Chapter 7, Problem 7.1P

A cylindrical cast-iron casting has an axial bole extending partway through the casting, as shown. Locate the center of gravity of the casting.

To determine

The location of the center of gravity of the casting.

Answer to Problem 7.1P

  $\overline{X}=2.68ft.$ .

Explanation of Solution

Given information:

Given a cylindrical cast iron casting having an axial hole extending pathway through the casting.

The centroid of a body can be found with two axes taken into consideration. As, the given figure is symmetrical about X-axis, so the centroid must lie on the X-axis. Now, by calculating the weight of the given member with reference to densities taken from U S Customary codes for typical average properties of some common materials. Thereby, in reference to the codes the weight of cast iron is taken as $450pcf$ .

Calculation:

Let W₁the weight of the whole cylinder and W₂ the weight of the empty space considering its density same as that of the cast iron.

  $\begin{array}{l}{w}_1=\pi r^{2}h\times 450pcf\\ w_1=3.14\times \frac{6}{12}\times \frac{6}{12}\times 5\times 450\\ w_1=1766.5lb.\\ w_2=\pi r^{2}h\times 450pcf\\ w_2=3.14\times \frac{3}{12}\times \frac{3}{12}\times 3\times 450\\ w_2=264.94lb.\\ W=w_1+w_2\\ W=1766.5lb+\left(-264.94\right)lb\\ W=1502.56lb.\end{array}$

Now we have,

  $\begin{array}{l}\overline{X}=\frac{{\displaystyle \sum wy}}{W}\\ \overline{X}=\frac{(w_1\times y_1)-\left(w_2\times y_2\right)}{W}\\ \overline{X}=\frac{\left(1766.5\times 2.5\right)-\left(264.94\times 1.5\right)}{1501.56}\\ \overline{X}=2.68ft.\end{array}$

Conclusion:

Therefore, the location of center of gravity of the casting is $\overline{X}=2.68ft.$ .