# Need help with torsion bar design

• posted
I need some torsion bars that will generate between 25 ft/lbs and 40 ft/lbs
of torque with 45 degrees of deflection. The bar must be between 10 and 12"
long, if possible. How do I calculate the diameter that the bar should be?
What material should I use? I do not have a heatreat facility available to
me, but I do have a more than adequate machine shop. Does anyone know a
Steve
• posted
These are not terribly difficult calculations, with all the formulas in Machinery Handbook. Start with 1/4 inch round bar. I don't know if you can get 45 degrees of deflection in that short a bar without exceeding the yield point, however, but that can be calculated also.
Alternately, you can find this out experimentally. Just take a 1/4 inch steel bar, clamp one end and apply 25 foot-pounds of torque. See how many degrees of deflection you get. Then choose a larger or smaller bar if needed. To check for the yield point, apply double the maximum torgue, and then see if the bar returns to the original point or takes a set. No calculations required.
I understand that springs are generally high carbon steel. All steel is equally stiff, only the yield point is different.
Richard
Steve Lusardi wrote:
• posted
Gut feel is that is a lot of torque and angle for such a short bar. Probably eliminates standard round steel bars :(
If you go with the lowest bar and longest length, you stand the best chance. Try 25 lb ft = 300 lb in and 11 inches effective length (need to leave some ends to grab) and you get a diameter of .247 inches with steel. Call it 0.25 inches.
Shear stress with that bar would be 97,800 lb/in^2. That puts you into at least 1085 to 1090 steel and requires careful design of the ends, like splines that are larger diameter than the bars. Closer to practical than I guessed but not a good home shop project.
You can make a bar that meets your requirements by stacking up thin rectangular bars. This is often done for torsion systems that require a high amount of windup and torque in a short length. I did a short search and didn't find any design guidelines online, but the basic approach is.
Design a rectangular shaped bar that will take the windup you want (45 degrees) and be within the stress limits for the material you are using. Calculate how stiff that rectangular bar is, then stack up bars until the total stack of bars meets your stiffness requirements.
The nice thing about stacked rectangular bars is that the end attachments are easy to make and if you need to adjust the stiffness of the torsion bar, all you need to do is add or eliminate bars and modify the end caps accordingly.
As a SWAG, you could probably build what you want by stacking up 0.375 x 0.062 rectangular stock with good properties. But someone needs to crunch some numbers first.
With careful design, you might be able to do it with stacked rectangular stock made of cold rolled 1020 or such. Just a semi-educated guess.
Dick
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I just happen to have an excel sheet that does this problem
thata= T*L/(J*G) Stress = T*D/(2J)
J= polar area moment of inertia solid rod = 3.14*D^4/32 G= torsion modulus of material steel 11.6E7 psi Al 3.9E6 psi 60-NiTinol 1-4 E6 psi depending on heat treat T= Torque in inlbs L length of rod inches D diameter of rod inches
25 ftlbs at 12 inchs with 45 deg windup.
Assuming steel rod I calculate a 0.25 dia rod will do it but it is going to be stressed at nearly 100Ksi in torsion. Not really practical.
Assuming Al rod a 0.33 OD rod 12 inch long with a stress of 44 ksi also a bit high to be practical.
If you can pay I can do it with 60-Nitinol but it will not be cheap. I dont have Ti properties handy but it might work.
Dan Clingman
Steve Lusardi wrote:
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Thanks Dan. With this I should be able to calculate a solution. After revisiting the design and applying some changes, I can fit an 18" length. That should distance the yield point into the safety zone. What is the steel you have referenced? 4340 has been suggested, but would it require heat treatment? Steve
• posted
Sounds like it would be a lot easier to use a rod with a coil spring wound around it. Much lower movement per unit length. Check out this site for some ballpark info http://home.earthl> I need some torsion bars that will generate between 25 ft/lbs and 40 ft/lbs
• posted
It's been pointed out that your bar is too short for the intended specs. An idea - I've no idea if it would be practical for you: The German Panther tank of WWII used torsion bar suspension and was similarly limited in the available length of the torsion bar. There, they took the bar from the wheel-bearing arm, straight across the hull to the other side where it was geared to another bar which ran straight back across the hull again. In effect, they doubled the length of the torsion bar this way, to the great improvement of the ride comfort of the tank.
Staale Sannerud
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The torsion bar springs that hold the trunk lid open on the 'Ford Falcon' (mid 70's) were 5/16" dia. & about 4ft. long. Search the wrecker's yards for something similar & do some experimenting.
Paul Cordell Nth. Qld. Australia.

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