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Mechanics of Materials: Deformation of a Hollow Cylinder

  1. Sep 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Hollow Circular tube of Length (L) 600 mm is compressed by forces P (axially).
    Outside diameter (d2) is 75 mm.
    Inside diameter (d1) is 63 mm.
    Modulus of Elasticity (E) is 73 GPa
    Poisson's ratio (v) is 0.33.
    axial strain (ε) is 781 x 10-6

    Find shortening of tube (δ). (This was calculated to be 0.469 mm)
    Find % change in cross sectional area. (Answer is -0.081%)
    Find volume change of the tube. (Answer is -207 mm3)

    2. Relevant equations
    δ = εL
    δ = L' - L
    A = 1/4 π d2
    v = -ε'/ε (ε' is lateral strain)
    σ = Eε = P/A (σ is stress)


    3. The attempt at a solution
    Using equation 1, the tube shortens 0.469 mm.

    I figured I should calculate the lateral strain (ε') first.
    Using equation 4, ε' = 258 x 10-6

    Then, using equation 1 (and 2) laterally for d2, I get d'2 is 75.02 mm.

    Similarly using equation 1 and 2 for d1, I get d'1 is 62.98 mm. (I assumed that the walls would expand which is why d1 to d'1 decreased).

    Now, calculating the area difference I used the following equation 3:
    A = 1/4 π (d12-d12) = 1300 mm2
    A' = 1/4 π (d'12-d'12) = 1304.5 mm2

    (A' - A) / A = 0.346% *Should be -0.081%*

    I also tried to calculate Volume change, but received the incorrect answer. I believe my issue with Volume is the same reason for missing Area. Please let me know what I'm missing.
     
  2. jcsd
  3. Sep 13, 2015 #2

    billy_joule

    User Avatar
    Science Advisor

    The second answer should ring alarm bells right off the bat. Things bulge during axial compression, not contract, so we expect an increase in CSA.
    I think your working is correct.


    fig%201.jpg
     
  4. Sep 13, 2015 #3
    What was the load P?

    Chet
     
  5. Sep 14, 2015 #4
    The two diameters should increase in the compressed state. They should each increase by 0.0258 % (as you showed). The cross sectional area should increase by twice this percent, or 0.0516%. So the length decreases by 0.0781%, and the area increases by 0.0516%. So the volume decreases by 0.0781-0.0516=0.0265%. This percent of the volume translates into volume decrease of 207 mm3.

    With an increase of 0.0258% for each of the diameters, what is the new cross sectional area? With a decrease of 0.0718% in length, what is the new length? What is the new volume? What is the decrease in volume? How does this compare with the 207 mm3?

    Chet
     
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