Plate Bending 1

[mdmiller2000]

I have attached a sketch of my application. I am trying to figure out how thin i can make the 2" plate in order to put the 10K pounds of force on the plate. There is no reason I cannot use gussets on top if that will help. The whole structure will be sitting on a platform (not suspended). I used a program while in college for this type of analysis but the name of it is escaping me right now. If you could point me in the right direction I would greatly appreciate it.

 

[metman]

Stress(S) = M/Z

M = P X L
P = 10,000Lbf
L = 24"

M = 240,000In-Llbf

Z = b X h^2/6

b = 24"
h = 2"

Z = 24" X (2")^2/6 = 16 In^3

S = M/Z = 240,000In-Lbf/16In^3 = 15,000 psi

Do not use thinner than 2" unless you add gussets then the calc method changes. 24,000 psi is a fair Yield Strength for "mild" Steel which gives you less than 2:1 factor of safety.

h = square root of (6PL/bS)

Therefore, if you use "enough" gussets, the thinning of 2" plate becomes directly proportional to the square root of the ratio L/24". L is the space between gussets.

It appears you will be depending on welded structure. 1,400 Lbf/in for 1/4" weld will give a healthy safety factor for mild steel.

 

[mdmiller2000]

Thank you very much. I understand I am looking at the stress of the plate. Where can I look up the stress per thickness of plate?

 

[corus]

Isn't the plate sat on a cylinder? Metman's formula is for a beam bending, ie. if it were rectangular. I'd consider it as a circular plate simply supported at its edges with a central load. You may also consider it as a circular plate built in at its edges as the supporting cylinder will give some rotational stiffness to the egdes, but not completely. The answer will lie somewhere between the two cases, but take the worst case. Look in Roark, formulas for flat plates, for examples.

corus

 

[metman]

The sketch seems to be rectangular (12" X 24"). If true then Stress is proportional to thickness squared. You can't look it up, you must calculate it.

If it is cylindrical, as corus suggests, then you need the formula for that condition to calculate it. Roark is an excellent source.

 

[corus]

In the diagram it refers to a dimension labeled Phi, which is commonly taken to be the diameter. Looking at it again though it must be referring to the plate thickness of the side parts, or is it?

Even if it were rectangular the formula by metman must be incorrect. If the force is P = 10,000 lbf then the moment from the reaction forces at the edges must be P/2 x L/2 for a simply supported beam, or PL/4. If as I said earlier, the side supports provide some rotational restraint then the moment will be PL/8 for edges fully built in at both the centre and at the edges. The true answer will be somewhere inbetween. I'd use PL/4 to size the plate thickness. If you use gussets then the moment will be effectively reduced but then I'd consider the bending moment at the edges for a built in beam to size the gussets and welds.

corus

 

[mdmiller2000]

The point where the 10Klbs is applied is a 3/4 bolt that is threaded into a 10klb Load Cell that measures the force applied. The load cell its self is about a 6" diameter and about 1.5" thick and is secured in 6 places with smaller bolts. This might make thing more complicated or easier.

 

[corus]

Assuming a point load will be a worst case for the bending of the plate.

corus