PHOTOELASTICITY -1
Aim: To determine the material fringe value of a circular disc specimen subjected to
compression.
Fig – Photoelasticity Apparatus
Theory:
Photoelasticity: It is an experimental method of stress analysis employing transparent plastic
models of a prototype and a field of polarized light produced by an instrument known as
"polariscope". The interaction of polarized light with stressed models produce characteristic
patterns, known as "Stress Patterns" from which the complete distribution of the principal
stresses in two dimensional as well as three dimensional problems can be determined.
Polarizer: A filter for converting randomly polarized light into plane polarized light; and which
is located immediately in front of the light source.
Analyzer: A Filter at the viewing end of the polariscope. The polarizer and analyzer are
optically identical, and are thus named to distinguish their respective functions.
Quarter wave plates: An optical element for inducing ¼ of a wavelength of relative
retardation between two spatially perpendicular component light rays. The transmission
polarizer contains two-quarter wave plates – one in conjunction with the polarizer, and one in
conjunction with the analyzer.
Plane polarized light: Light characterized by having the transverse vibrations limited to a
single plane for a ray; or to parallel planes for a beam.
Circularly polarized light: Light in which the plane of polarization rotates continuously with
propagation of the light ray. Circularly polarized is produced when the vibration of two
component rays of equal amplitude are perpendicular to each other on space and 1/4 of a
wavelength out of phase.
Fringe: Generically, colored or black lines, bands, or areas forming the photoelastic pattern,
and representing loci of constant difference in principal stresses. Specifically, in the case of
isochromatic fringes, the tint-of-passage in white light, or the centers of the dark bands in
monochromatic light.
Fringe order (N): ordinal numerical designation assigned to an isochromatic fringe. N equals
the magnitude of the birefringence at a point (expressed in fringes), and is proportional to the
difference in principal stresses at that point.
Isochromatic: A fringe (single-colored in white light, essentially black in monochromatic light)
representing a locus of constant difference in principal stresses (x - y) as the locus of constant
maximum stress, max.
Isoclinic: A black line representing a locus of constant directions of principal stresses, or in
other words, constant inclinations of the principal axes with respect to an arbitrary reference
axis.
Monochromatic light: Light of only one wavelength.
Principal Stresses: The algebraically maximum and minimum normal stresses at a point. The
principal stresses coincide in direction with the principal strains in isotropic materials having
the same elastic properties in all directions.
Description of Apparatus:
Polariscope is a precision optical instrument for performing quantitative stress measurements.
The basic model consists of a polarizing assembly, analyzing assembly, mechanical drive
coupling system for control of all four filters. All components are mounted on a common base
frame. The polarizer, analyzer, and quarter wave plates are glass-laminated construction. They
are mounted in Aluminum rings, and rotate on ball bearings. The rotation is indicated on a
precision engraved dial. In use, a photoelastic model is placed in the polariscope and when
forces are applied, a colorful fringe pattern results. This pattern reveals a visible picture of the
stress distribution over the whole area of the model.
The basic polariscope provides the capability for the following types of analysis and
measurement.
 Overall assessment of nominal stress magnitude and gradients.
 The directions of the principal stresses
 The magnitude of sign of the tangential stresses along free (unloading) boundaries and in
regions where the state of stress is uniaxial.
 The magnitude of the difference in principal stresses in a biaxial stress state.
Procedure:
(i)
Switch 'ON' the monochromatic source and allow it to attain full intensity.
(ii)
Align the axis of the 'Polarizer' 'Analyzer' and 'Quarter wave plates' to obtain either a dark
or bright field.
(iii)
Fix the specimen in the loading frame and bring the horizontal arm in position by means
of balancing weights.
(iv)
Load the specimen suitably and determine the fringe order by Tardy's* method of
compensation.
(v)
Increase the load in equal steps and note the corresponding fringe order.
(vi)
Plot a graph of load vs. fringe order (F Vs. n)
Diagram:
Observations:
1.
2.
3.
4.
Diameter of the circular disc d = _____ mm
Thickness of the circular disc h = _____ mm
Length of fulcrum to centre of loading pan l1 = _____mm
Length of fulcrum to centre of specimen l2 = _____ mm
Tabular column:
Load (FL)
Sl.
No.
Kg
N
Load on Model
‘F’, N
Fringe order
(n)
1.
2.
3.
4.
Calculations:
1.
Load Ratio r = l1/l2
2.
Load on model: F = FL * r
3.
Slope = F / n
4.
5.
N/fringe
(from Graph)
Material fringe value: (f)
f = 8/D (F / n) N/mm-fringe
Model fringe value: (fm)
fm = 8P/Dh
N/mm2 -fringe
Slope
(F/n)
Material
Fringe
Value
f
Model
Fringe
Value
Fm
PHOTOELASTICITY -2
Aim: To determine the material fringe value of a photoelastic material using a simply supported
beam specimen subjected to pure bending.
Theory:
In photoelastic analysis the stress distribution in a complex model is sought for applied load.
To obtain the stress distribution we should first know the material fringe value. The material
fringe values of photoelastic materials vary with the supplier, the batch of resin, the temperature
and the age. For this reason it is always necessary to calibrate each sheet of photoelastic
material at the time of test. Accurate determination of material fringe value can be obtained by
using
1.
Tensile specimen
2. Disc specimen
The circular polariscope setup is used in this experiment.
supported beam loaded as shown in figure.
3. Beam specimen
The test specimen is a simply
Description of Apparatus:
Polariscope is a precision optical instrument for performing quantitative stress measurements.
The basic model consists of a polarizing assembly, analyzing assembly, mechanical drive
coupling system for control of all four filters. All components are mounted on a common base
frame. The polarizer, analyzer, and quarter wave plates are glass-laminated construction. They
are mounted in Aluminum rings, and rotate on ball bearings. The rotation is indicated on a
precision engraved dial. In use, a photoelastic model is placed in the polariscope and when
forces are applied, a colorful fringe pattern results. This pattern reveals a visible picture of the
stress distribution over the whole area of the model.
The basic polariscope provides the capability for the following types of analysis and
measurement.
 Overall assessment of nominal stress magnitude and gradients.
 The directions of the principal stresses
 The magnitude of sign of the tangential stresses along free (unloading) boundaries and in
regions where the state of stress is uniaxial.
 The magnitude of the difference in principal stresses in a biaxial stress state.
Procedure:
(vii) Switch 'ON' the monochromatic source and allow it to attain full intensity.
(viii) Align the axis of the 'Polarizer' 'Analyzer' and 'Quarter wave plates' to obtain either a dark
or bright field.
(ix)
Fix the specimen in the loading frame and bring the horizontal arm in position by means
of balancing weights.
(x)
Load the specimen suitably and determine the fringe order at the extreme fibres by
adopting Tardy’s method of compensation.
(xi)
Increase the load in equal steps of 20 N and note the corresponding fringe order.
