Question

What is the retraction force (in lbf) of a 4 inch diameter cylinder with 2 inch rod and pressure of 1500psi?

Answer 1

The area of the load of cylinder can be given as,

$A=\pi(d^2_2-d^2_1)$

Plug in the known values,

$\begin{gathered} A=(3.14)((4in)^2-(2in)^2) \\ =(3.14)(16in^2-4in^2) \\ =(3.14)(12in^2) \\ =37.68in^2 \end{gathered}$

The retraction force acting on the cylinder is,

$F=PA$

Substitute the known values,

$\begin{gathered} F=(1500psi)(37.68in^2)(\frac{1\text{ lbf}}{1p\text{ s}in^2}) \\ =56520\text{ lbf} \end{gathered}$

Thus, the force exerted on the cylinder is 56520 lbf.