Explanation Given information: The length of the aluminum rod is denoted by L. The diameter ( d ) of the aluminum rod is 200 mm . The eccentricity ( e ) of the loading is 5 mm . The load applied on the aluminum rod is denoted by P . The one end of the aluminum rod is fixed and the other end is free. The Young’s modulus of the aluminum rod, E al = 72 GPa . The yield stress of the aluminum rod, σ Y = 410 MPa . Calculation: Calculate the cross-sectional area ( A ) of the circular rod using the relation: A = π 4 d 2 (1) Substitute 200 mm for d in Equation (1): A = π 4 ( 200 ) 2 = 31415.92 mm 2 × 1 m 2 10 6 mm 2 = 0.03141592 m 2 Calculate the moment of inertia using the relation: I = π 64 d 4 (2) Substitute 200 mm for d in Equation (2): I = π 64 ( 200 ) 4 = 78539816.33 mm 2 × 1 m 4 10 12 mm 4 = 78.539 × 10 − 6 m 4 Calculate the value of radius of gyration ( r ) using a relation: r = I A (3) Substitute 78.539 × 10 − 6 m 4 for I and 0.03141592 m 2 for A in Equation (3). r = 78.539 × 10 − 6 m 4 0.03141592 m 2 = 0.0025 m 2 = 0.05 m Thealuminum rod is fixed at one end and free at the other end, thus the value of effective length factor ( K ) is 2. Consider the buckling criteria: Calculate the length of the rod based on the buckling criteria. Show the expression for Euler’s formula as follows: P cr = π 2 E I ( K L ) 2 (4) 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 72 GPa for E , 78.539 × 10 − 6 m 4 for I , 200 kN for P cr , and 2 for K in Equation (4). 200 kN = π 2 ( 72 GPa ) ( 78.54 × 10 − 6 m 4 ) ( 2 L ) 2 200 kN × ( 1 , 000 N 1 kN ) = π 2 ( 72 GPa × 10 9 N m 2 1 GPa ) ( 78.54 × 10 − 4 m 4 ) ( 2 L ) 2 200 × 10 3 N = π 2 × 72 × 10 9 N m 2 × ( 78.54 × 10 − 6 m 4 ) ( 2 L ) 2 4 L 2 = 72 × 10 9 π 2 N m 2 × ( 78.54 × 10 − 6 m 4 ) 200 × 10 3 N 4 L 2 = 55 , 811.4 × 10 3 N ⋅ m 2 200 × 10 3 N 4 L 2 = 279

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 13.51P

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The greatest allowable length of the rod for without buckling.

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Chapter 13.5, Problem 13.50P

Chapter 13.5, Problem 13.52P

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Chapter 13 Solutions

Mechanics of Materials (10th Edition)

Ch. 13.3 - A 50-in long steel rod has a diameter of 1 in....Ch. 13.3 - A 12-ft wooden rectangular column has the...Ch. 13.3 - The A992 steel column can be considered pinned at...Ch. 13.3 - A steel pipe is fixed supported at its ends. If it...Ch. 13.3 - Determine the maximum force P that can be...Ch. 13.3 - The A992 steel rod BC has a diameter of 50 mm and...Ch. 13.3 - Determine the critical buckling load for the...Ch. 13.3 - The column consists of a rigid member that is...Ch. 13.3 - The aircraft link is made from an A992 steel rod....Ch. 13.3 - Rigid bars AB and BC are pin connected at B. If...

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