6. windmill gear assembly

Group Members:
PC Mubashar Sharif (2009-462)
NS Abdullah Bin Masood (2009-204)
NS Haider Iqbal (2009- )
Dated: 20 January 2014
Windmill
Gear
Assembly
Design of Machine
Elements
Sir Raja Amer
Azeem

Design of Machine elements
1
Wind Mill
Actually the idea of our project is to design different components used in the wind mill pump which
includes shaft on which the fan is mounted, bearings in which shaft is mounted and after that gears and
the process which is usually used for joining of two components on a single point/place.
A brief introduction:
Wind energy is the actually the conversion of wind
energy into a some kind of useful form of energy,
such as using wind turbines to make electricity, wind
mills for mechanical power, wind pumps for pumping
water or drainage, or sails to propel ships as
described in the applications.
It works when wind hits the windmills blade it spins
the rotor and the gearbox and the mechanic box
activate the rotor and the wires inside take all the
wind and make it into energy and the electricity poles
carry the electricity and powers and gives 1000
houses in an hour or so.
Here we are describing about the windmill pump.
Windmill for pumping the water:
Actually windmill pump is a device that pumps the
water from different surfaces. Water is pumped to
the surface when the pump rod raises the piston.
When piston check valve closes and holds the water
above the piston. As the piston rises, water is moved
up the pipe towards the surface. Water is also drawn
into the lower section of the pump cylinder through a
screen and the lower check valve. When the pump
rod reverses and begins to descend, the lower check
valve closes and the piston check valve opens
allowing the water in the cylinder to pass through the
piston check valve and become trapped above the
piston when the check valve closes. This cycle is
constantly repeated as the wind wheel turns, operating the reciprocating mechanism in the gearbox,
which operates the pump rod and pump

2
The gear assembly we are trying to design and fabricate is an assembly which has 2 gears of different
diameters and number of teeth, fixed mounted on the two different shafts of same geometry. The two
gears are meshed in such a way with a certain input on one shaft you get a different speed (rpm) and
torque on the other. For this purpose, the two shafts have been mounted on two same bearing (since
both of the shafts are same) and fixed in a support.
In our project, following are the major components that need to be designed:
- Gears
- Shafts
- Bearings
- Welding
We are a group of three members and have divided the tasks such that each member is responsible for
one component. The name of member and his respective job/task is mentioned below:
Abdullah Bin Masood: Designing and calculations of Gears
Mubashar Sharif: Designing and calculations of Shafts
Haider Iqbal: Calculations and selection of Bearings
First of all the calculations for the shaft have been made and finalized and then before fabrication of
shaft we went to buy the bearings of the appropriate size that could fit mount on our shaft but after
moving in the market we found out that the bearings are not available in every, every single size we
design. There are in fact, some certain specific sizes of the bearings. Same happened to our case. Our
shaft was basically much smaller than the bearings size available so we had to revise our design and with
everything same but the Factor of safety increased significantly due to over-designing of the shaft.
However, in order to keep the report brief and to the point, only finalized calculations have been
included.

3
Design Layout/Modeling:

4
Wind Power Calculation:
Shaft is basically a mechanical component that transmits power.
We know that from wind the available energy to the pump is
𝑷 𝒂𝒗𝒂𝒊𝒍𝒂𝒗𝒍𝒆=
𝟏
𝟐
ƿɛ𝑨𝑽 𝟑
𝑉 𝑤𝑖𝑛𝑑=20𝑚/𝑠𝑒𝑐
Diameter of shaft=1cm
𝜌 = 1.225𝑘𝑔/𝑚3
𝜀 = 0.59
By putting values in above equation we get
𝑃𝑎𝑣𝑎𝑖𝑙𝑎𝑣𝑙𝑒= 190.9562 𝑤𝑎𝑡𝑡𝑠
Assumptions
Blade tip speed = 1/√2 ×wind speed
V=rω
by putting values we get
ω=1308.917 rev/min
N=108
(for infinite cycles assumption)
Calculations of Shafts:
Shaft is basically a mechanical component that transmits power. In our assembly two shafts are being
used. Since both are same therefore we will make calculations for one shaft only.
Assuming the following constraints/given condition for our system:
Power to be transmitted: 𝑃𝑎𝑣𝑎𝑖𝑙𝑎𝑣𝑙𝑒= 190.95 𝑤𝑎𝑡𝑡𝑠
Angular velocity: ω=1308.917 rev/min
ω=136.97 rad/sec

