Evaluating the capacity of steel plate

  • Thread starter Fugro
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  • #1
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Hello All.

I'm trying to evaluate a design based on first principles.

I have 2 trusses made from RHS (simple A frame) between which I wish to tech-screw a sheet of 2.9 mm galvanised sheet.

Loading will primarily be from material (i.e. soil to a max depth of 400 mm) being placed on the steel sheet.

I intend to analyse this setup like a beam - assume a UDL over the span, calculate reactions provided by the trusses, determine shear and bending moment diagrams.

at the point of maximum bending moment, i will evaluate the stress (σ=My/I), and hope that it is less than the yield stress of the sheet.

Am I on the right track? Does analysing steel sheets have any special rules or idiosyncracies?

Any feedback would be greatly appreciated.
 

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  • #2
nvn
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Fugro: I am currently having some trouble understanding your structure configuration, free-body diagram (FBD), and boundary conditions (BCs). Would you be able to post dimensioned diagrams?
 
  • #3
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there you go, hopefully this helps.

My steel sheet is to be assumed fully loaded (i.e a uniform pressure on it from the soil).

Is it reasonable for me to treat this steel sheet member as a simply supported beam, and evaluate it like that? (i.e. max bending moment in centre, then evaluate as σ=M/Z)

I can tell that it would not be a good structure from experience (and it looks flimsy and will deflect too much) so will need some cross members going from truss to truss. I just want to prove that this is the case but don't know if these are reasonable assumptions.

ρεαℂε out
 

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  • #4
nvn
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Fugro: Yes, you can conservatively analyse your steel sheet as a simply-supported beam. Do you know the density of your soil? I think saturated soil density can go up to about rho = 1920 kg/m^3. In your case, for a soil depth of d = 400 mm, and a strip (beam) width of b = 100 mm, I think this would be a UDL of w = rho*g*cos(alpha)*d*b = (1920 kg/m^3)(9.81 m/s^2)*cos(6.843 deg)*[m^3/(1000 mm)^3](400 mm)(100 mm) = 0.7480 N/mm. Do you know the allowable tensile stress, Sta, of your steel? Is it Sta = 160 MPa?
 
  • #5
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Thank you nvn.

I shall procede with this analysis, using Sta of 150 MPa - thank you for your help.
 
  • #6
nvn
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Fugro: Good Sta value. An idiosyncrasy for thin sheet is nonlinear deflection and stress (called stress stiffening, or nonlinear effect of membrane stresses), due to large deflection, which can have a major effect on the results. Unfortunately, to my knowledge, Roark does not contain a stress-stiffening solution for your particular BCs (square plate, say, b = 1235 mm, two opposite sides simply supported, two sides free). I think it contains only simply supported (held in-plane, free to rotate) on all four sides, which, for your scenario, would appear to give centre deflection y = 5.133 mm (?), and centre stress sigma = 22.61 MPa (?). Keep in mind, however, four sides simply supported is probably radically different from two sides simply supported, two sides free. I just currently do not know how different. And I currently do not know how different your scenario is from a simply-supported beam, if significantly different.
 
  • #7
AlephZero
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Roark does not contain a stress-stiffening solution for your particular BCs (square plate, say, b = 1235 mm, two opposite sides simply supported, two sides free).
In any case, that is not the situation you have. If there is no connection along the "ridge" of the roof, the trusses will twist and pull in torwards each other at the ridge as the plates bend, but the opposite sides will behave differently, because the corner mounting points will (presumably) stay a fixed distance apart.

A conservative analysis would be to assume the plates are simply supported beams, and keep the deflections less than the thickness of the plate so the nonlinear effect (i.e. the plate stretches as it bends) is small.

A better design would be to support all 4 sides of each plate anyway, and then you can use the nonlinear formulas in Roark to do a more realistic analysis. You also need to check you have enough fixing screws to withstand the tension in the plate when it is stretched as well as bent.
 
  • #8
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Hey!

I put some angle membersin between the two trusses at 400mm intervals. Using Roark's I was able to analyse this as a simply supported plate (i.e the area between the two trusses and two spans). I am able to determine y_max =3.6mm in the center and σ_max= 73.2 MPa from Roarks.

I intend to screw the sheet down over the length of the spans and the trusses. Having never used this text, I am wondering if this turns it into a rectangular plate, all edges fixed.

Alphzero you mention evaluating tension - I am confident through experience that I'll use enough fasteners to hold the sheet down, but out of interest - how would I evaluate the tension that the uniform load would induce at the edges of the sheet?

Thanks again for your help
 
  • #9
AlephZero
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You can get an estimate of the tension like this.

Assume the plate at the mid-span is bent into a circular arc.

Call the length between the supports L and the deflection in the center d.

If you draw a picture and call the radius of the arc R, you get a right angled triangle with sides R, L/2, and (R-d). So R^2 = L^2/4 + (R-d)^2 which gives

R = (L^2/4 + d^2) / 2d

If the angle subtended by the arc is 2 theta, you have

sin theta = (L/2) / R

And the stretched length of the plate will be

2 R theta

Putting your numbers in, L = 400, d = 3.6
R = (L^2/4 + d^2) / 2d = 5557 mm
theta = arcsin 200 / 5557 = 0.0360 radians
stretched length = 2 x 0.0360 x 5557 = 400.10 mm
So the tensile strain in the plate at the mid point is about .025%
Which give a tensile stress of about 50MPa at the mid point of the edge, tapering off to 0 at the corners.

That is the same order of magnitude the bending stress. You need to convert the tension into the shear force per screw, depending on the screw spacing.

This is all approximate, but it should be the right order of magnitude. Since you have stiffened up the sheet so the deflection is about the same size as the thickness of the sheet, it isn't going to be a huge effect, unless your design didn't have any factor of safety anyway!

Note, you may need to repeat this for L = 1200mm. I'm not quite sure where you put your stiffneners!
 
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  • #10
nvn
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Using Roark's, I was able to analyse this as a simply supported plate .... I am able to determine y_max = 3.6 mm in the center, and sigma_max = 73.2 MPa.
Fugro: I am not obtaining your same answers yet, so I would like to know how you computed your above answers. What applied pressure are you applying? The applied pressure is q = 0.007 480 MPa, right? (If you are using a different pressure, could you show how you computed your pressure?) Using your assumption of simply supported on all four sides (for a typical, linear plate), I am currently obtaining the following, from Roark. Could you show how you computed your answers?

a = 1220 mm, b = 400 mm, t = 2.9 mm, a/b = 3.05
alpha = 0.1338, beta = 0.7148
q = 0.007 480 MPa, E = 200 000 MPa
y_max = alpha*q*(b^4)/(E*t^3) = 0.1338(0.007 480)(400^4)/[(200 000)(2.9^3)] = 5.253 mm
sigma_max = beta*q*(b/t)^2 = 0.7148(0.007 480)(400/2.9)^2 = 101.72 MPa
 

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