muhammad jawhar

1. 1
Gear Box
Name: Muhammad Jawhar Anwar
Class: 4 Stage –A
Subject: Design Project
Department: Mechanical and
Mechatronics
Supervised by : DR. Buland
College of
Engineering

2. 2
Salahaddin
University-Erbil
Academic Year
2021-2022

3. 3
Specifications Uni
ts
value
Power to be
delivered
Hp 27.6
Input speed rpm 1992
Output speed
Range ±3
rpm 103
Height in 24
Width x Length in 14 x 14
Gear and bearing
life
Hou
r
(s)
>13000
I/P and O/P shafts
extension
in 4
I/P and O/P shafts
orientation
In-line (reverted
gearbox)
Shock level Usually low and
occasional moderate
shocks
Chapter1
Speed, Torque, and Gear Ratios
• Calculation of the Number of Teeth for each gear
𝑒 =
𝜔5
𝜔2
𝑒 =
103
1992
=
1
19.4
𝑒 =
1
19.4
=
𝑁2
𝑁3
𝑁4
𝑁5
𝑁2
𝑁3
=
𝑁4
𝑁5
= √
1
19.4
=
1
4.4
` 𝑁𝑝 =
2𝑘
(1+2𝑚) sin2 𝜙
(𝑚 + √𝑚2 + (1 + 2𝑚)sin2 𝜙)

4. 4
𝑁𝑝 =
2 ∗ 1
(1 + 2(4.4))sin2 20
(4.4 + √(4.4)2 + (1 + 2(4.4))sin2 20)
𝑁𝑝 = 15.53 ≈ 16 teeth
𝑁2 = 𝑁4 = 16 teeth
𝑁3 = 4.4(𝑁2) = 70.4 ≈ 70 teeth
𝜔5 = (
𝑁2
𝑁3
)(
𝑁4
𝑁5
)𝜔2
𝜔5 = (
16
70
)(
16
70
)∗ 1992 = 104.07rpm
100 < 𝜔5 < 106
𝜔3 = 𝜔4 = (
16
70
) ∗ 1992 = 455.3 rpm
𝑇2 =
𝐻
𝜔2
𝑇2 = (
27.6ℎ𝑝
1992𝑟𝑝𝑚
)(550
𝑓𝑡 − 𝑙𝑏/𝑠
ℎ𝑝
)(
1𝑟𝑒𝑣
2𝜋𝑟𝑎𝑑
)(60
𝑠
𝑚𝑖𝑛
)
𝑇2 =
27.6∗ 550 ∗ 60
1992∗ 2𝜋
𝑇2 = 72.81𝑙𝑏𝑓.𝑓𝑡
𝑇3 = 𝑇2 (
𝜔2
𝜔3
)………… …. . . eq 9
𝑇3 = 72.81∗ (
1992
455.3
) = 318.55lbf.ft
𝑇5 = 𝑇2 (
𝜔2
𝜔5
)
𝑇5 = 72.81∗ (
1992
104.07
) = 1393.65lbf.ft
𝑃min =
(𝑁3 +
𝑁2
2
+
𝑁5
2
+ 2)
𝑌 − (clearances + wall thickness)
…… ……. . Eq. (11)

5. 5
Clearances + wall thickness = 1.5in
𝑃min =
(70 +
16
2
+
70
2
+ 2)
24 − (1.5)
= 5.
11 teeth
in
round down to 𝑃min = 5 teeth /in
𝑑2 = 𝑑4 =
𝑁2
𝑃
=
16
5
= 3.2in
𝑑3 = 𝑑5 =
𝑁3
𝑃
=
70
5
= 14in
𝑉23 =
𝜋𝑑2𝜔2
12
…………… .. 𝐸𝑞12
𝑉23 =
𝜋(3.2)(1992)
12
𝑉23 = 1668ft/min
𝑉
45 =
𝜋𝑑5𝜔5
12
𝑉
45 =
𝜋(14)(104.07)
12
𝑉
45 = 381.24ft/min
𝑊
23
𝑡
= 33000
𝐻
𝑉23
……………… . 𝐸𝑞. 13
𝑊
23
𝑡
= 33000
27.6
1668
= 546lbf
𝑊
45
𝑡
= 33000
H
𝑉
45
𝑊
45
𝑡
= 33000
27.6
381.24
= 2389lbf
Gear N 𝝎 T d V W
# rpm Ibf.ft in Ft//min Ibf
2 16 1992 72.81 3.2 1668 546
3 70 455.3 318.55 14 1668 546
4 16 455.3 318.55 3.2 381.24 2389
5 70 104.07 1393.65 14 381.24 2389

