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Mechanics of Materials - Shear and Compression

  1. Dec 10, 2011 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    http://img62.imageshack.us/img62/9815/mechmatt.jpg [Broken]
    P = 235 kN
    d = 50mm (diameter of the axis)
    t = 20 mm
    t1 = 12 mm
    Safety coeffecient = 2.0
    σy = 320 MPa


    Check the axis for strength (shear and compression)

    2. Relevant equations

    http://img202.imageshack.us/img202/8345/formulas2.jpg [Broken]


    3. The attempt at a solution

    http://img714.imageshack.us/img714/210/gzira.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 11, 2011 #2

    I like Serena

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    I do not know what all your symbols mean, but your calculation for the shear stress and tensile strength is correct.

    I do see that your relevant formula mentions 0.5/0.6, while you apply 1.5/2.0.
    Is that intentional?
    TBH, I do not know what this factor is. Do you?
     
    Last edited: Dec 11, 2011
  4. Dec 14, 2011 #3

    Femme_physics

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    Sorry it took me a while to reply, I had a materials strength lab reports to give.

    I was hoping you'd tell me that :smile:

    Well, so my answer is correct? At any rate, let's interpret the symbols.


    Ts [N/mm^2] = max shearing strength in the part
    Fmax [N] – Max shearing force
    As = Area of shearing
    [T] = allowable shearing strength
    Safe factory

    Sigma/c [N/mm^2] – Max compression strength in the part
    Fmax = max compression force
    A/c = area of the compression
    [Sigma/c] [Mpa] – allowable compression strength
    safely coeffecient
     
  5. Dec 14, 2011 #4

    I like Serena

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    How did the material strengths lab report go?


    Well, I didn't study material sciences, so there are a few technical aspects that escape me...

    What is [itex]d \cdot t_{min}[/itex]?
    And your formulas appear to contain [itex]lc[/itex] instead of [itex]/c[/itex]...?
    What are [itex][\sigma_t][/itex] and [itex]\sigma_y[/itex]?
    What's the difference between shearing strength and compression strength?

    I would expect the extra factor to be some type of rule-of-thumb factor, but I have no clue for what...
     
  6. Dec 15, 2011 #5

    Femme_physics

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    Great! I copied it from the smartest guy in class (specifically said "guy" :wink: ) so it have better turned out great. I just don't like to give out reports as much as I like to solve exercises.

    I don't see d x t (min)

    As far as Ic instead of /c, I might have miscopied! Not sure whether it's "I" or "/"

    Sigma y is "yield strength"
    Sigma t is (I think) "tensile strength". There's appearing a relation between that and compression or shearing strength.
    pushpull.gif

    :smile:
     
    Last edited: Dec 15, 2011
  7. Dec 16, 2011 #6

    I like Serena

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    I'm certain it would turn out even better if you let the smartest person in class do it! :smile:


    There it is!
    http://img202.imageshack.us/img202/8345/formulas2.jpg [Broken]


    Okay... I'll just accept that for now...


    Aha!
     
    Last edited by a moderator: May 5, 2017
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