CANKAYA UNIVERSITY FACULTY OF

ENGINEERING MECHATRONICS ENGINEERING

SERTAÇ ŞİMŞEK
201717204
09.01.2023

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 TABLE OF CONTENTS

1. What is Differential?................................................................4
1.1What are the functions of the differential?.........................4
1.2 OPERATION OF THE DIFFERENTIAL……………….5
2.BEARİNG SELECTİON………………………………………6
2.1 SCHAEFFLER KHM88649-HM88610………………..6
2.2 SCHAEFFLER K25877-25821………………………..7
3.DESIGN OF SHAFTS………………………………………..8-11
4.STATIC DESIGN OF SHAFTS………………………………..12
4.1 Stress Concentratio……………………………………..….13-15
4.2 Equivalent Stress…………………………………………..15-17
5.TECHNICAL DRAWING DIMENSIONS …………………….17
6.Conclusıon………………………………………………………18
7.Referance……………………………………………………….18

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1.What is Differential?

The differential directs the torque output from the gearbox to the two axle shafts. For rear wheel
drive vehicles, the differential breaks the motion 90 degrees and transmits it to the wheels. The
differential is there so that the right and left wheels can spin at different speeds. On curves and
rough roads, the right and left wheels must travel at different speeds as they travel different
amounts. Otherwise, the wheels will slip or spin, in this case the road grip will decrease and the
tires will wear out. Differential in its most basic definition; It is a gearbox that transmits the
movement it receives from the gearbox to the axle shafts.

1.1What are the functions of the differential?

*Transferring the rotation movement it receives from the gearbox to the wheels.

*Ensuring that the wheels inside and outside the bend can rotate at different speeds while the
vehicle is turning.

*Increasing the torque coming from the gearbox and reducing the number of revolutions while
transmitting the movement to the wheels

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1.2 OPERATION OF THE DIFFERENTIAL

The number of revolutions and torque produced in the engine is changed in the gearbox according
to the choice of the driver; The torque can be decreased by increasing the speed, or the speed can be
decreased and the torque increased. After this adjustment, the torque and speed values of the engine
are changed one last time in the differential. This change in the differential is constant, that is, its
amount cannot be adjusted, it is called the final gear ratio.

In the differential, a gear ratio is formed between the pinion gear that drives the bevel gear and the
bevel gear. The rotating pinion gear is small and the rotating bevel gear is larger, that is, the small
gear rotates the large gear. In passenger vehicles, this gear ratio in the differential is approximately
i=3. Thanks to this final gear ratio in the differential, the speed decreases and the torque (traction
force) increases. That is, if the shaft is rotating at 3000 rpm, the bevel gear will rotate at 1000 rpm,
the left and right axle shafts will also rotate at a speed of 1000 rpm (when the vehicle is going
straight).

In straight driving: Bevel gear speed = (Right axle speed + left axle speed) /2

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2.BEARİNG SELECTİON

Tapered roller bearings have very high radial-load capacity and high axial-load
capacity in one direction. For this reason, tapered roller bearings are often used in
pairs, mounted in opposite directions, to cater for high radial loads and axial loads
in either direction. Tapered roller bearings are often used as in the
wheel bearings of cars and heavy trucks.

2.1 SCHAEFFLER KHM88649-HM88610

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2.2 SCHAEFFLER K25877-25821

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3.DESIGN OF SHAFTS

Material: AISI 4140

ka = surface condition modification factor

I chose a floor surface because we are going to practice tight fitting.

𝑘a = 𝑎𝑆utb
ka = 1.58 x (951)-0.085
ka = 0.88

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 kb = size modification factor

d:30.2mm

kb=1.24𝑑−0.107

kb=0.86

kc = load modification factor

Kc=0.85

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 kd = temperature modification factor

Since the working temperature of the shaft I designed is 50 degrees on average kd=1.008

ke = reliability factor

𝑘e = 1 − 0.08𝑧a
ke = 1 – 0.08 x 2.326
ke = 0.814

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 kf = miscellaneous-effects modification factor

kf=1

Se’ = 0.5 x 951 = 475.5 MPa

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4.STATIC DESIGN OF SHAFTS

Maximum torque is 160 N/m and 3000rpm

160
Fz= 0,055 = 5.8kN
2

Fy=5.8×cos30=5.022kN

∑ F = (−53 x 103 ) + Ra – (65.5× 103 ) - (5.022× 103 ) = 0

Ra=123,52 kN

T=F x r
r = d/2
r =30.2/2
r = 15.1

T= 5.02×cos30×0.0151=0,65kN/m

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4.1 Stress Concentratio

D = 35 mm
d = 30.2 mm
r = 2.5 mm
D / d = 35 / 30.2 = 1.15 mm
r / d = 2.5 / 30.2 = 0.082 mm

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Looking Kt in figure

Kt=1.6 Bending

Looking Kt in figüre

Kt=1.4 Torsion

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4.2 Equivalent Stress

Bending √𝑎=0.246-3.08(10-3)(951x106)+1.51(10-5)(951x106)2-2.67(10-8)(951x106)3

√𝑎 =0.346mm

Torsion √𝑎=0.190-2.51(10-3)(951x106)+1.35(10-5)(951x106)2x2.67(10-8)(951x106)3

√𝑎=0.325mm
1
q= √0,346
1+
√2,5

q ≌ 0.9

1
qs= √0,325
1+
√2,5

qs ≌ 0.9

Kf = 1 + 0.9(1.6 – 1)
Kf = 1.54 ( Bending )
Kfs = 1 + 0.9(1.4 – 1)
Kf = 1.26 (Torsion)

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(BENDING)

M.c 32.c
σa = Kf = Kf
I 𝜋𝑑3

32x4.52
σa = 1.54 3
𝜋(30.2𝑥10−3)

σa = 25.74 MPa
(TORSION)
16.T
 = Kf 𝜋𝑑3
16x 0,65
 = 1.26 3
𝜋(30,2𝑥10−3)

 = 15.14 MPa

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𝜎𝑎 𝜏𝑚 1
+ =
𝑆𝑒 𝑆𝑢𝑡 𝑛
𝜎𝑎 = 25.74 Mpa
𝜏𝑚 = 15.14 MPa
𝑆𝑢𝑡 = 951 MPa
𝑆𝑒 = 251 MPa

25.74 15.14 1
+ =
251 951 𝑛

n=8.44

5.TECHNICAL DRAWING DIMENSIONS

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6.Conclusion

In this project, I tried to create a design within the information given at the university. I hope I
was able to complete the project. I paid attention to Shaft design techniques.
I searched for suitable bearings on the internet for the design. While solving the problems, I usually
got help from the tables. But I learned a lot from this project. This is a great contribution for me. It
would be better if a professional touch my design. From my point of view this is a good start. There
were also application areas where I had difficulty in the project. For example, the torsion and
bending values were difficult to find, but I was able to find the results of my research. I hope you
prepared a good homework.

7.Referance

• https://www.e-rulman.com/ara.aspx
• shigley mechanical engineering design solutions 9th edition
• https://medias.schaeffler.de/tr
• https://www.traceparts.com/tr

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