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.
To determine the required diameter D for a log to function as a beam carrying a uniformly distributed load, we need to consider the maximum bending stress (Fb) and the maximum shear stress (Fv) that the log can withstand.
First, let's calculate the maximum bending stress (Fb) in the log. The bending stress is given by the formula:
Fb = (M * c) / (I * e)
Where:
M is the moment applied to the log (which is the product of the uniformly distributed load and the length of the beam)
c is the distance from the neutral axis to the point of interest (which is half the diameter of the log)
I is the moment of inertia of the log (which is pi/64 * D^4, where D is the diameter of the log)
e is the modulus of elasticity of the log material (which we can assume is a constant value)
In this case, the moment (M) is equal to the product of the uniformly distributed load (W) and the length of the beam (L):
M = W * L
Given that W = 400#/ft and L = 32 feet, we can calculate the moment:
M = 400#/ft * 32 ft
Next, we need to calculate the maximum shear stress (Fv) in the log. The shear stress is given by the formula:
Fv = V / A
Where:
V is the shear force applied to the log (which is the product of the uniformly distributed load and the length of the beam divided by 2)
A is the cross-sectional area of the log (which is pi/4 * D^2, where D is the diameter of the log)
Given that W = 400#/ft and L = 32 feet, we can calculate the shear force and the cross-sectional area:
V = (W * L) / 2
A = pi/4 * D^2
Finally, we need to consider the maximum values for Fb and Fv. In this case, Fb should not exceed 1,200 psi and Fv should not exceed 100 psi.
Using the calculated values for M, c, I, e, V, and A, we can rearrange the formulas for Fb and Fv to solve for D. We can use trial and error or an iterative calculation method to find the required diameter D that satisfies both the bending stress and shear stress conditions.
Alternatively, you can use available software or online calculators specifically designed for beam design, where you can input the values and quickly get the required diameter D.