Torsion bar used to support weight
Torsion bar used to support weight
(OP)
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.
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.





RE: Torsion bar used to support weight
My guess would be that there is another mechanical support somewhere in the system. Otherwise when you remove the load (people and retaining straps) the thing would would snap shut like a mouse trap and there is no way they would make you pull that hard to open it. The torsion bar is probably only to facilitate easy closing of the system.
RE: Torsion bar used to support weight
RE: Torsion bar used to support weight
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RE: Torsion bar used to support weight
if you're using solid steel bar, weight might also be a factor !? (i'd suggest tube)
4140 seems to max out at 180 ksi (Ftu), which has a Fsu = 108 ksi.
i looked up "flippac" ... couldn't see what you were trying to say ... sketch pls
RE: Torsion bar used to support weight
Rb1957 you are right, there may be a bending stress, however the bar is being run through the hinging device so the weight of the steel torsion bar will have no effect on the actual torque of the system however it may be a significant factor in determining how much shear stress the torsion bar can handle. I will have to check that.
I have attached a sketch to give you a better idea.
RE: Torsion bar used to support weight
RE: Torsion bar used to support weight
"On the human scale, the laws of Newtonian Physics are non-negotiable"
RE: Torsion bar used to support weight
RE: Torsion bar used to support weight
From doing research the supports on the front are more to hold the flippac down when no load is in the lid section as well as preventing the thing from shaking in the wind, however they do not seem to be a significant load bearing support.
RE: Torsion bar used to support weight
and then the torque bar is attached to something (the rest of the world) since it has to take the torque from the lid and put it somewhere.
RE: Torsion bar used to support weight
The way it works is that the passenger side of the torsion bar is the point where is it grounded. This is because is if fixed to the part of the hinge that is attached to the canopy part of the flippac. On the drivers side of the torsion bar, the bar is fixed to the lid. They two other support have some type of bushing that lets the bar rotate. So this mean the passenger side of the torsion bar does not rotate, but the drivers side does.
Another thing to consider is that when the torsion bar is just supporting the weight of the lid (about 100 lb) the angle of twist is 1.359 rad which is about 78 degrees. The lid supports are then used to hold the flippac at 90 degree when only the weight of the lid is effecting the torsion bar.
Here is a sketch of what I mean.
RE: Torsion bar used to support weight
To get a better idea of what this looks like here is a piture of a broken flippac. As you can see distributing the load from the torsion bar can be quite a problem.
RE: Torsion bar used to support weight
i wonder if the structure is at rest (unloaded) with the lid down, and you have to work (pull on some cable) to close the thing ? does it "spring" open or closed ?
there's going to be some "honking" big fastener reacting the torque (with a couple only the diameter of the bar apart) ?
RE: Torsion bar used to support weight
When the torsion bar is in it neutral position (angle of twist of zero degrees), the lid is straight up. This way the torsion bar twists from a range of 90 degrees to -90 degrees (not ideal for torsion bars... but thats why we include safety factors!!) the idea is that this prevents the lid from slamming down into the top of your truck as well as slamming down when you are closing the lid.
To help open and close the system a two foot long rod is attached to the free end of the torsion bar. The operator can then get the needed leverage with the help of the torsion bar to open and close the lid easily. When completely closed or open releasing the latched or hood support rods should cause the lid to spring open and the far end should not raise higher than 18 inches.
To get a better idea I suggest watching this youtube video
http://www.youtube.com/watch?v=ydQ8S-1X64s
I also attached more drawings!
And yes after figuring out this torsion bar the next step is designing fastener. I agree they will have to be quite beefy!
RE: Torsion bar used to support weight
you're using a lid loaded with two people as the loadcase (creating the torque in the bar); but aren't there stops that the lid is resting on (in the down position) ?
if the operator is pulling the the lid down, then it'll spring back (once he releases his pull). i think the torque in the bar depends on the load that the "stay down" latches carry. i suspect that the "payload" of the lid (the 400 lbs) is reacted on the stops, and the torque in the bar is alot less than you've calc'd. i suspect that the latches load up about double the weight of the lid (ie the torque due to the latch load is about double the torque due to the weight), so that when the latches are releasd there is enough energy in the torsion bar to pull the weight up.
RE: Torsion bar used to support weight
I do think your on to something with the stops in the down position, however I looking at different pictures of flippac I can not seem to figure out where the stops are located (could possibly be a cable system running through the tent, once the cable was in full tension it could act as a stop).
The torque on the bar that I calculated earlier was based on the torsion bar being the only force supporting the weight and load. I have changed my analyse of considering the torsion to only facilitate the opening and closing of the lid rather than supporting any load (this is what the flippac really appears to be doing).
I have reworked the problem changing a couple things such as the length of the torsion bar (now 62") I will post an image to show my work. Results are giving a maximum shear of 93168 psi at 90 degrees which gives a safety factor of 1.35 rather than 1.5.
RE: Torsion bar used to support weight
The blue line represents the torque in the torsion bar and the green line represents the moment created by the mass. As you can see the two line intersect. At this point the torque in the bar is greater than the moment caused by the mass, therefore the mass lid will stop at this point.
RE: Torsion bar used to support weight
if the lid starts vertically, and is pulled down and held down, then released; it will spring up because the force to pull it down has been removed.
no?
RE: Torsion bar used to support weight
Now when then latches are released the torque is greater than the moment created by the weight of the lid so the torsion bar will spring the lid back up to 1.4776 radian (the point where the moment created by the weight equals the torque in the torsion bar). At this point the operator then attaches the lever arm and applies force to lift the lid up to the neutral point. Although the moment created by the mass is greater than the torque, there will still be torque in the bar which is creating an assisted lift system.
Does this make more sense now?
RE: Torsion bar used to support weight
On another note:
How do you intend to close the lid?
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RE: Torsion bar used to support weight
RE: Torsion bar used to support weight
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support 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 ?)
RE: Torsion bar used to support weight
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?
RE: Torsion bar used to support weight
but the bar is functioning as a torsion spring, what do you gain ? (or am i missing something ??)
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support weight
RE: Torsion bar used to support weight
from thick wall tubes you should expect to lose some strength correct?
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support 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
RE: Torsion bar used to support weight
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?
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support weight
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).
RE: Torsion bar used to support weight
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
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support weight
90° is a lot for a torsion bar, unless it is very long. You might try using multiple torsion elements (springs or bars).
RE: Torsion bar used to support weight
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 !).
RE: Torsion bar used to support weight
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...??
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support weight
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)
RE: Torsion bar used to support weight
RE: Torsion bar used to support weight
as for creep, i guess only time will tell !?
RE: Torsion bar used to support weight
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.
RE: Torsion bar used to support weight
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).
RE: Torsion bar used to support weight
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.