machine design ,gear box design mahamad jawhar.pdf

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
Salahaddin University-Erbil
Academic Year 2021-2022

2
Specifications Units 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 Hour
(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
Chapter2
Speed, Torque, and Gear Ratios
• Calculation of the Number of Teeth for each gear
√
` ( )
(𝑚 √𝑚 ( 𝑚) )
( ( ))
. √( ) ( ( )) )

3
( )
( ) ( )
( ) ( )
( )
(
𝑝
𝑟𝑝𝑚
) (
𝑡 𝑙 𝑠
𝑝
) (
𝑟 𝑣
𝑟
) .
𝑠
𝑚 𝑛
/
𝑙 𝑡
( )
( )
( )
( )
. /
( )
( )

4
. /
( )
𝑞
( )( )
( )( )
𝑞
Gear N 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

5
Chapter 3
√ ( )
√𝑙 𝑛 for steel pinion and steel mating gear
𝑡 𝑡 𝑛 𝑛𝑡 𝑙 𝑡𝑟 𝑛𝑠𝑚 𝑡𝑡 𝑙𝑜
is the overload factor. Usually =1 (no shocks)
also see table 3.2.
is the size factor =1 (as suggested by AGMA)
(
√
)
( ) 𝑞 ( )
( ) 𝑞 ( )
( )
( )
(
√
)
size factor
( ) 𝑞 ( )
because is uncrowned teeth
. / 𝑞 for 1<F<17in.

6
( ) . / 𝑞 ( )
. /
Round up to F= 3 in
(
( )
( ))
We choice commercial condition because Qv =7
( )
( ) ( ) ( )
( )
( ( ) ( ))

7
𝑚
(
𝑚
𝑚
) ( )
( )
( )
√
√ ( ) ( )
( ) ( ) ( )
𝑡 𝑛 ( )
rev
Using fig 3.3 to find

8
So
Using table 3-4 the type of steel gear
𝑟 𝑡 𝑟 𝑜𝑟
𝑛
Bending of gear 4

10
Then we use fig 3.5 L= rev
psi
𝑛

11
Wear of gear 5
√
√ ( ) ( )
Choosing grade 1 through hardened steel to 250HB
Using fig 3.6 at HB=250 to find
𝑛

12
Bending of gear 5
N5=70 , then J=0.415
Choosing the same material grade1
Using fig 3.7 at HB=250
St= 77.3 HB+ 12 800 psi
𝑛

13
Wear of gear 2
Using eq 15
(
√
)
Gear 2&3 is lower than 4&5 therefore
Select F=2.25 in.
( )
(
( )
( ))
Using eq 20
( ) ( ) ( )
( )
( ( ) ( ))
Using eq 14
√
√ ( ) ( )
Using eq 24

14
rev
Use fig 3.3
Use table 3.4 & ant try grade 1, flame hardened steel
𝑛
Bending of gear 2
N2=16 teeth then
J=0.27
Using fig 3.5 as rev ,
Using eq 26
Using the same material from table 3.5 flame hardened steel
Using eq 27
𝑛

15
Wear of gear 3
rev
√
√ ( ) ( )
Choosing grade 1 through hardened steel to 200HB
Using fig 3.6 then
𝑛
Bending of gear 3
N3=70 then J=0.415 from fig 3.4
Fig 3.5
Using the same material grade 1

16
𝑛
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
5 2 Through-
hardened
to 250HB
109600 32125 14 3.0

17
Chapter 4
Shaft Design
Determination of the axial location of gears and bearings
Figure 4. 1 Shaft layout for the reverted gearbox indicating the estimated axial distances between
the gears and bearings, distances are in inches.
The face width of each gear is known
2 = 3 = 1.5 𝑛.
Similarly
4 = 5 = 2.0 𝑛.

18
Figure 4. 2 Counter shaft diameters, shoulders and the axial locations of the gears and bearing.
From the previous chapter
23𝑡 = 546 𝑙
45𝑡 = -2389 𝑙
𝑟 = 𝑡 tan … … … … … … … … … 𝑞. 28
Therefore
23𝑟 = 23𝑡 tan
23𝑟 = 546 tan 20
23𝑟 = −198.72 𝑙
Similarly 45𝑟 = 45𝑡 tan
= tan =2389 tan20 = -869.5 ibf
The next step is to select a suitable material for the shaft followed by a rough estimation of
the shaft diameters at different axial locations. This could be achieved by determination of

19
the fatigue and static stress capacity based on the infinite life of the shaft and a minimum
safety factor of 1.5.
Force analysis
• Determination of reaction forces on bearings.
x-z plane
𝑥 𝑥 𝑥
( ) ( )
( )
𝑙
𝑙
x-y plane
𝑥 𝑥 𝑥
( ) ( )
( )
lbf
lbf

20
0 0
3822 3822
0 0
0
500
1000
1500
2000
2500
3000
3500
4000
4500
y
[in.]
x [in.]
torque diagram

21
101 101
647 647
-1742 -1742
0
-2000
-1500
-1000
-500
0
500
1000
y
[in.]
x [in.]
shear force diagram
0
202
3275
3922
2180
0
0
500
1000
1500
2000
2500
3000
3500
4000
4500
y
[in.]
x [in.]
moment diagram(x -z)

22
354 354
155.28 155.25
-714.22 -714.22
0
-800
-600
-400
-200
0
200
400
600
y
[in.]
x [in.]
shear force diagram(x -y)
0
708
1446
887
0
0
200
400
600
800
1000
1200
1400
1600
y
[in.]
x [in.]
moment diagram(x -y)
0
736
3580
4236
2354
0
0
500
1000
1500
2000
2500
3000
3500
4000
4500
y
[in.]
x [in.]

23
Figure 4. 3 Free body diagram, torque, shear force and bending moment diagrams for the shaft.
• The torque diagram
T3= *
T3= 546 *
= 3822 ibf .in
T3= *
T3= - 2389 * = -3822 ibf .in
The Shear force and Bending moment
diagrams x-y plane
𝐺 = = 354 𝑙
𝐺 = -
𝐺 = 354 – 198.72
𝐺 = 155.28 𝑙
= 𝐺 −
= 155.28 – 869.5
= −714.22 Ibf =
=0

24
𝑥
( )
𝑙 in
( ) ( )
( ) ( )
𝑙 𝑛
( ) ( ) ( )
( ) ( ) ( )
𝑙 𝑛
x-z plane
𝐺 = = 101 𝑙
𝐺 = +
𝐺 = 101 + 546
𝐺 = 647 𝑙
= 𝐺 −
= 647 – 2389
= −1742 ibf =
𝑥
𝑙 𝑛
( ) ( )
( ) ( )
𝑙 𝑛
( ) ( ) ( )
( ) ( ) ( )
𝑙 𝑛

