Explanation Given information: The eccentric load on the strut is denoted by P. The length ( L ) of the strut is 3 m . The width ( b ) of the cross-section of the strut is 50 mm . The depth ( d ) of the cross-section of the strut is 100 mm . The eccentricity ( e ) of the loading is 150 mm . The ends of the strut are pinned. Calculation: Consider the Young’s modulus ( E ) of the 2014-T6 aluminum alloy is 73.1 × 10 3 N mm 2 . Consider the yield stress ( σ Y ) of the 2014-T6 aluminum alloy is 414 MPa . Calculate the cross-sectional area ( A ) of the rectangular strut using the relation: A = b d (1) Substitute 50 mm for b , 100 mm for d Equation (1): A = 50 × 100 = 5 , 000 mm 2 Calculate the moment of inertia ( I y ) using the relation: I y = 1 12 d b 3 (2) Substitute 50 mm for b , 100 mm for d Equation (2): I y = 1 12 × 100 × 50 3 = 1.0416 × 10 6 mm 4 Calculate the moment of inertia ( I x ) using the relation: I x = 1 12 b d 3 (3) Substitute 50 mm for b , 100 mm for d Equation (3): I x = 1 12 × 50 × 100 3 = 4.166 × 10 6 mm 4 Calculate the value of radius of gyration ( r x ) using a relation: r x = I x A (4) Substitute 4.166 × 10 6 mm 4 for I x and 5 , 000 mm 2 for A in Equation (4). r x = 4.166 × 10 6 5 , 000 = 833.2 = 28.865 mm Consider the buckling of strut about y-y axis: The ends of the strut are pinned, thus the value of effective length factor ( K ) is 1. Calculate the critical load on the strut based on the buckling criteria. Show the expression for Euler’s formula as follows: P cr = π 2 E I ( K L ) 2 (5) Here, E is the Young’s modulus, I is the moment of inertia, K is the effective length factor, L is the length of the member. Substitute 73.1 × 10 3 N mm 2 for E , 1.0416 × 10 6 mm 4 for I , 3 m for L and 1 for K in Equation (5). P cr = π 2 ( 73.1 × 10 3 ) ( 1.0416 × 10 6 ) ( 1 × 3 m × 1 , 000 mm 1 m ) 2 = 751.481 × 10 9 9 × 10 6 = 83.497 × 10 3 N × 1 kN 1000 N = 83.5 kN Thus, the critical load based on the buckling about y-y axis is 83.5 kN _ . Check the validity of using Euler’s formula: Euler’s formula is valid only when the critical stress ( σ cr ) is less than the yield stress ( σ Y )

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 13.5, Problem 13.65P

To determine

The value of maximum allowable eccentric load P on the strut for buckling and yielding not to take place.

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Chapter 13.5, Problem 13.64P

Chapter 13.5, Problem 13.66P

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Chapter 13 Solutions

Mechanics of Materials (10th Edition)

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