Hydraulic Calculations

In this case, I think a drawing would help. Draw it and scan it to a jpg file then you can upload it with the little camera icon.

The force created by any spring is expressed as F=kx where k is the spring stiffness in N/mm and x is in mm. Once you have the force of the spring, then multiply that by the area that it is acting against to get pressure. So FxA where the units of F are N and area is mm² giving pressure in N/mm² .

Having said all that, I am not really sure that I really understood your question and I have probably written what you already knew.

Hi Steve S, Thanks for your quick reply. I have a picture but I cant find this camera icon, any ideas as how to attach it??? Regards Samurai7

Here you go....

Samurai7,

Torque wrench you say...

I work for Hydratight and we design and manufacture torque wrenches and tensioning systems, for the oil and gas industry mainly but we consider any request to be honest. Therefore I am interested in seeing more of the concept you have.

As another poster further down has said, a drawing would be handy in this situation so that we can comment further. You have come to the right place for help on this subject as there are many on here who can point you in the right direction. You may not get the answer given to you straight away but general pointers in the direction you should be looking especially seeing as it is a 'study' question if you catch my drift. It would be wrong for anyone to give you the answer straight away as there would be no learning involved there. Be assured though that you should be able to get the answers you are looking for....

...a drawing would help though :-)

Kind Regards

Kev Brown

Hi, Seventh Samurai!

Welcome to the CR4 crew! Great to have you aboard!

Your torque wrench idea sounds interesting.

Maybe Kev brown at Hydratight (or somebody else there!) can help you patent and market it if it's a new concept.

All the best,

Mark

Thanks to you all who have responded so quickly to my question!

I now have the tools (thanks to Steve S) to provide a basic picture for you to see exactly where I am having trouble with my design.

I hope that this picture can shine a brighter light on my issue for you. Just a reminder though, it is the pressure felt on the inner walls that I am struggling to find an equation for.

I look forward to your advice,

Samurai7.

I'm fumbling about, here, but can't it be seen like this?

If that represents a (200mm diameter) cylinder with a piston, the pressure must be = force / area, so

P = F / A

F = 1kg x 9.8 = 9.8N

A = pi x 0.1 x 0.1 ≈ 0.0314 m²

So P = 9.8 / 0.0314 ≈ 312Nm^-2 - which is the same in every direction (assuming the height is small enough to ignore the weight of the fluid!).

For your app, you just need the area of the 'piston' and the "force from bolt tightening".

(I'm very tired, so if I'm talking a load of BS, please will someone put me straight?)

[Edit - guess you've got to add atmospheric pressure to that, because (from my sums) with no weight present, you'd have a prefect vacuum. Hey! maybe I'm onto something !]

Thanks again to all of you that have added there little bit to help me!!!! So have I got this right: Area of a cylinder is (A=2 x pi x r Sq + 2 x pi r x h) Sq =squared And if this is right, for my equation; r = 2.5 mm and h =254 mm Then I get the answer: 4029.1 mm or 4.0291 m Now I could work this figure into the area of a circle as JohnDG kindly provded, which gets me 169.2 N m. Now does this mean that that is the pressure felt all along the cylinder i.e handle or just the piston ends? Confused, Samurai7

Piston will continue moving until it reaches equalibrium. At that time, force will be

equal throughout cylinder.

Hi, Domino!

Welcome to CR4!!

Have you picked up on a couple of points over which I'm still uncertain?

How is the wrench actually used to measure the torque on an object (eg nut on a stud)?

Where, if anywhere, does the "slip" occur once the wrench hits the desired torquing level?

How is the hydraulic fluid affected by the change in pressure? What is it supposed to do to the spring, and where and how is a readout made as to the desired or achieved torque level?

How is the hydraulic fluid contained to prevent leak-by between the wrench and the spring?

I follow that if a torque wrench were to have a spring all the way down the handle, it could possibly be replaced by a fluid, but what would be the point or usefulness/ advantage/disadvantage of this approach?

There are probably all kinds of similar machines for comparison. I'm just not familiar with this application, so any clarification would be useful for my understanding of how it would actually work. I just haven't visualized the process yet.

Can any of you give me a hand here?

