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PART - B
Unit I Force Analysis
1. A single cylinder, single acting, four stroke gas engine develops 20 kW at 300 r.p.m. The
work done by the gases during the expansion stroke is three times the work done on the
gases during the compression stroke, the work done during the suction and exhaust strokes
being negligible. If the total fluctuation of speed is not to exceed 2 per cent of the mean
speed and the turning moment diagram during
compression and expansion is assumed to be triangular in
shape. Find the moment of inertia of the flywheel.
Given: P=20kW;N=300rpm; =2N/60=31.442rad/s
since the total fluctuation of speed (1-2) is not to
exceed 2 percent of the mean speed (),
Therefore, 1-2=4%; CS=1-2/=0.04
n=N/2=300/2=150; Workdone/cycle=P*60/n=8000N-m
Max. Fluctuation Energy E=I2CS=255.2Kg-m2
2. (a) Derive the equation of forces on the reciprocating parts of an engine, neglecting weight
of the connecting rod.

Velocity of the piston:

Acceleration of the piston:
Accelerating force or inertia force of the reciprocating parts:
(b) What is turning moment diagram and draw its for four stroke IC engine?
A turning moment diagram for a 4 stroke cycle IC engine is shown in figure. We know that
in a 4 stroke cycle IC engine, there is one working stroke after the crank has turned
through two revolutions i.e.7200 (or 4 radians)

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3. The torque delivered by a two-stroke engine is represented by T = (1000+300 sin 2 500

cos 2 ) N-m.where is the angle turned by the crank from the inner dead centre. the
engine speed is 250 rpm.the mass of the flywheel is 400 kg and radius of gyration 400 mm.
determine (i) the power developed,(ii) the total percentage fluctuation of speed,(iii) the
angular acceleration of flywheel when the crank has rotated through an angle of 60 from
the inner dead centre. (iv) The maximum angular acceleration and retardation of the
flywheel.
Given: T = (1000+300 sin 2 500 cos 2 ) N-m: N=250rpm; m=400kg; k=0.4m; =600;
=2*250/60=26.18rad/s;
Hint:Tmean=Workdone per cycle/Crank angle per rev=976.13N-m; 1=29.5102=119.50;
Power Developed P= Tmean* =25.56kW; Max. Fluctuation of Energy=E=mk22cs;
cs=1.33%; Angular acceleration when =600 ; =7.965rad/s2 ; when 2=149.040 ; TTmean=583N-m; 2=329.04; T-Tmean=583N-m=-583.1N-m; Max or Min= T-Tmean/T=9.11rad/s2
4. A vertical petrol engine 150 mm diameter and 200 mm stroke has a connecting rod 350 mm
long. The mass of the piston is 1.6 kg and the engine speed is 1800 rpm.on the expansion
stroke with crank angle 30 from the top dead centre, the gas pressure is 750
kN/m2.Determine the net thrust on the engine.
Given: D=150=0.15m; L=200mm=0.2m; Radius of the crank; r=L/2=0.1m; Connecting rod
length l=0.35m
m=1.6kg; p=750kN/m2
N=1800rpm; Angular velocity =188.49rad/s, Crank angle =300; Gas pressure p=750kN/m2
Piston Force Fp=p*area of the piston= 13253.59N; Fi=Inertia force= -(mass of
piston)*Accleration of piston
Net thrust for vertical engine is given by F=FP+FiW= 7534.396N
5. In a slider crank mechanism, the length of the crank and connecting rod are150 mm and
600 mm respectively. The crank position is 60 from inner dead centre. The crank shaft
speed is 450 r.p.m. clockwise. Determine 1. Velocity and acceleration of the slider, 2.
Velocity and acceleration of point D on the connecting rod which is 150 mm from crank pin
C, and 3. angular velocity and angular acceleration of the connecting rod.
Given : OC = 150 mm = 0.15m ; PC = 600 mm = 0.6 m ; CD = 150 mm = 0.15 m ; N = 450
r.p.m. or
= 2 450/60 = 47.13 rad/s
Hint:
1. Velocity & Accceleration of the slider:
Vp=*OM=6.834m/s; ap=2*NO=124.4m/s2
2. Velocity and acceleration of point D on the connecting rod:
VD=*OD1=6.834m/s; aD=2*OD22=266.55m/s2
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3. Angular velocity and angular acceleration of
the connecting rod:
PC= Vpc/PC=6.127rad/s; PC=atPC/PC=481.27rad/s2
6. A vertical double acting steam engine develops 75 kW at 250 rpm.the maximum fluctuation
of energy is 30 percent of the work done per stroke. The maximum and minimum speeds
are not to vary more than 1% on either side of the mean speed. Find the mass of the
flywheel required if the radius of gyration is 0.6 meters.
Given: Power=75kW; N=250rpm; 1-2=1% =0.01; k=0.6
Cs= 1-2/ =0.01; Maximum fluctuation of energy, E = Work done per cycle CE
We know that E =mk2 2 Cs; Mass of the flywheel=547kg
7. The lengths of crank and connecting rod of a horizontal reciprocating engine are 200 mm
and 1 meter respectively. The crank is rotating at 400 rpm.when the crank has turned
through 30 from the inner dead centre. The difference of pressure between cover and
piston rod is 0.4 N/mm2.if the mass of the reciprocating parts is 100 kg and cylinder bore is
0.4 meters, then calculate: (i) inertia force, (ii) force on piston, (iii) piston effort, (iv) thrust
on the sides of the cylinder walls, (v) thrust in the connecting rod, and (vi) crank effort.
Given: r=0.2m; l=1m;N=400rpm or =41.88rad/s: =300; p1-p2=0.4N/mm2; m=100kg;
D=0.4m;n=l/r=5

(i)Inertia Force(Fi): Fi=-m*a [-ve sign is due to the fact that inertia force opposes the accelerating
force]
a=acceleration of the piston whish is given as:
=338.86m/s2 ; Therefore Fi=-m*a=-33886N
(ii) Force of the piston: FP=p*area of the piston; p=p1-p2=0.4 N/mm2; FP=50265N
(iii) Piston effort: F=Fi+FP=16379N
(iv) Thrust on the sides of the cylinder walls:
The thrust on the sides of cylinder walls(or normal reaction), FN is given as; FN=Ftan
=Angle made by connecting rod with line of stroke, the value of in terms of is given as

(v) Thrust in the connecting rod:FC=F/cos=16461.5N

(vi) Crank effort (FT) or Tangential Force: FT=FCsin(+)=9615N
8. The radius of gyration of a fly wheel is 1 meter and the fluctuation of speed is not to exceed
1% of the mean speed of the flywheel. If the mass of the flywheel is 3340 kg and the steam
engine develops 150 kW at 135 rpm, then find (i) maximum fluctuation of energy and (ii)
coefficient of fluctuation of energy.
Given: k=1m; fluctuation of speed=1% of mean speed or 2- 1=1% of or 2- 1/ =0.01 or
coefficient of fluctuation of speed, KS=0.01; m=3340kg; P=150Kw; N=135rpm; =14.137rad/s
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=6675.13Nm; KE=Max.Fluctuation of energy/Workdone per cycle