25
The total bending moment diagram
It could be found by superposition of the moments in the x-y plane and x-z plane.
𝐺 =√( ) ( )
=736 Ibf.in
Similarly MJ=√( ) ( )
= 4236Ibf . 𝑛
MK=2354 lbf.in
The shoulder at point I
The bending moment is the highest at the shoulder (point I), the torque is high and
there is a stress concentration as well.
In the x-y plane
( ) ( )
( ) ( )
𝑙 𝑛
In the x-z plane
( ) ( )
( ) ( )
𝑙 in
Combined

26
√( ) ( )
= 3580 ibf. 𝑛
At point I, is alternating moment usually referred to as and defined by
𝑚 𝑥 𝑚 𝑛
While the midrange moment is defined as
𝑚 𝑥 𝑚 𝑛
The midrange torque is
𝑚 = 3822 𝑙 . 𝑛 𝑚 = = 0

27
Estimation of the stress concentrations
For bearings
= 1.2~1.5
Where
𝐷 𝑠 𝑡 𝑙 𝑟 𝑟 𝑚 𝑡 𝑟 𝑜 𝑡 𝑠 𝑡 𝑡 𝑡 𝑠 𝑜𝑢𝑙 𝑟, 𝑛.
𝑠 𝑡 𝑠𝑚 𝑙𝑙 𝑟 𝑚 𝑡 𝑟 𝑜 𝑡 𝑠 𝑡 𝑡 𝑡 𝑠 𝑜𝑢𝑙 𝑟, 𝑛.
The upper limit of 1.5 could be used as a first estimation and a worst-case scenario.
The fillet radius should also be sized to coincide with that of a standard bearing or gear.
= 0.02~0.06
Where
𝑟 𝑠 𝑡 𝑙𝑙 𝑡 𝑟 𝑢𝑠, 𝑛.
𝑠 𝑡 𝑟 𝑛 𝑜𝑟 𝑜𝑟 𝑡 𝑠𝑚 𝑙𝑙 𝑟 𝑠 𝑡 𝑚 𝑡 𝑟 𝑡 𝑡 𝑠 𝑜𝑢𝑙 𝑟, 𝑛.
The Marin equation for the determination of the endurance limit is
= 𝑘 𝑘 𝑘𝑐𝑘 𝑘 𝑘 ′
… … … … … … . . 𝑞. 29
Table 4. 1 Parameters for Marin surface modification factor ka.

28
Table 4. 2
A 1040 CD steel is chosen for its low price with 𝑺𝒖𝒕 = 85 𝒌𝒑𝒔𝒊, Sy = 71 kpsi. From
table 4.2.
therefore, 𝑛 𝑟 𝑙 𝑠𝑡 𝑛 𝑡 𝑙 4.1
= 2.7 𝑘𝑝𝑠 = −0.265
Then
𝑘 = 2.7(85)−0.265
= 0.831

29
Let
𝑘 = 0.9 (𝑡𝑜 𝑐 𝑐𝑘 𝑙 𝑡 𝑟 − 𝑜𝑛 𝑤 𝑛 𝑠 𝑘𝑛𝑜𝑤𝑛)
𝑘𝑐 = 𝑘 = 𝑘 = 1
And
′
= 0.5 𝑢𝑡 𝑜𝑟 𝑢𝑡 ≤ 200 𝑘𝑝𝑠
′
= 100 𝑘𝑝𝑠 𝑜𝑟 𝑢𝑡 > 200 𝑘𝑝𝑠
Using eq. 29
= 0.831 0.9 0.5 85
= 31.78 𝑘𝑝𝑠
Using DE-Goodman criterion to make the first estimate for the smaller diameter at the
shoulder
√ * √ (𝑘 ) (𝑘 𝑠 ) √ (𝑘 𝑚) (𝑘 𝑠 𝑚) +
Assuming shoulder-fillet well-rounded or
r/d=0.1 then From table 4.3
𝑡 = 1.7
𝑡𝑠 = 1.5
Assume
= 𝑡 𝑛 𝑠 = 𝑡𝑠
with

30
√ { √ }
Since the Goodman criterion is conservative, then it possible to select the next standard
size below 1.65 which is
=1
= 1.625 𝑛.
A typical diameter ratio at the shoulder is
= 1.2
Therefore,
𝐷 = 1.2 1.625 = 1.95 𝑛.
Round up to 𝐷 = 2.0 𝑛.
A nominal 2.0 in. CD shaft is used
Checking the acceptability of the estimates
𝐷
Assume fillet radius to diameter ratio
𝑟
= 0.1
Therefore,
𝑟 = 0.1 1.625 = 0.16 𝑛.

31
Using 𝑡𝑜 𝑛 𝑡 𝑠𝑡𝑟 𝑠𝑠 𝑐𝑜𝑛𝑐 𝑛𝑡𝑟 𝑡 𝑜𝑛 𝑘𝑡
𝑡 = 1.6
Using fig.4.6 to find the sensitivity factor q.
Then
𝑞 = 0.82
= 1 + ( 𝑡 − 1) … … … … . . 𝑞. 33
= 1 + 0.82(1.6 − 1)
=1.49
Using fig. 4.7 for (r/d) =0.1 and (D/d) =1.23
then
𝑡𝑠 = 1.35
Using fig. 4.8 for r=1.6 in. then
𝑞𝑠 = 0.95
𝑠 = 1 + 𝑞𝑠 ( 𝑡𝑠 − 1) … … … … … 𝑞. 34
𝑠 = 1 + 0.95(1.35 − 1)
𝑠 = 1.33
= 0.831 (𝑛𝑜𝑡 𝑐 𝑛 )
Kb=( ) 𝑜𝑟 0.11 < ≤ 2 𝑛
Using eq. 35
Kb=( )

32
Using eq. 29
= 0.831 0.835 0.5 85
= 29.49 𝑘𝑝𝑠
The Von Mises stresses for rotating round, solid shafts, neglecting axial loads are given by:
=( ( ) 0( ) ( ) 1
( )
𝑝𝑠
=( ( ) ⁄
0( ) ( ) 1
* ( ) +
√
( )
𝑛
𝑛
𝑛
Check for Yielding

33
The von Mises maximum stress is :
𝑚 𝑥′= [( 𝑚 + )2 + 3( 𝑚 + )2]1⁄2
*(
𝑘 ( 𝑚 )
) (
𝑘 𝑠( 𝑚 )
) +
To check for yielding, Eq. 30 is compared to the yield strength, then
𝑛𝑦 =
Note: for a quick conservative check, the sum of alternating stress and the midrange
stress is always greater or equal to the maximum stress therefore, the results will be
conservative.
Therefore,
𝑛𝑦
𝑠𝑦 𝑠𝑦
𝑛𝑦 = 3

