**Actions: **
Add Comment

I am building a cart out of rectangular steel tubing that is 6 feet long. It will rest on casters on each end. It will carry a load of about 600 lbs concentrated within 22 inches of one of the ends. I am using the formula D = P x l^3 / 48EI to calculate the deflection which assumes the load is concentrated in the middle however my load is concentrated within 22 inches of the end. How do I adjust the calculation for this.

Hi, kevmc:

For clarity, is the 600lb load concentrated within 22 inches of only one of the ends, or is it equally concentrated between points at that distance from both ends?

Either way the deflection will be less than what would occur with loading concentrated at the midpoint of the beam.

regards, mathtalk

The load is concentrated from one of the ends to 22 inches from that end. There is very little load if any from 22 inches to the other end

here is a simple drawing.

LLLLL

XXXXXXXXXXXXXXXXXXXX

L is the load and X is the length of the steel beem. I should also clarify that there will be two lengths of steel tube so I think the load for each beam wold only be 300 LBS. I also posted a sketchup drawing at http://files.me.com/kevcoleman/3t6nt1 if you want to look at what I am building. The load, a table saw will rest on the four triangular gussets.

You are right that the load for each beam would only be 300 lbs. when the cart is level. If the cart were tipped toward one beam (or the other), the load would increase on the lower side (and decrease on the upper side). So for engineering purposes you might want to build in a modest margin for the effects of tipping (though if the angle were to reach 45 degrees you'll probably have bigger problems than beam deflection!).

Similarly you can modify you diagram to indicate the loading being concentrated at the four "corner" gussets:

L L

XXXXXXXXXXXXXXXXXXXX

with a load of 150 lbs. at each corner.

Now the leftmost load is nearly inline with the beam's support, so it will contribute almost no deflection. Again we probably want a modest overestimate to account for angling the table downward toward the interior gusset (if the table saw cart were being moved), but in most practical operation one wants a table saw perfectly leveled.

Have you looked at the demo download that redhoss suggested?

[DTWARE Engineering Software: DTBeam version 1.0]

http://dtware.com/

Redhoss (formerly redhoss-ga) is our resident expert on beams and engineering, so I recommend you invite him to post that as the official answer. If you're looking for more than just the computed answer, I'll be happy to add a few words about the history of mathematics involved.

regards, mathtalk

I will plug your numbers into the beam program and give you the resulting deflection if you will tell us the wall thickness of the 3x2 tubing.

That explanation helps alot. redhoss, I dont necessarily need a 3 X2 tube, in fact I wouldnt mind a slighly smaller one so long as it doesnt deflect much more than probably 1/16 at the center.

Also, the thickness is not a contraint. from the looks of available tubing, it is probably 3/16 or could be 1/4. I originally selected 3X2 tubing in 3/16 inch thickness because using the center loaded formula showed that would result in a deflection of around 1/16 inch which is acceptable to me. I am thinking however based on your guys feedback that given the load is at the end, I may be vastly overestimating the size of tubing I will need to limit the deflection.

I did try to use the beam software mentioned but I am not an engineer so I had a hard time figuring out all the inputs.

I ran 3x2x.125 D = .029 in.

2.5x 1.5x .10 D = .065 in.

Something to think about is the cost-benefit of using an over-engineered tubing. One may have an occasional heavy piece to cut on the table saw cart. Mahogany is not (despite its reputation) the densest of commercially valuable woods, but it makes something of a benchmark. A cubic meter of mahogany weighs something like 1200 lbs. You'd probably never cut something that big (and heavy), but you might want to give the weight of working pieces a thought.

That's on the benefit side. The other question is how much less in price or work time will you save by using lighter steel tubing? I'd guess you can find a 12 foot or longer piece of the 3" x 2" x 3/16" steel tubing you considered for $50 or less. So there may be limited savings to going with a lighter gauge (or weaker cross-section).

regards, mathtalk

**Actions: **
Add Comment

Frequently Asked Questions | Terms & Conditions | Disclaimer | Privacy Policy | Contact Us | Spread the word!

Fri 23 Jun 2017 - 7:07 pm UTC - © 2017 Uclue Ltd

redhoss

Researcher

15 Jul 2010 13:15 UTCThu 15 Jul 2010 - 1:15 pm UTC

I downloaded the demo version of a beam program from dtware.com. It is easy to use and will solve your problem.