Max Load on a plate 1

[ttuterry]

Here is the situation: a coworker has asked me to figure out the maximum loading that a piece of 1/4" aluminum plate will hold. This plate is to go between two large beams on a haul trailer. They just want to stand on the plate and don't want to fall through. I have looked up the properties of the aluminum but I can't figure out how to calculate the max load. Can anyone help me out?? Thanks

 

[COEngineeer]

I read your question slowly but I still dont understand what you are trying to do with the plate. Explain it a little more? You dont want the beam to fall through?

 

[ttuterry]

Sorry i didn't explain it well enough...the plate is to go over an open space between the two beams, and there are going to be people standing on the piece of plate

 

[ttuterry]

some more info...the plate overlaps each beam by 1.5"

 

[ACtrafficengr]

what is the span?

Are the people a live load or standing still?

How many?

What factor of safety are you giving it?



"...students of traffic are beginning to realize the false economy of mechanically controlled traffic, and hand work by trained officers will again prevail." - Wm. Phelps Eno, ca. 1928

"I'm searching for the questions, so my answers will make sense." - Stephen Brust

 

[ttuterry]

The span between the two beams is 36" and the people are probably going to just be standing still or barely be moving, as they are just unscrewing large bolts from a container. The plate is 36" wide and 33" long, so it isn't very big. I have tried to use the equation stress=P/A but my answer seems way too large. I assume the factor of safety will have to be large, as safety is the priority when you work for the government! But ACFtrafficengr, i did think about all those questions before you posed them, I am just having trouble relating it all...

 

[ACtrafficengr]

You may be overthinking this. Got any 2x10's lying around?

"...students of traffic are beginning to realize the false economy of mechanically controlled traffic, and hand work by trained officers will again prevail." - Wm. Phelps Eno, ca. 1928

"I'm searching for the questions, so my answers will make sense." - Stephen Brust

 

[swearingen]

My take on this:

At 33" wide, I'd say to design it for no more than two people at 250# each. Assume that the 500 pounds is divided into 4 point loads along the center of the span. To simplify this case, split the plate into 4 individual plates, each holding 125# in the center. Also, I'm assuming simple supports and 2011 T6 aluminum (fy=24.5ksi, E=10300ksi).

I = bh^3/12 where b is the width and h is the thickness.
I = 8.25*t^3/12 = 0.6875*t^3
y = 1/2*t = t/2
M = Moment = P*L/4 where P is the load and L is the span
M = .125*33/4 = 1.03k-in
Sigma = Stress in plate = M*y/I
Sigma = (1.03*t/2)/(0.6875*t^3)
Sigma = 0.75/t^2

If you take the plate to yield at 24.5ksi, you'd need:

Sigma = 24.5 = 0.75/t^2, t = .175in

Deflections:

Delta = P*L^3/(48*E*I)
Delta = .125*33^3/(48*10300*0.6875*.175^3) = 2.5" !!!
That's a problem.

To keep deflections decent, use a deflection criteria of L/360 = .092" and recalculate the thickness required:

Delta = 0.092 = .125*33^3/(48*10300*0.6875*t^3)
t = 0.46"

I'd use a 1/2" plate and call it a day...

 

[ttuterry]

Swearingen, thanks for putting it all together for me...i had all those equations on the paper just couldn't put it together...one thing though, when you use L/360 as your deflection criteria, your fy is not even considered to calculate the new thickness? i don't quite understand this...

 

[ttuterry]

another thing...maybe i have typed your deflection equation wrong in excel or something swearingen, but when i put in a larger load, the deflection gets smaller...how can this be??

 

[CarlB]

You must have a boo-boo:

Delta = P*L^3/(48*E*I)


Delta goes up if P or L increases, or if E or I decreases.

 

[swearingen]

Deflection in the elastic range is dependent on the modulus of elasticity and not the yield strength, which is why Fy does not appear in the deflection equation. To put it another way, the yield strength tells you what stress is required to yield it, while the modulus of elasticity tells how much it stretches (deflects) on your way to yield.

 

[GeoPaveTraffic]

ttuterry,

Beams are often controlled by deflection, not stress. That is why the equations are written the way they are. If you want to check Fy, just plug you deflection into the Sigma = 0.75/t^2 equation.

I believe I got that right, been a long time since I did much structures.

 

[DRC1]

For Deflection try Delta = (P*L^3)/(48*E*I)