Question 1071643: 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 85 pounds per square inch if the diameter is 9 inches and the thickness is 0.80 inch.
I am very unsure where this falls under or what to do first?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let P = internal pressuer
let D = internal diameter
let T = thickness
let S = stress
let K = constant of variation
variation formula would be:
S = K * P * D / T
you need to solve for K based on what you know and then you can use K to solve for what you don't know.
you know that:
S = 100 pounds per square inch when P = 25 pounds per square inch and D = 5 inches and T = .75 inches.
the variation formula of S = K * P * D / T becomes:
100 = K * 25 * 5 / .75
solve for K to get:
K = (100 * .75) / (25 * 5) = .6
now that you know the value of K, you can solve your problem.
you are given that:
P = 85
D = 9
T = .8
you know that K = .6.
since it is the constant of variation, it never changes, once it is originally determined for the particular problem on hand.
formula of S = K * P * D / T becomes:
S = .6 * 85 * 9 / .8
solve for S to get S = 573.75 pounds per square inch.
here's a reference on variation type problems and how to solve them.
http://www.purplemath.com/modules/variatn.htm
|
|
|
