posted by Jon on .
A log of diameter D is available to be used as a beam carrying a uniformly distributed load of 400#/ft over a length of 32 feet. Determine the required diameter D necessary if Fb=1,200psi and Fv=100psi.
What are Fb and Fv? Is one of them the rupture or yield stress?
What factor of safety do you wish to use?
Fb=allowable bending stress and Fv=unit shearing stress , I just need to find the minimum diameter allowable to meet those conditions
That 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
ref: 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.