View Full Version : Torque or force to bend tubing?

01-11-2010, 08:50 PM
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.

01-11-2010, 09:14 PM
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.

01-11-2010, 09:22 PM
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.

01-11-2010, 09:29 PM
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

01-11-2010, 10:05 PM
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):


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.

George Barnes
01-11-2010, 10:15 PM
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.


01-11-2010, 10:55 PM
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

01-12-2010, 12:00 AM
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.

01-12-2010, 01:21 AM
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

01-12-2010, 01:27 AM
I found the Pines web site. Here is the link

01-12-2010, 06:40 AM
Checking the specifications on their pipe benders is very instructive. The model 2 will bend up to 2" sched 80 pipe and comes standard with a 15 horsepower motor and an option for 24 hp. Granted that they are production machines but it does give an idea of what is required to do the job. Even if you trade off speed for reduced horsepower it doesn't reduce the forces involved.

I have a 20 ton log splitter and I am running it with a 15 hp Briggs. It has a 4 inch cylinder and when it is working at max effort the entire frame springs visibly. The frame is a large wide flange H beam about 8" web x 6" flanges.

Bending 2" x .125 chrome moly pipe is a pretty ambitious goal. To achieve it is going to require some heavy equipment and a careful analysis to eliminate any possible weakness. The one main thing to remember is that leverage is your friend.

The geometry is also important. Bending steel requires maximum effort to start the bend. Once the crystals are sliding the force required drops significantly until work hardening begins. That happens near the ultimate yield strength of the material and the force required suddenly skyrockets just before the material fails.

The other very important factor is the speed of the bend. The slower the bend the greater the force required throughout the entire bend cycle. It also makes a difference to the amount of bend and the radius of bend that can be achieved. When the material is bent quickly it tends to flow more easily as temperatures at the crystal boundaries are higher. This is easily seen with my picket twister. For maximum twist I bend by hand since I can twist the handles much faster than the motor drive. Using the motor I can get about 10 to 12 twists per foot on 1/2" square bar. By hand at twice the twisting speed I can get nearly 20 turns before failure.

The ductility of steel is very dependent on the deformation rate. That's why explosive forming is able to produce shapes that can't be attained by other forming methods.

01-12-2010, 09:15 AM
I've seen quite a few benders that use roller chain and sprockets so it's nothing wild eyed.

If I were building it though I would use two strands of roller chain and two sets of sprockets.

Separate the two strands far enough apart so the cylinder can fit between them.

Use a double acting single rod cylinder,but fix the rod to the machine frame and let the cylinder move with the chain.Doing this you get the maximum force from your cylinder on the power stroke.

The base of the cylinder would be a heavy steel block which both strands of chain would pin to and close the loops.

With two strands of chain smaller chain can be used.Two strands of #60 will carry more load than one strand of #80.Also smaller teeth means more of them around a given circumference.More teeth generally means more ability to transmit torque in a chain drive system.

You won't have much problem with stretch,forklift mast chains don't stretch much after all.

01-12-2010, 09:31 AM
A forklift is a different situation entirely since there isn't a stick/slip effect but I'll take your word that it is workable on a bender. I still don't like it though.

01-12-2010, 10:08 AM
Hope those chains don't snap.. that + spring of tube = chain whiping out at mach 3 towards whoever is standing in front of the machine.

01-12-2010, 08:42 PM
As far as chain breaking Burden Surplus has #160 4 layers 10' for $70.00. Cheep enough to over build with it but they don't have the sprocket. There's one on Gov. Surplus. I bet you cant find it right now probably under 6' of snow.
Hum links won't work.

Evan you mention 20 ton on your log splitter with 15HP. As I under stand it you can cut the HP in half if you use a electric motor. Is that correct? If so the Pines one gives 15HP as standard and 25HP up grade. So does that mean there using more like 40 tons.

01-12-2010, 10:59 PM
how about using the load chain of a lift truck?, i might be talking crap but its the first thing popped into the cranium, fairly sure you can get the strength needed
oops sorry that's what you were already talking about, my apologies

Mike Burdick
01-13-2010, 12:04 AM

Here's something for you to look over...

Pro Tools offer plans for a tube bender - not a mandrel bender like you posted but this will do most jobs! If you phone them they'll email you the plans for free.

