Question

What is the retraction force (in lbf) of an 8 inch diameter cylinder with a 6 inch rod and pressure of 600psi?

Answer 1

We are asked to determine the retraction force on a cylinder. A diagram of the problem is the following:

To determine the retraction force we will use the fact that the pressure is defined as:

$P=\frac{F}{A}$

Where:

$\begin{gathered} P=\text{ pressure} \\ F=\text{ Force} \\ A=\text{ }Area \end{gathered}$

To solve for the force we will multiply both sides by "A":

$PA=F$

The area is given by the following:

We have the circle of the cylinder and the circle formed by the rod. Therefore, the area in contact with the pressure is equivalent to subtracting the area of the 6 inches circle from the area of the 8 inches circle like this:

$A=A_8-A_6$

The area of a circle is given by:

$A=\pi\frac{D^2}{4}$

Where "D" is the diameter. Replacing in the formula for the area;

$A=\frac{\pi D^2_8}{4}-\frac{\pi D^2_6}{4}$

Replacing the diameters:

$A=\frac{\pi(8in)^2}{4}-\frac{\pi(6in)^2^{}_{}}{4}$

Solving the operations:

$A=22in^2$

Replacing in the formula for the force we get:

$(300\frac{lb_f}{in^2})(22in^2)=F$

Solving the operations:

$6600lb_f=F$

Therefore, the retraction force is 6600 lbf.