Torsion Bar to support Tow Bar

[cheddarbob56]

I'm new to the site and so far have found alot of good info. I am working on designing a torsion bar to hold a tow bar of a trailer up. The tow bar pivots about two pins and I need a torsion bar set up to hold the tow bar in a level position when not hooked up. What I'm needing is a little help calculating the size of my torsion bar. I have it modeled as a hex shaft. The coupling is 7" ahead of the pivot (could be moved if need be) and the weight of the tow bar is 505 lbs. at 52" from the pivot. I have two torsion bars modeled up held at the ends and free in the middle coming together to one link. The bars are 12" long each. Could somebody help me with material type and how to determine if I have a large enough bar. To me it seems small but my calculations say I have enough. Oh yes I need about 20 degrees range of motion, 10 up and 10 down from level. I would like it level and then an additional pulling force could be applied to gain the 10 degrees up travel. Any help would be greatly appreciated. Thanks in advance.
Also I've uploaded a couple of PDF's to see what I'm working on. Might be more clear.

 

[desertfox]

Hi cheddarbob56

There is only one pdf not two, however if I understand correctly the hexagon shaft terminates with two bannana shape links onto a rod with eyes, so is the torsion bar pushing upwards?
If so, the ratio of forces from the pivot point ie 52" * 505lb divided by the distance from the above pivot point to the distance to the vertical rod with eyes and that should get you your vertical force on the bannana shaped link.
That force then multiplied by the distance to your torsion bar will give you the torsion on the spring.

desertfox

 

[cheddarbob56]

desertfox - Thanks for the quick response. Yes the torsion bar is pushing the tow bar up to hold it in the air at a level position. The hexagon shaft terminates at the banana links like you assumed and that distance from my pivot to the rod eye link is 7" so I came up with a force of 3751 lbs (vertical) on the rod eye link so each banana link would take half of that. Now that said how would I calculate my torsion bar size correctly and then determine what material it is to be made of?

 

[desertfox]

Hi cheddarbob56

Well the standard torsional equation will be a good starting point.
I assume from your earlier post that the torsion bar is clamped centrally against rotation and the rotation take place at each end.
So the formula you need is the standard one ie:-

[τ]= T*r/J

where [τ]= shear stress

r = radius of shaft

J= polar second moment of area

T= torque on shaft

also you will need:- for the stiffness and deflection:-

T/[θ] = G*J/l

where G= modulus of rigidity
[θ] = angular defl in radians
l= length of shaft

other symbols as before.

Now the one difficulty you have is the hexagon shaft because it J value is not as easy to find as a circular section however you could approximate it by assuming a circular section of a diameter equal to the across flats dimension.

desertfox

 

[cheddarbob56]

So can you use any material type for a torsion bar or does as long as it falls into range with the stress and deflection you need? Or does it have to be some sort of spring steel or something of that sort?

Thanks again desertfox for the response. I remember this stuff but its been a long while since messing with spring stuff

 

[desertfox]

hi Again

Not sure how good your torsion bar will be if you pivot in the middle usually one end is fixed and the other turns, you could do with considering a torsion spring or two springs. basically both halfs of your shaft want to twist in the same direction at the centre which isen't ideal.
I'll have a nose in some of my books.

desertfox

 

[cheddarbob56]

The way I have it set up now I actually have two torsion bars which are fixed at the ends and about 12" long with the other end free to rotate (the end the banana links are connected to are seperate torsion bars and both rotate at that end, they are just supported by a round bushing which the hex will be machined round at that end to fit into the round bushing) if that all makes sense.

 

[desertfox]

Hi cheddarbob56

Ah I see you have two seperate springs okay that sounds better, so now you use the formla I provided earlier, you have your load from your earlier calculation and you know your deflection +/_ 10 degrees.
You will need to deflect each spring a little it first to give it some pre-load then your final deflection needs to meet your desired load.
You need to check the shear stress you calculate with the allowable shear stress for the material your using, a rough rule of thumb is the max shear stress is about 50% of tensile strength.

good luck

desertfox

desertfox

 

[cheddarbob56]

Thanks alot desertfox....you are a great help. I'll see what I come up with. Thanks again.

 

[desertfox]

hi cheddarbob56

Thinking further I am wondering about the movement of the tow bar doesn't it move through 90 degrees? if thats the case the torsion bar will have to do the same movement so your +/_ 10 degrees is out of the window.

desertfox

 

[cheddarbob56]

Desertfox - the tow bar would only go from where it is located down to the ground which around 10 degrees down and then it would probably only range up about 10 from level.

 

[desertfox]

hi cheddarbob56

Ah I thought it might be put vertical when not in use how you doing with the calcs, if you upload them I'll have a look at them.

desertfox

 

[cheddarbob56]

Desertfox - I came up with a torsion stress of 357,706 psi....way to high so I did a little change of design and ran the torsion bar in line with the draw bar so I could get it about 30" long which gave me a shaft of 1" dia and a torsion stress of 59,438 psi so was thinking of using a 4140 alloy with a tensile of 148,000 psi and that would give me 20 degrees of motion I do believe. If I get time I'll upload some calcs for ya to check over.

 

[desertfox]

Hi cheddarbob56

Thanks be interesting to see, if the shaft is 30" long whats it deflection like?

desertfox

 

[cheddarbob56]

In using the T/? = G*J/l equation I set that for 20 degrees and with my new force and shorter lever arm I have a torque of 11,670 in lbs from which I determined to use a 1" dia. bar based on the above equation.

 

[desertfox]

hi cheddarbob56

I used your torque from the last post, assumed a 1" dia shaft and I agree with your stress but deflection I get to be 18.57 degrees but I assume it isn't that critical.

desertfox

 

[cheddarbob56]

Thanks desertfox....I am going through the calculations again to add a little safety factor in there. You are right the deflection isn't hugely critical. I'm hoping to be around 20 degrees but +/- 1 to 2 degrees won't hurt anything.