Question

What is the retraction force (in lbf) of a 24 inch diameter cylinder with a 12 inch rod and pressure of 1500 psi?

Answer 1

ANSWER:

1356480 lbf

STEP-BY-STEP EXPLANATION:

We have the formula to calculate the force is the following:

$\begin{gathered} F=P\cdot A \\ P=\text{ pressure = 1500 psi} \\ A=\text{area of cylinder} \\ F=\text{ retraction force} \end{gathered}$

First we calculate the value of the area, like this:

$\begin{gathered} A=2\cdot\pi\cdot r\cdot h \\ r=\frac{d}{2}=\frac{24}{2}=12 \\ \text{ replacing} \\ A=2\cdot3.14\cdot12\cdot12 \\ A=904.32in^2 \end{gathered}$

We replace and calculate the force:

$\begin{gathered} \text{psi = }\frac{\text{lbf}}{in^2} \\ F=1500\cdot904.32 \\ F=1356480\text{ lbf} \end{gathered}$