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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

For the state of stress shown, it is known that the normal and shearing stresses are directed as shown and that σ_x= 14 ksi, σ_y= 9 ksi, and σmin= 5 ksi. Determine (a) the orientation of the principal planes, (b) the principal stress σmax, (c) the maximum in-plane shearing stress

Expert Solution

Step 1

Given Data:

The stress in x-direction is $σ_{x} = 14 ksi$ .
The stress in y-direction is $σ_{y} = 9 ksi$ .
The minimum stress is $σ_{m i n} = 5 ksi$ .

The expression of minimum stress is given as,

$\begin{array}{rcl} σ_{m i n} & = & \frac{σ_{x} + σ_{y}}{2} - \sqrt{{(\frac{σ_{x} - σ_{y}}{2})}^{2} + τ_{x y}^{2}} \\ 5 ksi & = & \frac{14 ksi + 9 ksi}{2} - \sqrt{{(\frac{14 ksi - 9 ksi}{2})}^{2} + τ_{x y}^{2}} \\ τ_{x y} & = & 6 ksi \end{array}$

Step 2

Part (a)

The expression of orientation of principal plane is,

$\begin{array}{rcl} \tan 2 θ_{P 1} & = & \frac{2 τ_{x y}}{σ_{x} - σ_{y}} \\ \tan 2 θ_{P 1} & = & \frac{2 \times 6 ksi}{14 ksi - 9 ksi} \\ θ_{P 1} & = & 33.7 ° \end{array}$

Another principal plane is at θ_P2 = 33.7° +90° =123.7°.

Part (b)

The maximum principal stress is,

$\begin{array}{rcl} σ_{x} + σ_{y} & = & σ_{m a x} + σ_{m i n} \\ 14 ksi + 9 ksi & = & σ_{m a x} + 5 ksi \\ σ_{m a x} & = & 18 ksi \end{array}$

Thus, the maximum stress is 18 ksi.

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