rotate

01-20-2010, 10:23 PM

I need a shaft that will not plastic deform with 400 ft-lbs of torque. Using low carbon steel like 1018, what's the minimum diameter that I can use. Is there a formula for this?

View Full Version : Torsional strength of shaft

rotate

01-20-2010, 10:23 PM

I need a shaft that will not plastic deform with 400 ft-lbs of torque. Using low carbon steel like 1018, what's the minimum diameter that I can use. Is there a formula for this?

mechanicalmagic

01-20-2010, 10:46 PM

ALL shafts will deform with a load, any load. If it returns to the original shape, that's elastic deformation, by definition. Looking up info.

Evan

01-20-2010, 10:50 PM

If it returns to it's original shape it's elastic deformation. If it doesn't then it is plastic deformation.

edit: Heh, you beat me to the correction by a minute. (while I was typing)

A drive shaft is a good example of a shaft designed for that sort of torque load.

edit: Heh, you beat me to the correction by a minute. (while I was typing)

A drive shaft is a good example of a shaft designed for that sort of torque load.

Carld

01-20-2010, 11:15 PM

Your talking about something like a torsion bar suspension. I don't think your going to get by with 1018, it will deform and not return to the original state.

You'll probably have to use a high carbon steel and heat treat it. Or maybe a 4000 series steel and heat treat it.

What every you use I would think it will have to be heat treated.

You'll probably have to use a high carbon steel and heat treat it. Or maybe a 4000 series steel and heat treat it.

What every you use I would think it will have to be heat treated.

camdigger

01-20-2010, 11:30 PM

There are several releveant equations.

DIY analysis here

http://me-wserver.mecheng.strath.ac.uk/group2006/groupa/design_driveshaft.htm

I'd go further, but it's been 25 years...:o

I feel old all of a sudden...

DIY analysis here

http://me-wserver.mecheng.strath.ac.uk/group2006/groupa/design_driveshaft.htm

I'd go further, but it's been 25 years...:o

I feel old all of a sudden...

oldtiffie

01-20-2010, 11:49 PM

I need a shaft that will not plastic deform with 400 ft-lbs of torque. Using low carbon steel like 1018, what's the minimum diameter that I can use. Is there a formula for this?

The point where elastic becomes plastic deformation is the yield point.

If you have a pre-determined or limiting diameter, I'd suggest a trip to your local steel merchant for a list of manufactures or distributors and web addresses and phone numbers and start looking and asking.

That will give you a range of options, costs and any pre- and post- heat-treatment requirements to meet your objectives and specifications.

The point where elastic becomes plastic deformation is the yield point.

If you have a pre-determined or limiting diameter, I'd suggest a trip to your local steel merchant for a list of manufactures or distributors and web addresses and phone numbers and start looking and asking.

That will give you a range of options, costs and any pre- and post- heat-treatment requirements to meet your objectives and specifications.

Herm Williams

01-21-2010, 12:08 AM

Depending on how much torque you need, you might consider a drive axelo from a junk yard. It can be machined with carbide. Very tough, but good material.

re

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rotate

01-21-2010, 12:10 AM

I don't have any dimensional constraints. I'm just trying to figure out a ball park figure. Is 1/2" good enough?

Carld

01-21-2010, 12:18 AM

I don't know how long you want it but you could use an old torsion bar from a car to make it out of. It may be to hard to machine though.

doctor demo

01-21-2010, 12:53 AM

I need a shaft that will not plastic deform with 400 ft-lbs of torque. Using low carbon steel like 1018, what's the minimum diameter that I can use. Is there a formula for this?

A 1'' diameter extension does not deform when used on My 600ft-lb torque wrench, so buy a cheap extension the proper length for Your ap.

Of course it probably is not 1018:) .

Steve

A 1'' diameter extension does not deform when used on My 600ft-lb torque wrench, so buy a cheap extension the proper length for Your ap.

Of course it probably is not 1018:) .

Steve

JoeCB

01-21-2010, 12:57 AM

If I have the picture... you want a mild steel (1018) shaft to resist a 400ft lb torque without plastic deformation. now assuming a bunch of things like; no bending stress, no keyways, static loading, no shock loading, nornal temperature...

using 36,000 psi yeld strength for mild steel, I come up with a shaft 0.88 inch dia. ... if you allow a 2X safety factor you need a 1.10 " shaft. So I would say a 1 inch shaft should do just fine.

Now it's been (quite) a while since I had do do this kind of thing for real so I would appreciate comments from the experts.

Joe B

using 36,000 psi yeld strength for mild steel, I come up with a shaft 0.88 inch dia. ... if you allow a 2X safety factor you need a 1.10 " shaft. So I would say a 1 inch shaft should do just fine.

Now it's been (quite) a while since I had do do this kind of thing for real so I would appreciate comments from the experts.

Joe B

Evan

01-21-2010, 02:04 AM

I'm just trying to figure out a ball park figure. Is 1/2" good enough?

No, not at all. I can twist 1/2" material into a pretzel with my picket twister using less than 100 lbs on the end of a 3 ft handle.

1 inch material will take it but will be pretty springy. If you don't want a lot of spring then go to 2" pipe or use a car drive shaft.

No, not at all. I can twist 1/2" material into a pretzel with my picket twister using less than 100 lbs on the end of a 3 ft handle.