5
Designing for the infinite life: N=108
(since shafts will be continuously
subject to fatigue in our case)
Diameter of shaft= 0.7 cm
The material chosen is CD 1020 steel
Following the iterations approach:
1 → P=Tω
Torque=T=190.95/136.97
T = 1.39 Nm
2 → Shear stress=S=16T/𝜋𝑑3
After putting values
S =20.65 Mpa
Now since,
3 →𝑺 𝒆=𝑲𝒂×𝑲𝒃×𝑲𝒄×𝑲𝒅×𝑲𝒆×𝑲𝒇×𝑺𝒆′
𝑆 𝑢𝑡 = 470𝑀𝑝𝑎 𝑆 𝑦=390Mpa
𝑆𝑒′
= 0.5𝑆 𝑢𝑡 = 0.5× 470 = 235𝑀𝑝𝑎
Ka=a𝑆 𝑢𝑡
𝑏
=4.51× 470−0.265
= 0.8832
Kb=1.24(1 × 10−2
)−0.1.07
=0.969
Kc=0.59 for torsion
Kd=Ke=Kf=1
Se=0.8832×0.969×0.59×235
Se=118.66Mpa
4→a=
[𝒇 × (𝑺 𝒖𝒕)]ˆ𝟐
𝑺 𝒆
f=0.9
b=
−1
3
× log(𝑓𝑆𝑢𝑡/𝑆𝑒)

6
After putting values we get
a=1507.91Mpa
b=-0.189655
𝑆𝑓=a𝑁 𝑏
Put values in above where N=108
cycles
We get
𝑆𝑓= 47.53 Mpa
This is the fatigue strength for the rotating shaft.
Now from the stress concentration factor chart using our fillet radius and diameter of shaft we
calculated Ks after which we calculated Kfs:
Where,
Notch radius=0.508 mm
Notch sensitivity=q(s)=0.64
Kt=2.2
Kfs=1+q (Kt-1)
→Kfs=1.76
Now
→Ƭmax=20.06 × 1.76 = 35.30 Mpa
Ƭa=35.30/2=17.65Mpa
Ƭm=17.65Mpa
From soderberg criteria:

7
Ƭa +Ƭm =
0.577𝑆𝑦
𝑛
Putting values we get;
n=
0.577×390
14.88
= 6
So,
Factor of safety = n= 6
Comment:
This factor of safety is reasonable and this design can easily sustain under given conditions.
Selection of Bearing:-
Here we have chosen deep groove ball bearing and not the roller bearing because roller bearing is used
for high speed applications but here we are dealing with small velocities.
As our machine‘s working service is 24hrs. So using Shigley’s Chap 11 (table 11-4)

8
Since, LD=55kh;
We have selected our bearing for 1 million cycles.
So,
Solution:
We have selected our bearing for a lifespan of 1 million cycles.
L10 = LRnR60 = 106
Cycles
F 𝐷 = mg
Desired life = ld= 55000 hrs
desired radial load = FD = 1.47N = 0.00147kN
L10 = LRnR60 = 106
Cycles
= 394.04N = 0.394kN

9
ω = nD = 500 rev
min⁄
C10 = FD(
LDnD60
LRnR60
)
1
a⁄
C10 = 0.00147 × (
55000 × 500 × 60
106
)
= 2.42kN
So using catalog provided by Timken catalog given in the book. Here, we have got the rating for the
bearing i.e. 2.42kN. Now if we see in the catalog, we get the nearest possible result i.e.
C10rated
= 5.07kN
For this, we have selected our bearing on the basis of our rated value from catalog Deep Groove Single
Row 02-Series.
Bore = 10mm
Outer Diameter= = 30mm
Width = 9mm
Fillet radius = 0.6mm
ds = 12.5mm
dH = 27mm
However, the bearings installed/used in our assembly are of the following characteristic value which was
nearest to the calculated one:
C10rated
= 6.89kN
For this, we have selected our bearing on the basis of our rated value from catalog Deep Groove Single
Row 02-Series.
Bore = 12mm
Outer Diameter = 32mm
Width = 10mm
Fillet radius = 0.6mm
ds = 14.5mm
dH = 28mm

