Torsion bar used to support weight 1

[DownHillHero]

I'm trying utilize a torsion bar to support a load as well as facilitate a flipping motion. A similar set up of this can be seen in a product known as a flippac and to start I was trying to figure out how the torsion bar in the flippac is able to support so much load.

The dimensions of the supported section of the flippac are about 80"x64".

I believe the torsion bar is made out of some kind of 4140 steel that has been heat treated and oil quenched. The shear modulus G=11600 ksi and yield tensile is 251600 psi. Since we will be dealing with shear stress, the shear yield is about 50% of the tensile, so 125800 psi. I have incorporated a safety factor of 1.5 giving an allowable shear stress of 83867 psi.

Here are the dimensions and constants I have been using to determine the torque in the torsion bar.

Torsion bar length = 54"

So far I have been using the formula
phi = (T*L)/(J*G)
where phi = angle of twist (rad)
T = torque (in*lb)
L = length of bar (54 in)
J = (pi/2)*r^4 (in^4)
G = 11600 ksi

At the point of interest, the load is about 500 lb which is acting 40" away from the hinging point. This results in a moment of 20000 in*lb, therefore the torque in the torsion bar must be 20000 in*lb to support the weight (the flippac has two support beams in the front, however 95% of the load for two people, about 400 lb plus the weight of the lid is supported only by the torsion bar). The angle of twist in the torsion bar at this point is 90 degrees = 1.57079 rad.

Using the equation above we can find the require radius of the torsion bar to allow for this motion to happen. From my calculation I'm getting r = 0.4407"

Next I used the formula t_max = (T*r)/J to find the maximum shear (t = shear). In this case the maximum shear is 148718 psi.... well above the yielding shear strength of 83867 psi.

Does anyone have any ideas on how flippac does it??? I am stumped.

 

[rb1957]

a 0.04" wall sounds ok, maybe check buckling of the tube wall ?
the neutral point angle reduced 'cause J reduced ? so design for twist angle rather than stress (stress is less critical) ... i'd keep an eye on both criteria (turn your back on one and it'll turn on you like a rabid dog !).

 

[DownHillHero]

Although a torsion bar seems like a fairly simple system actually designing one to do what you want proves to be difficult.

For some of the current calculations I have been making to get results that are within an allowable shear stress I am having to extend my torsion bar to as much as 62" (quite a long bar....) The main problem I am having is that I may be limited to a torsion bar no longer that 54". From calculations I have done I don't think there is any way to design a torsion bar that will give me the proper torque as well as the 90 degree angle of twist at this length.

Any idea on how to in corporate a torsion bar and springs? If a torsion bar is still used with some combination of springs, the problem I see is having to potentially disengage the torsion bar after it reach a certain angle of twist (say about 80 degrees). Would something like that even be possible...??

 

[DownHillHero]

Alright rb1957 starting to get some better results with your suggestion.

What I have been doing is specifying my outside diameter of the rod. For the current calculation I am using an outside diameter of 0.375". Solving for the inner diameter from the angle of twist formula with the max torque (4657 in*lb) I get an inner diameter of 0.3092" (seems like the thickness is very small now t=0.0678" about 1/16 of an inch!) with a bar that is 62" long this will give me a stress of 104.5 ksi which would have a safety factor of 1.2 compared to the 125 Fsu.

One thing that I am concerned with is that we are considering Fsu when really the shear yield should be considered, the bar needs to be within the completely elastic region or else it will have creep and become plastically deformed which would disable the operating function of the torsion bar.

 

[rb1957]

.0678" is actually 2t, no?

FWIW, i'm getting a much higher stress, for torque = 4500in.lbs, stress = 450 ksi ...
OD = 0.375", ID = 0.25" (making a 5' long tube like that might be tricky)

 

[DownHillHero]

Sorry for the mistake on the last post the .375 is the out radius and the .3092 is the inner radius. should give you closer to the 104.5 ksi that I calculated before, however I'm still unsure about if this value will cause creep in the torsion bar it all depends on the shear yielding.

 

[rb1957]

yeah, that's better.

as for creep, i guess only time will tell !?

 

[DownHillHero]

From doing the analysis it has become more obvious why a lot of the flippac torsion bars actually break...

Do you think there would be a way to disengage the torsion bar after a certain point (say 80 degree angle of twist) and possibly catch the lid with some other type of system such as a compression spring?

The problem I see is that if you were to disengage the torsion bar it would go back to its neutral position, and then you would have to preload the bar to be able to close and open the lid... could be interesting to think about because that way the torsion bar would be less likely to experience creep or failure.

 

[rb1957]

don't like the idea of disengaging the torque tube when it's under load !

you could install a counterweight as a temporary thing, only fitted during opening and closing.

you could have a "cunning" shock absorber built into the tube. if you filled the tube with oil, and dragged a piston along the tube, controlled by a cable. putting in more thought than i've time for, i think you could get a mechanism that'll slow the descent of the lid, likely both sides of zero. again, without much thought, it looks as though it'd also function as a lift "drag" ... increasing the lift force.

thinking about how this works ... as i understand it, it falls down to some angle under gravity, for small angles the torque due to weight exceeds the internal torque the tube generates as a result of it's twist. then the operator pulls it down a little to latch it down. so releasing the latches means the tube is twisted more than it wants to be (to react the weight) so it'd raise itslef (but only up to the balance point, and the operator has to lift it beyond this point. so you want something to absorb the potential released as the lid descends (right now we're using strain energy in the torsion bar).

 

[DownHillHero]

Yes disengaging the torsion bar when its under that amount torque does sound like a bad idea.

You are right with the description. The reason why it becomes tricky is because you need something to absorb the potential released as the lid descends in both directions, both opening and closing. The torsion bar effectively does this however is prone to failure, makes sense from some of the calculation we have been doing.

Perhaps a good solution would be to flip to the side rather than over the front, this would significantly reduce the moment arm created by the mass of the lid, also if a torsion bar is used then the length of the bar could be much longer allowing to reach that 90 degree angle of twist much easier than for the short bar.