Plate Analysis 1

[LIGWY]

I have a 3/8" plate that is span over a 26'x6' area and is supported every 4'3". The circumference is welded. I have a uniform load for the area at 2,200 lbs/ft or a 515 lbs/ft^2 and I also have a point load that is 4,985 lbs (there are two of them at 3' apart)

How do I go about analysis a plate for moment shear and deflection?

The formulas would really help and how the K or C factors are found.

 

[Ussuri]

Could you use yield line analysis?

 

[miecz]

The obvious approach is to locate beams under the concentrated loads and design the plate as a one-way continuous beam with a uniform load. If you can't locate beams under the concentrated loads, I think you need to do a finite element analysis. Alternately, you could assume some effective width equal to the width of your load plus some multiple of the plate thickness, like 6t to each side. However, I know of no source that recommends what that multiplier should be, and whatever it is, the result should be small compared to the width of the load, so I would ignore it and call the effective width the width of the load. I rather doubt that a 3/8" plate will carry your concentrated loads, unless you do a finite element analysis.

 

[archeng59]

If you have access to the book, Roark's Formulas for Stress & Strain, there are equations for plates having various fixities around the edges. I've used that occasionally for checking plates. if nothing else, check the plate as a 12" strip, either continuous or simple span between support beams. Section properties based on 12" width. If the support beams are equally spaced, you can calculate statical moments and shears for continous spans with uniformly distributed loads or equally spaced concentrated loads using the co-efficients in ASD manual (p. 2-312) or LRFD manual (p. 4-206).

 

[LIGWY]

I did try to do a 12" strip but my moment of inertia is so small it is showing great deflection. Because is it not b=12 h=3/8"? But that is how I have calculated the loads.

Yes I have looked into both manuals on thos pages The load being placed is a small skid steer if that helps at all.

This system has been in service for 20+ years and there are no signs of any problems.

 

[LIGWY]

I have bits and pieces of the equations for a plate with various fixed ends but can not get my hands on the actual equation.

I have information for a psi loading that k*wl^4/Et^3

k is the factor from the plate edge I tried .0834 from a question but I do not know how this applies the plate edge supports
w - load
l- longest lenght
E - 29000 ksi
t - thicknes

I got a large deflcetion that was not realistic

 

[aggman]

From the information above you will be conservative to design it as one-way. Well you are almost always conservative to design it as one-way but this assumption is reasonable for your plate aspect ratio. I am getting a stress around yield and deflection around 1 1/3 times plate thickness just with the uniform load. Thus the small defelection theory is not valid for your plate, nor would "normal structural program" FEA be valid and you need to do a large deflection theory analysis. I would research the yield line approach as it would be most appropriate and use it for your design or use a plate that would generate deflections that are around 1/2 plate thickness to validate the small deflection theory.

 

[LIGWY]

To be honest this is the first time I have had to design a plate like this and I am uncertian in regards to a large deflection theory analysis.

 

[aggman]

It’s the same theory that holds true for your standard beam analysis. The part of the theory that says we assume small deflections and that plane sections remain plane after deflecting. For plates the beam "small deflection" theory validity diminishes as the deflection exceeds 1/2 the plate thickness. At that point your plate will not "fail" but it will begin redistributing the load as membrane forces. Deflections will become larger at this point and can cause serviceability issues. I suggest you head to your local library and pick up a book on plates and shells. Otherwise you need to maintain the small deflection assumptions and go on with your design. Plate theory is not simple by any means. In fact it's quite a bit more complex than beam theory and numerical solutions are the more appropriate method of analysis.

 

[JStephen]

Roark includes some information on large deflection of circular plates, not sure about rectangular- check.

If you assume zero stiffness, you can calculate allowable loads which will be much higher- think Roark has that as well, with 1-dimensional loading. It assumes the ends are fixed for tensile forces, so you have to have adequate supports at those points.

Don't figure the loader as a "point" load, figure some way to use the actual area, or an estimate of it.

 

[LIGWY]

I am still concern with what exactly my b for the deflection when using
for the uniform load and I calculate I = bh^3.12 isn't my b 12"? or is it 6'? if the plate spans or transfers load at the 4.33" supports?
5wl^4/384 EI

 

[aggman]

Assuming that you are welding the plate to flanges of wide flange beams and you are looking at one-way bending then I would use the following formula's.

Assume fixed ends: (Weld on both sides of flange fixes plate)

Mmax = 1/12 * w * L^2 (L = 4'-3") (Case 15 ASD manual)
w = unit load (typically psi or psf)

Deflection = (w * L^4) / (384 * E * I) (Case 15 ASD manual)
I = (1/12) * bi * t^3
bi = effective plate width (typically 1" or 12")
t = plate thickness

If your plate does not generate fixation at the continuous supports then I would use continuous span formulas. A lot of people like to use (1/10 * w * L^2) for moment and (.0069 * w * L^4 / (E * I)) for deflection. This is the three span condition.

