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International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072
© 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1977
Design And Analysis Of Vibration Exciter
JAWAZ ALAM1, Mr. V. SUNIL2, Mr. S.UDAYA BHASKAR3
1p.G. Scholar,Machine Design, Al-Habeeb Colleg Of Engineering Technology (J,N.T.U.H)
2asst.Prof.M.Tech. Mech. Dept., Al-Habeeb Colleg Of Engineering Technology (J.N.T.U.H)
3Asso. prof, M.Tech (Ph.D), Mech. Dept., Al-Habeeb Colleg Of Engineering Technology (J.N.T.U.H)
---------------------------------------------------------------------------***----------------------------------------------------------------
Abstract :- A body is said to be in vibration whenithasto &
fro motion. Vibration is of mainly three type which are Free
vibration, Damped vibration and Forced vibration. The
Mechanical Vibration Exciter worked on forced vibratory
motion. Design and Analysis of Mechanical VibrationExciter
which produce mechanical vibratory motion to provide
forced vibration to a specimen on which modal analysis and
testing is to be performed. In the design & analysis of a
mechanical vibration exciter, it contains cam and follower
mechanism used to generate uniaxial vibrations. Theexciter
is designed to produce displacement by the given range of
frequency which is to be provided. The analysis and
construction of working device, its important parts and the
results obtainedfromFast FourierTransform (FFT)analyser
are described here.
Key Words: Design ,Analysis, S.H.M Cam, Snail Cam,
Follower, Shaft, Spring.
1. INTRODUCTION
A body is said to be in vibration when it has to & fro motion.
Vibration is of mainly three type which are Free vibration,
Damped vibration and Forced vibration.
The Mechanical Vibration Exciter worked on forced
vibratory motion. Design and Analysis of Mechanical
Vibration Exciter which produce mechanical vibratory
motion to provide forced vibration to a specimen on which
modal analysis and testing is to be performed. In the design
& analysis of a mechanical vibration exciter, it contains cam
and follower mechanism used to generate uniaxial
vibrations. The exciter is designed to produce displacement
by the given range of frequency which is to be provided. The
analysis and construction of working device, its important
parts and the results obtained from Fast Fourier
Transform (FFT) analyser are described here.
Fig 1:- Vibration Exciter With F.F.T
2. PROBLEM DEFINITION
The Mechanical vibration exciters can prosecute only a
single type of output like Simple Harmonic Motion ie S.H.M.
These usual cam profiles cannot provide shocks and sudden
jerks. V.Ryan has mentioned a model of snail or drop cam
which can be used forproviding a sudden jerk motion.Soitis
intended to design a vibration exciter with various output by
using cams of different profiles i.e S.H. M and SNAIL cam
profile. Hence the problem definition is
“Designanddevelopmentofmechanicalvibrationexciter
with multiple outputs”
3. SCOPE OF THE PROJECT
The scope of our project work includes design ofmechanical
vibration exciter by means of cam and follower
mechanism.There are two types of motions will be provided
using S.H.M. profile and drop profile of cam. The scope also
includes selection of suitable materials, fabrication and
testing of the exciter using FFT analyzer. In order to design
vibration exciter, its components like cam, follower, shaft
and springs have to be designed, which are discussed.
4. DESIGN:-
4.1 DESIGN CALCULATIONS:-The design calculations for
shaft, cams and follower, springs and key are given below.
These calculations were done with the help of design criteria
mentioned in the reference book and the design data book.
4.1.1. SHAFT :-
The design power for shaft (P) is taken as 2kW considering
factor of safety.
The motor speed (N) is taken to be 1000rpm.
Data:-
Let,
Power of shaft [P] = 2 KW
Motor speed [N] =1000 r.p.m
So,
Torque [T] = P / (2π N /60 )
T = 2000 / ( 2 × 22 × 100/ 60 )
T = 19.1 Nm .… (1)
The Eq. (1) will give the torque acting on the shaft.
