Chapter 61, Problem 33A

Find the volume of a hollow machined steel piece shown. Round the answer to 1 decimal place.

To determine

The volume of machined piece shown in the figure.

Answer to Problem 33A

The volume of machined piece is $\text{V = 234}\text{.474}{\text{cm}}^{\text{3}}$

Explanation of Solution

Given information:

  

Concept used:

The piece being machined is shown in the figure. The shape of the piece is of frustum of a cone and a thorough a hollow frustum shape has been machined from the piece.

The volume of the given piece can be calculated by subtracting the volume of hollow frustum from the volume of frustum of cone.

The volume of the frustum of right circular cone is

  $\text{V =}\frac{\text{1}}{\text{3}}\pi \text{h}\left({\text{R}}^2\text{+}{\text{r}}^2\text{+Rr}\right)$

Where, h = height, and R = radius of larger Base and r = radius of smaller base.

Calculation:

  ${\text{V}}_F\text{= }\frac{\text{1}}{\text{3}}\text{×3}\text{.14×}\text{h}\text{×}\left({\left(\frac{\text{D}}{\text{2}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{d}}{\text{2}}\right)}^{\text{2}}\text{+}\frac{\text{D}}{\text{2}}\text{×}\frac{\text{d}}{\text{2}}\right)$

  $\begin{array}{l}{\text{V}}_F\text{ = }\frac{\text{1}}{\text{3}}\text{×3}\text{.14×}5.\text{06×}\left({\left(\frac{11.32}{2}\right)}^{\text{2}}\text{+}{\left(\frac{9.28}{2}\right)}^{\text{2}}\text{+}\frac{11.32}{2}\text{×}\frac{9.28}{2}\right)\\ {\text{V}}_F\text{ =}\frac{\text{1}}{\text{3}}\text{×3}\text{.14×5}\text{.06 }\text{× }\left(\text{32}\text{.0356 +21}\text{.5296 +26}\text{.2624}\right)\\ {\text{V}}_F\text{ =}\frac{\text{1}}{\text{3}}\text{×3}\text{.14×5}\text{.06×}79.8276\\ {\text{V}}_F\text{ = 422}\text{.777}{\text{cm}}^{\text{3}}\\ \end{array}$

Now, the volume of the thorough cut hollow frustum can be calculated from the same formula.

  ${\text{V}}_H\text{= }\frac{\text{1}}{\text{3}}\text{×3}\text{.14×}\text{h}\text{×}\left({\left(\frac{\text{D}}{\text{2}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{d}}{\text{2}}\right)}^{\text{2}}\text{+}\frac{\text{D}}{\text{2}}\text{×}\frac{\text{d}}{\text{2}}\right)$

  $\begin{array}{l}{\text{V}}_H\text{ = }\frac{\text{1}}{\text{3}}\text{×3}\text{.14×}5.\text{06×}\left({\left(\frac{7.88}{2}\right)}^{\text{2}}\text{+}{\left(\frac{5.84}{2}\right)}^{\text{2}}\text{+}\frac{7.88}{2}\text{×}\frac{5.84}{2}\right)\\ {\text{V}}_H\text{ =}\frac{\text{1}}{\text{3}}\text{×3}\text{.14×5}\text{.06 }\text{× }\left(\text{15}\text{.5236 +8}\text{.5264 +11}\text{.5048}\right)\\ {\text{V}}_H\text{ =}\frac{\text{1}}{\text{3}}\text{×3}\text{.14×5}\text{.06×}35.5548\\ {\text{V}}_H\text{ = 188}\text{.303}{\text{cm}}^{\text{3}}\\ \end{array}$

Now, the volume of the given piece

  $\begin{array}{l}{\text{V =V}}_{\text{F}}{\text{-V}}_{\text{H}}\\ \text{V = 422}\text{.777-188}\text{.303}\\ \text{V = 234}\text{.474}{\text{cm}}^{\text{3}}\end{array}$

Conclusion:

Thus, the volume of hollow machined piece is $\text{V = 234}\text{.474}{\text{cm}}^{\text{3}}$.