34
The Keyway
The keyway is a groove made in the shaft to fix the gear mounted on the shaft in
position. This keyway affects the diameter of the shaft and reduces its strength. The
keyway extends to a point just to the right of point (I).
From the total bending moment diagram, let the moment at the end of keyway close
to point (I) is
= 3 750𝑙 . 𝑛
Assume that at the bottom of the keyway
= 0.02 ( 𝑠𝑡 𝑛 𝑟 𝑝𝑟𝑜𝑝𝑜𝑟𝑡 𝑜𝑛)
Therefore,
𝑟
𝑟 .
Using fig. 4.5
𝑡 = 2.4
And using fig. 4.6
𝑞 = 0.69
Using eq.33
= 1 + ( 𝑡 − 1)
= 1 + 0.69(2.4 − 1)
= 1.966
Using fig. 4.7 𝑡𝑠 = 2.2
Fig 4.8 gives9 𝑞𝑠 = 0.9
Using eq.34 𝑠 = 1 + 𝑞𝑠 ( 𝑡𝑠 − 1)
𝑠 = 1 + 0.9(2.2 − 1)
𝑠 = 2

35
( )
𝑝𝑠
* ( ) +
√
( )
=15722 psi
𝑛
𝑛
𝑛
Note: the results show that the keyway is more critical than the shoulder. Therefore, to
overcome the conundrum, it is recommended to change the shaft material to a higher
strength or to change its diameter to a larger size.
Trying
steel 1050 CD with 𝑺𝒖𝒕 = 𝟏𝟎𝟎 𝒌𝒑𝒔𝒊, 𝑺𝒚 = 𝟖𝟒 𝒌𝒑𝒔𝒊. From table 4.2.
and recalculating the factors affected by 𝑢𝑡.
Using eq.30
𝑘 = 𝑢𝑡
𝑘 = 2.7(100)−0.265
= 0.797
Using eq. 29
= 0.797 0.835 0.5 100

36
= 33.275 𝑘𝑝𝑠
Fig. 4.6 gives
𝑞 = 0.73
= 1 + ( 𝑡 − 1)
= 1 + 0.73(2.4 − 1)
= 2.022
( )
𝑝𝑠
𝑛
𝑛
𝑛
This value is quite close to the required 1.5, and since the Goodman equations are
conservative therefore, the value of 𝑛 = 1.433 is acceptable.

37
Inspecting point K The retaining ring groove
At point (K) a flat bottom groove exists to accommodate the retaining ring. This point
is characterized by:
𝑡 𝑠 𝑣 𝑟𝑦 𝑜𝑟 𝑙 𝑡 𝑜𝑡𝑡𝑜𝑚 𝑟𝑜𝑜𝑣 𝑠
= 0 𝑠 𝑛 𝑐 𝑡 𝑦 𝑡 𝑡𝑜𝑟𝑞𝑢 𝑟 𝑚
= 2354 𝑙 . 𝑛 𝑠 𝑛 𝑐 𝑡 𝑦 𝑡 𝑚𝑜𝑚 𝑛𝑡 𝑟 𝑚
𝑚 = = 𝑚 = 0
In order to check how critical point (K) is
Using table 4.3 to estimate the stress concentration factor as follows:
= 𝑡 = 5.0
′ = =
( )( )
( )
𝑝𝑠
𝑛
This quick check revealed a low safety factor.
For a detailed determination of point (K) specs, standard retaining ring specs could be
obtained from related tables and going online to (www.globalspec.com)
Using the shaft diameter of 1.625 in. to get
Groove width
= 0.068 𝑛.
Groove depth
𝑡 = 0.048 𝑛.
Fillet radius at bottom of the groove
𝑟 = 𝑛.
Therefore,

38
𝑟
𝑡
/𝑡=1.42
= 𝐷 − 2𝑡
= 1.625 − 2 0.048
= 1.529
𝑛 Using fig. 4.9 for flat groove on a round shaft
𝑡 = 4.3
Fig. 4.6 yields
𝑞 = 0.63
= 1 + ( 𝑡 − 1)
= 1 + 0.63(4.3 − 1)
= 3.09
′ = =
( )( )
( )
𝑝𝑠
𝑛
• Determination of D6
Since
𝐷
= 1.2~1.5
Then d=
Therefore
𝐷6 = = 1.354 𝑛
Round up to the next standard size
Therefore 𝐷6 = 1.4 𝑛

39
Inspecting point M
Point (M) is characterized by being a shoulder to axially hold the bearing. Only
bending is present due to a small moment. However, the cross-section area is small
too, which results in a high stress concentration especially the fillet radius is small to
accommodate the bearing.
Finding the Moment at point M
In the x-y plane
= 354 (10.25-0.75) –198.72(10.25-2.75)-
869.5(10.25-8.5)
=351 ibf . 𝑛
In the x-z plane
= 101(10.25-0.75)+546(10.25-2.75)-2389(10.25-
8.5)
=874 ibf. 𝑛
Combined
=√( ) ( )
=942 ibf . 𝑛
Therefore
= 942 𝑙 . 𝑛
And
𝑚 = 𝑚 = = 0
Using table 4.3 for a quick estimation of Kt for a sharp fillet shoulder where
= 0.02

40
And
𝑡 = 2.7
Also
= 1 𝑛.
Therefore
𝑟 =d*0.02= 1 *0.02=0.02
Using fig. 4.6
𝑞 = 0.7
= 1 + ( 𝑡 − 1)
= 1 + 0.7(2.7 − 1)
= 2.19
′ = =
( )( )
( )
𝑝𝑠
𝑛
This value of (nf) is satisfactory and should be checked as the bearing is selected.
As the shaft diameters at the critical positions were determined and by taking the
shoulders heights, to retain the gears and bearings, in consideration, other shaft
diameters could be estimated. These diameters will be checked when the bearings are
selected.
𝐷1 = 𝐷7 = 1 𝑛.
𝐷2 = 𝐷6 = 1.4 𝑛.
𝐷3 = 𝐷5 = 1.625 𝑛.
𝐷4 = 2.0 𝑛.

41
The rigidity of the shaft Deflection of the shaft
Any deflection in the shaft could have detrimental consequences on the mounted gears
and bearings with high noise levels due to the vibrations. Deflection analyses require a
full knowledge of the shaft geometry and dimensions. Table 4.4 exhibits rough
guidelines for the deflections and the maximum slopes under particular elements
mounted on the shaft. Fillets, grooves, and keyways do not have a tangible effect on
shaft deflection
Table 4. 4 Typical maximum ranges for slopes and transvers deflections.
The deflection is usually determined by double integration of the moment equation
= … … … … … … … 𝑞. 40
M= 𝑥 〈𝑥 𝑥 〉 𝑐〈𝑥 𝑥 〉 〈𝑥 𝑥 〉 0 𝑥
At the shoulder

42
Y P P
Shoulder
X
RA RB
A D G H I J M B
n
Previous sectio
Next section
Figure 4. 11 A stepped shaft subjected to concentrated loads.
Divide the moment equation by the second moment of area (I) of the previous section
of the shaft. The equation becomes.
 At the distance (𝑥 ) where the shoulder exists, add a step change in ( ⁄ ) as
follows:
𝑠𝑡 𝑝 . /
𝑠𝑡 𝑝 ( ) ( )
 At the distance (𝑥 ) where the shoulder exists, add a ramp change in ( ⁄ ) as
follows:
𝑠𝑙𝑜𝑝 𝑚 [
. /
𝑥
]
𝑚 [
. / . /
𝑥
]
( )
𝑛
( )
𝑛

