Max Load on a plate
Max Load on a plate
(OP)
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





RE: Max Load on a plate
RE: Max Load on a plate
RE: Max Load on a plate
RE: Max Load on a plate
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
RE: Max Load on a plate
RE: Max Load on a plate
"...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
RE: Max Load on a plate
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...
RE: Max Load on a plate
RE: Max Load on a plate
RE: Max Load on a plate
Delta = P*L^3/(48*E*I)
Delta goes up if P or L increases, or if E or I decreases.
RE: Max Load on a plate
RE: Max Load on a plate
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.
RE: Max Load on a plate