Explanation Given information: The effective length of the column is L e = 2.4 m . The allowable yield strength of the steel is σ Y = 250 MPa . The modulus of elasticity of the steel is E = 200 GPa . The centric load acting in the column is P = 180 kN . Calculation: Consider the thickness of the angle section as 9.5 mm. Refer to Appendix C “Properties of Rolled-Steel Shapes” in the textbook. For L 89 × 64 × 9.5 mm angle section; The cross sectional area of the angle ( A ) is A = 1 , 360 mm 2 . The minimum moment of inertia is I = I y = 0.454 × 10 6 mm 4 . The minimum radius of gyration is r = r y = 18.2 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 effective length ( L e ) and the minimum radius of gyration ( r ) as follows; L e r = 2.4 m × 1,000 mm 1 m 18.2 = 131.87 < L r = 133.22 Find the effective stress ( σ e ) using the equation. σ e = π 2 E ( L e / r y ) 2 Substitute 200 GPa for E and 131.87 for L e / r y . σ e = π 2 × 200 GPa × 1,000 MPa 1 GPa ( 131.87 ) 2 = 113.514 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 113.514 MPa for σ e . σ cr = [ 0.658 250 113.514 ] 250 = 99.45 MPa Calculate the allowable stress ( σ a l l ) using the relation. σ a l l = σ cr 1.67 Substitute 99.45 MPa for σ cr . σ a l l = 99.45 1.67 = 59.55 MPa Calculate the allowable load ( P all ) using the equation. P all = σ all A Substitute 59.55 MPa for σ all and 2 ( 1 , 360 ) mm 2 for A . P all = 59.55 × 2 ( 1 , 360 ) = 162 × 10 3 N × 1 kN 1,000 N = 162 kN The centric load is greater than the allowable load. Hence, the design is unsafe

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Author: BEER

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 10.3, Problem 83P

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Chapter 10.3, Problem 82P

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Chapter 10 Solutions

EBK MECHANICS OF MATERIALS

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