10
Conclusion:
We have selected our bearing using standard catalog provided by Timken, but unfortunately this type of
bearing was not available so we have to change our bearing according to the availability. So, in our
project single side shielded bearing No. Z-0009 has been used.
Gear designing:
Assumptions:
The assumptions taken for our gears are as under:
Failure acts only the yielding point. We have used here spur gear because they are for low loads only.
Actually Spur Gears are the most common type of gears. They are mostly used where the application is
of rotary type while maintaining constant torque and speed.
Material selection:
The main problem while designing was that either we will get our required material or not. Here the
material used for the manufacturing of gears was mainly dependent on working conditions like wear,
noise etc. So, here we have used annealed steel for our working.
The Gear:
Pressure angle = θ = 20
Number of teeth =N= 70
Face width = b = 8 𝑚𝑚
Power = 𝑃 = 192.95 watt = 0.258 hp
n = 1308 rev/min=136.97 rad /sec
T = P/n =190.95/136.97=1.39 N-m
Pitch diameter= Dp = 70 mm=0.07 m
Diameteral pitch = Pd = number of teeth/pitch dia
Pd=70/63.5

11
Pd = 1.102
teeth
mm
Module=m=
Dp
number of teeth
=63.5/70=0.90 mm
Circular pitch=p= (π*Dp)/N= (π*m)=2.84 mm
Now there will be many terms to obtain as shown in following figures

12
Transmitted force = Ft =
2T
Dp
= 43.77 N
Normal force = Fn = Ft tan θ = 15.93 N
Resultant force = Fr =
Ft
cos θ
= 46.58 N
Surface speed = Vm =
πDpn
12
=2.27 m/sec
Now , we use Lewis equation
Fs =
SnYb
Pd
Where
Fs = Allowable dynamic bending force
Y = Lewis form factor
The material used is 4140 annealed steel
Sut = 655 MPa
Sn = 0.5 Sut = 327.5 MPa
Y = 0.429 (from table for 70 teeths)
Fs =
SnYb
Pd
= 1.019 kN
Dynamic force = Fd =
600 + Vm
600
Ft
= 46.76 N
Design Method:
Strength of gear tooth should be greater than the dynamic force
Fs ≥ Fd
Including factor of safety Nsf:
Fs
Nsf
≥ Fd
So , we have

13
Fs
Fd
≥ Nsf
𝐂𝐨𝐦𝐦𝐞𝐧𝐭 𝐨𝐧 𝐝𝐞𝐬𝐢𝐠𝐧
21 is the factor of safety for designing of big dia gear. This shows that our gear is over designed.
This is a huge value and logically we should review our design but in our case neither the cost nor the
size matter that much.
Pinion:
Pressure angle = θ = 20
Number of teeth = 18
Face width = b = 7.6 mm
Power = 192.95 watt = 0.258 hp
N1/N2 = n2/n1
N2=number of teeth of gear
N1=number of teeth of pinion
n2=speed of gear
So,
n1=speed of pinion=5199 rev/min=544.43 rad/sec
n = 544.43 rad/sec
N1/N2 = T2/T1
N2=number of teeth of gear
N1=number of teeth of pinion
T2=torque of gear
So,
T1=torque of pinion=0.18 N-m
Pitch diameter= Dp = 18 mm

14
Diameteral pitch = Pd = number of teeth/pitch dia
Pd =
18
18
= 1
teeth
mm
Module=m=
Dp
number of teeth
=18/18=1 mm
Circular pitch=p= (π*Dp)/N= (π*m)=3.14 mm
Transmitted force = Ft =
2T
Dp
= 20 N
Normal force = Fn = Ft tan θ = 7.27 N
Resultant force = Fr =
Ft
cos θ
= 21.2 N
Surface speed = Vm =
πDpn
12
= 2.56
m
sec
Now , we use Lewis equation
Fs =
SnYb
Pd
Where
Fs = Allowable dynamic bending force
Y = Lewis form factor
The material used is 4140 annealed steel
Sut = 655 MPa
Sn = 0.5 Sut = 327.5 MPa
Y = 0.309 (from table for 18 teeths)
Fs =
SnYb
Pd
= 769.1 N
Dynamic force = Fd =
600 + Vm
600
Ft = 21.29 N
Design Method:
Strength of gear tooth should be greater than the dynamic force
Fs ≥ Fd

