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So i have a little diagram to visualize:

So lets say i have a 20 mm plate connected with two M8 bolts, and i apply a 30kN force in the direction shown in picture. Question is - would the force be equally distributed between bolts - i.e., each bolt will have a 15kN of shear stress? Or each bolt will still take full stress of 30kN?

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  • $\begingroup$ if you are actually bolting things together some of the load is carried by friction between the clamped surfaces as well. For a safe/conservative design you may want to assume one bolt carries all the load though. $\endgroup$
    – agentp
    Commented Nov 28, 2017 at 21:53
  • $\begingroup$ Well it's pretty hard to calculate how much of the force there will be.. I want to do a towing hitch in my car - so i'm just taking an average 1500kg car, doubling that weight for safety measures, and i get about 30kN. I just need to calculate if two M8 bolts would hold occasional towing. Original towing hitch was actually bolted on the center of the car with two M8 bolts(bolted to frame, and then the frame was bolted to the car on both sides of the car), so i guess bolts hold. But to be on the safe side, i still need to be sure. $\endgroup$
    – waitwhat
    Commented Nov 29, 2017 at 17:48
  • $\begingroup$ You need a class II hitch. mechanics.stackexchange.com might be a better place for this question. $\endgroup$
    – agentp
    Commented Nov 29, 2017 at 19:34

3 Answers 3

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Assuming a material like cold rolled steel: The steel will yield and deform slightly at small points of very high stress , equalizing the load/stress. So the bolts will have very similar loads. This plastic deformation and cold work is what makes steel so useful or "forgiving".

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  • $\begingroup$ The obvious prerequisite is precision - the steel yielding slightly before both bolts engage in equal measure. $\endgroup$
    – SF.
    Commented Nov 29, 2017 at 15:08
  • $\begingroup$ Thanks for an answer, didn't that about the steel. $\endgroup$
    – waitwhat
    Commented Nov 29, 2017 at 17:51
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As others have mentioned, the load will be distributed between the bolts. How they will be distributed is more complicated, but it is usually assumed to be an equal division between the bolts (in this case), as SolarMike mentioned. This assumption is made easier by the cold work explained by blacksmith37.

From an analytical standpoint, all that can be inferred is that the sum of the forces acting on the bolts must be equal to the applied force of 30 kN. This equality must be satisfied.

To see why, don't look at the bolts, but at the plate. It suffers the effects of three different forces: the applied 30 kN force at the bottom (let's call it $P$), and the reactions of each of the two bolts (let's call them $R_A$ and $R_B$). If the sum of all of these forces (let's call it $F$) were not zero

$$F = P + R_A + R_B \neq 0$$

then that would imply that there is a net force applied to the plate. And as we know from Newton's Second Law of Motion, force equals mass times acceleration. If there's a non-null net force applied to the plate, then that means that the plate will need to be accelerating!

Obviously, if the applied load is too large and the bolts snap, then the applied load will no longer be balanced, there will indeed be a net force applied to the plate, and the plate will indeed accelerate in the direction of the force.

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  • $\begingroup$ Thanks for an answer. If force is mostly equally distributed, i just need to calculate if two M8 bolts would hold about 30kN. $\endgroup$
    – waitwhat
    Commented Nov 29, 2017 at 17:45
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Are the holes / bolts a machined fit? If not then one is most likely taking more stress than the other.

However, when engineers do this type of analysis they take the load and divide it equally between the bolts.

A similar result is obtained when you have one bolt and three plates ie the middle plate is sandwiched...

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  • $\begingroup$ Nah, holes will be drilled by me, so they wont be centered perfectly. So it will be likely that one bolt will take more than the other, but the stress is still reduced, so that's good. Thanks for an answer. $\endgroup$
    – waitwhat
    Commented Nov 28, 2017 at 19:02

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