Computation of a gearbox for a wind power plant

POLITECNICO DI TORINO
Master of science in mechanical engineering
Computation of a Gearbox for a Wind power plant
Professors:
Teresa Berruti
Muzio Gola
Candidates:
Pietro Galli 220205
Marco Merlotti 217196
Antonio Russo 214173
Spring semester 2015

Index
Index 1
1. Design of a Gearbox for a wind power plant 1 2
1.1 Static Design 5
1.1.1 Design of shaft 1 5
1.1.2 Design of bearings 7
1.1.3 Internal stresses 11
1.1.4 Safety factor 8
1.1.5 Design of shaft 2 9
1.2 Fatigue 12
1.2.1 Shaft 1 section 3 12
1.3 Rolling bearings 23
1.3.1 Roller bearings C and D 23
1.3.2 Bearing E 23
1.3.3 Rating life_bearing D 24
1.3.3 Rating life_bearing E 27
1.4 Spur gears 29
1.4.1 Computation of transmission ratio, pt and b 29
1.4.2 Tooth bending strength 31
1.4.3 Tooth surface fatigue strength 32

2
1. Technical Report 1
The object of the report is the computation of the gearbox of the wind power plant shown below.

3
The gearbox works as a speed increaser and the increment consists of two following stages:
1) Planetary gearing;
2) wheel-pinion stage.
A detailed representation of the gearbox is represented in figure:

5
The data given are the ones reported in table:
Given data
Dimensions mm 1730*2748*2100
Operating data
Input power P (kW) 1150
Input speed ni (rpm) 24,5
Gear data
Total gear reduction i 41,63
Gears (second stage)
Pressure angle α° 20
Helical teeth spiral angle β° 15
Bearings and bore diameter (d)
Bearings A and B - series 32318 J2 d (mm) 90
Bearings C and D - series NCF 2952 V d (mm) 260
Suggested material for the shafts
Quenched and tempered steel 36NiCrMo16 UNI EN10083
Rm,N (Mpa) 1250
ReH,N (Mpa) 1050
σD-1 (Mpa) 565
1.1 Static Design
The aim of this section is to study the wheel-pinion stage and, in particular, to perform the static
calculation of the output shaft 1 and the hollow shaft 2.
1.1.1 Design of shaft 1
The input torque from the rotor can be computed as:
𝑀𝑡,𝐼𝑁 =
60 ∗ 1000 ∗ 𝑃
2 ∗ 𝜋 ∗ 𝑛𝑖
= 4.5 ∗ 102
𝑘𝑁 ∗ 𝑚
and the output one to the main generator, assuming gear efficiency equal to 1, results to be:
𝑀𝑡,𝑂𝑈𝑇 =
𝑀𝑡,𝐼𝑁
𝑖
= 1.1 ∗ 10 𝑘𝑁 ∗ 𝑚

6
In order to compute the forces acting on shaft 1 and shaft 2, the dimension of the pitch radius is
computed by means of a proportion between the input dimension (bore diameter of bearings A and
B) and the one directly measured in the drawing.
A list of the evaluated dimensions for the shaft 1 is reported in figure 1.5:
Shaft 1
Then it is possible to evaluate the forces acting between the pinion 1 and the wheel 1, which are
displayed in the figure 1.6:
Forces acting on wheel 1
The forces acting on wheel 1 are evaluated and the results are following:
𝐾𝑝 =
𝑀𝑡,𝑂𝑈𝑇
𝑟1
= 1.8 ∗ 102
𝑘𝑁
𝐾𝑎 = 𝐾𝑝 ∗ tan 𝛽 = 4.8 ∗ 10 𝑘𝑁
Pitch radius will be indicated as r1

7
𝐾 𝑛 = 𝐾𝑝 ∗
tan 𝛼
cos 𝛽
= 6.7 ∗ 10 𝑘𝑁
1.1.2 Design of bearings
The designed roller bearings are the ones reported: the point of application of the forces applied by
the bearings depend on the inclination of the rolls of the bearings and it can be computed as follows:
Bearings A and B
Here is a sketch of the dimensional layout of shaft 1:
Unloaded shaft 1
The forces acting on shaft 1 are reported in the planes XZ and YZ, respectively (for simplicity forces
and moments which do not contribute to neither vertical or horizontal nor to bending equilibrium are
omitted):

8
Forces acting in plane XZ
Forces acting in plane YZ
The forces are evaluated through static balances of forces:
𝑉𝐴 + 𝑉𝐵 − 𝐾 𝑛1 = 0
−𝐾 𝑛1 ∗ 𝑙 𝑔 + 2 ∗ 𝑙 𝑔 ∗ 𝑉𝐵 + 𝐾𝑎1 ∗ 𝑟1 = 0
𝐾𝑝1 ∗ 𝑙 𝑔 − 2 ∗ 𝐻 𝐵 ∗ 𝑙 𝐺 = 0
𝐻𝐴 + 𝐻 𝐵 − 𝐾𝑝1 = 0
The values of the unknowns forces resulting from the linear system are then evaluated:
𝑉𝐴 = 4.4 ∗ 10 𝑘𝑁
𝐻𝐴 = 9.0 ∗ 10 𝑘𝑁
𝑉𝐵 = 2.4 ∗ 10 𝑘𝑁
𝐻 𝐵 = 9.0 ∗ 10 𝑘𝑁

9
The axial forces on the bearings are evaluated according to the direction of 𝐾𝑎; to identify the right
case among the ones proposed in each figure the radial forces on each bearings A and B are evaluated
and case 2b results to be the one which is valid in the case proposed.

10
Axial loading according to bearing arrangements
The evaluated axial forces are:
𝐴 𝐴 = 7.5 ∗ 10 𝑘𝑁
𝐴 𝐵 = 2.7 ∗ 10 𝑘𝑁
The right-end sided bearing (B) is loaded by a radial force which causes an axial one due to its oblique
mounting. The left-end sided bearing (A), instead, is the one which carries both the thrusts of the B
bearing and of the pinion-wheel contact.

