Explanation Given information: The hollow circular tube is made of 2014-T6 aluminum. The outer diameter ( d o ) of the hollow circular tube is 150 mm . The inner diameter ( d i ) of the hollow circular tube is 100 mm . The tube is pinned at both the ends. The length ( L ) of the tube is 6 m . Calculation: Consider the Young’s modulus ( E ) of the material is 73.1 × 10 9 MPa . Show the moment of inertia of the hollow circular tube as follows: I = π ( d o 4 − d i 4 ) 64 Substitute 150 mm for d o and 100 mm for d i . I = π ( 150 4 − 100 4 ) 64 = π ( 50625 × 10 4 − 10 , 000 × 10 4 ) 64 = 40625 × 10 4 π 64 = 634.7656 × 10 4 π mm 4 × ( 1 m 4 10 − 12 mm 4 ) = 634.7656 × 10 − 8 π m 4 Show the expression for Euler’s formula as shown below: P cr = π 2 E I ( K L ) 2 (1) Here, P cr is the maximum load on the material just before it began to buckle, E is the Young’s modulus of the material, I is the least moment of inertia of the material cross-sectional area, L is the length of the material and K is the effective-length factor. The value of effective-length factor for rod, K = 1 , since the both ends are pinned. Substitute 1 for K , 73.1 × 10 9 MPa for E , 634.7656 × 10 − 8 π for I , and 6 m for L in Equation (1). P c r = π 2 × 73.1 × 10 9 × 634.7656 × 10 − 8 π ( 1 × 6 ) 2 = π 2 × 73

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ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Chapter 13.3, Problem 13.7P

To determine

the largest load applied on the column without causing buckling.

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Chapter 13.3, Problem 13.6P

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Mechanics Of Materials, Si Edition

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