Explanation Given information: A wide-flange W10 × 45 of A992 steel acts as a column. The length of the column is L = 24 ft . The maximum eccentric load applied on the column is denoted by P . The allowable bending stress, ( σ b ) allow = 15 ksi . 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 W10 × 45 section as follows: The area ( A ) of the W10 × 45 section is 13.3 in . 2 The radius of gyration ( r y ) of the W10 × 45 section is 2.01 in . The radius of gyration ( r x ) of the W10 × 45 section is 4.32 in . The moment of inertia, ( I y ) = 53.4 in . 4 . The thickness of the web of the section, b f = 8.02 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 4.32 in . for r x . ( K L r ) x = ( 2 × 24 ft × 12 in . 1 ft 4.32 in . ) = 576 4.32 = 133.33 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 2.01 in . for r y . ( K L r ) y = ( 0.7 × 24 ft × 12 in . 1 ft 2.01 in . ) = 201.6 2.01 = 100.3 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 = 50 ksi . Substitute 29 × 10 3 ksi for E and 50 ksi for σ Y in Equation (1). ( K L r ) c = 2 π 2 × 29 × 10 3 50 = 58 , 000 π 2 50 = 107 Show the condition for the column to be a long column as follows: ( K L r ) c ≤ ( K L r ) x ≤ 200 (2) Substitute 133

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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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 13.114P

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To find: the value of maximum eccentric force P using interaction formula.

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Chapter 13.7, Problem 13.113P

Chapter 13.7, Problem 13.115P

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