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# Tau Max Decrease as the Diameter Increases?

08/25/2017 8:26 AM

This is a bit confusing to me;

Why does tau max decrease as the radius of a solid shaft increases?

20mm solid shaft = ((1255*0.02)/(3.1415927/2)/0.02^4))/1000000 = 100MegaPascal for tau max

40mm solid shaft = ((1255*0.04)/(3.1415927/2)/0.04^4))/1000000 = 12.5MegaPascal for tau max

100mm solid shaft = ((1255*0.1)/(3.1415927/2)/0.1^4))/1000000 = 798958 Pascals

Why does tau max decrease as the diameter of the shaft increases and the torque remains the same?

Doesn't the force increase as the distance increase as well?

Pathfinder Tags: force tau max torque
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#1

### Re: tau max decrease as the diameter increases?

08/25/2017 8:30 AM

Because in those calculations, the answer goes down with the reciprocal of the cube of the diameter of the shaft.

<...a bit confusing...> It would be helpful to the forum if <...tau max...> were defined as a Google search brings up the name of a German-Norwegian writer, editor, and publisher.

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#3

### Re: tau max decrease as the diameter increases?

08/25/2017 8:43 AM

Knowing how SEO works maybe if I link tau max like I just did; once the page gets indexed your wish will be a reality.

On the hand of the question though;

if the yield strength of the material is 500MPa. Let's say I use 50% of the yield strength as my acceptable limit;

tau_acceptable = sigma_y/2

as long as tau max is < tau_acceptable then without unaccounted for shocks; the material shouldn't fail?

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#5

### Re: tau max decrease as the diameter increases?

08/25/2017 8:48 AM

On those figures it has a factor of safety of 2.5 or so.

Has it been tested yet? CR4 doesn't enter into making money-back guarantees...

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#2

### Re: Tau Max Decrease as the Diameter Increases?

08/25/2017 8:37 AM
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### Re: Tau Max Decrease as the Diameter Increases?

08/25/2017 8:46 AM

The postulate [the maximum shear stress to be found within the shaft at a given fixed operating torque as the diameter varies with the reciprocal of the cube of the diameter of the shaft] leaves nothing in the way of <...a bit confusing...> here. So in principle, the larger the shaft, the easier it can withstand the applied torque without failure. That makes sense.

So the nature of any remaining <...bit confusing...> needs to be presented.

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#6

### Re: Tau Max Decrease as the Diameter Increases?

08/25/2017 12:26 PM

Think of shear as sort of an analogue to pressure. For pressure, the same force spread over a larger area results in a lower pressure.

Welcome to the world of extensive units (weight, size, volume, etc.), and intensive units (pressure, shear, concentration of a solute).

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#7

### Re: Tau Max Decrease as the Diameter Increases?

08/25/2017 12:28 PM

Think of it in terms of lever arm. When you tighten a bolt, the longer the wrench, the less force you have to exert for a given amount of torque.

When you apply a torque to a shaft, the shaft has to generate the opposite torque. A larger shaft has a larger lever arm (radius of shaft), and so for a given torque, the maximum torsion is less for a larger shaft.

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#8

### Re: Tau Max Decrease as the Diameter Increases?

09/23/2017 8:30 AM

Dear Mr.marlons

It is very simple. The relation between torque, shear stress and diameter is linked by the formula

T = (π/4) x ( D^3) x Fs

Where T = torque, D = dia of Shaft, Fs = shear stress

Therefore Fs = T/(π/4) x ( D^3)

Therefore when D increases in the denominator, value of (D^3) increases in the denominator, hence Fs will come down. In simple terms Fs = k/ D^3 since T/(π/4) remains constant for a given calculation.

Dhayanandhan.S

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#9

### Re: Tau Max Decrease as the Diameter Increases?

09/27/2017 6:12 PM

This seem like the best place to ask since you at least use the formula that I am referncing.

With this material and it's properties:

Hardness, Brinell201 - 269201 - 269
Hardness, Rockwell C13.8 - 27.613.8 - 27.6
Tensile Strength, Ultimate

686 MPa

99600 psi

Tensile Strength, Yield

490 MPa

71100 psi

Elongation at Break17 %17 %
Reduction of Area45 %45 %
Modulus of Elasticity

205 GPa

29700 ksi

Typical steel
Poissons Ratio0.290.29Typical steel
Machinability55 %55 %Based on AISI 1212 steel as 100% machinability
Shear Modulus

80.0 GPa

11600 ksi

Typical steel
Impact8.08.0kg(f)/cm²

T = (π/4) x ( D^3) x Fs

I know the Torque: 1.255x10^6Nmm to solve for the theoretical shaft diameter I would use the Shear Modulus from the above material property sheet as Fs?

Doing substitution the formula becomes:

1.255x10^6N/((π/32)*D*4) = 80/(D/2)

Then solving for D will give me the size of the shaft without a safety factor. Is that correct? If not where do I get Fs?

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