Shaft diameter for a given torque ?

[picclock]

Hi

Can anyone advise me, or point me to a solution for the minimum shaft diameter I would need to transmit a torque of 9 Nm or around 80 lb.inch (6-7 ft.lb). I presume the unsupported length comes into it somewhere. which in this case is around 6". This is a kind of intermediate shaft and is supported by bearings at either end.  I was planning to use unhardened silver steel. I have found some internet info (http://www.engineeringtoolbox.com/torsion-shafts-d_947.html) which goes through lots of calculations and comes out with an angle of deflection (twist) figure, but what is a 'safe' angle of deflection? 

Any help much appreciated

Best Regards

Picclock

[RotarySMP]

That formula will give you the shear stress. The material data sheet will give you the allowable stress. The usage will determine the factor safety you need between the two.

[picclock]

@ RotarySMP
I came across another site that said that 1 thou of movement over 20x the diameter was a good way to go. So I tried applying it to one of the formula in the example section and it came up with nonsense. I then tried working through the example and it doesn't give the answer they say. From the site :

The angular deflection of the shaft can be calculated as

θ = L · T / (Ip ·G)

  = L · T / ( (π · D4 / 32) · G)

  = 1 (m) · 1000 (Nm) / (π · (0.05 (m))^4 / 32) · 79 (GPa)

  = 0.021 (radians)

You can see that the second line from the bottom has a numerator of 1000 and a denominator of ~4.8x10^-5 giving an answer of over 2x10^7 so how you derive 0.021 radians from this. ( I also wrote to the site editor). I have edited the formula to show that the 0.05 value is to the power of 4 as this did not copy correctly.

Hence the reason for trying to seek further advice from the forum.

Best Regards

picclock

[David Jupp]

You have your calculation of denominator wrong.

Maybe you missed out the 10^9 in GPa ?

[picclock]

@David Jupp
Hi David - The formula is the one posted on the engineeringtoolbox website - no mention of an extra 10^9. If the error is simply that (and in truth the figures are the same) I have it cracked and can sort it myself. Thanks for having a look at it for me.
Best Regards

picclock

[Lew_Merrick_PE]

The angle of twist in a shaft driven by a torque is: Theta = (Torque X Length)/(Polar Moment of Inertia X Shear Modulus) -- all being in consistent units with Theta being provided in Radians.  My mnemonic for this is: Twist = Tom Locke/John of Ghent as T = Torque, L = Length, J = Polar Moment of Inertia, and G = Shear Modulus in American mechanical engineering shorthand.

The shear stress in a shaft is: S(shear) = Torque/Polar Moment of Inertia -- all being in consistent units.

Polar moment of inertia for a circle (i.e. round shaft) is: J = pi X (D^4)/64 -- where J is the Polar Moment of Inertia and D is the diameter of the shaft.

Assuming that silver steel is actually AISI O1 tool steel, then your allowable shearing stress (no safety margin) is a minimum of 53,750 psi and its Shear Modulus is 23,840,000 psi.  If it is AISI W1 tool steel, then your allowable shearing stress (no safety margin) is a minimum of 57,000 psi and its Shear Modulus is 23,840,000 psi.

Does this help?

[picclock]

@ Lew
Interesting mnemonic - I really had hoped to avoid formula like these, however the editor of toolbox, Tom emailed me to say that the 10^9 figure was missing and he's since updated the site. So at least I can crunch the numbers now and make some sense of the results.

It also occurred to me that I might have been better off measuring the twist by putting a load on an arm at a given radius to the shaft and then measuring the deflection of a dial gauge at a suitable distance. I suspect this may have been faster and given a good result for a given shaft.

I was hoping to use a size of shaft for which I have other components, but the forces are too large and the angular twist too great for my std size so I will have to get the parts needed for a larger diameter.   Modulus of rigidity for the stubbs silver steel I have is 80GPa according to the data.

Many thanks for your time and sharing the information.

Best regards

picclock

[Lew_Merrick_PE]

Picklock -- 80 GPa is the Bulk Modulus, not the Shear Modulus (which would be a nominal 160 GPa for most tool steels).  I do not think in Pascals as the mesh is too small for convenient memorization.

[picclock]

Hi Lew - I freely admit that I am out of my depth here, however http://www.engineeringtoolbox.com/modulus-rigidity-d_946.html gives 79 for structural steel.  The silver steel data I obtained from http://www.engineeringsupplies.co.uk/silver-steel-tech-data-i-53.html , which says its around 80.

I think I might actually test and measure angular deflection (twist).

However, I'm on holiday for the next couple of weeks so I doubt it will be before I return.

Thanks for your assistance

best regards

picclock

[Lew_Merrick_PE]

Picclock.

The primary modulii are:  (1) Tensile (Young's) Modulus (E) is the amount of elongation in tension or delongation in compression of a material as measured in (lbs/inē)/(in/in) ((N/mē)/(m/m)) where the (in/in) or (m/m) value tends to be ignored as they are a divide by one factor.  It amazes me how often otherwise experienced engineers forget that this is an extension/compression ratio more properly cast as a spring rate. (2) Shear Modulus (G) is the amount of deformation induced by a shear load applied to a surface of a material as measured in (lbs/inē)/(in/in) ((N/mē)/(m/m) and is related to the Tensile Modulus as: G = E/2*(1+nu) where nu is the Poisson's Ratio for the material that relates longitudinal strain to transverse strain and is about .3 for most steels.  (3) The Bulk Modulus, also called the Modulus of Rigidity, (K) is the relationship between longitudinal strain and bi-directional transverse strain in a material.  It is measured in (lbs/inē)/(inē/inē) ((N/mē/(mē/mē)) and is related to the Tensile Modulus as: K = E/3*(1-2*nu).

Actual measurements of Shear Modulus and Bulk Modulus vary slightly from their mathematical definitions because energy stored in metallic crystal lattices are not exact values consistent in all conditions.  This is why metallurgists have added several hundred packing types to metallic crystal lattice structures over the past 40 or so years.  When I had my introduction to metallurgy course back in the 1970's, advanced lattice structure considerations were left for post-graduate level study.  We worked with only half-a-dozen lattice structures.  That (over)simplification would get you flunked out of a modern introduction to metallurgy course today.

I admit to being uncomfortable with Newton derived units.  Their "mesh" is so small that they do not (IMO) give appropriate feedback to the designer.  When I headed up the mechanical design department for the development of (2nd generation) automotive airbag restraint systems, engineers from Germany and Japan (who had never used anything but Pascals) were catastrophically failing pressure vessels regularly because of powers of 10 errors -- something that disappeared when we forced them to work in lbs/inē units.  The "mesh" of psi/ksi/msi values provides a feedback missing from the Pa/kPa/MPa/GPa "spread."

Also, engineeringtoolbox.com, like Wikipedia, has provided some interesting values I have had to correct in work done by the (new)internet generation and should be verified before use.  I had to redo some thermodynamic/fluidics work performed by a PhD candidate engineer a few years back because he downloaded "value" that turned out to be (shall we say) highly aggressive for a NASA development effort.  Real values tied to defined and verified standards are the only ones that should be used for anything with the potential to injure people.  I say this as somebody who had the same e-mail address from 1975 through 1999.