The document discusses the design of flywheels. Flywheels store kinetic energy and are used to reduce power fluctuations in engines and machines. They have a heavy rotating rim connected to a central hub by several arms. Flywheels can be made of cast iron due to its ability to absorb vibrations. The stresses in flywheels include tensile stresses from centrifugal force and bending stresses from the arms resisting torque fluctuations. Proper design of the rim, arms, and materials is needed to ensure flywheels withstand the stresses during high-speed rotation.

1. DESIGN OF FLYWHEEL

2. Introduction
• Flywheel is a heavy rotating body that acts as
a reservoir of energy. Energy is stored in the
form of kinetic energy.
• Flywheel is extensively used in applications
like “Punching Press’ (power is supplied at
constant rate)and I/C Engines” (Power is
generated at variable rate)

3. Functions
• To store and release energy when needed
during the work cycle
• To reduce the power capacity of electric
motor
• To reduce the amplitude of speed
fluctuations.

4. Flywheel &Governor
Flywheel
• Reduces the unavoidable
fluctuation of speed ,arising
from fluctuations of turning
moment on the crankshaft.
• It neither has influence on
mean speed nor it
maintains a constant speed.
• Always remains in operation
when the engine is running
Governor
• Controls the mean speed of
the engine by varying the
fuel supply to the engine.
• It has no influence on cyclic
speed fluctuations.
• Only operates when the
engine is running away from
its mean speed.

5. Flywheel
• It may not be used if the
cyclic fluctuations of energy
output are small or
negligible.
• Energy stored is in the form
of kinetic energy and is
100% convertible to work
without friction.
Governor
• It is essential for all types of
engines to adjust the fuel
supply as per the demand.
• It involves frictional losses

6. Materials for flywheel
• It is generally made of cast iron, as it has
following advantages;
1) C.I Flywheels are cheapest
2) C.I Flywheels can be given any complex shape
without involving machining operations.
3) Excellent ability to damp vibrations
However it has poor tensile strength, and its
failure is sudden and total.

7. Mass density of flywheel materials
Material Mass density kg/m3 ρ
FG150 7050
FG200 7100
FG220 7150
FG260 7200
FG300 7250
Carbon Steels 7800

8. Construction

9. Split flywheel

10. • Arms have elliptical c/s. In small flywheels, the
arms are replaced by a solid web.
• In large flywheels stresses are induced in arms
due to casting process. Heavy concentration of
mass at the rim and at the hub results in
unequal cooling rates, which results in cooling
stresses.
• These are taken care by using split type
construction

11. Torque analysis
A flywheel mounted on a relatively stiff shaft is
shown.
When the flywheel absorbs energy its speed
increases and vice versa.
This reduces speed fluctuations.
Considering the equilibrium of torques,
I (dω/dt) = Ti – To ----(1)

13. where;
I = Mass moment of inertia of flywheel(kg-m2)
Ti = Driving or input torque (N-m).
To = Load of output torque (N-m)
ω = Angular velocity of the shaft (rad/s)
When (Ti – To ) is positive then the flywheel is
accelerated .
When (Ti – To ) is negative then the flywheel is
retarded.

14. Coefficient of fluctuation of speed (Cs)
• It is defined as the ratio of maximum
fluctuation of speed to the mean speed.
Cs = (Maximum fluctuation of speed) /mean
speed.
Cs = (ωmax – ωmin)/ω ----- (2)
ω = average or mean angular velocity given by
ω = (ωmax + ωmin)/2 --------(3)

15. • Similarly
Cs = (nmax – nmin)/n ----- (4)
n = (nmax + nmin)/2
Cs = 2(ωmax - ωmin)/(ωmax + ωmin) ---(5)
Cs =2(nmax – nmin)/(nmax + nmin) ---(6)

16. • The reciprocal of coefficient of fluctuation is
called as coefficient of steadiness. It is denoted
by letter ‘m’
• m = 1/ Cs
= ω/ (ωmax – ωmin)
= n/ (nmax – nmin)

17. Type of equipment Cs
Punching, shearing and forming
presses
0.200
Compressor ( belt driven) 0.120
Compressor ( gear driven) 0.020
Machine Tools 0.025
Reciprocating Pumps 0.040
Geared Drives 0.020
Internal Combustion engines 0.030
D.C generators (Direct drive) 0.010
A.C generators (Direct drive) 0.005

18. Fluctuation of energy in flywheel

19. • The turning moment diagram for a multi-
cylinder engine is shown above.
• The intercepted areas between the torque
developed by the engine and the mean torque
line is taken in order from one end are-
+a1, -a2, +a3, -a4, +a5, -a6
Let the energy in the flywheel at A = U