Workdone per cycle=Tmean*= Tmean*2; Tmean= P/=10610.45Nm
Workdone per cycle=10610.45*2=66667.42Nm/cycle

9. (i) Deduce the expression for the inertia force in the reciprocating force neglecting the
weight of the connecting rod. Solution: Refer 2(a) Problem in Part-B
(ii) A vertical petrol engine with cylinder of 150 mm diameter and 200 mm stroke has a
connecting rod of 350 mm long. The mass of the piston is 1.6 kg and the engine speed is 1800
rpm.on the expansion stroke with crank angle 30 from TDC, the gas pressure is 750
kPa.Determine the net thrust on the piston.
Given: D=0.15m; r=l/2=0.1m; l=0.35m; mk=1.6kg; N=1800rpm; =300; Fgas Pr=750*103N/m2
By Analytical Method: Net Thrust on the piston FP=Fgas pr+WR-FI
=5736N
Fgas pressure force = Fgas pr * Area = 13,254N; Weight of piston WR=15.7N
Net Thrust on the piston FP=7534N
10. A single cylinder vertical engine has a bore of 100 mm and a stroke of 120 mm has a
connecting rod of 250mm long. The mass of the piston is 1.1kg. The speed is 2000rpm. On
the expansion stroke, with a crank at 20 from top dead center, the gas pressure is
700kN/mm2. Determine (i) Net force acting on the piston (ii) Resultant load on the gudgeon
pin (iii) Thrust on the cylinder walls, and (iv) Speed above which, other things remaining
the same, the gudgeon pin load would be reversed in direction.
Given: D=0.1m; L=0.12m; r=L/2=0.06m; l=0.25m; mR=1.1kg;
=209.5rad/s; =200; p=700kN/m2
1. Net force on the piston: FL=5.5kN; 3.Net Force=2256.8N
2. Inertia
Force

N=2000rpm;

3. Resultant load on the gudgeon pin: FQ=FP/cos=2265N

4. Thrust on the cylinder walls: FN= FP tan=185.5N
5. Speed, above which the gudgeon pin load would be reversed in the direction
Corresponding speed in rpm.N1>2606rpm
11. The slider of a slider crank mechanism is subjected to a force of 3000 N.the crank is at 60
position. The length of the crank and connecting rod are 100 mm and 300 mm respectively.
Determine the driving torque on the crank.
Given: F=3000N; r=length of the crank=0.1m; l=length of the connecting rod=0.3m;
n=l/r=0.3/0.1=3; sin=sin/n;
Torque on the crank
12. In a reciprocating engine mechanism, if the crank and the connecting rod are 300 mm and 1

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m long respectively and the crank rotates at a constant speed of 200 rpm.determine
analytically: (i) the crank angle at which the maximum velocity occurs, and (ii) the
maximum velocity of the piston (iii) derive the relevant equations.
Given : r = 300 mm = 0.3 m ; l = 1 m ; N = 200 r.p.m. or = 2 200/60 = 20.95 rad/s
1. Crank angle at which the maximum velocity occurs, n=l/r=3.33

2. Maximum velocity of the piston; vp(max)=6.54m/s

13. A vertical double acting steam engine has a cylinder 300 mm diameter and 450 mm stroke
and runs at 200 rpm.the reciprocating parts has a mass of 225 kg and the piston rod is 50
mm diameter. The connecting rod is 1.2 m long. When the crank has turned through 125
from the top dead centre the steam pressure above the piston is 30 kN/m 2 and below the
piston is 1.45 kN/m2.Calculate: (i) Crank pin effort, and (ii) Effective turning moment on
the crank shaft.
Given: D=0.3m; L=0.45m; r=L/2=0.225m; N=200rpm; =20.95rad/s; mR=225kg; d=0.05m;
l=1.2m; =1250
p1=30kN/m2; p2=1.5kN/m2

Effective turning moment of the crankshaft

14. The turning moment diagram for a petrol engine is drawn to the following scales : Turning
moment, 1 mm = 5 N-m crank angle, 1 mm = 1. The turning moment diagram repeats itself
at every half revolution of the engine and the areas above and below the mean turning
moment line taken in order are 295, 685, 40, 340, 960, 270 mm2. The rotating parts are
equivalent to a mass of 36 kg at a radius of gyration of 150 mm. Determine the coefficient of
fluctuation of speed when the engine runs at 1800 r.p.m.
Given: m=36kg; k=0.15m; N=1800rpm; =188.52rad/s
E = Maximum Energy Minimum Energy=985mm2
E =mk22CS
985=36*0.12*188.522* CS
CS= 0.3%

15. A connecting rod of an I.C. engine has a mass of 2 kg and the distance between the centre of
gudgeon pin and centre of crank pin is 250 mm. The C.G. falls at a point 100 mm from the
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gudgeon pin along the line of centres. The radius of gyration about an axis through the C.G.
perpendicular to the plane of rotation is 110 mm. Find the equivalent dynamical system if
only one of the masses is located at gudgeon pin. If the connecting rod is replaced by two
masses, one at the gudgeon pin and the other at the crank pin and the angular acceleration
of the rod is
23 000 rad/s 2 clockwise. Determine the correction couple applied to
the system to reduce it to a dynamically equivalent system.
Given: m=2kg; l=0.25m; l1=0.1m; kG=0.11m; =23000rad/s2
(a) Equivalent Dynamical System: m1=l2m/l1+l2=1.1kg; m2=l1m/l1+l2=0.9kg
(b) Correction Couple: T=m(k12-kg2) =133.4N-m
Unit II Balancing of Masses
1. Four masses A, B, C and D revolve at equal radii and are equally spaced along a shaft. The
mass B is 7 kg and the radii of C and D make angles of 90 and 240 respectively with the
radius of B. Find the magnitude of the masses A, C and D and the angular position of A so
that the system may be completely balanced. (Nov / Dec 2012)
Key : Given: mA = 7 kg ; C = 90 with B; D = 240 with B
Plane
Mass (m) kg
Radius (r) m
Cent.force /2 Distance from Couple
/2
(m.r) kg-m
R.P (l) m
(m.r.l) kg-m2
A
mA
rA
mA rA
1
mA rA
B
mB
rB
mB rB
1
0
C
mC
rC
mC rC
1
mC rC
D
mD
rD
mD rD
1
mD rD
Ans. 5 kg ; 6 kg ; 4.67 kg ; 205 from mass B in anticlockwise direction