1. V.F.D
International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072
© 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1978
∑ Fy = 0
Ra+ Rf=1000 N
∑Ma =0
500 × 160 + 500×240 = Rf ×400
Rf =500 N
Ra= 500N
2. V.B.M.D
Mc=Md=80000 N-mm
Following fig give shear force and bending moment diagram
of the shaft,
Fig 2:- S.F.D & B.M.D Of Force Acting On Shaft
Let,
kt = kb =2
then,
Mv =
Mt = 164.5 * 103 Nmm ……………………………………(2)
The Eq. (2) will give the equivalent moment acting on the
shaft .
Now, d3
[τ] = = 81 N/mm2
∴ d = 22 mm
Let
d = 30 mm
Thus, the diameter of shaft is 30 mm.
4.1.2. CAM AND FOLLOWER (SHM MOTION):
Generally,
Mechanical Exciters provide displacement upto 30mm.
For our project ,
we have taken 10 mm as maximum rise of the cam.
Base circle radius is taken as 40mm and the width as 20mm.
Following figure shows the forces acting on cam and the
symbols used for various dimensions of the cam and the
follower.
Fig 3 :- Force Acting On Cam And Follower
Given Data:-
Maximum lift,hm= 10mm
Or, rb = 40mm
Or, b = 20mm
Or, N =1000 rpm
Or, F = 4000 N
Or, σyt = 400 N/mm2
Or, Ѳb = Ѳr =150o
Or, Ѳd = 600
Design of cam: -
rp = 40 + 10 = 50 mm
International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072
© 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1979
ω = = 105 rad/s ...(4)
The Eq. (4) will gives the value of angular velocity of cam.
v = (sin( ) ) ….(5)
The Eq. (5) will give the value of linear velocity of cam.
Now,
At = 00 & = we will get v = 0
and at = , v = vmax
so, v = = 0.63 m/s
ar =( )2 × ×ω2 = 79.38 m / sec2
y = hm = 10 mm
Fm = Fmax = 400 N
Now,
rc =
= 53.7 mm …(6)
The Eq. (6) will give radius of curvature of cam surface.
Now,
For contact stress between cam and follower,
σy= 189.8
=222.75 N/mm2 which is less than 1130 …(7)
The Eq. (7) will give contact stresses between cam and
follower.
Thus, the induced stress value is less thanthedesignvalue of
contact stress.
Follower body:-
The dimensions of follower body are based on the
dimensions of the roller, which has widthanddiameterof 20
mm.
D = 46 mm
Thread: - M 46 ×2
Design of Roller pin:-
The Design of Roller pin is to withstand the double shear
failure
[τ] = 45 N/mm2
Following equation will give the value for diameter of roller
pin
Fmax= 2π ( [τ] =400N (8)
dp = 3mm
Let, dp = 8mm
Thus, the value of roller pin is 8 mm by considering our
factor of safety.
Figures 4& 5 shows the cam and follower models prepared
on Autodesk Inventor as per the design calculations & its
behaviour shown in graph 1.
Fig 4 :- Inventor Model of SHM Cam
Fig 5 :- Inventor Model of Follower
International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072
© 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1980
Blue – displacement , Yellow – velocity , Red- acceleration
Graph 1:- Graphs for SHM cam
4.1.3 CAM and Follower (SNAIL CAM) :
After designing the SHM profile cam,
Now drop cam is designed, for which base circle radius is
taken as 40mm and the width as 20mm of cam and Its
maximum rise is 10mm.
Given Data:-
Maximum lift,hm= 10mm
Or, rb = 40mm
Or, b = 20mm
Or, N =1000 rpm
Or, F = 400 N
Or, σyt = 400 N/mm2
Or, Ѳb = 2700
Or , Ѳr = 0o
Or, Ѳd = 900
Design of cam: -
rp = 40 + 10 = 50 mm
ω = = 105 rad/s ..(9)
The Eq. (9) will gives the value of angular velocity of cam.
v = (sin( ) ) …….(10)
The Eq. (10) will give the value of linear velocity of cam
Now,
At = 00 & = we will get v = 0
so, v = = 0.35 m/s
ar =( )2 × × 2 = 24.5 m / sec2
y = hm = 10 mm
Fm = Fmax = 400 N
Now,
rc =
= 62.31 mm …(11)
The Eq. (11) will give radius of curvature of cam surface.
