Steel Plate bending 1

[jmen68]

I am trying to convince myself that I am using a formula from the AISC manual correctly.

The project is to design a large counterweight for a device, and I am designing it as a steel plate box welded together and filled with concrete.

Here is the problem:
I want to design the counterweight walls to a deflection standard of L/360 so 0.09 inches. The depth of the box is 33 inches and is about 20 ft wide. It will have a bottom brace and a top brace while placing concrete.

My assumptions:
I am planning to use 1/2" plate, E=29x10^6, weight of concrete 150#/cf

My calculation:
I am simplifying this down to a simple beam, so the beam will be 33" long, 12" wide, and 0.5" thk. with bending in the 0.5" thk. direction. My concrete will be at full liquid head so the beam will be loaded with a continuous load increasing to one side (a triangle) with the top (left if you did a traditional beam drawing horizontally on a sheet of paper) having a support and 0 load, and the bottom (right) having 412.5 lbs. ...aka W would be 150 lbs/lf...
Looking at the AISC manual for delta_max or any publication online the formula for a simply supported beam with a concentrated load increasing to one side is: delta_max= 0.013[(wL^3)/(EI)]
My I=(bh^3)/12 = (12"x0.5^3)/12 = 0.125in^4
sub all into delta_max eq. = 0.013((150*33^3)/(29,000,000*0.125*12)) and I get 0.0016 in.

Here is my problem, 0.0016 inches my world isn't even recognizable to any equipment I own so basically there is no deflection. My intuition is just not allowing me to believe that there will be no deflection in this plate.

Can anyone double check my work/tell me that my assumption is incorrect that this is a simple beam problem. I understand that it is simplifying it to a degree but I am at a loss with the number the formula give me.

Thanks

 

[TehMightyEngineer]

Wait, is the plate spanning 20 ft or 33 inches? I'm confused about the dimensions.

Maine EIT, Civil/Structural.

 

[jmen68]

To answer your question 33" tall.

But oh jeez, I guess I should check the deflection in the horizontal direction as well. However, I believe I am ok there, the box will be braced during the pour with kickers and vertical walers to secure the top at a spacing of about 24 inches. so the deflection I am anticipating is the form bulging between the walers. Here is where I am at with the drawing now.

http://imgur.com/ySNWDev

I am going to put the round bar in there regardless of the deflection, because the elders are demanding it, haha.

 

[JStephen]

I'm not sure where the load is coming from. Is 150 pcf your concrete weight? Then pressure at 33" deep will be 412 psf, and total load W on the beam will be 0.5*412.5*2.75' = 567 lb. (Note that in the beam formula you are using, W is the total triangular load, not the load intensity). That gives me 0.073".

I'm assuming that when you pour, the steel plate is forming a 33" high tank, with a stiffener at the top and bottom, so the plate is just spanning between the stiffeners.

 

[TehMightyEngineer]

While I agree that it appears you shouldn't have much deflection horizontally (20 ft span) you definitely have a complicated plate setup. You probably should have a structural engineer take a harder look at it if it's that critical. Your math appears to be correct and your assumption of a simple span seems reasonable but you really have to take a look at the long span direction as well as the deflection of the walers and the kickers.

Maine EIT, Civil/Structural.

 

[TehMightyEngineer]

Ah yes, as JStephen pointed out your load does appear to be off.

Maine EIT, Civil/Structural.

 

[Agent666]

I dare say your steel plate formwork would be stiffer than any similar timber formwork so problem solved!

Just put a few more welded bars between the walls and incorporate them in the Concrete if you are concerned. Alternatively weld a series of horizontal/vertical angle to the plate would stiffen it considerably.

Something else to consider is the deflection/distortion from welding it all up is likely to be greater than the deflection from the concrete. Similar if you were galvanising the item.

