Capital university of science and technology
Assignment
Mechanical Vibration Lab
Design and model analysis to find the critical speed of shaft
Hussain Muhayyoddin
BME163110
April 27, 2020
Submit to
Eng. Awais Liquat
Department of Mechanical Engineering
Contents
Introduction .................................................................................................................................................. 3
Calculation .................................................................................................................................................... 3
For diameter 0.004m ................................................................................................................................ 4
Theoretical data ............................................................................................................................................ 5
Ansys Data..................................................................................................................................................... 5
Ansys results ................................................................................................................................................. 6
For data A: .......................................................................................................................................... 12
For data B:........................................................................................................................................... 12
Conclusion: .................................................................................................................................................. 14
Introduction
All rotating shafts, even in the absence of external load, deflect during rotation. The magnitude of
the deflection depends upon stiffness of the shaft and its supports, the total mass of shaft and
attached part, the unbalance of the mass with respect to the axis of rotation, and the amount of
damping in the system. As long as the deflections are minimal, the shaft can still operate
satisfactorily. However, as the speeds increase, the shafts become unstable. Thus, there is the need
to determine these critical speeds at which instability sets in. With the deflection, now considered
as a function of speed, only the lowest (first) and occasionally and the second are of interest to the
designer, the others will usually be so high as to be well out of range of the operating speed At the
first critical speed, the shaft will bend to the simplest shape possible. At the second critical speed
it will bend to the second simplest shape possible etc. For example, a shaft supported at its end and
having a large (compared to shaft) masses attached, will bend according to the configurations
shown in Figures 1 and Figure 2 at the first and second critical speed respectively.
Figure 1 Loading Configuration at the first critical speed Figure 2 A variant of the loading configuration at the first critical speed
In this project we first find out the frequencies of our shaft on ANSYS>workbench>Modal that is
fixed from both of the ends. After analysis on Ansys using different lengths and diameters of the
shaft we get the different values of frequency of 1st and 2nd mode and then compare it with the
experimental results.
Calculation
Following data, we have given to solve theoretical calculation from paper.
𝜋 1 𝑔∗𝐸∗𝐼
𝑓𝑛 = [𝑁 + ]2 √
4 2 𝑊 ∗ 𝐿4
In which,
fn= Natural Frequency, Hz
N = Mode number i.e. 1, 2, 3…
I = Moment of inertia of shaft
E = Young’s Modulus (GPA)
W = Weight of shaft per meter length
L = Length of shaft (m)
g = Acceleration due Gravity (m/s2)
The shaft is made up of stainless steel and its material properties are:
PROPERTIES VALUE UNIT
Young’s Modulus 207 GPa
Density 7860 Kg/𝑚3
Poisson’s ratio 0.3 -
Ultimate Tensile Strength 1350 MPa
Yield strength 1200 MPa
Now first solve, when keeping length same and changing diameter of shaft
For diameter 0.004m
For Mode one:
Diameter = D =0.004m
Length = L =1m
Young’s Modulus (GPA) = E =200×109
Moment of inertia of shaft = I = 1.26×10-11
Acceleration due Gravity = g = 9.81m/s2
𝜋 1 𝑔∗𝐸∗𝐼
𝑓𝑛 = [𝑁 + ]2 √
4 2 𝑊 ∗ 𝐿4
Now by putting values in formula
π 1 9.81×200×10∗9×1.26×10∗−11
fn= 4 [ N +2 ]2 √ 0..96×1∗4
fn =3.534*5.074
fn =17.93Hz
For Mode two:
Diameter = D =0.004m
Length = L =1m
Young’s Modulus (GPA) = E =200×109
Moment of inertia of shaft = I = 1.26×10-11
Acceleration due Gravity = g = 9.81m/s2
𝜋 1 𝑔∗𝐸∗𝐼
𝑓𝑛 = [𝑁 + ]2 √
4 2 𝑊 ∗ 𝐿4
Now by putting values in formula
π 1 9.81×200×10∗9×1.26×10∗−11
fn= 4 [ 3 +2 ]2 √ 0..96×1∗4
fn =9.817*5.074
fn =49.81Hz
similarly, for remaining calculation results are given as
Theoretical data
Table A
Sr. No. Shaft Diameter First Mode (Hz) Second Mode (Hz)
Length(m)
1 1 0.004m 17.93 49.81
2 1 0.006m 26.82 74.25
3 1 0.008m 35.52 98.63
Similarly calculated natural frequency for different diameters and length=0.9m
Table B
Sr. No Shaft length Diameter First Mode (Hz) Second Mode (Hz)
(m) (m)
1 0.9 0.004 22.20 61.24
2 0.9 0.006 33.14 91.42
3 0.9 0.008 44.48 122.21
Ansys Data
Table A
Sr. No. Shaft Diameter First Mode (Hz) Second Mode (Hz)
Length(m)
1 1 0.004m 17.181 47.356
2 1 0.006m 27.408 75.434
3 1 0.008m 36.667 101.06
Similarly natural frequency for different diameters and length=0.9m
Table B
Sr. No Shaft length Diameter First Mode (Hz) Second Mode (Hz)
(m) (m)
1 0.9 0.004 21.212 58.465
2 0.9 0.006 33.838 93.249
3 0.9 0.008 45.281 124.76
Ansys results
For data A
Figure 3 First mode (D=0.004m, L=1m, 17.181Hz)
Figure 4 Second mode (D=0.004m, L=1m, 47.356Hz)
Figure 5 First mode (D=0.006m, L=1m, 27.408Hz)
Figure 6 Second mode (D=0.006m, L=1m, 75.534Hz)
Figure 7 First mode (D=0.08m, L=1m, 36.677Hz)
Figure 8 Second mode (D=0.008m, L=1m, 101.06Hz)
For data B
Figure 9 First mode (D=0.004m, L=0.9m, 21.212Hz)
Figure 10 Second mode (D=0.004m, L=0.9m, 60.843Hz)
Figure 11 First mode (D=0.006m, L=0.9m, 33.838Hz)
Figure 12 Second mode (D=0.006m, L=0.9m, 93.249Hz)
Figure 13 First mode (D=0.008m, L=0.9m, 45.289Hz)
Figure 14 Second mode (D=0.08m, L=0.9m, 124.76Hz)
From above theoretical and Ansys values.
For data A:
(1):
First Mode:
17.93 − 17.181
∗ 100 = 𝟒. 𝟏𝟕𝟕%
17.93
Second Mode:
49.81 − 47.356
∗ 100 = 𝟒. 𝟗𝟐%
49.81
(2):
First Mode:
27.408 − 26.82
∗ 100 = 𝟐. 𝟏𝟒%
27.408
Second Mode:
75.534 − 74.25
∗ 100 = 𝟏. 𝟔𝟗%
75.534
(3):
First Mode:
36.667 − 36.564
∗ 100 = 𝟎. 𝟐𝟖%
36.667
Second Mode:
101.06 − 98.63
∗ 100 = 𝟐. 𝟒%
101.06
For data B:
(1):
First Mode:
22.20 − 21.212
∗ 100 = 𝟒. 𝟔𝟓%
21.212
Second Mode:
61.24 − 58.465
∗ 100 = 𝟒. 𝟓𝟑%
61.24
(2):
First Mode:
33.838 − 33.14
∗ 100 = 𝟐. 𝟎𝟔%
33.838
Second Mode:
93.249 − 91.42
∗ 100 = 𝟏. 𝟗𝟔%
93.249
(3):
First Mode:
45.281 − 44.48
∗ 100 = 𝟏. 𝟕𝟔%
45.281
Second Mode:
124.76 − 122.21
∗ 100 = 𝟐. 𝟎𝟒%
124.76
Table A3
Sr. No. Shaft Diameter %error
Length(m)
First Mode Second Mode
1 1 0.004m 4.177 4.92
2 1 0.006m 2.14 1.69
3 1 0.008m 0.28 2.4
Table B3
Sr. No. Shaft Diameter %error
Length(m)
First Mode Second Mode
1 0.9 0.004m 4.65 4.53
2 0.9 0.006m 2.06 1.96
3 0.9 0.008m 1.76 2.04
Conclusion:
From above data, there is no so much error between experimental result and ANSYS result these
errors can also be reduced if we increase the number of nodes during analysis on ANSYS and
solve it that would be time taking for lower end pc but if we use higher-end pc and then solve it
we will get approx. accurate results. In other case error may occur due to human error. All the
deformations approximately same as research paper.