Mark

A spring has a certain amount of yield, give, flex, per length. The longer that it is, the more give that it has. Kind of like adding resistors to a circuit, to give the circuit more resistors, more paths to take, lowers the resistance of that circuit. To make a spring stronger, cut out coils and make it shorter. This yield is measurable and repeatable which makes it good for a repeatable job like checkimg torque values. When a spring is used, just multiply it's length by it's value per length. If the spring is bent, as in this case, all values will change and we no longer know what we have. This is where hydraulics is wonderful. A solid column of liquid is fairly incompressable, so it goes around corners so well. Like the brakes in your car, the brake lines are bent whichever way that they need to be. The hydraulic fluid simply acts like liquid steel. Like the master and slave cylinders in your car, the solid steel brake lines screw into these and rubber cups prevent fluid from leaking out. As in your brakes, air is compressible and ruins the hydraulics. An air bleeder would be necessary.

Two types of torque wrenches that are comonly used are torsion bar, where the stationary bar compares against a flexing bar, and a spring and ball where you here a click when the spring equals the torque value and a ball "clicks". There is no actual "slippage". You can ignore the click and contine tightening if you choose.

The torque wrench has marks to let you know the values and should be calibrated every so often.

Hi again, I see I may have not divulged enough information for you all, well, in my case the force is not the problem, I require the pressure felt within the body of the wrench caused by the compression of the hydraulic fluid between the piston or bolt end and the spring. As to how I will get the torque directed onto the piston is my little secret, needless to say if my calculations prove correct, I will be venturing down the patent route. I can say this however, the prime objective for me is to design a torque wrench that is not of a straight or vertical design i.e. Curved, and one which must work in a very limited space (If you have seen close up the thrust nozzles on the RAF Harrier that allows it to hover, then it needs to fit under the curve of the nozzle.). I need the hydraulic fluid to act as a go between along the body, this is due to the curve having an adverse effect on a spring. However this addition of fluid brings it own issues, one of which is the pressure it will inevitably produce within the body under compression. And yes, I do know that Hydraulic fluid compresses very very little, but I need to find the pressure to work out a material strong enough to cope and for health and safety regulations etc! I realize that this would normally be a straight forward hydraulic piston equation, however all pistons that I know of are straight! Unfortunately this is how the various engineering maths books show them to be. Even my tutor is lacking in the knowledge of my issue! I have seen formulas that deal with pressure felt from a fluid on a curved surface. Only these all tend to show a bed of fluid as opposed to contained fluid such as a tube, none of which show a curved plus a straight surface combined! You can now hopefully see how frustrating this is to me, as I really have nowhere more to look for help! Even the torque wrench manufacturers that I have contacted have either sent me a kind of buyers guide to one of their products or have told me that the information that I seek is privy to them and them alone. Please can anyone shed a tiny bit of light on the subject, Samurai7

You're right, I don't get it. A piston doesn't have to be straight. It can be round like a ball or flat like a dime. It needs to prevent a seal from going down the tube past it. The cylinder or tube should be concentric to keep the piston from cocking over in the bore. Otherwise, the pressure inside the tube isn't going to be that great. I don't know what the torque spec is on a Harrier jet nozzle, though.

Hi Domino, When I said piston, I meant the whole component as in it's cylinder. Sorry for any confusion. As for the Harrier, I've been working on them for the last 6 years so don't worry about that. I know all that I need to know about the aircraft side of things such as the torque rate for the bolts! What I cant get my head around is this damn cylinder pressure!!!! I am open to any suggestions at this point!!!!! Samurai7

Never mind bent or straight, never mind volume - whatever the shape & size - the pressure (psi, Nm^-2 or whatever) will be the same everywhere inside the fluid-filled space - so it will act on the walls (where you've drawn your circle in #7) with whatever pressure you calculate for the piston.

Hi again JohnDG, I will keep this short as I'm guessing that people are getting annoyed with this thread!!! Your saying that the Maximum pressure that the hydraulic fluid achieve will be 42 N/m (this being the max permissible torque allowed) form the spring at max compression and the force created by the torque of the bolt when at maximum tightness also 42 N/m, or will that be 84 N/m? Samurai7

Hi, Samurai7,

You're mixing up units, here. Pressure must always have units of force/area. The N/m you're talking about is a unit of torque.

The pressure throughout the fluid will be given by e.g. the force exerted by the spring, divided by the area of the piston attached to the spring.

It's difficult (for me) to relate the pressure in the fluid to torque at the nut without detail of how the pressure at the top end of the 'tube' is converted to a force and applied to the 'wrench' (UK 'spanner') part of the tool, which engages with the nut.

Hi JohnDG, Basically the nut end will rotate clockwise, causing it's screw thread to rotate into the body forcing the fluid to act directly onto the spring (at the handle) which is already compressed to a predetermined length / force rating! Then when the torque at the nut end surpasses that of the spring it will cause the spring to compress even more the upper piston will separate allowing for slip! I hope this clears things up slightly. Samurai7