Explanation Given information: The effective length of the steel column is L e = 5.8 m . The magnitude of the axial load is P = 296 kN . The eccentricity of the load in steel column is e = 125 mm . The allowable yield stress of the steel column is σ Y = 250 MPa . The modulus of elasticity of the steel column is E = 200 GPa . The allowable stress in bending is ( σ all ) bending = 150 MPa . Calculation: Consider the W shaped section of W 200 × 52 ; Refer to Appendix C “Properties of Rolled-Steel Shapes” in the textbook. For W 200 × 52 shape; The cross sectional area is A = 6 , 650 mm 2 . The section modulus in x -axis is S x = 511 × 10 3 mm 3 . The minimum radius of gyration is r = r y = 51.6 mm . Find the slenderness ratio L r using the equation. L r = 4.71 E σ Y Here, the modulus of elasticity of the material is E and the allowable yield strength is σ Y Substitute 200 GPa for E and 250 MPa for σ Y : L r = 4.71 200 GPa × 1,000 MPa 1 GPa 250 = 133.22 Find the ratio of the effective length to the minimum radius of gyration. L e r = 5.8 m × 1,000 mm 1 m 51.6 = 112.403 < L r = 133.22 Find the effective stress ( σ e ) using the equation. σ e = π 2 E ( L e / r ) 2 Substitute 200 GPa for E and 112.403 for L e / r . σ e = π 2 × 200 GPa × 1 ,000 MPa 1 GPa ( 112.403 ) 2 = 156.233 MPa Find the critical stress ( σ cr ) using the relation. σ cr = [ 0.658 σ Y σ e ] σ Y Substitute 250 MPa for σ Y and 156.233 MPa for σ e . σ cr = [ 0.658 250 156.233 ] 250 = 127.959 MPa Find the allowable stress ( σ all ) centric due to centric load using the relation. ( σ all ) centric = σ cr 1.67 Substitute 127.959 MPa for σ cr . ( σ all ) c e n t r i c = 127.959 1.67 = 76.622 MPa Find the maximum moment ( M ) using the relation. M = P e Here, the allowable load is P and the eccentricity of the load is e . Substitute 296 kN for P and 125 mm for e . M = 296 kN × 1,000 N 1 kN × 125 = 37 , 000 , 000 N-mm Check the allowable stress using the centric and bending equation. P / A ( σ all ) centric + M / S x ( σ all ) bending ≤ 1 Substitute 296 kN for P , 6 , 650 mm 2 for A , 76.622 MPa for ( σ all ) centric , 37,000,000 N-mm for M , 511 × 10 3 mm 3 for S x , and 150 MPa for ( σ all ) bending . ( 296 kN × 1,000 N 1 kN 6 , 650 ) 76

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Chapter 10.4, Problem 115P

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Chapter 10.4, Problem 114P

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Chapter 10 Solutions

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Choose the wide-flange shape of 200 mm nominal depth.

Solution Summary: The author analyzes the W shaped section of W200times 52 and the allowable yield stress of the steel column.

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Choose the wide-flange shape of 200 mm nominal depth.

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Chapter 10 Solutions