11
1.1.3 Internal stresses
Once the forces at the bearings and at the contact pinion 1-wheel 1 are known; the internal stresses
of shaft1 are evaluated at each cross section as function of the z coordinate:
 Section 1: 𝟎 < 𝒛 < 𝒍 𝒄𝑨 + 𝒍 𝒑𝑨
(Mout is exchanged by the shaft through a mechanical tab; a simplifying hypothesis has been done:
it has been assumed that this moment is due to a force couple applied in the middle of the tab)
𝑁 = 0 N
𝑀 𝑏
𝑥𝑧
= 0 Nm
𝑇 𝑥𝑧
= 0 N
𝑀 𝑏
𝑦𝑧
= 0 Nm
𝑀𝑡 = −𝑀𝑡,𝑂𝑈𝑇
𝑇 𝑦𝑧
= 0 N
 Section 2: 𝐥 𝐜𝐀 + 𝐥 𝐩𝐀 < 𝐳 < 𝐥 𝐜𝐀 + 𝐥 𝐩𝐀 + 𝐥 𝐆
𝑁 = −𝐴 𝐴
𝑀 𝑏
𝑥𝑧
= −𝑉𝐴 ∗ (𝑧 − 𝑙 𝑐𝐴 − 𝑙 𝑝𝐴)
𝑇 𝑥𝑧
= −𝑉𝐴
𝑀𝑡 = −𝑀𝑡,𝑂𝑈𝑇
𝑇 𝑦𝑧
= 𝐻𝐴
𝑀 𝑏
𝑦𝑧
= 𝐻𝐴 ∗ (𝑧 − 𝑙 𝑐𝐴 − 𝑙 𝑝𝐴)

2
 Section 3 : 𝒍 𝒄𝑨 + 𝒍 𝒑𝑨 + 𝒍 𝑮 < 𝒛 < 𝒍 𝒄𝑨 + 𝒍 𝒑𝑨 + 𝟐 ∗ 𝒍 𝑮
𝑁 = 𝐾𝐴 − 𝐴 𝐴
𝑀 𝑏
𝑥𝑧
= −𝑉𝐴 ∗ (𝑧 − 𝑙 𝑐𝐴 − 𝑙 𝑝𝐴) + 𝐾𝑛 ∗ (𝑧 − 𝑙 𝑐𝐴 − 𝑙 𝑝𝐴 − 𝑙 𝐺)
+ 𝐾𝐴 ∗ 𝑟1
𝑇 𝑥𝑧
= −𝑉𝐴 + 𝐾 𝑛
𝑀𝑡 = 𝐾𝑝 ∗ 𝑟 − 𝑀𝑡,𝑂𝑈𝑇
𝑇 𝑦𝑧
= 𝐻𝐴 − 𝐾𝑝
𝑀 𝑏
𝑦𝑧
= 𝐻𝐴 ∗ (𝑧 − 𝑙 𝑐𝐴 − 𝑙 𝑝𝐴) − 𝐾𝑝 ∗ (𝑧 − 𝑙 𝑐𝐴
− 𝑙 𝑝𝐴 − 𝑙 𝐺)
 Section 4: 𝟎 < 𝒍𝒕𝒐𝒕 − 𝒛 < 𝒍𝒕𝒐𝒕 − (𝒍 𝒄𝑨 + 𝟐 ∗ 𝒍 𝑮)
𝑁 = 0 N
𝑀 𝑏
𝑥𝑧
= 0 Nm
𝑇 𝑥𝑧
= 0 N
𝑀𝑡 = 0 Nm
𝑇 𝑦𝑧
= 0 N
𝑀 𝑏
𝑦𝑧
= 0 Nm
By means of the equations evaluated at each cross section axial load, bending and torsion moments
as well as the area, flexural strength modulus Wb and the torsion strength modulus Wt are plotted
as a function of z. They show the follow trends, computed by using Excel software:

2
0.0E+00
2.0E+03
4.0E+03
6.0E+03
8.0E+03
1.0E+04
1.2E+04
1.4E+04
0 50 100 150 200 250 300 350 400 450 500 550 600 650
mm^2
z [mm]
Area
0.0E+00
5.0E+04
1.0E+05
1.5E+05
2.0E+05
2.5E+05
3.0E+05
3.5E+05
4.0E+05
4.5E+05
0 50 100 150 200 250 300 350 400 450 500 550 600 650
mm^3
z [mm]
Wt

3
0.0E+00
5.0E+04
1.0E+05
1.5E+05
2.0E+05
2.5E+05
3.0E+05
3.5E+05
4.0E+05
4.5E+05
0 50 100 150 200 250 300 350 400 450 500 550 600 650
mm^3
z [mm]
Wb
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-9.0E+01
-8.0E+01
-7.0E+01
-6.0E+01
-5.0E+01
-4.0E+01
-3.0E+01
-2.0E+01
-1.0E+01
0.0E+00
1.0E+01
0 50 100 150 200 250 300 350 400 450 500 550 600 650
kN
z [mm]
N Area

4
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-7.0E+00
-6.0E+00
-5.0E+00
-4.0E+00
-3.0E+00
-2.0E+00
-1.0E+00
0.0E+00
1.0E+00
0 50 100 150 200 250 300 350 400 450 500 550 600 650
kN*m
z [mm]
Mb (x-z) Area
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-2.0E+00
0.0E+00
2.0E+00
4.0E+00
6.0E+00
8.0E+00
1.0E+01
1.2E+01
1.4E+01
0 50 100 150 200 250 300 350 400 450 500 550 600 650
kN*m
z [mm]
Mb (y-z) Area

5
Once forces and moments are known, the normal and tangential stresses are computed by means of
the following formulas:
𝜎 𝑁 =
𝑁
𝐴
𝜎𝑏 =
𝑀 𝑏
𝑊 𝑏
𝜏 =
𝑀𝑡
𝑊𝑡
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-2.0E+00
0.0E+00
2.0E+00
4.0E+00
6.0E+00
8.0E+00
1.0E+01
1.2E+01
1.4E+01
1.6E+01
0 50 100 150 200 250 300 350 400 450 500 550 600 650
kN*m
z [mm]
Mb tot Area
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-1.4E+01
-1.2E+01
-1.0E+01
-8.0E+00
-6.0E+00
-4.0E+00
-2.0E+00
0.0E+00
2.0E+00
0 50 100 150 200 250 300 350 400 450 500 550 600 650
kN*m
z [mm]
Mt Area

6
The trend of the stresses is displayed in the following diagrams:
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-1.4E+01
-1.2E+01
-1.0E+01
-8.0E+00
-6.0E+00
-4.0E+00
-2.0E+00
0.0E+00
2.0E+00
0 50 100 150 200 250 300 350 400 450 500 550 600 650
MPa
z [mm]
sigmaN Area
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-5.0E+01
0.0E+00
5.0E+01
1.0E+02
1.5E+02
2.0E+02
2.5E+02
0 50 100 150 200 250 300 350 400 450 500 550 600 650
MPa
z [mm]
sigmab Area

7
Consequently we can calculate and plot equivalent stresses:
 according to Tresca:
𝜎𝑒𝑞 = √(𝜎 𝑁 + 𝜎𝑏)2 + 4𝜏2
 according to Von Mises:
𝜎𝑒𝑞 = √(𝜎 𝑁 + 𝜎𝑏)2 + 3𝜏2
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-1.4E+02
-1.2E+02
-1.0E+02
-8.0E+01
-6.0E+01
-4.0E+01
-2.0E+01
0.0E+00
2.0E+01
0 50 100 150 200 250 300 350 400 450 500 550 600 650
MPa
z [mm]
tau Area
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-5.0E+01
0.0E+00
5.0E+01
1.0E+02
1.5E+02
2.0E+02
2.5E+02
3.0E+02
0 50 100 150 200 250 300 350 400 450 500 550 600 650
MPa
z [mm]
sigma eq Tresca Area