43
Xz plane
Shoulder H
( ) 𝑙 𝑛
( ) ( )
( ) ( ) 𝑙 𝑛
( ) 𝑙 𝑛
𝑠𝑙𝑜𝑝 𝑙 𝑛
( ) 𝑙 𝑛
( ) 𝑙 𝑛
𝑚 𝑠𝑙𝑜𝑝 𝑠𝑙𝑜𝑝
𝑚 𝑙 𝑛
𝑠𝑡 𝑝 ( ) ( )
𝑠𝑡 𝑝 𝑙 𝑛
Shoulder I
( ) 𝑙 𝑛
( ) 𝑙 𝑛

44
𝑚 𝑙 𝑛
𝑠𝑡 𝑝 ( ) ( )
Units Poin
t G
point H Sectio
n GH
sectio
n HI
point I point J Section IJ
on
AH
on HI on HI on IB
Wt
lb 546 2389
M Lb.in 202 687.25 3275 3922
I In4
0.3
42
0.34
2
0.78
5
0.78
5
0.34
2
0.342
M/I Lb/in3
590 200
8
875.
14
4170 9570 1146
1
Δ(M/I
)
Lb/in3
-1133.19 1436 3338 5400 2046.5
x In. 2 2.75 6.75 7.75
Δx In. 0.75 4 1
Δ(M/I
)/ Δx
Lb/in4
1890
.7
823.
8
1890.6
Δm Lb/in4
-1067 1067
Step
Slope

45
The moment equation
( ) 〈𝑥 〉 〈𝑥 〉
𝑥 〈 〉 . / 〈𝑥 〉 𝑚 〈 〉 〈 〉
. / 〈 〉 𝑚 〈𝑥 〉
𝑥 〈 〉 〈𝑥 〉 〈 〉
〈 〉 〈 〉 〈𝑥 〉
But
0
590
2008
875.14
4,170
9570
11461
2739.06
0
0
2000
4000
6000
8000
10000
12000
14000
0 2 4 6 8 10 12
M/I
xz
plane

46
𝑦
𝑥
Double integration yields
𝑦
𝑥
𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉
𝑦 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 𝑥
Boundary conditions
X=0 , y=0 therefore C2 =0
X=l , y=0
And substituting
E=30Mpsi for steel
𝑦
𝑥
, 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 -
𝑦 , 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 𝑥-
The deflection
𝑦 , 𝑥 𝑥- 𝑜𝑟 𝑥
𝑦 , 𝑥 〈𝑥 〉 𝑥- 𝑜𝑟 𝑥
𝑦 , 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
𝑥- 𝑜𝑟 𝑥

47
𝑦 , 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 𝑥- 𝑜𝑟 𝑥
𝑦 , 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉
𝑥- 𝑜𝑟 𝑥
The slope
, 𝑥 - 𝑜𝑟 𝑥
, 𝑥 〈𝑥 〉 - 𝑜𝑟 𝑥
𝑦
𝑥
, 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 -
𝑜𝑟 𝑥
𝑦
𝑥
, 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 -
𝑜𝑟 𝑥
𝑦
𝑥
, 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 -
𝑜𝑟 𝑥
x
[in.]
Point Fittings y
[in.]
dy/dx
[deg.]
0 A Left bearing 0 0.0221
2 G Gear 3 - 0.00076 0.0210
2.75 H Shoulder H - 0.00102
6.75 I Shoulder I - 0.00184
7.75 J Gear 4 - 0.00167 0.0203
10 B Right
Bearing
0 0.0671

48
𝑠𝑡 𝑝 . /
𝑠𝑡 𝑝 ( ) ( )
 At the distance (𝑥 ) where the shoulder exists, add a ramp change in ( ⁄ ) as
follows:
𝑠𝑙𝑜𝑝 𝑚 [
. /
𝑥
]
0
-0.00076
-0.00102
-0.00184
-0.00167
0
-0.0025
-0.002
-0.0015
-0.001
-0.0005
0
y
[in.]
x [in.]
Deflection

49
𝑚 [
. / . /
𝑥
]
( )
𝑛
( )
𝑛
X-Y plane Shoulder H
( ) 𝑙 𝑛
( ) ( )
( ) ( ) 𝑙 𝑛
( ) 𝑙 𝑛
( ) 𝑙 𝑛
( ) 𝑙 𝑛
𝑚 𝑙 𝑛

50
𝑠𝑡 𝑝 ( ) ( )
Shoulder I
( ) 𝑙 𝑛
( ) 𝑙 𝑛
𝑚 𝑙 𝑛
𝑠𝑡 𝑝 ( ) ( ) 𝑠𝑡 𝑝 𝑙
Units Point
G
point H Sectio
n GH
sectio
n HI
point I point J Section IJ
on
AH
on HI on HI on IB
Wr
lb 199 870
M Lb.in 708 824.5 1446 1601
I In4
0.34
2
0.34
2
0.78
5
0.785 0.342 0.342
M/I Lb/in3
206
9
240
9
1050.
5
1841 4225.
6
4678.
5
Δ(M/I) Lb/in3
-1360.45 1436 3338 2384.6 453
x In. 2 2.75 6.75 7.75
Δx In. 0.75 4 1
Δ(M/I)
/ Δx
Lb/in4
453.3 199 453
Δm Lb/in4
-255 255
Step
Slope

51
The moment equation
( ) 〈𝑥 〉 〈𝑥 〉
𝑥 〈 〉 . / 〈𝑥 〉 𝑚 〈 〉 〈 〉
. / 〈 〉 𝑚 〈𝑥 〉
𝑥 〈 〉 〈𝑥 〉 〈 〉
0
2069
2409
1050.55
1,841
4225.6
4678.5
1129.1117
0
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
0 2 4 6 8 10 12
M/I
xy
plane

52
〈 〉 〈 〉 〈𝑥 〉
But
𝑦
𝑥
Double integration yields
𝑦
𝑥
𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉
𝑦 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 𝑥
Boundary conditions
X=0 , y=0 therefore C2 =0
X=l , y=0
And substituting
E=30Mpsi for steel
𝑦
𝑥
, 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 -
𝑦 , 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 𝑥-
The deflection

53
𝑦 , 𝑥 𝑥- 𝑜𝑟 𝑥
𝑦 , 𝑥 〈𝑥 〉 𝑥- 𝑜𝑟 𝑥
𝑦 , 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
𝑥- 𝑜𝑟 𝑥
𝑦 , 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 𝑥- 𝑜𝑟 𝑥
𝑦 , 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 𝑥- 𝑜𝑟 𝑥 10
The slope
, 𝑥 - 𝑜𝑟 𝑥
, 𝑥 〈𝑥 〉 - 𝑜𝑟 𝑥
𝑦
𝑥
, 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 -
𝑜𝑟 𝑥
𝑦
𝑥
, 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 -
𝑜𝑟 𝑥
𝑦
𝑥
, 𝑥 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉 〈𝑥 〉
〈𝑥 〉 〈𝑥 〉 -
𝑜𝑟 𝑥

