Tau Max Decrease as the Diameter Increases?

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.

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?

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...

https://en.wikipedia.org/wiki/Shear_stress

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.

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).

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.

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

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:

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?