8
1.1.4 Safety factor
From the previous plot we can get the safety factors by using both Tresca and Von Mises criteria:
 the maximum equivalent stress according to Tresca is obtained for z equal to 404 mm and
it’s 245,5 MPa;
 the maximum equivalent stress according to Von Mises is obtained for z equal to 404 mm
and it’s 233,5 MPa.
Finally, since specimens used to test this material according FKM standard have diameter
Deff,N,m=Deff,N,p=16mm≠deff for every section of the shaft, coefficients KA (anisotropy factor) and Kd
(technological size factor) have to be considered in order to correct the values of yield strength and
the ultimate tensile strength:
 𝐾𝐴 = 1
It has been assumed that anisotropy eventually present is negligible for the shaft.
𝐾𝑑,𝑚 =
1 − 0,7686 ∗ 𝑎 𝑑,𝑚 ∗ log(
𝑑 𝑒𝑓𝑓
7,5
)
1 − 0,7686 ∗ 𝑎 𝑑,𝑚 ∗ log(
𝑑 𝑒𝑓𝑓,𝑁
7,5
)
0.00E+00
2.00E+03
4.00E+03
6.00E+03
8.00E+03
1.00E+04
1.20E+04
1.40E+04
-5.0E+01
0.0E+00
5.0E+01
1.0E+02
1.5E+02
2.0E+02
2.5E+02
0 50 100 150 200 250 300 350 400 450 500 550 600 650
MPa
z [mm]
sigma eq Von Mises Area

9
𝐾𝑑,𝑃 =
1 − 0,7686 ∗ 𝑎 𝑑,𝑃 ∗ log(
𝑑 𝑒𝑓𝑓
7,5
)
1 − 0,7686 ∗ 𝑎 𝑑,𝑃 ∗ log(
𝑑 𝑒𝑓𝑓,𝑁
7,5
)
where ad is found in tables:
a d,m 0,28
a d,P 0,32
So Rm and Rp are evaluated for each deff as follow:
𝑅 𝑚 = 𝑅 𝑚,𝑁 ∗ 𝐾𝐴 ∗ 𝐾𝑑,𝑚
𝑅 𝑒 = 𝑅 𝑒,𝑁 ∗ 𝐾𝐴 ∗ 𝐾𝑑,𝑃
So the safety factor has been computed for each section by using as 𝜎𝑒𝑞 both the ones evaluated by
Tresca and with Von Mises:
SF =
𝑅 𝑒𝐻
𝜎𝑒𝑞
Finally as safety factor of the whole shaft the minimum one has been taken; in particular by using
Tresca criterion it is equal to 3.42 whereas by using Von Mises criterion it is equal to 3.59 (it is
remarkable to notice that the safety factor calculated with Tresca is lower and this is in accordance
with the more conservative kind of failure criterion).
1.1.5 Design of shaft 2
In order to compute the forces acting on shaft 2, the dimension of the pitch radius of the wheel 1, all
the diameters and lengths are computed by means of a proportion between the input dimension (bore
diameter of bearings C and D) and the one directly measured in the drawing.
A list of the evaluated dimensions for the shaft2 is reported in figures below:

10
Shaft 2
Then it is possible to evaluate the forces acting between the pinion 1 and the wheel 1, which are equal
and opposite to the ones computed for shaft1:

11
Forces acting on wheel 1
𝐾𝑝 = 1.8 ∗ 102
𝑘𝑁 ; 𝐾𝑎 = 𝐾𝑝 ∗ tan 𝛽 = 4.8 ∗ 10 𝑘𝑁 ; 𝐾 𝑛 = 𝐾𝑝 ∗
tan 𝛼
cos 𝛽
= 6.7 ∗ 10 𝑘𝑁
The designed roller bearings are the ones reported in figure 1.7: the point of application of the forces
applied by the bearings is supposed to be located at the centre of the length of the bearings:
Bearings C and D

12
Here is a sketch of the dimensional layout of shaft 2:
Unloaded shaft 2
The forces acting on shaft 2 are reported in the planes XZ and YZ, respectively:
Forces acting in plane XZ
LbC/2+Lg
Ls

13
Figure 1.10: Forces acting in plane YZ
The forces are evaluated through static balances of forces:
𝑉𝐷 − 𝑉𝐶 − 𝐾 𝑛 = 0
𝐾 𝑛 ∗ (
𝑙 𝑏𝐶
2
+ 𝑙 𝐺) − (2 ∗ 𝑙 𝐺 + 𝑙 𝑏𝐶) ∗ 𝑉𝐷 + 𝐾𝑎 ∗ 𝑟2 = 0
𝐾𝑝2 ∗ (
𝑙 𝑏𝐶
2
+ 𝑙 𝐺) − 𝐻 𝐷 ∗ (2 ∗ 𝑙 𝐺 + 𝑙 𝑏𝐶) = 0
𝐻 𝐶 + 𝐻 𝐷 − 𝐾𝑝2 = 0
The values of the unknown forces resulting from the linear system are then evaluated:
𝑉𝐶 = 2.4 ∗ 10 𝑘𝑁
𝐻 𝐶 = 8.9 ∗ 10 𝑘𝑁
𝑉𝐷 = 9.2 ∗ 10 𝑘𝑁
𝐻 𝐷 = 8.9 ∗ 10 𝑘𝑁
The axial forces on the bearings are evaluated according to the direction of 𝐾𝑎; since it is directed
towards the D bearings, only this is loaded (in particular, even if usually bearings mounted with this
LbC/2+Lg

14
kind of configuration are not allowed to carry an axial load, a special model of bearings is taken into
account in order to provide even this kind of request). So by equilibrium equation it can be found
that:
𝐴 𝐶 = 0 𝑘𝑁
𝐴 𝐷 = 4.8 ∗ 10 𝑘𝑁
Once the forces at the bearings and at the contact pinion 1-wheel 1 are known; the internal stresses
of shaft2 are evaluated at each cross section as function of the z coordinate:
 Section 1: 𝟎 < 𝒛 <
𝒍 𝒃
𝟐
𝑁 = 0 N
𝑇 𝑥𝑧
= 0 N
𝑀 𝑏
𝑥𝑧
= 0 Nm
𝑀𝑡 = 0 Nm
𝑀 𝑏
𝑦𝑧
= 0 Nm
𝑇 𝑦𝑧
= 0 N
 Section 2:
𝒍 𝒃
𝟐
< 𝒛 < 𝒍 𝒃 + 𝒍 𝑮

