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Average sheer stress

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data
    The wooden device is pulled apart by force P, determine the axial force P applied. The average normal stress is 2 ksi. Find the average shear stress along a-a.

    http://i60.tinypic.com/2vt1h8y.png

    2. Relevant equations
    τ = V/A
    σ = P/A


    3. The attempt at a solution

    2 ksi = P/A

    A = (2)*(4)*(2) = 16

    P = 16 * 2 = 32 kip

    For average sheer stress, I attempt to do the problem but I end up with 2 ksi again, and I don't think that's right.
    τ = V/A

    (32/2)/(2*4) = 2 ksi
     
  2. jcsd
  3. May 8, 2015 #2

    SteamKing

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    You're just going around in circles with this calculation. :frown:

    If the average normal stress is 2 ksi, over which area would this be measured on the gizmo shown in the figure?

    Hint: The load P is not putting a normal axial stress on the cross section a-a. Where, then, will the axial stress be applied?
     
  4. May 8, 2015 #3
    Over the edges of the block where it hooks in with the device?
     
  5. May 8, 2015 #4

    SteamKing

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    Is the stress there a normal (i.e. perpendicular) stress or is it shear stress?

    You need to learn the difference between an axial stress and a shear stress.

    axial-stressstrain-curve-modulus-of-elasticity-2-638.jpg

    shear-stress-strain-curve-modulus-of-rigidity-100103039-3-638.jpg
     
  6. May 8, 2015 #5
    The stress would be perpendicular to where a-a is, right?

    I can't ascertain these concepts in words from my head so I am having trouble explaining what's confusing me.
     
  7. May 8, 2015 #6

    SteamKing

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    Look at the diagram of the part carefully.

    The plane a-a is oriented so that it runs parallel to the direction of the applied load P. Since a-a is parallel to a-a, the load P cannot produce a normal stress of 2 ksi there.

    However, there is another location on the part at which P can produce a normal stress. Remember, the plane at this location must be perpendicular to the load P.

    Refer to the graphic entitled What is Axial Stress? in Post #4 for reference. Compare to how the member is loaded in this graphic to how the part is loaded in the link here:

    http://i60.tinypic.com/2vt1h8y.png

    Care to guess where this location is?
     
  8. May 8, 2015 #7
    So axial stress would be the stress on the object because of the two forces pulling, it would be on the places the middle part is in contact with the object being pulled?
    If that makes sense to you.
     
  9. May 8, 2015 #8

    SteamKing

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    Yes, it would. :wink:

    Now what is the cross sectional area of the middle of the part? Is it different from the area at plane a-a?
     
  10. May 8, 2015 #9
    Plane a-a looks to be 4 x 2

    but the plane we're looking for the area doesn't appear to have a measurement to me? it would be 2 * something for one contact area. Unless we can just assume it's 1 inch also because the middle measurement is 1 inch and we assume it's equal measurement for those? Unless i'm missing something.
     
  11. May 8, 2015 #10

    SteamKing

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    It's 2" x 1" because those are the dimensions shown for the middle portion of the part. They're very clearly indicated on the sketch.
     
  12. May 8, 2015 #11
    But if we want the area of the part of the wood in contact with the part pulling, how do I assume from the diagram those are 1" because the middle is 1 inch"?
    The axial force is acting on the two parts in contact, right?
     
  13. May 8, 2015 #12

    SteamKing

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    Take it slow here.

    Yes, the axial force is acting on the jaws which hold the wood piece. However, the greatest axial stress is not going to be located at the point of contact between the jaws and the wooden piece; it's going to be where the cross sectional area of the piece normal to the force P is the smallest, namely in the middle between the two jaws. That's where the wooden piece measures 2" x 1" as shown on the picture.

    The problem statement is asking you first to find out what P is based on the average normal stress being 2 ksi. Once you have determined the value for P, then you are asked to calculate the average shear stress this force produces on plane a-a.

    There's two distinct parts to this problem, and you must solve the first part in order to solve the second part.
     
  14. May 8, 2015 #13
    Ohh I get it,I think because of double shear on this, the two edge pieces on the jaw will half the force on each, but the middle piece will have the full force so we use that for this stress? I hope that's what you meant.
     
  15. May 9, 2015 #14

    SteamKing

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    No, that's not the reason. You're jumping ahead of yourself, again, making things more compliicated than they are.

    Forget the shear and the shear stress for now, while P is determined.

    The problem statement says the in order to determine P, we should use the geometry of the wooden piece and the fact that the average normal stress is 2 ksi. Not the average shear stress, but the average normal stress. The normal stress will not occur near plane a-a or anywhere near the ends of the wooden piece where is is held by the jaws pulling on it with force P. We've established that the normal stress will occur in the long narrow piece of wood connecting the two ends which are held by the jaws.

    Look, as much as I've enjoyed explaining this problem to you, it's time for you to show some revised work toward a solution.

    First, what is the value of P? Answer this question, and then we can go on to find out what the average shear stress in plane a-a is.
     
  16. May 9, 2015 #15
    Sorry, I'm just trying to understand why...if I don't understand why then I can't actually continue with problems in the future...

    σ = P/A

    σ = 2 ksi
    A = (2 * 1) = 2 in^2

    2 ksi = P/(2 * 1)
    P = 4 kips
     
  17. May 9, 2015 #16

    SteamKing

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    This looks good.

    Now, you can continue and answer the second part of the problem, namely, what is the average shear stress on the plane a-a? Here, it's OK to consider double shear.
     
  18. May 9, 2015 #17
    So each little cube is (4*2) = 8 in^2

    so τ = V/A

    τ = (4 kips)/ (8) = 1/2 kips
    Dividing by two for double sheet 1/4 kips ?
     
  19. May 9, 2015 #18

    SteamKing

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    Since you're calculating shear stress, the units would be ...?
     
  20. May 9, 2015 #19
    Whoops, should be in ksi.
     
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