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School of Engineering (Mechanical and Automotive)
Major Assessment Course: Mechanics of Machines (MIET1077)
Experiment: Six Bar Linkage Mechanism
Course Coordinator: Prof. Firoz Alam ([email protected])
Student Information Family Name Given Name
Pace Samuel
Student Number: S3659265
Due Date: 25/10/20
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Contents Summary ........................................................................................................................................................................................................... 2
Objectives .......................................................................................................................................................................................................... 2
Related Theory .................................................................................................................................................................................................. 2
Calculating Lengths of Links .............................................................................................................................................................................. 3
Graphical Position Analysis ................................................................................................................................................................................ 4
Graphical Velocity Analysis ................................................................................................................................................................................ 5
Complete Vector Polygon ............................................................................................................................................................................ 7
Graphical Velocity Results ............................................................................................................................................................................ 7
Graphical Acceleration Analysis ........................................................................................................................................................................ 8
Complete Vector Polygon .......................................................................................................................................................................... 15
Graphical Acceleration Results .................................................................................................................................................................. 16
Graphical Instantaneous Centre Analysis ........................................................................................................................................................ 16
Comlete Instantaneous Centre Diagram .................................................................................................................................................... 22
Results for Instantaneous Centre of Zero Velocity Method ....................................................................................................................... 23
Analytical Method for Position, Velocity and Acceleration ............................................................................................................................. 23
Position Analysis ........................................................................................................................................................................................ 23
Part A - Crank ........................................................................................................................................................................................ 23
Part B β Coupler & Rocker .................................................................................................................................................................... 24
Part C β Slider ....................................................................................................................................................................................... 29
Velocity Analysis ........................................................................................................................................................................................ 30
Part A - Crank ........................................................................................................................................................................................ 30
Part B β Coupler & Rocker .................................................................................................................................................................... 30
Part C - Slider ........................................................................................................................................................................................ 32
Acceleration Analysis ................................................................................................................................................................................. 34
Part A β Crank ....................................................................................................................................................................................... 34
Part B β Coupler & Rocker .................................................................................................................................................................... 34
Part C β Slider ....................................................................................................................................................................................... 37
Summary of Analytical Values .................................................................................................................................................................... 38
Dynamic Forces ............................................................................................................................................................................................... 39
Equations of Motion .................................................................................................................................................................................. 39
Equation Matrix ......................................................................................................................................................................................... 45
Working Model Simulation .............................................................................................................................................................................. 45
Balancing Strategy Development ...................................................................................................................................................................... 0
Results ............................................................................................................................................................................................................... 1
Discussion .......................................................................................................................................................................................................... 2
Conclusion ......................................................................................................................................................................................................... 3
References ........................................................................................................................................................................................................ 3
Attachments ...................................................................................................................................................................................................... 3
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Summary This report details the methods for calculating the position, velocity and acceleration of a 6 bar linkage mechanism. The graphical method of calculating the position, velocity, acceleration and instantaneous centres of zero velocity use vector polygons and position diagrams to calculate and display graphically the magnitude and direction of each component which can be measured to provide an accurate result. The values are then cross checked using the analytical method for calculating each component. A simulation was made using Working Model to provide further analysis and verification of the results. Each of these methods provided results that would be used to calculate the dynamic forces of the system and generate a balancing strategy for reducing the vibration in the system. From this report, the positions, velocity, and accelerations of the linkages have been accurately calculated and verified using multiple methods of calculating the results.
Objectives The objectives of this major assessment are to:
β’ Apply the understanding of Mechanics of Machines course content β’ Calculate position, velocity, acceleration, and instantaneous centres of the system using
graphical methods β’ Calculate position, velocity and acceleration using analytical methods β’ Develop and analyse a working model simulation to evaluate the outputs β’ Calculate the dynamic forces on the system and develop a balancing strategy development
Related Theory A linkage mechanism has a grounded link, driver link (crank) and a slave link (rocker). The grounded (link g) link throughout the whole motion of the system will always have no velocity or acceleration. This link is the link which the system revolves around and defines the joints which the links rotate around. The crank (a and Ξ±) is what drives the motion of the system and determines the limits of motion for each link and joint. The rocker (b and Ξ²) has its motion driven by the crank and that motion is dependent on the size of the crank. The floating link (link f) has complex motion dependant on the crank and rocker and is what connects them together.1
Figure 1 4 Bar Linkage Mechanism Source: http://dynref.engr.illinois.edu/aml.html
Many different linkage mechanisms can be seen in our daily life, some examples can be seen below.
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Figure 2 Bike Pedalling Source:
http://dynref.engr.illinois.edu/aml.html
Figure 3 Knee Joint Source
:http://dynref.engr.illinois.edu/aml.html
Figure 4 Suspension with Watt's linkage Source: http://dynref.engr.illinois.edu/aml.html
These different linkages seen in our daily lives are made up of different types of 4 bar linkage mechanisms and can have the same principles applied to finding out their function and ranges of movement.
