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Homework Help: Expansion difficulty with clearance

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data

    An 80mm aluminium alloy piston fits into a cast iron cylinder with a uniform radial clearance of 0.75mm at 15C temp.the piston and cylinder reach 120C.

    1. Determine radial clearance at 120C
    2.what is diametral clearance 120C
    3.What is change in cross sectional area of piston.

    I am given coefficients of linear expansion for both metals.
    aluminium alloy: 19x10^-6/C
    iron: 10x10^-6/C

    2. Relevant equations

    I know how to use the formula for linear expansion and area expansion.

    3. The attempt at a solution

    Part 1 is what gives me trouble.
    I am not sure what radial clearance means.I think it means the gap between the piston and the iron cylinder.
    In that case, I can use the formula to find how much the piston expands.
    I need to find the expansion of the piston and the expansion of the metal cylinder and take these two answers away from 0.75mm to find the new radial clearance.

    My problem is: How do I find the expansion of the iron cylinder?

    Part 2:
    If I have found the radial clearance, then I can multiply it by 2 and that will give me the diametral clearance?

    Part 3:
    Change in cross-sectional area:

    jjust fill in the formula for this using original cross-sectional area to find new area
  2. jcsd
  3. Aug 25, 2010 #2


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    Staff Emeritus
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    Homework Helper

    If you know how to find the expansion of the aluminum piston, you use the same method (Except that you use the thermal expansion coefficient of iron, not aluminum.)
    Don't know if that is what you are asking, post back if you meant to ask something else.
  4. Aug 25, 2010 #3
    I dont know what the original thickness of the iron is, so i cant figure out by how much it expands
  5. Aug 25, 2010 #4
    Is the 0.75mm the figure i need to use for the iron?
  6. Aug 25, 2010 #5


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    Staff Emeritus
    Science Advisor
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    The iron has a radius that is 0.75 mm larger than that of the aluminum.
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