Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 6 - Mechanical Properties of Metals - Questions and Problems - Page 207: 6.1

Answer

$$\sigma_{x} = \sigma \cos^2 \theta$$

Work Step by Step

In the x-direction, the exerted force will be $$F_{x}=F \cos \theta$$ Also, the area in the x-direction $A=A_x \cos \theta$ The mechanical stress equals the force divided by the area $A$ and in x-direction will be $$\sigma_{x}=\frac{F_{x}}{A_{x}}$$ Now, substitute the terms of $A_x$ and $F_x$ to get the stress in the form $$\sigma_{x}=\dfrac{F_{x}}{A_{x}} = \dfrac{F \cos \theta}{(A)/( \cos \theta)} = \dfrac{F}{A} \cos^2 \theta = \boxed{ \sigma \cos^2 \theta} $$ Where $\sigma = \dfrac{F}{A}$
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