20. • Then from fig we have
• Energy at B = U + a1
• Energy at C = U + a1 – a2
• Energy at D = U + a1 – a2 + a3
• Energy at E = U + a1 – a2 + a3 –a4
• Energy at F = U + a1 – a2 + a3 –a4 – a5
• Energy at G = U + a1 – a2 + a3 –a4 – a5 +a6 = U

21. • Let the maximum and minimum energy occurs
at points E and B
Maximum energy in the flywheel
= U + a1 – a2 + a3 – a4
Minimum energy in the flywheel
= U + a1

22. • Maximum fluctuation of energy (U0) is defined
as the difference between the maximum
kinetic energy and minimum kinetic energy in
the cycle.
• (U0) = UE – UB
= (U + a1) – (U + a1 – a2 + a3 – a4 )
= + a2 - a3 + a4

23. Coefficient of fluctuation of energy (Ce)
• It is defined as the maximum fluctuation of
energy to the work done per cycle. It is denoted
by Ce
• Ce = (maximum fluctuation of energy)/(Work done per cycle)
• = U0 / work done per cycle
Work done per cycle = area below mean torque line
form 0 to 360 degrees
Work done per cycle = 2πTm for 2 stroke engine.
Work done per cycle = 4πTm for 4 stroke engine.

24. Type of engine Ce
Single cylinder, double acting steam
engine
0.21
Cross-compound steam engine 0.096
Single cylinder, four-stroke petrol
engine
1.93
Four cylinder, four-stroke petrol engine 0.066
Six cylinder, four-stroke petrol engine 0.031

25. Solid Disk Flywheel
• The simple type of flywheel is a solid circular
disk. The mass moment of inertia of this disk
is given by ;
I = (mR2 /2) - - - - (a)
Where
I = mass moment of inertia of disk(kg-m2)
m = mass of disk (kg)
R = outer radius of disk (m)

26. The mass of the disk is given by;
m = π R2 t ρ ----- (b)
t = thickness of disk (m)
ρ = mass density of flywheel material (kg/m3)
From (a) & (b)
I = (π/2) ρtR4 - ---- (c)

27. Fig: Solid Disk Flywheel

28. • There are two principal stresses in the rotating
disk – (i) tangential stress (σt) & (ii) radial
stress (σr)
(σt) = (ρv2 )/106 x [μ +3]/8 {1- (3μ+1)/(μ+3)(r/R)2 }
(σr) = (ρv2 )/106 x [μ +3]/8 {1-(r/R)2}
μ = Poisson’s ratio (0.3 for steel and 0.27 for cast iron)
v = peripheral velocity (m/s) (Rω)
(σt) max = (σr) max = (ρv2 )/106 x [μ +3]/8 at (r = 0)

29. Rimmed flywheel

30. Construction of rimmed flywheel
Rimmed flywheel consists of following
components
1) Rim (outer ring) 2) Hub (Inner hub)
3) Arms (4 to 6)
Arms possesses elliptical cross-section for increased
stress absorption, while the rim has rectangular
c/s.
It is difficult to determine the mass moment of
inertia of rimmed flywheel ,due to its
complicated shape

31. Design of rimmed flywheel
• Analysis of flywheel is made by using any one
of the following assumptions
• 1) the spokes, the hub, and the shaft do not
contribute any moment of inertia, entire
moment of inertia is due to the rim alone
• 2) the effect of shaft, hub, spokes is to
contribute 10% while the arm contributes 90%
of the required moment of inertia.

32. • The mass moment of inertia of the rim is given by
Ir = C x I
Ir = Mass moment of inertia of the rim about its axis
of rotation
C = Factor of mass moment of inertia
= 1 (for assumption no.1)
= 0.9 (for assumption no.2)
I = Mass moment of inertia of the entire rimmed
flywheel

33. Let b,t = width and thickness of the rim
D = Mean diameter of flywheel
R/Rm = Mean radius of flywheel,
ρ = Density of flywheel material,
a1,b1 = elliptical dimensions (mm)
mr = mass of the rim of flywheel
kr = radius of gyration of the rim about its axis
of rotation

34. • The mass moment of inertia of the rim is given
by Ir = mr kr
2
• For a rim kr = Rm hence ;
Ir = mr Rm
2
Mass of rim = volume of rim X mass density
= (frontal circular surface area) X (width) X ρ
mr = (2πRm t) X (b) X (ρ) ; hence Ir is given as
Ir = 2 π Rm
3 t b ρ

35. Stresses in rimmed flywheel
The following types of stresses are induced in the
rim of a flywheel:
1. Tensile stress due to centrifugal force,
2.Tensile bending stress caused by the restraint
of the arms,
3. The shrinkage stresses due to unequal rate
of cooling of casting. It is taken care of by a
factor of safety.