2
𝑁 = −𝐴 𝑐
𝑀 𝑏
𝑥𝑧
= −𝑉𝐶 ∗ (𝑧 −
𝑙 𝐵
2
)
𝑇 𝑥𝑧
= −𝑉𝐶
𝑀𝑡 = 0
𝑇 𝑦𝑧
= −𝐻 𝐶
𝑀 𝑏
𝑦𝑧
= −𝐻 𝐶 ∗ (𝑧 −
𝑙 𝐵
2
)
 Section 3: 𝒍 𝒃 + 𝒍 𝑮 < 𝒛 <
3𝒍 𝒃
2
+ 2𝒍 𝑮
𝑁 = −𝐾𝑎 − 𝐴 𝐶
𝑀 𝑏
𝑥𝑧
= −𝑉𝐶 ∗ (𝑧 −
𝑙 𝑏
2
) − 𝐾𝑛 ∗ (𝑧 − l 𝑏𝐶 − 𝑙 𝐺) + 𝐾𝑎 ∗ 𝑟2
𝑇 𝑥𝑧
= −𝑉𝐶 − 𝐾 𝑛
𝑀𝑡 = 𝐾𝑝 ∗ 𝑟
𝑇 𝑦𝑧
= −𝐻 𝐶 + 𝐾𝑝
𝑀 𝑏
𝑦𝑧
= −𝐻𝐶 ∗ (𝑧 −
𝑙 𝑏
2
) + 𝐾𝑝 ∗ (𝑧 − 𝑙 𝑏 − 𝑙 𝐺)
 Section 4:
𝟑∗𝒍 𝒃
2
+ 2𝒍 𝑮 < 𝒛 < 𝑙 𝑡𝑜𝑡 −
𝑙 𝑠
2
𝑁 = −𝐾𝑎 − 𝐴 𝐶 + 𝐴 𝐷
𝑀 𝑏
𝑥𝑧
= −𝑉𝐶 ∗ (𝑧 −
𝑙 𝑏
2
) − 𝐾 𝑛 ∗ (𝑧 − 𝑙 𝑏 − 𝑙 𝐺) + 𝐾𝑎 ∗ 𝑟1 + 𝑉𝐷 ∗ (𝑧 −
3 ∗ 𝑙 𝑏
2
− 2 ∗ 𝑙 𝐺)
𝑇 𝑥𝑧
= −𝑉𝐶 − 𝐾 𝑛 + 𝑉𝐷
𝑀𝑡 = 𝐾𝑝 ∗ 𝑟

2
𝑇 𝑦𝑧
= −𝐻 𝐶 + 𝐾𝑝 − 𝐻 𝐷
𝑀 𝑏
𝑦𝑧
= −𝐻 𝐶 ∗ (𝑧 −
𝑙 𝑏
2
) + 𝐾𝑝 ∗ (𝑧 − 𝑙 𝑏 − 𝑙 𝐺) − 𝐻 𝐷 ∗ (𝑧 −
3 ∗ 𝑙 𝑏
2
− 2 ∗ 𝑙 𝐺)
 Section 5: 𝒍𝒕𝒐𝒕 > 𝒛 > 𝒍𝒕𝒐𝒕 −
𝑙 𝑠
2
(MS is exchanged by the shaft through a splined coupling joint: a simplifying hypothesis has been
done: it has been assumed that this moment is due to a force couple applied in the middle of the joint)
𝑁 = 0 N
𝑀 𝑏
𝑥𝑧
= 0 Nm
𝑇 𝑥𝑧
= 0 N
𝑀𝑡 = 0 Nm
𝑇 𝑦𝑧
= 0 N
𝑀 𝑏
𝑦𝑧
= 0 Nm
By means of the equations evaluated at each cross section axial load, bending and torsion moments
as well as the area, flexural strength modulus Wb and the torsion strength modulus Wt are plotted
as a function of z. They show the follow trends, computed by using Excel software:

2
0.0E+00
1.0E+04
2.0E+04
3.0E+04
4.0E+04
5.0E+04
6.0E+04
7.0E+04
0 50 100 150 200 250 300 350 400 450 500 550
mm^2
z [mm]
Area
0.0E+00
1.0E+06
2.0E+06
3.0E+06
4.0E+06
5.0E+06
6.0E+06
0 50 100 150 200 250 300 350 400 450 500 550
mm^3
z [mm]
Wt

3
0.0E+00
1.0E+06
2.0E+06
3.0E+06
4.0E+06
5.0E+06
6.0E+06
0 50 100 150 200 250 300 350 400 450 500 550
mm^3
z [mm]
Wb
0
10000
20000
30000
40000
50000
60000
70000
-6.0E+01
-5.0E+01
-4.0E+01
-3.0E+01
-2.0E+01
-1.0E+01
0.0E+00
1.0E+01
0 50 100 150 200 250 300 350 400 450 500 550
N
z [mm]
N Area

4
0
10000
20000
30000
40000
50000
60000
70000
-1.0E+01
-5.0E+00
0.0E+00
5.0E+00
1.0E+01
1.5E+01
2.0E+01
0 50 100 150 200 250 300 350 400 450 500 550
kNm
z [mm]
Mb(x-z) Area
0
10000
20000
30000
40000
50000
60000
70000
-1.8E+01
-1.6E+01
-1.4E+01
-1.2E+01
-1.0E+01
-8.0E+00
-6.0E+00
-4.0E+00
-2.0E+00
0.0E+00
2.0E+00
0 50 100 150 200 250 300 350 400 450 500 550
kNm
z [mm]
Mb(y-z) Area

5
0
10000
20000
30000
40000
50000
60000
70000
-5.0E+00
0.0E+00
5.0E+00
1.0E+01
1.5E+01
2.0E+01
2.5E+01
0 50 100 150 200 250 300 350 400 450 500 550
kNm
z [mm]
Mbtot Area

6
Once forces and moments are known, the normal and tangential stresses are computed by means of
the following formulas:
𝜎 𝑁 =
𝑁
𝐴
𝜎𝑏 =
𝑀 𝑏
𝑊 𝑏
𝜏 =
𝑀𝑡
𝑊𝑡
0
10000
20000
30000
40000
50000
60000
70000
-2.0E+01
-1.0E+01
0.0E+00
1.0E+01
2.0E+01
3.0E+01
4.0E+01
5.0E+01
6.0E+01
7.0E+01
8.0E+01
9.0E+01
0 50 100 150 200 250 300 350 400 450 500 550
kNm
z [mm]
Mt Area

7
The trend of the stresses is displayed in the following diagrams:
0
10000
20000
30000
40000
50000
60000
70000
-1.4E+00
-1.2E+00
-1.0E+00
-8.0E-01
-6.0E-01
-4.0E-01
-2.0E-01
0.0E+00
2.0E-01
0 50 100 150 200 250 300 350 400 450 500 550
MPa
z [mm]
SigmaN Area
0
10000
20000
30000
40000
50000
60000
70000
-2.0E+03
0.0E+00
2.0E+03
4.0E+03
6.0E+03
8.0E+03
1.0E+04
1.2E+04
1.4E+04
1.6E+04
0 50 100 150 200 250 300 350 400 450 500 550
MPa
z [mm]
σb Area

8
0
10000
20000
30000
40000
50000
60000
70000
-5.0E+00
0.0E+00
5.0E+00
1.0E+01
1.5E+01
2.0E+01
2.5E+01
3.0E+01
3.5E+01
4.0E+01
0 50 100 150 200 250 300 350 400 450 500 550
MPa
z [mm]
τ Area

9
Consequently we can calculate and plot equivalent stresses:
 according to Tresca:
𝜎𝑒𝑞 = √(𝜎 𝑁 + 𝜎𝑏)2 + 4𝜏2
0
10000
20000
30000
40000
50000
60000
70000
-1.0E+01
0.0E+00
1.0E+01
2.0E+01
3.0E+01
4.0E+01
5.0E+01
6.0E+01
7.0E+01
8.0E+01
0 50 100 150 200 250 300 350 400 450 500 550
MPa
z [mm]
σeq Tresca Area

10
 according to Von Mises:
𝜎𝑒𝑞 = √(𝜎 𝑁 + 𝜎𝑏)2 + 3𝜏2
By the previous plot we can get the safety factors by using both Tresca and Von Mises criteria:
 the maximum equivalent stress by Tresca is obtained for z=430-492mm and it’s 62 MPa;
 the maximum equivalent stress by Von Mises is obtained for z=430-492mm and it’s 54MPa;
Now, as for shaft 1, since specimens used to test this material according FKM standard have diameter
Deff,N,m=Deff,N,p=16mm≠deff for every section of the shaft, coefficients KA (anisotropy factor) and Kd
(technological size factor) have to be considered:
KA=1(it has been assumed that anisotropy eventually present is negligible for the shaft)
𝐾𝑑, 𝑚 =
1 − 0,7686 ∗ 𝑎𝑑, 𝑚 ∗ log(
𝑑𝑒𝑓𝑓
7,5
)
1 − 0,7686 ∗ 𝑎𝑑, 𝑚 ∗ log(
𝑑𝑒𝑓𝑓, 𝑁
7,5
)
0
10000
20000
30000
40000
50000
60000
70000
-1.0E+01
0.0E+00
1.0E+01
2.0E+01
3.0E+01
4.0E+01
5.0E+01
6.0E+01
7.0E+01
0 50 100 150 200 250 300 350 400 450 500 550
MPa
z [mm]
σeq Von Mises Area

11
𝐾𝑑, 𝑝 =
1 − 0,7686 ∗ 𝑎𝑑, 𝑝 ∗ log(
𝑑𝑒𝑓𝑓
7,5
)
1 − 0,7686 ∗ 𝑎𝑑, 𝑝 ∗ log(
7,5
)
a d,m 0,28
a d,p 0,32
So Rm and Rp are evaluated for each deff as follow:
𝑅𝑚 = 𝑅𝑚, 𝑁 ∗ 𝐾𝐴 ∗ 𝐾𝑑, 𝑚
𝑅𝑒 = 𝑅𝑒, 𝑁 ∗ 𝐾𝐴 ∗ 𝐾𝑑, 𝑝
So the safety factor has been computed for each section by using as 𝜎𝑒𝑞 both the one evaluated by
Tresca and with Von Mises:
𝑆𝑐 =
𝑅 𝑒𝐻
𝜎𝑒𝑞
Finally as safety factor of the whole shaft the minimum one has been taken; in particular by using
Tresca criterion it is equal to 14 whereas by using Von Mises criterion it is equal to 16 (it is remarkable
to notice that the safety factor calculated with Tresca is lower and this is in accordance with the more
conservative kind of failure criterion).

12
1.2 Fatigue
1.2.1 Shaft 1 section 3
From the Table 1.4 we can get values:
Rm,N(Mpa) 1250
Re,N(Mpa) 1050
σD-1 (Mpa) 565
Now, since specimens used to test this material according FKM standard have diameter
Deff,N,m=Deff,N,p=16mm≠deff=90 mm , coefficients KA (anisotropy factor) and Kd (technological size
factor)have to be considered:
KA=1 (it has been assumed that anisotropy eventually present is negligible for the shaft)
𝐾𝑑, 𝑚 =
1 − 0,7686 ∗ 𝑎𝑑, 𝑚 ∗ log(
𝑑𝑒𝑓𝑓
7,5
)
1 − 0,7686 ∗ 𝑎𝑑, 𝑚 ∗ log(
7,5
)
𝐾𝑑, 𝑝 =
1 − 0,7686 ∗ 𝑎𝑑, 𝑝 ∗ log(
𝑑𝑒𝑓𝑓
7,5
)
1 − 0,7686 ∗ 𝑎𝑑, 𝑝 ∗ log(
7,5
)
a d,m 0,28 Kd,m 0,90
a d,p 0,32 kd,p 0,88
So Rm and Rp are calculated:
𝑅𝑚 = 𝑅𝑚, 𝑁 ∗ 𝐾𝐴 ∗ 𝐾𝑑, 𝑚=1120 MPa;
𝑅𝑒 = 𝑅𝑒, 𝑁 ∗ 𝐾𝐴 ∗ 𝐾𝑑, 𝑝=920 MPa.
Now from table σD-1
tc
can be read in order to obtain σc,D-1 of the component:
σ 𝑐,𝐷−1 = σ 𝐷−1 ∗ bo ∗ Kv ∗ Ks

13
where bo can be found in graph according to roughness and Rm and Kv=Ks=1 because no further
information are given.
Finally the value of the admissible value of fatigue limit σc,D-1/1,5 is computed (NB: Safety factor has
been taken equal to 1,5 because according to FKM fatigue life survival probability is 97,5% and in
case of failure no relevant consequence will occur).
In table all values are shown:
σD-1 565 Mpa bo 0,92
σc,D-1 520 Mpa Kv 1
Ks 1
𝝈 𝒄,𝑫−𝟏
𝟏, 𝟓
350 Mpa
Next we can compute σm,nom and σa,nom by applied loads(taken from previous calculation in the static
verification):
Bending moment σb,nom= 66 MPa
Torsion moment τnom= 0 Mpa
Axial load σn,nom= -4 Mpa
Then through gradient factors nσ and nτ, σa,eq,eff and σm,eq,eff are computed, by taking into account that
we are dealing with a shoulder (type E series 1, so from tables notch radius r=2,5mm):
 first of all Xσ and Xτ are computed for the load. In particular:
- for axial load Xσ(d)=0 ; Xσ(r)=
2,3
𝑟
=0,92;
- for bending moment Xσ(d)=
2
𝑑
=0,02; Xσ(r)=
2,3
𝑟
=0,92;
- for torsion moment Xτ(d)=
2
𝑑
=0,02; Xτ(r)=
1,15
𝑟
=0,46;
 then nσ and nτ are evaluated according to X
- if X≤0,1 mm-1
n = 1 + X ∗ 10−(𝑎−0,5+
𝑅𝑚
𝑏
)

