Bangladesh Army University of Science and Technology (BAUST)
Department of Mechanical Engineering
Strength of Materials and Mechanics of Machinery Lab
List of Experiments
S/
L
No
i.
Investigate the influence of arc of contact in belt-pulley friction.
Page
Numb
er
02-03
ii
Determination of critical speed of a shaft with single mass.
04-06
iii
Experimental verification of Gyroscopic Law’s.
07-08
iv
Determination of frequency of spring vibration by using two degree freedom
vibration with similar spring.
09-14
v
Determination of frequency of spring vibration by using two degree freedom
vibration with different spring.
13-20
Experiment Name
Page 1 of 20
No. of Experiment: 01
Name of the Experiment: Investigate the influence of arc of contact in belt-pulley friction.
Objectives:
1. To know about influence of arc of contact between belt and pulley friction.
Required Apparatus:
1.
2.
3.
4.
Belt Friction Apparatus
Rope
Spring Balance
Allen key
Theory: Influence of arc of contact: When a rope or belt slides over a pulley, the friction
between the rope and the pulley produces differing rope forces in the two ends. The
difference between the rope forces corresponds to the friction force at the pulley.
The belt or rope friction is a function of the coefficient of friction and the arc of contract .
The ratio of the two rope forces is calculated by way of Eytelwein’s rope friction equation.
=
Experiment Procedure:
1.
2.
3.
4.
5.
6.
Set the rope with pulley.
Loosen clamping lever at support, lower pulley and slacken off belt.
Pull out locking pin at support and set locking pin at the hole of 30°.
Adjust both ends force 50N.
Rotate the pulley handle.
When the rope slides over the pulley keep rotating the handle and record rope
forces in the two ends.
7. Now set the locking pin at 45°, 60°, 90°, 120°, 150° and 180° and measure the
forces in same way as 30°.
8. Record all the forces and compare experimental and theoretical values.
Page 2 of 20
Data Table:
Angle ∝
(degrees)
Angle ∝ (radiant)
Force F1 (N)
Force F2 (N)
Ratio F1/F2,
Experimental
Ratio
30°
60°
90°
120°
150°
180°
0.52
1.05
1.57
2.09
2.62
3.14
e , Calculation
Discussion:
Page 3 of 20
Experiment No.: 02
Name of the Experiment: Determination of critical speed of a shaft with single mass.
Objectives:
1) To know about critical speed of shafts..
2) To know how to determine critical speed of a shaft with single mass.
Required Apparatus:
1) Whirling Shaft Apparatus
2) 6 mm diameter a shaft
3) 1 kg mass
Theory: In this experiment the oscillation behavior of an elastic shaft will be studied.
Phenomena such as resonance and self-centering will be particularly clear in the over-critical
range.
In order to prevent influences from the gyrostatic effect, a symmetrical rotor with a centered
mass disc will be considered. In this case, the mass disc does not carry out tilting motions, but
instead runs through only pure translations in addition to its rotation around the axis of
rotation.
First the expected Critical Speed is calculated. In order to show the influence of the distances
between bearings on the critical speed, the test will be conducted for two rotor lengths of 300
mm and 450 mm.
The inherent circular frequency is calculated from the rigidity c of the shaft and the mass m
of the disc
ω° =
With it, this equation follows for the critical speed
n
. ω° =
=
.
The rigidity of the shaft is calculated for the Centre position of the disc to
c=
!
With E = 210000 N/mm and the geometrical moment of inertia
I=
#$%
=
#. %
= 63.61 mm
Page 4 of 20
Experimental Procedure:
1) Move rear bearing into desired position. Marking and scale make adjusting
easier.
2) Clamp bearing in place.
3) Release screw of superfluous second mass disc, release related safety bearing
with clamping lever and move safety bearing and position in centre between two
bearings.
4) Clamp safety bearing in place.
5) Adjust axial clearance of mass disc.
6) Carefully fix mass disc in place with clamping screw.
7) Rotate shaft by hand and ensure proper running.
8) Cover rotor with protective cover and lock quick-release catches.
9) Set speed potentiometer 1 on switchbox to zero.
10) Set switch for speed adjustment to potentiometer 1.
11) Switch on motor.
12) Run up rotor very slowly up to resonance point with potentiometer 1.
13) As soon as rotor contacts safety bearing, read speed n off speed counter and
record.
14) Reduce speed again.
15) Set speed with potentiometer which is guaranteed to lie above speed (approx.
500 rpm above speed n ). In this case, one potentiometer rotation corresponds to
approx. 300 rpm.
16) Switch specified rotor as quickly as possible and run through resonance point.
17) Now lower speed slowly with potentiometer 2 in over-critical range until rotor is
near resonance point and contact safety bearing.
18) Read off speed n' and record.
Calculation:
n
=
()*(+
Page 5 of 20
Data Table:
Comparison of measured and calculated critical speed
Distance
between
bearings L in
mm
Speed from
below n in
rpm
Speed from
above n in
rpm
Critical speed
Critical speed
in rpm
in rpm
n
Discussion:
Page 6 of 20
n
No. of Experiment: 03
Name of the Experiment: Experimental verification of Gyroscopic Law’s.
Objectives:
1. To know about Gyroscopic Law’s.
Theory: The mass of the slider weight (m = 65.6 g), the acceleration due to gravity g, and the
radius r of the slider weight produce the moment Mw dictated by the balance bar.:
Mw = m.g.r = 0.0656 kg.9.81 m/s2. r = 0.06435N. r
This moment Mw is counteracted by the gyroscopic moment, causing the balance bar to be
lifted to the horizontal position.
The theoretical gyroscope moment Mk is calculated from the rotational speed of the frame nF,
the rotational speed of the gyro ne and the mass moment of inertia of the gyro Jz (Jz = 375
cm2g) as follows:
M k = , F , e Jz =
#
nF .
#
ne .0.0000375 kg m2
Required Apparatus:
1. Gyroscope
2. Slide Calipers
3. Allen key
Experimental Procedure:
1.
2.
3.
4.
5.
6.
Place the protective hood in the retaining ring.
Turn the two speed potentiometers to zero.
Switch on the meter for the gyro.
With the speed potentiometer run up to the desired rotational speed.
Switch on the motor for the frame.
With the speed potentiometer increase the rotational speed until the balance bar is
horizontally aligned.
7. Make a note of both rotational speeds.
Page 7 of 20
Data Table:
Radius r
in m
Moment
Mw in Nm
Rotational
speed of
gyro ne in
rpm
0.025
0.050
0.075
0.095
Discussion:
Page 8 of 20
Rotational
speed of
Frame nF
in rpm
Moment
Mk in Nm
Deviation
in %
No. of Experiment: 04
Name of the Experiment: Determination of frequency of spring vibration by using two
degree freedom vibration with similar spring.
Objectives:
1. To know about the two degree of freedom vibration analyzer system.
2. To observe the similarities and dissimilarities between the mathematical value and
experimental value.
3. To know about frequency measurement by using Tachometer with similar spring.
Required Apparatus:
1.
2.
3.
4.
5.
Two Degree Freedom Vibration Analyzer
1 kg mass
Compression Spring
Steel wire
Tachometer
Theory:
Two Degree of Freedom: Some dynamic systems that require two independent coordinates,
or degrees of freedom, to describe their motion, are called “two degree of freedom systems”.
Degrees of freedom may or may not be in the same coordinate direction.
Number of degrees of freedom of the system = Number of masses in the system - Number of
possible types of motion of each masses.
Figure: Two Degree of Freedom
Page 9 of 20
Theoretical calculation:
Data Table:
No of obs.
Loads
Length of the
Length of the
Extension
(gm)
spring without
spring with load,
load, ./ (cm)
.0 (cm)
L = .0 -./
(cm)
1
200
3.4
3.5
0.1
2
250
3.4
4
0.6
3
300
3.4
4.3
0.9
4
350
3.4
4.6
1.2
5
400
3.4
5
1.6
Table 1: Spring constant measurement
Figure: Load vs extension graph
From load vs extension graph,
X axis, m = dm = 260 gm = 0.26 kg
Y axis, l = dl = 0.9 cm = 0.009m
Page 10 of 20
1
Slope =
.
=
2
.
m/kg
= 0.03461 m/kg
Slope 7 =
81'9:
= 28.8934 kg/m
S = g - slope 7
= 9.81 - 28.8934 ( BCDE 7 /D)
H
= 283.44 I
[1 kgms 7 = 1 N]
Let, Similar spring constant and mass.
