%.
P where The aluminum W 310 x 202 (wide flange) column is subjected to an eccentric axial load P as shown. The maximum compressive stress in the column is given by the so-called secant formula: L o max 0 [1+% sec (√)) ở = P/A = average stress A = 25 800 mm² = cross-sectional area of the column e = 85 mm = eccentricity of the load c = 170 mm = half depth of the column r = 142 mm = radius of gyration of the cross section L = 7100 mm = length of the column E = 71 x 10⁹ Pa = modulus of elasticity Determine the maximum load P that the column can carry if the maximum stress is not to exceed 120 × 10^6 Pa. Terminate the computation when Σa < 0.0005%.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Methods in solving Non-linear equations
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