Explanation Given information: A wide-flange W12 × 50 of A36 steel acts as a column. The length of the column is L = 24 ft . The eccentric load ( P ) applied on the column is 15 kip . The allowable bending stress, ( σ b ) allow = 15 ksi . Show the AISC equation of Sec 13.6 as follows: σ allow = [ 1 − 1 2 ( K L r ( K L r ) c ) 2 ] σ Y [ 5 3 + 3 8 ( K L r ( K L r ) c ) − 1 8 ( K L r ( K L r ) c ) 3 ] Here, σ allow is the allowable stress, σ Y is the yield stress, E is the Young’s modulus, and ( K L r ) is the slenderness ratio. Show the interaction formula as follows: ( P A ) ( σ a ) allow + ( M c A r 2 ) ( σ b ) allow ≤ 1 Here, ( σ a ) allow is the allowable axial stress, P is the force, ( σ b ) allow is the allowable bending stress, A is the cross-sectional area, c is the distance between the centroid of member and outer fiber, M is the moment, and r is the radius of gyration. Calculation: Refer the Appendix B. Show the geometric properties of W12 × 50 section as follows: The area ( A ) of the W12 × 50 section is 14.7 in . 2 The least radius of gyration ( r y ) of the W12 × 50 section is 1.96 in . The moment of inertia, ( I y ) = 56.3 in . 4 . The thickness of the web of the section, b f = 8.08 in . Show the expression for slenderness ratio as follows: Slenderness ratio = ( K L r ) Calculate the value of ( K L r ) x as follows: The value of K x is 2 as the bottom of the column is fixed and top is free. Substitute 2 for K x , 24 ft for L , and 5.18 in . for r x . ( K L r ) x = ( 2 × 24 ft × 12 in . 1 ft 5.18 in . ) = 576 5.18 = 111.2 Calculate the value of ( K L r ) y as follows: The value of K y is 0.7 as the bottom of the column is fixed and top is pinned. Substitute 0.7 for K y , 24 ft for L , and 1.96 in . for r y . ( K L r ) y = ( 0.7 × 24 ft × 12 in . 1 ft 1.96 in . ) = 201.6 1.96 = 102.86 Show the expression for ( K L r ) c as follows: ( K L r ) c = 2 π 2 E σ Y (1) Here, K is the effective length factor, L is the length of the column, and ( r ) is the radius of gyration, E is the Young’s modulus, and σ Y is the yield stress of the material. Consider the Young’s modulus of the steel is E = 29 × 10 3 ksi . Consider the yield stress of the steel is σ Y = 36 ksi . Substitute 29 × 10 3 ksi for E and 36 ksi for σ Y in Equation (1). ( K L r ) c = 2 π 2 × 29 × 10 3 36 = 58 , 000 π 2 36 = 126 Show the condition for the column to be a long column as follows: ( K L r ) c ≤ ( K L r ) x ≤ 200 (2) Substitute 111

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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 13.7, Problem 13.116P

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To find: whether the column is adequate to support the loading.

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Chapter 13.7, Problem 13.115P

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Mechanics of Materials

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