Explanation Given information: A wide-flange W12 × 50 of A36 steel acts as a column. The length of the column is L = 10 ft . The column is fixed at the bottom and pinned at the top. The eccentric load applied on the column, P . 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 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 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 depth of the section, d = 12.19 in . The thickness of the web of the section, b f = 8.08 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 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 Calculate the value of ( K L r y ) as follows: The value of effective length factor, K = 2 , since one end fixed and another end free. Substitute 2 for K, 10 ft for L, and 1.96 in . for r y . ( K L r y ) = ( 2 × 10 ft × 12 in . 1 ft 1.96 in . ) = 122.44 Show the condition for the column to be a long column as follows: ( K L r ) c ≤ ( K L r y ) ≤ 200 (2) Substitute 42.857 for ( K L r y ) and 126 for ( K L r ) c in Equation (2)

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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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 13.110P

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

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Chapter 13.7, Problem 13.109P

Chapter 13.7, Problem 13.111P

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A 20-ft-long column is made of aluminum alloy 2014-T6. If it is pinned at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and Eq. 13–30.

Determine the maximum allowable eccentric load P that can be applied to the wood column. The column is fixed at its base and free at its top. Use the NFPA equations of Sec. 13.6 and Eq. 13–30.

Chapter 13 Solutions

Mechanics of Materials (10th Edition)

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

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