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Shear stress of a plate

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2.
    2. Relevant equations
    σ = V / A

    3. The attempt at a solution
    The first thing I did was rearrange the formula to isolate for V.
    I got the area which (A=πr2) gave me 314.15mm2. I change that to m and then I wne ahead and calculate the force which gave me 110.11 MN

    But the answer is 549.8 kN ( they multiplied 20x25xπ to get the area)

    Not sure where I am going wrong.
     
  2. jcsd
  3. Nov 13, 2014 #2

    SteamKing

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    You have to be careful which area you use for A. When the punch punches a hole in a plate, the shear force is not applied perpendicular to the plate. It's the circular punching which shears from the rest of the plate.

    If you make a sketch looking at the edge of the plate, you can see quite clearly what the shear area must be.
     
  4. Nov 13, 2014 #3
    The problem shows me a picture and yes I do see that I use include the thickness of the plate in there but how can I get the value?
    The solutions says to multiply pi x 20 x25 but I dont understand where that formula comes from or I get the same value by multiplying 2 x pi x 10 x 25 but again not sure what formula that is
     
  5. Nov 13, 2014 #4

    SteamKing

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    It's the circumference of the punched hole multiplied by the thickness of the plate. ;)
     
  6. Nov 13, 2014 #5
    so multiplying circumference by height gives area as well :S ?
     
  7. Nov 13, 2014 #6

    SteamKing

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    Obviously. Look, shearing takes place when two surfaces want to slide over one another. The circumference of the piece being punched out slides over the circumference of the hole left in the plate. Both the plate and the punch out obviously have the same thickness, so the only other property you need to calculate the shear area is the circumference of the hole left in the plate.
     
  8. Nov 13, 2014 #7
    thank you for the explanation :)
     
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