(xii) After completing the experiments the specimen is removed from the corresponding value
of and the light source is switched off.
(xiii) Plot a graph of load vs. fringe order (P Vs n)
(xiv) The fringe order is given by the equation n = 1 ± (/180)
Where  is the angle through which the analyser must be rotated to obtain extinction at the edge
of the specimen. Fringe order for the different loads applied is noted. The results are obtained
and ‘P’ v/s ‘n’ curve is plotted and the slope is obtained.
Diagram :
Observations :
Length
l1
Length
l2
Distance
X1
Distance
X2
Depth of the specimen
Thickness of the specimen
Load on the specimen Feff
= ____ mm
= ____ mm
= ____ mm
= ____ mm
= ____ mm
= ____ mm
= F * (l1/l2), N
Tabular column :
Sl.
No.
1.
2.
3.
4.
5.
Load
Kg
N
Effective Load,
Feff, N
Fringe Order, n
Material Fringe
value ‘f’
Model
Fringe
value, fm
Calculations:
Section Modulus
Z
= td2/6, mm3
Bending moment on the specimen
Mb
= [F/2] * [(X1-X2)/2] N-mm
Bending stress
b
= Mb/Z MPa
Maximum Bending Stress
b max = M/Z = n*f / t
 F 
= 
 n  slope
Slope (from the graph F Vs n)
3( X 1  X 2 )  F 
 n 
2d 2
slope
Material Fringe value
f 
Model fringe value
fm 
3( X 1  X 2 )  F 
 n 
2d 2t
slope
PHOTOELASTICITY -3
Aim: To determine the stress concentration factor for a flat plate with a circular hole, subjected
to a tensile load.
Theory:
In photoelastic analysis the stress distribution in a complex model is sought for applied load. To
obtain the stress distribution we should first know the material fringe value. The material fringe
values of photoelastic materials vary with the supplier, the batch of resin, the temperature and the
age. For this reason it is always necessary to calibrate each sheet of photoelastic material at the
time of test.
Accurate determination of material fringe value can be obtained by using the circular polariscope
setup is used in this experiment. The test specimen is a flat plate with a circular hole made of
photoelastic material.
Definitions:
Photoelasticity: It is an experimental method of stress analysis employing transparent plastic
models of a prototype and a field of polarized light produced by an instrument known as
"polariscope". The interaction of polarized light with stressed models produce characteristic
patterns, known as "Stress Patterns" from which the complete distribution of the principal
stresses in two dimensional as well as three dimensional problems can be determined.
Polarizer: A filter for converting randomly polarized light into plane polarized light; and which
is located immediately in front of the light source.
Analyzer: A Filter at the viewing end of the polariscope. The polarizer and analyzer are
optically identical, and are thus named to distinguish their respective functions.
Quarter wave plates: An optical element for inducing ¼ of a wavelength of relative
retardation between two spatially perpendicular component light rays. The transmission
polarizer contains two-quarter wave plates – one in conjunction with the polarizer, and one in
conjunction with the analyzer.
Plane polarized light: Light characterized by having the transverse vibrations limited to a
single plane for a ray; or to parallel planes for a beam.
Circularly polarized light: Light in which the plane of polarization rotates continuously with
propagation of the light ray. Circularly polarized is produced when the vibration of two
component rays of equal amplitude are perpendicular to each other on space and 1/4 of a
wavelength out of phase.
Fringe: Generically, colored or black lines, bands, or areas forming the photoelastic pattern,
and representing loci of constant difference in principal stresses. Specifically, in the case of
isochromatic fringes, the tint-of-passage in white light, or the centers of the dark bands in
monochromatic light.
Fringe order (N): ordinal numerical designation assigned to an isochromatic fringe. N equals
the magnitude of the birefringence at a point (expressed in fringes), and is proportional to the
difference in principal stresses at that point.
Isochromatic: A fringe (single-colored in white light, essentially black in monochromatic light)
representing a locus of constant difference in principal stresses (x - y) as the locus of constant
maximum stress, max.
Isoclinic: A black line representing a locus of constant directions of principal stresses, or in
other words, constant inclinations of the principal axes with respect to an arbitrary reference
axis.
Monochromatic light: Light of only one wavelength.
Principal Stresses: The algebraically maximum and minimum normal stresses at a point. The
principal stresses coincide in direction with the principal strains in isotropic materials having
the same elastic properties in all directions.
Stress concentration: A stress concentration (often called stress raisers or stress risers) is a
location in an object where stress is concentrated. An object is strongest when force is evenly
distributed over its area, so a reduction in area, e.g. caused by a crack, results in a localized
increase in stress. A material can fail, via a propagating crack, when a concentrated stress
exceeds the material's theoretical cohesive strength. The real fracture strength of a material is
always lower than the theoretical value because most materials contain small cracks or
contaminants (especially foreign particles) that concentrate stress.
Stress concentration factor: A theoretical factor Kt expressing the ratio of the greatest stress in
the region of stress concentration to the corresponding nominal stress.
Description of Apparatus:
Polariscope is a precision optical instrument for performing quantitative stress measurements.
The basic model consists of a polarizing assembly, analyzing assembly, mechanical drive
coupling system for control of all four filters. All components are mounted on a common base
frame. The polarizer, analyzer, and quarter wave plates are glass-laminated construction. They
are mounted in Aluminum rings, and rotate on ball bearings. The rotation is indicated on a
precision engraved dial. In use, a photoelastic model is placed in the polariscope and when
forces are applied, a colorful fringe pattern results. This pattern reveals a visible picture of the
stress distribution over the whole area of the model.
The basic polariscope provides the capability for the following types of analysis and
measurement.
 Overall assessment of nominal stress magnitude and gradients.
 The directions of the principal stresses
 The magnitude of sign of the tangential stresses along free (unloading) boundaries and in
regions where the state of stress is uniaxial.
 The magnitude of the difference in principal stresses in a biaxial stress state.
Procedure:
(xv) Switch 'ON' the monochromatic source and allow it to attain full intensity.
(xvi) Align the axis of the 'Polarizer' 'Analyzer' and 'Quarter wave plates' to obtain either a dark
or bright field.
(xvii) Fix the specimen in the loading frame properly with the help of shackles and bring the
horizontal arm into position
(xviii) Load the specimen suitably and determine the fringe order at the extreme fibres by
adopting Tardy’s method of compensation.
(xix) Increase the load in equal steps of 5 Kg and note the corresponding fringe order.
(xx) After completing the experiments the specimen is removed from the corresponding value
of and the light source is switched off.
(xxi) Plot a graph of load vs fringe order (P Vs N)
(xxii) Determine the stress concentration factor
Diagram:
Observations:
Width of the specimen
w = ____mm
Diameter of the circular hole d = ____mm
Thickness of the specimen
t = ____mm
Length
l1 = ____mm
Length
l2 = ____mm
Tabular column:
Sl.
No.
1
2
3
4
5
Load
FL
Kg
N
Effective
Load, F
N
Angle

Fringe order
x
n
Peak
Stress
p MPa
Normal
Stress
n MPa
SCF
Kt
= p /n
Calculations:
Effective load
F = FL * (l1/l2)
Fringe order
n = x + ( / 180)
Normal stress
n = F / (w-d) t
Peak Stress or Max. Stress p = (n f)/t
Stress concentration factor Kt= p /n
(10 to 12 N/mm/sec)
Porter Governor
Aim: To conduct experiment on Porter governor & to determine the frictional resistance at the
sleeve, centrifugal forces on the governor balls & draw the controlling force diagram.
Apparatus: Porter Governor, Tachometer, Measuring tape.
Observations:
1. Mass of governor fly balls m=------ Kg
2. Mass of sleeve M= ---------- Kg
3. Length of upper links = Length of lower links L= ------
meters
4. Offset of links from axis of rotation y = -------- meters
5. Initial distance between top & bottom pivots H= --------meters
h
L
C

2
mw r
S
H
r
L
M
Sleeve motion
y
Schematic diagram of Porter Governor
Tabular Column:
Trial Speed Sleeve Distance Distance
No
‘N’
Lift
'c'
's'
rpm
‘x’
meters
meters
meters
Radius
of
rotation
'r'
meters
Angle
Height of Friction
governor
al
''
degrees 'h' meters Force 'f'
Newton
Centrifugal Effort
force
‘E’
'Fc'
Newton
Newton
1
2
3
4
5
Specimen calculations:
H x 
1. Distance c     metres
 2 2
2. Distance s 
L
2
 c 2  meters
3. Radius of rotation r   y  s  metres where y=offset of links
c 
4. Angle   sin 1  
L 
5. Height of governor h  r tan 
6. Frictional force
we know that the equilibrium speed of a porter governor
with equal link lengths & equal offset of the upper &
lower links is given by
 mg  ( Mg  f )  895
N2  

mg

 h
N 2 mgh
Hence f=
 (m  M ) g
895
7. Effort of the governor is the mean force exerted at the sleeve for a fractional change in
speed. E=0.01(Mg+mg+f) for 1% change in speed.
8. Power =Effort x Sleeve lift=E x x=--------Nm
2N
9. Angular velocity w 
rad/sec
60
10. Centrifugal force or controlling force Fc=mwr Newton
Power
P, Nm
centrifugal force Fc
Radius of rotation r
Controlling force curve
Proell Governor
Aim: To conduct experiment on Proell governor & to determine the frictional resistance at the
sleeve, centrifugal forces on the governor balls & draw the controlling force diagram.
Apparatus: Proell Governor, Tachometer, Measuring tape.
Observations:
6. Mass of governor fly balls m=------ Kg
7. Mass of sleeve M= ---------- Kg
8. Length of upper links = L = ------
meters
9. Offset of links from axis of rotation y = -------- meters
10. Extension of links which carry rotating masses e = ----- meters
11. Initial distance between top & bottom pivots H= --------meters
Pivot
2
mw r
h
L
C

e
S
r
M
Sleeve motion
x
Pivot
y
Schematic diagram of Proell Governor
Tabular Column:
Trial
No
Speed
‘N’
Rpm
Sleeve
Lift ‘x’
meters
Distan Distance Radius
ce 'c' 's'
of
meters meters
rotation
'r'
meters
Angle
Height of Frictional
governor
Force 'f'
''
Newton
degrees 'h' meters
Centrifugal
force
'Fc' Newton
Effort
‘E’
Newton
1
2
3
4
5
Specimen calculations:
H x 
11. Distance c     metres
 2 2
12. Distance s 
L
2
 c 2  meters
13. Radius of rotation r   y  s  metres where y=offset of links
c 
14. Angle   sin 1  
L 
15. Height of governor h  r tan 
16. Frictional force
As equilibrium speed of a proell governor with equal offset of links is
c  mg  ( Mg  f )  895


(c  e) 
mg
 h
where e=length of extension links
N2 
N 2 mgh  c  e 

  (m  M ) g
895  c 
17. Effort of the governor is the mean force exerted at the sleeve for a fractional change in
speed. E=0.01(Mg+mg+f) for 1% change in speed.
Hence f=
18. Power =Effort x Sleeve lift=Ex=--------Nm
2N
19. Angular velocity w 
rad/sec
60
20. Centrifugal force or controlling force Fc=mwr Newton
Power
P, Nm
centrifugal force Fc
Radius of rotation r
Controlling force curve
Hartnell Governor
Aim: To conduct experiment on Hartnell (Spring controlled) governor & to determine the
frictional resistance at the sleeve, centrifugal forces on the governor balls & draw the controlling
force diagram.
Apparatus: Hartnell Governor, Tachometer, Measuring tape.
It is a spring type governor and much more sensitive then porter and proell governor. In this
governor the balls are controlled by spring which is mounted on the spindle axis.
It consists of the casing in which the pre-compressed spring is housed so as to apply the force on the sleeve.
There are two bell crank levers and each of it carries a ball at one of the end and the roller at
another end, both are fitted on the frame of the casing. The casing along with spring and frame
rotates about the spindle axis.
When the speed of the governor increase then the balls fly out away from the axis, bell crank
lever moves on the pivot and then the roller end lifts the sleeve against force.
This movement is then transferred to the throttle of the engine. We can also adjust the force just
by tightening it.
In this a represents the vertical arm of the bell crank lever and b represents the horizontal arm of
the bell crank lever, W is the weight of the sleeve, w is the weight of the ball and s is the spring
force exerted on sleeve. For deriving the relation we take half of the combine weight exerted by
sleeve and spring.
Fly ball
r1
r
2
F=m
c w r
a
Ball arm
a
(S+f)
2
b
x
b
Sleeve arm
x =sleeve lift (or compression of spring)
Fulcrum
Taking moments about the fulcrum ,
(S  f )
Fc  a 
 b,
(i) where
2
Fc =centrifugal force on the rotating masses=mw 2 r Newton
S= Spring force exerted on the sleeve
 S=stiffness of spring  compression of spring (or sleeve lift)=k  x
Rearranging the terms in the equation (i),
a
frictional force at the sleeve f=2Fc    ( S ) Newton
b
Observations:
12. Mass of governor fly balls m=------ Kg
13. Stiffness of the spring, k= 3 kg/cm= 2943N/m
14. Length of vertical (ball arm) link = a= ------ meters
15. Length of horizontal (sleeve arm) link = b= ---------meters
16. Initial radius of rotation r1=--------- meters
Tabular Column:
Tria
l
No
Speed
‘N’
rpm
Sleeve
Lift ‘x’
meters
Spring
Radius of Centrifugal
force ‘S’ rotation ‘r’ force
Newton
meters
'Fc' Newton
Frictional
Force 'f'
Newton
Effort
‘E’
Newton
Power
P, Nm
1
2
3
4
5
Specimen calculations:
2 N
rad/sec
60
22. Spring force S=Stiffness x sleeve lift = k  x N
a
23. Radius of rotation r  x    r1 metres where r1=Initial distance between the governor
b
masses & the spindle axis (Stationary condition)
24. Centrifugal force Fc= mw 2 r Newton where m= mass of governor fly balls.
a
25. frictional force at the sleeve f=2Fc    ( S ) Newton
b
26. Effort of the governor is the mean force exerted at the sleeve for a fractional change in
speed. E=0.01(Mg+S+f) for 1% change in speed.
21. Angular velocity of the governor spindle w 
27. Power =Effort x Sleeve lift= E  x =--------Nm
centrifugal force Fc
Radius of rotation r
Controlling force curve
GYROSCOPE
Aim: To conduct an experiment on motorized gyroscope and to determine;
(i)
Gyroscopic couple
(ii) Applied couple and
(iii) To demonstrate the effect of gyroscopic couple using right hand screw rule
Observations:
1. Mass of the spinning disc M= ------------ kg
2. Diameter of the disc D= ----------- meters
3. Distance of the weights added from the fulcrum of the motor x= -------meters
Tabular column:
Trial Weight
No
added,
'W'
Newton
RPM
of the
disc
'N'
Angular Angle Time
velocity ''
taken
t
of spin deg
(for
precession
'w'
of '') sec
rad/sec
Angular
Gyroscopic Applied
velocity of couple Cg, couple
precession Nm
Ca , Nm
'wp' rad/sec
1
2
3
4
Specimen calculations:
1.
2.
3.
4.
5.
MD 2
=
kg-m2
Mass moment of inertia of the disc I 
8
2 N
=
rad/sec
Angular velocity of spin w 
60
 
w

=
rad/sec
Angular velocity of precession p
t 180
Nm
Gyroscopic couple Cg = I  w  wp =
Nm
Applied couple Ca = W  X =
a'
C
g
C
Ca
Ca
O
g
a
Characteristic curve
Spin Vector diagram
As the disc is spinning counterclockwise when viewed from the front, the applied
couple acts so as to tilt the spin vector to oa'. The reactive couple tends to rotate the
entire system clockwise when viewed from the top, which is evident from the
Right hand screw rule.
FULCRUM
DISC
WEIGHT
W
X
DYNAMIC BALANCING
Aim: To conduct an experiment on dynamic balancing machine and to determine;
(iv) Unknown angular settings between unbalanced masses
(v) Un known radius of rotation
(vi) To demonstrate the effect of rotating unbalance before & after balancing
Apparatus used: Dynamic balancing machine, Vernier caliper, Rotating masses,
Allen key, Spanners, etc.
Observations:
4. Rotating unbalance masses m=66 gms=0.066 kg
5. Diameter of the slotted disc D= 175mm
6. Diameter of the unbalance masses=30 mm
7. Radius of rotation of the masses in extreme position,
X = (radius of slotted disc-radius of unbalance discs) = (175-30)/2= 72.5mm
Problem 1
Determine the relative angular settings & the radius of mass A for complete
dynamic balance, given the masses in the planes A,B, C & D are all equal to 66
gms. Radii of rotation of masses in planes B, C & D are 72.5 mm, 60 mm 50
mm respectively. The axial distance between the planes A-B, B-C, & C-D is 100
mm each.
R.P
A
r
B
C
D
A
100mm
100mm
100mm
Tabular column: (Problem 1)
Plane
Mass M, Radius of Centrifugal
Distance
Couple
*A(Reference
Plane)
B
C
D
kg
rotation
‘r’ mm
0.066
rA
force ‘Mr’ from
‘MrL’ kgkg-mm
reference
mm2
plane
‘L’,
mm
0.066rA
0
0
0.066
0.066
0.066
72.5
60
50
4.785
3.96
3.3
100
200
300
478.5
792
990
D
2
Scale: 100 kg-mm=1cm
Scale: 1 kg-mm =1 cm
 D =232 deg
990
792
3.3
C
478.5
B
0.066rA
3.96
 C =81 deg
4.785
Couple Polygon
From couple polygon find the
relative angular settings between B, C & D
Answers:
A
 B =0 deg
 C =81 deg
 D =232 deg
Force Polygon
From couple polygon find the
unknown radius of rotation & angle of
mass in plane A
Answers:  A =201
r A =55 mm
The masse may then be arranged as per the calculations and checked for
complete dynamic balance.
ANSWERS FOR PROBLEM 1
C
0
201
0
232
0
81
B
A
D
Relative Angular Positions obtained from graphical solution
Given Data: rB=72.5 mm, rC=60 mm, rD=50 mm
Distance between planes =100 mm

Answers: rA=55 mm, CD
Problem 2:
A rotating shaft carries four equal masses A, B, C & D of 66 gm each which
are placed in planes as shown in fig. The radius of rotation of mass B is 70 mm
and the angle between B & C is 900 and between B & D is 2400. Find the radii
of rotation of masses A, C & D and also the angular position of mass A for
complete dynamic balance. Check for balance graphically.
R.P
A
B
C
C
D
0
240
rA
0
90
B
100mm
50mm
100mm
D
Tabular column:
Sl.No Plane Mass
kg
1
2
3
4
A
B
C
D
0.066
0.066
0.066
0.066
M, Radius of Centrifugal
rotation ‘r’ force ‘Mr’
mm
kg-mm
rA
0.066rA
70
4.62
rC
0.066rc
rD
0.066rD
Distance from Couple
reference plane ‘MrL’
‘L’, mm
mm2
0
0
100
462
200
13.2rC
300
19.8rD
kg-
240 deg
2
Scale: 100 kg-mm=1cm
Scale: 1 kg-mm =1 cm
240 deg
13.2 r
c
3.1
19.8rD
90°
462
B
 A 0.066rA
 C =90 deg
4.62
=0 deg
Couple Polygon
Force Polygon
Fromcouple polygon find the
unknown radius of rotation of masses
C&D
Answers:
4.026
From couple polygon find the
relative angular settings betweenA &B
rA =51 mm
 A =214 deg
rC =61 mm
r D =47 mm
From the closure of the couple polygon, we get rc=61 mm rD=47 mm.
Now, in the force table, 0.066rc=0.066x61=4.026 Kg-mm
And 0.066rD=0.066x47=3.1 kg-mm.
Substituting the above values & drawing the force polygon we obtain the direction
of mass A & its radius of rotation.
Ans: rA=51 mm, rC=61 mm rD=47 mmA=2140
*********
Critical Speed of Shafts
Aim: To determine the critical speeds of a simply supported beam theoretically & to compare
with the experimental values along with mode shapes.
Apparatus: Steel rods of diameters 6mm & 8 mm, Photosensitive Tachometer, Critical speed
test kit.
Observations:
1. Elastic modulus of steel E= 207 GPa
2. Length of rod L=1.05 meters
3. Density of steel = 7800 kg/m3
4. Rod diameter 6mm and 8 mm 
Shaft
Diameter
‘d’ meters
Mass per
unit
length’m’
kg/m
Mode
Number
6 x10-3
8 x10-3
Theoretical
Critical
frequency (Hz)
Experimental
Frequency
(Hz)
I
I
Mode
Number
Theoretical
Critical
frequency
(Hz)
II
II
Specimen Calculations:
1. Mass per unit length of shaft =density x cross sectional area
d 2
m= 
kg/m
4
2.
Theoretical critical frequency =
3.
Theoretical critical frequency =
2.45
EI
mL4
7.95
for I mode
EI
for mode II
mL4
Critical speed ( rpm )
4. Experimental frequency =
Hz
60
Mode Shapes
I mode
Node
II mode
Experimental
Critical
frequency (Hz)
Spring Mass System (Undamped vibrations)
Aim: To determine the natural frequency of vibrations experimentally & compare with the theoretical
value.
Apparatus: Spring, Weight, Scale, Graph sheet.
Observations:
1. Free length of spring lo= -------meters
2. Mass attached with hanger = ------kg
3. final length after attaching the mass l = -------- meters
4. Diameter of recorder drum dr= ------mm
5. Rotational speed of the drum n= 6 rpm.
Tabular column (Undamped Vibration)
Sl.No
Theoretical frequency
fn (the) Hz
Distance ‘H’ per
cycle from graph,
mm
Experimental frequency
fn (exp) Hz
% Error
Specimen calculations:
1. Static deflection of spring = (length of spring after attaching mass – free length of spring)
= (l – l0) = ------- meters.
2. Stiffness of the spring k 
mg

=-----------N/m where g =9.81m/sec2
k
g
=---------- radians/sec
(OR w n 
rad/sec)

m
w
1 k
4. Frequency in Hertz fn=
or n =-----------cycles per sec (CPS) or Hz
2
2 m
1
5. Experimental natural frequency f n (exp)  = where
T
H
T    where H=distance measured on the graph per cycle, mm
V 
  dr  n
and V=linear velocity of the drum=
, mm/sec
60
and n=rpm of the recorder drum =6 rpm.
3. Circular frequency wn=
V 
Hence f n (exp)   
H
 f n (the ) f n (t exp) 
 100

f
n
(
the
)


6. % error = 
H
Undamped Vibrations
k

k
m
k
m
x
Undamped Free vibrations
Spring Mass System (Damped vibrations)
Aim: To determine the logarithmic decrement, damping ratio & damping coefficient for the spring mass
system with viscous damping.
Observations:
1. Free length of spring lo= -------meters
2. Mass attached with hanger = ------kg
3. final length after attaching the mass l = -------- meters
x1
x2
x
3
Under damped vibrations
(Viscous Damping)
Tabular column (Damped Vibration)
Sl.No
Static
deflection 
meters
Ratio of
Successive
amplitudes
Logarithmic
decrement

Damping ratio

Damping coefficient
C
Frequency of Damped
vibrations fd
Specimen calculations:
(1) Static deflection of spring = (length of spring after attaching mass – free length of spring)
= (l – l0) = ------- meters.
(2) Circular frequency w n 
(3) Frequency in Hertz
fn=
1
2
g

rad/sec
w
g
or n =-----------cycles per sec (CPS) or Hz
2

(4) Ratio of successive amplitudes 
xn
=
xn 1
x 
x 
i.e  1  or  2  and so on from graph
 x2 
 x3 
 x 
(5) Logarthmic decrement   ln  n  
 xn 1 
 2 

  Damping ratio  =
(6) But   

2
2
 1  2 
4  


(7)Frequency of damped vibration f d  f n 1   2 =
(8) Damping coefficent C    2mwn 
N / m / sec
Hz
k
C
m
x
Damped Free vibrations
Torsional Pendulum (Torsional vibrations-Undamped)
Aim: To determine the natural frequency of Torsional vibrations experimentally & compare with the
theoretical value.
Apparatus: Torsional pendulum, Scale, Graph sheet & marker (sketch) pen.
Observations:
6. Diameter of the wire or shaft d =--------meters
7. Length of wire L =---------meters
8. Rigidity modulus of wire material (Steel) G = 84 GPa= 84x109 N/m2
9. Mass of the disc M= -------- kg
10. Diameter of the disc D= ------ meters
11. Diameter of the recorder drum dr = --------- mm
12. Rotational Speed of the drum n=6 rpm.
Tabular column (Undamped Vibration)
Sl.No
Theoretical frequency
fn (the) Hz
Distance ‘H’ per
cycle from graph,
mm
Experimental frequency
fn (exp) Hz
% Error
Specimen calculations:
MD 2
kg-m2 = ----8
 d4
7. Polar moment of inertia of the shaft J 
=-----------m4
32
6. Mass moment of inertia of the disc I 
8. Natural frequency of Torsional vibration (Theoretical ) f n (the ) 
9. Circular frequency w n  2 f n =---------- radians/sec
7. Experimental natural frequency f n (exp) 
1
2
1
= where
T
H
T    where H=distance measured on the graph per cycle, mm
V 
  dr  n
and V=linear velocity of the drum=
, mm/sec
60
and n=rpm of the recorder drum =6 rpm.
V 
Hence f n (exp)   
H
GJ
= -------- Hz
IL
 f n (the ) f n (t exp) 
 100

f
n
(
the
)


