Chapter 4, Problem 73P

To determine

The magnitude of the shear strain rate about point $P$ in the $xy$ plane is given by : ${\epsilon }_{xy}=\frac{1}{2}\ge \left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)$.

Explanation of Solution

The shear strain rate is half of the rate of decrease of the angle between two perpendicular lines which intersect at the point. Initially at time $t_1$, line $a$ and $b$ are perpendicular and intersect at point $P$. The angle between line $a$ and $b$ is ${\alpha }_{a-b}$ at time $t_2$.

The following figure shows the fluid element at time $t_2$.

  

Figure-(1)

At time $t_2$, the point $A$ moves a distance of $\left(u+\frac{\partial u}{\partial x}dx\right)dt$ to right and $\left(v+\frac{\partial v}{\partial x}dx\right)dt$ to up and the point $B$ moves a distance of $\left(u+\frac{\partial u}{\partial y}dy\right)dt$ to the right and $\left(v+\frac{\partial v}{\partial y}dy\right)dt$ up. The point $P$ moves a distance of $udt$ to right and $vdt$ up.

Refer to Figure-(I), to obtain the values of the horizontal distance between point $P^{\prime }$ and $A^{\prime }$ as $dx+\frac{\partial u}{\partial x}dxdt$, the vertical distance between point $P^{\prime }$ and $A^{\prime }$ as $\frac{\partial v}{\partial x}dxdt$, the horizontal distance between point $P^{\prime }$ and $B^{\prime }$ as $-\frac{\partial u}{\partial y}dydt$ and the vertical distance between point $P^{\prime }$ and $B^{\prime }$ as $dy+\frac{\partial v}{\partial y}dydt$.

Write the expression for the angle $\alpha$ at point $A$

  $\begin{array}{l}\tan {\alpha }_a=\frac{\frac{\partial v}{\partial x}dxdt}{dx+\frac{\partial u}{\partial x}dxdt}\\ {\alpha }_a={\tan }^{-1}\left(\frac{\frac{\partial v}{\partial x}dxdt}{dx+\frac{\partial u}{\partial x}dxdt}\right)\end{array}$

The distance $dx+\frac{\partial u}{\partial x}dxdt$ is approximate to $dx$.

  $\begin{array}{c}{\alpha }_a={\tan }^{-1}\left(\frac{\frac{\partial v}{\partial x}dxdt}{dx}\right)\\ ={\tan }^{-1}\left(\frac{\partial v}{\partial x}dt\right)\\ \approx \frac{\partial v}{\partial x}dt\end{array}$

Write the expression for the angle $\alpha$ at point $B$

  $\begin{array}{l}\tan {\alpha }_b=\frac{-\frac{\partial u}{\partial y}dydt}{dy+\frac{\partial v}{\partial y}dydt}\\ {\alpha }_b={\tan }^{-1}\left(\frac{-\frac{\partial u}{\partial y}dydt}{dy+\frac{\partial v}{\partial y}dydt}\right)\end{array}$

The distance $dy+\frac{\partial v}{\partial y}dydt$ is approximate to $dy$.

  $\begin{array}{c}{\alpha }_b={\tan }^{-1}\left(\frac{-\frac{\partial u}{\partial u}dudt}{du}\right)\\ ={\tan }^{-1}\left(-\frac{\partial u}{\partial y}dt\right)\\ \approx -\frac{\partial u}{\partial y}dt\end{array}$

Write the expression for the angle ${\alpha }_{a-b}$ at time $t_1$.

  ${\left({\alpha }_{a-b}\right)}_{t_1}=\frac{\pi }{2}$

Refer to figure (I), to obtain the value of ${\left({\alpha }_{a-b}\right)}_{t_2}$ as ${\alpha }_b+\frac{\pi }{2}-{\alpha }_a$ at time $t_2$.

  ${\left({\alpha }_{a-b}\right)}_{t_2}={\alpha }_b+\frac{\pi }{2}-{\alpha }_a$   ...... (I)

Substitute $\frac{\partial v}{\partial x}dt$ for ${\alpha }_a$ and $-\frac{\partial u}{\partial y}dt$ for ${\alpha }_b$ in Equation (I).

  ${\left({\alpha }_{a-b}\right)}_{t_2}=-\frac{\partial u}{\partial y}dt+\frac{\pi }{2}-\frac{\partial v}{\partial x}dt$

Write the expression for the shear strain rate.

  $\begin{array}{c}{\epsilon }_{xy}=-\frac{1}{2}\frac{d}{dt}{\alpha }_{a-b}\\ {\epsilon }_{xy}=-\frac{1}{2}\frac{d}{dt}\left({\left({\alpha }_{a-b}\right)}_{t_2}-{\left({\alpha }_{a-b}\right)}_{t_1}\right)\end{array}$   ...... (II)

Substitute $-\frac{\partial u}{\partial y}dt+\frac{\pi }{2}-\frac{\partial v}{\partial x}dt$ for ${\left({\alpha }_{a-b}\right)}_{t_2}$ and $\frac{\pi }{2}$ for ${\left({\alpha }_{a-b}\right)}_{t_1}$ in Equation (II).

  $\begin{array}{c}{\epsilon }_{xy}=-\frac{1}{2}\frac{d}{dt}\left(-\frac{\partial u}{\partial y}dt+\frac{\pi }{2}-\frac{\partial v}{\partial x}dt-\frac{\pi }{2}\right)\\ {\epsilon }_{xy}=-\frac{1}{2}\frac{d}{dt}\left(-\frac{\partial u}{\partial y}dt-\frac{\partial v}{\partial x}dt\right)\\ {\epsilon }_{xy}=\frac{1}{2}\frac{d}{dt}\left(\frac{\partial u}{\partial y}dt+\frac{\partial v}{\partial x}dt\right)\end{array}$