Please be aware that the plans use their dies etc. so it won't save all that much money but it does give one an idea of what's needed. One can always duplicate the dies if one has the tools to do so.

Here's the link - look down the page and you'll see the offer.


01-13-2010, 05:44 AM
As I under stand it you can cut the HP in half if you use a electric motor. Is that correct? If so the Pines one gives 15HP as standard and 25HP up grade. So does that mean there using more like 40 tons.

Those are production benders where time is money. They are fast machines. Since horsepower is force over time the shorter the time the greater the horsepower requirement.

01-13-2010, 07:04 AM
xtreme cycle.. xcycle? they posted a pdf on a model 3 bender drawings.. as mine is a clone I made on the milling machine off another I had (sold for child support payment).. On the back plate, I added to the length to sit the cylinder in the middle of the shoe.

I have the pdf if you need it. They post the center line radius of bending shoes, you can calculate cutting a tube-radius in the shoe. THEY are on sale right now on ebay, I think I saw them for $95.. now.. it'd be hard to find a chunk of steel and the lathe time for that. IF you must thou? I rigged a radius cutter from a 4" rotary table from HF and a lantern toolpost on a cross slide on it. I had to cut slow or chatter would result.

My bender has a 120 volt greenlee pump. (like on a conduit bender or knockout, I bought five of them at auction, they sold on ebay for $500 each except this one). and a two inch by 24" cylinder, it makes about 110degrees without moving the pins.. a 90 is mostly what you need, but I do 180's sometimes, cut them and weld them into a circle for a english wheel or? handwheel. All this is mounted on a 30x30 steel fab table.. sets off with a pin, all the shoes mounted on a rack, has a storage box for pins and lower shoes.. protractor-etc... This has wheels on one end to allow wheeling around like a wheel-barrow, but sit firmly when sat down and not roll around as you adjust tube to bend positions.

My bender will bend 2"x.25 wall tube. No groaning.

Scrounging? well a chevrolet power steering pump so they tell me works at 1200psi.. so.. when I rig the huge hydraulic pump off the pulley on the wrecker project here, I'll tap a 1200psi regulator to the steering and hydraulic brakes booster. I have saw them power steering pumps used for all kinds of stuff.

To make it more easy? I turned the model3 clone up on it's edge, let the pipe ride across a trolley with a old cracked lathe chuck on it with a 0.68rpm gearbox and 3phase motor. This created a second and third motion axis. OR? a U-bolt, piece of angle bolted firmly and a $19 craftsman digital level will do the secondary indexing, if you have someone to hold it in place. At one time all this had a plc running it with a digital readout. Mine has a crane-cams degree wheel on the lever that moves the shoe now.

Don't scrimp on the software, spend the $59 on bend-tech basic package if you can.. I got the full design package here on a laptop.

A three roll bender like G Barnes made works too.. Also on angle and strap if you do the rolls for it.

I have pictures somewhere..
Found a picture, lil plc is now in Canada staying with a HSM friend. All painted a non-intrusive bronze color.. quiet ain't it? FUng Sweigh?? Ha..

ONE MORE GOOD DEED? if you need the plans, pm me your email where you can recieve a pretty large pdf file.

George Barnes
01-13-2010, 06:06 PM
Sorry for the delay in posting. Had to sneak off to Tulsa for one night of the Chili Bowl Nationals. (Midget Race Cars)

I have actually seen two mandrel benders that appeared to be self built. One for sure and the other one I would bet on it.

I used to go down to Wichita to a company that did a lot of fabrication for the aircraft industry and they had 3 or 4 big Pines benders and rack upon rack of tooling for them. I remember being shown a piece of 4" x .028" stainless that was bent on a 4" C.L.R. with no kinks. WOW! Anyway, over in the corner and covered in dust was what appeared to me to be a home built version of a mandrel machine. No castings, all weldaments using a lot of structural shapes. I never did ask but I would about bet that that machine might have been the tool that started the company on the road to what it became. Just a guess on my part though.

Here is a drawing of the back side. It used the double acting cylinder mounted to the beam on one side and a solid bar on the other side that would slide horizontally in mounts and the chain on each end traversed on the sprockets. Just changing lineal motion to rotary. This bar would serve as a mounting for stops that would use a micro switch to activate a solenoid valve to control degree of bend.


The clamping action for the clamp die and the follow block were over center slides, no hydraulics other than the main cylinder.