6. 6
Chapter 3
𝜎𝑐 = 𝐶𝑝√𝑊𝑡𝐾𝑜𝐾𝑣𝐾𝑠
𝐾𝑚
𝑑𝑝𝐹
𝐶𝑓
𝐼
… …………. Eq. (14)
𝐶𝑝 = 2300 √𝑙𝑏𝑓/𝑖𝑛2 for steel pinion and steel mating
gear
𝑊𝑡
= 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑙𝑜𝑎𝑑
𝐾𝑜 = 1 is the overload factor. Usually =1
.....................(no shocks) also see
table 3.2.
𝐾𝑠 = 1 is the size factor =1 (as suggested

7. 7
by AGMA)
𝐶𝑓 = 1
𝐾𝑣 = (
𝐴 + √𝑉
𝐴
)
𝐵
………… .Eq. 15
𝐴 = 50 + 56(1− 𝐵) ………… 𝐸𝑞. (16)
𝐵 = 0.25(12 − 𝑄𝑣)2/3
……… .. 𝐸𝑞. (17)
𝑄𝑣 = 7
𝐵 = 0.25(12− 7)2/3
= 0.731
𝐴 = 50 + 56(1− 0.731) = 65.1
𝐾𝑣 = (
65.1+ √381.24
65.1
)
0.731
= 1.21
𝐾𝑠 = 1 size factor
𝐾𝑚 = 𝐶𝑚𝑓 = 1 + 𝐶𝑚𝑐(𝐶𝑝𝑓𝐶𝑝𝑚 + 𝐶𝑚𝑎𝐶𝑒 )…… ……. . 𝐸𝑞. (18)
𝐶𝑚𝑐 = 1 because is uncrowned teeth
𝐶𝑝𝑓 = (
𝐹
10𝑑
− 0.0375+ 0.0125𝐹)…… …. . 𝐸𝑞19 for 1<F<17in.
𝐹 = (3 ∼ 5)(
𝜋
𝑃
)… …………. 𝐸𝑞. (21)
𝐹 = 4 (
𝜋
5
) = 2.512in
Round up to F= 3 in
𝐶𝑝𝑓 = (
3
10(3.2)
− 0.0375+ 0.0125(3)) = 0.094 𝐶𝑝𝑚 = 1
𝐶𝑚𝑎 = 𝐴 + 𝐵𝐹 + 𝐶𝐹2

8. 8
We choice commercial condition because Qv =7
𝐴 = 0.127,𝐵 = 0.0158,𝐶 = −0.930(10)−4
𝐶𝑚𝑎 = 0.127+ 3(0.0158) − 0.930(10)−4
∗ (3)2
𝐶𝑚𝑎 = 0.17
𝐶𝑒 = 1
𝐶𝑓 = 1
𝐾𝑚 = 1 + 𝐶𝑚𝑐 (𝐶𝑝𝑓𝐶𝑝𝑚 + 𝐶𝑚𝑎𝐶𝑒)
𝐾𝑚 = 1 + 1 ∗ (0.094(1) + 0.17(1))
𝐾𝑚 = 1.26
𝐼 =
cos 𝜙sin 𝜙
2𝑚𝑁
(
𝑚𝐺
𝑚𝐺 + 1
)…………… … Eq. (22)
𝐼 =
cos 20sin20
2(1)
(
4.375
4.375 + 1
) = 0.1308
𝜎𝑐 = 𝐶𝑝√𝑊𝑡𝐾𝑜𝐾𝑣𝐾𝑠
𝐾𝑚
𝑑𝑝𝐹
𝐶𝑓
𝐼
𝜎𝑐 = 2300√2389∗ 1 ∗ 1.21∗ 1 ∗ (
1.26
3.2 ∗ 3
)(
1
0.1308
) = 123872psi

9. 9
𝜎𝑐, all = (
𝑆𝑐
𝑆𝐻
)(
𝑍𝑁 𝐶𝐻
𝐾𝑇𝐾𝑅
)…………. . Eq. (23)
𝐿4 = 𝑡 ∗ 𝑛4 ……… …. . Eq. (24)
𝐿4 = 13000 ∗ 60 ∗ 455 = 3.5 ∗ 108
rev
Using fig 3.3 to find 𝑍𝑁
𝑍𝑁 = 0.9
𝐾𝑇 = 𝐾𝑅 = 𝐶𝐻 = 1
𝑆𝐻 = 1.2 design factor
So
𝑆𝑐 =
𝜎𝑐𝑆𝐻
𝑍𝑁
=
1.2∗ 123872
0.9
= 165163psi

10. 10
Using table 3-4 the type of steel gear
𝑆𝑐 = 165163 is grade 1 carbuzied and hardened 𝑔𝑒𝑎𝑟, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝜎𝑐,𝑎𝑙𝑙 = 𝑆𝑐𝑍𝑁 = 180000∗ 0.9 = 162000psi
𝑛𝑐 =
𝜎𝑐,𝑎𝑙𝑙
𝜎𝑐
=
162000
123872
= 1.1.31
Bending of gear 4
𝜎 = 𝑊𝑡
𝐾𝑜𝐾𝑣𝐾𝑠
𝑃𝑑
𝐹
𝐾𝑚𝐾𝐵
𝐽
……………. Eq. 26

11. 11
Using fig 3.4
𝑁4 = 16 teeth, then 𝐽 = 0.27
𝑊𝑡
= 2389lbf
𝐾𝑣 = 1.21
𝐾𝑜 = 1
𝐾𝑠 = 1
𝐾𝑚 = 1.26
𝐹 = 3in.
𝐾𝐵 = 1
𝑃𝑑 = 5 teeth /in
𝜎 = 2389∗ 1.21∗ (
5
3
)(
1.26
0.27
) = 22483.psi

12. 12
Then we use fig 3.5 L=3.5 ∗ 108
rev
𝑌𝑁 = 0.95

13. 13
𝑆𝑡 = 55000 psi
𝜎𝑎𝑙𝑙 = 𝑆𝑡𝑌𝑁 = 55000∗ 0.95 = 52250psi
𝑛 =
𝜎𝑎𝑙𝑙
𝜎
=
52250
22483
= 2.32
Wear of gear 5
𝜎𝑐 = 𝐶𝑝√𝑊𝑡𝐾𝑜𝐾𝑣𝐾𝑠
𝐾𝑚
𝑑𝑝𝐹
𝐶𝑓
𝐼
𝜎𝑐 = 2300√2389∗ 1 ∗ 1.21∗ 1 ∗ (
1.26
14 ∗ 3
)(
1
0.1308
) = 59222psi
𝐿5 = 13000∗ 60 ∗ 104 = 8.11 ∗ 107
rev
𝑍𝑁 = 1.0
Choosing grade 1 through hardened steel to 250HB

14. 14
Using fig 3.6 at HB=250 to find 𝑆𝑐 = 109600psi
𝑛𝑐 =
𝜎𝑐,𝑎𝑙𝑙
𝜎𝑐
=
109600
59222
= 1.85
Bending of gear 5
𝜎 = 𝑊𝑡
𝐾𝑜𝐾𝑣𝐾𝑠
𝑃𝑑
𝐹
𝐾𝑚𝐾𝐵
𝐽
N5=70 , then J=0.415
𝜎 = 2389∗ 1 ∗ 1.21 ∗ 1 ∗
5
3
∗
1.26
0.415
= 14628psi
Choosing the same material grade1
Using fig 3.7 at HB=250
St= 77.3 HB+ 12 800 psi
𝑆𝑡 = 32125
𝑌𝑁 = 0.97

15. 15
𝜎𝑎𝑙𝑙 = 𝑆𝑡𝑌𝑁 = 32125∗ 0.97 = 31161
𝑛 =
𝜎𝑎𝑙𝑙
𝜎
=
31161
14628
= 2.1
Wear of gear 2
𝑉23 = 1668
ft
min
𝑊
23
𝑡
= 546 ibf
Using eq 15
𝐾𝑣 = (
65.1+ √1668
65.1
)
0.731
= 1.42
Gear 2&3 is lower than 4&5 therefore
Select F=2.25 in.
𝐶𝑝𝑓 = (
𝐹
10𝑑
− 0.0375+ 0.0125𝐹)
𝐶𝑝𝑓 = (
2.25
10(3.2)
− 0.0375 + 0.0125(2.25)) = 0.06
Using eq 20
𝐶𝑚𝑎 = 0.127+ 2.25(0.0158) − 0.930(10)−4
∗ (2.25)2
= 0.16
𝐾𝑚 = 1 + 𝐶𝑚𝑐 (𝐶𝑝𝑓𝐶𝑝𝑚 + 𝐶𝑚𝑎𝐶𝑒)
𝐾𝑚 = 1 + 1 ∗ (0.06(1) + 0.16(1))
𝐾𝑚 = 1.22
Using eq 14

16. 16
𝜎𝑐 = 𝐶𝑝√𝑊𝑡𝐾𝑜𝐾𝑣𝐾𝑠
𝐾𝑚
𝑑𝑝𝐹
𝐶𝑓
𝐼
𝜎𝑐 = 2300√546∗ 1 ∗ 1.42 ∗ 1 ∗ (
1.22
3.2 ∗ 2.25
)(
1
0.1308
) = 72892psi
Using eq 24
𝐿2 = 13000 ∗ 60 ∗ 1992 = 1.55∗ 109
rev
Use fig 3.3
𝑍𝑁 = 0.85
Use table 3.4 & ant try grade 1, flame hardened steel
𝑆𝑐 = 170 000psi
𝑛𝑐 =
𝜎𝑐,𝑎𝑙𝑙
𝜎𝑐
=
170000∗ 0.85
72892
= 1.98
Bending of gear 2
N2=16 teeth then
J=0.27
Using fig 3.5 as 𝐿2 = 1.55∗ 109
rev , 𝑌𝑁 = 0.86
Using eq 26
𝜎 = 𝑊𝑡
𝐾𝑜𝐾𝑣𝐾𝑠
𝑃𝑑
𝐹
𝐾𝑚𝐾𝐵
𝐽
𝜎 = 546∗ 1 ∗ 1.42 ∗ 1 ∗
5
2.25
∗
1.22
0.27
= 7976psi
Using the same material from table 3.5 flame hardened
steel

17. 17
𝑆𝑡 = 45000psi
Using eq 27
𝜎𝑎𝑙𝑙 = 𝑆𝑡𝑌𝑁 = 45000∗ 0.88 = 39600psi
𝑛 =
𝜎𝑎𝑙𝑙
𝜎
=
39600
7976
= 4.96
Wear of gear 3
𝐿3 = 13000 ∗ 60 ∗ 455 = 3.55∗ 108
rev
𝑍𝑁 = 0.9
𝜎𝑐 = 𝐶𝑝√𝑊𝑡𝐾𝑜𝐾𝑣𝐾𝑠
𝐾𝑚
𝑑𝑝𝐹
𝐶𝑓
𝐼
𝜎𝑐 = 2300√546∗ 1 ∗ 1.42 ∗ 1 ∗ (
1.22
14 ∗ 2.25
)(
1
0.1308
) = 34845psi
Choosing grade 1 through hardened steel to 200HB
Using fig 3.6 then
𝑆𝐶 = 90000psi

18. 18
𝑛𝑐 =
𝜎𝑐,𝑎𝑙𝑙
𝜎𝑐
=
90000 ∗ 0.9
34845
= 2.3
Bending of gear 3
N3=70 then J=0.415 from fig 3.4
𝜎 = 𝑊𝑡
𝐾𝑜𝐾𝑣𝐾𝑠
𝑃𝑑
𝐹
𝐾𝑚𝐾𝐵
𝐽
𝜎 = 546 ∗ 1 ∗ 1.42 ∗ 1 ∗
5
2.25
∗
1.22
0.415
= 5065psi
Fig 3.5
𝑌𝑁 = 0.9
Using the same material grade 1
𝑆𝐶 = 28000psi
𝑛 =
𝜎𝑎𝑙𝑙
𝜎
=
28000 ∗ 0.9
5065
= 4.97
Gear Material
grade
Treatment Wear
stress
psi
Bending
stress
psi
d in F in
2 1 Flame-
hardened
170000 45000 3.2 2.25
3 1 Through-
hardened
to 200HB
90000 28000 14 2.25
4 3 Carburized
and
hardened
180000 55000 3.2 3.0

19. 19
5 2 Through-
hardened
to 250HB
109600 32125 14 3.0