Linear force > Angular torque > Mechanical advantage

[Travis McWilliams]

Guys, I'm trying to figure some stuff out, but I'm stumped. I need to figure the what my output would be.

I'm applying linear force via hydraulic cylinder. The cylinder will turn a "gear" and in turn turn another. The "gear" is not necessarily a gear, it and the hydraulic cylinder will mate with each stroke of the cylinder. It will push it to be simple. The "gear" will mesh with the other final gear.

"gear" = 2"
"final" = 3"

Single Acting Cylinder
Cylinder surface area = 3"
Cylinder Stroke = 0.75"

Hydraulic Pump Pressure = 10,000psi max
Pump flow = 60 in3/min
Pump HP = 1.7

So I'd like to know what the max torque output would be @10,000psi.
I'd also like to know how to figure this on my own. I know how to do a little bit, but I'm not even sure if I"m doing it correctly. I'm trying to teach myself this stuff, but all the places I'm looking aren't exactly related to my scenario. Thanks guys, any help is greatly appreciated!

 

[Vedward]

Sounds interesting. How is the connecting rod from the cylinder connected to the gear?

What are the dimensions of the rod to the center of the gear?

With this information we can calculate your questions, but there may be some inefficiency and friction factors which will need to be estimated.

Is a sketch available?

V. Eddy

 

[Travis McWilliams]

Let me get a sketch together, with dimensions.

 

[Travis McWilliams]

Ok, so the CAD program I use is still under development and is very buggy. I did a quick model, The cylinder is not to proportion. The cad program crashed before I could save any of my individual models, so this is the best I can do at the moment. However, the cylinder will push the two bars on "gear" The end is the cylinder has two "U" shaped tips which would push and allow "gear" to rotate and still push at the same time. "gear" and cylinder should be lined up to perform the best mechanical advantage. Nothing is concrete here, I'm just trying to get an idea of what I'm looking at. The way I have it now is:

"gear" = 2" Diameter x 1.25" height - 5 vertical bars with a 0.30 diameter
"final output" = 3" x 1" height
I know my cylinder stroke will depend on the distance it takes to travel 1 bar at a time and still be able to retract freely and engage freely once more. I'm not too worried about that at this point. If I need more or less bars, stroke length, or a different pushing mechanism that is fine.

I'm needing the cylinder to be as small as possible, that is my main focus, and it needs to be able to put out 8,000 ft/lbs of torque to the "final output"

There will be quite a bit of friction, but 8,000 ft/lbs should overcome that. I don't know a whole lot about adding friction into the equation. I guess you would have a friction coefficient to degrade the torque rating? I have no clue on how to even find the friction coefficient for my application. 0.20?

If I left any other needed information out please let me know. Thanks again!

 

[Travis McWilliams]

Also, everything is a variable except, "final gear" It must be 3"

And the cylinder can push at any angle. I was thinking at a 90 degree angle would be the most efficient and allow the most mechanical advantage...

 

[CWatters]

How do you ensure the cylinder pushes on a rod and not a gap between rods?

 

[Travis McWilliams]

Everything will be all buttoned down. "final" will not move on its own.

 

[Travis McWilliams]

Anyone?

 

[Nidum]

You will probably get better responses if you explain what you are trying to achieve ie what does your mechanism actually do ?

 

[Travis McWilliams]

It will be used to torque a bolt. 1 7/8" stud - 2 15/16" nut.

 

[JBA]

I have seen a hydraulic cylinder powered assembly used for tightening subsea flange stud/nuts that consisted of a simple box end wrench design with a clevis/pin connection between the end of the wrench arm and the cylinder rod end. Of course, the cylinder had to be retracted and the wrench reset on the nut after about each 60 degree rotation of the wrench.

From viewing your diagram, I suspect you are intending to achieve a ratcheting motion reset the cylinder against each succeeding cross bar on the first gear on each stroke to keep from having to reset the "wrench" on the nut after each stroke. Are you using two gears due to the space restriction around the nut?

 

[Travis McWilliams]

yes, there are plenty of hydraulic torque wrenches on the market which are much simpler than what I'm doing here. however, I just really would like to figure out how to solve this, it's nagging at me!

 

[JBA]

Please see my edited version of my last post.

 

[Travis McWilliams]

JBA, correct on the confined space issue. But, no ratcheting. I have left out other major details which would clearly explain my design. I'm just nervous about sharing those details, if you know what I mean...