14
- if 0,1mm-1
≤X≤1 mm-1
n = 1 + √𝑋 ∗ 10−(𝑎+
𝑅𝑚
𝑏
)
nσ(r) axial 1,12
nσ(d) bending 1,35
nσ(r) bending 1,12
nτ(d) torsion 1
nτ(r) torsion 1,08
So stress concentration factors Kt are evaluated by using approximated formulas related to
experimental diagrams( considering that D/d=1,14 where D and d are respectively the biggest and the
smallest diameters of the shaft across the notch):
Axial Kt,n = 2,21
Bending Kt,b = 2,24
Torsion Kt,t = 1,43
Hence final stress concentration factors Kf are computed:
𝐾𝑓 =
𝐾𝑡
𝑛 𝜎 𝑛 𝜏
Axial Kf,n = 1,98 (NB: in this case nσ(d)=1 )
Bending Kf,b = 1,48
Torsion Kf,t = 1,32
Finally effective applied stress is evaluated σm, eff = σm, nom ∗ Kf;
Axial σn,eff = -8,5 MPa
Bending σb,eff = 98 MPa
Torsion τeff = 0 MPa
At this point σm,eq,eff and σa,eq,eff are computed by considering that alternated stress is due only to
bending stresses and axial and torsional stress provides the constant contribution:
𝜎 𝑚,𝑒𝑞,𝑒𝑓𝑓 = √σn, eff2 + 3𝜏, eff2 = 8.5 MPa
σa,eq,eff= σb,eff =98 MPa

15
Now according to FKM, admissible area is limited on the upper part by the minimum of 2 lines:
(yield line): if σm,eq,eff≥ 0 𝜎 𝑎 + 𝜎 𝑚 = 𝑅 𝑒;
if σm,eq,eff< 0 𝜎 𝑎 − 𝜎 𝑚 = 𝑅 𝑒;
𝜎 𝐷 + 𝑀 𝜎 ∗ 𝜎 𝑚 = 𝜎 𝐷−1
where 𝑀 𝜎 coefficient depends on the material and is equal to:
𝑀 𝜎 = 𝑎 𝑀 ∗ 103
𝑅 𝑚 + 𝑏 𝑚
Values in table:
aM 0,35 Mσ 0,29
bM -0,10 alfa 1,59
In the plot are shown the three lines correspondent to σD, σc,D and
𝝈 𝒄,𝑫
𝟏,𝟓
and the two lines of Yield and
𝑌𝑖𝑒𝑙𝑑
𝟏,𝟓
. In order to have infinite life the point defined by (σm,eq,eff
;σa,eq,eff) is required to be inside the permissible area bounded by the curves of FKM fatigue limit
𝝈 𝒄,𝑫
𝟏,𝟓
and of elastic limit
𝑌𝑖𝑒𝑙𝑑
𝟏,𝟓
both divided by the safety factor; from the plot it can be seen that shaft
has actually infinite life.
-600 -400 -200 0 200 400 600 800
0
100
200
300
400
500
600
700
800
900
1000
0
100
200
300
400
500
600
700
800
-600 -400 -200 0 200 400 600 800
σD σc,D σc,D/1,5 σa,eq Yield Yield/1,5

16
For shaft1_section 4, shaft2_section 1 and 2, we apply the same method already used for shaft
1_section 3; since procedure is exactly the same, only tables and graphs with the main results will be
reported.
d eff,N(mm) 16,00 mm KA 1,00
d eff(mm) 90,00 mm
a d,m 0,28 Kd,m 0,90
a d,p 0,32 kd,p 0,88
Rm,N(Mpa) 1250,00 Mpa Rm 1120
Re,N(Mpa) 1050,00 Mpa Re 924
aM 0,35 Mσ 0,29
bM -0,10 alfa 1,59
σD-1 565 Mpa bo 0,92
Ks 1
σc,D-1/1,5 347 Mpa
Bending moment σb,nom= 69 MPa
Torsion moment τnom= -75 Mpa
Axial load σn,nom= -12 Mpa
[type E series 1, so from tables notch radius r=2,5mm]
2,3
𝑟
=0,92;

17
2
𝑑
=0,02; Xσ(r)=
2,3
𝑟
=0,92;
2
𝑑
=0,02; Xτ(r)=
1,15
𝑟
=0,46;
nσ(r) axial 1,12
nσ(d) bending 1,35
nσ(r) bending 1,12
nτ(d) torsion 1
nτ(r) torsion 1
Axial Kt,n = 2,21
Bending Kt,b = 2,24
Torsion Kt,t = 1,43
Bending Kf,b = 1,48
Torsion Kf,t = 1,32
Axial σn,eff = -23 MPa
Torsion τeff = -100 MPa
 𝜎 𝑚,𝑒𝑞,𝑒𝑓𝑓 = √σn, eff2 + 3𝜏, eff2=174 MPa

18
d eff,N(mm) 16 mm KA 1,00
d int 124
d eff(mm) 260 mm
a d,m 0,28 Kd,m 0,85
a d,p 0,32 kd,p 0,83
Rm,N(Mpa) 1250 Mpa Rm 1070
Re,N(Mpa) 1050 Mpa Re 870
aM 0,35 Mσ 0,27
bM -0,10 alfa 1,78
σD-1 565 Mpa bo 0,92
Ks 1
0
100
200
300
400
500
600
700
800
900
1000
-600 -400 -200 0 200 400 600 800
σa,eq σD σc,D σc,D/1,5 Yield Yield/1,5

19
σc,D-1/1,5 347 Mpa
Bending moment σb,nom= 1,6 MPa
Axial load σn,nom= 0 Mpa
[type F series 1, so from tables notch radius r=4mm]
2,3
𝑟
=0,58;
2
𝑑
=0,01; Xσ(r)=
2,3
𝑟
=0,58;
2
𝑑
=0,01; Xτ(r)=
1,15
𝑟
=0,29;
nσ(r) axial 1,10
nσ(d) bending 1,23
nσ(r) bending 1,10
nτ(d) torsion 1,00
nτ(r) torsion 1,07
Axial Kt,n = 2,53
Bending Kt,b = 2,58
Torsion Kt,t = 1,55
Bending Kf,b = 1,91
Torsion Kf,t = 1,45
Axial σn,eff = 0 MPa

20
Torsion τeff = 0 MPa
d eff,N(mm) 16 mm KA 1,00
d int 124
d eff(mm) 260 mm
a d,m 0,28 Kd,m 0,85
a d,p 0,32 kd,p 0,83
Rm,N(Mpa) 1250 Mpa Rm 1070
Re,N(Mpa) 1050 Mpa Re 870
aM 0,35 Mσ 0,27
bM -0,10 alfa 1,78
0
100
200
300
400
500
600
700
800
900
1000
-600 -400 -200 0 200 400 600 800
σD σc,D σc,D/1,5 Yield σa,eq Yield/1,5

