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.

 

[Walterke]

about 1.4776 radians

Mind your decimals!

On another note:
How do you intend to close the lid?

NX 7.5.5.4 with Teamcenter 8 on win7 64
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Nvidia Quadro 2000

 

[DownHillHero]

Closing of the lid will be very similar to opening the lid. Since the torsion bar is neutral when it is vertical the response will be the same whether the lid is being opened to the 90 degree position or closed the the -90 degree position. In both cases the weight of the lid will cause the bar to twist to a certain point and then the operator will have to pull the lid down the rest of the way and latch it.

 

[rb1957]

yes, but i don't see the torque bar "assisiting" the lift. i see the operator doing twice the work ... his applied torque is reducing bothe the weight torque from the lid and the reaction torque to the vehicle. the advantage the operator has is that his torque arm is bigger than weight (so he can probably lift 100 lbs weight for 50 lbs effort). it sounds to me as though this torsion bar is more a "drop" control device.

 

[DownHillHero]

I see what your saying, and yes his applied torque will reduce both the weight torque and the reaction torque however since the torsion bar wants to return to it neutral position would it not help with the lift?

Heres a little graph I made, what it shows is the force the must be applied by the operator on a 2 foot long moment arm. This is done by taking the difference between the torque created by the mass and subtracting the torque in the torsion bar then dividing by the 24 in long lever arm. From the graph you can see that the operator never has to apply more than 35 lb of force to keep the lid at a static position. If the torsion bar was not there then the operator would have to apply a much greater force, therefore the torsion bar is helping assist the lift?

on the graph the y-axis is the force and the x-axis is the angle of twist in radians.

 

[rb1957]

i was thinking of a free body of the lid ... i thought the torque applied (raising the lid) would be reacted by reducing the torque for the weight ...

talking this through ...
initially the lid is at rest, some 1.47rad down. if the operator applied a force, it'll be reacted by the bar (at the tie-downs to the vehicle) so effectively he's applying a torque ... this'll raise the lid, reducing the weight torque (about the bar). the static balance is achived when net torque on the bar (weight torque - operator torque) causes the bar to twist (less) ...

weight torque = Wd*cos(theta)

theta = torque*L/GJ (or something like)

nett torque = Wd*cos(theta) - Pr ... initially assume the operator load is tangential, then fine tune (knowing that the operator is at a fixed point on the ground.

doesn't look like the operator is doing double the work, but i still don't think the torsion bar is helping (though i do see strain energy being released by the torsion bar, so that suggests it is helping ?)

 

[DownHillHero]

Yes I am also using those same type of equations with the free body diagrams to determine the torques. I think the stain energy is what is helping with the lift, maybe not helping is the correct term, but more reducing the amount of work and force the operator as to do.

Considering the system more do you think it would be possible to get the same result with a torsional spring rather than a torsion bar. My first question would be if there are even torsional springs with only using maybe 4 spring that would be attached to the hinges. This would make it so you didn't have to have a bar run across the entire front. One thing I see being a problem is that torsional spring are not linear?.... maybe this could actually be helpful in supporting more of a load rather than just the weight?

 

[rb1957]

i think a spring would be harder to support at the vehicle attach points ... harder, not impossible.

but the bar is functioning as a torsion spring, what do you gain ? (or am i missing something ??)

 

[DownHillHero]

I also agree using a spring would be more difficult with attaching points because you can simply spline the end of the torsion bar and it would work great.

The benefits of not having to use a torsional bar would be to eliminate the need to have the bar extending all away across the front of the unit.
Also since we are going from 90 degrees to -90 degrees the torsional bar could be more subject to sudden failure (not sure how torsional springs are with 90 degree to -90 degree bending)

I'm really just trying to explore all possible options to get the response I need so I am open to ideas and suggestions.

 

[rb1957]

if your torsion stress (closed or open) is higher than yield, i'd anticipate some creep happening, ie some loosening (and tightening !?) of the tie-down latches (as the bar premantly sets one way or the other). i'd expect solid bar would be somewhat heavy (and inefficient) ... a thick wall tube would be "better".

 

[DownHillHero]

Yes creep would occur so the main goal would be to keep the shear stress under the point of yielding to prevent that. From the calculation I was doing with the solid torsion bar I was getting a maximum shear stress of about 99000 psi and the yielding shear for the material I selected was 125800 psi so creep may not be a problem, however since the torsion bar will always be under torque while closed the bar will set.

from thick wall tubes you should expect to lose some strength correct?

 

[TheTick]

I've done dozens of similar mechanisms, though on a smaller scale.

The first thing you want to do is graph the torque required to hold the lid stationary at a variety of positions (e.g. 0° to 180°). Torque from weight will go to zero as the CG approaches the 90° position and max out when CG is at 0 and 180. (Excel works great for this).