2. The following particulars relate to an outside cylinder of uncoupled locomotive: Revolving

mass per cylinder = 300kg; Reciprocating mass per cylinder = 450 kg; Length of each crank
= 350 mm; Distance between wheels = 1.6 m; Distance between cylinder centers = 1.9 m;
Diameter of driving wheels = 2m; Radius of balancing mass = 0.8m; angle between the
cranks = 90. If the whole of the revolving mass and 2/3 of the reciprocating masses are to
be balanced in planes of driving wheels, determine;
Magnitude and direction of the balance masses, speed at which the wheel will lift
off the
rails when the load on each driving wheel is 35 KN, and Swaying couple at
speed
arrived in (ii) above. (Dec 2013)
Key:
parts to be balanced per cylinder at the crank pin, m = mB = mC = m1 + c.m2

Plane

Mass (m) kg

Radius (r) m

A
B
C
D

mA
mB
mC
mD

rA
rB
rC
rD

Fluctuation in rail pressure or hammer blow =

We know that maximum variation of tractive effort
We know that maximum swaying couple = a(1-

Cent.force /2
(m.r) kg-m
mA rA
mB rB
mC rC
mD rD

Distance from
R.P (l) m
1
1
1
1

Couple
/2
(m.r.l) kg-m2
mA rA
0
mC rC
mD rD

B. .b
c)/(2)1/2 x m2

3. The cranks are 3 cylinder locomotive are set at 120. The reciprocating masses are 450 kg
for the inside cylinder and 390 kg for each outside cylinder. The pitch of the cylinder is 1.2
m and the stroke of each piston 500 mm. The planes of rotation of the balance masses are
960 mm from the inside cylinder. If 40% of the reciprocating masses are to be balanced,
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determine:The magnitude and the position of the balancing masses required at a radial
distance of 500 mm; and The hammer blow per wheel when the axle rotates at 350 rpm.
Key:
1.
2.
3.
4.
5.

Since 40% of the reciprocating masses are to be balanced, therefore mass of the reciprocating
parts to be balanced for each outside cylinder, mA = mC = c mO
mass of the reciprocating parts to be balanced for inside cylinder, mB = c m1
Table
hammer blow = B. .b

4. A 4 cylinder engine has the two outer cranks as 120 to each other and their reciprocating
masses are each 400 kg. The distance between the planes of rotation of adjacent cranks are
400mm, 700mm, 700mm and 500mm. Find the reciprocating mass and the relative angular
position for each of the inner cranks, if the engine is to be in completely balance. Also find
the maximum unbalanced secondary force, if the length of each crank is 350 mm, the length
of each connecting rod 1.7m and the engine speed 500 rpm. (Nov / Dec 2012)
Key: Given : m1 = m4 = 400 kg ; r = 300 mm = 0.3 m ; l = 1.2 m ; N = 240 r.p.m.
Plane
Mass (m) kg
Radius (r) m
Cent.force /2 Distance from Couple
/2
(m.r) kg-m
R.P (l) m
(m.r.l) kg-m2
A
mA
rA
mA rA
1
mA rA
B
mB
rB
mB rB
1
0
C
mC
rC
mC rC
1
mC rC
D
mD
rD
mD rD
1
mD rD
5. A 4 cylinder vertical engine has cranks 150 mm long. The planes of rotation of first, second
and fourth cranks are 400 mm, 200 mm and 200 mm respectively from the third crank and
their respective masses are 50kg, 60kg, and 50 kg respectively. Find the mass of the
reciprocating mass for the third cylinder and the relative angular positions of the cranks in
order that the engine may be in computer primary balance.
Key:Given r1 = r2 = r3 = r4 = 150 mm = 0.15 m ; m1 = 50 kg ; m2 = 60 kg ;m4 = 50 kg
Plane
Mass (m) kg
Radius (r) m
Cent.force /2 Distance from Couple
/2
(m.r) kg-m
R.P (l) m
(m.r.l) kg-m2
A
mA
rA
mA rA
1
mA rA
B
mB
rB
mB rB
1
mB rB
C (R.P)
mC
rC
mC rC
0
0
D
mD
rD
mD rD
1
mD rD
Ans: 2 = 160, 4 = 26 m3 = 60 kg
6. A 3 cylinder radial engine driven by a common crank has the cylinders spaced at 120. The
stroke is 125 mm; the length of the connecting rod is 225 mm and the reciprocating mass
per cylinder 2 kg. Calculate the primary and secondary forces at crank shaft speed of 1200
rpm. (Dec 2013)
Key: Given : L = 125 mm ; l = 225 mm; m = 2 kg ; N = 1200 r.p.m.
1. Maximum Primary Force = 3m/2 x 2 r
2. Maximum Secondary force = 2m/2(2 2)(r/4n).
7. The reciprocating mass per cylinder in a 60 V-twin engine is 1.5 kg. The stroke is 100 mm
for each cylinder. If the engine runs at 1800 rpm, determine the maximum and minimum
values of the primary forces and find out the corresponding crank position.
Key: = 30, m = 1.5 kg, l = 100 mm ; N = 1800 r.p.m.
Maximum and minimum values of primary forces = m/2 x 2 r (9cos2+sin2)1/2

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Maximum and minimum values of secondary forces

8. The firing order of a six cylinder, vertical, four stroke, in-line engine is 1-4-2-6-3-5. The
piston stroke is 80 mm and length of each connecting rod is 180 mm. the
pitch distances
between the cylinder centre lines are 80 mm, 80 mm, 120 mm, 80
mm and 80 mm
respectively. The reciprocating mass per cylinder is 1.2 kg and
the engine speed is
2400 rpm. Determine the out-of-balance primary and
secondary forces and couples
on the engine taking a plane mid-way between the
cylinders 3 and 4 as the reference
plane.
Key: Given : L = 80 mm or r = L / 2 = 40 mm = 0.04 m ; l = 180 mm ; m = 1.2 kg ; N = 2400 r.p.m.
Plane

Mass (m) kg

Radius (r) m

1
2
3
4
5
6

1.2
1.2
1.2
1.2
1.2
1.2

0.04
0.04
0.04
0.04
0.04
0.04

Cent.force /2
(m.r) kg-m
0.04
0.04
0.04
0.04
0.04
0.04

Distance from
R.P (l) m
l1
l2
l3
l4
l5
l6

Couple
/2
(m.r.l) kg-m2
0.04l1
0.04l2
0.04l3
0.04l4
0.04l5
0.04l6

Draw force polygon and couple polygon.

