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A double shear butt joint is shown below in the diagram 4 plates are held together by two bolts. The applied force is 42KN. Determine the shear stress within the joint

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$$\begin{align} a &= 2 \frac{\pi}{4} d^2 \\ &= 113.0973355 \\ \frac{42}{2} &= 21 \\ \frac {21}{\pi/4\cdot12} &= 20.05352283\text{ kN/mm}^2 \end{align}$$

Is this correct?

Kyle Anderson
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    not sure what X is in you first formula. Also looks like you may have mixed up your order of operations. Break your answer down into steps. in a double shear connection, your bolt experiences 1/2 the load of the single member at each side of the single piece. load is 42 kN /2. You know stress is a force over an area so then take the load experienced by the bolt at 1 face and divide it by the cross sectional area of the bolt. Assuming you use mm for the bolt diameter your answer will be in kN/mm^2. 1 kN/mm^2 = 1 GPa. – Forward Ed Feb 13 '20 at 19:55
  • @ForwardEd i have changed the question to what you have said. Is this correct now? New to this and trying to learn about it. I have understood the Single Shear, but the double shear and got me a little confused – Kyle Anderson Feb 14 '20 at 07:08
  • you are doubling the area of your bolt, but you should only be looking at one interface at a time. You bolt needs to have sufficient strength on either side of the single plate. you check one side then you check the other. I'd say 99% of the time one side will be equal to the other so you really only need to check the one side. – Forward Ed Feb 14 '20 at 14:31
  • So $$ a = \frac{\pi}{4} d^2 = 113.0973355$$ I have changed my answer. Is this correct now? I haven't doubled the bolt diameter. – Kyle Anderson Feb 16 '20 at 14:23
  • I get a value of 185 MPa on the bolt at each slip plane – Forward Ed Feb 23 '20 at 03:12
  • you have your order of operation screwed up. you are doing 21/pi/4*12. What you need to be doing is 21/(pi/4*12) – Forward Ed Feb 24 '20 at 00:44

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