When a shaft is transmitting a power from driver equipment to the driven equipment, It is necessary to calculate the shaft diameter from the torque based on the maximum torque that can be transferred through the shaft or the maximum amount of twist in the shaft. Otherwise, the shaft may not satisfy the functionality of the equipment.

# Inputs that we need to calculate shaft diameter

As we know Torsional equation (Read more about the torsional equation here)

Where

**τ** = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).

** r** = Radius of the shaft.

**T** = Twisting Moment or Torque.

**J** = Polar moment of inertia.

**C** = Modulus of rigidity for the shaft material.

** l **= Length of the shaft.

**θ** = Angle of twist in radians on a length.

From Torsion Equation we can consider

It can be written as

Polar moment of inertia will be calculated from the following equation.

Polar moment of inertia for a circular cross-section is

From this, the maximum torque that can be transferred thru the shaft can be written as

From the above equation, we can calculate the unknown factors. In our case, the shaft diameter is the unknown factor.

So we need the following input data

- The maximum torque that would be transferred thru the shaft (We can calculate torque from the Power and the speed also)
- The torsional shear stress of the material

Let’s take a sample problem to calculate the shaft diameter from the torque.

# Sample Problem to calculate the shaft diameter from the torque

**Problem: A shaft is transmitting 200 kW at 1200RPM. Find a suitable diameter for the shaft, if the maximum torque transmitted exceeds the mean by 30%. Take maximum allowable shear stress as 70 MPa.**

**Answer:**

Given data

Power(P) = 200kW = 200 x 10^{3 }W

Speed (N) = 1200RPM

Operating Torque (T_{Operating}) = ?

Maximum torque that can be operated on the shaft = 30% to the operating torque

T_{Max} = 1.3 x T_{Operating}

We have power(P) = (2 π N T)/60 = (2 x 3.14 x 1200 x T)/60

200 x 10^{3 }= 125.6 x T

T = 1592.36 N-m

Operating Torque (T_{Operating}) = 1592.36 N-m

Where T_{Max} = 1.3 x T_{Operating}

T_{Max} = 1.3 x 1592.36

T_{Max} = 2070.06 N-m

Maximum torque that can be operated on the shaft (T_{Max}) = 2070.06 N-m

From this maximum operating torque, we can find the shaft diameter with above equation

2070.06 x 10^{3 }N-mm = (70Mpa (N-mm^{2}) x π x d^{3})16

d^{3 }= 150687.075 mm

d = 53.19 mm

The required shaft diameter will be a 53 mm shaft

# Conclusion

Here we only consider the maximum torque that can be transferred through the shaft as the known factor to calculate the shaft diameter. We can also find the shaft diameter with the maximum angular twist that can be allowable in the shaft by considering the shaft stiffness(C) and the torque relation(Given below) from the torsional equation.

Try the following exercise problem and post your answer in the comment box below! Let’s see who can post the right answer.

**Exercise problem: A shaft is transmitting 97.5 kW at 180 r.p.m. If the allowable shear stress in the material is 60 MPa, find the suitable diameter for the shaft. The shaft is not to twist more than 1° in a length of 3 metres. Take C = 80 GPa.**

Dane says

Not sure

Pankaj N Bid says

How to calculate shaft diameter, when subjected to torsion and bending moment. Know Power is 20 kW and rpm is 400. Material of shaft is AISI 4140.

alex says

d3 = 150687.075 mm

d = 53.19 mm

How have you got 53.19 out of d3, shouldn’t the answer be 438.01 ?

ø = 2 x √(A / π)

Sundar Dannana says

My friend, it is simply the (150687.075)^(1/3)

Devender Narang says

shaft dia = 103.098 mm

Sundar Dannana says

Yes, correct Devender Narang.

Good job