Size of the Shaft for Torque

Torque(T, in N.m)/Moment of area (J, in m⁴⁾= Shear stress (in MPa) /Outer fibre radius(R, in m).

Power = Torque x angular velocity

What application are you going to use this shaft? Which stress is 45 MPa? Is it a hollow or solid shaft?

This shaft will be used to rotate a heavy object which will be pivoted at the shaft center

and having a weight of 2500n its solid shaft.Its rotated at 1rps.

Anyway I believe you do know the basics of strength of materials or machine designs or any theory of failures. If not, I think much more comprehensive textbooks, journals, manufacture- and other publications are available and should be consulted.

Whatever theory you apply, it is clear that the maximum resultant bending moment and the resultant maximum torque must be first determined.

The bending moment at any section on a shaft is the algebraic sum of moments of the forces to the one side of the section. So you need to know the length of your shaft, its supports and the location of a heavy object on your shaft.

Shafts may be subjected to various combinations of axial, tensile and compressive loads as well as torsion and bending and these may be appied under gradual, sudden, shock and cyclic conditions. So you need to know the type of load on your shaft in order to consider i.e shock and fatigue load factors, etc.

The limiting ruling section(LRS), refers to the maximum value of the external diameter of a shaft for which the listed stress is applicable. As the cross sectional dimensions of steel increases, the strength decreases, so it is therefore required to reflect this tendency. Consult BS 970: Part 1: Check mechanical properties of a selection of steels

Legal aspects are covered under OSH Act. As shaft is such a common part of driven machinery, all regulations must be complied with.

For example if you consider only twisting and bending:

The polar second moment area of the shaft cross section (J) in m⁴

J = (π/32) x d⁴

Where: d = diameter of a solid shaft in metres.

If for example you are using Guest's or Tresca's maximum shear stress theory of failure:

If the maximum shear stress generated in the material of a shaft by action of combined bending and twisting, is equal to a shear stress generated by a twisting moment(torque) acting singly, the latter twisting moment is referred to as equivalent torque (T_e).

T/J = τ/R

Therefore: (1) T_e = (π τ_max d³)/16

(2) T_e = (M² + T²)^1/2 , Where: M= maximum bending moment and T= maximim torque

Incase combined shock and fatigue factors for bending (K_b) and torsion (K_t) are considered,then T_e = [ (K_bM)² + (K_tT)²]^1/2 .....................................(3)

By equating (1) to (2) or(3) , suitable shaft diameter for solid shafts may be calculated.

This is just an example, you may calculate according to any theory of failures popular in morden technology.I hope these basic principles will help.

It's easy! I assume you mean 25555 N.m torque (judging by the low speed 30 rpm this seems more likely than N.mm). Stress is MPa, not MPa/mm². (MPa = MN/m² = N/mm²)

Torsional modulus Z = torque/stress 567890 mm³. Assuming it's a solid shaft Z = pi.r³ where r = shaft radius. D = 2.r comes to 113.1 mm. If it's a hollow shaft it's a bit more complicated, torque/stress = I/r_o = pi.(r_o⁴ - r_i⁴)/r_o, using an obvious notation where I = 2nd moment of area. You then need to choose suitable r_o and r_i.

Shaft speed doesn't come into diameter calc, but at speed given power (= torque x speed in radians/sec) = 80.3 kW.

I made a mistake in earlier posting, specially embarrassing after saying it's easy! But at least it's better than the error being pointed out by somebody else.

For a solid shaft, torsional Z = pi/2.r³ not pi.r³ and D comes to 142.5 mm.

For a hollow shaft torque/stress = I/r_o = pi/2.(r_o⁴ - r_i⁴)/r_o. E.g. if Do = 160 mm, Di = 118 mm.

Cheers.... Codey