FMD – Exam solution 22/02/2021
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FUNDAMENTALS OF MACHINE DESIGN (A.Y. 2020-2021)
WRITTEN TEXT OF MACHINE DESIGN (22ND FEBRUARY 2021)

Figure 1. Gearbox and differential of a car

Figure 2. Simplified representation of the primary shaft of the gearbox
Z
Y
INPUT TORQUE
l
b
A B
5th gear 4th gear
Z
Y
a
FMD – Exam solution 22/02/2021
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Figure 3: sketch of the forces acting on the shaft
1. With reference to the figure, what is the value of FT when gear 4 is engaged [N]?
4 = 4 = 4 2? = 200 ? 103144 2? = 2778
2. With reference to the figure, what is the value of FR when gear 4 is engaged [N]?
4 = 4 ? = 2778 ? tan 20° = 1011
3. With reference to the figure, what is the value of FT when gear 5 is engaged [N]?
5 = 5 = 5 2? = 200 ? 103162 2? = 2469
4. With reference to the figure, what is the value of FR when gear 5 is engaged [N]?
5 = 5 ? = 2469 ? tan 20° = 899
5. With reference to the figure, what is the value of the horizontal force FHA4 acting on the bearing A
when gear 4 is engaged [N]?
4 = 4 ? ? = 2778 ? 166 ? 70166 = 1606
6. With reference to the figure, what is the value of the horizontal force FVA4 acting on the bearing A
when gear 4 is engaged [N]?
4 = 4 ? ? = 1011 ? 166 ? 70166 = 585
7. With reference to the figure, what is the value of the horizontal force FHA5 acting on the bearing A
when gear 5 is engaged [N]?
5 = 5 ? ? = 2469 ? 166 ? 17166 = 2216
8. With reference to the figure, what is the value of the horizontal force FVA5 acting on the bearing A
when gear 5 is engaged [N]?
A
B
FVA FHA
FHBFVBFT
FR
a or b
l
Z
YX
rp
FMD – Exam solution 22/02/2021
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5 = 5 ? ? = 899 ? 166 ? 17166 = 807
9. What is the resultant force RA acting on the bearing A for the most loaded condition (between fourth
and fifth gear on) [N]?

5 = ?52 + 52 = ?22162 + 8072 = 2359
10. What is the resultant force RB acting on the bearing B for the most loaded condition (between fourth
and fifth gear on) [N]?

4 = 4 ? = 2778 ? 70166 = 1171
4 = 4 ? = 1011 ? 70166 = 426
4 = ?42 + 42 = ?11712 + 4262 = 1247
11. What is the minimum dynamic load coefficient Cmin requested for the most loaded ball bearing,
considering both the service in fourth and fifth gear, to ensure a basic rating life of 5·107 revolutions
[N]? Assume a reliability of 95% and a coefficient askf = 1.
The most loaded ball bearing is the bearing A.
Considering the operating conditions in the 4th and 5th gear it is possible to evaluate the equivalent load on
the roller bearings:
= ?? ?


= ?0.25 ? 17103 + 0.45 ? 235933 = 1927 N
Evaluation of the minimum basic dynamic load rating:
10 = 1 ? ? ??
A
FVA
FHA
Z
X
RA
B
FVB
FHB
Z
X
RB
FMD – Exam solution 22/02/2021
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= ? ? 101 ?
Considering the reliability of 95% (a1 = 0.62) and a basic rating life of L10 = 50 million cycles:
= 1927 ? ? 500.62 ? 13 = 8.3 kN
Point B: spur gears
Furthermore, it is requested to design the gear module of the fourth gear according to the Hertz procedure.
In particular, it is requested to calculate: the coefficient Cp, the coefficient I, the torque C used to design
the module. Assume E = 206 GPa, a surface hardness of both gears equal to 670 HB. Assume a number of
teeth z1 = 48, z2 = 48 and a pressure angle α = 20°. The transmission ratio i = 1.0.
1. What is the value of the module m [mm]?
= 2 ?

=

= __4
_4 = 14448 = 3
2. What is the value of the coefficient Cp [√]?
= 0.564 ? ? 11 ? 2

+ 11 ? 2

= 0.564 ? ? 11 ? 0.32206000 + 11 ? 0.32206000 = 189.75 √
3. What is the value of the coefficient I?
= sin ? cos2 ? 1 ?1 + 1 ? = sin 20° ? cos 20°2 ? 1 1?1 + 1 1? = 0.08035
4. What is the value of the adopted torque T [Nm]?
= = 200
5. What is the value of S0 [MPa]?
0 = 2.8 ? (?) ? 69 = 2.8 ? 670 ? 69 = 1807
6. What is the value of λ (gear width factor) given by the Hertz theory?
3 = 2
0,2 ? 2 ? 2 ? ?
= ? 2
0,2 ? 2 ? 2 ? ? 3
0, = 0? = 18073 = 602
= 189.7526022 ? 2 ? 200000482 ? 32 ? 0.08035 = 7.9