Explanation Given information: The copper hollow circular tube acts as a column. The ends of the column are pinned. The outer diameter ( d o ) of the tube is 35 mm . The thickness ( t ) of the tube is 7 mm . The Young’s modulus of the copper, E c u = 120 GPa . The yield stress of the copper is σ Y = 750 MPa . The length of the copper tube is 2 m . The load on the column is denoted by P . The eccentricity ( e ) of the loading is 14 mm . Calculation: Consider the inner diameter of the tubular column as d i . Calculate the inner diameter of the tubular column using the relation: d i = d o − 2 t (1) Substitute 35 mm for d o and 7 mm for t in Equation (1). d i = 35 − 2 × 7 = 35 − 14 = 21 mm Calculate the cross-sectional area ( A ) of the circular hollowtube using the relation: A = π 4 [ ( d o ) 2 − ( d i ) 2 ] (2) Substitute 35 mm for d o , 21 mm for d i in Equation (2): A = π 4 ( 35 2 − 21 2 ) = π 4 ( 1225 − 441 ) = 615.75 mm 2 × ( 1 m 2 10 6 mm 2 ) = 0.61575 × 10 − 3 m 2 Calculate the moment of inertia ( I ) using the relation: I = π 64 [ ( d 0 ) 4 − ( d i ) 4 ] (3) Substitute 35 mm for d o , 21 mm for d i in Equation (3): I = π 64 [ ( 35 ) 4 − ( 21 ) 4 ] = π 64 [ 1 , 500 , 625 − 194 , 481 ] = π 64 [ 1 , 306 , 144 ] = 64115.19 mm 4 × ( 1 m 4 10 12 mm 4 ) I = 64.1152 × 10 − 9 m 4 Calculate the value of radius of gyration ( r ) using a relation: r = I A (4) Substitute 64.1152 × 10 − 9 m 4 for I and 0.61575 × 10 − 3 m 2 for A in Equation (4). r = 64.1152 × 10 − 9 m 4 0.61575 × 10 − 3 m 2 = 1.04125 × 10 − 4 m 2 = 0.01020 m Consider the buckling of column: The coppercolumn is pinned, thus the value of effective length factor ( K ) is 1. The Young’s modulus ( E ) of the copper is 120 GPa . Show the unit conversion of the Young’s modulus as follows: E = 120 GPa × ( 10 9 N m 2 1 GPa ) = 120 × 10 9 N m 2 Show the expression for Euler’s formula as follows: P cr = π 2 E I y ( K L ) y 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 2 m for L , 120 × 10 9 N m 2 for E , 64.1152 × 10 − 9 m 4 for I , and 1 for K in Equation (5). P c r = π 2 ( 120 × 10 9 ) ( 64.1152 × 10 − 9 ) ( 1 × 2 ) 2 = 75935 4 = 18 , 983.75 N × 1 kN 1 , 000 N = 18.983 kN Thus, the load on the column based on the buckling criteria is 18.983 kN _ . Check the validity of Euler’s formula: Consider the yielding stress of the steel, σ Y = 50 ksi . Calculate the critical stress using the relation; σ cr = P cr A (6) Substitute 18.983 kN for P cr and 0.61575 × 10 − 3 m 2 for A in Equation (6)

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 13.5, Problem 13.59P

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The load supported by the column without failure.

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Chapter 13.5, Problem 13.58P

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Chapter 13 Solutions

Mechanics of Materials (10th Edition)

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The load supported by the column without failure.

Solution Summary: The author explains how the copper hollow circular tube acts as a column. The outer diameter of the tube is 35mm and the thickness (t).

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The load supported by the column without failure.

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Chapter 13 Solutions