Now,
For contact stress between cam and follower,
σy= 189.8
= 218.14 N/mm2 which is less than 1130 N/mm2.(12)
The Eq. (12) will give contact stresses between cam and
follower.
Thus, the induced stress value is less thanthedesignvalue of
contact stress.
Follower body:-
The dimensions of follower body are based on the
dimensions of the roller, which has widthanddiameterof 20
mm.
d = 46 mm
Thread: -
M 46 ×2
Design of Roller pin:-
The Design of Roller pin is to withstand the double shear
failure
[τ] = 45 N/mm2
Following equation will give the value for diameter of roller
pin
Fmax= 2π ( [τ] =400N
dp = 3mm
Let, dp = 8mm
Thus, the value of roller pin is 8 mm by considering our
factor of safety..
Figures 6 & 7 shows the cam and follower models prepared
on Autodesk Inventor as per the design calculations & its
behaviour shown in graph 2.
International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072
© 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1981
Fig 6 :- Inventor Model of Snail Drop Cam
Fig 7 :- Inventor Model of Follower
Blue – displacement , Yellow – velocity , Red- acceleration
Graph 2 :- Graphs For Snail Cam
4.1.4 Spring :
The spring design is based on fatigue loading
Fmax = 400 N
δmax = 10 mm
c = 6
Material: ASTM A228
G = 79.3 GPa
Sut = 1800N/mm2
[ τ ] = 504 N/mm2
Now,
WAHL STRESS FACTOR Kw = ( =1.2525 …….(14)
Above equation will give the value of Wahl stress factor
1.2525.
For the value of wire diameter of the spring.
τ = Kw (
d =4 mm
D = 24 mm
Hence, the coil diameter is 24 mm and wire diameter is 4
mm.
Number of coil k = = 40 N/mm = ……(16)
so, n (active) = 7
and , n, (total ) = n +2 = 9
Thus, the total number of turns is 9.
Eq. (16) will give the value of solid length of spring.
LS = ( n+2 ) d = 36 m …(17 )
Below Eq. will give the value of free length of spring.
Lf = Ls+ max + (n, -1 ) ×1 = 54 mm …(18 )
Below Eq. will give the value of pitch of spring.
Lf = pn + 2d ….( 19 )
p = 6.6 mm
Thus the value of pitch is 6.6mm.
4.1.5 Key :
Key will be subjected to shear stresses and crushing
stresses, So it is designed in such a way to withstand these
stresses .
ds = 30 mm
width of key w = … (20)
w =
so, w = 7.5 mm
The width of key is 7.5 mm.
Now, Height of key
h = = (21)
h = 5 mm
So, The height of key is 5 mm.
L = 30 mm
International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072
© 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1982
Below Eq.will give the value of force acting on key .
F = ... (22)
F = = 3184 N
[τ] = 45 N/mm2
σcr = 135 N/mm2
Now, the actual value of shear stress,
[τ] = ...(23)
[τ] = 14.15 N/mm2
And, the actual value of crushing stress,
σcr = …(24)
σcr = 21.23 N/mm2
Thus, the actual crushing stress is less than the permissible
value of crushing stress.
Thus, the dimensions of the keys are in safe limit.
So design shaft, cams and follower, key and springs isdone
and their calculated stress values are within the specified
limit of stress.
5. Results:-
The mechanism used to operate one type of input at one
time in vibration exciter having two different type of cam
profile on same shaft is represented in fig.6.1 and 6.2 as
SHM input and jerk input. It’s been shown in both the
figures, two types of cams consider type A and type B.
Where type A has jerk type motion and those cams are
mounted on the shaft at the outer ends, where as type B has
SHM type motion and those cams are mounted on the shaft
at the central portion. The output is measured by using FFT
analyzer i.e. output/amplitudeinfrequencydomain.ForJerk
motion and SHM motion, output is shown in Figure 6.3 and
6.4 respectively.
Fig 6.1 :- Schematic Diagram For Jerk Motion
Fig 6.2 :- Schematic Diagram For S.H.M. Motion
Fig 6.3 :- Output For Jerk Motion
For jerk type, the input is given as 1 cm jerk withperiodic up
and down movement ,but outputrecordedinfigure6.3actas
2 cm maximum in upward direction and -1.6 cm in
downward direction.