UNIT III FREE VIBRATIONS
PART B
1. (a) A machine of weighs 18 kg and is supported on springs and dashpots. The total stiffness
of the springs is 12 N/mm and damping is 0.2 N/mm/s the system is initially at rest and a
velocity of 120 mm/s is imparted to the mass. Determine (1) the displacement and velocity of
mass as a function of time (2) the displacement and velocity after 0.4s.
(b) Describe the types of vibrations with simple sketch. (Dec 2013)
HINT:
(a) Given Data:
Mass m = 18 kg,
Initial Velocity = 120 mm/s = 0.12 m/s
Stiffness s = 12 N/mm = 12000 N/m
Damping coefficient c = 0.2 N/mm/s = 200 N/m/s

Step 1:
Find the Circular Natural frequency n = Sqrt [s/m]
Find the Damped Natural frequency d = Sqrt [n2 a2] where a = c/2m
Step 2:
The General Equation for Displacement of Damped vibration in terms of time is
X = C e-at sin (dt + )
Intially at rest t =0, Displacement X=0
Substitute X = 0 and t = 0 in the above equation and thus = 0,
Hence X becomes
X = C e-at sin dt
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Step 3:
Differentiate above equation to get velocity V
V = dX/dt = C [e-atd cosdt - ae-at sin dt] ..(differentiate by u.v method)
Now Substitute Initial velocity of V = 0.12 m/s at time t = 0
Hence Find the constant C = 4.76 mm.
Step 4:
a)Displacement and Velocity in terms of t:
Substitute C,a,d in
X = C e-at sin dt and
V = C [e-atd cosdt - ae-at sin dt]
Step 5:
a)Displacement and Velocity after 0.4 seconds:
Substitute t=0.4 sec in above equation to get X and V
b) Types of Free Vibrations
The following three types of free vibrations are important from the subject point of view :
1. Longitudinal vibrations, 2. Transverse vibrations, and 3. Torsional vibrations.

2. Derive the expression for the natural frequency of free transverse or longitudinal vibrations
by using any two methods.
HINT:
1. Equilibrium Method

Consider a constraint (i.e. spring) of negligible mass in an unstrained position, as shown in

Figure
Let s = Stiffness of the constraint. It is the force required to produce unit displacement in the
direction of vibration. It is usually expressed in N/m.
m = Mass of the body suspended from the constraint in kg,
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W = Weight of the body in newtons = m.g,
= Static deflection of the spring in metres due to weight W newtons, and
x = Displacement given to the body by the external force, in metres.

Restoring force =W - s ( + x) =W - s. - s.x

= s.- s. - s. x = -s.x
(since W = s.) . . . (i)
And
Accelerating force = Mass Acceleration

. . . (Taking upward force as negative)

Equating Restoring force and accelerating force

---(iii)

----(iv)

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2.

Energy Method:

The time period and the natural frequency may be obtained as discussed in the previous method.
3. A shaft of 100 mm diameter and 1 m long is fixed at one end and other end carries a
flywheel of mass 1 tonne. Taking youngs modulus for the shaft material as 200 GN/m 2; find
the natural frequency of longitudinal and transverse vibrations.
HINT:
Step 1:
Determine the cross sectional area of the shaft A = (/4) d2
Determine the Moment of Inertia I = (/64) d4
Step 2:
Natural Frequency of Longitudinal vibrations
Using the below formula, find fn which is natural frequency of longitudinal vibration
Static deflection of cantilever beam

Natural frequency of longitudinal vibration

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Step 3:
Natural Frequency of Transverse vibrations
Using the below formula, find fn which is natural frequency of transverse vibration
Static deflection of cantilever beam for transverse vibrations is

Natural frequency of transverse vibration

4. A flywheel is mounted on a vertical shaft as shown in figure. The both ends of the shaft are
fixed and its diameter is 50 mm. The flywheel has a mass of 500 kg. Find the natural
frequency of free longitudinal and transverse vibrations. Take E = 200GN/m2.

HINT
Step 1:
Determine the cross sectional area of the shaft A = (/4) d2
Determine the Moment of Inertia I = (/64) d4
Step 2:
Let m1 be the mass of flywheel carried by length l1 and
mm1 be the mass of flywheel carried by length l2
Extension of length l1 is

Compression of length 12 is

Equating Extension length and compression length find m1.

Step 3:
Static Deflection from extension length l1 is

Find the natural frequency using

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Step 4:
To determine natural frequency of transverse vibration, find using the below formula.

where a and b are the distance of flywheel from ends

Find the natural frequency using

5.

A shaft 1.5 m long is supported by two short bearings and carries two wheels each of 50 kg
mass. One wheel is situated at the centre of the shaft and other at a distance of 0.4 m from
the centre towards right. The shaft is hollow of external diameter 75 mm and inner
diameter 37.5 mm. The density of the shaft material is 8000 kg/m3. The youngs modulus for
the shaft material is 200 GN/m2. Find the frequency of free transverse vibration.
Step 1
Determine the Moment of Inertia

Step 2
Density of given material is 8000 kg/m3
Find mass of shaft per unit length using

Step 3
General expression for Static deflection [for simply supported beam] due to weight W is given by

Determine the static deflection due load W1 and W2. (Both 50 kg located at different places as specified in
the questions) using

Note: The value of a and b changes for both 1 and 2 as per the position given in questions
Step 4
General expression for Static deflection [for simply supported beam] due to self weight (uniformly
distributed load)
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Find w by using w = msg

Step 5
Determine Natural frequency of transverse vibration using

6. A Steel bar 25 mm wide and 50 mm deep is freely supported at two points 1m apart and
carries a mass of 200 kg in the middle of the bar. Neglecting the mass of the bar, find the
frequency of free transverse vibration. If an additional mass of 200 kg is distributed
uniformly over the length of the shaft, what will be the frequency of free transverse
vibration? Take E = 200 GN/m2.
Step 1:
Determine the Moment of Inertia using I = bd3/ 12
Step 2:
Determine the Static Deflection [for simply supported beam with pt load at centre] using

Step 3:
Find the natural frequency of transverse vibration neglecting the self weight using

Step 4:
General expression for Static deflection [for simply supported beam] due to self weight (uniformly
distributed load)

Determine w by using w = msg. [ms = Area X Length X Density ] [Take Density of steel = 7700 kg/m3
Step 5:
Find the Natural frequency of transverse vibration considering the self weight using
fn = 0.4985 / Sqrt [ + s/1.27]

7. A vertical steel shaft 15 mm diameter is held in long bearings 100 cm apart and carries at its
middle a disc weighing 147.15N. The eccentricity of the centre of gravity of the disc from the
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centre of the rotor is 0.3mm. The modulus of elasticity for the shaft material is 19.6 X 106
N/cm2, and the permissible stress is 6867 N/cm2. Determine :(i) the critical speed of the shaft
(ii) The range of speed over which it is unsafe to run the shaft. Neglect the weight of the
shaft.
HINT:
Step 1:
Determine the moment of Inertia

Step 2:
Critical speed of shaft Nc
Determine Static deflection with point load at middle [consider as fixed at both ends for long bearings]

Determine natural frequency fn as

Critical speed Nc in r.p.s is same as natural frequency in Hz, therefore

Nc = fn X60 rpm
Step 3:
To determine range of speed:
`
Additional Dynamic Load during shaft rotation can be determined by relations

Also

and
is the permissible stress (given in the question)
Find m1 using above relation [Equating both M values]
Additional deflection due to load [W1= m1g] is

Also

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Nc is known to us, Equate the value of y and find N value [taking positive and negative equations] to
get the range of speed N1 and N2.

