# Finding force load paths and reaction forces from ground

1. Jul 12, 2013

### engineer365

I am looking to find the reaction forces on the bottom of the rectangular shape I have included (sorry for the crude diagram). I have a moment on the top right beam (390600 lb/in) and a distributed load of 160 lb/in across the entire surface of the top of the rectangular prism. I am looking for the reaction forces coming from the bottom of the rectangle to size the proper supports. I am not sure where to make the section cuts and how to include the distributed load on top of the top surface along with the moment on the top right side which comes from an overhanging beam. Any input is greatly appreciated.

Thanks

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2. Jul 13, 2013

### Simon Bridge

Looks like a 40' box (square cross-section) that continues another 10' as an L-beam.

https://www.physicsforums.com/attachment.php?attachmentid=60233&d=1373639416

... is the box hollow? (i.e. a square tube with two sides cut away in the last 10')
... how many supports, what kind, and where are they supposed to go?
... is the moment about the indicated line (i.e. end of the L-beam, about the end-horizontal)?
... will the pivot for the moment also act as a support?

Look for the center of mass.
A distributed load can be replaced by a single load having equivalent effect - shifts effective center of mass.

A support under the com will stop it from falling. Supports off center mean that the com will produce a torque about the support and you'll need another one.

A support on the far left will have to pull to counter the moment.

3. Jul 13, 2013

### engineer365

First off, I apologize the box is in inches, I somehow over saw that when I was making the sketch. The entire box is only square tubing which I am assuming to be point beams. There is a flat plate along the top in which the load is distributed. The supports will be casters so essentially rollers not pinned supports. The moment comes from a another beam attached on that indicated line directly in the middle of the horizontal line on the top. There will be a total of four supports on the corners of the entire box so there wont be one on the bottom where the beam meets 3/4 through the 50" beam. I also forgot to label the length of the bottom beam which is 38". Thank you and any other questions I am glad to answer in order to figure out my dilemma.

4. Jul 13, 2013

### Simon Bridge

ζ
The actual units don't matter to the analysis ... I take it English is not your first language? I suspect the term you are looking for is "overlooked" ... "oversaw" means that you were person in charge of others who did the actual drawing.

Ah - each line represents a beam made out of square tubing - uniform crossection? Got it.
So it's a frame-box, with a kinda frame-shelf hanging off the right-hand end?

So we are looking at some sort of mobile platform?

You need to be clear about this: moments should be drawn about the axis of rotation. If it is just that there is a weight (from the attached beam) that acts there, then it is a load.

Then there will be moments about the left and right ends of the box, about the castor axles.

You were roughly to scale then :)
I am not unclear what the dilemma is - the procedure is still the same: work out the center of mass and the center for the distributed load to figure the reaction forces at each corner.

The wheels on the right will need to support more load than the wheels to the left, and the far wheels will have to support a little more than the near ones.

5. Jul 13, 2013

### Simon Bridge

Something like this:

... only you don't have the BB' and B'E' beams.
I've put cones for the position of castors and a wire mesh for the load surface.
The red parts are behind the blue parts. (JIC)

For your purpose you'll want pretty heavy-duty castors - this should not be too difficult.

Aside: it does not look like you have enough rigidity in the frame - it will flex and wobble lots even with angle bracing, but that's not your question.. Have you built this yet?

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6. Jul 14, 2013

### nvn

There is no wheel at point E', according to post 3. That wheel is instead located at point F'.

engineer365: Your given problem is statically indeterminate. There is no place you can place a section cut to solve it using statics.

However, to simplify the problem, we could say reaction force Rd' = 0. This is similar to saying the wheel at point D' is almost lifting off the ground. Then, the problem becomes statically determinate, and gives the following vertical reaction forces (lbf) at the wheels at points D, D', E, and F'. Notice, your uniformly-distributed load is assumed to act over a length of a2 = 50 (the entire table top); if this assumption (P1) is wrong, let us know.

a1 = 38, a2 = 50, b1 = 20, h1 = 17.25
P1 = 160*a2 = 160*50 = 8000

Rd = -6279
Rd' = 0
Re = 10280
Rf' = 4000​

Notice, reaction force Rd is negative. This means your wheel at point D is lifting off the ground. Therefore, your table is currently unstable, meaning it will tip over.

