Explanation Given information: The load on the wooden column is denoted by P. The length ( L ) of the column is 10 ft . The Young’s modulus ( E w ) of the wooden column is 1.8 × 10 3 ksi . The yield stress ( σ Y ) of the wooden column is 8 ksi . The width ( b ) of the column is 10 in . The depth ( d ) of the column is 4 in . The eccentricity ( e ) of the loading is 5 in . The load applied on the wooden column is denoted by P . The ends of the wooden column are pinned for movement about x-y axis. The bottom end of the column is fixed and top end free for movement about y-y axis. Calculation: Calculate the cross-sectional area ( A ) of the rectangular column using the relation: A = b d (1) Substitute 10 in . for b and 4 in . for d Equation (1): A = 10 × 4 = 40 m 2 Calculate the moment of inertia ( I y ) using the relation: I y = 1 12 d b 3 (2) Substitute 10 in . for b and 4 in . for d Equation (2): I y = 1 12 × 4 × 10 3 = 333.33 in . 4 Calculate the moment of inertia ( I x ) using the relation: I x = 1 12 b d 3 (3) Substitute 10 in . for b and 4 in . for d Equation (3): I x = 1 12 × 10 × 4 3 = 53.33 in . 4 Calculate the value of radius of gyration ( r y ) using a relation: r y = I y A (4) Substitute 333.33 in . 4 for I y and 40 in . 2 for A in Equation (4). r y = 333.33 m 4 40 m 2 = 8.33325 m 2 = 2.886 m Consider the buckling of column about x-x axis: The wooden column is pinned, thus the value of effective length factor ( K ) is 1. Calculate the length of the column based on the buckling criteria. Show the expression for Euler’s formula as follows: P cr = π 2 E I ( K L ) 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 1.8 × 10 3 ksi for E , 53 .33in . 4 for I , 10 ft for L and 1 for K in Equation (5). P cr = π 2 ( 1.8 × 10 3 ksi ) ( 53.33 in . 4 ) ( 1 × 10 ft × 12 in

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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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 13.54P

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The maximum allowable eccentric load P on the column without buckling or yielding.

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Chapter 13.5, Problem 13.53P

Chapter 13.5, Problem 13.55P

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Assume that the wood column is pinned top and bottom for movement about the x-y axis, and ixed at the bottom and free at the top for movement about the y-y axis. Determine the maximum eccentric load P that canbe applied without causing the column to buckle or yield. Ew = 1.8(103) ksi, sY = 8 ksi.

The wood column is pinned at its base and top. If L = 5 ft, determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.8(103) ksi, sY = 8 ksi.

The brass rod is ixed at one end and free at the other end. If the eccentric load P = 200 kN is applied, determine the greatest allowable length L of the rod so that it does not buckle or yield. Ebr = 101 GPa, sY = 69 MPa.

Chapter 13 Solutions

Mechanics of Materials (10th Edition)

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The maximum allowable eccentric load P on the column without buckling or yielding.

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