Explanation Given information: The wide flange section W8 × 67 made up of A36 steel is used as a column. The area A for wide flange section W8 × 67 is 19.7 in . 2 . The least moment of inertia of the wide flange section is I y = 88.6 in . 4 . The Young’s modulus for A36 steel is E = 29 × 10 3 ksi . The value of yield stress for A36 steel, σ Y = 36 ksi . The length of the column is L = 25 ft . The column is fixed at the bottom and free at the top end. Calculation: 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 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. The value of effective-length factor for the column, K = 2 , since bottom end is fixed and top end is free. Substitute 2 for K , 29 × 10 3 ksi for E , 88.6 in . 4 for I , and 25 ft for L in Equation (1). P cr = π 2 ( 29 × 10 3 ksi ) ( 88

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Author: HIBBELER

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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 13.3, Problem 13.15P

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The largest axial force P, that the column can support without buckling.

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Chapter 13.3, Problem 13.14P

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Chapter 13 Solutions

Mechanics of Materials

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The largest axial force P , that the column can support without buckling.

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