shear stress plane

[Anonymous]

A simpel pin, OD10 has two areas of 3mm each that are subject to 1800N in total (900N each). 

another area of 3mm on each side is supported. 

If I calculate shear average of 1800N/pi*r^2, I get 22,91MPa.

I just cant figure out ; 

a) what plane in nastran or inv stress fea that is correct for shear stress in my situation? 

b) what is acceptable stress of a pin 10mm od, s235 steel? 0,7x tensile?

c) Why dont i get 22,91 in stress in nastran? or aprox.

[John_Holtz]

Hi @Anonymous 

1. Based on your orientation, you want to look at shear XZ. See my image below.
2. You need to determine what factor of safety is required and the shear strength of the material.
3. The real situation is much more complex than the simple formula of force/area, but the results are in that range. The variation is cause by bending in the pin and the discontinuity of the constraint and load. Also, your mesh looks to be too coarse. (Even my mesh is probably too coarse.)

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[Roelof.Feijen]

Small addition to point c.

The effect of the Poisson ration is also not taken into account in your hand calculations, FEA does, unless it’s value is zero of course.