Calculating Lengths of Links To begin the following values were given to begin calculating the links.
OA (m) 0.2 B (deg) 12 ΞΈ1 (deg) 6 n2 (rpm) 480
The following lengths are calculated using the provided equations
π»π» =ππππ4
=0.24
= 0.05ππ
ππππ = πΆπΆπΆπΆ = 4 β ππππ = 4 β 0.2 = 0.8ππ
The following values are left to be calculated through the simultaneous equations provided.
πππΆπΆ = ?
πππ΄π΄ = ?
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π΄π΄ππ = ?
Trial and error was used to find the following values for the unknowns.
πππΆπΆ = 0.3ππ
πππ΄π΄ = 0.5ππ
π΄π΄ππ = 0.6ππ
These values were checked against the equations to test their validity.
ππππ + πππ΄π΄ < ππππ + π΄π΄ππ
0.2 + 0.5 < 0.8 + 0.6
0.7 < 1.4
ππππ + ππππ < πππ΄π΄ + πππ΄π΄
0.2 + 0.8 < 0.5 + 0.6
1.0 < 1.1
Since both these statements are true, the system may continue with the following values used for each of the links.
Graphical Position Analysis
Figure 5 Graphical Position Analysis
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The positions of 16 different angles of ππ2 and the subsequent positions of each of the other positions. The paths are πππ΄π΄, πππ΅π΅, πππΆπΆ and πππ·π· with the limits of the motion for B being π΄π΄πΏπΏ and π΄π΄π π and the limits of D being πΆπΆπΏπΏ and πΆπΆπ π .
Graphical Velocity Analysis To begin the graphical velocity analysis, the velocity of point A is calculated as it has a known rotational speed and direction and so can be found.
ππ2 = 480 ππππππ =48060
ππππππ = 8 ππππππ = 8 β 2ππ ππππππ ππβ = 50.26 ππππππ ππβ
πππ΄π΄ = πππ΄π΄πππππ΄π΄ = 50.26 β 200 = 10053.1ππππ/ππ
The direction of πππ΄π΄ is tangential to the position circle and the velocity is going anticlockwise.
The next component to be found is πππ΅π΅ which can be found using the following equation
πππ΅π΅β₯OE = πππ΄π΄β₯OA + πππ΅π΅/π΄π΄β₯BE
This equation produces the following vector polygon.
Figure 6 Finding VB
The same technique is used to find πππΆπΆ , but there are 2 equations which need to be used. πππΆπΆ = πππ΄π΄β₯OA + πππΆπΆ/π΄π΄β₯AC
πππΆπΆ = πππ΅π΅β₯BE + πππΆπΆ/π΅π΅β₯BC
Drawing out the known components and the relative components, the following vector polygon is found.
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Figure 7 Finding VC
The lines can then be connected to find the relative velocities and the point which they intersect is the arrowhead of πππΆπΆ .
Figure 8 Graphically Solving for VC
Now that πππΆπΆ is known, πππ·π· can be found using it in the following equation.
πππ·π·||ππβππ = πππΆπΆ + πππ·π·/πΆπΆβ₯CD
The following vector polygon is produced, and the connection points define the vectors of the velocities.
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Figure 9 Finding VD
Figure 10 Solving for VD
Complete Vector Polygon Now that all the individual components are found, the entire velocity vector polygon can be seen below.
Figure 11 Velocity Vector Polygon
The program which the vector polygon was drawn in had a scale which was used at πππ΄π΄ of 1 point = 100mm/s. Using this scale, the following magnitudes and angles for the rest of the vectors can be solved.
Graphical Velocity Results Width (pts) Height (pts) Length (pts) Velocity (mm/s) Velocity (m/s)
πππ΄π΄/ππ 72.649 61.685 95.30434 9530.434 9.530434
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πππΆπΆ/ππ 35.039 45.354 57.31245 5731.245 5.731245
πππΆπΆ 52.023 4.911 52.25429 5225.429 5.225429
πππ΄π΄ 14.413 11.42 18.38888 1838.888 1.838888
πππ΄π΄/πΆπΆ 37.674 16.331 41.06132 4106.132 4.106132
πππ΄π΄/πΆπΆ 54.92 0 54.92 5492 5.492
πππΆπΆ/πΆπΆ 2.896 4.911 5.701293 570.1293 0.570129 Table 1 Graphical Velocity Analysis Results
Graphical Acceleration Analysis To begin the graphical acceleration analysis, first the known values are calculated to calibrate the scale and distance. The following equation is used to find the magnitude of πππ΄π΄ππ.
πππ΄π΄ππ =πππ΄π΄2
πππ΄π΄=
(πππ΄π΄πππππ΄π΄)2
πππ΄π΄=
(50.265 β 200)2
200= 505323.6 ππππ/ππ2
Since it is known that the direction is perpendicular to link OA, πππ΄π΄ππ can be sketched as follows. Since the velocity is known to be constant, the tangential acceleration of the support is known to be 0. Because there is no tangential component to the acceleration, the normal component of the acceleration is equal to the overall acceleration of the link.
