Explanation Calculation: Show the expression of strain distribution as follows: ε = − ε m y c = − ε m u Here, y is distance of the point considered from the neutral surface, ε m is normal strain, and c is the maximum value of y . Show the expression of bending couple as follows: M = − ∫ − c c y σ b d y = 2 b ∫ 0 c y | σ | d y = 2 b c 2 ( ∫ 0 1 u | σ | d u ) (1) Show the value of u as follows: u = ε ε m (2) Substitute K σ n for ε and K σ m for ε m in Equation (2). u = K σ n K σ m = ( σ σ m ) n | σ | = σ m u 1 n Substitute σ m u 1 n for | σ | in Equation (1)

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 4.6, Problem 98P

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Show the maximum stress in prismatic bar is σm=1+2n3nMcI.

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Chapter 4.6, Problem 97P

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Mechanics of Materials, 7th Edition

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Show the maximum stress in prismatic bar is σ m = 1 + 2 n 3 n M c I .

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Show the maximum stress in prismatic bar is σm=1+2n3nMcI.

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