Explanation Given information: A wide-flange W8 × 15 of A992 steel acts as a column. The length of the column is L = 10 ft . The both ends of the column are pinned. The eccentric load applied on the column, P . Show the AISC equation of Sec 13.6 as follows: σ allow = 12 π 2 E 23 ( K L r ) 2 Here, σ allow is the allowable stress, E is the Young’s modulus, and ( K L r ) is the slenderness ratio. Show the expression for Equation 13-30 as follows: σ max = P A + M c I Here, σ max is the maximum compressive stress, P is the load, A is the area, M is the moment, c is the distance between the centroid of member and outer fibre, and I is the moment of inertia. Calculation: Refer to the Appendix B. Show the geometric properties of W8 × 15 section as follows: The area ( A ) of the W8 × 15 section is 4.44 in . 2 The radius of gyration ( r x ) of the W8 × 15 section is 3.29 in . The radius of gyration ( r y ) of the W8 × 15 section is 0.876 in . The moment of inertia, ( I x ) = 48 in . 4 . The depth of the section, d = 8.11 in . 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 A992 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 Calculate the value of ( K L r y ) as follows: The value of effective length factor, K = 1 , since the ends are pinned. Substitute 1 for K, 10 ft for L, and 0.876 in . for r y . ( K L r y ) = ( 1 × 10 ft × 12 in . 1 ft 0.876 in . ) = 136

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Buckling of columns

The column is described as a vertical member of the structure that carries an axial compressive load, and buckling is a lateral deformation of the column under the application of load.

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Chapter 13.7, Problem 13.107P

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To find: the largest eccentric load P that can be applied to the column.

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Chapter 13.6, Problem 13.106P

Chapter 13.7, Problem 13.108P

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Chapter 13 Solutions

Mechanics of Materials

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