Given that,
D = 1 kg [1, 2, 3, 4 …]
D = 1 kg [1, 2, 3, 4 …]
Spring constant L = L = L = 283.44 N/m [10, 20, 30, 40 …………………∞ ]
Then equation of motion of D is,
D
$N OP
$1N
= - s x + E (R - R ) ………………..(1)
Then equation of motion of m is,
D
S N TN
SU N
= - E ( R - R ) – E R ……………….. (2)
Page 11 of 20
Assume,
R = V cos ωt
R = V cos ωt
Then equation (1) becomes,
−D V , Cos ωt = −E V cos ωt + E (V cos ωt − V cos ωt)
i.e.: XD , − (E + E )ZV = −E V ……………………… (3)
And equation (2) becomes,
−D V , Cos ,t = − E (V \]E,^ − V \]E,^) − E V cos,^
i.e.: XD , − (E + E ZV = −E V
Eliminating V and V between equations (3) and (4),
D , − (E + E )
E
=
E
D , − (E + E )
From which,
⇨ ω −`
(8P *8N )
⇨ω −`
.
N
*
+
.
(8N *8! )
N
aω
.
+
*
+
.
8P 8N * 8N 8! *8! bP
P
N
=0
aω +
(283.44 - 283.44) + (283.44 - 283.44 ) + (283.44 - 283.44)
= 0
1-1
⇨ ω − 1133.76, + 241014.7008 = 0
⇨ω =
.g ±i(
.g )N 7 - -
.g
Page 12 of 20
⇨ω =
.g ±j
.
Now taking (+) value and negative (-) value,
ω = (1133.76 + 566.88)/2
ω = (1133.76 − 566.88)/2
= 850.32
= 283.44
So, , = 29.160
So, , = 10.835
Now, Frequency n =
=
lP
Now, Frequency n =
2.
=
= 4.641 Hz
lN
.
j
= 2.68 Hz.
Experiment Procedure:
1. Take two similar mass and three springs of similar / dissimilar size, constant value
and adjusting to the point.
2. Select the spring according to the spring constant measured by formula and graph
paper.
3. Take the mass and spring setup in horizontally.
4. Two thin brake wires will be attached on the two mass from the support stand.
5. A tachometer is set in front of the mass and an indicated sticker is attached to the
mass body.
6. Simply turn the spring manually and it will vibrate as like two degree freedom during
that time.
7. Press the tachometer button, holding for one minute.
8. Take the r.p,m value and calculate the frequency by using mathematical formula.
9. Compare among the theoretical and practical data and find out the error.
Page 13 of 20
Experimental calculation:
From the tachometer get r.p.m (revolution per minute) value.
.9.
So, Frequency =
-
= (?) r.p.s (revolution per second) or Hz
Now,
Measured r. p. m = 220, r. p. m = 217
Frequency ( n ) =
=
.9. P
-
Frequency (n )=
=
-
= 1.83 r.p.s or Hz.
.9. N
g
-
= 1.81 r.p.s or Hz
Data Table:
Spring constant
(N /m )
L = 283.44
L = 359.341
L = 105.107
Theoretical
frequency (opq )
(Hz)
n = 4.64
Experimental
(Hz)frequency (ors )
(Hz)
n = 1.83
n = 2.68
n = 1.81
Discussion:
Page 14 of 20
Error =
opq 7ors
opq
60.56
32.46
- /tt%
No. of Experiment: 05
Name of the Experiment: Determination of frequency of spring vibration by using two
degree freedom vibration with different spring.
Objectives:
4. To know about the two degree of freedom vibration analyzer system.
5. To observe the similarities and dissimilarities between the mathematical value and
experimental value.
6. To know about frequency measurement by using Tachometer with different spring.
Required Apparatus:
6. Two Degree Freedom Vibration Analyzer
7. 1 kg mass
8. Compression Spring
9. Steel wire
10. Tachometer
Theory:
Two Degree of Freedom: Some dynamic systems that require two independent coordinates,
or degrees of freedom, to describe their motion, are called “two degree of freedom systems”.
Degrees of freedom may or may not be in the same coordinate direction.
Number of degrees of freedom of the system = Number of masses in the system - Number of
possible types of motion of each masses.
Figure: Two Degree of Freedom
Page 15 of 20
Theoretical calculation:
Data Table:
No of obs.