8. A vertical shaft 25 mm diameter and 0.75 m long, is mounted in long bearings and carries a
pulley of mass 10 kg midway between the bearings. The centre of the pulley is 0.5 mm from
the axis of the shaft. Find (a) the Whirling Speed (b) the bending stress in the shaft, when it
is rotating at 1700 rpm
HINT:
Step 1:
Determine the Moment of Inertia of shaft

Step 2:
Determine the static deflection of the shaft [considered as fixed on both ends for long bearings]

Determine the natural frequency

Natural frequency of transverse vibration in Hz is same as Critical speed in rpm, Hence

Nc = fn X60 rpm
Step 3:
To find Max Bending Stress if N = 1700 rpm

and
Find W1 in terms of from the above expression.
Step 4:
Determine y using [in terms of , as W1 is in terms of ]

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Also find y using

[the value of e is the distance of pulley from centre]

Equating both y, we get Bending Stress .

9. The disc of a torsional pendulum has a moment of inertia of 600 kg-cm2 and is immersed in
a viscous fluid. The brass shaft attached to it is of 10 cm diameter and 40 cm long. When the
pendulum is vibrating, the observed on the same side of the rest position for successive
cycles are 9 degree, 6 degree and 4 degree. Determine (a) Logarithmic decrement (b)
damping force at unit velocity and (c) periodic time of vibration. Assume for the brass shaft,
G = 4.4 X 1010 N/m2. What would the frequency be if the disc is removed from the viscous
fluid?
HINT:
Step 1:
Determine the Circular Natural frequency

Step 2:
Determine the Damping Coefficient C using below method
x1 = 9 deg, x2= 6 deg and x3 = 4 deg
Convert x1,x2 and x3 in radians
We know that

Also
[x1/x3] = [x1/x2][x2/x3] = [x1/x2]2
or
[x1/x2] = [x1/x3]1/2
Also we know that

Where a = c/2m
From the above expression,
Determine the value of c
c is the Damping force per unit velocity [in N/m/s]
Step 3:
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Determine cc using cc = 2mn
Determine Logarithmic Decrement using

Step 4:
Time period tp = 1/fn
For Damped vibration fn = d/2 where d = Sqrt [n2-a2] and a = c/2m
Determine Periodic time of vibration tp from the above expression
Step 5:
For Undamped vibration [if disc is removed from viscous fluid]
Determine natural frequency using
fn = n/2

10. A vibrating system consists of a mass 0f 8 kg, spring of stiffness 5.6 N/mm and a dashpot of
damping coefficient of 40 N/m/s. Find (a) damping factor (b) logarithmic decrement (c)
ratio of two consecutive amplitudes.
HINT:
Step 1:
a) Determine the critical damping coefficient cc

Damping factor = c/cc [c is given as damping coefficient]

Step 2:
b) Determine the Logarithmic Decrement

Step 3:
Determine the ratio of two consecutive amplitudes

11. A centrifugal pump is driven through a pair of spur wheels from an oil engine. The pump
runs at 4 times the speed of the engine. The shaft from the engine flywheel to the gear is 75
mm diameter and 1.2 m long, while that from the pinion to pump is 50 mm diameter and
400 mm long. The moments of inertia are as follows: flywheel = 1000kg-m2; pinion = 10kg-

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m2; and pump impeller = 40kg-m2. Find the natural frequencies of torsional oscillations.
Take C = 84 GN/m2.
HINT:
Step 1:
Determine the mass moment of equivalent rotor B
IB = IB/G2
Additional length of equivalent shaft is
l3 = G2.l2 [d1/d2]4
Total length of equivalent shaft l = l1 + l3
Step 2:
lAIA = lBIB
Determine lA from the above expression
Polar Moment of Inertia for equivalent shaft is
J = [/32]d14
Step 3:
Determine the Natural frequency of torsional vibration
fn = [1/2]sqrt[C.J/lAIA]

12. A 4 cylinder engine and flywheel coupled to a propeller are approximated to a 3 rotor

system in which the engine is equivalent to a rotor of moment of inertia 800 kg-m2, the
flywheel to a second rotor of 320 kg-m2 and the propeller to a third rotor of 20 kg-m2. The
first and second rotors are being connected by a 25 mm diameter and 2m long shaft.
Neglecting the inertia of the shaft and taking its modulus of rigidity as 80 GN/m 2,
determine; (i) natural frequencies of free torsional vibrations and (ii) the positions of the
nodes.
HINT:
Step 1:
Determine the torsionaly equivalent shaft using
l = l1 + l2 [d1/d2]4

Step 2:
Determine lA and lC using
lAIA = lCIC
and
1/lCIC = 1/IB [ 1/(l1 lA) + 1/(l3 lC) ]

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Locate lA and lC in the drawing as shown, lA and lC will have two values [First value for single node and
second value of two node system]
Step 3:
Determine the Polar Moment of Inertia J = [ / 32 ]d4
Natural frequency of torsional vibration for single node/two node system is
fn1 = [1/2]. Sqrt [ CJ/lAIA] {Substituting two diff values of lA]
Step 4:
Position of Nodes:
From the value of lC of two node system, the corresponding value of lC from original system of propeller is
= lC [d2/d1]4, where lC { Value of two node system of lC]

13. The mass of a single degree damped vibrating system is 7.5 kg and makes 24 free

oscillations in 14 seconds when disturbed from its equilibrium position. The amplitude of
vibration reduces to 0.25 of its initial value after five oscillations. Determine: 1. stiffness of
the spring, 2. logarithmic decrement, and 3. damping factor, i.e. the ratio of the system
damping to critical damping.
Step 1:
From 24 oscillations in 14 seconds, determine fn Natural frequency
fn = 24/14
Circular Natural frequency n = 2 fn
From the above, Stiffness of spring can be determined using

[since m and n are known]

Step 2:
Determine logarithmic decrement using
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Logarithmic Decrement is

Step 3:
Damping factor: = c/cc
Determine critical damping coefficient from below expression
Determine Damping coefficient c from below expression (as is known)

Find damping factor using c/cc

14. Derive an expression for the frequency of free torsional vibrations for a shaft fixed at one

end and carrying a load on the free end.