54
x
[in.]
Point Fittings y
[in.]
dy/dx
[deg.]
0 A Left bearing 0 0.0340
2 G Gear 3 - 0.000114 0.0301
2.75 H Shoulder H - 0.001680
6.75 I Shoulder I - 0.003053
7.75 J Gear 4 - 0.005927 0.00746
10 B Right
Bearing
0 0.00249
0 -0.000114
-0.00168
-0.003053
-0.005927
0
-0.007
-0.006
-0.005
-0.004
-0.003
-0.002
-0.001
0
0.001
y
[in.]
x [in.]
Deflection

55
The final bearing slope results show
slope
Bearings xz plane
[Deg.]
xy plane
[Deg.]
Total
[Deg.]
Total [rad.]
Left 0.0221 0.0340 0.056 0.000977
Right 0.0671 0.00249 0.068 0.001186

56
Shaft Design for Gear 2
23𝑡 = - 546 𝑙
𝑟 = 𝑡 tan … … … … … … … … … 𝑞. 28
Therefore
23𝑟 = 23𝑡 tan
23𝑟 = 546 tan 20
23𝑟 = 199 𝑙 q
Force analysis
x-z plane
𝑥 𝑥
( ) ( )
𝑙
𝑙
x-y plane
𝑥 𝑥
( ) ( )
lbf
lbf

57
873 873
0
0
100
200
300
400
500
600
700
800
900
1000
y
[in.]
x [in.]
torque diagram (
246 246
301 301
0
0
50
100
150
200
250
300
350
y
[in.]
x [in.]
shear force diagram

58
0
112.168
490
264
0
0
100
200
300
400
500
600
y
[in.]
x [in.]
89 89
-110 -110
0
-150
-100
-50
0
50
100
y
[in.]
x [in.]
0
282.069
490
0
0
100
200
300
400
500
600
y
[in.]
x [in.]

59
The Shear force and Bending moment diagrams
x-y plane
𝐺 = = 89 𝑙
𝐺 = -
𝐺 = 89 – 199
𝐺 = -110 𝑙
𝑥
( )
𝑙 in
( ) 𝑟
( )
( ) ( )
x-z plane
𝐺 = = -246 𝑙
𝐺 = +
𝐺 = 245 + 546
0
119.366
564.279
515.172
0
0
100
200
300
400
500
600
y
[in.]
x [in.]

60
𝐺 =791
𝑥
( )
𝑙 in
( ) 𝑟
( )
( ) ( )
𝐺 =√( ) ( )
=521 Ibf.in
The shoulder at point H
The bending moment is the highest at the shoulder (point I), the torque is high and
there is a stress concentration as well.
In the x-y plane
( ) ( )
( ) ( )
𝑙 𝑛
In the x-z plane
( ) ( )
( ) ( )
𝑙 in
Combined
H √( ) ( )

61
H= 281 ibf. 𝑛
Torque of shaft gear2
𝑛
For bearings
= 1.2~1.5
= 0.02~0.06
.
𝑘 = 𝑢𝑡 … … … … … … … … 𝑞. 30
Using the 1030 CD steel is chosen for its low price with 𝑺𝒖𝒕 = 76 𝒌𝒑𝒔𝒊, Sy = 64 kpsi.
From table 4.2.
From table 4-1 use = 2.7 𝑘𝑝𝑠 = −0.265
Then
𝑘 = 2.7(76)−0.265
= 0.8569
Let
𝑘𝑐 = 𝑘 = 𝑘 = 1
And
′
= 0.5 𝑢𝑡 𝑜𝑟 𝑢𝑡 ≤ 200 𝑘𝑝𝑠

62
Using eq. 29 = 𝑘 𝑘 𝑘𝑐𝑘 𝑘 𝑘 ′
= 0.8569 0.9 0.5 76
= 29.305 𝑘𝑝𝑠
shoulder
Assuming shoulder-fillet well-rounded or r/d=0.1
then From table 4.3
𝑡 = 1.7
𝑡𝑠 = 1.5
Assume
= 𝑡 𝑛 𝑠 = 𝑡𝑠 𝑚 = = 0
The Goodman reduces to
( * √ (𝑘 ) (𝑘 𝑠 ) √ (𝑘 𝑚) (𝑘 𝑠 𝑚) +)
√
( )
{ √ }
From using the standard size
= 0.938 𝑛.
= 1.22
Therefore,
𝐷 = 1.22 0.938 = 1.145 𝑛.

63
𝐷
= 0.1
Therefore,
𝑟 = 0.1 0.938 = 0.0938 𝑛.
Using from table 4-5 get 𝑡 = 1.63
Then 𝑞 = 0.78
= 1 + ( 𝑡 − 1) … … … … . . 𝑞. 33
= 1 + 0.78(1.63 − 1)
=1.49
then
𝑡𝑠 = 1.39
𝑞𝑠 = 0.61
𝑠 = 1 + 𝑞𝑠 ( 𝑡𝑠 − 1) … … … … … 𝑞. 34
𝑠 = 1 + 0.61(1.39 − 1)
𝑠 = 1.2379
= 0.8569 (𝑛𝑜𝑡 𝑐 𝑛 )

64
Kb=( ) 𝑜𝑟 0.11 < ≤ 2 𝑛
Kb= ( )
= 0.885 0.8569 0.5 76 = 28823 𝑘𝑝𝑠
=( ( ) 0( ) ( ) 1
( )
𝑝𝑠
=( ( ) ⁄
0( ) ( ) 1
* ( ) +
√
( )
𝑛
𝑛
𝑛

65
Check for Yielding
𝑚 𝑥′= [( 𝑚 + )2 + 3( 𝑚 + )2]1⁄2
*(
𝑘 ( 𝑚 )
) (
𝑘 𝑠( 𝑚 )
) +
𝑛𝑦
𝑠𝑦 𝑠𝑦
= 3.9
The Keyway
From the total bending moment diagram, let the moment at the end of keyway close to
point (I) is
= 390 . 𝑛
Therefore,
𝑟 =0.02*0.938
𝑟 =0.01876
Using fig. 4.5
𝑡 = 2.49
And using fig. 4.6
𝑞 = 0.58

66
= 1 + ( 𝑡 − 1)
= 1 + 0.58(2.49− 1)
= 1.8642
Using fig. 4.7 𝑡𝑠 = 2.1
Fig 4.8 gives 𝑞𝑠 = 0.74
Using eq.34 𝑠 = 1 + 𝑞𝑠 ( 𝑡𝑠 − 1)
𝑠 = 1 + 0.74(2.1− 1)
𝑠 = 1.814
( )
𝑝𝑠
* ( ) +
√
( )
= 16935 psi
𝑛
𝑛
𝑛