I've wished a thousand times that I had taken some pictures and measurements on that machine. The drawing is a pretty close representation of how I remember it appearing.

The second machine was one that a fellow built himself to use in his sideline building dragster chassis. He built his dies from aluminum as he was not doing thousands of bends. On his, he just used a single acting cylinder in a pull mode and had a couple of large garage door springs to do the returning after the bend. He used a single strand chain of about a !" pitch which I believe is a #80 size. For extracting the mandrel, he had a threaded rod on the end of the mandrel rod and a nut. He used an impact wrench and a socket to pull it out.

This machine would not be well suited for production work as it was pretty slow and required a lot of fiddling but it produced good results. Thin wall 4130 is not the easiest thing in the world to bend.

I'm sure that a rotary draw bending machine should be built an entirely different way and use totally different materials, but I might someday just have to try it this way and If I do, I'll relay my findings. I tend to use readily available parts and operate on the old racer axiom...If it don't break, it's too heavy!

Guess that I should finish the drawings first

01-13-2010, 08:32 PM
Drawings ? We don't need No stinking drawings.....:cool:

Well I did plan on buying the dies. Probably RMD but their not the cheapest. your drawings are purity close to being what I had envisioned. If you look at the bender on Gov Surplus looks like I beam to me and not to much cast stuff.

George Barnes
01-13-2010, 11:30 PM
Have you got a link to the government surplus item that you mentioned? I'm interested in ideas, not purchasing stuff.

This stuff is so much easier to make in CAD than in metal. Makes me lazy!

01-14-2010, 08:51 AM
I will try the link last time they didn't work from their site. It's(govliquation.com) Go to the right column and click machinery then left column to bending and forming machines. What CAD program are you using? I down loaded Mach3 and Dolphin to use on the plasma table.

01-14-2010, 09:31 AM
Yup,there's the roller chain photo 14-


George Barnes
01-15-2010, 11:09 PM
Alright, I think that I will have a crack at figuring my question out mathematically.

My present setup is a 1-1/2 h.p., 1725 r.p.m. electric motor that drives a vane pump through a 2 to 1 reduction timing belt drive. The vane pump will show that it is producing just about 1500 p.s.i. if the cylinder is topped out. It will extend the cylinder its full 36" stroke in 20 seconds with no load. The cylinder (2-1/2" bore) has an area of 4.91 sq. in. and with the 36" stroke that gives 176.76 cu. in. in 20 seconds or 530.28 cu. in. per minute. That converts to 2.295 g.p.m. (530.28/231)

From my first post, it was shown that I had 5325.72 lbs of force that were being applied to the pin that was 8.75" from the pivot. This accomplished what was needed to bend the tube. Based on the reply from the OFN, the middle of the travel is where the rating would need to be determined (when the arm is 90 to the cylinders centerline). Since torque is rated in ft. lbs. , I think that I need to convert the 8.75" to .729'. Torque is T=F x D so it follows that T= 5325.72 x .729 or 3882.45 ft. lbs of torque. If that is correct, it's nice to know but I'm not sure what it does for me.

The sprocket in the drawing has a P.D. of 7.2" or a radius of 3.6". That would be the arm length that would be used for the application of the force that the hydraulic cylinder would input. I think that the arm length would remain the same through the bending process. The ratio of the two arm lengths is 2.43 to 1 (8.75/3.6) so it follows that the force applied to the sprocket would need to be 12,941.5 lbs.. (5325.72 x 2.43)

The hydraulic cylinder shown in the drawing is a 4" bore with a 1-1/4" rod. The area of the piston is 11.33 sq. in. after deducting the rods area. Since the pump in the present arrangement is capable of producing 1500 p.s.i. , a force of 16,995 lbs. is available to do the bending. In my present setup, the pump doesn't have to work as hard as this one would require but it should be capable of doing that. If my math is correct, it looks to me like there should be plenty of power to bend the range of tube that I am interested in doing.

The other thing that would be involved would be the speed of the bending process. The 36" of extension when applied to an arm that is 19" long gives 142.66 of radial travel. To get the same radial travel on the 7.2" sprocket would require 8.96" of stroke on the 4" cylinder. 101.52 cu. in. (101.52 x 8.96) converts to .439 g.p.m.. 5.23 times as fast according to my calculations. Probably need to slow it down some.

