Lecture 27
Example Problem
Thermal Strain
Given:
A 4 innch diameter (a = 3.17 in2), 25 ft long standard steel steam pipe is installed at 60 degrees F. The pipe temperature increases to 220 F when in use.
Determine:
the elongation of the pipe
Solution:
First, find the equation necessary to solve for delta L due to thermal expansion.
delta L = (alpha)(delta T)(L)
Find the coefficient of thermal expansion from the lecture table for steel (alpha = 0.000 0065 in per degree F) and the temperature variation (220 - 60 = 160 degrees F). Now replace the variables in the equation above to solve for delta L.
(0.000 0065 in/F)(160F)(25 ft)(12in/ft)
Delta L (elongation) = 0.312 in
Thus, the total deformation would be 0.312 inches and the overall length would be 300.312 inches.
Given:
assume in the above example that the pipe had been anchored "rigidly" in two "immovable" concrete buttresses, so that the pipe could not elongate.
Determine:
the stress in the pipe is caused by this temperature change.
Solution:
This now becomes a stress/strain problem rather than a thermal stress problem. The key to solving this problem is realizing that the stress caused by the fixed supports is effectively the same as a pipe with an original length of 300.312 in that had been
shortened to 25 feet or 300 inches. Start with E = stress / strain and solve for stress.
stress = (E)(strain)
= (E)(delta L/L)
= (29,000,000 psi)(0.312 inches)/(300 inches)
= 30,160 psi.
this exceeds the acceptable 24 000 psi limit.......
Determine:
What force would this exert on buttresses?
Solution:
Knowing that stess is equal to the force divided by the cross-sectional area, turn this around to get: Force = (stress)(area).
(30160 psi)(3.17in2) = 95,600 poinds of force. (quite a large force!!!)
These problems should illustrate the incredible force that a variation in termperature can exert upon a structure.
Copyright © 1995, 1996, 1997 by Chris H. Luebkeman and Donald Peting