15
Including factor of safety Nsf
Fs
Nsf
≥ Fd
So , we have
Fs
Fd
≥ Nsf
This comes out to be 36
So, the factor of safety is 36
𝐂𝐨𝐦𝐦𝐞𝐧𝐭 𝐨𝐧 𝐝𝐞𝐬𝐢𝐠𝐧:
36 is the factor of safety for designing of big dia gear. This shows that our gear is over designed.
This is a huge value and logically we should review our design but in our case neither the cost nor the
size matter that much.
Bending and Wear of Gear and Pinion:
And
(Dp)P = 18 mm=0.018m
(Dp)G = 70 mm=0.07 m
Vm =
πDpn
12
= 2.56
m
sec
Wt= 74 N
as
A = 50 + 56(1 − B) =60.08
B = 0.25(12 − Qv)2/3 = 0.82
Qv=6

16
So
Kv=1.29 m/sec
From following table
Lewis factor= Y=0.309 (for pinion )
Y=0.439( for gear )
From this equation
(Ks)P= 1.03
(Ks)G= 1.05
Cmc = 1 (for uncrowned teeth )

17
Cpf = 0.019
Cpm=1
Cma=0.15
Ce=1
Now as
Km= 1.23
Assuming constant gear thickness Kb=1

18
Speed ratio =Ng/Np
Mg=70/18= 3.88
Pinion cycles =10^8
Gear cycles =10^8/Mg=10^8/3.88
(YN)P=0.977
(YN)G=0.996
For grade 1 steel
HBP = 240 and HBG = 200
From following figure
(St)P = 216 MPa
(St)G =190 MPa
Pinion tooth Bending:
[(σ)P]B=23.7 MPa

19
[(SF)P ]B= 2.5
Gear tooth bending:
[(σ)G]B=23 .7 Mpa
[(SF)G]B= 3.6
Pinion tooth Wear:
[(σ)P]W=205 MPa
[(SF)P]W=1.4
Gear tooth Wear:
[(σ)G}W=207 Mpa
[(SF)G]W=1.36
Result of analysis :
For the pinion, we compare [(SF)P ]B with [(SF)P]W ^2, or 2.5 with =1.96, so the
threat in the pinion is from wear.
For the gear, we compare [(SF)G]B with [[(SF)G]W]^2
, or
3.6 with 1.85 , so the threat in the gear is also from wear.

20
Final Results:
Above calculations show that our design is over designed, therefore, our designed gears assembly can
withstand much larger conditions than the given constraints. Although it will be costly and heavier in
weight and size as well, but in our case, these are not the parameters of design success.
Applications:
Windmills are generally used to harness power of wind to perform mechanical work as well as
generation of electricity. Applications of windmills are as follows –
o Large Scale Electric Power Generation- Huge numbers of windmills are installed on a single site
called as wind farms to produce large scale electric power from wind. Small Scale Electric Power
Generation- Single windmill can be installed in remote locations to produce electricity, where
connection to the general electric network is not available
o .Pumping Water for Irrigation- Windmill is used for pumping of water from the well for
agricultural purposes.
o Milling Grain- Windmills are used to grind grain, cut wood and supply water in some locations
where electricity is not available.
o Windmills in Warfare- Initially, windmills were used as watchtowers because they were
constructed on the tops of hills or mountains.
o The energy produced by windmill does not generate any pollution, which is an infinite source of
energy.
o As windmill generates energy from wind it does not require fuel.
o It does not create greenhouse gases.
o It is also used for driving all sorts of mechanical devices for manufacturing.
o Windmill is generally used to convert wind energy into rotational energy with the help of blades.
o Energy produce by windmill is cheap.
o It is a renewable energy source.
o When wind energy is produced it does not create any ill effect on environment.
o Wind power is the cheapest source of energy that is available in current arena.
o The operating cost of windmill is very low.
o Wind energy provides jobs in production, maintenance, research and installations of windmills,
as windmills are becoming popular.
o Windmills produce electricity at 5 cents kilowatt per hour.
o Current windmills with 2 or 3 thin blades connected to a long pole type base add a more
modern aesthetic to land.

6. windmill gear assembly

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