Well, I don't know it this has been done correctly or not. Feel free to correct and point out any errors.

01-15-2010, 11:44 PM
Well, I don't know it this has been done correctly or not. Feel free to correct and point out any errors.

How much horsepower is really available? The motor, if it is a decent high efficiency type will put out about 80 to 90 percent of it's rated input. If it is an ordinary electric motor it is more like 70 to 80 percent. It could be as low as 50% for a cheap import. Then you need to make a stab at the mechanical inefficiencies and losses in the system. It's important that the pump maximum output match the actual volume required by the system in operation. If it doesn't the efficiency losses go up drastically. Under best case conditions overall system efficiency will be around 25 to 30 percent of input power.

You need to take that into account in your calculations to determine how much force is really available and how fast it can be applied. If the pump cannot move the cylinder at a rate equal to its maximum output rating then oil with be flowing through the bypass and contributing nothing to the effort produced.

01-16-2010, 08:30 AM
SPEED is the last thing a HSM'er needs.. being able to HIT the mark (degree) is it..

(ya racer)

My unit takes 30-40 seconds to bend a 90 degree bend, that gives my old reflexes time to catch it where I want it. I keep saying I will put a "adjustable stop limit in".. but.. alas.. hate to modify something I rarely use these days.

That being said, I got a modified design for the G-Barnes 3 roll bender I'd like someone to draw up in 3d just like you did there George.. It is a unique design not produced anywhere. I even figured a way to leave the encoders and computer off it and it faithfully reproduce the "variable rate bend".

01-16-2010, 09:27 AM
Speed is important. The faster the bend the easier it is to do the bend and the better the metal takes the shape. A fast bend produces higher temperatures at the slip planes of the crystals and increases the momentary ductility of the metal.

George Barnes
01-16-2010, 04:49 PM
Evan, I'm guessing that the math and the application of the math must be alright as you have not pointed out any errors.

I guess that to simplify what I am trying to determine is if I were to build a new machine, how close would my old power combination be to doing the job? I know that it will do everything that I want to do on the present machine. The example of the 2" x .120" wall tube is a very realistic sample. I have bent 2" x .250" wall D.O.M. and 1-3/4" x .188" wall 4130 on this machine, but I don't want to do very much of it as this old girl moans and groans when I do those. I bend bunches of 1-3/4" x .120" wall on a 3-1/2" C.L.R. without any problem.

With all of the problems that you point out, It would seem to me that they would be present in the old unit too. Lets say that I just removed the old motor, pump, drive, control valve, etc. and used them on the new machine. The new machine would have ball bearings on the pivot points instead of the pin in a hole and arm sliding on arm situation that now exists. I'd think that should be better.

I realize that I would be applying the force on a shorter lever length, but would be using a larger cylinder and more of the pumps capability to accomplish the goal. If I've done the math right, it looks like that should work.

The gallonage figures don't particularly bother me as increased speed would be nice when doing big runs of parts. A flow control valve could be used to slow things down some. I used to have one on the old machine but I got to the point of having better manual dexterity on the control valve for stopping the degree of bend when determining spring back. Once determined, I set a micro switch and it controls a solenoid valve in the extension side of the cylinder. The bends are exactly the same with any difference coming from the material itself.


I guess that the only way to find if this will work is too build it...in metal.

George Barnes
01-16-2010, 04:52 PM
David, Is this what you want?


01-16-2010, 09:44 PM
The factors I pointed out aren't problems George. They are just constraints that need to be identified and quantified to ensure a successful design. I really don't know how much loss of efficiency you can expect but it is significant. Hydraulic systems in general have high losses because of friction both mechanical and fluid. Some of those losses show up as heat in the hydraulic fluid.

In general, the greater the loads and the pressures the higher the losses. The higher the fluid flow velocity the greater the losses. It isn't linear either.

01-16-2010, 10:33 PM
Dang George how long did that ring roller take you to draw up? Look's Good! Went to scrap yard to day. I didn't see any H beam that I thought would work. Just standard I beam. The Idea was brought up to make the H beam. Weld 3 8"X1" plates together. Gives me a headache just thinking of how bad the plates would twist from welding. And then straighten them out. :mad:

01-16-2010, 10:35 PM
There is plans posted on the net, metalillness.com site in documents section.. Thanks to George.. I made a pdf from his post..

Madman, he did some beautiful rollers..