On just about every low clearance hydraulic torque wrench, double acting cylinders are used. They use more of a bar leverage mechanism to achieve mechanical advantage. Ratcheting is produced from a triangle with teeth which engages the "final gear" with each stroke and disengages on retract. So no resetting wrench with each torque. I know these tools from top to bottom, and they are too big (width wise) some ratcheting systems are better than others, and the ones with decent ratchet systems are way too big. I use these tools in very low clearance areas, blow out preventers on drilling rigs.

 

[Travis McWilliams]

Also, I believe I know which sub sea torque tool you're refferring to. Did It have a "rooster tail" extruding from the hex and a cylinder attached to an arm which went around another bolt a couple of bolts from the bolt being torqued?

If so, those are old school wrenches, quite dangerous, and inefficient due to the fact that the point of contact with the rooster tail varies. They will break almost anything and are extremely easy to maintain.

 

[JBA]

That was one design used on one sub sea construction site.

I understand your desire to keep critical elements of your design secure; but, to get any help on torque leverage, at least some basic layout of the design is required. One issue on applying the torque by pushing on the edge of a rotating disc or arm is that (as you stated above) the actual torque is reduced with the angle of rotation away from a 90 degree orientation of the load point to the center of rotation. As a result, it is hard to insure a given desired torque will be applied at the final tightening point of the nut.
With regard to the torque ratio of two gears, the amount of torque applied from one gear to another as shown in your figure is the inverse ratio of the gears diameters, i.e. if a given torque is applied to a 4 inch gear, the torque from a driven 2 inch gear will be 1/2 of that applied to the larger gear. Alternatively, if you apply a force to the edge of the first gear, as shown, then that gear is the equivalent of an equal arm length center pivot lever pushing on the edge of the second gear and the resulting torque of the second gear on its shaft is simply the amount of force applied to the edge of the first gear x the radius of the second gear.

 

[Travis McWilliams]

Ok wow, that is way more simple than I thought. So, it should be:
Force from cylinder x the diameter of first gear = first gear output
and then
First gear output x radius of second gear = final output torque

f x d = op
op x r = t
Is that right? Thank you very much for helping me, I greatly appreciate it.

 

[JBA]

Actually f = op
op x r (of the second disc) = t (of the second disc)

Remember that I said that the first disc was like a lever with equal arms and a center pivot, ie, a teeter totter, so a force on one end of the lever results in an equal but opposite direction force on the other end of the lever.

 

[JBA]

The diameter of the first disc has no effect on the force applied to the second disc. The relative diameter of the discs only effects the ratio of the amount of angle of rotation between the two discs, ie, if the diameter of the first gear/disc is twice that of the second gear/disc then 30 degrees of rotation of the first gear/disc will result in 60 degrees of rotation of the second gear/disc.

 

[Travis McWilliams]

I get it, ahh that was my first thought and then I started overthinking again. The force applied on the second gear would have a constant mechanical advantage... spoke too soon

 

[Travis McWilliams]

So, I have a hydraulic cylinder in which the base is an ellipse. It measures 1.4375 by 2.0625. I find the area with (a x b x pi) which = 2.33. Then I find the output force with ( PSI x Area ) which = 23,300. Then I would multiply 23,300 by the radius (1.5) of second gear, which I'm getting 34,950 ft/lbs torque, right? That would be the magnitude torque rating? Then I would have to account for friction, which I guess I need to do a lot more studying on...

I've read in I believe it was API Specification 16A, that the friction coefficient with graphite and heavy oil lubricant is 0.13. Could I use that as my friction factor?

 

[JBA]

Your calculation looks OK except you forgot to divide the gear radius by 12 to get the radius in ft. Additionally, a std 2.80" O.D. gear (see the below) has pitch diameter of 2.40 in. so the correct load radius is 1..2/12 = .10 ft, and your torque (ft-lbs) = 23,300 x .10 = 2,330 ft-lbs. (still in the range you require for the above bolt assuming you use a fluoropolymer coated bolt friction factor of .07).

The major problem is that small diameter gears have small teeth; for example, a std 2.80" O.D., 5 pitch, 12 tooth, 2.40 P.D. x 2.5 in width spur gear, of 4140 hardened steel with a tensile strength of 156,000 psi has a maximum safe tooth load of only 6,370 lbs, well below the load you need for your torque requirement.

So, as you said, you still have a bit of work to do.

For reference see the below sites:
http://www.botlanta.org/converters/dale-calc/gear.html
http://www.woodcousa.com/17D-TorqueChart.htm

For my basic gear size data see Pg 20 of:
http://www.bostongear.com/pdf/upload/lit/P-1482-BG_pg005-050.pdf