 

[dhengr]

Jmen68:
You can see it and imagine it, we can’t see it from here, and your description isn’t much help either. The first step in trying to get some help is to describe your problem well enough so that someone who isn’t looking at the same thing you are can understand what you are really trying to do, and what you are talking about. Get on the stick, if you want to do meaningful engineering, and want a meaningful discussion here. Look at your own description, and ask yourself..., if I had never seen this problem before, nor seen the equipment involved, have I given all the info. needed to fully describe the situation and the problem. Why do you want to know the deflection at about 1/3rd up the 33" side? Aren’t you much more worried about how the top lip, 20' long, will deflect? Is there a reason that you couldn’t form this conc. mass just like you would a 33" deep, by 20' square spread footing?

You have a 33" high, 20' wide and how long, a box with a bottom and four sides made out of .5" steel plate? It acts as a containment, and will be filled with conc. to act as a counter weight? How is it finally supported in its end use? You should put a cage of rebar in this conc. beam and you should do something to bond the steel box and conc. mass together; shear studs, ties across the conc. mass from steel side to steel side, or some such. Is this counter weight being build right on the structure of your equipment, and what is that?

 

[jmen68]

Really I am not trying to be a jerk here or offend any users or mods but I am just going to answer some questions here.
To dhengr,
"You can see it and imagine it, we can’t see it from here, and your description isn’t much help either. The first step in trying to get some help is to describe your problem well enough so that someone who isn’t looking at the same thing you are can understand what you are really trying to do, and what you are talking about."
-I put a picture in my second post, forgot to include it in my first

"Get on the stick"
-...really?

"Look at your own description, and ask yourself..., if I had never seen this problem before, nor seen the equipment involved, have I given all the info. needed to fully describe the situation and the problem. Why do you want to know the deflection at about 1/3rd up the 33" side? "
-Again I posted a picture, sorry it was late. I believe I gave enough info about the problem, I was just trying to determine if I was using the correct formula/using it correctly.

"Aren’t you much more worried about how the top lip, 20' long, will deflect? Is there a reason that you couldn’t form this conc. mass just like you would a 33" deep, by 20' square spread footing?"
-I will be kicking it back to a solid foundation at two foot spacing, I am just concerned about deflection from liquid head of concrete not the top rim.

"You have a 33" high, 20' wide and how long, a box with a bottom and four sides made out of .5" steel plate? It acts as a containment, and will be filled with conc. to act as a counter weight? How is it finally supported in its end use? You should put a cage of rebar in this conc. beam and you should do something to bond the steel box and conc. mass together; shear studs, ties across the conc. mass from steel side to steel side, or some such. Is this counter weight being build right on the structure of your equipment, and what is that?"
-not trying to be that guy here, but none of that info is needed for my question. But if you would like to know, I don't need a rebar cage, I do not need this to be structural concrete, it could be blue and spalling like nothing you have ever seen before, its just a cheap means to an end of needing extra weight.

 

[jmen68]

JStephen- I noticed that the formula for this beam had a capital W, where the other uniform loads where dentoed as a lower case w. Is there anywhere you can point in the manual that defines this? I believe you, however I would also like to prove it to myself.
Thanks

 

[BAretired]

W is the total load on the beam, i.e. W = 412.5*2.75/2 = 567#

Then, Δ_max = 0.01304WL³/EI = 0.0733".

BA

 

[JStephen]

I'm looking at the green AISC-ASD book. The page prior to the beam formulas is "Nomenclature" and defines W there as "Total load on beam, kips". (They also use E in KSI, so the K-factor cancels out).

 

[racookpe1978]

I'm going to echo the above request: Sketch your idea and scan the sketch. I can't picture it either. How is the steel & concrete weight going to be attached/suspended to the supports?

If, like a trebuchet or catapult weight - this is a dynamic (even if unofficial) assembly - you need to ensure it is reasonably safe under acceleration/deceleration!

 

[jmen68]

I have designed the frame to support the weight already. This isn't my concern of the question, it is basically a formwork question. I have provided a link to a graphic from my drawing above. Thanks for your concern though.

 

[TehMightyEngineer]

Wow, love how everyone is completely missing the image jmen posted.

Maine EIT, Civil/Structural.

 

[racookpe1978]

Sorry boss - I still don't see the link or image file. 8<(

 

[racookpe1978]

Oh. under the imgur file ...