Chapter 8, Problem 8.129RP

To determine

The resultant force of the pressure and the coordinates of its point of application.

Answer to Problem 8.129RP

The resultant force is $F=\left(\frac{2P_{0}ab}{\pi }\right)$ and co-ordinates of point of application are $\overline{x}=\frac{2a}{3}\text{, }\overline{y}=2b$

Explanation of Solution

Given Information:

The pressure acting on the rectangular plate varies as, $p=p_0\frac{x}{a}\cos \frac{\pi y}{2b}$ where $p_0$ is a constant.

Calculation:

Consider the following figure:

The pressure varies as $p=p_0\frac{x}{a}\cos \frac{\pi y}{2b}$.

Assume a small area element of area dxdy at coordinate (x,y). The resultant force on this small area is given by

  $dF=pdxdy=\left(p_0\frac{x}{a}\cos \frac{\pi y}{2b}\right)dxdy$

Integrate above to get total force as

  $\begin{array}{l}F={\displaystyle \underset{0}{\overset{a}{\int }}{\displaystyle \underset{-b}{\overset{+b}{\int }}\left(p_0\frac{x}{a}\cos \frac{\pi y}{2b}\right)dxdy}}\\ F={\displaystyle \underset{0}{\overset{a}{\int }}xdx{\displaystyle \underset{-b}{\overset{+b}{\int }}\frac{p_0}{a}\left(\cos \frac{\pi y}{2b}\right)dy}}\\ F={\left(\frac{x^2}{2}\right)}^a{}_0{\displaystyle \underset{-b}{\overset{+b}{\int }}\frac{p_0}{a}\left(\cos \frac{\pi y}{2b}\right)dy}\\ F=\left(\frac{a^2}{2}\times \frac{p_0}{a}\right)\left({\sin \frac{\pi y}{2b}|}_{-b}^{+b}\left(\frac{2b}{\pi }\right)\right)\\ F=\left(\frac{P_{0}ab}{\pi }\right)\left(\sin \frac{\pi b}{2b}-\left(\sin \frac{\pi \left(-b\right)}{2b}\right)\right)\\ F=\left(\frac{2P_{0}ab}{\pi }\right)\end{array}$

So the resultant force is $F=\left(\frac{2P_{0}ab}{\pi }\right)$.

Coordinates (x,y) is the location of line of action of small force dF. The moments about origin due to this small force are $d{Q}_x=ydF\text{ and }d{Q}_y=xdF.$

The x-coordinate of location of line of action of total resultant force is given by

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \underset{}{\overset{}{\int }}d{Q}_y}}{{\displaystyle \underset{}{\overset{}{\int }}dF}}=\frac{{\displaystyle \underset{}{\overset{}{\int }}xdF}}{F}=\frac{{\displaystyle \underset{0}{\overset{a}{\int }}{\displaystyle \underset{-b}{\overset{+b}{\int }}x{p}_0\frac{x}{a}\cos \frac{\pi y}{2b}dxdy}}}{\frac{2P_{0}ab}{\pi }}\text{=}\frac{{\displaystyle \underset{0}{\overset{a}{\int }}\frac{P_0}{a}{x}^2{\displaystyle \underset{-b}{\overset{b}{\int }}\cos \frac{\pi y}{2b}dxdy}}}{\frac{2P_{0}ab}{\pi }}\\ \overline{x}=\frac{\left(\frac{P_0}{a}\times \frac{a^3}{3}\right)\times \left(\frac{2b}{\pi }\right)\left(2\right)}{\frac{2P_{0}ab}{\pi }}\\ \overline{x}=\frac{2a}{3}\end{array}$

The y-coordinate of location of line of action of total resultant force is given by

  $\begin{array}{l}\overline{y}=\frac{{\displaystyle \underset{}{\overset{}{\int }}d{Q}_x}}{{\displaystyle \underset{}{\overset{}{\int }}dF}}=\frac{{\displaystyle \underset{}{\overset{}{\int }}ydF}}{F}=\frac{{\displaystyle \underset{0}{\overset{a}{\int }}{\displaystyle \underset{-b}{\overset{+b}{\int }}y{p}_0\frac{x}{a}\cos \frac{\pi y}{2b}dxdy}}}{\frac{2P_{0}ab}{\pi }}\text{=}\frac{{\displaystyle \underset{0}{\overset{a}{\int }}\frac{P_0}{a}x{\displaystyle \underset{-b}{\overset{b}{\int }}y\cos \frac{\pi y}{2b}dxdy}}}{\frac{2P_{0}ab}{\pi }}\\ \overline{y}=\frac{\left(\frac{P_0}{a}\times \frac{a^2}{2}\right)\times {\left(\left(\frac{2b}{\pi }\right)\left(y\right)\left(\sin \left(\frac{\pi y}{2b}\right)\right)\right)}_{-b}^{+b}-{\displaystyle \underset{-b}{\overset{+b}{\int }}\left(\frac{2b}{\pi }\right)\left(\sin \frac{\pi y}{2b}\right)dy}}{\frac{2P_{0}ab}{\pi }}\\ \overline{y}=\frac{\left(\frac{P_{0}a}{2}\right)\times \left(\left(\frac{2b}{\pi }\right)\left(2b\right)\left(2\right)\right)-{\left(\frac{2b}{\pi }\right)}^2{\left(\cos \frac{\pi y}{2b}\right)}_{-b}^{+b}}{\frac{2P_{0}ab}{\pi }}\\ \overline{y}=2b\end{array}$

Conclusion:

Therefore, the resultant force is $F=\left(\frac{2P_{0}ab}{\pi }\right)$ and co-ordinates of point of application are $\overline{x}=\frac{2a}{3}\text{, }\overline{y}=2b$.