Fig 6.4 :- Output For S.H.M Motion
International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072
© 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1983
6. Analysis :
The graphs obtained by FFT analyzer with the help of
accelerometer are slightly out of range due to following
reasons-
• Insufficient damping as the setup is not fixedtotheground.
• Required contact stresses between follower roller and
cams cannot be measured.
• Friction between roller and roller pin is causing slight
slipping of the roller initially.
Based on results obtained, the conclusion of the project is
discussed in next chapter.
7. CONCLUSION:
In previous topic, the results obtained fromtheFFTanalyzer
are discussed. The time vs amplitude graph shows that both
the types of output waveforms—SHM and sudden jerk, are
achieved with the cam and follower mechanism. The
vibration exciter can be used for rpm value of the motor up
to 1000 rpm. The variation in the amplitude indicates that
the damping provided by external weights are insufficient.
By fixing the frame to the ground or by using vibration
dampening pads, sufficient damping can be provided, which
will give better accuracy. Use of ball bearing as roller in the
roller follower will reducethefrictionbetweenthe roller and
cams and also reduce the stress on the roller pin. This will
increase the life of both the roller and the cams.
References
1. Refrence Books And Journal
 S.S. RATTAN “THEORY OF MACHINE” Third Edition
Pg 209-247 & Pg 615-674
 Robert L. Norton “Machine Design An Integrated
Approach” Second Edition Pg 546-566 & Pg -826.
 R.S. Khurmi “Theory Of Machine” Third Edition.
 R.S. Khurmi& J.K. Gupta, “A Text Book of Machine
Design”. Pg 509-557,820-884.[9]
 PSG, “Design Data Book”, p4.13, p.7.21, p7.22,
p7.100, p7.110, p7.113-p7.116.
 V. Ryan, “Snail/Drop Cams”, World Association of
Technology Teachers, 2009.[8]
2. REFRENCE URL
 http://www.globalspec.com/learnmore/sensors_tr
ansducers_detectors/acceleration_vibration_sensin
g/shakers_vibration_shock_testing[1]
 http://engineering.myindialist.com/2015/long-
answer-questions-on-vibrations/
 https://www.mechanical.in/types-vibration/

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Design and Analysis of Vibration Exciter

  • 1. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1977 Design And Analysis Of Vibration Exciter JAWAZ ALAM1, Mr. V. SUNIL2, Mr. S.UDAYA BHASKAR3 1p.G. Scholar,Machine Design, Al-Habeeb Colleg Of Engineering Technology (J,N.T.U.H) 2asst.Prof.M.Tech. Mech. Dept., Al-Habeeb Colleg Of Engineering Technology (J.N.T.U.H) 3Asso. prof, M.Tech (Ph.D), Mech. Dept., Al-Habeeb Colleg Of Engineering Technology (J.N.T.U.H) ---------------------------------------------------------------------------***---------------------------------------------------------------- Abstract :- A body is said to be in vibration whenithasto & fro motion. Vibration is of mainly three type which are Free vibration, Damped vibration and Forced vibration. The Mechanical Vibration Exciter worked on forced vibratory motion. Design and Analysis of Mechanical VibrationExciter which produce mechanical vibratory motion to provide forced vibration to a specimen on which modal analysis and testing is to be performed. In the design & analysis of a mechanical vibration exciter, it contains cam and follower mechanism used to generate uniaxial vibrations. Theexciter is designed to produce displacement by the given range of frequency which is to be provided. The analysis and construction of working device, its important parts and the results obtainedfromFast FourierTransform (FFT)analyser are described here. Key Words: Design ,Analysis, S.H.M Cam, Snail Cam, Follower, Shaft, Spring. 