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the stress in the material of a pipe subject to internal pressure varies jointly with the pressure and the internal pressure and the pressure and the internal diameter of the pipe and inversely with the thickness of the pipe. The stress is 100 lbs/ per in^2 when the diameter is 5 inches, thickness is .75 in, and the internal pressure is 25 lbs/ in^2. find the stress when the internal pressure is 65lbs inches^2 if the diameter is 9 inches and the thickness is .8 inch

Answer :

We are given that the stress on a pipe varies directly with the internal pressure and internal diameter and inversely with the thickness. This means that the relationship between the variables is as follows:

[tex]S=k\frac{PD}{e}[/tex]

Where:

[tex]\begin{gathered} S=\text{ stress} \\ P=\text{ internal pressure} \\ D=\text{ internal diameter} \\ e=\text{ thickness} \\ k=\text{ constant of proportionality} \end{gathered}[/tex]

Now, we determine the value of "k" using the data provided:

[tex]\begin{gathered} S=100psi \\ D=5in \\ e=0.75in \\ P=25psi \end{gathered}[/tex]

Now, we plug in the data:

[tex]100psi=k(\frac{(25psi)(5in)}{0.75in})[/tex]

Now, we solve for "k". First, we solve the operations:

[tex]100=k(166.6)[/tex]

Now, we divide both sides by 166.6:

[tex]\begin{gathered} \frac{100}{166.6}=k \\ \\ 0.6=k \end{gathered}[/tex]

Therefore, the function is:

[tex]S=(0.6)\frac{PD}{e}[/tex]

Now, we determine the value of "S" for the following data:

[tex]\begin{gathered} P=65psi \\ D=9in \\ e=0.8in \end{gathered}[/tex]

Now, we plug in the data:

[tex]S=(0.6)(\frac{(65psi)(9in)}{0.8in})[/tex]

Solving the operations:

[tex]S=438.75psi[/tex]

Therefore, the stress is 438.750 pounds per square inch.

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