Chapter 8, Problem 8.48P

To determine

x and y centroidal coordinates

Answer to Problem 8.48P

The centroidal coordinate $\left(\overline{x},\overline{y}\right)$ of the volume is $\left[\frac{2a\left(2h_1+h_2\right)}{3\left(3h_1+h_2\right)},\frac{2b\left(2h_1+h_2\right)}{3\left(3h_1+h_2\right)}\right]$.

Explanation of Solution

Given information:

We use double integration method to find the x and y centroidal coordinates of a volume.

  $\begin{array}{l}\overline{x}=\frac{Q_{yz}}{V}=\frac{{\displaystyle \underset{V}{\int }xdV}}{{\displaystyle \underset{V}{\int }dV}}\\ \overline{y}=\frac{Q_{xz}}{V}=\frac{{\displaystyle \underset{V}{\int }ydV}}{{\displaystyle \underset{V}{\int }dV}}\end{array}$

Calculation:

The differential volume $dV$ of element shown

  $\begin{array}{l}dV=zdxdy\\ =\left[h_1-\left(h_1-h_2\right)\frac{xy}{ab}\right]dxdy\end{array}$

Assume the centroidal coordinates $\left(\overline{x_{el}},\overline{y_{el}}\right)$ of element as:

  $\begin{array}{l}\overline{x_{el}}=x\\ \overline{y_{el}}=y\end{array}$

The volume $V$ of the element

  $\begin{array}{l}V={\displaystyle \underset{V}{\int }dV}\\ ={\displaystyle \underset{0}{\overset{b}{\int }}{\displaystyle \underset{0}{\overset{a}{\int }}\left[h_1-\left(h_1-h_2\right)\frac{xy}{ab}\right]dxdy}}\\ ={\displaystyle \underset{0}{\overset{b}{\int }}\left[h_{1}a-\left(h_1-h_2\right)\frac{ay}{2b}\right]dxdy}\\ =h_{1}ab-\left(h_1-h_2\right)\frac{ab}{4}\\ =\left(3h_1+h_2\right)\frac{ab}{4}\end{array}$

The first moment $Q_{yz}$ of element

  $\begin{array}{l}{Q}_{yz}={\displaystyle \underset{V}{\int }\overline{x_{el}}dV}\\ ={\displaystyle \underset{0}{\overset{b}{\int }}{\displaystyle \underset{0}{\overset{a}{\int }}\left[h_{1}x-\left(h_1-h_2\right)\frac{x^{2}y}{ab}\right]dxdy}}\\ ={\displaystyle \underset{0}{\overset{b}{\int }}\left[h_1\frac{a^2}{2}-\left(h_1-h_2\right)\frac{a^{2}y}{3b}\right]dxdy}\\ =h_1\frac{a^{2}b}{2}-\left(h_1-h_2\right)\frac{a^{2}b}{6}\\ =\left(2h_1+h_2\right)\frac{a^{2}b}{6}\end{array}$

The centroidal x coordinate

  $\overline{x}=\frac{Q_{yz}}{V}=\frac{\left(2h_1+h_2\right)\frac{a^{2}b}{6}}{\left(3h_1+h_2\right)\frac{ab}{4}}=\frac{2a\left(2h_1+h_2\right)}{3\left(3h_1+h_2\right)}$

Similarly, the centroidal y coordinate

  $\overline{y}=\frac{Q_{xz}}{V}=\frac{2b\left(2h_1+h_2\right)}{3\left(3h_1+h_2\right)}$

Conclusion:

The centroidal coordinate $\left(\overline{x},\overline{y}\right)$ of the volume is $\left[\frac{2a\left(2h_1+h_2\right)}{3\left(3h_1+h_2\right)},\frac{2b\left(2h_1+h_2\right)}{3\left(3h_1+h_2\right)}\right]$.