Solving Torque Problems - Metrologist Seeking Advice

  • Thread starter drdodge
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In summary: So, your piece of steel would not be able to handle the torque, even if it was hardened to HY-100 specification.
  • #1
drdodge
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Good afternoon all,

As some of you may remember, I am a metrologist (of 35 years). Mainly physical stuff. Well, I have a new job at a great place doing pressure, force, torque, etc.

My problem is:
I need to connect "female to female" 3/4 square drive socket size. (I could used 1 inch square, too)
This would be used to hang my torque arm on the transducers for calibration and checks.

I can not fine much available in the marketplace. The question is can I buy a piece of 3/4" square material that will handle up to 1000 ft lbs of torque?

I do not know what material specs that I would be looking for, so some education would also be very helpful

thanks,

dr
 
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  • #2
The short answer is 'no', 3/4" square material will not be strong enough to resist a torque of 1000 ft-lbs.

I don't know what your torque arm looks like now, but I think a re-design of your apparatus is in order.
 
  • #3
This setup is a commercially purchased unit. (CDI inc.) We have some of these male to male units already. (1/2" square, 1/4" square) They are magnetic as a drillbit. (so probably not a weird alloy) Seems that it may be a hardened material, looking at the broken one. The 3/4" and 1" drives on the wrenches themselves are able to handle the large torque, and many 250 ft lb wrenches use a 1/2" drive.

dr
 
  • #4
You are certainly more familiar with your apparatus than I, and you seem to have experience dealing with your equipment supplier. I suggest you contact them directly and see if they can provide a solution.
 
  • #5
What would be the way to determine where a piece of steel will yield to the torque?
A formula would help, along with a bit of explanation as to how torque yield is calculated.
Torque adapters from these companies can be very expensive. If the adapter is just a piece of steel heat treated/hardened to a specific spec, I can have 3-5 of the adapters made local for the cost of one adapter.

And in the coarse of time these adapters do wear out.

thanks so far.

dr
 
  • #6
Torsional stress calculations for various cross sections are given in this link:
http://www.roymech.co.uk/Useful_Tables/Torsion/Torsion.html

The max. shear stress calculated should be less than the max. allowable shear stress for the material. For mild steel, the max. all. shear stress is 0.4*yield stress. For example, a typical mild steel A36 has a min. yield of 36 ksi, so the max. all. shear stress would be 0.4*36 = 14.4 ksi.

Good luck!
 
  • #7
thanks for the link to formulas

how does the hardness/heat treating effect things?

I have some further info. on the adapter I have. (~1" x 1" x 2.375")
good to at least 1000 ft lbs, being as it was used previous years to cal the standards.
I threw it on the balance to get mass, and got out the calipers to measure its volumn.
the adapter that I have has a density of aprox. 7.60 g/cc (+/-0.05 g/cc)

seems that it may be some sort of sintered metal product, as that is the only place I see densities like that come up on a google search
that could explain the excessive costs of these adapters.

hhhmmm...

dr
 
  • #8
Steels have densities ranging from 7.75 to 8.05 g/cc. After all, steel is mostly iron (7.87 g/cc).

http://en.wikipedia.org/wiki/Steel

Heat treatment can be used to raise the elastic limit and the ultimate strength of steel.

Sintering is a manufacturing process, where the piece is made by using heat and pressure to turn a powdered material into a solid product. The best tools would be forged, and that process can be quite expensive.
 
  • #9
so is it realistic to be able to heat treat a 3/4" piece of steel to handle the force, or would it, even with the best possible tolerances, still fall short? (without exotic alloys)

I agree about the forged steel being the best, but that may again, explain the "priceyness" of these adapters.

It is interesting that torque is one of the measurement disiplines that has not progressed to the state that time, electric, mass, and temperature have. Lots of room for inovative new products to be developed. (I have a couple ideas myself...lol)

again, thanks for the input, it really helps

dr
 
  • #10
Most people like to avoid putting large torques on structures unless absolutely necessary, and the preferred section is circular because it greatly simplifies the analysis, it eliminates stress concentrations, and it reduces the possibility of crack formation.

When I calculated the max. shear stress on a 3/4" square piece with 1000 ft.-lb applied, I got a figure in excess of 136,000 psi. The minimum yield stress for the material would have to be at least 2.5 times that figure. The strongest steel which is regularly used is HY-100 (to make submarine pressure hulls), and that has a min. yield stress of about 100,000 psi.

Now, you might be able to lay your hands on some of that material, but it ain't cheap and because of the special tempering process, it's hard to work and you must use approved procedures to weld it lest you destroy the material's high strength.

Like I said in previous posts, I don't know anything about your apparatus. Perhaps 1000 foot-pounds of torque is an over-estimation of the true load. Perhaps your apparatus can be re-designed or re-arranged to eliminate, or at least reduce, this load. That is going to be something for you to investigate. But as the raw numbers stand now, I don't think you can design 3/4" stock to withstand this load.
 
  • #11
I understand how, as engineering a specific project, limiting torque is very important. I have the interesting job of creating it. I have, over the years of calibrating torque and force, have seen some stuff that just plain scares me. Being as we are renovating the torque/force lab I really want to eliminate all those problem areas. If you go to some labs that calibrate torque, they will have a bucket of the most twisted, cracked and marginally safe stuff known to man. One of my goals here is to eliminate that completely. That is the reason I have been trying to get my head around the math side of it. I really do appreciate the input. There are not many sources that you can get info from without paying for it, or getting answers you don't trust.

That is why I prefer to come here to get answers.

Thanks steamking (interesting name, by the way)

dr
 
  • #12
The torsion of non-circular sections is pretty advanced strength of materials, which is why most engineers tend to shy away from it. The mathematics of calculating the torsional stresses is usually not covered in undergraduate engineering courses, and only approximate solutions are presented. To fully analyze such problems requires the aid of computers and knowledge of the finite element method.
 

Related to Solving Torque Problems - Metrologist Seeking Advice

1. What is torque and why is it important in metrology?

Torque is a measure of the force that causes an object to rotate around an axis. In metrology, torque is important because it is used to measure the amount of force needed to rotate or tighten a part, which is crucial in ensuring precision and accuracy in measurements.

2. How do I calculate torque in a problem?

The formula for calculating torque is torque = force x distance. This means that to calculate torque, you need to know the amount of force being applied and the distance from the axis of rotation. It is important to use consistent units of measurement, such as Newtons for force and meters for distance.

3. What are some common challenges when solving torque problems in metrology?

One common challenge is determining the correct direction of the torque. This can be especially tricky when dealing with multiple forces acting on a single object. Another challenge is identifying the correct axis of rotation, as this can greatly affect the calculated torque.

4. What are some tips for solving torque problems accurately?

One tip is to draw a diagram of the problem to visualize the forces and their directions. This can help in identifying the correct direction of the torque and the axis of rotation. It is also important to pay attention to units and conversions to ensure accurate calculations.

5. How do I know if my solution to a torque problem is correct?

You can check your solution by using the equation torque = force x distance and plugging in your calculated values. The unit for torque is Newton-meters (N·m), so make sure your answer has the correct units. You can also double-check your calculations and make sure they align with the given information in the problem.

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