Shaft Diameter and Shaft Critical Speed

By.Ahmad Suhendra
The size of the diameter of the shaft is very influential at critical speed. If the shaft is rotating close to or exceed the critical speed, the shaft becomes very fast vibrate and cause the bearings become damaged, so the correct practice should be working below the shaft critical speed.

Example calculations.
A Pelton turbine disc has a weight of 30 kg , and planned to have a nominal rotational speed is 1500 rpm and Power = 100 kilowatts. Shaft material (solid cylinder shaft) to be used is a standard steel, with the number: AISI-SAE 1040, which has Ultimate tensile stress (UTS) 90.000 psi and Young's modulus (E) 29 10^6 psi .
What is the diameter (D,mm) of the shaft is used?
If the distance between the bearings and disc is L= 50 mm (see picture),
what is the critical rotational speed (Nc)?
Did diameter shaft that is used safely?

Given :
N = 1500 rpm
P =100 kW
m = 30 kg
UTS = 90,000 psi (620.5 N/mm^2)
E = 29*10^6 psi (199948 N/mm^2)

I.) Torsional or twisting moment (T, N.mm).
T = 9.55 10^6 P/N................................(1)
= 9.55 10^6 * 100 / 1500
= 636667 N.mm

II.) Ultimate shear stress (UST, N/mm^2).
UST = 0.75 *UTS........................................(2)
= 0.75 * 620.5
= 465 N/mm^2

The allowable stresses (Ss) that are generally used in practice are:
4000 psi ( 28 N/mm^2) for main power-transmitting shafts (Ss < UST).

III.) Polar Section Modulus (Zp, mm^3).
Zp =T / Ss............................................(3)
= 636667/28
=22738 mm^3
and,
for circular solid shaft is :
Zp = (phi D^3) / 16................................(4)
D^3 = Zp *16 /phi
= 22738 * 16 / 3.14
D = 115863^(1/3)
= 49 mm or 2 inchi. (Shaft diameter)

IV.) Polar moment of inertia (I ,mm^4) of a circular solid shaft.
I=1/4* R^4.............................................(5)
I = phi D^4 / 64
= 3.14 ( 49^4)/64
=282836 mm^4

V.) Critical speed.
shaft weight is negligible.
E =199948 N/mm^2
I=282836 mm^4
L=50 mm
m=30 kg

Nc =[ 3 * E * I / (m *L^3)]^0.5 / (2*phi)..............(6)
= [ 3 * 199948 * 282836 / (30 * 50^3)] ^0.5 / ( 6.28)
= 45242^0.5 / 6.28
=213 /6.28
=34 rev/s
=2040 rpm

Nc > N, 1500 rpm (OK, can be used safely).

N(1500) / Nc(2040) = 0.735
0.735 < 0.8 (subcritical operation)

Note:
lower than 0.8 * Critical speed - subcritical operation
higher than 1.25 * Critical speed - above critical operation


Ref:
http://www.tribology-abc.com/calculators/uts.htm
http://www.roymech.co.uk/Useful_Tables/Matter/shear_tensile.htm
http://www.engineeringtoolbox.com/torsion-shafts-d_947.html
http://www.roymech.co.uk/Useful_Tables/Drive/Shaft_Critical_Speed.html
http://jimnevins.net/Stuff/Machinery%27s%20%20handbook%2026th%20Edition/MH26/yc.pdf