Question

The stress in the material of a pipe subject to internal pressure varies jointly with the internal pressure and the internal diameter of the pipe and inversely with the thickness of the pipe. The stress is 100 pounds per square inch when the diameter is 5​ inches, the thickness is 0.75​ inch, and the internal pressure is 25 pounds per square inch. Find the stress when the internal pressure is 15 pounds per square inch if the diameter is 2 inches and the thickness is 0.65 inch.

Answer 1

Answer:

The answer is 27.69 [psi]

Explanation:

(Barinly´s editor seem to not be working properly so please see attachment)

We know that:

• $σ≈\frac{P_{in} D_{in}}{t}$

Where:

• σ => Stress in the material
• Pin => Internal Pressure
• Din => Internal diameter
• t => Thickness

Since we know that thre Stress is directly proportional to Pin and Din but inversely proportional to thickness, we need to add a proportionality constant "k" to our equation to make it complete:

• $σ=\frac{kP_{in} D_{in}}{t}$

We also know that the Stress is 100 [psi] when Pin is 25 [psi], Din is 5 [in] and thickness is .75 [in], using this values we can solve for k:

• $k=\frac{σt}{P_{in}D_{in} }$

thus k = .6

Now all we need to know is use the same equation but using the new parameters: Pin = 15 [psi], Din = 2 [in] and t= .65 [in]

• $σ2 = \frac{(.6)(15)(2)}{.65}$

The stress is 27.69 [psi]