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Homework Help: Question on deformations of members under axial loading

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A 250-mm long aluminum tube of 36mm outer diameter and 28mm inner diameter may be closed at both ends by means of single threaded screw on covers of 1.5mm pitch. With one cover screwed on tight, a solid brass rod of 25mm diameter is placed inside the tube and the second cover is screwed on. Since the rod is slightly longer than the tube it is observed that the cover must be forced against the rod by rotating it one-quarter turn before it can be tightly closed. Determine the deformation of the rod and of the tube.

    2. Relevant equationsI have no problem calculating the elongations of each part, but wonder in the end why they add up to 3.75E-4 (the total deformation of the system as found from the threaded screw) when I thought the outer tube is in tension and the inner rod is in compression. Wouldn't this mean the total deformation of the system would be the difference between the two values? Our professor says no, but I am confused as to why.

    Any help is appreciated/if you have more info as to the solution let me know.

    3. The attempt at a solution

    With other given values I calculated delta rod=0.1325mm and delta tube=0.1325, which adds up to delta total=3.75E-4 (found by multiplying .25 turn* the pitch of 1.5mm)
  2. jcsd
  3. Jun 26, 2010 #2
    Can you explain your statement: "delta rod=0.1325mm and delta tube=0.1325, which adds up to delta total=3.75E-4"? The final length of both the tube and the rod is 250.1325 mm? Or not?
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