36. 1.Tensile stress due to the centrifugal force
• Let b = Width of rim,
t = Thickness of rim,
A = Cross-sectional area of rim = b × t,
D = Mean diameter of flywheel
R = Mean radius of flywheel,
ρ = Density of flywheel material,
ω = Angular speed of flywheel,
v = Linear velocity of flywheel, and
σt = Tensile or hoop stress.

37. Fig. (a) Flywheel

38. Fig.(b) C/S of a flywheel arm

39. Consider a small element of the rim as shown
shaded in Fig. b. Let it subtends an angle δθ
at the centre of the flywheel.
Volume of the small element = A.R.δθ
Mass of the small element, dm = Volume ×
Density
= A.R.δθ.ρ = ρ.A.R.δθ

40. centrifugal force on the element, dF = dm.ω2.R
= ρ.A.R.δθ.ω2.R
= ρ.A.R2.ω2.δθ
Vertical component of dF = dF.sin θ
= ρ.A.R2.ω2.δθ sin θ

41. Total vertical bursting force across the rim
diameter X-Y
= ρ.A.R2.ω2
= ρ.A.R2.ω2
= 2 ρ.A.R2.ω2 ----- (1)

43. • This vertical force is resisted by a force of 2P,
such that
2P = 2σt × A ----- (2)
From (1) & (2) we have
2 ρ.A.R2.ω2 = 2σt × A
σt = ρ.R2.ω2 = ρ v2
Is the tensile stress due to centrifugal force

44. 2. Tensile bending stress caused by restraint of the arms

45. • The tensile bending stress in the rim due to
the restraint of the arms is based on the
assumption that each portion of the rim
between a pair of arms behaves like a beam
fixed at both ends and uniformly loaded, as
shown in Fig. above, such that length between
fixed ends, as
l =

46. = (2 π R)/n where n is number of arms
{UDL (w) per metre length} = {Centrifugal Force
between the pair of arms}
w = b.t.ρ.ω2.R N/m
Maximum bending moment is given by
M = (w l2 )/12
M = (b.t.ρ.ω2.R)/12 x {(2 π R)/n}2

47. Section modulus Z = bt2/6
σb = M/Z
= (b.t.ρ.ω2.R)/12 x {(2 π R)/n}2 x 6/bt2
σb = (19.74 ρ.ω2.R3 ) n2t = (19.74 ρ.v2.R) n2t
In practical conditions both stresses co-exist, so
G.Lanza suggested that the resultant stress in
the rim is (¾)th of σt & (¼)th of σb

48. • Resultant stress in the rim is given as
σ = (3/4)σt + (1/4) σb
σ = ρv2[(3/4) + (4.935.R)/n2t ]

49. Stresses in flywheel arms

50. The following stresses are induced in the arms
of a flywheel.
1. Tensile stress due to centrifugal force acting
on the rim.
2. Bending stress due to the torque transmitted
from the rim to the shaft or vice versa.
Considering each one separately

51. 1. Tensile stress due to centrifugal force acting on the rim
• Due to the centrifugal force acting on the rim,
the arms will be subjected to direct tensile
stress whose magnitude is same as discussed
in the previous article.
∴ Tensile stress in the arms, is σt1 = (3/4)σt
= ¾ ρ v2

52. 2.Bending stress due to the torque transmitted
Due to the torque transmitted from the rim to
the shaft or vice versa, the arms will be
subjected to bending, In order to find out the
maximum bending moment on the arms, it
may be assumed as a cantilever beam fixed at
the hub and carrying a concentrated load at
the free end of the rim as shown in Fig. above

53. Let Mt = Maximum torque transmitted by the
shaft,
R = Mean radius of the rim,
r = Radius of the hub,
n = Number of arms, and
Z = Section modulus for the cross-section of
arms.

54. • Load at the mean radius of the rim is
F = Mt/R
Load on each arm = Mt/(R.n)
Maximum bending moment on the arm at the
hub M = {Mt/(R.n)}[R-r]
• Bending stress in arms,
σb1 = M/Z = {(Mt / R n) (R – r)}/Z
Total tensile stress on the arms = σ = σt1 +σb1