Calculating Weight of Roll Cage Tubing: Density & Cost

AI Thread Summary
To calculate the weight of roll cage tubing, both solid and hollow, one must determine the volume using the formula for the respective shapes and then multiply by the material's density. For hollow tubes, the volume is calculated using the outer and inner diameters, while for solid tubes, the inner diameter is zero. The total cost can be estimated by multiplying the weight by the price per pound and adding labor and other material costs. Bent tubes can be treated as equivalent length segments for these calculations. For material costs, local steel tube merchants typically provide pricing tables online, which can be useful for estimates.
OutCell
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Hi,

How do i calculate the weight of a roll cage tubing, solid or hollow, by knowing the material density?

I am trying to estimate the cost of the cage by calculating the weight. Is that possible? Or should i look at estimating the roll cage from a different point?

Would really appreciate any ideas

Thanks
 
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Well, volume*density will give you the overall mass. So if you know the volume of all of the tubing used and you know the price per pound of the material, then all you have to do is account for time for labor and other materials required for construction.
 
OutCell said:
Hi,

How do i calculate the weight of a roll cage tubing, solid or hollow, by knowing the material density?

I am trying to estimate the cost of the cage by calculating the weight. Is that possible? Or should i look at estimating the roll cage from a different point?

Would really appreciate any ideas

Thanks

For a hollow tube:

V = (d_o^2 - d_i^2) \frac{\pi}{4} \cdot L

where,

do = outside diameter (ft)
di = inside diameter (ft)
L = length (ft)

Then multiply by the density in pounds per cubic foot to find the weight in pounds. Multiply by the cost per pound for the total cost of the tube.

If it is solid, then di is zero. Everything else is the same. Of course you'll need to sum all of the tube segments.

Plus like Fred mentioned, there is a labor cost and a cost for any other componets to the cage.

Hope this helps.

CS
 
FredGarvin said:
Well, volume*density will give you the overall mass. So if you know the volume of all of the tubing used and you know the price per pound of the material, then all you have to do is account for time for labor and other materials required for construction.

Thanks for your time and answer i really appreciate it. This is exactly what i was trying to confirm :)
 
stewartcs said:
For a hollow tube:

V = (d_o^2 - d_i^2) \frac{\pi}{4} \cdot L

where,

do = outside diameter (ft)
di = inside diameter (ft)
L = length (ft)

Then multiply by the density in pounds per cubic foot to find the weight in pounds. Multiply by the cost per pound for the total cost of the tube.

If it is solid, then di is zero. Everything else is the same. Of course you'll need to sum all of the tube segments.

Plus like Fred mentioned, there is a labor cost and a cost for any other componets to the cage.

Hope this helps.

CS

Excellent. Thanks a million, this is what i was looking for exactly. Does a bent tube act as something other than a segment or should i just consider it as a segment and add it up?
Any good sites where i can find a good estimate of steel materials cost? In English pounds if any :)
 
Last edited:
OutCell said:
...Does a bent tube act as something other than a segment or should i just consider it as a segment and add it up?

Just add it in as an equivalent length segment.

CS
 
stewartcs said:
Just add it in as an equivalent length segment.

CS

Thanks CS
 
OutCell said:
Any good sites where i can find a good estimate of steel materials cost? In English pounds if any :)

Google search the name of your local steel tube merchant. They more than likeley have tables on the net that give kG per meter or pound/foot for all the different sizes and grades. We have several catalouges in the office here with all that infomation and more.

The merchant may also have posted their RETAIL rates for purchase although these are far higher than if you send them a nice material schedule and ask for a quote. If you make sure your schedule lists how many of each size and grade you require and also purchase in standard lengths so the mechant doesn't have to do your leg work then they are also more likly to offer a better price. My general rule "The more you nag the more it costs" and this applies to anything.
 
engineroom said:
Google search the name of your local steel tube merchant. They more than likeley have tables on the net that give kG per meter or pound/foot for all the different sizes and grades. We have several catalouges in the office here with all that infomation and more.

The merchant may also have posted their RETAIL rates for purchase although these are far higher than if you send them a nice material schedule and ask for a quote. If you make sure your schedule lists how many of each size and grade you require and also purchase in standard lengths so the mechant doesn't have to do your leg work then they are also more likly to offer a better price. My general rule "The more you nag the more it costs" and this applies to anything.

Thanks. I will send them an email and explain why i need the costs. I am actually not buying the material, i only need the cost so i can calculate an estimate for my report/project in university.

Appreciate your help and time, thanks :)
 
  • #10
How would you calculate the volume of a square and hexagonal tube (hollow/solid)?? Would it be a complicated method?
 
  • #11
OutCell said:
How would you calculate the volume of a square and hexagonal tube (hollow/solid)?? Would it be a complicated method?

The same concept is used.

For example the area of a square is just Length times Width (since it is square it is L times L). Hence...

V = (d_o^2 - d_i^2) \cdot L

For a hexagon the volume is,

V = \frac{3 \sqrt{3}}{2} \cdot (t_o^2 - t_i^2) \cdot L

where,

to = the edge length of the outer hexagon
ti = the edge length of the inner hexagon

Everything else is the same.

CS
 
  • #12
stewartcs said:
The same concept is used.

For example the area of a square is just Length times Width (since it is square it is L times L). Hence...

V = (d_o^2 - d_i^2) \cdot L

For a hexagon the volume is,

V = \frac{3 \sqrt{3}}{2} \cdot (t_o^2 - t_i^2) \cdot L

where,

to = the edge length of the outer hexagon
ti = the edge length of the inner hexagon

Everything else is the same.

CS

Waw, i guessed the square right but your way of the hexagon is easier than what i found :) i greatly appreciate your help
V= 〖[((3 √3)/2 x l^2)〗_out- 〖((3 √3)/2 x l^2)〗_in] x L , this was my hexagon way :P
 
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