# How to determine the max load capcity of a steel square tube

A square tube beam, 4in x 4in x .1875in thick.
Length= 36in
Cross sectional Area= 2.859in^2
A36 steel

How do you determine its max load capacity?

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Max load capacity for what type of loading? Point load at the center when the ends are rigidly secured? Distributed load over the length of the beam when one end is fixed and the other is cantilevered? Bearing strength? Tensile Strength?

Max load capacity for what type of loading? Point load at the center when the ends are rigidly secured? Distributed load over the length of the beam when one end is fixed and the other is cantilevered? Bearing strength? Tensile Strength?
Its a sawhorse type of setup. with each of the legs secured to the floor.

Baluncore
2019 Award
1. The worst case situation will be when a load is applied to the point at the centre of the upper surface of the beam. The top surface will be punched through, or will collapse and pull in the side walls. As the upper tube surface approaches the lower surface, the tube will bend because separation of the top, (compression), and the bottom, (tension), faces has been significantly reduced.

2. The next worst situation will be when a point load is applied to only one upper edge, near the middle of the tube. A dent in the edge will form. The same sort of failure as case 1 will then occur, as the loaded sidewall bends along its midline and so reduces the top to bottom separation.

3. The third situation will be when the load is shared by both sidewalls, on a line across the top of the beam. That will carry twice the load of 2, except that any imbalance will reduce it back to case 2.

4. If the load is distributed evenly over the entire top surface then that is a different situation again.

1. The worst case situation will be when a load is applied to the point at the centre of the upper surface of the beam. The top surface will be punched through, or will collapse and pull in the side walls. As the upper tube surface approaches the lower surface, the tube will bend because separation of the top, (compression), and the bottom, (tension), faces has been significantly reduced.

2. The next worst situation will be when a point load is applied to only one upper edge, near the middle of the tube. A dent in the edge will form. The same sort of failure as case 1 will then occur, as the loaded sidewall bends along its midline and so reduces the top to bottom separation.

3. The third situation will be when the load is shared by both sidewalls, on a line across the top of the beam. That will carry twice the load of 2, except that any imbalance will reduce it back to case 2.

4. If the load is distributed evenly over the entire top surface then that is a different situation again.

the load will be applied evenly to the face that is marked green in the picture. (the entire top face)

Determining the allowable uniform load that can be applied to a beam

Determining the allowable uniform load that can be applied to an A36, HSS 4x4x3/16 steel square tube. Theres a 30,000lb weight being applied evenly across the top face in a downward direction. The square tube is supported equally at both ends.

I just wanted to know if i did this correctly. Iv done problems like this before but never that involved a square tube. I didnt know if i could still apply the same method as for an I-beam or if there was a compltelty different way it needed to be done. Any help would be awesome!

A=2.859in^2
S=3.45in^3
d=4in
t= .375in
wt=9.42lb/ft

M=Sσ
=(3.45in^3)(24,000lb/in^2) = 82,000lb*in=6,900lb*ft

w=M8/L^2
=(8)(6,900ft*lb)/(3ft)^2 = 6133.3lb/ft

Vmax=(wL/2)+(P/2)
=[(9.42lb/ft)(3ft)/(2)]+(30,000lb)/(2) = 15,015.13lb

τavg= Vmax/dt
=(15,014.13lb)/(4in)(.375in) = 10009.42 lb/in^2

Since 10009.42<14000, therefore Wallow = 6133.3lb/ft