8. % error = 
Experimental Graph obtained for Undamped Torsional vibration
H
Undamped Vibrations
d
L
I
Undamped Torsional vibrations
Torsional Pendulum (Damped vibrations)
Aim: To determine the logarithmic decrement, damping ratio & damping coefficient for the Torsional
pendulum with viscous damping.
Observations: Diameter of the wire or shaft: d =--------meters
1. Length of wire L =---------meters
2. Rigidity modulus of wire material (Steel) G = 84 GPa= 84x109 N/m2
3. Mass of the disc M= -------- kg
4. Diameter of the disc D= ------ meters
5. Diameter of the recorder drum dr = --------- mm
6. Rotational Speed of the drum n=6 rpm.
x1
x2
x
3
Under damped vibrations
(Viscous Damping)
Tabular column (Undamped Vibration)
Sl.No
Theoretical
frequency fn
Hz
Ratio of
Successive
amplitudes
Logarithmic
decrement

Damping ratio

Damping coefficient
C
Frequency of Damped
vibrations fd
Specimen calculations:
(4) Frequency in Hertz
fn=
1
2
GJ
IL
= ----------cycles per sec (CPS) or Hz
(4) Ratio of successive amplitudes 
xn
=
xn 1
x 
x 
i.e  1  or  2  and so on from graph
 x2 
 x3 
 x 
(5) Logarthmic decrement   ln  n  
 xn 1 
 2 

  Damping ratio  =
(6) But   

2
2
 1  2 
4





(7)Frequency of damped vibration f d  f n 1   2 =
(8) Damping coefficent C    2 I wn 
N / m / sec
Hz
d
L
Oil
I
Damped Torsional vibrations
STRAIN ROSETTES
Aim: To conduct experiment on a member subjected to combined bending and
torsion using Rectangular & Delta Rosettes and to determine;
(vii) Principal Strains & their orientations
(viii) Principal stresses
Apparatus required: Strain rosettes, Cantilever shaft with torsion arm, Weights,
Wheat stone bridge circuit, and Strain indicator.
Observations:
8. Elastic modulus of the material of the beam E=207 GPa
9. Poisson's ratio 
10.Type of strain gauges: Electrical resistance type : Rectangular & Delta
arrangement
Tabular column: (Rectangular Rosette)
Tria
l No
1
2
3
Load,
Newton
Micro
Strain
''
x(10-6)
Micro
Strain
''
x(10-6)
Micro
Strain
'C'
x(10-6)
Major
Principal
strain
1
Minor
Principal
strain
2
Major
Principal
stress
1
Minor
Principa
l stress
2
1
2
Specimen calculations:
6. Principal Strains;
  A  C
 2
1,2  
 1

 2
 A   C    2 B   A   C 
2
2

+ve sign for 1 (major principal strain) and – ve sign for 2 (Minor principal
strain)
(Note: if any of the individual strain values are negative, substitute them
as it is)
2. Principal Stresses;
1 
E  1  2 

2 
1   2 
Mpa
2 
E   2 1 


1  2  2 
Mpa
Where, 1 and 2 are the major and minor principal stresses.
1
1  2 B   A   C 
tan

=
3. Principal angle  2
  A  C

deg
and
 1   2 
Maximum
shear
stress


max

 = --------------------Mpa
4.
 2 
Tabular column: (Delta Rosette)
Tria
l No
Load,
Newton
Micro
Strain
''
x(10-6)
Micro
Strain
''
x(10-6)
Micro
Strain
'C'
x(10-6)
Major
Principal
strain
1
Minor
Principal
strain
2
Major
Principal
stress
1
Minor
Principa
l stress
2
1
2
1
2
3
Specimen calculations:
1. Principal Strains;
1,2 
1
2
 A   B   C  
3
3
  A   B     B  C    C   A 
2
2
2
+ve sign for 1 (major principal strain) and – ve sign for 2 (Minor principal
strain)
(Note: if any of the individual strain values are negative, substitute them
as it is)
2. Principal Stresses;
1 
E  1  2 


1  2  2 
2 
E   2 1 


1  2  2 
Where, 1 and 2 are the major and minor principal stresses.
1 1  3( C   B )
3. Principal angle  2 tan  2    
 A B C
and

 =

deg
 1   2 
Maximum
shear
stress


max

 = --------------------MPa
4.
 2 