×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Max Load on a plate

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

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?

RE: Max Load on a plate

(OP)
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

RE: Max Load on a plate

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

RE: Max Load on a plate

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

RE: Max Load on a plate

(OP)
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...

RE: Max Load on a plate

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

RE: Max Load on a plate

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...

RE: Max Load on a plate

(OP)
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...

RE: Max Load on a plate

(OP)
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??

RE: Max Load on a plate

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.

RE: Max Load on a plate

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.

RE: Max Load on a plate

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.

RE: Max Load on a plate

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

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources