Theory and Design of mechanical
system
Topic: 2 stage cylindrical-bevel speed reducer
Group: G2
1. 111910138 Michael Susngi
2. 111910139 Pranav Jadhav
3. 111910140 Rahul Jadhav
4. 111910141 Ravindra Shinde
Team members
● Problem Statement
“Design a suitable gearbox for box shipping conveyor which used to carry 3 boxes of 50kg each
at a speed of 28m/min.”
• Specifications :
1. Load acting on belt: 3 x 50kg
2. Speed of conveyor
3. Diameter of roller
4. Weight of conveyor belt
5. Friction coefficient between belt & roller
• Assumptions :
1. Material is assumed to be isotropic and homogeneous.
2. All types of vibrations are neglected.
3. Thermal stresses are neglected.
4. Gradual and uniform loading condition.
150kg
28m/min
200mm
50kg
0.67
(Reference : https://www.rubbercal.com/industrial-rubber/heavy-black-conveyor-belt.html & https://www.engineersedge.com/coeffients_of_friction.htm )
● Design Requirements
1. Nominal operating conditions of industrialized locations
2. Input and output shafts STD. size for typical couplings
3. Usually low shock levels, occasional moderate shock
4. Continuous operation
5. Low maintenance
6. Competitive cost
• Significance of gear drive:
1. It is a positive drive, and the velocity ratio remains constant.
2. The center distance between the shafts is relatively small, which results in compact
construction.
3. It can transmit very large power, which is beyond the range of belt or chain drives.
4. It can transmit motion at very low velocity, which is not possible with the belt drives.
5. The efficiency of gear drives is very high, even up to 99 per cent in case of spur gears.
6. A provision can be made in the gearbox for gear shifting, thus changing the velocity
ratio over a wide range.
• Types of gear:
1. Spur gear
1. Easy to design and manufacture.
2. Suitable for low torque
3. Make noise at high-speed operation
2. Bevel gear
1. Angular torque transmission.
2. Not desirable for High-speed reduction.
3. Compact design and highly efficient
4. Helical gear
3. Worm and Worm wheel
1. Large speed reduction.
2. Large increase in torque.
3. Self-Locking Mechanism.
4. Less efficient
1. Gradual teeth engagement
2. More durable
3. Less noise
4. Produce large thrust force due to helix angle
• Types of gear:
• Selection of gearbox
2 stage cylindrical bevel speed reducer.
a) In our case motor is in parallel to the conveyor
belt. so, we have to transmit power in
perpendicular direction. So perfect solution is
bevel gear.
b) But along with that we have to achieve high
reduction ratio of 30. so, we also used spur
gears to obtain the desired reduction.
• Orthographic views
with notations:
PART NO. PART NAME TYPE QTY
1 Driving spur gear Custom 1
2 Driven spur pinion Custom 1
3 Driving bevel pinion Custom 1
4 Driven bevel gear Custom 1
5 Input shaft Custom 1
6 Output shaft Custom 1
7 Intermediate shaft Custom 1
8 Input shaft Bearings STD. 2
9 Output shaft Bearings STD. 2
10 Intermediate shaft
Bearings
STD. 2
11 Upper casing Custom 1
12 Lower casing Custom 1
• Part list :
PART NO. PART NAME TYPE QTY
13 Casing Nut STD. 4
14 Casing bolt STD. 4
15 Foundation bolt STD. 4
16 Bearing cap bolt STD. 20
17 Input shaft oil seal STD. 1
18 Output shaft oil seal STD. 1
19 Banjo bolt STD. 1
20 Input shaft key STD. 1
21 Output shaft key STD. 1
22 Intermediate shaft
key 1
STD. 1
23 Intermediate shaft
key 2
STD. 1
24 Retaining ring STD. 3
25 Bearing cap Custom 5
• Selection of motor
i) Power required to convey boxes = µN x V
= 0.67 x (150+50) x 9.81 x
28
60
= 613.5 watt
ii) To encounter different losses in the gearbox and couplings 20% overload factor is
considered.
Thus, total power required = 613.36 + 613.36 x 20% = 736.03 watt
iii) Reasons for using Induction motor
1. Useful in low-speed operation
2. Frequent maintenance not needed as like DC
(Due to carbon brushes)
3. Suitable in operations where high starting
torque is not required such as in conveyors
The standard motor available in induction motor catalogue to obtain required power is of
0.75kW(1HP)
(Reference: https://3.imimg.com/data3/QM/MP/MY-3212916/three-phase-induction-motors.pdf, Table no. 3, pg. No. 3)
“3-phase Squirel Cage Induction Motor, 50Hz, 0.75kW(1HP)”
• Selection of motor
Electrical Specification Mechanical Specification
Number of poles : 8 Speed : 690 RPM
Full load current : 2A Efficiency : 70.0%
Operating voltage : 415V ± 10% Output : 0.75 kW
Power factor : 0.73 Frame : 100L
Name of components Abbreviation
used
Input shaft P
Driving Bevel Pinion G1
Driven Bevel gear G2
Intermediate Shaft Q
Driving Spur pinion S1
Driven Spur gear S2
Output shaft R
Input Shaft Bearing B1
Output Shaft Bearing B2
Intermediate Shaft Bearing B3
• Gear train Layout with
abbreviations used
MOTOR
ROLLER
• Power flow diagram
Motor
Coupling
1 P
G1
G2
Q
S1 S2 R
ROLLER
Coupling
2
• Force analysis of whole system
(3D)
• Gear ratio calculations
1. After selection of motor, Input speed to gearbox = 690 rpm
2. Speed at output of gearbox is calculated as;
Velocity of roller = radius of roller x angular velocity of roller
V = r x (
2π𝑁
60
) ; ∴ 𝑁 = 𝑉 ×60
2𝜋𝑟
=
28×60
60×2𝜋×0.2
= 23.3 𝑟𝑝𝑚
Thus, Output speed of gearbox = 23.3 rpm
3. That is, we have to reduce speed from 690rpm to 23.3rpm in 2 stages
4. After doing several permutations and combinations, we get the suitable configuration of,
5:1 in 1st stage (in bevel pair) ∴ Speed of shaft Q =
690
5
= 138 rpm &
6:1 in 2nd stage (in spur pair) ∴ Speed of shaft R =
138
6
= 23 rpm
Shaft Type Speed (RPM)
P 690
Q 138
R 23
• Bevel pair design
i) Part No. is 3,4 and part name is Driving bevel pinion and Driven bevel gear
ii) Function: Transmits power in perpendicular direction and also provides speed reduction
iii) Free body diagram
Material Syt
(𝑘𝑔𝑓/𝑚𝑚2
)
Sut
(𝑘𝑔𝑓/𝑚𝑚2
)
Pinion 40 Ni 1 Cr 1
Mo 28
88 110
Wheel C 55 Mn75 46 76
iv) For reduction ratio of 5:1 material selected are;
(Refer: PSG data book, Table 5, pg.no. 8.4 & PSG data book, pg.no. 1.12 ,1.15)
• Bevel pair design
v) Two orthographic views
SEC. F. V. SEC. R. H. V.
vi) Design calculations:
Let, rpm = 695 , module : 3 , reduction
ratio i = 5 , Number of teeth on pinion
Zp = 20
vii) Verifying Design Against Beam Strength:
viii) Verifying Design Against Wear Strength
ix) Final dimensions:
Notations Meaning Values
𝑚 module 3
𝛼 Pressure angle 20˚
𝑍𝑝 Num of teeth on pinion 20
𝑍𝐺 Num of teeth on gear 100
𝐷𝑝 p.c.d. of pinion (G1) 60 mm
𝐷𝐺 p.c.d. of gear (G2) 300mm
𝐴𝑜 Cone distance 153 mm
𝑏 Face width 38.25 mm
𝛾 Pitch angle 11.30˚
• Spur pair design
i) Part No. is 1,2 and part name is Driving spur pinion and Driven spur gear
ii) Function: Transmits power in parallel shafts and speed reduction
iii) Free body diagram
Material Syt
(𝑘𝑔𝑓/𝑚𝑚2
)
Sut
(𝑘𝑔𝑓/𝑚𝑚2
)
Pinion (S1) 40 Ni 1 Cr 1
Mo 28
88 110
Wheel (S2) C 35 Mn75 46 76
iv) For reduction ratio of 6:1 material selected are;
(Refer: PSG data book, Table 5, pg.no. 8.4 & PSG data book, pg.no. 1.9)
• Spur pair design
v) Two orthographic views
vi) Design calculations (Beam & wear Strength)
vi) Design calculations (Beam & wear Strength)
vi) Final dimensions of Spur gear
Notations Meaning Values
𝑚 module 3
𝛼 Pressure angle 20˚
ℎ Tooth depth (both) 6.75 mm
𝑍1 Num of teeth (pinion) 18
𝑍2 Num of teeth (Gear ) 108
𝐷1 P.C.D. of pinion (G1) 54 mm
𝐷2 P.C.D. of gear (G2) 324 mm
𝐷𝑎 Addendum Diameter(Gear) 330 mm
𝑑𝑎 Addendum Diameter(pinion) 60 mm
𝐷𝑏 Base diameter(Gear) 316.5 mm
𝑑𝑏 Base diameter (pinion) 46.5 mm
• Design of shafts
1. Intermediate shaft
i) Part no. is 7 and part name is intermediate shaft
ii) Design of based on ASME code
iii) Function: transmission of torque and power
iv) The material selected for shaft is Plain carbon steel with grade 40C8 because,
this steel is used for making shafts, crankshaft, automobile axle
(Refer: PSG design data book, pg.no. 1.9, 1.10)
v) Properties of material:
1. Sut = 580N/mm2
2. Syt = 330N/mm2
vi) Allowable stress calculation:
Ʈmax = 0.30xSyt = 99N/mm2 or Ʈmax = 0.18xSut = 104.4N/mm2
hence Ʈmax = 99N/mm2 … (which is less)
Also, shaft is keyed for mounting of gears,
∴ Ʈmax = 0.75x99 = 74.25N/mm2
vii) FBD & BMD of shaft in respective plane
∴ Maximum bending moment at S1 = 44384.182 + 126856.842 = 134,397.22 Nmm
viii) Thus, Mb = 134397.22 Nmm & Mt = 77887 Nmm
ix) Take Kb = Kt = 1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334)
x) By maximum shear stress theory,
τ𝑚𝑎𝑥 =
16
ߨ𝑑3 (𝑘𝑏 .
𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 =
16
ߨ𝑑3 (1.5 × 134397.22)2+(1.5 × 77887)2
∴ 𝑑 = 25.19𝑚𝑚
But we haven’t considered the weight of gears while calculating diameter of shaft, hence we
consider the shaft of diameter d = 35mm
xi) Finally,
Shaft dia. for bearing mounting d 35mm
Shaft dia. for gear mounting 1.1d 38.5mm
Shaft dia. for resting gear on one side 1.2d 42mm
xii) Checking stress concentration at critical sections
As there are steps/shoulders in shaft ;
there has to be a stress concentration.
To reduce this stress concentration fillet is
provided
From BMD it is clear that on spur gear(S1)
side bending moments are higher.
Hence critical sections X-X’ and Y-Y’ are
labelled in adjoining figure
We have to find stress(𝜎)at these sections and then comparing with Endurance limit (Se )
If σ > Se ,then we have to change the dependent dimension such as 1.1d,0.1d,1.2d and
if σ < Se , then we can say that Design is safe and correct.
A) At section X-X’ for Bending and Torsion
B) At section Y-Y’ for Bending and Torsion
• Selection of bearings
1. Intermediate shaft bearings
i) Part No. 3 and part name is Intermediate shaft bearing
ii) Selected type of bearing for given application is Taper roller bearing, because it
has following advantages-
It can carry combine radial and axial loads
(Refer: PSG Data book, pg.no. 4.1)
The construction, which involves two bearings
with their fronts facing each other, is called
‘face-to-face’ or direct mounting.
(Refer: DOME by Bhandari, Fig. 15.11,case 2(a)
iii) Expected bearing life in hrs. L10h = 25000hrs
Machines for use 8hrs/day ad fully utilized – E.g. cranes for bulk goods, ventilating fans,
countershaft of gearboxes, the recommended life is from 20000 to 30000h
(Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5)
iv) L10 =
60𝑛L10h
106 =
60×138×25000
106 = 202.5 millions revolution
since, intermediate shaft speed is running at 138 RPM
v) Trial 1 Tentatively, we select Bearing 32207A. (Refer: PSG design DATA, pg.no. 4.25)
vi) Bearing A:
FrA = 1156.152N and FaA = 926.8N
FaA
FrA
=
926.8
1156.152
= 0.8 > e = 0.37
P =Equivalent bearing load = X.Fr + Y.Fa = 0.4x1156.152+ 1.6x926.8 = 1945.34N
C =dynamic capacity = P(𝐿10)0.3
x (Load factor) = 1945.34 (202.5)0.3
(1.5) = 14355.37N
Hence, Bearing 32207A (C = 57600N) is suitable. (Refer: PSG design DATA, pg.no. 4.4 & 4.25)
vii) Similarly for Bearing B:
FrB = 2378.689N and FaB = 743.34N
Fa𝐵
Fr𝐵
=
743.34
2378.689
= 0.31 < e = 0.37
P = Equivalent bearing load = X.Fr + Y.Fa = 1x1156.152+ 0x926.8 = 1156.152N
C = dynamic capacity = P(𝑳𝟏𝟎)𝟎.𝟑
x (Load factor) = 1156.152 (202.5)0.3
(1.5) = 8531.66N
Hence, Bearing 32207A (C = 57600N) is suitable. (Refer: PSG design DATA, pg.no. 4.4 & 4.25)
vii) The dimensions of bearing
• Selection of Keys for gear mounting on intermediate shaft
i) Part No. is 22 and Part Name is Intermediate shaft key 1.
ii) Function: Mounting bevel gear (G2) on intermediate shaft
iii) TWO orthographic views of key iv) FBD of key
v) The martial selected for Key is plain carbon steel with grade C50 ,Because-
This steel is suitable for making keys, shaft, cylinder, and machined components requiring
wear resistance (Refer: PSG design data book, Page No. 1.9 & 1.10)
vii) Properties of C50 :
1. Sut = 660 N/mm2
2. Syt = 380 N/mm2
viii) Factor of safety chosen is 1.5 because, key failure is more economical than gear or shaft
failure in any extreme conditions.
ix) Allowable stresses:
1. 𝜎 =
380
1.5
= 253.33N/mm2 and 2. 𝜏 =
0.5×380
1.5
= 126.67N/mm2
x) Shaft diameter for gear mounting = 1.1d = 38.5mm ; The dimensions of parallel key obtained
from table is bxh = 12x8 and keyway depth is 6mm (Refer: PSG design data book, Page No. 5.24)
xi) The effective length is found using empirical relations available for key
∴ l = 1.5×38.5 = 58mm < 84mm bxhxl = 12x8x58
xii) Checking for induced stresses;
𝜏 =
2𝑇
𝑏ℎ𝑙
=
2×77887
12×8×𝟓𝟖
= 28N/mm2 < 126.67N/mm2
Also, 𝜎 = 2 𝜏 = 56N/mm2 < 253.33N/mm2
Hence, design is safe
i) Part No. is 23 and Part Name is Intermediate shaft key 2.
iv) Take l = 25mm and check for induced stresses;
𝜏 =
2𝑇
𝑏ℎ𝑙
=
2×77887
12×8×𝟐𝟓
= 64.90N/mm2 < 126.67N/mm2
Also, 𝜎 = 2 𝜏 = 129.81N/mm2 < 253.33N/mm2
iii) Orthographic views , FBD, Material for key , Fos are same as earlier case
Also dia. for gear mounting is same. Hence bxh = 12x8
only thing changes is length of key which should be < 27mm
ii) Function: Mounting spur gear (S1) on intermediate shaft
v) Hence, our design is safe
bxhxl = 12x8x25
• Selection of circlips (External type)
i) Part No. is 26 and Part Name is circlip 1
ii) Function: Locking bevel gear (G2) in its position on intermediate shaft
iii) Detailed views with design notations
iv) Commonly used steel for circlip is C65 (same as used in springs and washers)
(Refer: PSG design DATA,
pg.No. 5.3)
v) Properties of C65:
1. Sut = 750 N/mm2
2. Syt = 430 N/mm2
(Refer: PSG design DATA, pg.No. 1.9 & 1.10)
vi) In our case d1 = 1.1d = 38.5mm and for this circlip selected from light series (A) is
designated as Circlip, Light A 38 (Refer: PSG design DATA, pg.no. 5.4)
Notati
on
Meaning Value (mm)
d1 Shaft dia. For gear mounting 38
d2 Groove diameter o shaft for circlip 36
d3 Unexpanded dia. Of circlip 35.2
d4 Expanded dia. Of circlip 50.6
d5 Hole dia. For expanding circlip 2
s Thickness of ring in axial direction 1.5
b Thickness of ring in radial direction 5.8
a Height of ears of ring for expansion 4.2
v) Design
dimension getting
from catalogue for
above mentioned
circlip are
vii) As per catalogue, Maximum axial load carry by these circlip = 29100N >> 926.8N
so we can use it without any worry
• Design of shafts
2. Output shaft
ii) Design of based on ASME code
i) Part no. is 6 and part name is output shaft
iii) Function: Transmission
of torque and power to
conveyor roller
iv) TWO orthographic
views
iv) The material selected for shaft is Plain carbon steel with grade 40C8 because,
this steel is used for making shafts, crankshaft, automobile axle
(Refer: PSG design data book, pg.no. 1.9, 1.10)
v) Properties of material:
1. Sut = 580N/mm2
2. Syt = 330N/mm2
vi) Allowable stress calculation:
Ʈmax = 0.30xSyt = 99N/mm2 or Ʈmax = 0.18xSut = 104.4N/mm2
hence Ʈmax = 99N/mm2 … (which is less)
Also, shaft is keyed for mounting of gears,
∴ Ʈmax = 0.75x99 = 74.25N/mm2
vii) FBD & BMD of shaft in respective plane
∴ Maximum bending moment at S2 = Mb = 79258.0642 + 216238.72 = 230,306.35 Nmm
Also, Mt =
60 ×𝑃
2𝜋𝑁
=
60 ×0.75×106
2𝜋×23
= 311390.11 Nmm
Considering load factor = 1.2
Hence, Mt = 1.2 x 311390.11 = 373668.13 Nmm
viii) Thus, Mb = 230306.35 Nmm & Mt = 373668.13 Nmm
ix) Take Kb = Kt =1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334)
x) By maximum shear stress theory,
τ𝑚𝑎𝑥 =
16
ߨ𝑑3
(𝑘𝑏 .
𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 =
16
ߨ𝑑3
(1.5 × 230306.35)2+(1.5 × 376668.13)2
∴ 𝑑 = 35.61𝑚𝑚
But we haven’t considered the weight of gears while calculating diameter of shaft, hence we
consider the shaft of diameter d = 44 mm
xi) Finally,
Shaft dia. for bearing mounting 0.8d & 1.1d 35.2mm & 48.4mm
Shaft dia. for gear mounting d 44mm
Shaft dia. for resting gear on one side 1.2d 52.8mm
xii) Checking stress concentration at critical sections
To reduce the stress concentration at critical sections (X-X’, Y-Y’ and Z-Z’) suitable fillet radius of
(0.09d) is provided. Where d is the diameter of shaft
We have to find bending (𝛔𝐛) and torsional (𝝉) stresses at these sections and then comparing
with Endurance limit (Se )
If σ > Se ,then we have to change the dependent dimension such as 1.1d,0.8d,1.2d and
if σ < Se , then we can say that Design is safe and correct.
0.09d 0.09d 0.09d
A) At section X-X’ for Bending and Torsion
B) At section Y-Y’ for Bending and Torsion
C) At section Z-Z’ for Bending and Torsion
d
For steel 40C8
Se’ = 0.5xSut = 290MPa
Se = Ka.Kb.Kc.Kd.Se’
Ka = 0.84 (machined & cold-drawn)
Kb = 0.85 (0.75<d<75mm)
Kc = 1 (100% reliability)
q = 0.95 (Notch radius = 4mm)
𝑟
𝑑
=
0.09𝑑
0.8𝑑
= 0.11 &
𝐷
𝑑
=
𝑑
0.8𝑑
= 1.25
Thus Kt for bending case
Kt = 1.65 Refer: DOME by bhandhari,
fig. 5.5)
Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.62
Kd = 𝐾𝑓
−1
= 0.62
Se = 0.84x0.85x1x0.62x290
Se = 128Mpa
Bending stress at Z-Z’ is
𝝈𝒃 =
𝟑𝟐 𝑴𝒃 𝒛
𝝅 (𝟎. 𝟖𝒅)𝟑
𝑀𝑏 𝑧 = 31987 Nmm
𝜎𝑏 =
32×31987
𝜋 (35.2)3 = 7.5MPa < Se
From BMD
Similarly Kt for torsional case
Kt = 1.4 (Refer: DOME by bhandhari, fig. 5.6)
Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.38
Kd = 𝐾𝑓
−1
= 0.72
Torsional shear stress at Z-Z’ is
𝝉 =
𝟏𝟔 𝑴𝒕
𝝅 (𝟎. 𝟖𝒅)𝟑
Se = 0.84x0.85x1x0.72x290
Se = 150Mpa
𝜏 =
16×373668.18
𝜋 (35.2)3 = 43.63MPa
𝜏 < Se
i.e. Design is safe at section Z-Z’
• Selection of bearings
2. Output shaft bearings
i) Part No. 9 and part name is Output shaft bearing
ii) Selected type of bearing for given application is Taper roller bearing, because it
has following advantages-
It can carry combine radial and axial loads
(Refer: PSG Data book, pg.no. 4.1)
iii) Radial load acting on bearings;
FrA = 440.322 + 1201.3262 = 1279.48N
FrB = 609.672 + 1663.3472 = 1771.56N
FrB > FrA and Ka = 0
It matches with load case 2(a): from fig 15.22 of DOME by bhandhari
And here also, we are going with face-to-face
configuration
Hence axial load acting on bearings are
given as
FaB =
0.5 𝐹𝑟𝐵
𝑌
=
0.5 ×1771.56
1.6
= 553.61N
and as Ka = 0 FaA = FaB
iv) Expected bearing life in hrs. L10h = 25000hrs
Machines for use 8hrs/day ad fully utilized – E.g.
cranes for bulk goods, ventilating fans, shafts of
gearboxes, the recommended life is from 20000
to 30000h
(Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5)
v) L10 =
60𝑛L10h
106 =
60×23×25000
106 = 34.5 millions revolution
since, output shaft speed is running at 23 RPM
vi) Diameter for bearing mounting is 0.8d = 35.2mm (Non-gear side)
Trial 1 For Bearing A : 32207A is selected (Refer: PSG design DATA, pg.no. 4.25)
C = dynamic capacity = PBx(𝐿10)0.3x (Load factor)
= 1397.56 x (34.5)0.3 x (1.5) = 6064.67N
Hence, Bearing 32207A (C = 57600N) is suitable.
𝑭𝒂𝑨
𝑭𝒓𝑨
=
𝟓𝟓𝟑. 𝟔𝟏
𝟏𝟐𝟕𝟗. 𝟒𝟖
= 𝟎. 𝟒𝟑𝟐𝟔 > 𝒆 = 𝟎. 𝟑𝟕
(Refer: PSG design DATA, pg.no. 4.4 & 4.25)
vii) Diameter for bearing mounting is 1.1d = 1.1x44 = 48.4mm (Gear side)
Trial 1 For Bearing B : 32210A is selected (Refer: PSG design DATA, pg.no. 4.25)
C = dynamic capacity = PBx(𝐿10)0.3x (Load factor)
= 1771.56 x (34.5)0.3
x (1.5) = 7687.63N
Hence, Bearing 32210A (C = 72600N) is suitable.
𝑭𝒂𝑩
𝑭𝒓𝑩
=
𝟓𝟓𝟑. 𝟔𝟏
𝟏𝟕𝟕𝟏. 𝟓𝟔
= 𝟎. 𝟑𝟏𝟐𝟒𝟗 < 𝒆 = 𝟎. 𝟑𝟕
(Refer: PSG design DATA, pg.no. 4.4 & 4.25)
FrB = 1771.56N and FaB = 553.61N
PB =Equivalent bearing load = X.FrB + Y.FaB = 1x1771.56+ 0x553.61 = 1771.56N
i) Part No. is 21 and Part Name is output shaft key
ii) Function: Mounting spur gear (S2) on output shaft
• Selection of Keys for gear mounting on Output shaft
iii) TWO orthographic views iv) FBD of key
v) On output shaft torque acting = 311390.106Nmm which is very high, So here we go with
higher grade of steel which is C60 (Refer: PSG design data book, Page No. 1.9 & 1.10)
This steel is suitable for making hardened screws and nuts, couplings , machined spindles
vii) Properties of C60 :
1. Sut = 750 N/mm2
2. Syt = 420 N/mm2
viii) Factor of safety chosen is 1.2 because, key failure is more economical than gear or shaft
failure in any extreme conditions.
ix) Allowable stresses:
1. 𝜎 =
420
1.2
= 350N/mm2 and 2. 𝜏 =
0.5×380
1.2
= 175N/mm2
x) Diameter of shaft for gear mounting = d = 44mm ; for these
The dimensions of parallel key obtained from table is bxh = 14x9 and keyway depth is 6.5mm
(Refer: PSG design data book, Page No. 5.24)
xi) Length of key is taken to be 30mm = width of S2
xii) Checking for allowable stresses:
𝜏 =
2𝑇
𝑏ℎ𝑙
=
2×311390.11
14×9×30
= 164.75N/mm2 < 175N/mm2
Also, 𝜎 = 2 𝜏 = 329.51N/mm2 < 350N/mm2 Hence, design is safe
bxhxl = 14x9x30
i) Part No. is 27 and Part Name is circlip 2
ii) Function: Locking spurl gear (S2) in its position on output shaft
iii) The diameter for S2 mounting is d = 44mm thus circlip selected from light
series A is
• Selection of circlips (External type)
Circlip, Light A 45
Notation Value (mm) Notation Value (mm)
d1 45 D5 2
d2 42.5 s 1.75
d3 41.5 b 6.7
d4 59.4 a 4.7
(Refer: PSG Design DATA,
pg.no. 5.3 & 5.4)
• Design of shafts
3. Input shaft
i) Part no. is 5 and part name is input shaft
ii) Design of
based on ASME
code
iii) Function:
Transmission of
torque and power
from motor to G1
iv) The material selected for shaft is Plain carbon steel with grade 40C8 because,
this steel is used for making shafts, crankshaft, automobile axle
(Refer: PSG design data book, pg.no. 1.9, 1.10)
v) Properties of material:
1. Sut = 580N/mm2
2. Syt = 330N/mm2
vi) Allowable stress calculation:
Ʈmax = 0.30xSyt = 99N/mm2 or Ʈmax = 0.18xSut = 104.4N/mm2
hence Ʈmax = 99N/mm2 … (which is less)
Also, shaft is keyed for mounting of gears,
∴ Ʈmax = 0.75x99 = 74.25N/mm2
∴ Maximum bending moment at bearing B = Mb = 29556.152 + 9447.752 = 31029.44Nmm
Also, Mt =
60 ×𝑃
2𝜋𝑁
=
60 ×0.75×106
2𝜋×𝟔𝟗𝟎
=10379.67Nmm
Considering load factor = 1.3
Hence, Mt = 1.3 x 10379.67 = 13793.5 Nmm
viii) Thus, Mb = 31029.44 Nmm & Mt = 12455.6 Nmm
ix) Take Kb = Kt = 1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334)
x) By maximum shear stress theory,
τ𝑚𝑎𝑥 =
16
ߨ𝑑3 (𝑘𝑏 .
𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 =
16
ߨ𝑑3 (1.5 × 31029.44)2+(1.5 × 13493.5)2
∴ 𝑑 = 15.1mm
But we haven’t considered the weights of gears and bearings while calculating diameter of
shaft, Also minimum dia. for which taper roller bearing is available is 30mm.
hence, we consider the shaft of diameter d = 30mm
xi) Finally, Shaft dia. for bearing mounting d 30mm
Shaft dia. For coupling 0.8d 24mm
Shaft dia. for gear mounting d 30mm
Shaft dia. for resting gear on one side 1.2d 36mm
xii) Checking stress concentration at critical sections
1. From BMD it is clear that ;
X-X’ is a critical section as bending moment is high over there
2. To reduce the stress concentration at critical sections (X-X’)
suitable fillet radius of (0.08d) is provided. Where d is the
diameter of shaft
3. We have to find bending (𝛔𝐛) and torsional (𝝉) stresses at
these sections and then comparing with Endurance limit (Se )
4. If σ > Se ,then we have to change the dependent dimension
such as 0.8d,1.2d or 0.08d and
if σ < Se , then we can say that Design is safe and correct.
A) At section X-X’ for Bending and Torsion
For steel 40C8
Se’ = 0.5xSut = 290MPa
Se = Ka.Kb.Kc.Kd.Se’
Ka = 0.84 (machined & cold-drawn)
Kb = 0.85 (0.75<d<75mm)
Kc = 1 (100% reliability)
q = 0.84 (Notch radius = 2.5)
Refer: DOME by Bhandari, fig. 5.23)
𝑟
𝑑
=
0.08𝑑
𝑑
= 0.08 &
𝐷
𝑑
=
1.2𝑑
𝑑
= 1.2
Thus Kt for bending case
Kt = 1.8
Refer: DOME by Bhandari, fig. 5.5)
Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.672
Kd = 𝐾𝑓
−1
= 0.598
Se = 0.84x0.85x1x0.598x290
Se = 123.84Mpa
Bending stress at X-X’ is
𝝈𝒃 =
𝟑𝟐 𝑴𝒃 𝒙
𝝅 (𝒅)𝟑
𝑀𝑏 𝑧 = 26466.32 Nmm
𝜎𝑏 =
32×26466.32
𝜋 (30)3 = 9.98MPa < Se
From BMD
Similarly Kt for torsional case
Kt = 1.45 (Refer: DOME by Bhandari, fig. 5.6)
Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.378
Kd = 𝐾𝑓
−1
= 0.7257
Torsional shear stress at Z-Z’ is
𝝉 =
𝟏𝟔 𝑴𝒕
𝝅 (𝒅)𝟑
Se = 0.84x0.85x1x0.7257x290
Se = 150.26Mpa
𝜏 =
16×13493.5
𝜋 (35.2)3 = 2.545MPa
𝜏 << Se
i.e. Design is safe at section X-X’
• Selection of bearings
3. Input shaft bearings
i) Part No. 8 and part name is Intput shaft bearing
ii) Selected type of bearing for given application is Taper roller bearing, because it
has following advantages-
It can carry combine radial and axial loads (Refer: PSG Data book, pg.no. 4.1)
iii) Radial load acting on bearings;
FrA = 347.722 + 111.152 = 365.052N
FrB = 861.742 + 294.62 = 910.71N
FrB > FrA and Ka = 36.7N
It matches with load case 2(a): from fig 15.22 of DOME by Bhandhari
And here also, we are going with face-to-face
configuration
Hence axial load acting on bearings are
given as
FaB =
0.5 𝐹𝑟𝐵
𝑌
=
0.5 ×910.71
1.6
= 284.6N
and as Ka = 36.7N FaA = FaB + Ka
FaA = 321.3N
iv) Expected bearing life in hrs. L10h = 25000hrs
Machines for use 8hrs/day ad fully utilized – E.g.
cranes for bulk goods, ventilating fans, shafts of
gearboxes, the recommended life is from 20000
to 30000h
(Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5)
v) L10 =
60𝑛L10h
106 =
60×690×25000
106 =
= 1035 millions revolution
Since, Input shaft speed is running at 690 RPM
vi) Diameter for bearing mounting is d = 30mm (Non-Gear side)
Trial 1 For Bearing A : 32206A is selected (Refer: PSG design DATA, pg.no. 4.25)
C = dynamic capacity = PAx(𝐿10)0.3
x (Load factor)
= 660.1 x (1035)0.3 x (1.5) = 7946.64N
Hence, Bearing 32206A (C = 43800N) is suitable.
𝑭𝒂𝑨
𝑭𝒓𝑨
=
𝟑𝟐𝟏. 𝟑
𝟑𝟔𝟓. 𝟎𝟓𝟐
= 𝟎. 𝟖𝟖 > 𝒆 = 𝟎. 𝟑𝟕
(Refer: PSG design DATA, pg.no. 4.4 & 4.25)
PA = Equivalent bearing load = X.FrA + Y.FaA
= 0.4x365.052 + 1.6x321.3 = 660.1N
FrA = 365.052N & FaA = 321.3N
vii) Diameter for bearing mounting is d = 30 mm (Gear side)
Trial 1 For Bearing B : 32206A is selected (Refer: PSG design DATA, pg.no. 4.25)
C = dynamic capacity = PBx(𝐿10)0.3x (Load factor)
= 910.7 x (1035)0.3 x (1.5) = 10963.5N
Hence, Bearing 32206A (C = 46800N) is suitable.
𝑭𝒂𝑩
𝑭𝒓𝑩
=
𝟐𝟖𝟒. 𝟔
𝟗𝟏𝟎. 𝟕𝟏
= 𝟎. 𝟑𝟏𝟐𝟓 < 𝒆 = 𝟎. 𝟑𝟕
(Refer: PSG design DATA, pg.no. 4.4 & 4.25)
FrB = 910.71N and FaB = 284.6N
PB = Equivalent bearing load = X.FrB + Y.FaB
= 1x1771.56+ 0x553.61 = 910.71N
• Locking arrangement for bearing A (of input shaft only)
Lock
washer
Ring Nut
iv) We have to find suitable Lock washer and Ring nut from manufactures
catalogue which totally depends on shaft dia. At these section
i) Part No. are 29,30 and Part name is Lock washer
and Ring nut respectively.
ii) Function: Locking bearing A from other end to
restricts it’s movement in axial direction
iii) The shaft diameter for lock washer and ring nut is
0.8d = 0.8x30 = 24mm
v) Finally, the standard elements select from PSG design data tables
Notations Values (mm)
h 23
d1 32
d2 42
a 5
b 5
t 1.25
N 13
(Refer: PSG design DATA, pg. No. 5.86)
Notations Values (mm)
d 25
D 38
b 7
s 4
t 2
M25 x 1.5
(Refer: PSG design DATA,
pg. No. 5.85)
i) Part No. is 20 and Part Name is Input shaft key
v) Check for induced stresses;
𝜏 =
2𝑇
𝑏ℎ𝑙
=
2×13493.5
8×7×𝟑𝟎
= 16.72N/mm2 << 126.67N/mm2
Also, 𝜎 = 2 𝜏 = 33.44N/mm2 << 253.33N/mm2
iv) Orthographic views , FBD, fos = 1.5 are same as earlier case
Diameter for gear mounting is d = 30mm, Hence bxh = 8x7 and keyway depth is 1.7mm
also, length of key = 𝑙 = 30 mm < 60mm
ii) Function: Mounting Bevel gear (G1) on input shaft
vi) Hence, our design is safe
bxhxl = 8x7x45
• Selection of Keys for gear mounting on Input shaft
iii) The martial selected for Key is plain carbon steel with grade C50
1. Sut = 660 N/mm2
2. Syt = 380 N/mm2
i) Part No. is 28 and Part Name is circlip 3
ii) Function: Locking bevel gear (G1) as well as bearing B
iii) The diameter for G1 mounting is d = 30mm thus circlip
selected from light series A is
• Selection of circlip (External type)
Circlip, Light A 30
(Refer: PSG Design DATA, pg.no. 5.3 & 5.4)
Notation Value (mm) Notation Value (mm)
d1 30 d5 2
d2 28.6 s 1.5
d3 27.9 b 3.5
d4 41 a 5
• Locking arrangement for G1 from open end
iv) Lock washer and Ring nut assembly is same as that of selected on slides 72 & 73
because mounting dia. is same in both the cases i.e 24mm
i) Part No. is 31 and Part name is Washer
ii) Function: Evenly distributing load from nut to larger area
iii) The shaft dia. for washer is 24mm, thus dimension
obtained from manufactures catalogue are
Washer
Lock washer &
Ring nut assembly
Notation Values (mm)
d 24
D 48
s 7
i) Part No. is 17 and Part Name is Input shaft oil seal
ii) Function: Prevents oil leakage from input end of gearbox
iii) Diameter for oil seal mounting is 0.8d = 0.8x30 = 24 mm
iv) Also Type A oil is selected for input shaft since torque acting on
it is quite low as well as it running at high speed
Type A : Stiffener (metal) ring is covered with rubber elastomer to tolerate thermal expansion
• Selection of Oil seal for Input shaft
v) Thus the oil seal selected for 24mm shaft dia. from catalogue
have dimension of
Notation Value
b 7
c (min) 0.3
Nominal housing diameter 35
(Refer: PSG Design DATA, pg. No. 5.11)
Note: Find fit allowance and tolerances for Type A seals , see table on pg. No. 5.112
i) Part No. is 18 and Part Name is Output shaft oil seal
ii) Function: Prevents oil leakage from input end of gearbox
iii) Diameter of shaft R for oil seal mounting is 1.1d = 0.8x44 = 48.4 mm
iv) Also Type B oil is selected for output shaft since torque acting on it is very high
compared to input shaft
Type B : Stiffener (metal) ring is at outer periphery enclosing elastomer material
• Selection of Oil seal for Output shaft
v) Thus the oil seal selected for 48.4mm shaft dia. from catalogue
have dimension of
Notation Value
b 8
c (min) 0.4
Nominal housing diameter 62
(Refer: PSG Design DATA, pg. No. 5.11)
Note: Find fit allowance and tolerances for Type B seals , see table on pg. No. 5.112
• Design of Casing
1. Lower casing
 FBD
 2 orthographic views with design notation
 Material
 Factor of safety
 Calculations
 Table of final dimensions
• Design of Casing
1. Upper casing
 FBD
 2 orthographic views with design notation
 Material
 Factor of safety
 Calculations
 Table of final dimensions
• Design of bearing caps
 FBD
 2 orthographic views with design notation
 Material
 Factor of safety
 Calculations
 Table of final dimensions
• Design of stuffing box
 FBD
 2 orthographic views with design notation
 Material
 Factor of safety
 Calculations
 Table of final dimensions
• Weight of gearbox
Total weight of gearbox
= 120 + 3.6 + 5.4 +
10 + 34.4 + 12 =
𝟏𝟖𝟔𝒌𝒈
• Selection of Eye Bolts
i) Part No. is 28 and Part Name is eye bolt
ii) Function: Lifting of gearbox using cranes/hoists
iii) Orthographic Views of eye bolt
iv) Load acting on eye bolt = Overall weight of
gearbox when lifted (Pw) = 186 x 10 = 1860N
iv) The material selected for shaft is Plain carbon steel with grade 40C8 because, this
steel is used for making shafts,bolts, crankshaft, automobile axle
(Refer: PSG design data book, pg.no. 1.9, 1.10)
1. Sut = 580/mm2
2. Syt = 300N/mm2
v) Factor of safety is taken to be 7, since as gearbox get lifted all the weight is come on eye
bolt
Allowable stress calculation: σ =
300
7
= 42.85MPa
vii) Critical mode of failure : Shearing of threads of eye bolt
σ =
4𝑃𝑤
𝜋𝑑𝑐
2 𝑑𝑐
2 =
4𝑃𝑤
𝜋σ
∴ 𝑑𝑐
2
=
4 × 1860
𝜋 × 42.85 𝑑𝑐 = 7.43𝑚𝑚
𝑑 =
𝑑𝑐
0.86
= 8.64mm
M10 x 0.75 is selected.
Hence,
• Selection of Nut-Bolt
for casing
i) Part No. is 33,34 and Part Name is
casing bolt & casing nut
(Hexagonal Nut and Bolt)
ii) Function: Joining of upper and
lower casings
iii) Critical mode of failure : Tensile
iv) Material : 30C8 and fos = 6
M8 x 0.5 is selected.
(Refer: PSG design DATA, pg.no.5.49 & 5.50 )
Such 6 Nut-bolt pair is needed
The same size of screw(below figure) are used nearby bearing region
M8 x 0.5
Screw is selected.
For 4 bearing semi-circular
regions we requiring 8 screw
(Refer: PSG design DATA, pg.no.5.61 & 5.62 )
Again, 6 same hexagonal socket head cap screw [M8 x 0.5] is used for attachment of
stuffing box to the casing
The nut-bolts, screws used to join two casing are intentionally selected of fine series
to ensure proper tightening and thus avoiding oil leakage
• Selection of foundation Bolts
i) Part No. is 32 and Part Name is Foundation bolt
ii) Function: Fixing/Mounting of gearbox on support
iii) Load acting on foundation bolt = Overall weight of gearbox when lifted
(Pw) = 186 x 10 = 1860N
iv) Critical mode of failure : Compression and shearing of threads
v) The material selected for shaft is Plain carbon steel with grade 40C8 because, this
steel is used for making shafts, bolts, crankshaft, automobile axle
(Refer: PSG design data book, pg.no. 1.9, 1.10)
1. Sut = 580/mm2
2. Syt = 300N/mm2
v) Factor of safety is taken to be 6, since as gearbox
get lifted all the weight is come on eye bolt
Allowable shear stress calculation: τ =
0.5×300
5
=
30MPa
𝑑𝑐
2 =
4𝑃𝑤
𝜋𝜏
𝑑𝑐 = 8.88𝑚𝑚
𝑑 =
𝑑𝑐
0.86
= 10.32mm
M12 x 1.25
The next standard bolt is
selected of,
𝜏 =
4𝑃𝑤
𝜋𝑑𝑐
2
∴ 𝑑𝑐
2
=
4 × 1860
𝜋 × 30
On dimensional constraint
basis the screw selected
for-
1. bearing cap attachment
to housing and
2. cover plate attachment
to stuffing box is of
M5 x 0.5
6 for each cap x 5 = 30
Hexagonal socket head
cap screw are used.

Gearbox design

  • 1.
    Theory and Designof mechanical system Topic: 2 stage cylindrical-bevel speed reducer Group: G2 1. 111910138 Michael Susngi 2. 111910139 Pranav Jadhav 3. 111910140 Rahul Jadhav 4. 111910141 Ravindra Shinde Team members
  • 2.
    ● Problem Statement “Designa suitable gearbox for box shipping conveyor which used to carry 3 boxes of 50kg each at a speed of 28m/min.”
  • 4.
    • Specifications : 1.Load acting on belt: 3 x 50kg 2. Speed of conveyor 3. Diameter of roller 4. Weight of conveyor belt 5. Friction coefficient between belt & roller • Assumptions : 1. Material is assumed to be isotropic and homogeneous. 2. All types of vibrations are neglected. 3. Thermal stresses are neglected. 4. Gradual and uniform loading condition. 150kg 28m/min 200mm 50kg 0.67 (Reference : https://www.rubbercal.com/industrial-rubber/heavy-black-conveyor-belt.html & https://www.engineersedge.com/coeffients_of_friction.htm )
  • 5.
    ● Design Requirements 1.Nominal operating conditions of industrialized locations 2. Input and output shafts STD. size for typical couplings 3. Usually low shock levels, occasional moderate shock 4. Continuous operation 5. Low maintenance 6. Competitive cost
  • 6.
    • Significance ofgear drive: 1. It is a positive drive, and the velocity ratio remains constant. 2. The center distance between the shafts is relatively small, which results in compact construction. 3. It can transmit very large power, which is beyond the range of belt or chain drives. 4. It can transmit motion at very low velocity, which is not possible with the belt drives. 5. The efficiency of gear drives is very high, even up to 99 per cent in case of spur gears. 6. A provision can be made in the gearbox for gear shifting, thus changing the velocity ratio over a wide range.
  • 7.
    • Types ofgear: 1. Spur gear 1. Easy to design and manufacture. 2. Suitable for low torque 3. Make noise at high-speed operation 2. Bevel gear 1. Angular torque transmission. 2. Not desirable for High-speed reduction. 3. Compact design and highly efficient
  • 8.
    4. Helical gear 3.Worm and Worm wheel 1. Large speed reduction. 2. Large increase in torque. 3. Self-Locking Mechanism. 4. Less efficient 1. Gradual teeth engagement 2. More durable 3. Less noise 4. Produce large thrust force due to helix angle • Types of gear:
  • 9.
    • Selection ofgearbox 2 stage cylindrical bevel speed reducer. a) In our case motor is in parallel to the conveyor belt. so, we have to transmit power in perpendicular direction. So perfect solution is bevel gear. b) But along with that we have to achieve high reduction ratio of 30. so, we also used spur gears to obtain the desired reduction.
  • 10.
  • 11.
    PART NO. PARTNAME TYPE QTY 1 Driving spur gear Custom 1 2 Driven spur pinion Custom 1 3 Driving bevel pinion Custom 1 4 Driven bevel gear Custom 1 5 Input shaft Custom 1 6 Output shaft Custom 1 7 Intermediate shaft Custom 1 8 Input shaft Bearings STD. 2 9 Output shaft Bearings STD. 2 10 Intermediate shaft Bearings STD. 2 11 Upper casing Custom 1 12 Lower casing Custom 1 • Part list : PART NO. PART NAME TYPE QTY 13 Casing Nut STD. 4 14 Casing bolt STD. 4 15 Foundation bolt STD. 4 16 Bearing cap bolt STD. 20 17 Input shaft oil seal STD. 1 18 Output shaft oil seal STD. 1 19 Banjo bolt STD. 1 20 Input shaft key STD. 1 21 Output shaft key STD. 1 22 Intermediate shaft key 1 STD. 1 23 Intermediate shaft key 2 STD. 1 24 Retaining ring STD. 3 25 Bearing cap Custom 5
  • 12.
    • Selection ofmotor i) Power required to convey boxes = µN x V = 0.67 x (150+50) x 9.81 x 28 60 = 613.5 watt ii) To encounter different losses in the gearbox and couplings 20% overload factor is considered. Thus, total power required = 613.36 + 613.36 x 20% = 736.03 watt iii) Reasons for using Induction motor 1. Useful in low-speed operation 2. Frequent maintenance not needed as like DC (Due to carbon brushes) 3. Suitable in operations where high starting torque is not required such as in conveyors
  • 13.
    The standard motoravailable in induction motor catalogue to obtain required power is of 0.75kW(1HP) (Reference: https://3.imimg.com/data3/QM/MP/MY-3212916/three-phase-induction-motors.pdf, Table no. 3, pg. No. 3) “3-phase Squirel Cage Induction Motor, 50Hz, 0.75kW(1HP)” • Selection of motor Electrical Specification Mechanical Specification Number of poles : 8 Speed : 690 RPM Full load current : 2A Efficiency : 70.0% Operating voltage : 415V ± 10% Output : 0.75 kW Power factor : 0.73 Frame : 100L
  • 14.
    Name of componentsAbbreviation used Input shaft P Driving Bevel Pinion G1 Driven Bevel gear G2 Intermediate Shaft Q Driving Spur pinion S1 Driven Spur gear S2 Output shaft R Input Shaft Bearing B1 Output Shaft Bearing B2 Intermediate Shaft Bearing B3 • Gear train Layout with abbreviations used
  • 15.
    MOTOR ROLLER • Power flowdiagram Motor Coupling 1 P G1 G2 Q S1 S2 R ROLLER Coupling 2
  • 16.
    • Force analysisof whole system (3D)
  • 17.
    • Gear ratiocalculations 1. After selection of motor, Input speed to gearbox = 690 rpm 2. Speed at output of gearbox is calculated as; Velocity of roller = radius of roller x angular velocity of roller V = r x ( 2π𝑁 60 ) ; ∴ 𝑁 = 𝑉 ×60 2𝜋𝑟 = 28×60 60×2𝜋×0.2 = 23.3 𝑟𝑝𝑚 Thus, Output speed of gearbox = 23.3 rpm 3. That is, we have to reduce speed from 690rpm to 23.3rpm in 2 stages 4. After doing several permutations and combinations, we get the suitable configuration of, 5:1 in 1st stage (in bevel pair) ∴ Speed of shaft Q = 690 5 = 138 rpm & 6:1 in 2nd stage (in spur pair) ∴ Speed of shaft R = 138 6 = 23 rpm Shaft Type Speed (RPM) P 690 Q 138 R 23
  • 18.
    • Bevel pairdesign i) Part No. is 3,4 and part name is Driving bevel pinion and Driven bevel gear ii) Function: Transmits power in perpendicular direction and also provides speed reduction iii) Free body diagram Material Syt (𝑘𝑔𝑓/𝑚𝑚2 ) Sut (𝑘𝑔𝑓/𝑚𝑚2 ) Pinion 40 Ni 1 Cr 1 Mo 28 88 110 Wheel C 55 Mn75 46 76 iv) For reduction ratio of 5:1 material selected are; (Refer: PSG data book, Table 5, pg.no. 8.4 & PSG data book, pg.no. 1.12 ,1.15)
  • 19.
    • Bevel pairdesign v) Two orthographic views SEC. F. V. SEC. R. H. V.
  • 20.
    vi) Design calculations: Let,rpm = 695 , module : 3 , reduction ratio i = 5 , Number of teeth on pinion Zp = 20
  • 23.
    vii) Verifying DesignAgainst Beam Strength:
  • 24.
    viii) Verifying DesignAgainst Wear Strength
  • 25.
    ix) Final dimensions: NotationsMeaning Values 𝑚 module 3 𝛼 Pressure angle 20˚ 𝑍𝑝 Num of teeth on pinion 20 𝑍𝐺 Num of teeth on gear 100 𝐷𝑝 p.c.d. of pinion (G1) 60 mm 𝐷𝐺 p.c.d. of gear (G2) 300mm 𝐴𝑜 Cone distance 153 mm 𝑏 Face width 38.25 mm 𝛾 Pitch angle 11.30˚
  • 26.
    • Spur pairdesign i) Part No. is 1,2 and part name is Driving spur pinion and Driven spur gear ii) Function: Transmits power in parallel shafts and speed reduction iii) Free body diagram Material Syt (𝑘𝑔𝑓/𝑚𝑚2 ) Sut (𝑘𝑔𝑓/𝑚𝑚2 ) Pinion (S1) 40 Ni 1 Cr 1 Mo 28 88 110 Wheel (S2) C 35 Mn75 46 76 iv) For reduction ratio of 6:1 material selected are; (Refer: PSG data book, Table 5, pg.no. 8.4 & PSG data book, pg.no. 1.9)
  • 27.
    • Spur pairdesign v) Two orthographic views
  • 28.
    vi) Design calculations(Beam & wear Strength)
  • 29.
    vi) Design calculations(Beam & wear Strength)
  • 30.
    vi) Final dimensionsof Spur gear Notations Meaning Values 𝑚 module 3 𝛼 Pressure angle 20˚ ℎ Tooth depth (both) 6.75 mm 𝑍1 Num of teeth (pinion) 18 𝑍2 Num of teeth (Gear ) 108 𝐷1 P.C.D. of pinion (G1) 54 mm 𝐷2 P.C.D. of gear (G2) 324 mm 𝐷𝑎 Addendum Diameter(Gear) 330 mm 𝑑𝑎 Addendum Diameter(pinion) 60 mm 𝐷𝑏 Base diameter(Gear) 316.5 mm 𝑑𝑏 Base diameter (pinion) 46.5 mm
  • 31.
    • Design ofshafts 1. Intermediate shaft i) Part no. is 7 and part name is intermediate shaft ii) Design of based on ASME code iii) Function: transmission of torque and power
  • 32.
    iv) The materialselected for shaft is Plain carbon steel with grade 40C8 because, this steel is used for making shafts, crankshaft, automobile axle (Refer: PSG design data book, pg.no. 1.9, 1.10) v) Properties of material: 1. Sut = 580N/mm2 2. Syt = 330N/mm2 vi) Allowable stress calculation: Ʈmax = 0.30xSyt = 99N/mm2 or Ʈmax = 0.18xSut = 104.4N/mm2 hence Ʈmax = 99N/mm2 … (which is less) Also, shaft is keyed for mounting of gears, ∴ Ʈmax = 0.75x99 = 74.25N/mm2
  • 33.
    vii) FBD &BMD of shaft in respective plane ∴ Maximum bending moment at S1 = 44384.182 + 126856.842 = 134,397.22 Nmm
  • 34.
    viii) Thus, Mb= 134397.22 Nmm & Mt = 77887 Nmm ix) Take Kb = Kt = 1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334) x) By maximum shear stress theory, τ𝑚𝑎𝑥 = 16 ߨ𝑑3 (𝑘𝑏 . 𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 = 16 ߨ𝑑3 (1.5 × 134397.22)2+(1.5 × 77887)2 ∴ 𝑑 = 25.19𝑚𝑚 But we haven’t considered the weight of gears while calculating diameter of shaft, hence we consider the shaft of diameter d = 35mm xi) Finally, Shaft dia. for bearing mounting d 35mm Shaft dia. for gear mounting 1.1d 38.5mm Shaft dia. for resting gear on one side 1.2d 42mm
  • 35.
    xii) Checking stressconcentration at critical sections As there are steps/shoulders in shaft ; there has to be a stress concentration. To reduce this stress concentration fillet is provided From BMD it is clear that on spur gear(S1) side bending moments are higher. Hence critical sections X-X’ and Y-Y’ are labelled in adjoining figure We have to find stress(𝜎)at these sections and then comparing with Endurance limit (Se ) If σ > Se ,then we have to change the dependent dimension such as 1.1d,0.1d,1.2d and if σ < Se , then we can say that Design is safe and correct.
  • 36.
    A) At sectionX-X’ for Bending and Torsion
  • 37.
    B) At sectionY-Y’ for Bending and Torsion
  • 38.
    • Selection ofbearings 1. Intermediate shaft bearings i) Part No. 3 and part name is Intermediate shaft bearing ii) Selected type of bearing for given application is Taper roller bearing, because it has following advantages- It can carry combine radial and axial loads (Refer: PSG Data book, pg.no. 4.1)
  • 39.
    The construction, whichinvolves two bearings with their fronts facing each other, is called ‘face-to-face’ or direct mounting. (Refer: DOME by Bhandari, Fig. 15.11,case 2(a)
  • 40.
    iii) Expected bearinglife in hrs. L10h = 25000hrs Machines for use 8hrs/day ad fully utilized – E.g. cranes for bulk goods, ventilating fans, countershaft of gearboxes, the recommended life is from 20000 to 30000h (Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5) iv) L10 = 60𝑛L10h 106 = 60×138×25000 106 = 202.5 millions revolution since, intermediate shaft speed is running at 138 RPM v) Trial 1 Tentatively, we select Bearing 32207A. (Refer: PSG design DATA, pg.no. 4.25) vi) Bearing A: FrA = 1156.152N and FaA = 926.8N FaA FrA = 926.8 1156.152 = 0.8 > e = 0.37 P =Equivalent bearing load = X.Fr + Y.Fa = 0.4x1156.152+ 1.6x926.8 = 1945.34N C =dynamic capacity = P(𝐿10)0.3 x (Load factor) = 1945.34 (202.5)0.3 (1.5) = 14355.37N Hence, Bearing 32207A (C = 57600N) is suitable. (Refer: PSG design DATA, pg.no. 4.4 & 4.25)
  • 41.
    vii) Similarly forBearing B: FrB = 2378.689N and FaB = 743.34N Fa𝐵 Fr𝐵 = 743.34 2378.689 = 0.31 < e = 0.37 P = Equivalent bearing load = X.Fr + Y.Fa = 1x1156.152+ 0x926.8 = 1156.152N C = dynamic capacity = P(𝑳𝟏𝟎)𝟎.𝟑 x (Load factor) = 1156.152 (202.5)0.3 (1.5) = 8531.66N Hence, Bearing 32207A (C = 57600N) is suitable. (Refer: PSG design DATA, pg.no. 4.4 & 4.25) vii) The dimensions of bearing
  • 42.
    • Selection ofKeys for gear mounting on intermediate shaft i) Part No. is 22 and Part Name is Intermediate shaft key 1. ii) Function: Mounting bevel gear (G2) on intermediate shaft iii) TWO orthographic views of key iv) FBD of key
  • 43.
    v) The martialselected for Key is plain carbon steel with grade C50 ,Because- This steel is suitable for making keys, shaft, cylinder, and machined components requiring wear resistance (Refer: PSG design data book, Page No. 1.9 & 1.10) vii) Properties of C50 : 1. Sut = 660 N/mm2 2. Syt = 380 N/mm2 viii) Factor of safety chosen is 1.5 because, key failure is more economical than gear or shaft failure in any extreme conditions. ix) Allowable stresses: 1. 𝜎 = 380 1.5 = 253.33N/mm2 and 2. 𝜏 = 0.5×380 1.5 = 126.67N/mm2 x) Shaft diameter for gear mounting = 1.1d = 38.5mm ; The dimensions of parallel key obtained from table is bxh = 12x8 and keyway depth is 6mm (Refer: PSG design data book, Page No. 5.24) xi) The effective length is found using empirical relations available for key ∴ l = 1.5×38.5 = 58mm < 84mm bxhxl = 12x8x58
  • 44.
    xii) Checking forinduced stresses; 𝜏 = 2𝑇 𝑏ℎ𝑙 = 2×77887 12×8×𝟓𝟖 = 28N/mm2 < 126.67N/mm2 Also, 𝜎 = 2 𝜏 = 56N/mm2 < 253.33N/mm2 Hence, design is safe i) Part No. is 23 and Part Name is Intermediate shaft key 2. iv) Take l = 25mm and check for induced stresses; 𝜏 = 2𝑇 𝑏ℎ𝑙 = 2×77887 12×8×𝟐𝟓 = 64.90N/mm2 < 126.67N/mm2 Also, 𝜎 = 2 𝜏 = 129.81N/mm2 < 253.33N/mm2 iii) Orthographic views , FBD, Material for key , Fos are same as earlier case Also dia. for gear mounting is same. Hence bxh = 12x8 only thing changes is length of key which should be < 27mm ii) Function: Mounting spur gear (S1) on intermediate shaft v) Hence, our design is safe bxhxl = 12x8x25
  • 45.
    • Selection ofcirclips (External type) i) Part No. is 26 and Part Name is circlip 1 ii) Function: Locking bevel gear (G2) in its position on intermediate shaft iii) Detailed views with design notations iv) Commonly used steel for circlip is C65 (same as used in springs and washers) (Refer: PSG design DATA, pg.No. 5.3) v) Properties of C65: 1. Sut = 750 N/mm2 2. Syt = 430 N/mm2 (Refer: PSG design DATA, pg.No. 1.9 & 1.10)
  • 46.
    vi) In ourcase d1 = 1.1d = 38.5mm and for this circlip selected from light series (A) is designated as Circlip, Light A 38 (Refer: PSG design DATA, pg.no. 5.4) Notati on Meaning Value (mm) d1 Shaft dia. For gear mounting 38 d2 Groove diameter o shaft for circlip 36 d3 Unexpanded dia. Of circlip 35.2 d4 Expanded dia. Of circlip 50.6 d5 Hole dia. For expanding circlip 2 s Thickness of ring in axial direction 1.5 b Thickness of ring in radial direction 5.8 a Height of ears of ring for expansion 4.2 v) Design dimension getting from catalogue for above mentioned circlip are vii) As per catalogue, Maximum axial load carry by these circlip = 29100N >> 926.8N so we can use it without any worry
  • 47.
    • Design ofshafts 2. Output shaft ii) Design of based on ASME code i) Part no. is 6 and part name is output shaft iii) Function: Transmission of torque and power to conveyor roller iv) TWO orthographic views
  • 48.
    iv) The materialselected for shaft is Plain carbon steel with grade 40C8 because, this steel is used for making shafts, crankshaft, automobile axle (Refer: PSG design data book, pg.no. 1.9, 1.10) v) Properties of material: 1. Sut = 580N/mm2 2. Syt = 330N/mm2 vi) Allowable stress calculation: Ʈmax = 0.30xSyt = 99N/mm2 or Ʈmax = 0.18xSut = 104.4N/mm2 hence Ʈmax = 99N/mm2 … (which is less) Also, shaft is keyed for mounting of gears, ∴ Ʈmax = 0.75x99 = 74.25N/mm2
  • 49.
    vii) FBD &BMD of shaft in respective plane ∴ Maximum bending moment at S2 = Mb = 79258.0642 + 216238.72 = 230,306.35 Nmm Also, Mt = 60 ×𝑃 2𝜋𝑁 = 60 ×0.75×106 2𝜋×23 = 311390.11 Nmm Considering load factor = 1.2 Hence, Mt = 1.2 x 311390.11 = 373668.13 Nmm
  • 50.
    viii) Thus, Mb= 230306.35 Nmm & Mt = 373668.13 Nmm ix) Take Kb = Kt =1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334) x) By maximum shear stress theory, τ𝑚𝑎𝑥 = 16 ߨ𝑑3 (𝑘𝑏 . 𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 = 16 ߨ𝑑3 (1.5 × 230306.35)2+(1.5 × 376668.13)2 ∴ 𝑑 = 35.61𝑚𝑚 But we haven’t considered the weight of gears while calculating diameter of shaft, hence we consider the shaft of diameter d = 44 mm xi) Finally, Shaft dia. for bearing mounting 0.8d & 1.1d 35.2mm & 48.4mm Shaft dia. for gear mounting d 44mm Shaft dia. for resting gear on one side 1.2d 52.8mm
  • 51.
    xii) Checking stressconcentration at critical sections To reduce the stress concentration at critical sections (X-X’, Y-Y’ and Z-Z’) suitable fillet radius of (0.09d) is provided. Where d is the diameter of shaft We have to find bending (𝛔𝐛) and torsional (𝝉) stresses at these sections and then comparing with Endurance limit (Se ) If σ > Se ,then we have to change the dependent dimension such as 1.1d,0.8d,1.2d and if σ < Se , then we can say that Design is safe and correct. 0.09d 0.09d 0.09d
  • 52.
    A) At sectionX-X’ for Bending and Torsion
  • 53.
    B) At sectionY-Y’ for Bending and Torsion
  • 54.
    C) At sectionZ-Z’ for Bending and Torsion d For steel 40C8 Se’ = 0.5xSut = 290MPa Se = Ka.Kb.Kc.Kd.Se’ Ka = 0.84 (machined & cold-drawn) Kb = 0.85 (0.75<d<75mm) Kc = 1 (100% reliability) q = 0.95 (Notch radius = 4mm) 𝑟 𝑑 = 0.09𝑑 0.8𝑑 = 0.11 & 𝐷 𝑑 = 𝑑 0.8𝑑 = 1.25 Thus Kt for bending case Kt = 1.65 Refer: DOME by bhandhari, fig. 5.5) Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.62 Kd = 𝐾𝑓 −1 = 0.62 Se = 0.84x0.85x1x0.62x290 Se = 128Mpa Bending stress at Z-Z’ is 𝝈𝒃 = 𝟑𝟐 𝑴𝒃 𝒛 𝝅 (𝟎. 𝟖𝒅)𝟑 𝑀𝑏 𝑧 = 31987 Nmm 𝜎𝑏 = 32×31987 𝜋 (35.2)3 = 7.5MPa < Se From BMD Similarly Kt for torsional case Kt = 1.4 (Refer: DOME by bhandhari, fig. 5.6) Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.38 Kd = 𝐾𝑓 −1 = 0.72 Torsional shear stress at Z-Z’ is 𝝉 = 𝟏𝟔 𝑴𝒕 𝝅 (𝟎. 𝟖𝒅)𝟑 Se = 0.84x0.85x1x0.72x290 Se = 150Mpa 𝜏 = 16×373668.18 𝜋 (35.2)3 = 43.63MPa 𝜏 < Se i.e. Design is safe at section Z-Z’
  • 55.
    • Selection ofbearings 2. Output shaft bearings i) Part No. 9 and part name is Output shaft bearing ii) Selected type of bearing for given application is Taper roller bearing, because it has following advantages- It can carry combine radial and axial loads (Refer: PSG Data book, pg.no. 4.1) iii) Radial load acting on bearings; FrA = 440.322 + 1201.3262 = 1279.48N FrB = 609.672 + 1663.3472 = 1771.56N FrB > FrA and Ka = 0 It matches with load case 2(a): from fig 15.22 of DOME by bhandhari
  • 56.
    And here also,we are going with face-to-face configuration Hence axial load acting on bearings are given as FaB = 0.5 𝐹𝑟𝐵 𝑌 = 0.5 ×1771.56 1.6 = 553.61N and as Ka = 0 FaA = FaB iv) Expected bearing life in hrs. L10h = 25000hrs Machines for use 8hrs/day ad fully utilized – E.g. cranes for bulk goods, ventilating fans, shafts of gearboxes, the recommended life is from 20000 to 30000h (Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5) v) L10 = 60𝑛L10h 106 = 60×23×25000 106 = 34.5 millions revolution since, output shaft speed is running at 23 RPM
  • 57.
    vi) Diameter forbearing mounting is 0.8d = 35.2mm (Non-gear side) Trial 1 For Bearing A : 32207A is selected (Refer: PSG design DATA, pg.no. 4.25) C = dynamic capacity = PBx(𝐿10)0.3x (Load factor) = 1397.56 x (34.5)0.3 x (1.5) = 6064.67N Hence, Bearing 32207A (C = 57600N) is suitable. 𝑭𝒂𝑨 𝑭𝒓𝑨 = 𝟓𝟓𝟑. 𝟔𝟏 𝟏𝟐𝟕𝟗. 𝟒𝟖 = 𝟎. 𝟒𝟑𝟐𝟔 > 𝒆 = 𝟎. 𝟑𝟕 (Refer: PSG design DATA, pg.no. 4.4 & 4.25)
  • 58.
    vii) Diameter forbearing mounting is 1.1d = 1.1x44 = 48.4mm (Gear side) Trial 1 For Bearing B : 32210A is selected (Refer: PSG design DATA, pg.no. 4.25) C = dynamic capacity = PBx(𝐿10)0.3x (Load factor) = 1771.56 x (34.5)0.3 x (1.5) = 7687.63N Hence, Bearing 32210A (C = 72600N) is suitable. 𝑭𝒂𝑩 𝑭𝒓𝑩 = 𝟓𝟓𝟑. 𝟔𝟏 𝟏𝟕𝟕𝟏. 𝟓𝟔 = 𝟎. 𝟑𝟏𝟐𝟒𝟗 < 𝒆 = 𝟎. 𝟑𝟕 (Refer: PSG design DATA, pg.no. 4.4 & 4.25) FrB = 1771.56N and FaB = 553.61N PB =Equivalent bearing load = X.FrB + Y.FaB = 1x1771.56+ 0x553.61 = 1771.56N
  • 59.
    i) Part No.is 21 and Part Name is output shaft key ii) Function: Mounting spur gear (S2) on output shaft • Selection of Keys for gear mounting on Output shaft iii) TWO orthographic views iv) FBD of key
  • 60.
    v) On outputshaft torque acting = 311390.106Nmm which is very high, So here we go with higher grade of steel which is C60 (Refer: PSG design data book, Page No. 1.9 & 1.10) This steel is suitable for making hardened screws and nuts, couplings , machined spindles vii) Properties of C60 : 1. Sut = 750 N/mm2 2. Syt = 420 N/mm2 viii) Factor of safety chosen is 1.2 because, key failure is more economical than gear or shaft failure in any extreme conditions. ix) Allowable stresses: 1. 𝜎 = 420 1.2 = 350N/mm2 and 2. 𝜏 = 0.5×380 1.2 = 175N/mm2 x) Diameter of shaft for gear mounting = d = 44mm ; for these The dimensions of parallel key obtained from table is bxh = 14x9 and keyway depth is 6.5mm (Refer: PSG design data book, Page No. 5.24)
  • 61.
    xi) Length ofkey is taken to be 30mm = width of S2 xii) Checking for allowable stresses: 𝜏 = 2𝑇 𝑏ℎ𝑙 = 2×311390.11 14×9×30 = 164.75N/mm2 < 175N/mm2 Also, 𝜎 = 2 𝜏 = 329.51N/mm2 < 350N/mm2 Hence, design is safe bxhxl = 14x9x30 i) Part No. is 27 and Part Name is circlip 2 ii) Function: Locking spurl gear (S2) in its position on output shaft iii) The diameter for S2 mounting is d = 44mm thus circlip selected from light series A is • Selection of circlips (External type) Circlip, Light A 45 Notation Value (mm) Notation Value (mm) d1 45 D5 2 d2 42.5 s 1.75 d3 41.5 b 6.7 d4 59.4 a 4.7 (Refer: PSG Design DATA, pg.no. 5.3 & 5.4)
  • 62.
    • Design ofshafts 3. Input shaft i) Part no. is 5 and part name is input shaft ii) Design of based on ASME code iii) Function: Transmission of torque and power from motor to G1
  • 63.
    iv) The materialselected for shaft is Plain carbon steel with grade 40C8 because, this steel is used for making shafts, crankshaft, automobile axle (Refer: PSG design data book, pg.no. 1.9, 1.10) v) Properties of material: 1. Sut = 580N/mm2 2. Syt = 330N/mm2 vi) Allowable stress calculation: Ʈmax = 0.30xSyt = 99N/mm2 or Ʈmax = 0.18xSut = 104.4N/mm2 hence Ʈmax = 99N/mm2 … (which is less) Also, shaft is keyed for mounting of gears, ∴ Ʈmax = 0.75x99 = 74.25N/mm2
  • 64.
    ∴ Maximum bendingmoment at bearing B = Mb = 29556.152 + 9447.752 = 31029.44Nmm Also, Mt = 60 ×𝑃 2𝜋𝑁 = 60 ×0.75×106 2𝜋×𝟔𝟗𝟎 =10379.67Nmm Considering load factor = 1.3 Hence, Mt = 1.3 x 10379.67 = 13793.5 Nmm
  • 65.
    viii) Thus, Mb= 31029.44 Nmm & Mt = 12455.6 Nmm ix) Take Kb = Kt = 1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334) x) By maximum shear stress theory, τ𝑚𝑎𝑥 = 16 ߨ𝑑3 (𝑘𝑏 . 𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 = 16 ߨ𝑑3 (1.5 × 31029.44)2+(1.5 × 13493.5)2 ∴ 𝑑 = 15.1mm But we haven’t considered the weights of gears and bearings while calculating diameter of shaft, Also minimum dia. for which taper roller bearing is available is 30mm. hence, we consider the shaft of diameter d = 30mm xi) Finally, Shaft dia. for bearing mounting d 30mm Shaft dia. For coupling 0.8d 24mm Shaft dia. for gear mounting d 30mm Shaft dia. for resting gear on one side 1.2d 36mm
  • 66.
    xii) Checking stressconcentration at critical sections 1. From BMD it is clear that ; X-X’ is a critical section as bending moment is high over there 2. To reduce the stress concentration at critical sections (X-X’) suitable fillet radius of (0.08d) is provided. Where d is the diameter of shaft 3. We have to find bending (𝛔𝐛) and torsional (𝝉) stresses at these sections and then comparing with Endurance limit (Se ) 4. If σ > Se ,then we have to change the dependent dimension such as 0.8d,1.2d or 0.08d and if σ < Se , then we can say that Design is safe and correct.
  • 67.
    A) At sectionX-X’ for Bending and Torsion For steel 40C8 Se’ = 0.5xSut = 290MPa Se = Ka.Kb.Kc.Kd.Se’ Ka = 0.84 (machined & cold-drawn) Kb = 0.85 (0.75<d<75mm) Kc = 1 (100% reliability) q = 0.84 (Notch radius = 2.5) Refer: DOME by Bhandari, fig. 5.23) 𝑟 𝑑 = 0.08𝑑 𝑑 = 0.08 & 𝐷 𝑑 = 1.2𝑑 𝑑 = 1.2 Thus Kt for bending case Kt = 1.8 Refer: DOME by Bhandari, fig. 5.5) Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.672 Kd = 𝐾𝑓 −1 = 0.598 Se = 0.84x0.85x1x0.598x290 Se = 123.84Mpa Bending stress at X-X’ is 𝝈𝒃 = 𝟑𝟐 𝑴𝒃 𝒙 𝝅 (𝒅)𝟑 𝑀𝑏 𝑧 = 26466.32 Nmm 𝜎𝑏 = 32×26466.32 𝜋 (30)3 = 9.98MPa < Se From BMD Similarly Kt for torsional case Kt = 1.45 (Refer: DOME by Bhandari, fig. 5.6) Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.378 Kd = 𝐾𝑓 −1 = 0.7257 Torsional shear stress at Z-Z’ is 𝝉 = 𝟏𝟔 𝑴𝒕 𝝅 (𝒅)𝟑 Se = 0.84x0.85x1x0.7257x290 Se = 150.26Mpa 𝜏 = 16×13493.5 𝜋 (35.2)3 = 2.545MPa 𝜏 << Se i.e. Design is safe at section X-X’
  • 68.
    • Selection ofbearings 3. Input shaft bearings i) Part No. 8 and part name is Intput shaft bearing ii) Selected type of bearing for given application is Taper roller bearing, because it has following advantages- It can carry combine radial and axial loads (Refer: PSG Data book, pg.no. 4.1) iii) Radial load acting on bearings; FrA = 347.722 + 111.152 = 365.052N FrB = 861.742 + 294.62 = 910.71N FrB > FrA and Ka = 36.7N It matches with load case 2(a): from fig 15.22 of DOME by Bhandhari
  • 69.
    And here also,we are going with face-to-face configuration Hence axial load acting on bearings are given as FaB = 0.5 𝐹𝑟𝐵 𝑌 = 0.5 ×910.71 1.6 = 284.6N and as Ka = 36.7N FaA = FaB + Ka FaA = 321.3N iv) Expected bearing life in hrs. L10h = 25000hrs Machines for use 8hrs/day ad fully utilized – E.g. cranes for bulk goods, ventilating fans, shafts of gearboxes, the recommended life is from 20000 to 30000h (Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5) v) L10 = 60𝑛L10h 106 = 60×690×25000 106 = = 1035 millions revolution Since, Input shaft speed is running at 690 RPM
  • 70.
    vi) Diameter forbearing mounting is d = 30mm (Non-Gear side) Trial 1 For Bearing A : 32206A is selected (Refer: PSG design DATA, pg.no. 4.25) C = dynamic capacity = PAx(𝐿10)0.3 x (Load factor) = 660.1 x (1035)0.3 x (1.5) = 7946.64N Hence, Bearing 32206A (C = 43800N) is suitable. 𝑭𝒂𝑨 𝑭𝒓𝑨 = 𝟑𝟐𝟏. 𝟑 𝟑𝟔𝟓. 𝟎𝟓𝟐 = 𝟎. 𝟖𝟖 > 𝒆 = 𝟎. 𝟑𝟕 (Refer: PSG design DATA, pg.no. 4.4 & 4.25) PA = Equivalent bearing load = X.FrA + Y.FaA = 0.4x365.052 + 1.6x321.3 = 660.1N FrA = 365.052N & FaA = 321.3N
  • 71.
    vii) Diameter forbearing mounting is d = 30 mm (Gear side) Trial 1 For Bearing B : 32206A is selected (Refer: PSG design DATA, pg.no. 4.25) C = dynamic capacity = PBx(𝐿10)0.3x (Load factor) = 910.7 x (1035)0.3 x (1.5) = 10963.5N Hence, Bearing 32206A (C = 46800N) is suitable. 𝑭𝒂𝑩 𝑭𝒓𝑩 = 𝟐𝟖𝟒. 𝟔 𝟗𝟏𝟎. 𝟕𝟏 = 𝟎. 𝟑𝟏𝟐𝟓 < 𝒆 = 𝟎. 𝟑𝟕 (Refer: PSG design DATA, pg.no. 4.4 & 4.25) FrB = 910.71N and FaB = 284.6N PB = Equivalent bearing load = X.FrB + Y.FaB = 1x1771.56+ 0x553.61 = 910.71N
  • 72.
    • Locking arrangementfor bearing A (of input shaft only) Lock washer Ring Nut iv) We have to find suitable Lock washer and Ring nut from manufactures catalogue which totally depends on shaft dia. At these section i) Part No. are 29,30 and Part name is Lock washer and Ring nut respectively. ii) Function: Locking bearing A from other end to restricts it’s movement in axial direction iii) The shaft diameter for lock washer and ring nut is 0.8d = 0.8x30 = 24mm
  • 73.
    v) Finally, thestandard elements select from PSG design data tables Notations Values (mm) h 23 d1 32 d2 42 a 5 b 5 t 1.25 N 13 (Refer: PSG design DATA, pg. No. 5.86) Notations Values (mm) d 25 D 38 b 7 s 4 t 2 M25 x 1.5 (Refer: PSG design DATA, pg. No. 5.85)
  • 74.
    i) Part No.is 20 and Part Name is Input shaft key v) Check for induced stresses; 𝜏 = 2𝑇 𝑏ℎ𝑙 = 2×13493.5 8×7×𝟑𝟎 = 16.72N/mm2 << 126.67N/mm2 Also, 𝜎 = 2 𝜏 = 33.44N/mm2 << 253.33N/mm2 iv) Orthographic views , FBD, fos = 1.5 are same as earlier case Diameter for gear mounting is d = 30mm, Hence bxh = 8x7 and keyway depth is 1.7mm also, length of key = 𝑙 = 30 mm < 60mm ii) Function: Mounting Bevel gear (G1) on input shaft vi) Hence, our design is safe bxhxl = 8x7x45 • Selection of Keys for gear mounting on Input shaft iii) The martial selected for Key is plain carbon steel with grade C50 1. Sut = 660 N/mm2 2. Syt = 380 N/mm2
  • 75.
    i) Part No.is 28 and Part Name is circlip 3 ii) Function: Locking bevel gear (G1) as well as bearing B iii) The diameter for G1 mounting is d = 30mm thus circlip selected from light series A is • Selection of circlip (External type) Circlip, Light A 30 (Refer: PSG Design DATA, pg.no. 5.3 & 5.4) Notation Value (mm) Notation Value (mm) d1 30 d5 2 d2 28.6 s 1.5 d3 27.9 b 3.5 d4 41 a 5
  • 76.
    • Locking arrangementfor G1 from open end iv) Lock washer and Ring nut assembly is same as that of selected on slides 72 & 73 because mounting dia. is same in both the cases i.e 24mm i) Part No. is 31 and Part name is Washer ii) Function: Evenly distributing load from nut to larger area iii) The shaft dia. for washer is 24mm, thus dimension obtained from manufactures catalogue are Washer Lock washer & Ring nut assembly Notation Values (mm) d 24 D 48 s 7
  • 77.
    i) Part No.is 17 and Part Name is Input shaft oil seal ii) Function: Prevents oil leakage from input end of gearbox iii) Diameter for oil seal mounting is 0.8d = 0.8x30 = 24 mm iv) Also Type A oil is selected for input shaft since torque acting on it is quite low as well as it running at high speed Type A : Stiffener (metal) ring is covered with rubber elastomer to tolerate thermal expansion • Selection of Oil seal for Input shaft v) Thus the oil seal selected for 24mm shaft dia. from catalogue have dimension of Notation Value b 7 c (min) 0.3 Nominal housing diameter 35 (Refer: PSG Design DATA, pg. No. 5.11) Note: Find fit allowance and tolerances for Type A seals , see table on pg. No. 5.112
  • 79.
    i) Part No.is 18 and Part Name is Output shaft oil seal ii) Function: Prevents oil leakage from input end of gearbox iii) Diameter of shaft R for oil seal mounting is 1.1d = 0.8x44 = 48.4 mm iv) Also Type B oil is selected for output shaft since torque acting on it is very high compared to input shaft Type B : Stiffener (metal) ring is at outer periphery enclosing elastomer material • Selection of Oil seal for Output shaft v) Thus the oil seal selected for 48.4mm shaft dia. from catalogue have dimension of Notation Value b 8 c (min) 0.4 Nominal housing diameter 62 (Refer: PSG Design DATA, pg. No. 5.11) Note: Find fit allowance and tolerances for Type B seals , see table on pg. No. 5.112
  • 80.
    • Design ofCasing 1. Lower casing  FBD  2 orthographic views with design notation  Material  Factor of safety  Calculations  Table of final dimensions
  • 81.
    • Design ofCasing 1. Upper casing  FBD  2 orthographic views with design notation  Material  Factor of safety  Calculations  Table of final dimensions
  • 82.
    • Design ofbearing caps  FBD  2 orthographic views with design notation  Material  Factor of safety  Calculations  Table of final dimensions
  • 83.
    • Design ofstuffing box  FBD  2 orthographic views with design notation  Material  Factor of safety  Calculations  Table of final dimensions
  • 84.
    • Weight ofgearbox Total weight of gearbox = 120 + 3.6 + 5.4 + 10 + 34.4 + 12 = 𝟏𝟖𝟔𝒌𝒈
  • 85.
    • Selection ofEye Bolts i) Part No. is 28 and Part Name is eye bolt ii) Function: Lifting of gearbox using cranes/hoists iii) Orthographic Views of eye bolt iv) Load acting on eye bolt = Overall weight of gearbox when lifted (Pw) = 186 x 10 = 1860N iv) The material selected for shaft is Plain carbon steel with grade 40C8 because, this steel is used for making shafts,bolts, crankshaft, automobile axle (Refer: PSG design data book, pg.no. 1.9, 1.10) 1. Sut = 580/mm2 2. Syt = 300N/mm2
  • 86.
    v) Factor ofsafety is taken to be 7, since as gearbox get lifted all the weight is come on eye bolt Allowable stress calculation: σ = 300 7 = 42.85MPa vii) Critical mode of failure : Shearing of threads of eye bolt σ = 4𝑃𝑤 𝜋𝑑𝑐 2 𝑑𝑐 2 = 4𝑃𝑤 𝜋σ ∴ 𝑑𝑐 2 = 4 × 1860 𝜋 × 42.85 𝑑𝑐 = 7.43𝑚𝑚 𝑑 = 𝑑𝑐 0.86 = 8.64mm M10 x 0.75 is selected. Hence,
  • 87.
    • Selection ofNut-Bolt for casing i) Part No. is 33,34 and Part Name is casing bolt & casing nut (Hexagonal Nut and Bolt) ii) Function: Joining of upper and lower casings iii) Critical mode of failure : Tensile iv) Material : 30C8 and fos = 6 M8 x 0.5 is selected. (Refer: PSG design DATA, pg.no.5.49 & 5.50 ) Such 6 Nut-bolt pair is needed
  • 88.
    The same sizeof screw(below figure) are used nearby bearing region M8 x 0.5 Screw is selected. For 4 bearing semi-circular regions we requiring 8 screw (Refer: PSG design DATA, pg.no.5.61 & 5.62 ) Again, 6 same hexagonal socket head cap screw [M8 x 0.5] is used for attachment of stuffing box to the casing The nut-bolts, screws used to join two casing are intentionally selected of fine series to ensure proper tightening and thus avoiding oil leakage
  • 89.
    • Selection offoundation Bolts i) Part No. is 32 and Part Name is Foundation bolt ii) Function: Fixing/Mounting of gearbox on support iii) Load acting on foundation bolt = Overall weight of gearbox when lifted (Pw) = 186 x 10 = 1860N iv) Critical mode of failure : Compression and shearing of threads v) The material selected for shaft is Plain carbon steel with grade 40C8 because, this steel is used for making shafts, bolts, crankshaft, automobile axle (Refer: PSG design data book, pg.no. 1.9, 1.10) 1. Sut = 580/mm2 2. Syt = 300N/mm2
  • 90.
    v) Factor ofsafety is taken to be 6, since as gearbox get lifted all the weight is come on eye bolt Allowable shear stress calculation: τ = 0.5×300 5 = 30MPa 𝑑𝑐 2 = 4𝑃𝑤 𝜋𝜏 𝑑𝑐 = 8.88𝑚𝑚 𝑑 = 𝑑𝑐 0.86 = 10.32mm M12 x 1.25 The next standard bolt is selected of, 𝜏 = 4𝑃𝑤 𝜋𝑑𝑐 2 ∴ 𝑑𝑐 2 = 4 × 1860 𝜋 × 30 On dimensional constraint basis the screw selected for- 1. bearing cap attachment to housing and 2. cover plate attachment to stuffing box is of M5 x 0.5 6 for each cap x 5 = 30 Hexagonal socket head cap screw are used.