Loads
Length of the
Length of the
Extension
(gm)
spring without
spring with load,
load, ./ (cm)
.0 (cm)
L = .0 -./
(cm)
1
200
3.4
3.5
0.1
2
250
3.4
4
0.6
3
300
3.4
4.3
0.9
4
350
3.4
4.6
1.2
5
400
3.4
5
1.6
Table 1: Spring constant measurement
Figure: Load vs extension graph
From load vs extension graph,
X axis, m = dm = 260 gm = 0.26 kg
Y axis, l = dl = 0.9 cm = 0.009m
Page 16 of 20
1
Slope =
=
.
.
2
m/kg
= 0.03461 m/kg
Slope 7 =
81'9:
= 28.8934 kg/m
S = g - slope 7
= 9.81 - 28.8934 ( kgms 7 /m)
= 283.44
v
[1 kgms 7 = 1 N]
Similarly From above,
Spring No
Spring Constant (N/m)
1
283.44
2
359.341
3
105.107
Let, Different spring constant and same mass.
Given that, D = 1 kg [1, 2, 3, 4 …]
D = 1 kg [1, 2, 3, 4 …]
Spring constant L = 283.44 N/m, L = 359.341 N/m, L = 105.107 N/m.
Then equation of motion of D is,
Page 17 of 20
D
S N TP
= - E R + E (R - R ) ………………..(1)
SU N
Then equation of motion of D is,
D
S N TN
= - E ( R - R ) – E R ……………….. (2)
SU N
Assume,
R = V cos ωt
R = V cos ωt
Then equation (1) becomes,
−D V , Cos ωt = −E V cos ωt + E (V cos ωt − V cos ωt)
i.e.: XD , − (E + E )ZV = −E V ……………………… (3)
And equation (2) becomes,
−D V , Cos , t = − E (V \]E,^ − V \]E,^) − E V cos,^
i.e.: XD , − (E + E ZV = −E V
Eliminating V and V between equations (3) and (4),
D , − (E + E )
E
=
E
D , − (E + E )
From which,
⇨ , −`
(wP *wN )
⇨, −`
.
IN
+
(wN *w! )
* j2.
IN
+
a,
j2.
+
*
j.
wP wN * wN w! *w! xP
IP IN
=0
g
a, +
(283.44 - 359.341) + (359.341 - 105.107 ) + (105.107 - 283.44)
= 0
1-1
Page 18 of 20
⇨ , − 1107.229, + 169412.3956 = 0
⇨ω =
⇨, =
g.
2 ± i(
g.
2)N 7 - -
-
2
. 2j
1107.229 ± 740.48
2
Now taking (+) value and negative (-) value,
ω = (1107.229 + 740.48)/2
ω = (1107.229 − 740.48) /2
= 923.8545
= 183.37
So, ω = 30.39
So, ω = 13.541
Now, Frequency n =
=
yP
#
Now, Frequency n =
. 2
yN
#
=
#
= 4.84 Hz
.j
#
= 2.30Hz
Experiment Procedure:
1. Take two similar mass and three springs of dissimilar size and adjusting to the point.
2. Select the spring according to the spring constant measured by formula and graph
paper.
3. Take the mass and spring setup in horizontally.
4. Two thin brake wires will be attached on the two mass from the support stand.
5. A tachometer is set in front of the mass and an indicated sticker is attached to the
mass body.
6. Simply turn the spring manually and it will vibrate as like two degree freedom during
that time.
7. Press the tachometer button, holding for one minute.
Page 19 of 20
8. Take the r.p,m value and calculate the frequency by using mathematical formula.
9. Compare among the theoretical and practical data and find out the error.
Experimental calculation:
From the tachometer get r.p.m (revolution per minute) value.
.9.
So, Frequency =
-
= (?) r.p.s (revolution per second) or Hz
Now,
Measured r. p. m = 424, r. p. m = 420
Frequency ( n ) =
=
.9. P
-
Frequency (n )=
=
-
= 2.25 r.p.s or Hz.
Spring constant
(N /m )
L = 283.44
L = 359.341
L = 105.107
.9. N
-
= 2.23 r.p.s or Hz
opq 7ors
Theoretical
frequency (opq )
(Hz)
Experimental
(Hz)frequency (ors )
(Hz)
n = 4.84
n = 2.25
53.51
n = 2.30
n = 2.23
3.04
Page 20 of 20
Error =
opq
- /tt%