Last edited: Jul 14, 2013
7. Jul 14, 2013

### Simon Bridge

There's also no point F ;) Lets see: the relevant part of post #3 was:
... Ahhh ... I confused "won't" and "entire box" huh?
No worries - easy enough to shift the cone :)

I agree there is not enough information to solve for the reaction forces just yet ... and it is not best resolved as a statics problem.

We know:
... this suggests a uniform load of 160lbs/in between AA' and BB' - it's com is right above the center of the box (being the volume ABED to A'B'E'D') ... so that won't destabilize the cart by itself.

The moment is problematic - the center of rotation is asserted to be CC' but I cannot think of a (sensible) way to build a cart so that will be the case ... what: is there a hinge there and something is pulling upwards from the other end of the cart? - anyway, the units are wrong.
I'd want to know where the moment comes from in more detail.
... see, that does not sound like something that produces a moment about CC'.
It sounds like a lever or an additional weight... either would more likely create a moment about the diagonal: EF' or something.

At face value:
160lbs/in x 38"= 6080lbs(force) ... which is 231040lbs-in moment about CC' ... not enough, by itself, to balance the given moment there (if we just take the units as a typo). The unbalanced moment looks like it will lift the cart off it's wheels, pivoting it about CC'.

Maybe I'm making a mistake and "the entire top of the box" means the whole 50" from AA' to CC'? That would make a bigger force closer to the pivot.
160lbs/in x 50" = 8000lbs ... which is 200000lbs-in moment. Not big enough either.

Supports would have to be placed to the right of F' to counter the unbalanced moment.
(Or tie down the other end or something.)

... and there is no mention of the weight of the cart itself.

Since the whole point is to size the proper supports, I suspect that OP has reason to believe that the frame will not lift off the ground like that: perhaps it has already been built and it is a matter of choosing wheels that are strong enough(?) ... we don't actually know.

Aside:
@engineer365: how to handle distributed loads (JIC):
https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=st&chap_sec=04.5&page=theory

Last edited: Jul 14, 2013
8. Jul 14, 2013

### nvn

Yes, I made a typographic mistake in post 6. Each F instead should be F'. Therefore, I corrected this now in post 6. Thanks for pointing that out.

If one attaches a cantilever beam to beam CC', as engineer365 described, then it will indeed apply a moment M1 to the midspan of CC', as engineer365 has drawn. M1 is given in post 1.

The cart self weight is perhaps included in the uniformly-distributed load, w = 160. If not, engineer365 could increase w slightly, to include it, if nonnegligible.

Perhaps w extends a distance a1, instead of a2. In post 6, I assume it extends a distance a2; and I asked engineer365 to let us know if this assumption is wrong.

Last edited: Jul 14, 2013
9. Jul 15, 2013

### Simon Bridge

Well said: in order -

I didn't think it was actually going to confuse anyone :)

To generate the moment about one end, the cantilever beam must be fixed at that end ... doesn't it?
I'm having trouble fixing a beam to the drawing in such a way that it will provide a moment where CC' is a pivot - I'd also need some way of fixing CC' in place wouldn't I?
OP can settle this one.

That's what I thought at first - it would be an odd way to do it though.
The "self-weight" is not uniform over the long axis of the cart, nor across the middle axis, but the distributed load is both ... you'd have to work out the total weight then divide it by the appropriate length that also keeps the com consistent.
Note: if the distributed load were provided by sheet steel - how thick is the sheet?

Anyway - there are things we need OP to tell us before knowing how to continue.
Engineer365: you got your ears on goodbuddy?

10. Jul 16, 2013

### nvn

Agreed. engineer365, are you sure your moment value given in post 1 is correct? Why are the moment units wrong in post 1? I.e., why are the units not lbf*inch or N*mm?