# Thread: Torque or force to bend tubing?

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## Torque or force to bend tubing?

The plasma table project is coming to a close and it's time to start gathering data and material for the next project. This will be a rotary tubing bender. The question is how much torque dus it take to bend tubing. The biggest I can think of bending is 2"X .125" chrome molly. I know there's a lot of variables. But what I'm looking for is a base line to get started with. So lets say bend 2" tube with a 12" lever. I really don't have any of the materials or hydraulics to gather. I have some 1.5" plate for the table and that's it. I don't like picking up stuff that doesn't fit my need at a given time. I like to leave it for the next guy that needs it. Except in the scrap yard I'm one of the few that gets to take stuff out of the scrap pile. And if it stays it will get melted.I will buy the shoes for the tubing and build the rest some thing like this.
http://www.vansantent.com/mandrel_bending_machines.htm

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I analyzed this by calculating the centre point load it would take to produce a 1" deflection of a 2 foot x 2" x .125 wall tubing supported at both ends. A 20 ton load produces a deflection of 1.18 inches. From my own experience this is about as much as can be applied without the member going into non linear buckling failure. I can't quantify that other than to say I have a pretty good feeling for these things.

20 tons will require a 4" diameter cylinder if run at usual hydraulic pressures of no more than 3000 psi.

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Found some info but how do I apply it. Do I use yield strength as when it deforms. Or tensile strength. I would think tensile is when it fails.
http://www.aedmotorsport.com/4130/RoundTube.htm

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Thanks Evan, I was thinking of at least a 4" cylinder myself. If I subtract for the cylinder rod. I come up with 17.2 tons. I think I'll bump up to 4.5" to 5" cylinder.

17.2 M T
16.2 US T
Last edited by jeremy13; 01-11-2010 at 08:38 PM.

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## Bazaars

If it were me, I'd have a look at local shops that do small pipe-work ("Schedule" as well as vehicle/bike exhaust work) and see what they have and do and get some "hands on" advice.

I'd probably seriously look at buying machine either from a "Trade" supplier or eBay etc.

Here is a quick selection from OZ - I've seen them cheaper in the "bazaars" (small tool shops):
https://www.machineryhouse.com.au/Pr...stockCode=P072

https://www.machineryhouse.com.au/Se...mageField.y=28

Use the "no GST" (Goods and Services Tax = 10%) cost.

AU\$1 ~ US\$0.90

I am sure they will be cheaper in the US.

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Here is a post that I made on the HAMB board a while back that might help you:

"I just bent this scrap piece of tubing on my Hossfeld bender and here is what I found. The tube is 2" x .120" HREW P & O material. The die set is a 6" CLR. I have a 36" stroke cylinder for doing 90° bends in one stroke that has a 2-1/2" bore. At the start of the bend (when the leverage is the lowest) my pressure gauge reads just at 500 PSI. As the cylinder extends, the pressure drops to a little over 400 PSI and then starts climbing as it approaches the 90° mark.

The area of the piston for a 2-1/2" cylinder is 4.909 Sq. In.. Thus the force exerted by the cylinder at 500 PSI is 2454.5 Lbs. The force is applied at a pivot pin that is 19" from the center pin. The pin that the wiper die is pivoted on is 8-3/4" from the center pin. Thus there is a ratio of 2.171 to 1 to add into the calculation. That gives a force of 5325.72 lbs at the wiper die pin.

Converting this to torque is where I get lost on this. Maybe one of the engineers on here can check my math and take it from here."

I didn't get any replies from any engineers about the question though.

I did repost the same info on the Off Road Fabrication board and got this reply from one of the members:

" At the start of the bend (when the leverage is the lowest) my pressure gauge reads just at 500 PSI. As the cylinder extends, the pressure drops to a little over 400 PSI and then starts climbing as it approaches the 90° mark. The area of the piston for a 2-1/2" cylinder is 4.909 Sq. In.. Thus the force exerted by the cylinder at 500 PSI is 2454.5 Lbs.

(His response) This is correct

The force is applied at a pivot pin that is 19" from the center pin. The pin that the wiper die is pivoted on is 8-3/4" from the center pin. Thus there is a ratio of 2.171 to 1 to add into the calculation. That gives a force of 5325.72 lbs at the wiper die pin.

(His Response) These measurements are not really useful without knowing the angle between the cylinder and the arm. or the wiper and the arm. Ignore these for now.

Converting this to torque is where I get lost on this. Maybe one of the engineers on here can check my math and take it from here."

(His Response) Torque is a function of the force and PERPENDICULAR distance. The distances and ratio you applied above give you incorrect results because you weren't dealing with perpendicular distances. You probably know from using a cheater bar or ratchet wrench that you have to pull perpendicular to the wrench handle. If you push along the wrench handle, no torque is generated. Also as you make the handle longer, less force is required to make more torque.

In the case of the bender, the same is true. Take a look at the picture below and you can get an idea of the torque the tube needed. Your force calculations above are spot on, but what you need to measure is the perpendicular distance (shown as "r") from the force of the ram ("F") to the center pin. You can do this by placing a 90degree square along the length of the ram and measuring to the center pin. This measurement will change as the bender moves through its range of motion, which is why the pressure in your cylinder also changes even if the torque to bend the tube may be roughly constant. Friction of course also accounts for some variation.

Part of the reason for wanting this information is that I have considered building a new bender that operates in a different manner. Kind of a poor man's mandrel machine.

Drawing is incomplete but you can get the idea of how it will work.

If I knew what amount of torque my old machine was applying to the tube on the largest material that I might encounter, I should be able to use that in sizing the sprockets and cylinder for a new machine. If I follow what you are saying, the radius of the sprocket would be the effective arm length that the cylinders force would be applied to and it would be constant due to the rotary movement of the sprocket.

Hope this isn't too confusing. I type very poorly, so I got lazy and just copied and pasted. I hope this might help and I'll be interested in seeing the replies that you get.

George

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Get out of my mind!! George

That is exactly what I want to build. But I nixed the double end cylinder and thought about a large spring on the other side of the I beam for the return. I don't think there is a need for power return but I might be wrong. The 12" I threw out there. more like a 12' diameter. Also the chain drive dos away with the ram getting closer as the bend progresses.I have no one thing I have to use so there is a lot of give and take. Cylinder to small, Use bigger sprocket and pump psi. to low, use bigger cylinder

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I sure don't like the idea of a chain drive. All chains have some stretch in operation. Combined with the friction in the dies on the tubing this could result in some pretty severe stick slip effects. I don't have a power bender but I do have a shop built strong arm machine that will bend up to 1/4" by 2" material to 90 dgerees. When you use such a short lever arm to apply the force it greatly amplifies the effect of even tiny amounts of stretch and flex in the system since you don't have the torque multiplication that a long lever arm gives.

This was something I really noticed when I built a power drive for my picket twister. It takes up to 500 ft lbs of torque to twist the heavier material and going to a 10 inch sprocket instead of a 3 foot radius pair of handles required some pretty serious gearing and a powerful motor. When a picket has been twisted to a high degree and I declutch the motor it springs back almost a half turn.

Basic rule is everything is made of rubber. The chain system is adding a lot more rubber.

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## Drive System

You could also think about a rack and pinion drive system. The cyclender could push the rack.

I have quite a bit of experience in tube bending. You are on the right track from the looks of your drawing.

A well known brand of bender of this type is "Pines" Google them you will find a lot of good info. for your bender and tooling

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## Pines

I found the Pines web site. Here is the link
http://www.pines-mfg.com/

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