# Mechanical Loading Problem

1. Feb 14, 2010

### Ragley

Hi

I'm a product designer, with little knowledge of mechanics. I've designed a product in which there is a captive Pin 2.5mm diameter x 12mm Long, made from Half-hard 316 Stainless Steel. The Pin is totally captivated for 4.75mm at each end, leaving an unsupported length of 2.5 mm in the centre. At this point, a load is applied. I can see that bending can't take place because of the captivation at each end, so with my limited knowledge it seems that the Pin will fail in Shear, What is the maximum load it can take before failure?

Can anyone help me please?

Thanks in anticipation
Ragley

2. Feb 14, 2010

### torquil

Is this equivalent to the following?:

Put two metal slabs on the table, on on top of the other. Then drill a vertical hole through both. Insert at metal rod. Then try to slide one metal slab to the side. There would be a force that attemps to cut the rod, assuming that the slabs are much harder than the rod.

Is this the same kind of force that will be acting on each sides of the central 2.7mm section of the pin? Maybe this is one possible idealization that makes it possible to estimate the strength, because these properties must be documented somewhere. Maybe somewhere related to metal cutting? At least, in this simplification, the strength will be proportional to the cross-sectional area of the pin.

This picture might be completely wrong if the load acting on the pin is not sufficiently uniform along the whole 2.75mm. In addition, there is the danger that it gives an over-estimate for the strength.

Torquil

3. Feb 14, 2010

### Q_Goest

Hi Ragley, welcome to the board. This really should go in the mechanical engineering forum, just FYI.

Shear strength is related to tensile and yield strength as given here:
http://www.roymech.co.uk/Useful_Tables/Matter/shear_tensile.htm

However, as torquil eludes to, whether or not this is truly in shear only is doubtful. As I understand you, the pin is 2.5 mm in diam and unsupported along a length of 2.5 mm. Bending stresses will be significant, so I'd suggest looking at both bending and shear then applying Mohr's circle to get the principal stresses. If you're unsure how to determine the bending moment, you should be able to find that in most texts or better yet, use http://www.roarksformulas.com/" [Broken].

Last edited by a moderator: May 4, 2017
4. Feb 14, 2010

### Ragley

Hi Torquil

Thanks for the reply.

Effectively, your understanding is correct. The device operates in a horizontal orientation, and is quite a simple device that is conected to a fishing swivel. It is the load applied through the swivel onto the Pin that will cause it to shear. It can bend very slightly, but with both ends captivated and clamped by hydraulic pressure when assembling, the swivel is trying to tear through the central 2.5mm of the Pin. As long as the Pin is stronger than the swivel, the swivel will fail first.

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