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• # Combined shear and bending on 16 dia pin

Discussion in 'The main mechanical design forum' started by moneytree1964, Aug 12, 2014.

1. ### moneytree1964New Member

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3. ### LochnagarWell-Known Member

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From the drawing you have posted, you need to assume the point of application of the 29000N force. You have assumed that it is applied at half the depth of the hole into which the shaft goes - which is not an unreasonable assumption as a first approximation. In reality, it will depend on a number of factors such as:-

1. Whether the part with the hole in it remains absolutely parallel - with the part with the pin in it.
2. The relative stiffness of the pin and the part with the hole in it.
3. Whether there is any deformation at the top of the hole when the pin makes contact with the hole, which will effect the assumed point of contact.
4. The type of fit the pin has in the hole, loose, clearance or interference.

In all probability, the bearing stress distribution in the hole, where the pin makes contact, will be parabolic, and so what you are trying to do is find the c of g of that parabola. Some people assume a triangular bearing stress distribution, and so the c of g of the triangle is 1/3 the hole depth. If it is a very deep hole, then you need to revise this "model" yet again.

It is hard for me to say, without knowing more about this problem, to say how these points above, will effect the assumed point of contact. You have assumed the point of contact as the depth of the hole divided by 2, (which is conservative), but it possibly could be the depth of the whole divided by 4. By assuming depth divided by 2, you have assumed a rectangular bearing stress distribution.

What I don't know, as you haven't provided any details on this, is how the pin is secured to its structure on the left hand side of your drawing, is it in a loose boss for example, and if it is then you need to assume a reaction point to the left of the plane "a" "b", which will change the problem yet again.

Hope this helps.