Explanation Given information: The load on the column is denoted by P . The Young’s modulus of the wooden column, E = 10 GPa . The yield stress of the wooden column, σ Y = 15 MPa . The width of the cross-sectional area of the column, b = 50 mm . The depth of the cross-sectional area of the column, d = 150 mm . The eccentricity ( e ) of the loading is 150 mm . The ends of the column are pinned. Calculation: Calculate the cross-sectional area ( A ) of the rectangular column using the relation: A = b d (1) Substitute 50 mm for b and 150 mm for d Equation (1): A = 50 × 150 = 7500 mm 2 × 1 m 2 10 6 mm 2 = 0.0075 m 2 Calculate the moment of inertia ( I y ) using the relation: I y = 1 12 d b 3 (2) Substitute 50 mm for b and 150 mm for d Equation (2): I y = 1 12 × 150 × 50 3 = 1 , 562 , 500 mm 4 × 1 m 4 10 12 mm 4 = 1.5625 × 10 − 6 m 4 Calculate the moment of inertia ( I x ) using the relation: I x = 1 12 b d 3 (3) Substitute 50 mm for b and 150 mm for d Equation (3): I x = 1 12 × 50 × 150 3 = 14 , 062 , 500 mm 4 × 1 m 4 10 12 mm 4 = 14.0625 × 10 − 6 m 4 Calculate the value of radius of gyration ( r x ) using a relation: r x = I x A (4) Substitute 14.0625 × 10 − 6 m 4 for I x and 0.0075 m 2 for A in Equation (4). r x = 14.0625 × 10 − 6 m 4 0.0075 m 2 = 1875 × 10 − 6 m 2 = 0.0433 m Consider the buckling of column about y-y axis: The wooden column is pinned, thus the value of effective length factor ( K ) is 1. Check the safety of the column based on the buckling criteria. 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 10 GPa for E , 1.5625 × 10 − 6 m 4 for I y , 10 ft for L , and 1 for K in Equation (5). P cr = π 2 ( 10 GPa ) ( 1.5625 × 10 − 6 m 4 ) ( 1 × 35 ) 2 = π 2 ( 10 GPa × 10 9 N m 2 1 GPa ) ( 1.5625 × 10 − 6 m 4 ) ( 1 × 3.5 m ) 2 = 15.421 × 10 4 12.25 N = 12 , 588.57 N × 1 kN 1 , 000 N P cr = 12.59 kN ≈ 12

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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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 13.56P

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The maximum eccentric load that can be supported by the column without yielding or buckling.

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please find Ix and Iy in mm4

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The maximum eccentric load that can be supported by the column without yielding or buckling.

Solution Summary: The author calculates the maximum eccentric load that can be supported by the column without yielding or buckling.

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