Answered: 80 ksi 90° 60 ksi 45 60°

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

The stress acting on two planes at a point is indicated. Determine the shear stress on plane a–a and the principal stresses at the point.

Expert Solution

Step 1

The component of 60 ksi in horizontal direction,

$\begin{array}{rcl} σ_{x} & = & (60 ksi) \sin 60 ° \\ = & 51.96 ksi \end{array}$

The component of 60 ksi in plane x-y is written as,

$\begin{array}{rcl} τ_{x y} & = & (60 ksi) \cos 60 ° \\ = & 30 ksi \end{array}$

The normal stress on a-a axis is written as,

$\begin{array}{rcl} σ_{a} & = & \frac{σ_{x} + σ_{y}}{2} + \frac{σ_{x} - σ_{y}}{2} \cos 2 θ + τ_{x y} \sin 2 θ \\ but, \\ σ_{a} & = & 80 ksi \\ θ & = & 45 ° \\ then, \\ 80 ksi & = & \frac{51.96 ksi + σ_{y}}{2} + \frac{51.96 ksi - σ_{y}}{2} c o s 2 (45 °) + (30 ksi) s i n 2 (45 °) \\ \frac{51.96 ksi + σ_{y}}{2} & = & 50 ksi \\ σ_{y} & = & 48.04 ksi \end{array}$

Step 2

The shear stress on plane a-a is written as,

$\begin{array}{rcl} τ_{a} & = & - (\frac{σ_{x} - σ_{y}}{2}) \sin 2 θ + τ_{x y} \cos 2 θ \\ = & - (\frac{51.96 ksi - 48.04 ksi}{2}) \sin 2 (45 °) + (30 ksi) \cos 2 (45 °) \\ = & - 1.96 ksi \end{array}$

Therefore, the shear tress on plane a-a is -1.96 ksi.

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