Hello ksauce, I am glad that we were able to understand each other.
Here is how I would go about solving your problem:
First we need to find the maximum bending moments. The appropriate
formula for the 40 inch direction would be for a simple beam with two
equal concentrated loads symmetrically placed. For the 20 inch
direction it would be for simple beam with a concentrated load at the
center. From my AISC manual:
M (40 in. direction) = Pa = 2,500# x 3 in = 7,500 in#
Where P is the load and a is the distance from the end.
M (20 in. direction) = Pl/4 = (2,500# x 20 in) / 4 = 12,500 in#
So, our section modulus must be calculated using the 12,500 in# value.
Next we need a resonable allowable bending stress. The yield strength
of the material you will be using will be about 36,000 psi. Using a
safety factor of 6, we will use s(y) = 6,000 psi.
The required section modulus is given by:
S = M / s(y) = 12,500 / 6,000 = 2.08 in^3
Even though the plates will be 20 x 40 I am afraid that the load is
applied so close to the end that we would have localized bending. I am
only going to consider 6 inches of plate (3 inches on each side of the
load).
S = b(d^3 - d(i)^3) /6d = 6(1.25^3 - 1^3) / 6(1.25) = .32 in^3
Which is not near enough. Let's try 2 inch sq. tube.
S = 6(2.25^3 - 2^3) / 6(2.25) = 1.5 in^3
I would put an extra piece of sq. tube directly under the load to
prevent buckling. The section modulus for 2x2x.065 is 0.31. Our total
would then be 1.5 + 0.62 = 2.12 in^3 which should work.
You are not adding much weight by going with 2 inch rather than 1 inch
sq. tubing and of course the 1/8 plate remains the same. I would add a
single tube down the center in the 40 inch direction also, if I were
making the dolly.
I hope you were able to follow this. If you have any questions, please
ask for a clarification.
Good luck with your beefy dolly, Redhoss |