Explanation Given information: A tube is made up A992-steel. The both ends of the tube are pinned The axial load applied on the tube, P = 150 kN . The inner diameter ( d i ) of the tube is 80 mm . The outer diameter ( c ) of the tube is 100 mm . Calculation: Consider the length of the tube is denoted by L . Calculate the area ( A ) of the hollow circular tube using the relation: A = π 4 ( d o 2 − d i 2 ) (1) Here, d o is outer diameter of the tube. Substitute 80 mm for d i and 100 mm for d o in Equation (1). A = π 4 ( 100 2 − 80 2 ) = π 4 ( 10 , 000 − 6 , 400 ) = 900 π mm 2 × 1 m 2 10 6 mm 2 = 0.9 π × 10 − 3 m 2 Calculate the moment of inertia ( I ) of the tube cross section using the relation: I = π 64 ( d o 4 − d i 4 ) (2) Substitute 80 mm for d i and 100 mm for d o in Equation (2). I = π 64 ( 100 4 − 80 4 ) = 92.25 π × 10 4 = 0.9225 π × 10 6 mm 4 × 1 m 4 10 12 mm 4 = 0.9225 π × 10 6 m 4 Calculate the radius of gyration ( r ) using the relation: r = I A (3) Substitute 0.9225 π × 10 6 m 4 for I, and 0.9 π × 10 − 3 m 2 for A in Equation (3). r = 0.9225 π × 10 6 0.9 π × 10 − 3 = 1.025 × 10 9 = 0.03202 m Calculate the slenderness ratio using the relation: Slenderness Ratio = K L r (4) Here, K is the effective length factor, L is the length of the member, and r is the radius of the gyration. The ends of the tube are pinned, then the value of K is 1. Substitute 2 for K and 0.03202 m for r in equation (4). Slenderness Ratio = 1 L 0.03202 = 31.23 L Assume the tube to be a long column. Show the expression for the allowable stress ( σ allow ) using AISC column design formula as follows: σ allow = 12 π 2 E 23 ( K L r ) 2 Modify the above expression as follows: σ allow = 12 π 2 E 23 ( K L r ) 2 P allow A = 12 π 2 E 23 ( K L r ) 2 ( K L r ) 2 = 12 π 2 E 23 ( P allow A ) K L r = 12 π 2 E 23 ( P allow A ) (5) Here, P allow is the allowable load, A is the area, E is the Young’s modulus, L is the length, K is the effective-length factor, and r is the radius of gyration, Consider the Young’s modulus of the steel is E = 200 GPa

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 13.6, Problem 13.90P

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the maximum length of the tube.

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Mechanics of Materials

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