1. INTRODUCTION A body is said to be in vibration when it has to & fro motion. Vibration is of mainly three type which are Free vibration, Damped vibration and Forced vibration. The Mechanical Vibration Exciter worked on forced vibratory motion. Design and Analysis of Mechanical Vibration Exciter which produce mechanical vibratory motion to provide forced vibration to a specimen on which modal analysis and testing is to be performed. In the design & analysis of a mechanical vibration exciter, it contains cam and follower mechanism used to generate uniaxial vibrations. The exciter is designed to produce displacement by the given range of frequency which is to be provided. The analysis and construction of working device, its important parts and the results obtained from Fast Fourier Transform (FFT) analyser are described here. Fig 1:- Vibration Exciter With F.F.T 2. PROBLEM DEFINITION The Mechanical vibration exciters can prosecute only a single type of output like Simple Harmonic Motion ie S.H.M. These usual cam profiles cannot provide shocks and sudden jerks. V.Ryan has mentioned a model of snail or drop cam which can be used forproviding a sudden jerk motion.Soitis intended to design a vibration exciter with various output by using cams of different profiles i.e S.H. M and SNAIL cam profile. Hence the problem definition is “Designanddevelopmentofmechanicalvibrationexciter with multiple outputs” 3. SCOPE OF THE PROJECT The scope of our project work includes design ofmechanical vibration exciter by means of cam and follower mechanism.There are two types of motions will be provided using S.H.M. profile and drop profile of cam. The scope also includes selection of suitable materials, fabrication and testing of the exciter using FFT analyzer. In order to design vibration exciter, its components like cam, follower, shaft and springs have to be designed, which are discussed. 4. DESIGN:- 4.1 DESIGN CALCULATIONS:-The design calculations for shaft, cams and follower, springs and key are given below. These calculations were done with the help of design criteria mentioned in the reference book and the design data book. 4.1.1. SHAFT :- The design power for shaft (P) is taken as 2kW considering factor of safety. The motor speed (N) is taken to be 1000rpm. Data:- Let, Power of shaft [P] = 2 KW Motor speed [N] =1000 r.p.m So, Torque [T] = P / (2π N /60 ) T = 2000 / ( 2 × 22 × 100/ 60 ) T = 19.1 Nm .… (1) The Eq. (1) will give the torque acting on the shaft. 1. V.F.D
  • 2. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1978 ∑ Fy = 0 Ra+ Rf=1000 N ∑Ma =0 500 × 160 + 500×240 = Rf ×400 Rf =500 N Ra= 500N 2. V.B.M.D Mc=Md=80000 N-mm Following fig give shear force and bending moment diagram of the shaft, Fig 2:- S.F.D & B.M.D Of Force Acting On Shaft Let, kt = kb =2 then, Mv = Mt = 164.5 * 103 Nmm ……………………………………(2) The Eq. (2) will give the equivalent moment acting on the shaft . Now, d3 [τ] = = 81 N/mm2 ∴ d = 22 mm Let d = 30 mm Thus, the diameter of shaft is 30 mm. 4.1.2. CAM AND FOLLOWER (SHM MOTION): Generally, Mechanical Exciters provide displacement upto 30mm. For our project , we have taken 10 mm as maximum rise of the cam. Base circle radius is taken as 40mm and the width as 20mm. Following figure shows the forces acting on cam and the symbols used for various dimensions of the cam and the follower. Fig 3 :- Force Acting On Cam And Follower Given Data:- Maximum lift,hm= 10mm Or, rb = 40mm Or, b = 20mm Or, N =1000 rpm Or, F = 4000 N Or, σyt = 400 N/mm2 Or, Ѳb = Ѳr =150o Or, Ѳd = 600 Design of cam: - rp = 40 + 10 = 50 mm
