need help w/torsion bar spring design

[tdw]

I'm looking for software and or calculations to design a 100 Lb. inch rate torsion bar spring approx. 18" long.
any help will be appreciated.

 

[desertfox]

Hi tdw

I have some formula here that should help you complete your design:-

(T/J)= (t/R)= (GQ/L)

where T= torque = 100lb.inch (in your case)
J= polar second moment of area for a circular
bar this equals 3.142*D^4/(32), for a tube this
equals 3.142*(Do^4-Di^4)/(32)

where D= diameter of bar
Do & Di equal the outer and inner diameter
of the tube.
t= shear stress
R= radius of bar or tube
G= Modulus of rigidity of the material(for steel this
is about 80GN/m^2 or 12*10^6 psi)
Q= angle of twist of the shaft in radians

L= length of bar or tube (18" in your case)


First of all select a suitable material and obtain a value
of safe working stress (ie (U.T.S./2.5) say).

Select a bar or tube that seems suitable and work out its
polar second moment of area, and then using the above work out the shear stress in the bar or tube and compare it with the safe stress ie:- rearranging the above formula we get

TR/J=t

compare t with your safe working stress.

Now to work out the deflectionof the bar or tube in radians
use this rearranged formula/

TL/GJ

Obviously if your value of shear stress for a given section is greater than the safe stress you have selected then you either need a bigger section or a higher strength material.
Note :- as your torque is high it might be better to go for a tube to keep the weight of the torsion bar down.Also bear in mind that the outer surface of any bar or tube carries the maximum stress therefore if you go for a solid bar of a large diameter you are carrying a lot of inefficient weight.

Hope this helps

Regards

Desertfox

 

[tdw]

Thanks Desert Fox;
Weight is not a factor.
Do I understand the safe stress (TR/J=t) must be multiplied by the total number of degrees being twisted, to determine shear stress?
Would you agree that safe stress for quenched and tempered (400 degree) SAE 1085 is approx 76,000 PSI?
Maxium twist is 22 degrees, can you recommend a material?

 

[desertfox]

Hi tdw


Yes an SAE1085 has a safe stress of about 76000 psi,however
this figure may vary depending on bar size.
No I think you mis-understand there is no need to multiply
the safe stress by the number of degrees being twisted, the shear stress is determined by the formula :-

TR/J= t

You just need to compare this value with the safe stress for the material and ensure that (t) is less than the working safe stress.Working on the assumption that you want a torsion bar and not a torsion spring and using the safe stress as above for an example :-

proceed as follows :-

T/J= GQ/L re-arrange to find J

TL/GQ = J

therefore 100x18/(12x10^6x22x2x3.142/360)=6.51x10^-5 inch^4

therefore TR/J = t now using t =76000


100xR/ 6.51x10^-5 = 76000

now transpose to find R

R = 76000 x 6.51x 10^-5/100 = 0.049476 inches

so the diameter would be 2 x 0.049476 = 0.098952 inches.


If your after a torsion spring ie:- a spring with a number of turns made from wire then the formula for these are different to the previous ie:-

k=M/Q=Ed^4/ (10.8DNa)

where k = spring rate
M = moment or torque (100lbin in your case)
Q = angular deflection in radians (22deg in your case
but need converting to radians)

D = mean coil dia

d= wire dia

Na= number of active turns

E= modulus of elasticity of the material 30x10^6 psi
for steel


stress for this spring would be given by

s= (32M/(3.142d^3))Kb

where Kb = stress correction factor (Wahl)

to proceed with this calculation I would need details of what the spring arms where like, size of shaft the spring would be mounted on etc.

Hope this is of some help I am not sure whether your after a torsion bar or spring which is why I have given these other formula. If you require further help just post in here and I'll see what I can do.

regards

desertfox

 

[tdw]

Thanks Desertfox;
First, let me clarify, we are working on a torsion bar spring, not a coil of any sort.
second, I understand;

T/J= GQ/L re-arrange to find J
TL/GQ = J (but I disagree with your result 6.51x 10^-5)
100x18/(12x10^6x22 x2x3.142/360)=6.51x10^-5 inch^4
shouldn't it be>>...

100x18/(12,000,000x22x2x3.142/360)= 3.91x10^-4

therefore TR/J = t now using t =76000


100xR/ 3.91x10^-4 = 76000

now transpose to find R

R = 76000 x 3.91x10^-4 /100 = .296896 inches

so the diameter would be 2 x .2969 = .5938 inches.

 

[desertfox]

Hi tdw

Yes my figure for J is incorrect it should be 3.91*10^-4.
I agree with your calculations also so your bar should be
around 0.5938" on diameter.

regards

desertfox

 

[tdw]

Thanks for all the help.
I feel better now!