Thread:

I-beam length / load calculation

If I have a 22 foot steel ibeam that is support on each end with the ibeam height being 5 inches and the "I" part being 3 inches, what is the max center load (ie. how much can I lift with it)? I have been to the web sites but the calculators get a little blurry after reading them.

what type of steel? (ksi) - How thick is the web?

You have to check both shear stress and bending stress.

I think I'm misunderstanding your measurments. You need these to solve the problem (besides density):

Originally Posted by Icy

what type of steel? (ksi) - How thick is the web?

You have to check both shear stress and bending stress.

Why don't you ask him what he is trying to lift instead of all the engineering questions?

Originally Posted by fourtznme

Why don't you ask him what he is trying to lift instead of all the engineering questions?

Why do you even post? It's obvious you fail at reading.

Even if he says an object and it's not remedial like an engine block I'd have to do the calculation.

Originally Posted by probablecause

what is the max center load (ie. how much can I lift with it)?

Originally Posted by Icy

what type of steel? (ksi) - How thick is the web?

You have to check both shear stress and bending stress.

My chart only has a 6'' x 9'' beam so I can't determine the most likely web thickness...

6" X 9"? Are you sure about that?

Originally Posted by probablecause

If I have a 22 foot steel ibeam that is support on each end with the ibeam height being 5 inches and the "I" part being 3 inches, what is the max center load (ie. how much can I lift with it)? I have been to the web sites but the calculators get a little blurry after reading them.

Thats an S5X10 American Standard Beam. It would help to know what you plans are. Having somebody go thru calcs ton find out you lifting 36" wide screens is a waste of time.
It not a real hefty beam if thats what your asking.

Originally Posted by Icy

Why do you even post? It's obvious you fail at reading.

Even if he says an object and it's not remedial like an engine block I'd have to do the calculation.

If you are as smart as you think you are, you would know the answer if you knew what he was lifting. Why get so technical.

Originally Posted by gn7

6" X 9"? Are you sure about that?

Thats an S5X10 American Standard Beam. It would help to know what you plans are. Having somebody go thru calcs ton find out you lifting 36" wide screens is a waste of time.
It not a real hefty beam if thats what your asking.

W6x9 is my smallest beam on my chart.

I thought he was saying his height was 5 and his width was 3. Now I get what he meant.

W3x5.7 is the smallest beam I could find quickly online and not lining up with his measurements I realized I mis-interrupted what he meant, not that it matters at this point. Plus i'm retarded and it's lbs/ft...

Originally Posted by fourtznme

If you are as smart as you think you are, you would know the answer if you knew what he was lifting. Why get so technical.

I know how to solve this problem. I don't deal with it on a day to day basis. I don't even know the common steel alloy used in this.

I prototype mechanical systems half time and deal with radar "things" the other half on a day to day basis.

I don't deal with loading anymore. I'm not going to tell the dude he can life a semi with it without checking it first and getting all the details. If someone more qualified that is in construction or civil engineering wants to step in and tell him the right answer that's fine by me.

Edit: Length also drives this problem especially when he's talking more than 6 feet. I could hurt someone not analyzing it. Sorry I don't live up to how smart you think I think I am. I'm not naive and I'm not going to kill the guy, someone he knows or hurt his equipment or the item he is lifting by trying to act "smart". I don't have the ego you think I do, you are really just that dumb.
______



Steel @ 36ksi yield strength - I'm pulling this number to be really safe, it's conservative.

60% of the yield strength is the general acceptable sheer. Lets say that the max sheer τmax = 36,000*0.6 = 21,000 psi

Shear Stress:

Max shear is in the X-axis (center of the flange)

τmax = V*Qcenter / I * tcenter

rewrite to

V = τmax * I * tcenter / Qcetner

We have everything but V (which is what we are solving for) and Qcenter. (Qcenter is the moment of area around the neutral axis)

Qcenter = Σy'A' Where y' is known as ybar, (it should have a bar on top of the y'), this is basically determining the center of mass for given geometry.

A' is the area of the two rectangles we are concerned about that make up the mass of the top part of the I-beam

Qcenter = [(3.004 * .326) * (2.5 - .326/2) + (.214 * (2.5 - .326) * ((2.5 - .326) / 2)]

Qceneter = 2.794

V = τmax * I * tcenter / Qcetner

V = 21000 * 12.3 * .214 / 2.794

V = 19,783.89Lb

Now you need max bending stress. I'm not going to type units out on everything so don't bother criticizing unless you want to do it yourself. This beam will fail in bending a lot sooner than shear.

Max Bending Stress, we are again looking at the center of the web. I'm assuming one side is a pin and the other is a roller. I do not know how to do anything else.

σmax= Moment*dcenter / I

What kind of bending moment are we getting based on the load of an unknown quantity will determine when the beam starts to fail.

Moment * dcenter = Moment * d / 2

d/2 = 2.5

Moment = Load / * Length / 2

Length = 264 inches (22 feet)

Moment = Load * 132

σmax = ksi * 2/3 - Generally accepted

σmax = 24,000psi

σmax= Moment*dcenter/I can be rewritten too:

σmax*I = M * dcenter

σmax*I/2.5 = Load*132

36,000*(2/3)*12.3/2.5/132 = Load

Load = 894.54lbs

Now compare the two. I knew this would fail in bending before I did it, but its always good to consider both.

V = 19,783.89Lb
VS
Load = 894.54lbs

After 894 lb, the beam will deform. It'll fail at some other greater loading. Adding gravitational forces to the load can cause issues as you are inducing a greater force.

I wouldn't lift more than 750 lbs with it. That's just me. Please do not accept my answers as 100%, I'm not a Licensed Professional Engineer and have not done one of these problems in over 4 years. I did not consider the weight of the beam itself in the calculations, this is negligible generally. If you want to be safe take another 55lbs off the answer above.

Being he wants "the max center load" with the info given I think he will never get the proper answer.

Originally Posted by Icy

L = 894.54lbs

After 894 lb, the beam will deform. It'll fail at some other greater loading. Adding gravitational forces to the load can cause issues as you are inducing a greater force.

I wouldn't lift more than 750 lbs with it. That's just me. Please do not accept my answers as 100%, I'm not a Licensed Professional Engineer and have not done one of these problems in over 4 years. I did not consider the weight of the beam itself in the calculations, this is negligible generally. If you want to be safe take another 55lbs off the answer above.

Answer I hope... Math above.

You'll get more out of the beam by shortening or putting the supports closer to the load. Only you know if you can do this based on the information provided. Length does not help you when dealing with bending moments.

Using 11 feet between supports will give 1789lbs.
Using 5.5 feet between supports will give 3578lbs.
Using 3 feet between supports will give 6,560lbs.
Using 1 inch you'll fail in shear (lol).

You get the point.

All this without seeing the beam.