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oldgit

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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 all right just so it does not permanently bend.*******************************************************************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 - b1d1^3/6db = 1"

d = 3"

b1 = 1 - 2x0.091 = 0.818

d1 = 3 - 2x0.091 = 2.818S = [(1 x 3^3) - (0.818 x 2.818^3)] / (6 x 3) = 0.483 in^3M (maximum bending moment) = [P (point load) x l (length)] / 4Solving for P:P = 4M/lM = 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 psiM = .55 x 30,000 x 0.483 = 7,969 in-lbP = 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

******************************************************************

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

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 all right just so it does not permanently bend.*******************************************************************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 - b1d1^3/6db = 1"

d = 3"

b1 = 1 - 2x0.091 = 0.818

d1 = 3 - 2x0.091 = 2.818S = [(1 x 3^3) - (0.818 x 2.818^3)] / (6 x 3) = 0.483 in^3M (maximum bending moment) = [P (point load) x l (length)] / 4Solving for P:P = 4M/lM = 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 psiM = .55 x 30,000 x 0.483 = 7,969 in-lbP = 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

******************************************************************

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

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