Let

Total Angle of Twist of shaft is equivalent to sum of angle of twist of each shaft.
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From this concept
T.l/C.J = [T.l1/C.J1] + [T.l2/C.J2] + [T.l3/C.J3]
Substituting value of J, J1, J2 and J3 in all equation and simplifying we get
Torsional equivalent shaft is
l = l1 + l2 [d1/d2]4 + l3 [d1/d3]4

Unit IV Forced Vibration

1. (a) Discuss the forcing due to support motion.
(b) What is meant by magnification factor in case of forced vibrations?
HINT:
a) Force due to support motion
The excitation of at the location of support is commonly known as base excitation

With the motion of the base denoted as y and the motion of the mass relative to the
intertial reference frame as x, the differential equation becomes

The solution can be written by

Steady state amplitude and phase form are given by

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b) Magnification factor
It is the ratio of maximum displacement of the forced vibration (xmax ) to the deflection
due to the static force F(xo). We have proved in the previous article that the maximum
displacement or the amplitude of forced vibration,

2. (a) Derive the relation for the displacement of mass from the equilibrium position of a
damped vibration system with harmonic forcing.
(b) Define the term vibration isolation.
HINT:
a) Harmonic forcing

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Diff equation is given by

Solution is
Complementary function is
Particular Integral is

The complete solution with harmonic function is

where
and

b) Vibration Isolation

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Transmissibility Ratio

3. A machine has a mass of 125 kg and unbalanced reciprocating mass 3 kg which moves
through a vertical stroke of 90 mm with SHM. The machine is mounted upon 5 springs.
Neglecting damping, calculate the combined stiffness of the spring in order that force
transmitted is 1/20th of the applied force, when the speed of the machine crank shaft is 1200
rpm. When the machine is actually supported on the springs, it is found that damping
reduces the amplitude of successive free vibration by 30%
Determine;
(1) Force transmitted to the foundation at 1200 rpm
(2) Force transmitted to the foundation at resonance.
HINT:
Step 1:
Determine the angular velocity [circular frequency] using the equation
= 2N/60
and
Determine the Eccentricity e = Stroke / 2
Step 2:
Determine circular natural frequency using transmissibility ratio.
Transmissibility ratio = +/- [ 1/ (1-r2)]
where r = /n
Since force transmitted is 1/20th of applied force = 1/20
In the transmissibility ratio equation put (1-r2) as (r2-1) to get positive root
Find combined stiffness using n = Sqrt [s/m]
Step 3:
To determine the Force transmitted to the foundation at 1200 rpm
Find frequency ratio r = /n
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Given that percentage of successive amplitude is 30%, hence X1 = 0.70 Xo
Logarithmic Decrement is
ln[Xo/X1] = 2c/Sqrt [cc2-c2]
Find critical damping coefficient from cc = 2mn,
Find the value of damping coefficient c from the above expression.

Actual value of transmissibility is

Step 4
The maximum unbalance force due to reciprocating parts is given by
F = mu2r where mu is mass of reciprocating part = 3 kg
Force transmitted to the foundation is
FT = F
Step 5
Force transmitted to the foundation at resonance,
At resonance = n
Hence the expression is reduced to

Find FT in same manner as above

[F = mu2r where mu is mass of reciprocating part = 3 kg
Force transmitted to the foundation is
FT = F ]
4. A vibrating system having a mass of 1.5 kg is suspended by a spring of stiffness 1200N/m
and it is put to harmonic excitation of 12 N. Assuming viscous damping, Determine, (1)
Resonant Frequency (2) Phase angle at resonance (3) Amplitude at resonance (4) Damped
frequency; Take c = 48 NS/m
HINT
Step 1
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Determine circular natural frequency

n = Sqrt [s/m] [ s & m are given]

At resonance = n
i) Find Resonant frequency using fn = n/2
Step 2
ii)Determine the phase angle at resonance using
[At resonance = n]

[c- given in the question]

Step 3
(iii) Amplitude at resonance
Max amplitude can be determined using below expression [Force F=12N,=n at resonance]

Step 4:
Damped frequency
fd = d/2
where
d = sqrt [n2- a2] and a = c/2m
Substitute and find frequency of damped vibration
5. A machine supported symmetrically on five springs, has a mass of 90 kg. The mass of the
reciprocating parts is 3 kg which moves through a vertical stroke of 90 mm with SHM.
Neglecting damping determine the combined stiffness of the springs so that force
transmitted to the foundation is 1/30th of impressed force. The machine crank shaft rotates
at 750 rpm. If the under actual working conditions the damping reduces the amplitude of
successive vibration by 25%, find:
(i) Force transmitted to the foundation at 900 rpm
(ii)Force transmitted to the foundation at resonance.
(iii)The amplitude of vibration at resonance
[PROCEDURE OF THIS PROBLEM AS SAME AS PROBLEM NO 3 EXCEPT AMPLITUDE
DETERMINATION]
Step 1:
Determine the angular velocity [circular frequency] using the equation
= 2N/60
and
Determine the Eccentricity e = Stroke / 2
Step 2:
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Determine circular natural frequency using transmissibility ratio.
Transmissibility ratio = +/- [ 1/ (1-r2)]
where r = /n
Since force transmitted is 1/30th of applied force = 1/30
In the transmissibility ratio equation put (1-r) as (r2-1) to get positive root
Find combined stiffness using n = Sqrt [s/m]
Step 3:
To determine the Force transmitted to the foundation at 900 rpm
Find frequency ratio r = /n
Given that percentage of successive amplitude is 25%, hence X1 = 0.75 Xo
Logarithmic Decrement is
ln[Xo/X1] = 2c/Sqrt [cc2-c2]
Find critical damping coefficient from cc = 2mn,
Find the value of damping coefficient c from the above expression.

Actual value of transmissibility is

Step 4
The maximum unbalance force due to reciprocating parts is given by
F = mu2r where mu is mass of reciprocating part = 3 kg
Force transmitted to the foundation is
FT = F
Step 5
Force transmitted to the foundation at resonance,
At resonance = n
Hence the expression is reduced to

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Find FT in same manner as above

[F = mu2r where mu is mass of reciprocating part = 3 kg
Force transmitted to the foundation is
FT = F ]
Step 6:
Determine the Amplitude using :
Amplitude = Force transmitted at the resonance / Combined Stiffness
6. A 75 kg machine is mounted on springs of stiffness K= 11.76 X 10 5 N/m with an assumed
damping factor of 0.2. A 2 kg piston within the machine has a reciprocating motion with a
stroke of 0.08 m and a speed of 3000 rpm. Assuming the motion of the piston to be
harmonic, determine the amplitude of vibration of the machine and the vibratory force
transmitted to the foundation.
Step 1:
Determine the angular velocity of unbalance force using
= 2N/60
Circular Natural frequency n = Sqrt [s/m] [s is stiffness which is denoted as k in question]
Eccentricity e = stroke / 2
Frequency ration r = /n
Step 2
Determine the exciting force F = mu.2.e [where mu mass of reciprocating part piston]
Max amplitude of vibration will be given by

Damping factor c/cc is given in question.

cc = 2mn
From this find c = [cc.][Damping factor]
Step 3:
Determine transmissibility ratio from the below expression

