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Design of Steel Plate 4

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CPENG78

Civil/Environmental
Sep 2, 2008
186
Can anyone point me to a link or documentation that shows design guidelines for a simply supported steel plate? I am applying a 10.4 kip load to a 10.7 inch wide by 52.2 inch long plate (1/2" thick). I am getting a quite larger bending stress than the allowable 66% (bending stress) and wondering if I'm missing something. Are there any allowable reductions when it comes to design of plates?
 
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I would be using 75% of yield as allowable for a flat plate, or Zx (plastic section modulus). AISC states this is allowed for weak axis bending of rectangular bars.

I don't know much about statistics, but I do know that if something has a 50-50 chance of going wrong, 9 times out of 10 it will.
 
You need to supply more information. Which way is it spanning (52.2 inch direction or 10.7 inch direction)? How is the load applied and over what area? Is deflection an issue?
One half inch plates are not really that good in bending. That's why they make so many steel shapes.
 
The 10.4 kip load is applied on a 10 in x 10 in square area in the middle of the plate. The span length is the 52.2 inch dimension.
 
In that case your deflection would be 9.5 inches, which I suspect is not acceptable.
You're also way beyond yield on the steel.
 
Jed,
In my research, I have come across some equations for plate design. However it wasn't clear to me as to what kind of streesses they were aimed at calculating. Do you know of any that are strictly used for design of such a plate in bending and shear? Thanks.
 
CPENG78,

I assume you do civil work, but not structural, as your question is elementary structural stuff.

For bending, M=PL/4 for concentrated load at centre span

Elastic section modulus S=bd^2/6

Bending stress f=M/S

Deflection = PL^3/48EI
 
A plate has the same formula as any other structural member that spans a distance. hokie66 has them summarized. I don't know what you're driving at, but in short, you can't span 52.2 inches with a 1/2 inch thick plate and carry 10.4 kips. The plate would exceed yield stress quickly and either fail or if the ends are adequately supported, act like a cable and support the load like a catenary.
 
Yes, the majority of my work is on the civil side, and I do have a background in structural design both through partially on my degree and through work experience in a structural company for 3 years. This experience however was a while back and yes I am aware of the basic and elementary structural equations.

The first reason for my post was to see if there were any assumptions/equations (since I don't regularly come across plate design) that I was missing other than the basic equations. Such as the post by Johnny.

It sounds like my results are matching those equations used by Hokie and results by Jed.

The second reason for posting my question in an elementary format was also to keep the question as simple as possible since my results are in disagreenment with a co-worker and I wanted to compare his assumptions as well.

Thank you everyone for your help. Greatly appreciated.
 
If you look at plate design the plate will be supported on 3 or all 4 sides to act like a plate, otherwise it is merely a beam. Due to your aspect ratio the plate will act largely like a beam in the short direction when supported on all 4 sides. Consequently I think your plate will only act as a plate when supported along 3 sides with one short edge free. The formulae of your research will then become clear
 
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