I am designing a wheeled dolly to support a load that rests on two
points (the edge of a steel square tube, so approximately 1/8" by 3").
The load is approximately 5000 pounds total. The dollies measure 1'8"
x 3'4" and the load points are on 34" centers. Because this dolly
needs to be easy to move, I am trying to keep it light. As such, I was
thinking of using a stressed-skin design utilizing a 1"x1" steel box
tube as the frame, and skinning it on both sides with plate steel.
Finally, on the side with the load points, a piece of plywood to help
prevent the load from sliding. My questions are: how do I calculate
the required gauge thickness for the box tube and the required
thickness for the plates? Ideally, I would like to use 18 or 16 gauge
tubing for the frame and 1/8" plate for the skins, but am not sure if
this will support the load.

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Request for Question Clarification by redhoss-ga on 10 Feb 2006 11:39 PST

I can help you with this if I can draw a sketch. I have tried and here
is what I think you are describing. I drew a 20" x 40" rectangle with
two point loads of 2,500# each located 3" from the 40" ends and
centered in the 20" direction. I assume the wheels are on the extreme
corners. Is this getting close.

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Clarification of Question by ksauce-ga on 10 Feb 2006 12:13 PST

Yes, you are correct in describing the diagram.

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

Exactly what I needed to know. Cheers!

I thought by 'beefy dolly', you meant a chubby Barbie!