What is the Maximum Load Capacity for a HSS 6x6x1/4in Steel Square Tube?

[blake92]

I have a HSS 6x6x1/4in steel square tube that is 63.5in long. I am trying to determine its maximum load when standing vertically on end (assume it is perfectly vertical and fixed at both ends).

I have calculated the following:

Moment of Inertia = 28.6in^4
radius of gyration = 2.34in
Cross sectional area = 5.75in^2
Modulus of elasticity = 29,000ksi

I then used the formula,

Pcr= (∏^2)*(E)(I)/(.5L)^2

when i do this though I get an answer near 7,000kips or 7,000,000 lbs! there's no way that can be right and i was wondering if anyone could help me figure this out! Thanks in advance!

 

[SteamKing]

I have a HSS 6x6x1/4in steel square tube that is 63.5in long. I am trying to determine its maximum load when standing vertically on end (assume it is perfectly vertical and fixed at both ends).

I have calculated the following:

Moment of Inertia = 28.6in^4
radius of gyration = 2.34in
Cross sectional area = 5.75in^2
Modulus of elasticity = 29,000ksi

I then used the formula,

Pcr= (∏^2)*(E)(I)/(.5L)^2

when i do this though I get an answer near 7,000kips or 7,000,000 lbs! there's no way that can be right and i was wondering if anyone could help me figure this out! Thanks in advance!

Actually, plugging in your numbers, the formula gives a critical buckling load closer to 8100 kips than 7000 kips.

However, if you check the compressive stress in the column at this critical buckling load, you get σ = 8120 / 5.75 = 1400 ksi, which is way more than the elastic limit of any steel which could be used to fabricate this column. (for example, regular A36 steel has a min. elastic yield stress of 36 ksi)

The Euler critical buckling load formula has limits of applicability, since the formula doesn't take into account the elastic strength of the material. The value of E changes once the material is stressed beyond the elastic limit, so you can't just plug in 29,000 ksi for steel and get reasonable results.

 

[blake92]

Actually, plugging in your numbers, the formula gives a critical buckling load closer to 8100 kips than 7000 kips.

However, if you check the compressive stress in the column at this critical buckling load, you get σ = 8120 / 5.75 = 1400 ksi, which is way more than the elastic limit of any steel which could be used to fabricate this column. (for example, regular A36 steel has a min. elastic yield stress of 36 ksi)

The Euler critical buckling load formula has limits of applicability, since the formula doesn't take into account the elastic strength of the material. The value of E changes once the material is stressed beyond the elastic limit, so you can't just plug in 29,000 ksi for steel and get reasonable results.

okay thanks, i was also able to find a table that shows the "Allowable Concentric Loads in Kips" i was then able to find the tubing listed there and it says for a 6ft section of the tubing that max load is 131 kips. Does that sound correct to you?

 

[SteamKing]

okay thanks, i was also able to find a table that shows the "Allowable Concentric Loads in Kips" i was then able to find the tubing listed there and it says for a 6ft section of the tubing that max load is 131 kips. Does that sound correct to you?

That load gives a compressive stress of about 22.8 ksi, which should be well below the elastic limit of even the most common steel. Of course, if there is even a slight eccentricity in the loading, then the max. allowable load on the column would decrease.

 

[blake92]

That load gives a compressive stress of about 22.8 ksi, which should be well below the elastic limit of even the most common steel. Of course, if there is even a slight eccentricity in the loading, then the max. allowable load on the column would decrease.

yeah according to the table it says Fy=46ksi for this HSS tubing. So does that mean the max allowable stress is 30.36 kips (46*.66) and then since this 22.8ksi is within that, everything is good?

 

[SteamKing]

yeah according to the table it says Fy=46ksi for this HSS tubing. So does that mean the max allowable stress is 30.36 kips (46*.66) and then since this 22.8ksi is within that, everything is good?

As long as the load is applied to the column without eccentricity, you should be good to go.

A lot will ride on how good the welds are which fix the ends of the column. If the ends start rotating under load, the situation could rapidly deteriorate.

 

[blake92]

As long as the load is applied to the column without eccentricity, you should be good to go.

A lot will ride on how good the welds are which fix the ends of the column. If the ends start rotating under load, the situation could rapidly deteriorate.

what exactly is eccentricity?

and i just need a close estimation on how much weight it can hold so as long as its close to what it can actually hold it should be fine. I don't need the exact number.

 

[Travis_King]

He means that that calculation you've done is for axial loading where the loading line of action is along the center axis/axes of the HSS. If there's more weight to one side, the equation you've done isn't accurate as there will be bending stresses in the beam which will increase the tendency to buckle.

 

[blake92]

He means that that calculation you've done is for axial loading where the loading line of action is along the center axis/axes of the HSS. If there's more weight to one side, the equation you've done isn't accurate as there will be bending stresses in the beam which will increase the tendency to buckle.

Im assuming there are no forces such as that but, the beam is conected to another identical beam with a cross member and the force is being applied stright down in the middle (kind of like a kid standing in the middle of monkey bars). There are also gussets on both sides though to help strengthen it.

 

[Chronos]

There are any number of civil engineering tables that provide load ratings for reinforced concrete. They include built in load imbalance allowances based on service conditions and real life experience. It would be unwise to meaningfully depart from these values.