To determine the maximum axial load applied on the column without causing buckling. Explanation Given information: Thetube is made of A992 steel. The outer width ( b ) of the tube is 4 in . The outer depth ( d ) of the tube is 4 in . The inner width ( b 1 ) of the tube is 3 in . The inner depth ( d 1 ) of the tube is 3 in . The tube is fixed at one end and free at another end. The length ( L ) of the tube is 15 ft . Calculation: Consider the Young’s modulus ( E ) of the material is 29 × 10 3 ksi . Show the moment of inertia of the tube as follows: I = b d 3 12 − b 1 d 1 3 12 Substitute 3 in . for b 1 , 3 in . for d 1 , 4 in . for b , 4 in . for d . I = 4 × 4 3 12 − 3 × 3 3 12 = 21.333 − 6.75 =14 .583 in . 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 = 2 , since one end is fixed and another end is pinned. Substitute 2 for K , 29 × 10 3 ksi for E , 14 .583 in . 4 for I , and 15 ft for L in Equation (1). P c r = π 2 × 29 × 10 3 × 14

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 13.3, Problem 13.21P

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the maximum axial load applied on the column without causing buckling.

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