67
Inspecting point F The retaining ring groove
At point (F) a flat bottom groove exists to accommodate the retaining ring. This point is
characterized by:
𝑚 = = 𝑚 = 0
= 𝑡 = 5.0
′ = =
( )( )
( )
𝑝𝑠
𝑛
This quick check revealed a low safety factor.
For a detailed determination of point (K) specs, standard retaining ring specs could be
obtained from related tables and going online to ( www.globalspec.com )
Groove width
= 0.046 𝑛.
Groove depth
𝑡 = 0.028 𝑛.
𝑟 = 𝑛.
Therefore,

68
𝑟
𝑡
𝑡
= 𝐷 − 2𝑡
= 0.938 − 2 0.028
= 0.882
𝑛 Using fig. 4.9 𝑡 = 3.57
Fig. 4.6 yields
𝑞 = 0.59
= 1 + ( 𝑡 − 1)
= 1 + 0.59(3.57 − 1)
= 2.51
′ = =
( )( )
( )
𝑝𝑠
𝑛 • Determination of D
Since
𝐷/d = 1.2~1.5
Then d=
Therefore
𝐷6 = = 0.781 Therefore

69
Inspecting point D
in the x-y plane
( ) ( )
( ) ( )
𝑙 𝑛
In the x-z plane
( ) ( )
( ) ( )
𝑙 in
Combined
=√( ) ( )
=777 ibf . 𝑛
Therefore
= 777 𝑙 . 𝑛
And
𝑚 = 𝑚 = = 0
= 0.02
And
𝑡 = 2.49
Also
= 0.750 𝑛.
Therefore
𝑟 =d*0.02= 0.750*0.02=0.015
Using fig. 4.6

70
𝑞 = 0.61
= 1 +0.61(2.49 − 1)
= 1.909
′ = =
( )( )
( )
𝑝𝑠
𝑛
𝐷1 = 0.750 𝑛.
𝐷2 = 0.781 𝑛.
𝐷3 = 0.938 𝑛.
𝐷4 = 1.19 n

71
Shaft Design for Gear 5
45𝑡 = 2389 𝑙
𝑟 = 𝑡 tan … … … … … … … … … 𝑞. 28
Therefore
45𝑟 = 23𝑡 tan
45𝑟 = 2389 tan 20
45𝑟 = 870 𝑙 .
Force analysis
x-z plane
𝑥 𝑥
( ) ( )
𝑙
𝑙
x-y plane

72
𝑥 𝑥
( ) ( )
lbf
lbf
16723 16723
0
0
2000
4000
6000
8000
10000
12000
14000
16000
18000
y
[in.]
x [in.]
torque diagram (
1303
395
0
0
200
400
600
800
1000
1200
1400
y
[in.]
x [in.]
shear force diagram

73
0
2443
0
0
0
500
1000
1500
2000
2500
3000
y
[in.]
x [in.]
395 395
-395 -395
0
-500
-400
-300
-200
-100
0
100
200
300
400
500
y
[in.]
x [in.]
0
891
0
0
100
200
300
400
500
600
700
800
900
1000
y
[in.]
x [in.]

74
Figure 4. 3 Freebody diagram, torque, shear force and bending moment diagrams for the shaft.
The Shear force and Bending moment diagrams
x-y plane
𝐺 = = 475 𝑙
𝐺 = -
𝐺 = 475 – 870
𝐺 = -395 𝑙
𝑥
( )
𝑙 in
( ) 𝑟
( )
( ) ( )
0
0
0
2600
0
0
500
1000
1500
2000
2500
3000
y
[in.]
x [in.]

75
x-z plane
𝑥
( )
𝑙 in
( ) 𝑡
( )
( ) ( )
𝐺 =√( ) ( )
=2600 Ibf.in
The shoulder at point I
In the x-y plane
( ) ( )
( ) ( )
𝑙 𝑛
In the x-z plane
( ) ( )
( ) ( )
𝑙 in
Combined
I √( ) ( )
I = 1213 ibf. 𝑛

76
𝑛
The midrange torque is 𝑚 = 16723 𝑙 . 𝑛
𝑚 = = 0
A 1030 CD steel is chosen for its low price with 𝑺𝒖𝒕 = 76 𝒌𝒑𝒔𝒊, Sy =64 kpsi. From
table 4.2.
therefore,
= 2.7 𝑘𝑝𝑠 = −0.265
Then
𝑘 = 2.7(76)−0.265
= 0.8569
Let
𝑘𝑐 = 𝑘 = 𝑘 = 1
′
= 0.5 𝑢𝑡 𝑜𝑟 𝑢𝑡 ≤ 200 𝑘𝑝𝑠
Using eq. 29 = 𝑘 𝑘 𝑘𝑐𝑘 𝑘 𝑘 ′
= 0.8569 0.9 0.5 76
= 29.305 𝑘𝑝𝑠
shoulder

77
√ * √ (𝑘 ) (𝑘 𝑠 ) √ (𝑘 𝑚) (𝑘 𝑠 𝑚) +
Assuming shoulder-fillet well-rounded or r/d=0.1
then From table 4.3
𝑡 = 1.7
𝑡𝑠 = 1.5
Assume
= 𝑡 𝑛 𝑠 = 𝑡𝑠
with
𝑚 = = 0
The Goodman reduces to
√
( )
{ √ }
Since the Goodman criterion is conservative, then it possible to select the next standard
size below 1.76 which is
=1.812 in
= 1.2
Therefore,
𝐷 = 1.2 1.812 = 2.17 𝑛.
A nominal 2.25 in. CD shaft is used

78
𝐷
𝑟
Therefore,
𝑟 = 0.1 1.825 = 0.1825 𝑛.
. Using using fig 4.5 to find stress concentration
𝑡=1.61
Then
𝑞 = 0.81
= 1 + ( 𝑡 − 1) … … … … . . 𝑞. 33
= 1 + 0.81(1.61− 1)
=1.5
then
𝑡𝑠 = 1.4
𝑞𝑠 = 0.95
𝑠 = 1 + 𝑞𝑠 ( 𝑡𝑠 − 1) … … … … … 𝑞. 34
𝑠 = 1 + 0.95(1.4 − 1)
𝑠 = 1.37
= 0.8569 (𝑛𝑜𝑡 𝑐 𝑛 )

79
Kb=( ) 𝑜𝑟 0.11 < ≤ 2 𝑛
Using eq. 35
Kb=( )
Using eq. 29
= 0.824 0.8569 0.5 76
= 26.83 𝑘𝑝𝑠
=( ( ) 0( ) ( ) 1
( )
𝑝𝑠
=( ( ) ⁄
0( ) ( ) 1
* ( ) +
√
( )
𝑛
𝑛
𝑛

