The project Computation of a Gearbox for a Wind Power Plant dealt with the study and the analysis of the Generator in order to provide data about the following characteristics
o Static Design of the gearbox:
Design of shaft 1
Design of bearings
Internal stresses
Safety factor
Design of shaft 2
o Fatigue analysis:
Shaft 1 section 3
Shaft 1 section 4
Shaft 2 section 1
Shaft 2 section 2
o Rolling bearings analysis, life and fatique:
Roller bearings C and D
Bearing E
Rating life_bearing D
Rating life_bearing E
o Spur gears analysis:
Computation of transmission ratio, pt and b
Tooth bending strength
Tooth surface fatigue strength
A collection of the PDF presentation published at the Clean Energy Council (CEC) Wind Forum 2016 in Melbourne. Key issues discussed were colocation of wind and solar, noise impacts, planning requirements across Australia and developments in the technology.
A collection of the PDF presentation published at the Clean Energy Council (CEC) Wind Forum 2016 in Melbourne. Key issues discussed were colocation of wind and solar, noise impacts, planning requirements across Australia and developments in the technology.
Frequency regulation of deregulated power system having grc integrated with r...eSAT Journals
Abstract This paper addresses the automatic generation control of deregulated multi area power system including one of the most important renewable energy resource viz. wind power plant. The interconnected two area power system is thermal-hydro system with two GENCOs in thermal area and one GENCO in hydro area. Generation rate constraint (GRC) is considered in all the GENCOs separately. Integral square error technique is used to optimize the gains of various integral controllers. It is seen that system response becomes poorer in terms of peak overshoots and settling time by considering GRC. DISCO participation matrix is chosen on the basis of open market strategy which is continuously changing. So, effect of changing DPM on dynamic responses is studied, following a step load perturbation. It is revealed that there is no effect of changing DPM on system’s dynamic responses. Keywords – Automatic generation control (AGC), deregulated power system, DISCO participation matrix (DPM), wind turbines, generation rate constraint (GRC).
With state-of-the-art hybrid technology that comes in the form of SolarMill that efficiently utilizes the potential of wind and sun energy in any climatic conditions, minimizing your carbon footprint becomes to breeze
Improved reactive power capability with grid connected doubly fed induction g...Uday Wankar
In the past, most national grid codes and standards did not require wind turbines to support the power system during a disturbance. For example during a grid fault or sudden drop in frequency wind turbines were tripped off the system. However, as the wind power penetration continues to increase, the interaction between the wind turbines and the power system has become more important. This is because, when all wind turbines would be disconnected in case of a grid failure, these renewable generators will, unlike conventional power plants, not be able to support the voltage and the frequency of the grid during and immediately following the grid failure. This would cause major problems for the systems stability.
Therefore, wind farms will have to continue to operate during system disturbances and support the network voltage and frequency. Network design codes are now being revised to reflect this new requirement. A special focus in this requirement is drawn to both the fault ride-through capability and the grid support capability. Fault ride-through capability addresses mainly the design of the wind turbine controller in such a way that the wind turbine is able to remain connected to the network during grid faults (e.g. short circuit faults). While grid support capability represents the wind turbine capability to assist the power system by supplying ancillary services, i.e. such as supplying reactive power, in order to help the grid voltage recovery during and just after the clearance of grid faults. Due to the partial-scale power converter, wind turbines based on the DFIG are very sensitive to grid disturbances, especially to voltage dips during grid faults.
Faults in the power system, even far away from the location of the turbine, can cause a voltage dip at the connection point of the wind turbine. The abrupt drop of the grid voltage will cause over-current in the rotor windings and over- voltage in the DC bus of the power converters. Without any protection, this will certainly lead to the destruction of the converters. In addition, it will also cause over-speeding of the wind turbine, which will threaten the safe operation of the turbine. Thus a lot of research works have been carried out on the LVRT ability of DFIG wind turbines under the grid fault. These LVRT strategies can be divided into two main types: the active method by improving control strategies, the passive scheme with additional hardware protective devices.
This report gives an overview of patenting activity around Doubly-fed Induction Generators (DFIG) used in the horizontal axis wind turbines for efficient power generation. Patents were categorized as per key DFIG technologies and analyzed for generating different trends within PatSeer Project.