21
σD-1 565 Mpa bo 0,92
Ks 1
σc,D-1/1,5 347 Mpa
Bending moment σb,nom= 4,2 MPa
Axial load σn,nom= 0 Mpa
[type F series 1, so from tables notch radius r=4mm]
 - for axial load Xσ(d)=0 ; Xσ(r)=
2,3
𝑟
=0,58;
 - for bending moment Xσ(d)=
2
𝑑
=0,01; Xσ(r)=
2,3
𝑟
=0,58;
 - for torsion moment Xτ(d)=
2
𝑑
=0,01; Xτ(r)=
1,15
𝑟
=0,29;
 nσ(r) axial 1,10
 nσ(d) bending 1,23
 nσ(r) bending 1,10
 nτ(d) torsion 1,00
 nτ(r) torsion 1,07
 Axial Kt,n = 2,53
 Bending Kt,b = 2,58
 Torsion Kt,t = 1,55
 Axial Kf,n = 2,31 (NB: in this case nσ(d)=1 )
 Bending Kf,b = 1,91

22
 Torsion Kf,t = 1,45
 Axial σn,eff = 0 MPa
 Bending σb,eff = 8 MPa
 Torsion τeff = 0 MPa
From the analysis it can be stated that each section verified has infinite life.
0
100
200
300
400
500
600
700
800
900
1000
-600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00
σa,eq σD σc,D σc,D/1,5 Yield Yield/1,5

23
1.3 Rolling bearings
1.3.1 Roller bearings C and D
In order to calculate the rating life (in hours) of the most loaded bearing between C and D, the
equivalent dynamic load P is computed.
For both the radial bearings the axial force Fa and the radial force Fr are evaluated:
Fa [kN] Fr [kN] Fa/Fr e
Bearing C 0 93 0 0.3
Bearing D 48 129 0.37 0.3
According to the relative values of Fa and Fr the equivalent dynamic load P is computed with the
formulas given in the SKF catalogue.
𝑃 = 0.92 ∗ 𝐹𝑟 + 0.4 ∗ 𝐹𝑎 𝑖𝑓
𝐹𝑎
𝐹𝑟
> 𝑒
𝑃 = 𝐹𝑟 𝑖𝑓
𝐹𝑎
𝐹𝑟
≤ 𝑒
The computed values are shown below and the highest dynamic load results to be applied on
bearing D.
PC=9.3*10 kN
PD=1.4*102
kN
1.3.2 Bearing E
The radial force acting on each bearing E is displayed in the following picture and it is evaluated
from the input torque:

24
𝑅 𝑒 =
𝑀𝑡,𝐼𝑁
𝑟𝑐
⁄
3
The equivalent dynamic load results to be:
𝑃 = 𝐹𝑟 𝑏𝑒𝑐𝑎𝑢𝑠𝑒
𝐹𝑎
𝐹𝑟
≤ 𝑒
Re [kN] rc [mm] P [kN]
4.6*10^2 324 4.6*10^2
1.3.3 Rating life_bearing D
The rating life is calculated as:
𝐿 𝑛𝑚 = 𝑎1 ∗ 𝑎 𝑆𝐾𝐹 ∗ (
𝐶
𝑃
)
𝑝
 aSKF depends on the viscosity ratio 𝑘 =
𝑣
𝑣1
, with v actual operating viscosity of the lubricant
(ISO VG 320) and v1 is the rated viscosity

25
In diagram 1 it is considered an operating temperature T=50°C; in diagram 2 v1 comes from the values
of the bearing bore diameter d and the bearing outer diameter D (see related figure in part1_static
design), and from the value of the angular speed of the shaft n2 (rpm).
The angular speed of the shaft 2 is evaluated in the following way:
𝜔 𝑜𝑢𝑡 = 𝜔𝑖𝑛 ∗ 𝑖
𝜔2 = 𝜔 𝑜𝑢𝑡 ∗
𝑟1
𝑟2
𝑛2 = 𝜔 𝑜𝑢𝑡 ∗
2 ∗ 𝜋
60
= 133 𝑟𝑝𝑚
𝑘 = 5.14
Due to the fact that the value of k>4 it is necessary to change the lubricant and to adopt one
guaranteeing a value of k=4.
Then it is possible to find the value of 𝑎 𝑆𝐾𝐹 from the following diagram, given the values of k just
found and the coefficient ηc (given as input datum) and Pu=143 kN (defined in the catalogue).

26
 a1 is the life adjustment factor for reliability and it is assumed to be equal to 1 for reliability
90%;
 C is the basic dynamic load rating (kN) from the datasheet of the bearing on the catalogue
and its value is 737 kN;
 p is the exponent of the life equation which is 10/3 for roller bearings
The value of rating life is evaluated and it results to be equal to:
𝐿 𝑛𝑚 = 1740 𝑚𝑖𝑙𝑙𝑖𝑜𝑛𝑠 𝑜𝑓 𝑐𝑦𝑐𝑙𝑒
Finally the value of operating hours is computed:
𝐿10ℎ =
𝐿 𝑛𝑚 ∗ 106
𝑛2 ∗ 60
= 218000 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 ℎ𝑜𝑢𝑟𝑠

27
1.3.3 Rating life_bearing E
Following the same procedure discussed in the previous section, the next results are obtained:
Bearing E
rc [mm] 323
rp [mm] 228
vc [m/s] 0,831
wp[rad/s] 3,64
np 35
R[kN] 4,6*102
A[kN] 0
e 0,35
Y1 1,9
P[kN] 4,6*102
d[mm] 190
D[mm] 400
dm [mm 295
T [°C] 50
v[mm^2/s] 185
v1[mm^2/s] 105
k 1,76
Pu [kN] 210
eta c 0,5
eta c*Pu/P 0,225
a SKF 1,5
a1 1
C [kN] 2120
p 3,33
Lnm 240
Lnh 1,2*105

28
The reference diagrams are displayed below:

29
In order to compute the value of the angular velocity related to the bearing E the following formulas
are used:
 𝜔𝑐 = 𝜔𝑖𝑛
 𝑣𝑐 = 𝜔𝑐 ∗ 𝑟𝑐
 𝜔 𝑝 =
𝑣 𝑐
𝑟 𝑝
1.4 Spur gears
1.4.1 Computation of transmission ratio, pt and b
To compute transmission ratio of stage two we referred to the following scheme and considerations:

30
𝑣𝑐 = 𝜔𝑐 (𝑟𝑝 + 𝑟𝑠)
Value of ωc is given as input data and it is equal to ωi = 24.5 rpm, consequently:
2𝑣𝑐 = 𝑣 𝐵 = 𝜔𝑠 𝑟𝑠 => 𝜔𝑠 =
2𝑣 𝑐
𝑟𝑠
= 198.6 𝑟𝑝𝑚
𝜔 𝑝 =
𝑣𝑐
𝑟𝑝
Transmission ratio of stage one can now be evaluated as:
𝑖 𝑠𝑡𝑎𝑔𝑒 𝐼 =
𝜔 𝑜𝑢𝑡
𝜔 𝑖𝑛
=
𝜔 𝑠
𝜔 𝑝
= 8.11
𝑖 𝑡𝑜𝑡 = 𝑖 𝑠𝑡𝑎𝑔𝑒 𝐼* 𝑖 𝑠𝑡𝑎𝑔𝑒 𝐼𝐼 => 𝑖 𝑠𝑡𝑎𝑔𝑒 𝐼𝐼 =
𝑖 𝑡𝑜𝑡
𝑖 𝑠𝑡𝑎𝑔𝑒 𝐼
= 5.14
Starting from the just evaluated transmission ratio, also equal to
𝑍2
𝑍1
with Z number of teeth, and by
knowing that since α = 20˚ the minimum number of teeth is = 17 ∗ 𝑐𝑜𝑠3
𝛽 = 15:
Z1 = 15
Z2 = 77
Values have been obtained in order to find a ratio close to 5.14.
Next step is to evaluate mn:
𝑝𝑡 =
𝑝 𝑛
𝑐𝑜𝑠𝛽
=
2𝜋𝑟1
𝑍1
=> 𝑝 𝑛 =
𝑝𝑡
𝑐𝑜𝑠𝛽
= 26 𝑚𝑚
𝑚 𝑡 =
𝑚 𝑛
𝑐𝑜𝑠𝛽
= 8.6 𝑚𝑚
𝑚 𝑛 =
𝑝 𝑛
𝜋
= 8.3 𝑚𝑚
An iterative procedure as been followed, due to the given constrain to choose a value of mn from the
table, to obtain the following results:
 r1 = 62.1 mm
 mn = 8 mm
 mt = 8.3 mm
 pt = 26.0 mm
Let’s now calculate the value of face width. By measuring it from the drawing and applying a
proportion factor related to the previous obtained values of radii, we get b = 190 mm .
Since b has to be at least equal to three or four times the circumferential pitch, the following
comparison has been made:
26 ∗ 4 = 104 < 190
Consequently requirement has been satisfied.

31
1.4.2 Tooth bending strength
Safety factor with respect to bending fatigue failure is computed for both the pinion and the wheel.
𝑤 𝐹𝑡 = 𝑤 + 𝐾𝐼 + 𝐾 𝑉 + 𝐾𝐹𝛼 + 𝐾𝐹𝛽
Given 𝑤 𝐹𝑡, we compute the tooth bending stress as:
𝜎 𝐹 =
𝑤 𝐹𝑡
𝑚 𝑛
∗ 𝑌𝑓 ∗ 𝑌𝜀 ∗ 𝑌𝛽
Unknown coefficients are computed according to the procedure suggested in the tutorial; results are
shown in the following table:
Safety factor is found with the following procedure:
𝑆 𝐹 𝑠ℎ𝑎𝑓𝑡1 =
𝜎 𝐹𝐷
𝜎 𝐹
𝑆 𝐹 𝑠ℎ𝑎𝑓𝑡2 =
𝜎 𝐹𝐷
𝜎 𝐹
The values of safety factors are computed for different materials: 34CrNiMo6 is the one with the
most similar composition of the material of the shaft. GGG 60 was chosen to verify the required
safety factor.
Ft N
W N/mm
Ki
Ke
fpe
Cv
εβ
Cβ
V
Kv
fE
qL
qLeff
ξα
KFα
KFβ
WFt
Zn
YF
Yε
Yβ
σF
σF GGG 60
S 2,472604 > 1,8
σF 34CrNiMo6
S 2,082193 > 1,8
320,00
shaft 2
173341,31
912,32
1,25
4,00
8,00
0,02
0,70
6,63
1,11
1,96
1,00
0,43
0,87
0,50
1,73
1095,05
85,44
2,21
0,58
0,88
153,68
380,00
24,00
Ft N
W N/mm
Ki
Ke
fpe
Cv
εβ
Cβ
V
Kv
fE
qL
qLeff
ξα
KFα
KFβ
WFt
Zn
YF
Yε
Yβ
σF
σFD GGG 60
S 1,768725 > 1,8
σFD 34CrNiMo6
S 1,489453 > 1,8
320,00
shaft 1
4,00
7,00
0,02
0,70
6,63
173341,31
912,32
1,25
3,09
1,96
0,87
1,00
1,11
0,50
1,73
20,00
0,42
1094,87
0,58
16,64
0,88
214,84
380,00

32
1.4.3 Tooth surface fatigue strength
Safety factor with respect to surface fatigue failure is computed for both the pinion and the wheel.
𝑤 𝐻𝑡 = 𝑤 + 𝐾𝐼 + 𝐾 𝑉 + 𝐾 𝐻𝛼 + 𝐾 𝐻𝛽
Given 𝑤 𝐻𝑡, we compute the Hertz stress as:
𝜎 𝐻 = √
𝑤 𝐻𝑡
𝑑1
∗
𝑢 + 1
𝑢
∗ 𝑍 𝑀 ∗ 𝑍𝜀 ∗ 𝑍 𝐻
Unknown coefficients are computed according to the procedure suggested in the tutorial; results are
shown in the following table:
Safety factor is found with the following procedure:
𝑆 𝐻 𝑠ℎ𝑎𝑓𝑡 1 =
𝜎 𝐻𝐷
𝜎 𝐻
𝑆 𝐻 𝑠ℎ𝑎𝑓𝑡 1 =
𝜎 𝐻𝐷
𝜎 𝐻
In this case, it wasn’t possible to find a value for ZM in the tables, so we chose to use the values of
the material selected before ( GGG 60 ).
Ft N
W N/mm
Ki
Ke
fpe
Cv
εβ
Cβ
V
Kv
fE
qL
qLeff
ξα
KHα
KHβ
Zε
WHt
Zh
ZM GGG 60
σH
σHd GGG 60
S 0,603802 > 1,4
shaft 1
912,32
1,25
4,00
173341,31
0,70
6,63
1,11
7,00
0,02
1,96
1,73
20,00
0,42
0,50
0,76
1,00
0,63
960,82
1,714546278
248
811,5237718
490
Ft N
W N/mm
Ki
Ke
fpe
Cv
εβ
Cβ
V
Kv
fE
qL
qLeff
ξα
KHα
KHβ
Zε
WHt
Zh
ZM GGG 60
σH
σHd GGG 60
S 1,355947 > 1,25
shaft 2
912,32
1,25
4,00
173341,31
0,70
6,63
1,11
8,00
0,02
1,96
1,73
24,00
0,43
0,50
0,77
1,00
0,63
978,01
1,714546278
248
361,3711439
490

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