On the same graph, draw applied torque from the torsion bar or spring at various angle positions. This will be a straight line. The slope of the line is determines by the spring rate ("k") and the height or y-intercept is determined by preload.

When you find a "k" and preload that you like, design your torsion bar accordingly.

 

[rb1957]

a tube will have less strength than a solid bar of the same OD. but in a solid bar, the inner material isn't stressed to the same level as the outer material. so a tube with a slightly larger OD can have the same stress as a solid bar, but less (significantly, about 50%, 80%) weight.

125 ksi Fty looks to be about 1/2 hard temper (about 160 ksi Ftu) ... 125 Ftu (1/4 hard), Fty = 100 ksi, may be more readily available (and cheaper), but you need to watch Fsu as well (yes, i know (as i suspect you know) in pure shear the shear stress = principal stress) ...
Ftu Fty Fsu
180 150 108
150 125 90
125 100 75
ok, these for 4130, but 4140 similar

 

[DownHillHero]

Thanks TheTick I have already done something very similar to what you are describing with matlab.

The graph ended up looking something like this.
I only considered an angle of twist from 0 to 90. At zero the lid is straight up and the torsion bar is neutral at this point (no pre-load). The response of the torque created by the mass will be the same regardless the direction the lid falls.

Similar to results you get from your mechanisms?

 

[MintJulep]

Any chance that the "grounded" end of the torsion bar isn't completely grounded?

As you note, at 0 degrees it isn't doing anything anyway.

So maybe there is some play allowed in the ground connection. No ground reaction for say 10 degrees on either side of 0. Cut down the range of twist from 90 to 80 degrees.

There is clearly a positive mechanical stop for the open position of the flipak.

 

[DownHillHero]

Thanks rb1957 you have been a big help in leading me to figure this thing out.
I have a quick question about where are you getting your values??

The values I am using are from matweb and the material is:
AISI 4140 Steel, normalized at 870°C (1600°F), reheated to 845°C (1550°F), oil quenched, 205°C (400°F) temper, 25 mm (1 in.) round

Ftu = 285 ksi
Fty = 251.6 ksi

since matweb does not give info about Fsu so I used maximum shear stress yield criterion which would give us Fsu = Fty/2 or Fsu = (sqrt(2)/3)*Fty if you use the octahedral shear stress yield criterion.

Just like that good old example they always use of applying torque to the piece of chalk giving that nice 45 degree fracture.

 

[rb1957]

no probs, that's what we're here for ...

AR-MMPDS-01 (our standard materials reference) available online
find MIL-HDBK-5 on-line

i'd caution using the ultra high strength temper (280 ksi), it tends to be very brittle (intolerant of scores, damage, etc; not much ductility)

i'd suggest 180 ksi as a maximum for good all round performance.

 

[DownHillHero]

Your selecting a steel with Ftu = 180 ksi?

If that is the case then Fty could be more around 160 ksi giving a Fsu around 80 ksi. I think to have a torsion bar thick enough with withstand the torque as well as the angle of twist would become much greater in length then I would want.
(This is why I suggested a torsion spring. It seems you might be able to get the same torque with two springs that have a 2 in. mean diameter with a 0.5" wire diameter and about 7 active coils. This would take up significantly less space).

 

[rb1957]

as previously posted Ftu = 180 ksi, Fsu = 125 ksi, and i'd use tube instead of bar.

how about a combination tube with a spring inside of it ? the tube would provide a nicer interface than spring alone, and provide a means of "hiding" the spring (from little hands, etc ...).

T = 4500 ft.lbs (i'm guessing units from your chart)
R = 2"
stress = T/(2*A*t) = (4500*12)/(2*3.14*t) ...
thk = 0.1", stress = 86 ksi

 

[DownHillHero]

That is a very interesting idea of placing a spring inside a tube to give you that extra torque that way you could reduce the amount of torque needed in the torsion bar.

From the chart the units are actually in*lbs from calculations the maximum torque would be 4657 in*lbs. I take it that R in the situation is the outside radius.. a 2" radius is much larger than what I would expect (4 in diameter tube seems very large) I would assume that R would not be more 0.5" giving a bar with a 1" diameter.

What I have done is assuming the value you stated and looking at the MIL-HDBK-5 the 4130 steel with Fsu = 125 ksi if i incorporate a safety factor of 1.5 the Fs_allowed = 83.333 ksi. Next I used the equation Fs = T*c/J where c is the outer diameter (I used 0.5") and J = (pi/2)(0.5^4 - c_i^4). Solving for the inner radius I am getting 0.4598"

if I now solve for the angle of twist for this bar I am getting 0.9386 radians, which is only about 54 degrees, a lot less than 90.

 

[TheTick]

Yes, my graphs were similar. In another twist, you can draw your spring curve between two points (e.g. "balance points" on the lid torque curve) and determine ideal rate and preload.

90° is a lot for a torsion bar, unless it is very long. You might try using multiple torsion elements (springs or bars).