Aircraft Structures for Engineering Students 5th Edition Megson Solutions ManualRigeler
Full donwload : http://alibabadownload.com/product/aircraft-structures-for-engineering-students-5th-edition-megson-solutions-manual/ Aircraft Structures for Engineering Students 5th Edition Megson Solutions Manual
EGME 306A The Beam Page 1 of 18 Group 2 EXPER.docxSALU18
EGME 306A The Beam
Page 1 of 18
Group 2
EXPERIMENT 3:The Beam
Group 2 Members:
Ahmed Shehab
Marvin Penaranda
Edwin Estrada
Chris May
Bader Alrwili
Paola Barcenas
Deadline Date: 10/23/2015
Submission Date: 10/23/2015
EGME 306A – UNIFIED LABORATORY
EGME 306A The Beam
Page 2 of 18
Group 2
Abstract (Bader):
The main objective for this experiment was to determine the stress, deflection, and strain of a supported beam
under loading, and to experimentally verify the beam stress and flexure formulas. Additionally, maximum
bending stress and maximum deflection were determined. To accomplish this, a 1018 steel I-beam with a strain
gage bonded to the underside was utilized in conjunction with a dial indicator to monitor beam deflection. In
order to determine the values for strain and deflection, the beam underwent testing utilizing the MTS Tensile
Testing machine, which applied a controlled, incrementally increasing load to the beam. This data was then
utilized along with calculations for the beams neutral axis, moment of inertia, and section modulus to determine
the required objective values. Final values of 12,150 psi for the maximum actual stress (vs. 12,784.8 psi for
theoretical stress), and 0.0138 in for the maximum actual deflection (vs. .0130 in for theoretical deflection)
correlated closely with each other, and successfully verify established beam stress and flexure formulas.
EGME 306A The Beam
Page 3 of 18
Group 2
Table of Contents:
List of Symbols and Units 4
Theory 5
Procedure and Experimental Set-up 8
Results 9
Sample Calculations and Error Analysis 12
Discussion and Conclusion 15
Bibliography 16
Appendix 17
EGME 306A The Beam
Page 4 of 18
Group 2
List of Symbols and Units (Chris):
List of Symbols and Units Name of variables (units) Units
𝜎 Stress psi
𝑃 Applied load lbf
𝐼 Moment of Inertia in.4
𝜀 Strain in/in
𝐿 Length of the bar in
Z Section Modulus of Beam in3
𝑐 Distance to Beam Neutral Axis in
𝐸 Modulus of Elasticity psi
EGME 306A The Beam
Page 5 of 18
Group 2
Theory (Edwin):
There are two main objectives for this experiment: to determine maximum bending stress values in
the beam and to determine the deflection in the beam. To help visualize this phenomena, imagine
cutting a section of a symmetrically loaded beam:
Now, examine diagrams of this section before (Fig. A) and after bending (Fig. B):
(Fig. A)
(Fig. B)
The main points to take away from the above diagrams are as follows: When the moment, M is applied
as shown in Fig. A, forces will be in compression near the top (positive moment) and in tension near
the bottom (negative moment). The effects from this moment are seen in Fig. B.
For determining max stress values, one concept to note is that our bending moment M can help
calculate bending stress. First, we rec
it contains the basic information about the shear force diagram which is the part of the Mechanics of solid. there many numerical solved and whivh will give you detaild idea in S.f.d.
Diganosis of pulley belt system faults using vibrations analysiskhalid muhsin
This work presents the dynamic response related to varying in indispensable parameters of pulley-belt system such as rotational speed, pulleys center distance, diameters ratios and type of belt (cross section). A pulley- belt system was manufactured, and the experimental results were obtained and analyzed accurately which reveal the effects of the above parameters on system working. Then, the pulley-belt system faults were studied using vibration analysis technique. A Selected faults were created like belts damage, belt resonance case, manipulation of initial tension in the belt, introducing belt slippage while running with the pulley by adding oil between the belt and the pulley, and catching the pulley damage then an experimental results were obtained and analyzed to find out how these types of faults appear in the manufactured system.
Manufactured system was assembled from a heavy steel structure, variable speed DC motor, speed measurement tool (tachometer), two shafts of 25 mm diameter, four pillow block ball bearings, V-belts and pulleys with different diameters (10, 15, 20, 25) cm. The fabricated system presents good mechanical characteristics, like gradually changing in rotational speed (500 to 2000) rpm, diameters ratios (1 to 2.5) and center distance (27 to 120) cm. Vibration sensors ADXL335 (3-axis accelerometer) were mounted on bearing brackets of drive and driven shafts and connected to Arduino type mega 2560 (microcontroller), which sends the data of vibration to the laptop in order to display it in Sigview software in the time and frequency domain band by FFT (fast fourier transform).
The results of FFT is used for the effect of each type of faults and dynamic response of changing parameters comparing with the healthy condition (FFT band) of the system.
Simulation of a small scale cogeneration system using a microturbinePietro Galli
The aim of the thesis was to develop an operating model of some sizes of Microturbines on the software eQuest.