80
Check for Yielding
𝑚 𝑥′= [( 𝑚 + )2 + 3( 𝑚 + )2]1⁄2
*(
𝑘 ( 𝑚 )
) (
𝑘 𝑠( 𝑚 )
) +
𝑛𝑦
= 1.73
The Keyway
= 1320 . 𝑛
Therefore,
𝑟 =0.02*1.812

81
𝑟 =0.0362
Using fig. 4.5
𝑡 = 2.5
And using fig. 4.6
𝑞 = 0.7
Using eq.33
= 1 + ( 𝑡 − 1)
= 1 + 0.7(2.5− 1)
= 2.03
Using fig. 4.7 𝑡𝑠 = 2
Fig 4.8 gives9 𝑞𝑠 = 0.86
Using eq.34 𝑠 = 1 + 𝑞𝑠 ( 𝑡𝑠 − 1)
𝑠 = 1 + 0.86(2.05− 1)
𝑠 = 1.9
( )
𝑝𝑠
* ( ) +

82
√
( )
=47135 psi
𝑛
𝑛
𝑛
Use the steel 1050 CD Sut =1000kpsi and Sy=84kpsi
Ka=2.7( ) =0.797
= 0.797 0.9 0.5 100
= 35865 k psi
q=0.75
= 1 + 0.75( 𝑡 − 1)
= 1 + 0.75(2.5− 1)
= 2.1
( )
𝑝𝑠
𝑛
𝑛
𝑛

83
Inspecting point K The retaining ring groove
At point (K) a flat bottom groove exists to accommodate the retaining ring. This point is
characterized by:
𝑚 = = 𝑚 = 0
= 𝑡 = 5.0
′ = =
( )( )
( )
𝑝𝑠
𝑛
low safety factor.
standard retaining ring specs could be obtained from related tables and going online to
(www.globalspec.com)
Groove width
= 0.068 𝑛.
Groove depth
𝑡 = 0.052 𝑛.
𝑟 = 𝑛.
Therefore,
𝑟
𝑡

84
𝑡
= 𝐷 − 2𝑡
= 1.812 – (2 0.052)
= 1.708
𝑛 Using fig.
4.9.
𝑡 = 4.48
Fig. 4.6 yields
𝑞 = 0.6
= 1 + ( 𝑡 − 1)
= 1 + 0.6(4.48 − 1)
= 3
′ = =
( )( )
( )
𝑝𝑠
𝑛
• Determination of D
Since
𝐷/d = 1.2~1.5
Then d=
Therefore
𝐷6 = = 1.750
Therefore 𝐷6 = 1.750 𝑛

85
Inspecting point m
Point (M) is characterized by being a shoulder to axially hold the bearing. Only bending
is present due to a small moment. However, the cross-section area is small too, which
results in a high stress concentration especially the fillet radius is small to accommodate
the bearing.
Finding the Moment at point d
n the x-y plane
( ) ( )
( ) ( )
𝑙 𝑛
In the x-z plane
( ) ( )
( ) ( )
𝑙 in
Combined
=√( ) ( )
=5024 ibf . 𝑛
Therefore
= 5024 𝑙 . 𝑛
And
𝑚 = 𝑚 = = 0
= 0.02
And
𝑡 = 2.39
Also
= 1.688 𝑛.

86
Therefore
𝑟 =d*0.02= 1.938*0.02=0.03876
Using fig. 4.6
𝑞 = 0.75
= 1 + 0.75(2.39 − 1)
= 2.0425
′ = =
( )( )
( )
𝑝𝑠
𝑛
𝐷1 = 𝐷7 = 1.688 𝑛.
𝐷2 = 𝐷6 = 1.750 𝑛.pll
𝐷3 = 𝐷5 = 1.812 𝑛.
𝐷4 = 2.25 n

87
chapter5
Bearing Selection
Bearings are selected based on the design requirements and the previous calculations for the
shaft and gears which are summarized below.
𝐺 𝑟 𝑛 𝑟 𝑛 𝑙 = 13 000 𝑜𝑢𝑟𝑠
𝑜𝑢𝑛𝑡 𝑟 𝑠 𝑡 𝑠𝑝 = 455 𝑟𝑝𝑚.
𝑠𝑡 𝑚 𝑡 𝑜𝑟 𝑠 𝑧 = 1 𝑛.
𝑠𝑡 𝑚 𝑡 𝑟 𝑛 𝑤 𝑡 = 1 𝑛.
𝑙 𝑙 𝑡𝑦 = 99 %
Right bearing reactions
𝑧 = 1742 𝑙 Rby = 714 RA= 1883𝑙 Left
bearing reactions
𝑧 = 101 𝑙 𝑦 = 354 𝑙 = 368
Due to the higher load on the right side of the shaft, the right bearing is considered more
critical.
Right bearing selection procedure:
The design dimensionless life measure for the bearing is determined by using eq. 5.3

88
=
Assuming a ball bearing with
= 3
And using eq. 5.2 [
( )( ) ⁄
]
= 1 𝑜𝑟 𝑠𝑡 𝑦 𝑙𝑜 𝑠.
[
( ) ⁄
]
10 = 22086ibf
10 = 22086 0.00444
10 = 98 𝑘
Checking this load with SKF bearing catalogue for ball bearings in the vicinity of 1 in.
diameter, as shown in table. 5.1. It sounds that the load is too high for a deep groove ball
bearing. Repeating the load calculation for a roller bearing with

89
[
( ) ⁄
]
10 = 17266
10 = 17266 0.00444
10 = 76.66 𝑘
Checking this load with SKF bearing catalogue for roller bearings in the vicinity of 1 in.
diameter, as shown in table. 5.2. The tables show that the following roller bearing matches
the load.
= 83𝑘
𝐷=30𝑚𝑚=1.188 in
𝑂𝐷 = 72 𝑚𝑚 = 2.834 𝑛.
= 27 𝑚𝑚 = 1.063 𝑛.

90
Left bearing selection
Choosing a ball bearing with
= 3
And using eq. 5.2
[
( )( ) ⁄
]
= 1 𝑜𝑟 𝑠𝑡 𝑦 𝑙𝑜 𝑠.
[
( ) ⁄
]
10 = 4316
10 = 4316 0.00444
10 = 19.16𝑘
Searching table 5.1 that gives the specifications of deep groove ball bearings in the proximity
of 1 in bore and applied load of 19.16kN.
The bearings in this category that matches these requirements is SKF 6305.
The left bearing specifications are
𝐷 = 25𝑚𝑚 = 1 𝑛.
𝑂𝐷 = 62 𝑚𝑚 ≈ 2.5 𝑛.
= 17 𝑚𝑚 = 0.67 𝑛.
= 23.4𝑘 = 5270 𝑙 .
𝑜𝑢𝑙 𝑟 𝑚 𝑡 𝑟 = 1.3~1.4 𝑛.
𝑚 𝑥 𝑚𝑢𝑚 𝑙𝑙 𝑡 𝑟 𝑢𝑠 = 0.08 𝑛.
The right bearing bore is 1.1811 in. which is slightly larger than the estimated one. However,
this has no effect of the design since the diameter of the next step of the shaft is 1.4 in. that
provide the required shoulder for the bearing.

