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If I know the tensile strength of a pin, how do I compute the total strength of multiple pins? That is, 2 objects are connected by multiple pins, evenly distributed, each with a height of $2mm$ and a cross sectional area of $3.33mm^2$. For some reason the answer of just multiplying the tensile strength of each pin by the number of pins, doesn't make 100% sense to me.

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The short answer is yes, it is that simple.

Think about it this way. The pins are connected to end plates of a given area. If all we're worried about is the tensile strength, we can place a one dimensional force on each end plate to put the system in pure tension. Now, split the system into a couple parts and do a force balance.

If we split it right down the middle, so that we're seeing the actual force on the top plate, and the internal force along each of the pins. These two forces have to be equal, otherwise the system would be in motion, meaning the pins have yielded or fractured.

Now, to calculate the stress in the pins, we simply divide the force applied across all the pins by the total area. Because in this case, the pins are all of even cross sectional area, the stress will be divided evenly between them. The implication of this is that none of the pins should fail before the others, and any difference in their tensile strength is due to manufacturing defects or variations, so we can just multiple the tensile strength of a single pin by the number of pins to get the overall tensile strength.

There is an important assumption here that the pins are distributed evenly on the plates. If the pins are lopsided, they're going to end up under some amount of bending, which complicates the stress calculations in them and adds to the stress that they see under the same applied load.

If we make the pins of variable cross-sectional area, then the load begins to concentrate in the larger pins, but it should still end up with the same stress in each pin because the load will distribute proportionally to the area, which is how we calculate stress anyway.

Where this gets more complex is if the pins are made of different materials or treated differently to change the tensile strength. This is essentially creating a composite material, and there we have to take into account volume fractions to calculate new yield and tensile strength numbers, but also consider which material fails first. For further reading I'd suggest chapter 3 in this MIT book (that is a 128 page pdf, beware if you're on mobile).

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    $\begingroup$ This assumes the top plate is infinitely stiff, correct? If it has similar flexibility as the pins, you wouldn't be guaranteed equal stress on different sized pins, right? (Not disagreeing with the assumption, just trying to clarify.) $\endgroup$ – Ethan48 Feb 10 '15 at 21:11
  • $\begingroup$ @Ethan48 Essentially, yes. If the plates are not significantly stiffer than the pins, you'll have to take into account their behavior in calculating how the force propagates through the system. I think that's a safe assumption though, if the total area of the pins is only a small fraction of the area of the plates. $\endgroup$ – Trevor Archibald Feb 10 '15 at 21:17
  • $\begingroup$ @Ethan48, & at Trevor, no this does not assume an infinitely stiff plate. It assumes a perfectly equal distribution of forces to all pins. Even an infinitely stiff plate, with unsymmetrical load placement (relative to the pins), will not distribute the load to all pins equally. When in doubt, make a free body diagram. $\endgroup$ – William S. Godfrey- S.E. Apr 5 '15 at 12:16
  • $\begingroup$ @WilliamS.Godfrey Sure, it assumes an infinitely stiff plate that is also constrained to a single degree of freedom. I think Trevor and I both figured that was a safe assumption based on the question. $\endgroup$ – Ethan48 Apr 6 '15 at 0:58
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    $\begingroup$ Yeah, the infinitely stiff (or even just much much stiffer than the pins) assumption might not be totally necessary, but it's a good, simple way to ignore any effects of the plate, and use it simply as a platform to transfer the load to the pins evenly. $\endgroup$ – Trevor Archibald Apr 6 '15 at 2:46

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