Then used the model developed to simulate the installation of a Cogeneration (CHP) plant to satisfy the heat and electric demand of a school. The final step was to consider the economics of the investment and choosing which size and use of the turbine was the most convenient for the building.
The key topics of the project were the following:
• Collecting experimental data about some size of Microturbines
• Development of the turbine model for the software eQuest
• Development of the building following the ASHRAE standard
• Simulation of some size of Cogeneration Plants
• Analysis of the results of the simulations
• Energy auditing of the building and analysis of the cost of the current and new plant
• Choice of the best solution for the building
The thesis was developed in collaboration between Politecnico di Torino and the University of Illinois at Chicago, during the course of the double degree program TOP-UIC
During the course, we had to develop some projects and assemblies on CAD using the Advaced GT&D dimensioning system to label the work.
In the attachments are presented some of the final assembly and dimensions given to the products
During the course, we had to develop some projects and assemblies on CAD using the Advaced GT&D dimensioning system to label the work.
In the attachments are presented some of the final assembly and dimensions given to the products
Study of a Gear-Rack and links System: Equations, configuration and CAD designPietro Galli
The study dealt with the Designing of a mechanism form a starting configuration. Development of equations (Study of the motion and of the velocities), CAD files and engineer drawings (Using Solidworks)
Design of a Handle Mechanism (Solidworks)Pietro Galli
The project was developped on Solidworks and dealt with the costruction of an Hadle mechanism, the study of its motion and the Finite Element Analysis (FEA) of the assembly.
Bachelor's Thesis: Use of CAD technologies to optimize the productivity of a ...Pietro Galli
The Thesis was developed during the Internship at Plurigest SRL. It deals with one of the projects done during that experience. In particular the aim of this was to provide a plant layout for a Pellet Heating system.
All the components of the plant were designed on Solidworks and AutoCAD. Then was performed a study of the layout of the components in order to provide best solution in terms of space used and functioning.
If you’d like discover more about this project in the attachment there is the pdf version of the thesis (Italian language) and some pictures. For more info please contact me
Plant layout of a Warehouse for a big toys companyPietro Galli
The project dealt with the definition of all the characteristics of a Warehouse using the principles of the Kaizen and of Lean manufacturing. The main points of the analysis were the following:
• Definition of the optimal geographical area were to build the Warehouse considering the connections among cities that will be served.
• Definition of the flux of the goods
• Definition of the Industrial shelving needed considering the dimension and frequency of movement of the goods.
• Definition of the plant layout of the Warehouse considering the principle of Lean manufacturing and the safety norms.
• Definition of the Forklift trucks needed
• Definition of the personnel needed
• Definition of the proper heating system
• Definition of the light system
The calculus were performed using Excel and the plant layout using Autocad and Solidworks
The project included the realization on Soliworks of some tools to realize through 3D printing. The components had to respect some important characteristics in order to be being 3D printable
Pictures and drawings of the assembly are reported
In this project was realized a gear mechanism able to move and its motion was studied
In the attachment you can find some of the video of its motion and the pictures of the gears
Solar Energy Equipment: Design of a solar plant for a buildingPietro Galli
The project dealt with the design of a solar plant heating-system for a building. The following analysis were performed:
• Data collection and preliminary design: (building characteristics and geographical information)
• Estimation of the how water demand
• Evaluation of the Irradiance:
• Computation of the volume of the hot storage tank and definition of its insulation
• Off-Design Calculations
• F-chart
• Thermal balance of the solar panel
• Heat exchanger analysis
• Exergy Analysis for each component of the plant (Collectors, Tank, Boiler)
• Economics and cost analysis
• Determination of the temperature profile of the thermal fluid along the pipe (Using MatLab)
NUMERICAL SIMULATIONS OF HEAT AND MASS TRANSFER IN CONDENSING HEAT EXCHANGERS...ssuser7dcef0
Power plants release a large amount of water vapor into the
atmosphere through the stack. The flue gas can be a potential
source for obtaining much needed cooling water for a power
plant. If a power plant could recover and reuse a portion of this
moisture, it could reduce its total cooling water intake
requirement. One of the most practical way to recover water
from flue gas is to use a condensing heat exchanger. The power
plant could also recover latent heat due to condensation as well
as sensible heat due to lowering the flue gas exit temperature.
Additionally, harmful acids released from the stack can be
reduced in a condensing heat exchanger by acid condensation. reduced in a condensing heat exchanger by acid condensation.