  • 3. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1979 ω = = 105 rad/s ...(4) The Eq. (4) will gives the value of angular velocity of cam. v = (sin( ) ) ….(5) The Eq. (5) will give the value of linear velocity of cam. Now, At = 00 & = we will get v = 0 and at = , v = vmax so, v = = 0.63 m/s ar =( )2 × ×ω2 = 79.38 m / sec2 y = hm = 10 mm Fm = Fmax = 400 N Now, rc = = 53.7 mm …(6) The Eq. (6) will give radius of curvature of cam surface. Now, For contact stress between cam and follower, σy= 189.8 =222.75 N/mm2 which is less than 1130 …(7) The Eq. (7) will give contact stresses between cam and follower. Thus, the induced stress value is less thanthedesignvalue of contact stress. Follower body:- The dimensions of follower body are based on the dimensions of the roller, which has widthanddiameterof 20 mm. D = 46 mm Thread: - M 46 ×2 Design of Roller pin:- The Design of Roller pin is to withstand the double shear failure [τ] = 45 N/mm2 Following equation will give the value for diameter of roller pin Fmax= 2π ( [τ] =400N (8) dp = 3mm Let, dp = 8mm Thus, the value of roller pin is 8 mm by considering our factor of safety. Figures 4& 5 shows the cam and follower models prepared on Autodesk Inventor as per the design calculations & its behaviour shown in graph 1. Fig 4 :- Inventor Model of SHM Cam Fig 5 :- Inventor Model of Follower
  • 4. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1980 Blue – displacement , Yellow – velocity , Red- acceleration Graph 1:- Graphs for SHM cam 4.1.3 CAM and Follower (SNAIL CAM) : After designing the SHM profile cam, Now drop cam is designed, for which base circle radius is taken as 40mm and the width as 20mm of cam and Its maximum rise is 10mm. Given Data:- Maximum lift,hm= 10mm Or, rb = 40mm Or, b = 20mm Or, N =1000 rpm Or, F = 400 N Or, σyt = 400 N/mm2 Or, Ѳb = 2700 Or , Ѳr = 0o Or, Ѳd = 900 Design of cam: - rp = 40 + 10 = 50 mm ω = = 105 rad/s ..(9) The Eq. (9) will gives the value of angular velocity of cam. v = (sin( ) ) …….(10) The Eq. (10) will give the value of linear velocity of cam Now, At = 00 & = we will get v = 0 so, v = = 0.35 m/s ar =( )2 × × 2 = 24.5 m / sec2 y = hm = 10 mm Fm = Fmax = 400 N Now, rc = = 62.31 mm …(11) The Eq. (11) will give radius of curvature of cam surface. Now, For contact stress between cam and follower, σy= 189.8 = 218.14 N/mm2 which is less than 1130 N/mm2.(12) The Eq. (12) will give contact stresses between cam and follower. Thus, the induced stress value is less thanthedesignvalue of contact stress. Follower body:- The dimensions of follower body are based on the dimensions of the roller, which has widthanddiameterof 20 mm. d = 46 mm Thread: - M 46 ×2 Design of Roller pin:- The Design of Roller pin is to withstand the double shear failure [τ] = 45 N/mm2 Following equation will give the value for diameter of roller pin Fmax= 2π ( [τ] =400N dp = 3mm Let, dp = 8mm Thus, the value of roller pin is 8 mm by considering our factor of safety.. Figures 6 & 7 shows the cam and follower models prepared on Autodesk Inventor as per the design calculations & its behaviour shown in graph 2.
