Strength of Materials - Ramps

1. Dec 14, 2008

oldgit

Hi

I wonder if you could help a really thick oldie with a problem, I found this:-

I need to build a set of ramps for my dump trailer. The ramps will be
6' long. I have and want to use 1" x 3" 11 gauge rectangular steel. If
I were to take one 6' piece of this material, stand it on its'
narrowest edge, support it at both ends and place a load at its'
center how much weight will it support before failing. A little
deflection is allright just so it does not permanently bend.

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Hello usajohnson, I think I can help you. I searched for properties of
1" x 3" 11 gauge rectangular steel tubing, but that is an odd size. We
will have to calculate the section modulus (excluding corner radius):

S = bd^3 - b1d1^3/6d

b = 1"
d = 3"
b1 = 1 - 2x0.091 = 0.818
d1 = 3 - 2x0.091 = 2.818

S = [(1 x 3^3) - (0.818 x 2.818^3)] / (6 x 3) = 0.483 in^3

M (maximum bending moment) = [P (point load) x l (length)] / 4

Solving for P:

P = 4M/l

M = s x S
Where:
s (allowable bending stress) = .55 x yield strength of steel
To be conservative we will assume that the steel you have is 30,000 psi

M = .55 x 30,000 x 0.483 = 7,969 in-lb

P = 4 x 7,969 / 72 in = 442#

So, you can say that the tubing will safely support at least 442#. The
tubing you have may actually be as high as 50,000 psi yield strength,
but we don't really know.

Please ask for a clarification if there is any of this that you don't understand.

Good luck with your project, Redhoss

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and was hoping you could help me understand! I'm trying to calculate for the thickness of steel plate, not box section! If I remove the -b1d1 from S to leave:
S = bd^3/6d
Is that OK? Or am I way off? (Please don't quote 'Integration math', I'm just too thick, never could get the hang of it!)
Any help would be much appreciated

Oldgit

Last edited: Dec 14, 2008
2. Dec 15, 2008

PhanthomJay

I am assuming, as Redhoss did, that you have a 1" wide x3" high x6' long piece of rectangular hollow tubing that has a wall thickness of 11 gauge, (or 0.091"). If that is a correct assumption, the calcs look good, although I didn't check the math. Are you saying that this is an incorrect assumption? Or are you worried about local failure of the top wall of the tube under the point load?

Last edited: Dec 15, 2008
3. Dec 15, 2008

mathmate

If the tubing is not a closed section, even if there is a slit with overlap, it will not work as calculated, since it could split open.
If it is an extruded closed tubing, the calculation assumptions should be OK.

4. Dec 16, 2008

oldgit

Hi Phanthom J, and Mathmate

Thanks for the reply, I'm afraid that my post has been misunderstood! The stuff in red is a post I found that used maths that was more to my level of understanding, but is for box tube section! I intended to use it to calculate for flat plate and was wondering if the removal of part of the equation ie:-

S = bd^3 - b1d1^3/6d (remove - b1d1^3) to leave

S = bd^3/6d

is the correct interpretation for flat plate, with of course the rest of the equation as is!

I'm trying to calculate the thickness of mild steel plate 12" x 12" x ? for a point load of 20tons. A friend has asked me to build them an olive press and I'd intended to use a 20Ton hydraulic bottle jack to press the olives between two 12" square plates! The olive paste would be in layers between a stack of rigid plastic squares (12" square), in turn between the steel plates! Any distortion of the steel plates is libel to fire the plastic sheets from the press causing injury to my friend, therefore the plates need to be thick enough not to bend under the full load of 20Tons!

I hope you can follow my rambling!

Thanks in advance for any reply's, but please keep the maths simple or my tiny mind my implode!

Oldgit

5. Dec 16, 2008

mathmate

I am sorry to disappoint you, but anything will bend under a load of 20 tons (or any load). It's a matter of how much deflection that can be tolerated.