Calculating PSI from Torque

[aldumoul]

The customer's print calls out 250 ft-lbs of torque mininum on a weld. This part is cylindrical via a progressive die. And this part is welded to complete the cylinder. A round bar (1 foot long, close enough) is inserted into the inside diameter of the part.... on the other end of the bar there is a 0.75" air cylinder pushing up on the bar. The cylinder radius block attached to it (to easily make the conversion) The part is constrained in a fixture. There was a gage made sometime ago... In which someone calculated 50 PSI=250 ft-lbs. I Disagree. I calculated appx 565 PSI. Obviously there are other factors such as the friction created by seals, rings, etc in the cylinder. Let's consider them negligable.

Torque=P(A)Dist
Cylinder area @0.75"=Appx 0.442..... pi*(.375^2)

So 250 ft-lbs=P(0.442)*1 ft
Divide both sides by 1 ft (ft cancels out)
Next, divide both sides (0.442)...=565 PSI

 

[DuncanM]

moon161 - turn the jig upside down so the worker can push down rather than lift? I'd be looking for at least a 3 foot bar if I had to apply 250lb.ft repeatedly.

aldumoul - Ever used a socket set? Ask your technicians if they have one you can look at. Perhaps you know them by some other name.

http://www.amazon.co.uk/Socket-Sets-Hand-Tools/s?ie=UTF8&rh=n:114584031&page=1

has some good images.

" But it isn't going to be a problem using the end opposite of the ratchet side?" Yes. Hopefully the images +/or seeing these things in the flesh will help.

 

[aldumoul]

Not sure how you can use a torque wrench the more I think about it. The bar is inserted into the part and is relatively loose fitting. What i was referring to in regards to using the butt end was-inserting it into the ID of the part. I will have to do some experimenting.

 

[DuncanM]

Take a 1" hex socket, 1/2" drive.

Weld the OD surface of it onto the end of your one foot bar in such an orientation that any ratchet or torque wrench when fitted to the socket acts as an extension to the length of your one foot bar, turning it into a 2/3/4 foot bar with a torque setting.

Then torque on your part = (torque setting on wrench/length of wrench) * (length of wrench + length of 'one foot bar')

 

[rb1957]

a torque is "just" a calibrated bar ... if you're going to do this with muscle power (as opposed to airpressure) i'd use a 2' bar (2 1' bars would work, double the torque wench reading)

 

[aldumoul]

rb1957: (2 1' bars would work, double the torque wench reading)

Wouldn't you want to divide the reading by 2? Also, When you hold a normal (stock) torque wrench - How does the torque wrench take in account how far away you are from the object in which the force is being applied?

 

[DuncanM]

The torque setting in a torque wrench is based on a fixed length - centreline of the socket to mid point in the handle somewhere. You can apply force x at the handle or 2x at the half way point, it'll still 'break' at the same applied torque.

 

[rb1957]

i think the torque wrench works by knowing the torque arm (it's essentially a force gauge). if it's calibrated for a 1' arm but you've added an extension piece, multiplying the torque by a factor of 2 then when the torque gauge is reading 1/2 your torque, you're applying the correct torque.

 

[aldumoul]

Looks like they took...

Force=50 PSI(0.75in^2/4)*pi
=22.089#

Of course F=P/A
They then calculated 22.089#/(0.75in^2/4)

That doesn't make any sense to me. Considering they do not put the Torque equation in equillibrium with the pressure. Trying to explain this to my boss who doesn't have an engineering degree hasn't been fun.

 

[rb1957]

"F=P/A" ? ... F =P*A

"they effed up the calc" should be sufficient. or a simple demostration ... use your set up to drive a torque wrench, supply 50psi and get the torque reading.

 

[aldumoul]

My bad F=P*A (you can see I demonstrated this correctly). If I start redesigning this I will attach a pic of the model.