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April 20, 2014

Posted by **Jon** on Wednesday, December 7, 2011 at 8:03am.

- Statics -
**drwls**, Wednesday, December 7, 2011 at 8:18amWhat are Fb and Fv? Is one of them the rupture or yield stress?

What factor of safety do you wish to use?

- Statics -
**Jon**, Wednesday, December 7, 2011 at 8:24amFb=allowable bending stress and Fv=unit shearing stress , I just need to find the minimum diameter allowable to meet those conditions

- Statics -
**drwls**, Wednesday, December 7, 2011 at 8:42amThat allowable shearing stress of 100 psi seems suspiciously low. There is usually a certain plane where the shearing stress is about half the tensile stress, so everything would fail in shear if the strength were that low.

Look up or calculate the location and value of the maximum bending moment, M.

For uniform loading at w weight/length, the maximum bending moment is

M = w L^2/8

I recommend using w = 33.33 lb/inch, with L in inches, if you are going to compute stresses in psi.

The maximum shear occurs at the ends, and is

V = w L/2

The maximum shear stress is

Fshear = V/(pi D^2/4)

The maximum bending stress is

Ftensile = M c/I

where c = D/2 and I = pi*D^4/64

- Statics (P.S.) -
**drwls**, Wednesday, December 7, 2011 at 8:53amref: Eshbach: Handbook of Engineering Fundamentals. College Edition, 1952.

What a great book! Mine cost me $6.40 in about 1957.

It was a pleasure to consult it again.

I am sure you will find the same formulas in many strength of materials texts and handbooks.

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