To determine The maximum load P it can carry without buckling. Explanation Given information: Factor of safety with respect to buckling is F . S . = 1.75 . The length of the column is 24 in. Calculation: The cross section is a rectangular of 1.5 in . × 0.5 in . . Modulus of elasticity of steel is 29,000 ksi. The yield stress of the steel is 102 ksi. Find the cross sectional area as shown below. A = b × d Here, b is the width of the cross section and d is the depth of the cross section Substitute 0.5 in. for b and 1.5 in. for d . A = 0.5 × 1.5 = 0.75 in 2 Find the moment of inertia about x axis as shown below. I x = b d 3 12 Substitute 0.5 in. for b and 1.5 in. for d . I x = 0.5 × 1.5 3 12 = 0.140625 in 4 Find the moment of inertia about y axis as shown below. I y = d b 3 12 Substitute 0.5 in. for b and 1.5 in. for d . I y = 1.5 × 0.5 3 12 = 0.015625 in 4 Both ends of the column are pinned along x-x axis. Hence, K = 1 . Both ends of the column are fixed along y-y axis. Hence, K = 0.5 . Apply Euler’s formula as shown below. P c r = π 2 E I ( K L ) 2 (1) Here, E is the modulus of elasticity, I is the moment of inertia, K is the effective length factor, and L is the unsupported length of the column. Find the critical buckling load with respect to x-x axis as shown below. Substitute 29,000 ksi for E , 0.140625 in 4 for I , 1 for K , and 24 in

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 13.3, Problem 13.24P

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The maximum load P it can carry without buckling.

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Chapter 13.3, Problem 13.23P

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