 

[LIGWY]

Thank You I will look at this

 

[LIGWY]

The detail I have has no fixation so I used the load as shown on 2-306 ASD for this and have 6" deflection

How should I handle the load from the skid steer. do I use a point load or distribute it out of the area it is touching the ground

 

[miecz]

Abutler

For a uniform load and small deflection plate theory, my old Mechanics of Materials textbook has the deflection at k*wb^4/Et^3, where b is the short dimension, not the long dimension. The k factor depends on your edge conditions and may be less than .0834, which is for pinned supports.

The textbook,(Seelye and Smith) says small deflection theory is OK up to several times the thickness of the plate. With fixed supports, I'm getting deflections of 1 (for uniform load)to 2 (For 5k load) times the plate thickness, so you may be OK with small deflection theory.

Also, I'm getting bending stresses over 40 ksi for the 5k load case.

 

[LIGWY]

What I am concluding to that the plate needs to be increased to 1" thick plate.

 

[JStephen]

Or consider grating as an alternative to the plate. Or put stiffeners on the underside of the plate.

 

[mrMikee]

For my own curiosity I solved this problem using the equations in the Gaylord and Gaylord text "Design of Steel Bins..."

a=72"
b=51"
p=3.58 psi

kyf=0.464 {eq6-53}

assuming A36 steel
Fb = 0.75Fy = 27 ksi

b/t = Sqrt[Fb/(kyf*p)] = 127.5 {eq6-55}

t = 51/127.5 = 0.4" reqd plate thickness

Rather than using a heavier plate throughout to handle the concentrated loads I would add a stiffener under these two loads if possible and if they don't move.

Regards,
-Mike

 

[LIGWY]

The load 4980 is a moving load.

When calculating the bending stress what formula do you use for plate?

 

[aggman]

Abutler,
I am going to get on a soap-box here and this in an answer to your other post. You need to go to your local library or university and get either Roark's formulas for stress and strain or another plate theory book. People have answered your questions for you short of just flat out doing the design for you. You are making a mistake in your calculations because your deflection is unreasonable. All of us have come up with the same numbers for the most part. In my first post I stated that I am getting a stress around yield. If you back through mrMikee's numbers using 3/8" plate you get plate stresses at 30,724 psi using plate theory. I generated about 33,000 using beam theory. Their are two problems with the design. One you have a concentrated "patch" load which needs to be distributed over some width of plate to get accurate results. You can get that from Roark's or any other plate book. You also need to watch your deflections. Roarks states the following as does Gaylord's bin design book that mrMikee references;

"When the deflection becomes larger than about 1/2 the thickness, as may occur in thin plates, the middle surface becomes appreciably strained and the stress cannot be ignored. This stress, called diaphragm stress, or direct stress, enables the plate to carry part of the load as a diaphram in direct tension. This tension may be balanced by radial tension at the edges if the edges are held or by circumferential compression if the edges are not horizontally restrained. In thin plates this circumferential compression may cause buckling.
When this condition of large deflection obtains, the plate is stiffer than indicated by the ordinary theory and the load deflection and load stress relations are nonlinear. Stresses for a given load are less and stresses for a given deflection are generally greater than the ordinary theory predicts."

Now not to sound harsh or rude but you have been trying to find out online how to solve this problem since last friday. You could have easily either gone to the library over the weekend or ordered a book from Amazon.com and received it next day air this morning. (I would recommend Roark's by the way) Instead you are still trying to find out online how to solve the problem. I have been a member of this forum for 2 1/2 years and never once has someone ever flat out solved a problem for someone on here. I again suggest you take the information we have gave you and get the required information from your local library or from a book-store. You don't seem to have a very good understanding of plate theory and because of that you are trying to design something you really don't understand. It's fine not to know how to do something, we all run into that all the time and that is why this forum is so popular, but you have to be able to take the time and research it. Think of it this way. Say you just go on with the design as is. Say everything is fine for 2 years, but suddenly one day the skid-steer falls through the floor and the operator dies. When you are sitting on the witness stand and the hot shot lawyer with the PH.d in structural engineering is wearing you out about your analysis I certainly wouldn't want to say, "Well I talked to some people on a forum on the internet about it and they thought it would be ok so I went on with it.". I realize thats like worst case but the point is that you take suggestions but the design in the end falls on your sholders and is for you to understand so you better have the information at your fingertips so you can make a descision about how it should be. It drives me crazy when people just "wing" it. I will now get down from my now and I hope you are not offended as that was not my intent.