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Bolted joints loaded in shear

  1. Mar 15, 2016 #1
    I am having trouble understanding the F used in these equations. Is it supposed to be the force F shown on the left, or is it F/2 or F/4?
     

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  3. Mar 15, 2016 #2

    haruspex

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    It must surely be the F in the diagram. This is shared over two bolts each side, so I expect to see F/2 in the equations. And indeed that is what we do see.
     
  4. Mar 15, 2016 #3
    So each bolt get F/4 and thats why the bearing stress equations are

    ##\frac{\frac{F}{4}}{\frac{t}{2} d} = \frac{F}{2td}##

    ?
     
  5. Mar 16, 2016 #4

    haruspex

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    I'm saying each bolt gets F/2, but I can't comment beyond that without knowing what t and d represent.
     
  6. Mar 16, 2016 #5
    for the bearing stress equations, d is supposed to be the bolts major diameter and t is the thickness of the thinnest plate
     
  7. Mar 16, 2016 #6

    haruspex

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    Ok,then it is the same F as in the diagram. Each bolt takes half the load, F/2. To get the bearing stress, divide by the area, td.
    To get the shear stress, you have to consider that there are two shear planes for each bolt, one at each side of the central plate (so distance t apart). Each of those has a shear force F/4, and an area πd2/4.
     
  8. Mar 16, 2016 #7
    but according to the definition they gave, shouldn't t be the thickness of the thinnest plate. in the question, its also mentioned that:
    "F/2 is transmitted by each of the splice plates, but since the areas of the splice plates are half those of the center bars, the stresses associated with the plates are the same. So for stresses associated with the plates, the force and areas used will be those of the center plate"

    is the shear force a max of F/4 only at these shear planes?
     
  9. Mar 16, 2016 #8

    haruspex

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    Sorry, can't decipher the meaning of that.
    How do you mean only in that question? At the shear planes as opposed to, where else?
     
  10. Mar 16, 2016 #9
    as opposed to anywhere else between the bolt head and the nut
     
  11. Mar 16, 2016 #10

    haruspex

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    There is no other shear plane between those.
     
  12. Mar 16, 2016 #11
    so shear only exists at the shear planes? doesn't it vary between the planes like this:
    http://postimg.org/image/3z13rz8kj/
     
  13. Mar 16, 2016 #12

    haruspex

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    I'll get this right in a minute....
    There are two shear planes at each bolt, but they are not adjacent to the head or nut. The shear planes are where the bolt passes from one plate to another.
     
  14. Mar 19, 2016 #13
    for the shear equations, it says the area is supposed to be the total cross sectional area of all the bolts in the group. So since each of the shear planes has a shear force F/4, and an area πd2/4 and since there are four bolts,

    so ## \frac{\frac{F}{4}}{4 ( \pi \frac{d^2}{4})} ## = ##\frac{F}{4 \pi d^2}##

    but that doesn't agree with the equation in the image
     
  15. Mar 19, 2016 #14

    haruspex

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    You are dividing by 4 twice over.
    You can think of it in many different ways:
    - a total force of F spread over four (half) bolts with a total area of πd2
    - a force of F/2 on each of two bolts, each with a shear plane area of πd2/2
    - a force of F/4 on each of four shear planes, each with an area of πd2/4
    All lead to the same answer.
     
  16. Mar 28, 2016 #15
    thanks. also, for bearing stress in the members, what is the reason for it being ##\frac{F}{2dt}## ? In the solutions, I see ##\frac{\frac{F}{4}}{\frac{t}{2} d} = \frac{F}{2td}## , but dont know why
     
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