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Tensile stress in perforated steel plate

  1. Apr 24, 2017 #1
    1. The problem statement, all variables and given/known data

    Two lengths of steel plate 100mm wide and 15mm thick are riveted together by a single 18mm diameter rivet. If the joint carries a tensile load of 8kN calculate:
    a) The shear stress in the rivet
    b) The tensile stress in the perforated plate

    2. Relevant equations

    stress = force / cross sectional area

    3. The attempt at a solution

    a) stress = force / cross sectional area
    = 8x10^3 / pi/4(0.018)^2
    = 31.43MPa

    This is the answer given in the textbook so I am satisfied it is correct.

    b) This is where I am stuck.
    The given answer is 6.51MPa.

    I thought maybe if I use 100mmX15mm (1500mm) as the area for the plate and subtract the area of the 18mm diameter rivet (pi/4(0.018)^2) I could divide the 8kN force (although this is the tensile load of the joint) by this area and get the stress for the plate.

    However if I do the above:

    8x10^3 / [(1.5x10^-3)-pi/4(0.018)^2]
    = 6.42MPa

    Where am I going wrong?
  2. jcsd
  3. Apr 24, 2017 #2


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    Staff: Mentor

    Those areas are not oriented in the same plane. One is perpendicular to the plate surface, the other parallel to the plate surface.

    What's the minimum cross sectional area of the plate? Note that would be the area of the section (so a slice through contiguous material).
  4. Apr 24, 2017 #3
  5. Apr 24, 2017 #4


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    Staff: Mentor

    Nope. Suppose you make your cross section through where the rivet is. How much plate material is sliced?
  6. Apr 24, 2017 #5
    Um if each plate is 100mm wide and 15mm thick then I would say 100mmX15mm - 2X15mmX18mm for the hole
    2(100x15)-(2)(15)(18) = 2460mm^2

    Not sure if I am visualizing this correctly..
  7. Apr 24, 2017 #6


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    Staff: Mentor

    Not sure why you're multiplying by two. You want to deal with only one plate.
  8. Apr 24, 2017 #7
    Okay so 100x15mm - 15x18mm = 1230mm^2

    Then stress = force / cross sectional area
    = 8x10^3 / 1.23x10^-3
    = 6.5MPa

    :woot: thank you
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