Explanation Given information: The two steel channels are laced together to form a bridge column. The length of the bridge column is L = 30 ft . The column is pinned at both the ends. The cross-sectional area of each channel is A = 3.10 in . 2 . The moment of inertia of each section are I x = 55.4 in . 4 and I y = 0.382 in . 4 . The centroid of the area is denoted by C . The distance between the centroid of the channel is denoted by d . Consider the buckling of the column about x − x and y ′ − y ′ occurs for the same load. The Young’s modulus for steel is E st = 29 × 10 3 ksi . The yield stress of the material is σ Y = 50 ksi . Calculation: Calculate the moment of inertia of the cross-section of column as follows: I x = 2 × ( moment of inertia of each channel ) = 2 × ( 55.4 ) = 110.84 in . 4 I y = 2 × ( moment of inertia of each channel ) + 2 × ( Area of each channel ) ( d 2 ) 2 = 2 × ( 0.382 ) + 2 × ( 3.10 ) ( d 2 ) 2 = 0.764 + 1.55 d 2 The column has to buckle about x − x and y ′ − y ′ at the same. Then, The value of I x and I y are Equal. I y = I x 0.764 + 1.55 d 2 = 110.84 1.55 d 2 = 110.84 − 0.764 d 2 = 110.84 − 0.764 1.55 d 2 = 71.016 d = ( 71.016 ) 1 2 d = 8.427 in . ≈ 8 .43 in . Thus, the distance between the centroid of the channel is 8 .43 in . _ Show the expression for Euler’s formula as shown below: P c r = π 2 E I ( K L ) 2 (1) Here, P c r is the maximum load on the material just before it begins 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

9th Edition

ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Chapter 13.3, Problem 13.14P

To determine

the distance between the centroid of the channel.

The value of critical load on the column.

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Chapter 13.3, Problem 13.13P

Chapter 13.3, Problem 13.15P

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the distance between the centroid of the channel. The value of critical load on the column.