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Mechanics - stress on rod of differing thickness and change in length

  1. Oct 21, 2011 #1
    1. The problem statement, all variables and given/known data

    a)where the greatest stresses are
    b) overall change in length


    That is meant to be a hollow cylinder on the bottom section inbetween the dotted lines

    E = 200 GPa

    2. Relevant equations

    σ = Fn / A

    ε = dl / lo = σ / E

    3. The attempt at a solution

    The book gives the answer as the top of the rod having the greatest stress = 7.64 MPa

    So working back from that answer the force on the top of the rod = 15 kN

    I assumed this was because -20+10-5 = -15 kN, but when I used -15 kN for the rest of the sections the stress was much higher due to the smaller area.

    ie. 15e3/[(pi 25 mm^2)] = 7.64 MPa (top section)

    15e3/[(pi 20 mm^2)] = 11.94 MPa (middle section)

    15e3/ [(pi 10mm^2)] = 47.75 MPa (bottom section)

    So my problem is that I don't know whether to include all the forces when calculating strain for each section or not..

    Last edited: Oct 21, 2011
  2. jcsd
  3. Oct 21, 2011 #2


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    mh1985: You did not show constraints on the rod yet. Therefore, we do not know where nor how the rod is attached to the outside world. By the way, regarding units, please see the following. You can edit your post by hitting the Edit button.

    1. Always leave a space between a numeric value and its following unit symbol. E.g., 10 kN, not 10kN. See the international standard for writing units (ISO 31-0).

    2. The unit symbol for kilonewton is spelled kN, not KN. K means kelvin; k means kilo. Always use correct capitalization of units. See NIST for the correct spelling of any unit symbol.

    3. MN/m^2 is called MPa. Always use the correct, special name for a unit. E.g., 7.64 MPa, not 7.64 MN/m^2. See the above links in item 1.

    4. Parentheses must be used to show grouping for division. E.g., 15e3/[pi*(625 mm^2)], not 15e3/pi 625 mm.
  4. Oct 21, 2011 #3

    thanks for pointing that out, I've made the changes now
    Last edited: Oct 21, 2011
  5. Oct 21, 2011 #4


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    In order to calculate the proper axial stress, you must draw a free body diagram for the section of the rod where you want to calculate stress. You cannot add up all of the applied loads and say that the sum of the loads acts anywhere along the length of the rod.
  6. Oct 21, 2011 #5


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    mh1985: The unit symbol for megapascal is spelled MPa, not Mpa. And you still need a space in 200 GPa.

    Also, the forum software currently has a bug and is erroneously changing some characters if you quote my text. Therefore, could you delete your quotation, in post 3, of my post, because it erroneously changed some spaces to asterisk, and could confuse others on how to write units correctly. Or else change the seven extraneous asterisks to spaces. There should be only one asterisk in my post, not eight asterisks. Thanks.
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