Technical Bulletin
TBN017.0/1998
CRITICAL SPEED OF SHAFTS
In general, the Rayleigh-
Ritz equation
All rotating shaft, even in overestimates and the
the absence of external Dunkerley equation
laod, deflect during rotation. underestimates the natural
The combined weight of a frequency.
shaft and wheel can cause
deflection that will create The equation illustrated
resonant vibration at certain below is the Rayleigh-Ritz
speeds, known as Critical equation, good practice
Speed.
The magnitude of
deflection depends
upon the followings :-
(a) stiffness of the
shaft and it’s support
(b) total mass of shaft
and attached parts
(c) unbalance of the
mass with respect to
the axis of rotation
(d) the amount of
damping in the system
Therefore, the
calculation of critical
speed for fan shaft is
necessary.
Critical Speed Equation
(Nc)
There are two method used
to calculate critical speed,
Rayleigh-Ritz and
Dunkerley Equation. Both
the Rayleigh-Ritz and
Dunkerley equation are an
approximations to the first
natural frequency of
vibration, which is assumed
to be nearly equal to the
critical speed of rotation.
The maximum static
deflection, δst, is obtained
by adding both the
maximum static deflection
suggests that the maximum of the rotating shaft and
operation speed should not the load.
exceed 75% of the critical
speed. (1) Maximum static
deflection on shaft
(δstI)
3
Criticalspeed,Nc= 0
π 1.1)
where :
g = gravity acceleration
(9.81 m/s2) δst = total δstI=
maximum static deflection
5
w
Critical speed depend upon
L
the magnitude or location of
the load or load carried by 3
the shaft, the length of the
shaft, its diameter and the
kind of bearing support. 3
4
Total Maximum Static
Deflection (δst) E
1-5
2.1 )
δst2=
1.2) W
L
3
δstI=
4
w 8
L E
I
3
8
2.2 )
E
I
(2) Maximum static
deflection on load 2 2
δst2=WB(L −B )
only (δst2)
3/2
9
3EIL
Technical Bulletin -
TBN017.0/1998
2.3)
2 2
δst2=WA(3L − 4A
) 24EI
2.4)
δst2=
W
L
3
3
E
I
where :
w = weight
of shaft, kg
W = weight
of wheel, kg
E = modulus of
elasticity, kg/m2
for shaft
C40=200 x108
kg/m2
I = moment of
inertia=πD4/64, m4 L =
length of shaft, m
Shaft Diameter Moment of inertia Weight per metre
D (mm) I (m4) (kg/m)
20 7.85 x 10-9 2.47
25 19.17 x 10-9 3.85
30 39.76 x 10-9 5.51
35 73.66 x 10-9 7.99
40 125.66 x 10-9 9.87
45 201.29 x 10-9 13.00
50 306.79 x 10-9 15.40
55 449.18 x 10-9 18.70
60 636.17 x 10-9 22.20
70 1178.59 x 10-9 30.20
Table I
2-5
384(200x108 )(125.66x10−9 )
=0.00018 m
(b) Deflection from load
Example 1 only (δst2)
Given the following
specifications, find the WA(3L − 4A2 )
2
δst2=
critical speed. --refer to Fig. 2.3
24EI
=7.5(0.205)[3(1.37)2
− 4(0.205)2 ]
24(200x108 )
(125.66x10−9 )
=0.000139 m
Model : KAT 15/15 S2
with 2-bearings
Diameter of shaft, D = 40 mm
Weight of wheel, W = 7.5 kg
Shaft length, L = 1.37 m
Length, A = 0.205 m
Moment of inertia, I = 125.66 x 10
Modulus of Elasticity, E= 200 x108
(C40)
Shaft weight, w=1.37 x 9.87
=13.52 kg ---- refer to Table I
(a) Deflection from
shaft weight only
(δst I)
3
δstI= 5wL ---------------- refer to Fig. 1.1
384EI
= 5(13.52)(1.37)
(d) Critical Speed
(Nc)
Technical Bulletin -
TBN017.0/1998
Nc= 30 g
π δst
(b) Total maximum =30 9.81
π 0.000319
static deflection
(δst)
=1675 rpm
δst = δst1 + δst2
Safety factor 25%,
= 0.00018 + therefore max.
0.000139 operation speed =
1675 x 0.75
= 0.000319 m = 1256
rpm
3-5
Check Critical Speed
For Long Span
Example 2
To check critical speed for
KAT 12/12 S3 with 2-
Length, A = 0.197 m
bearing, one side of the
Length, L = 1.114 m
bearing overhung.
Shaft weight,w = 8.9 kg
(a) Deflection from shaft
weight (δst I)
δstI=
5
Diameter of shaft, D = 35 mm w
Weight of wheel, W = 5.4 kg L
Moment of inertia, I = 73.66 x 10 3
Modulus of Elasticity, E=
200 x108 kg/m2 3
( 8
C 4
4 E
I
0)
= 5(8.9)(1.114)3
384(200x108 )
(73.66x10−9 )
=0.000109 m
Technical Bulletin -
TBN017.0/1998
(b) Deflection from load
only (δst2)
2
δst2=WA(3L −
4A2 )
24EI
=5.4(0.197)
[3(1.114)2 −
40(0.197)2
24(200x108 )
(73.66x10−9 )
=0.000107 m
(c) Total
maximum static
deflection (δst)
δst =δstI+δst2
=0.000109+0.0001
07
=0.000216m
(d) Critical speed for long
span (Nc)
Nc= 30 g
π δst
=30 9.81
π 0.000216
=2035rpm
Safety Factor 25%,
therefore max.
operation speed =
2035 rpm x 0.75
= 1526 rpm
4-5
WA
δst2= 3EI3
= 5.4(0.5215)3
3(200x108 )
(73.66x10
−9
)
Check Critical Speed =0.000173 m
For Overhung
(b) Total
maximum static
deflection (δst)
δst
=δstI+δst
Length, A = 2
0.5215 =0.00005
Length, L = 5+0.0001
0.534 m
73
shaft
weight,w = =0.00022
4.27 kg 8m
(a) Deflection from
shaft weight only
(δstI)
δstI=
w
L
3
8
E
I
= 4.27(0.534)3
8(200x108 )
(73.66x10−9 )
=0.000055 m
(b) Deflection from load
only (δst2)
Safety factor 25%,
max. operation speed
= 1980 x 0.75 = 1485
Technical Bulletin - rpm
TBN017.0/1998
Conclusion
(d) Critical Speed at
overhung (Nc) Long Span
Critical Speed = 2035 rpm
Max. operation speed = 1526 rpm
Nc = 30 g
π δst Overhung
= 30 9.81 Critical Speed = 1980 rpm
Max. operation speed = 1485 rpm
π
0.000228 Therefore, the max.
operation speed for this
KAT 12/12 S3 should be
= 1980rpm according to the overhung,
ie. whichever lesser, which
is = 1485 rpm
5-5