91
Bearing of Gear 2
Left bearing Selection
𝐺 𝑟 𝑛 𝑟 𝑛 𝑙 = 13 000
High counter shaft speed =1992 rpm.
𝑠𝑡 𝑚 𝑡 𝑜𝑟 𝑠 𝑧 = 0.750 𝑛.
𝑙 𝑙 𝑡𝑦 = 99 %
𝑧 = 245 𝑦 = 89 = 260 𝑙
=
Assuming a ball bearing with
= 3
[
( ) ⁄
]
10 = 5491.946
10 = 4988 0.00444
10 = 22.14 𝑘
At Table 5.1 ( C= 26 kN)

92
ID = 25mm
OD = 62mm
W = 17mm
Right bearing Selection
B𝑧 =301 i B𝑦 = 110 B= 320 𝑙
Assuming a roller bearing with
=
=10/3
[
( ) ⁄
]
10 = 4569 𝑙
10 = 4569 0.00444
10 = 20.28 𝑘

93
At Table 5.2 ( C = 22.9 kN)
ID = 17mm
OD = 62mm
W = 17mm
Bearing of Gear 5
Left bearing Selection
𝐺 𝑟 𝑛 𝑟 𝑛 𝑙 = 13 000
High counter shaft speed =104
𝑠𝑡 𝑚 𝑡 𝑜𝑟 𝑠 𝑧 = 1.688 𝑛.
𝑙 𝑙 𝑡𝑦 = 99 %
𝑧 = 1086 𝑦 = 475 = 1185 𝑙
=
= 10/3

94
[
( ) ⁄
]
10 = 6975
10 = 6975 0.00444
10 = 31 𝑘
At Table 5.2 ( C= 44.6 kN)
ID = 45mm
OD =75 mm
W = 16mm
Right bearing Selection
B𝑧 = 1086 B𝑦 = 395 𝑙 B = 1155 𝑙
= 10/3
[
( ) ⁄
]
10 = 6798 𝑙
10 = 6798 0.00444
10 = 30.18 𝑘
At Table 5.2 ( C = 44.6 kN)
ID = 45mm
OD = 75mm
W = 16mm

95
Chapter 6
Key and Keyway Design
For Gear4 & Gear 3.
From the previous data
Bore [in.] Hub length [in.] Torque lbf.in. Safety factor
Gear 3 1.625 1.5
3822 2
Gear 4 1.625 2.0
Using table 6.3 for a square key
Shaft diameter w h Keyway depth
1.625 1 5/8 3/8 3/8 3/16
Choosing key material of 1020 CD steel with 𝑺𝒚 = 𝟓𝟕 𝒌𝒑𝒔𝒊.
The lateral force on the key at the surface of the shaft is:
=
The area prone to bearing stress is
𝑙 𝑙
The bearing stress is

96
𝑙
But 𝑛
𝑛 𝑙
𝑙
𝑛
𝑦
𝑛
Retaining Rings
The retaining ring is selected based on the diameter of the shaft at which the groove is
required. The selected ring specs are associated with ring groove dimensions which are
supposed to commensurate with the designed dimensions. Table 6.4 is a typical sample of
retaining ring specifications.
The summary of the retaining rings to retain the gears and bearings are listed in the table
below:
Dimension Both Gears Tolerance Left
Bearing
Tolerance Right
Bearing
Tolerance
Shaft Dia. 1.625 1.023 1.375
Groove Dia. 1.529 ± 0.005 0.961 ± 0.004 1.291 ± 0.004
Groove Width 0.068
+
0.004
− 0.000
0.046
+ 0.004
− 0.000 0.056
+ 0.004
− 0.000
Groove Depth 0.048 0.031 0.042
Max. groove
fillet radius
0.010 0.010 0.010
Mini. Edge
margin
0.144 0.105 0.105
Allowable axial
thrust
14 800 lbf 6 000
lbf
9700 lbf
All dimensions are in inches.
h
w
F

97
For Gear 2
Gear 2 0.938 1.5 873 2
0.938 7.5/8 1/4 1/4 1/8
Choosing key material of 1020 CD steel with 𝑺𝒚 = 𝟓𝟕 𝒌𝒑𝒔𝒊.
=
𝑙 𝑙
𝑙
But 𝑛
𝑛 𝑙
𝑙
𝑛
𝑦
𝑙 𝑛

98
For Gear 5
Gear 2 1.812 2 16723 2
1.812 1 13/16 1/2 1/2 1/4
Choosing key material of 1045 CD steel with 𝑺𝒚 = 7𝟕 𝒌𝒑𝒔𝒊.
=
𝑙 𝑙
𝑙
But 𝑛
𝑛
𝑙
𝑛
𝑦
𝑛

99
Chapter 7
Eg:
Hexagonal-head bolts and nuts are used to keep the head of a pressure vessel tight with the
vessel. The joint is identical of 0.75 in. thickness for the vessel and its head. They are made
of grade 25 cast-iron. The separating force is 36 k lbf. The bolt specs are:
𝑛 𝑛
( ) 𝐷 𝑡 𝑟𝑚 𝑛 𝑘 ,𝑘𝑚, 𝑛 .
( ) 𝑛 𝑡 𝑛𝑢𝑚 𝑟 𝑜 𝑜𝑙𝑡𝑠 𝑟 𝑞𝑢 𝑟 . 𝑙𝑜 𝑐𝑡𝑜𝑟 𝑠 2 𝑛 𝑡 𝑜𝑙𝑡𝑠 𝑚 𝑦
𝑟 𝑢𝑠 𝑤 𝑛 𝑡 𝑗𝑜 𝑛𝑡 𝑠 𝑡 𝑘 𝑛 𝑝 𝑟𝑡.
Solution
𝑙 𝑛
To find the bolt length, nut thickness should be
determined from table 7.3

100
𝑡 𝑛
Adding two threads after the nut to the total length of the bolt therefore the length of these two
threads is
( )
𝑙 𝑡
𝑛
L=2.5 in
𝑛
The length of the threaded part is
𝑛
The length of the unthreaded part is
𝑙 𝑛
The length of the threaded part in the grip is
𝑙 𝑙 𝑙 𝑛
Using fig.(4.7)
𝑛
𝑛
𝑙 𝑙
𝑙 𝑛

101
For No 25-cast iron wall
Select =14 𝑝𝑠
For two identical members in the grip
(
𝑙
𝑙
)
( )
𝑙 𝑛
Using table 7.5 to find
/in
𝑟𝑟𝑜𝑟
The joint stiffness
Using table 7.6 to obtain the minimum proof strength of grade 5 SAE bolt
𝑘𝑙
𝑛
(
𝑛
)
Rearrangement gives
𝑛

102
Roundup to 6 bolts and check for n
𝑛
The value of n is higher than the required value (2)
The tightening torque is
(my first design is end, At the end of my project i use my data and make a gear
box drawing picture .)

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