1 inch material will take it but will be pretty springy. If you don't want a lot of spring then go to 2" pipe or use a car drive shaft.

darryl

01-21-2010, 04:32 AM

Just my estimation, but my thoughts are that a 1 inch mild steel shaft would do it, and a 1 1/4 diameter shaft would give a pretty good safety margin. I think you'd be twisting a 3/4 inch shaft at 400 in/lbs.

rotate

01-21-2010, 11:30 AM

Thanks everyone. I think I'll go with 1" since 400 ft lbs requirement already had some safety factor built-in.

camdigger

01-21-2010, 11:43 AM

Why solid?

Tubes (AKA pipe) is much more efficient for torque transmission, lighter, and stiffer to resist whirl.

There's a reason drive shafts in cars and the like are thin walled tubes.

Tubes (AKA pipe) is much more efficient for torque transmission, lighter, and stiffer to resist whirl.

There's a reason drive shafts in cars and the like are thin walled tubes.

jungle_geo

01-21-2010, 03:43 PM

I agree closely with JoeCB and have included the calculation below:

Sy = 36 kpsi for steel

assuming we go all the way to the point of yield to determine the minimum diameter:

Sy <= TD/2J

Where:

T : torque

D : diameter of shaft

J : polar moment of inertia which is equal to PI*D^4 / 32

Substituting everything in algebraically, we come up with:

Sy <= 16T/PI*D^3

Solving for D:

D >= (16*T/PI*Sy) ^ (1/3)

For your parameters:

T = 400 ft-lbs * 12 in/ft = 4800 in-lbs

D >= ((16*4800)/(PI*36000)) ^ (1/3)

D >= 0.879 inches

Assuming you go with a 1 inch shaft, you would get about 3.1 degrees of twist per foot of length at 400 ft-lbs, so it won't be very springy at this load assuming you aren't building a 10 foot long shaft.

Applying a safety factor makes it more interesting; both the torque and material have variability. I would figure how much the torque might overshoot nominal (400 ft-lbs) and then use that as my upper bound and apply a safety factor on that.

Also as JoeCB stressed, this is for a non-interrupted shaft. Keyways, flats, etc all add additional knockdown factors and you should then also consider cyclic loading if this shaft is going to see more than a few thousand cycles since such features exhibit crack propagation in a cyclic environment. A great book for safety factors and cyclic loading (a cook book essentially) is Mechanical Engineering Design by Shigley (ISBN-10: 0072921935). Excellent desk reference.

I hope these equations will help should you change a parameter and need to recalculate your diameter or at least for your own edification.

Sy = 36 kpsi for steel

assuming we go all the way to the point of yield to determine the minimum diameter:

Sy <= TD/2J

Where:

T : torque

D : diameter of shaft

J : polar moment of inertia which is equal to PI*D^4 / 32

Substituting everything in algebraically, we come up with:

Sy <= 16T/PI*D^3

Solving for D:

D >= (16*T/PI*Sy) ^ (1/3)

For your parameters:

T = 400 ft-lbs * 12 in/ft = 4800 in-lbs

D >= ((16*4800)/(PI*36000)) ^ (1/3)

D >= 0.879 inches

Assuming you go with a 1 inch shaft, you would get about 3.1 degrees of twist per foot of length at 400 ft-lbs, so it won't be very springy at this load assuming you aren't building a 10 foot long shaft.

Applying a safety factor makes it more interesting; both the torque and material have variability. I would figure how much the torque might overshoot nominal (400 ft-lbs) and then use that as my upper bound and apply a safety factor on that.

Also as JoeCB stressed, this is for a non-interrupted shaft. Keyways, flats, etc all add additional knockdown factors and you should then also consider cyclic loading if this shaft is going to see more than a few thousand cycles since such features exhibit crack propagation in a cyclic environment. A great book for safety factors and cyclic loading (a cook book essentially) is Mechanical Engineering Design by Shigley (ISBN-10: 0072921935). Excellent desk reference.

I hope these equations will help should you change a parameter and need to recalculate your diameter or at least for your own edification.

Evan

01-21-2010, 04:14 PM

Note that 400 ft lbs already includes a safety factor (how much is unstated). The stated material is 1018 which has a minimum yield strength of around 53ksi if cold drawn and 44 ksi if annealed. It can be hardened to RC 42 in thin sections (<3-4") by heating and quenching water which will improve the mechanical properties considerably. It isn't necessary to draw back the temper.

It also work hardens to a high degree which has a large effect on the plastic yield strength. If it has been work hardened it needs to either be normalized or allowed to age for a few days.

[edit] I forgot to mention that the yield point of most 10 series steels in particular is very sensitive to the applied strain rate. This has implications if the shaft is to operate at high rpm or with sharp shock loads. The faster the applied strain rate the lower the yield point.

It also work hardens to a high degree which has a large effect on the plastic yield strength. If it has been work hardened it needs to either be normalized or allowed to age for a few days.

[edit] I forgot to mention that the yield point of most 10 series steels in particular is very sensitive to the applied strain rate. This has implications if the shaft is to operate at high rpm or with sharp shock loads. The faster the applied strain rate the lower the yield point.

davidh

01-21-2010, 04:38 PM

look up "torque sticks" for tightening truck wheels with an impact wrench. thats exactly what they do, and they come in many different torque specs.