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Also = FT/F
Since F is known, determine FT from the above expression, which is force transmitted to the
foundation.
7. A body of mass 70 kg is suspended from a spring which deflects 2 cm under the load. It is
subjected to a damping effect adjusted to a value of 0.23 times that required for critical
damping. Find the natural frequency of the un-damped and damped vibrations and ratio of
successive amplitudes of damped vibrations. If the body is subjected to a periodic
disturbing force of 700 N and of frequency equal to 0.78 times the natural frequency, find
the amplitude of forced vibrations and the phase difference with respect to the disturbing
force.
HINT
Step 1
Determine the natural frequency of free vibration [without damping] using
fn = 0.4985/Sqrt[] where Deflection of spring due to load = 20cm = 0.2 m[given]
Damping coefficient c = 0.23 cc [given- damping coeff is 0.23 times of critical damping coeff]
Circular natural frequency n = 2fn
Critical Damping coefficient cc = 2mn
Determine damping coefficient c from the above expressions
Step 2
Natural frequency of undamped vibration is fn.
Natural frequency of damped vibration
fd = d/2
where d = Sqrt [n2- a2] and a = c/2m
Step 3
For forced vibration [excitation] in the same setup
Frequency f = 0.78fn [given in question as frequency is 0.78 times of natural frequency]
Hence = 2f
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Max amplitude of forced vibration can be determined by

Phase difference can be determined by

Determine s by using n = sqrt [s/m], [since n and m is known, s can be determined]

8. The support of a spring mass system is vibrating with amplitude of 6 mm and a frequency
of 1200 cycles/min. If a mass is 95 kg and the spring has a stiffness of 1950 N/m, determine
the amplitude of vibration of the mass. If a damping factor of 0.2 is include, what would be
the amplitude?
Step 1:
Determine the frequency of the support = 2f
Determine the natural frequency n = Sqrt [s/m]
Step 2:
Amplitude of Support Y = 6mm = 0.006m
Amplitude of forced vibration due to excitation of support is
Xmax = Y. [sqrt{s2+(c2)}2] / [sqrt { (s m2)2 +(c)2]
Intially without damping, damping coefficient c = 0
Step 3:
If a damping factor of 0.2 is included, c/cc = 0.2
Determine cc = 2mn and find xmax with same above expression
Xmax = Y. [sqrt{s2+(c2)}2] / [sqrt { (s m2)2 +(c)2]

9. A machine has a mass of 100 kg and unbalanced reciprocating parts of mass 2 kg which
move through a vertical stroke of 80 mm with SHM. The machine is mounted on 4 springs,
symmetrically arranged with respect to center of the mass, in such a way that the machine
has one degree of freedom and can undergo vertical displacements only. Neglecting
damping, calculate the combined stiffness of the spring in order that the force transmitted
to the foundation is 1/25th of the applied force, when the speed of the rotation of machine
crank shaft is 1000 rpm. When the machine is actually supported on the springs it is found
that the damping reduces the amplitude of successive five vibrations by 25%. Find:(i)The

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force transmitted to the foundation at 1000 rpm; (ii)The force transmitted to the foundation
at resonance; (iii)
The amplitude of the forced vibration of the machine at resonance.
[SIMILAR MODEL TO PROBLEM NO 3 and 5. Refer Hint of Problem No. 5]
Step 1:
Determine the angular velocity [circular frequency] using the equation
= 2N/60
and
Determine the Eccentricity e = Stroke / 2
Step 2:
Determine circular natural frequency using transmissibility ratio.
Transmissibility ratio = +/- [ 1/ (1-r2)]
where r = /n
Since force transmitted is 1/25th of applied force = 1/25
In the transmissibility ratio equation put (1-r) as (r2-1) to get positive root
Find combined stiffness using n = Sqrt [s/m]
Step 3:
To determine the Force transmitted to the foundation at 1000 rpm
Find frequency ratio r = /n
Given that percentage of successive amplitude is 25%, hence X1 = 0.75 Xo
Logarithmic Decrement is
ln[Xo/X1] = 2c/Sqrt [cc2-c2]
Find critical damping coefficient from cc = 2mn,
Find the value of damping coefficient c from the above expression.

Actual value of transmissibility is

Step 4
The maximum unbalance force due to reciprocating parts is given by
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F = mu2r where mu is mass of reciprocating part = 3 kg
Force transmitted to the foundation is
FT = F
Step 5
Force transmitted to the foundation at resonance,
At resonance = n
Hence the expression is reduced to

Find FT in same manner as above

[F = mu2r where mu is mass of reciprocating part = 2 kg
Force transmitted to the foundation is
FT = F ]
Step 6:
Determine the Amplitude using :
Amplitude = Force transmitted at the resonance / Combined Stiffness

10. A machine of mass 75 kg is mounted on springs of stiffness 1200 kN/m and with an assumed

damping factor of 0.2. A piston within the machine of mass 2 kg has a reciprocating motion
with a stroke of 80 mm and a speed of 3000 cycles/min. Assuming the motion to be simple
harmonic, find: 1. the amplitude of motion of the machine, 2. Its phase angle with respect to
the exciting force, 3. the force transmitted to the foundation, and 4. the phase angle of
transmitted force with respect to the exciting force.
Step 1
Determine the angular velocity of unbalanced force = 2 N/60
Determine the circular natural frequency n = sqrt [s/m]
Eccentricity e = stroke / s
Unbalanced excited force F = mu.2.e [where mu is the mass of the reciprocating parts]
Step 2
1) Amplitude of motion of machine

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Step 3
2) Phase lag with respect to excited force

Step 4
3) Force transmitted FT to the foundation
Determine transmissibility ratio using

Find FT from = FT/F

Step 5
4) Phase lag of transmitted force with respect to applied force
It can be determined by
= tan -1 [ 2 (c/cc)(/n)]
11. A single cylinder engine of total mass 200 kg is to be mounted on an elastic support which
permits vibratory movement in vertical direction only. The mass of the piston is 3.5 kg and
has a vertical reciprocating motion which may be assumed simple harmonic with a stroke of
150 mm. It is desired that the maximum vibratory force transmitted through the elastic
support to the foundation shall be 600 N when the engine speed is 800 rpm and less than
this at all high speeds. Find: (i) the necessary stiffness of the elastic support, and the
amplitude of vibration at 800 rpm, and (ii) if the engine speed is reduced below 800 rpm at
what speed will the transmitted force again becomes 600N.
Step 1
Determine the unbalanced force on the piston
F = mu2e [where e = stroke /2 and mu is the mass of reciprocating parts]
Step 2
Max vibratory force transmitted to the foundation will be given by
FT = [Stiffness of elastic support].[Max amplitude of vibration]

Since FT is given in the question, determine s from the above expression

Step 3
Determine the Maximum Amplitude of vibration using
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Step 4
To find the speed at which the transmitted force again becomes 600 N
Find Disturbing force F = mu12e [ Keep the value in terms of 1]
Step 5
Since the engine speed is reduced, find amplitude with same above concept [as given formulae]

Force transmitted FT = [Stiffness]. xmax [FT = 600 N known value]

