 View Question
Q: How to design beefy dolly. ( Answered ,   1 Comment ) Question
 Subject: How to design beefy dolly. Category: Miscellaneous Asked by: ksauce-ga List Price: \$50.00 Posted: 10 Feb 2006 10:36 PST Expires: 12 Mar 2006 10:36 PST Question ID: 444212
 ```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.``` 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.``` Clarification of Question by ksauce-ga on 10 Feb 2006 12:13 PST `Yes, you are correct in describing the diagram.` Subject: Re: How to design beefy dolly. Answered By: redhoss-ga on 10 Feb 2006 20:35 PST Rated: ```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```
 ksauce-ga rated this answer: `Exactly what I needed to know. Cheers!` `I thought by 'beefy dolly', you meant a chubby Barbie!` 