# How to calculate the diameter of a transmission shaft from RPM & Torque

If I know the torque and rpm that the shaft need to operate at, can I calculate the diameter of the shaft?

A motor operating at 60 rpm applying 1,000 Nm of force on a shaft that has a modulus of rigidity of 50 GPa and is 10 cm in length.

Can I calculate the diameter of the shaft so that it would not fail?

The tensile strength is:

Tensile Strength, Ultimate  686 MPa
Tensile Strength, Yield     490 MPa


This will be a horizontal shaft, the length isn't final yet but I am estimating that it will be 2,000 mm long and supported at 200 mm from each end.

There shouldn't be any unforseen shocks being applied to the shaft. It will be transmitting the power from a 15 HP motor that's connected via a coupling on one end.

• Can you clarify the question: 1) application and background of why you are asking this particular question. (So that answer will contain all your concerns) 2) a sketch of the problem 3) what sort of instantaneous forces will act on the shaft ? 4) will it be looking up in the sky or horizontal? 5) how will the force (torque) transfer happen between shaft and load, what interface? 10cm sounds small... Sep 21 '17 at 13:32

The elastic modulus would give you how much it would twist in use, you would need the tensile strength: taken from engineering toolbox http://www.engineeringtoolbox.com/torsion-shafts-d_947.html :

τ = T r / J                              (1)

where

τ = shear stress (Pa, psi)

T = twisting moment (Nm, in lb)

r = distance from center to stressed surface in the given position (m, in)

J = Polar Moment of Inertia of Area (m4, in4)

Diameter of a Solid Shaft
Diameter of a solid shaft can calculated by the formula

D = 1.72 (Tmax / τmax)^1/3                            (4)


this should answer your question with regard to the minimum theoretical value, but of course in real work applications you would need to consider safety factors and deflection (which you can find calculations for on the aforementioned link).

• In this equation Tmax would be the 1000Nm, I'd be solving for τmax, where do you get the number 1.72 from? Also why divide by three? The polar moment of inertia for a circular shaft is something different. Sep 21 '17 at 7:59
• @user1610950 follow the link and it derives the equations... Sep 21 '17 at 8:09
• Fixed the power Sep 21 '17 at 8:52
• Depending on the application (school exercise, theoretical question or engineering application) a factor of safety should be included in the answer. This is the minimum diameter, for ideal cases (ideal material, ideal manufacturing, etc). Sep 21 '17 at 9:42
• Are you saying that all that I need to do is use the modulus of elasticity as τmax in the formula to get the diameter of the shaft? Sep 21 '17 at 13:19

Given:

$T\ =\ 1000\ Nm^2$, $G\ =\ 50 GPa$, $L\ =\ 10\ cm\ =\ 100\ mm$

The torque has the wrong units, please double check.