Determine 1 from the above expression and hence find N1- that is speed required
12. An industrial machine weighing 445 kg is supported on a spring with a static deflection of
0.5 cm. If the machine has rotating imbalance of 25 kg-cm, determine the force transmitted
at 1200 rpm and the dynamic amplitude at that speed.
Step 1
Determine the Angular velocity of the unbalance force
= 2N/60
Determine the natural frequency n = sqrt [s/m] = sqrt [g/]
Since is known to us {given in the question as 0.5 cm}
Hence find the stiffness of spring s

Step 2
Determine the exciting force F = [mue]2
{Rotating Imbalance in the machine is given as 0.25 kgcm which is the quantity [mue]
Step 3
Force transmitted at 1200 rpm, without damper
= FT/F = +/- [1 / {1- [/n]2}]
Since /n > 1, the equation can be written as
= 1/[[/n]2 1]
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Since can be found using the above, find FT which is force transmitted without damper
at 1200 rpm
Step 4
Amplitude of vibration without damper
Since /n > 1
Xmax = [F/s] / [ 1 (/n)2]

13. What do you understand by transmissibility? Describe the method of finding the
transmissibility ratio from unbalanced machine supported with foundation.
The ratio of the force transmitted (FT) to the force applied (F) is known as the isolation factor or
transmissibility ratio of the spring support.
We have discussed above that the force transmitted to the foundation consists of the following
two forces :
1. Spring force or elastic force which is equal to s. xmax, and
2. Damping force which is equal to c. xmax.

Unit V Mechanisms for Control

1. What is the effect of gyroscopic couple on the stability of a two wheel vehicle taking a turn?
(Dec 2013)

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2. (a) Explain the function of a proell governor with the help of a neat sketch. Derive the
relationship among the various forces acting on the link.

(b) What are centrifugal governors? How do they differ from inertia governors? (Dec 2013)
The centrifugal governors are based on the balancing of centrifugal force on the rotating balls
by an equal and opposite radial force, known as the controlling force
3. The mass of each ball of a proell governor is 7.5 kg and the load on the sleeve is 80 kg.each
of the arms is 300 mm long. The upper arms are pivoted on the axis of rotation whereas the
lower arms are pivoted to links of 40 mm from the axis of rotation. The extensions of the
lower arms to which the balls are attached are 100 mm long and are parallel to the
governor axis at the minimum radius. determine the equilibrium speeds corresponding to
extreme radii of 180 mm and 240 mm
Speed

T = 223 N and N = 165.3 rpm

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4. Calculate the range of speed of a porter governor which has equal arms of each 200 mm
long and pivoted on the axis of rotation. the mass of each ball is 4 kg and the central mass of
the sleeve is 20 kg.the radius of rotation of the ball is 100 mm when the governor begins to
lift and 130 mm when the governor is at maximum speed.
Height of the governor
Speed of the governor

5. Find the angle of inclination with respect to the vertical for a two wheeler having the
following details negotiating a turn of radius 50 m: combined mass of vehicle with rider =
250 kg; Centre of gravity with rider in vertical position = 0.6 m; Moment of inertia of
flywheel = 0.3 kgm2; M.I of each road wheel = 1 kg-m2; Speed of engine is five times that of
road wheels and in same direction; Vehicle speed = 90km/h; wheel diameter = 600 mm.

6. A porter governor has two balls each of mass 3 kg and a central load of mass 15 kg.the arms
are all 200 mm long pivoted on the axis. The maximum and minimum radii of rotation are
160 mm and 120 mm respectively. Find the range of speed.
Height of the governor
Speed of the governor

7. A hartnell governor having a central sleeve spring and two right angled bell crank levers
operates between 290 rpm and 310 rpm for a sleeve lift of 15 mm.the sleeve and ball arms
are 80 mm and 120 mm respectively. The levers are pivoted at 120 mm from the governor
axis and mass of each ball is 2.5 kg.the ball arms are parallel at lowest equilibrium speed.
Determine (i) loads on the spring at maximum and minimum speeds, and (ii) stiffness of the
spring.

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8. In a Hartnell governor the lengths of ball and sleeve arms of a bell crank lever are 120 mm and
100 mm respectively. the fulcrum of the bell crank lever is located at 140 mm from the governor
axis each governor ball is 4 kg.the governor runs at 5 rps with ball arms vertical and sleeve arms
horizontal the sleeve movement is found to be 10 mm (upwards) for an increase of speed of
4%.find (i) maximum speed if the total sleeve movement is limited to 20 mm (ii) the spring
stiffness (iii) sensitiveness of governor (iv) required spring stiffness for isochronous at 300 rpm.
(Nov/Dec 2012)

9. A porter governor has equal arms each 250 mm long and pivoted on the axis of rotation each ball
has a mass of 5 kg and the mass of central load on the sleeve is 25 kg.the radius of rotation of the
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ball is 150 mm and the governor begins to lift and 200 mm.when the governor is maximum
speed. find the minimum and maximum speeds and range of speed of governor

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10.
a. In a spring controlled governor, the curve of the controlling force is a straight line. When
balls are 400 mm apart, the controlling force is 1200 N and when 200 mm apart, the
controlling force is 450 N.at what speed will the governor run when the balls are 250 mm
apart? What initial tension on the spring would be required for isochronisms and what
would then be the speed? Take masses of each ball to be 10 kg.
We know that for the stability of the spring controlled governors, the controlling force (FC) is
expressed in the form
* FC = a.r b . . . (i)
When r = r1 = 100 mm = 0.1 m, then
450 = a 0.1 b = 0.1 a b . . .(ii)
and when r = r2 = 200 mm = 0.2 m, then
1200 = a 0.2 b = 0.2 a b . . . (iii)
From equations (ii) and (iii), we find that
a = 7500, and b = 300

11. A ship is propelled by a turbine rotor which has a mass of 5 tonnes and a speed of 2100 rpm.the
rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction when viewed from the
stern. Find the gyroscopic effect in the following conditions: (i) the ship sails at a speed of 30
km/hr and steers to the left in curve having 60 m radius; (ii) the ship pitches 6 above and 6
below the horizontal position. The bow is descending with its maximum velocity. the motion due
to pitching is simple harmonic and a periodic time is 20 seconds.(iii) the ship rolls and at a certain
instant it has an angular velocity of 0.03 rad/sec clockwise when viewed from stern. (Nov/Dec
2012)

12. A four wheeled motor car weighing 2000 kg has height of C.G of 600 mm above ground level.
the engine parts and transmission are equivalent to a flywheel of 80 kg with radius of gyration
150 mm and their axis coincide with longitudinal axis of the vertical the car negotiates a curve
(turning right) of 60 m radius at 72 kmph with over all gear ratio 4:1.the radius of road wheel is
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300 mm and moment of inertia is 3 kg-m2.assuming wheel track as 1.5 m, weight distribution as
50:50.determine reaction at each wheel.

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