Condensation of vapors in flue gas is a complicated
phenomenon since heat and mass transfer of water vapor and
various acids simultaneously occur in the presence of noncondensable
gases such as nitrogen and oxygen. Design of a
condenser depends on the knowledge and understanding of the
heat and mass transfer processes. A computer program for
numerical simulations of water (H2O) and sulfuric acid (H2SO4)
condensation in a flue gas condensing heat exchanger was
developed using MATLAB. Governing equations based on
mass and energy balances for the system were derived to
predict variables such as flue gas exit temperature, cooling
water outlet temperature, mole fraction and condensation rates
of water and sulfuric acid vapors. The equations were solved
using an iterative solution technique with calculations of heat
and mass transfer coefficients and physical properties.
HEAP SORT ILLUSTRATED WITH HEAPIFY, BUILD HEAP FOR DYNAMIC ARRAYS.
Heap sort is a comparison-based sorting technique based on Binary Heap data structure. It is similar to the selection sort where we first find the minimum element and place the minimum element at the beginning. Repeat the same process for the remaining elements.
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Hierarchical Digital Twin of a Naval Power SystemKerry Sado
A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.
NO1 Uk best vashikaran specialist in delhi vashikaran baba near me online vas...Amil Baba Dawood bangali
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Computation of a gearbox for a wind power plant
1. 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
2. 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
1.2.2 Shaft 1 section 4 16
1.2.3 Shaft 2 section 1 18
1.2.4 Shaft 2 section 2 20
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
3. 2
1. Technical Report 1
The object of the report is the computation of the gearbox of the wind power plant shown below.
4. 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:
6. 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
7. 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
8. 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):
9. 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
10. 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.
11. 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.
12. 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)
13. 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:
17. 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
18. 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
19. 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
20. 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
21. 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:
22. 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:
23. 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
24. 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
25. 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
26. 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
28. 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:
36. 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
37. 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
38. 11
𝐾𝑑, 𝑝 =
1 − 0,7686 ∗ 𝑎𝑑, 𝑝 ∗ log(
𝑑𝑒𝑓𝑓
7,5
)
1 − 0,7686 ∗ 𝑎𝑑, 𝑝 ∗ log(
𝑑𝑒𝑓𝑓, 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:
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).
39. 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(
𝑑𝑒𝑓𝑓, 𝑁
7,5
)
𝐾𝑑, 𝑝 =
1 − 0,7686 ∗ 𝑎𝑑, 𝑝 ∗ log(
𝑑𝑒𝑓𝑓
7,5
)
1 − 0,7686 ∗ 𝑎𝑑, 𝑝 ∗ log(
𝑑𝑒𝑓𝑓, 𝑁
7,5
)
where ad is found in tables:
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
40. 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
)
41. 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
42. 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
43. 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.
1.2.2 Shaft 1 section 4
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
σc,D-1 520 Mpa Kv 1
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]
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;
45. 18
1.2.3 Shaft 2 section 1
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
σc,D-1 520 Mpa Kv 1
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
46. 19
σc,D-1/1,5 347 Mpa
Bending moment σb,nom= 1,6 MPa
Torsion moment τnom= 0 Mpa
Axial load σn,nom= 0 Mpa
[type F series 1, so from tables notch radius r=4mm]
first of all Xσ and Xτ are computed for the load. In particular:
- for axial load Xσ(d)=0 ; Xσ(r)=
2,3
𝑟
=0,58;
- 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
Torsion Kf,t = 1,45
Axial σn,eff = 0 MPa
Bending σb,eff = 3 MPa
47. 20
Torsion τeff = 0 MPa
𝜎 𝑚,𝑒𝑞,𝑒𝑓𝑓 = √σn, eff2 + 3𝜏, eff2=0 MPa
σa,eq,eff= σb,eff =3 MPa
1.2.4 Shaft 2 section 2
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
48. 21
σD-1 565 Mpa bo 0,92
σc,D-1 520 Mpa Kv 1
Ks 1
σc,D-1/1,5 347 Mpa
Bending moment σb,nom= 4,2 MPa
Torsion moment τnom= 0 Mpa
Axial load σn,nom= 0 Mpa
[type F series 1, so from tables notch radius r=4mm]
first of all Xσ and Xτ are computed for the load. In particular:
- for axial load Xσ(d)=0 ; Xσ(r)=
2,3
𝑟
=0,58;
- 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
49. 22
Torsion Kf,t = 1,45
Axial σn,eff = 0 MPa
Bending σb,eff = 8 MPa
Torsion τeff = 0 MPa
𝜎 𝑚,𝑒𝑞,𝑒𝑓𝑓 = √σn, eff2 + 3𝜏, eff2=0 MPa
σa,eq,eff= σb,eff =8 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
50. 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:
51. 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
52. 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).
53. 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
54. 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
56. 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:
57. 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.
58. 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
59. 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