Calculating new torque on a modified wrench

  • Thread starter Testarossa
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  • #1
I am a Nuclear Reactor Mechanic in the US Navy, and I have run into a disagreement between my peers and I on the proper use of torque wrenches. The situation is simple... (NOTE: these numbers are arbitrary for consistency) You have a dial type torque wrench with a 1 foot handle and you apply 6NM of force to the end of the handle (assume proper counter-force from here on in) and you torque the stud to 100ftlbs. Now you take another torque wrench with a 2 foot handle and you only have to apply 3NM to the end of the wrench to obtain 100ftlbs on the stud. now the dilemma.... if you were to place your hand 1 foot in on the 2 foot torque wrench, would you still apply 6NM to obtain 100ftlbs on the stud? (Assume the wrench is a solid piece of steel alloy that does not give and has a 100% energy transfer efficiency) My theory is, the force applied 1 foot in is still applied, somewhat, to the extra 1foot on the handle, and with 6NM applied to the 1foot position you might get more like 105 or 110ftlbs of torque applied to the stud instead of 100ftlbs.
 

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  • #2
tiny-tim
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Hi Testarossa! :smile:

It's good to know that our nuclear reactors are being powered up by good old imperial-measure wrenches! :biggrin:

INow you take another torque wrench with a 2 foot handle and you only have to apply 3NM to the end of the wrench to obtain 100ftlbs on the stud. now the dilemma.... if you were to place your hand 1 foot in on the 2 foot torque wrench, would you still apply 6NM to obtain 100ftlbs on the stud? (Assume the wrench is a solid piece of steel alloy that does not give and has a 100% energy transfer efficiency) My theory is, the force applied 1 foot in is still applied, somewhat, to the extra 1foot on the handle, and with 6NM applied to the 1foot position you might get more like 105 or 110ftlbs of torque applied to the stud instead of 100ftlbs.

I have to agree with your peers :redface:

the torque (moment) about the centre is always the force "dot" the distance to the centre …

the presence of extra mass either closer or further from the centre than the point of application of the force cannot make any difference.
 
  • #3
I think you may have misunderstood my question, and yes I realized my units were wrong after I typed it. First, forget that you are even using a "Torque wrench" because it doesn’t matter what the gage reads or what the wrench breaks at, in fact using the click or breakaway type wrench would void the argument altogether. So if anything assume you are using a dial type torque wrench that only SHOWS you how much torque you are applying. What I’m saying is, if it took 3N of force at the end of the 1foot wrench to apply 100NM of torque to the bolt, what would happen if you applied 3N of force halfway in on a 2foot wrench, would you still torque the bolt to exactly 100NM or would it be slightly more or possibly less? My question has nothing to do with the added mass of the wrench but simply the fact that if you apply the force in the middle of a 2foot piece of steel, the force isn’t only transferred down to the bolt itself, but some of the force is transferred and possibly multiplied by the extra end of the handle. Maybe to clarify..... on submarines our screws connected to what we call a "Quill shaft". Basically our engines are connected to reduction gears (like a transmission with only 3rd gear) and then the gear attaches to the shaft and then the screw..., but because we need to produce a large amount of torque and don’t have room for a larger shaft we make a "Quill" design where the shaft actually extends through both sides of the reduction gears increasing its effective length and therefore increasing torque produced by a given power output of the main engines. Because the force of the engine is being applied across the entire length of the shaft not just from the point that the gears attach to the shaft down to the screw. The reason all or most of the force is transferred to the screw is because the water is providing resistance, and as we all know "Every action has an equal and opposite reaction". So I would imagine the same would apply to using a longer wrench and holding it closer to the base, you would still apply more torque than desired, maybe not as much as holding it at the end, but still more than using a wrench that was smaller.
 
  • #4
tiny-tim
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Hi Testarossa! :smile:
… on submarines our screws connected to what we call a "Quill shaft".

Basically our engines are connected to reduction gears (like a transmission with only 3rd gear) and then the gear attaches to the shaft and then the screw..., but because we need to produce a large amount of torque and don’t have room for a larger shaft we make a "Quill" design where the shaft actually extends through both sides of the reduction gears increasing its effective length and therefore increasing torque produced by a given power output of the main engines.

Because the force of the engine is being applied across the entire length of the shaft not just from the point that the gears attach to the shaft down to the screw.

The reason all or most of the force is transferred to the screw is because the water is providing resistance, and as we all know "Every action has an equal and opposite reaction".

So I would imagine the same would apply to using a longer wrench and holding it closer to the base, you would still apply more torque than desired, maybe not as much as holding it at the end, but still more than using a wrench that was smaller.

Sorry, I know almost nothing about quill shafts.

I thought a quill shaft was just an ordinary shaft that could move in and out a little, with the power determined only by the reduction gear ratio (ie the number of cogs), and having nothing to do with the length of the shaft. :confused:

Could you possibly link to a diagram that shows how a submarine quill shaft operates?
 
  • #5
549
28
Torque is Force times Distance (the distance being the line perpendicular to the line of force that goes through the centre of rotation).
If you have a 1M torque wrench and apply IN at the end of the handle you'll get 1NM
If you apply 1N halfway down the handle you'll get 1/2 NM. (If it is the click type set to 1NM to get it to click you need to apply 1N at the end or 2N halfway down).
What you describe is what I would call a Torque Converter basically a reduction gearbox with a socket drive and two handles one which is jammed against something solid and the other pulled. The length of the handle doesn't matter it is where on the handle that you apply the force that gives the torque.
 
  • #6
I will have to recheck my facts about the quill shaft and get back to you. Unfortunately the design is still considered classified since we are still using it today. So I cannot post pictures at the moment, but there may be an older design I can show. I had always believed the same about how torque is only applied from the point of force to the object at the end, until I was introduced to this shaft design. I may have very well completely misunderstood how the design works. I will find out and clarify my response as soon as I can.
 
  • #7
867
61
length of lever x force = torque so...

2 ft. lever x 50 lb force = 100 ft lb of torque
1 ft. lever x 100 lb force = 100 ft lb of torque.

When you say you are using a dial type torque wrench do you mean
[PLAIN]http://www.acy.com.sg/images/J6121NMF.jpg [Broken]
or
[PLAIN]http://www.hydraulicpumpsmotors.com/wp-content/uploads/2010/10/torque-wrenches.jpg [Broken]

The second type relies on the amount which the main beam bends when applying torque from the handle. If you were to apply your force half way up the handle then the other half of the beam would not be bending and you would apply a much greater torque before the indicator reads 100 ft lb.

I'm not sure how the first type works. It may or may not have similar shortcomings.
 
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