  • 5. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1981 Fig 6 :- Inventor Model of Snail Drop Cam Fig 7 :- Inventor Model of Follower Blue – displacement , Yellow – velocity , Red- acceleration Graph 2 :- Graphs For Snail Cam 4.1.4 Spring : The spring design is based on fatigue loading Fmax = 400 N δmax = 10 mm c = 6 Material: ASTM A228 G = 79.3 GPa Sut = 1800N/mm2 [ τ ] = 504 N/mm2 Now, WAHL STRESS FACTOR Kw = ( =1.2525 …….(14) Above equation will give the value of Wahl stress factor 1.2525. For the value of wire diameter of the spring. τ = Kw ( d =4 mm D = 24 mm Hence, the coil diameter is 24 mm and wire diameter is 4 mm. Number of coil k = = 40 N/mm = ……(16) so, n (active) = 7 and , n, (total ) = n +2 = 9 Thus, the total number of turns is 9. Eq. (16) will give the value of solid length of spring. LS = ( n+2 ) d = 36 m …(17 ) Below Eq. will give the value of free length of spring. Lf = Ls+ max + (n, -1 ) ×1 = 54 mm …(18 ) Below Eq. will give the value of pitch of spring. Lf = pn + 2d ….( 19 ) p = 6.6 mm Thus the value of pitch is 6.6mm. 4.1.5 Key : Key will be subjected to shear stresses and crushing stresses, So it is designed in such a way to withstand these stresses . ds = 30 mm width of key w = … (20) w = so, w = 7.5 mm The width of key is 7.5 mm. Now, Height of key h = = (21) h = 5 mm So, The height of key is 5 mm. L = 30 mm
  • 6. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1982 Below Eq.will give the value of force acting on key . F = ... (22) F = = 3184 N [τ] = 45 N/mm2 σcr = 135 N/mm2 Now, the actual value of shear stress, [τ] = ...(23) [τ] = 14.15 N/mm2 And, the actual value of crushing stress, σcr = …(24) σcr = 21.23 N/mm2 Thus, the actual crushing stress is less than the permissible value of crushing stress. Thus, the dimensions of the keys are in safe limit. So design shaft, cams and follower, key and springs isdone and their calculated stress values are within the specified limit of stress. 5. Results:- The mechanism used to operate one type of input at one time in vibration exciter having two different type of cam profile on same shaft is represented in fig.6.1 and 6.2 as SHM input and jerk input. It’s been shown in both the figures, two types of cams consider type A and type B. Where type A has jerk type motion and those cams are mounted on the shaft at the outer ends, where as type B has SHM type motion and those cams are mounted on the shaft at the central portion. The output is measured by using FFT analyzer i.e. output/amplitudeinfrequencydomain.ForJerk motion and SHM motion, output is shown in Figure 6.3 and 6.4 respectively. Fig 6.1 :- Schematic Diagram For Jerk Motion Fig 6.2 :- Schematic Diagram For S.H.M. Motion Fig 6.3 :- Output For Jerk Motion For jerk type, the input is given as 1 cm jerk withperiodic up and down movement ,but outputrecordedinfigure6.3actas 2 cm maximum in upward direction and -1.6 cm in downward direction. Fig 6.4 :- Output For S.H.M Motion
  • 7. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 08 | Aug -2017 www.irjet.net p-ISSN: 2395-0072 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 1983 6. Analysis : The graphs obtained by FFT analyzer with the help of accelerometer are slightly out of range due to following reasons- • Insufficient damping as the setup is not fixedtotheground. • Required contact stresses between follower roller and cams cannot be measured. • Friction between roller and roller pin is causing slight slipping of the roller initially. Based on results obtained, the conclusion of the project is discussed in next chapter. 7. CONCLUSION: In previous topic, the results obtained fromtheFFTanalyzer are discussed. The time vs amplitude graph shows that both the types of output waveforms—SHM and sudden jerk, are achieved with the cam and follower mechanism. The vibration exciter can be used for rpm value of the motor up to 1000 rpm. The variation in the amplitude indicates that the damping provided by external weights are insufficient. By fixing the frame to the ground or by using vibration dampening pads, sufficient damping can be provided, which will give better accuracy. Use of ball bearing as roller in the roller follower will reducethefrictionbetweenthe roller and cams and also reduce the stress on the roller pin. This will increase the life of both the roller and the cams. References 1. Refrence Books And Journal  S.S. RATTAN “THEORY OF MACHINE” Third Edition Pg 209-247 & Pg 615-674  Robert L. Norton “Machine Design An Integrated Approach” Second Edition Pg 546-566 & Pg -826.  R.S. Khurmi “Theory Of Machine” Third Edition.  R.S. Khurmi& J.K. Gupta, “A Text Book of Machine Design”. Pg 509-557,820-884.[9]  PSG, “Design Data Book”, p4.13, p.7.21, p7.22, p7.100, p7.110, p7.113-p7.116.  V. Ryan, “Snail/Drop Cams”, World Association of Technology Teachers, 2009.[8] 2. REFRENCE URL  http://www.globalspec.com/learnmore/sensors_tr ansducers_detectors/acceleration_vibration_sensin g/shakers_vibration_shock_testing[1]  http://engineering.myindialist.com/2015/long- answer-questions-on-vibrations/  https://www.mechanical.in/types-vibration/