πππ΄π΄ = πππ΄π΄ππ||πππ΄π΄ + πππ΄π΄π‘π‘β₯πππ΄π΄
πππ΄π΄π‘π‘β₯πππ΄π΄ = 0
Figure 12 Calibrating Scale with aA
Using the following graphic, the directions for each component of the accelerations can be found.
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Figure 13 Directions of Acceleration Components
The first component to be found is πππ΅π΅/π΄π΄ππ using the following calculation.
πππ΅π΅/π΄π΄ππ =
πππ΅π΅/π΄π΄2
πππ΅π΅/π΄π΄=
9530.4342
500= 181658.3 ππππ/ππ2
This is graphed as follows with πππ΅π΅/π΄π΄ππ off the end of πππ΄π΄ and the line of πππ΅π΅/π΄π΄
π‘π‘ being tangential.
πππ΅π΅/π΄π΄ = πππ΅π΅/π΄π΄ππ
||π΄π΄π΅π΅+ πππ΅π΅/π΄π΄
π‘π‘β₯π΄π΄π΅π΅
Figure 14 Finding aB/A
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The next part to be calculated is πππ΅π΅ππ using the following equation.
πππ΅π΅ππ =πππ΅π΅2
πππ΅π΅=
1838.882
600= 5635.85 ππππ/ππ2
πππ΅π΅ππ is graphed from the origin with the πππ΅π΅π‘π‘ being tangential.
πππ΅π΅ = πππ΅π΅ππ||π΄π΄π΅π΅ + πππ΅π΅π‘π‘β₯π΄π΄π΅π΅
Figure 15 Finding aB
Since πππ΅π΅π‘π‘ and πππ΅π΅/π΄π΄π‘π‘ intersection defines their lengths, the values can be sketched accordingly.
πππ΅π΅ππ||π΄π΄π΅π΅ + πππ΅π΅π‘π‘β₯π΄π΄π΅π΅ = πππ΅π΅/π΄π΄ππ
||π΄π΄π΅π΅+ πππ΅π΅/π΄π΄
π‘π‘β₯π΄π΄π΅π΅
Figure 16 Solving for aB and aB/A
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From this graph, the connection between the vector between the tangential and normal components of the vector can be resolved to find the full vectors for πππ΅π΅ and πππ΅π΅/π΄π΄.
Figure 17 Finding aB and aB/A
Now the acceleration at C is to be calculated starting with πππΆπΆ/π΄π΄. First πππΆπΆ/π΄π΄ππ is to be calculated using
the following.
πππΆπΆ/π΄π΄ππ =
πππΆπΆ/π΄π΄2
πππ΄π΄πΆπΆ=
5731.242
300= 109490.7 ππππ/ππ2
πππΆπΆ/π΄π΄ππ can be sketched from πππ΄π΄ with πππΆπΆ/π΄π΄
π‘π‘ being perpendicular to the normal component as follows.
πππΆπΆ/π΄π΄ = πππΆπΆ/π΄π΄ππ
||π΄π΄πΆπΆ+ πππΆπΆ/π΄π΄
π‘π‘β₯π΄π΄πΆπΆ
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Figure 18 Finding aC/A
The same principle can be applied for finding πππΆπΆ/π΅π΅ππ and πππΆπΆ/π΅π΅
π‘π‘ as follows.
πππΆπΆ/π΅π΅ππ =
πππΆπΆ/π΅π΅2
πππ΅π΅πΆπΆ=
4106.132
215.77= 78141.1 ππππ/ππ2
With πππΆπΆ/π΅π΅ππ coming from the end of πππ΅π΅ and πππΆπΆ/π΅π΅
π‘π‘ being perpendicular to πππΆπΆ/π΅π΅ππ, the following
vector polygon can be drawn.
πππΆπΆ/π΅π΅ = πππΆπΆ/π΅π΅ππ
||π΅π΅πΆπΆ+ πππΆπΆ/π΅π΅
π‘π‘β₯π΅π΅πΆπΆ
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Figure 19 Finding aC/B
From here, the πππΆπΆ/π΄π΄ and πππΆπΆ/π΅π΅ values are combined to find the tangential components using the following equation and vector polygon.
πππΆπΆ/π΅π΅ππ + πππΆπΆ/π΅π΅
π‘π‘ = πππΆπΆ/π΄π΄ππ + πππΆπΆ/π΄π΄
π‘π‘
Figure 20 Solving for aC/B and aC/A
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The lines can be connected accordingly and the vector component of πππΆπΆ/π΅π΅, πππΆπΆ/π΄π΄ and πππΆπΆ can be found as follows.
Figure 21 Calculating aC/B and aC/A
The final acceleration to be found is πππ·π· which required πππΆπΆ and since the directions are known the following vector polygon can be created using the same principles as before.
πππ·π·/πΆπΆππ =
πππ·π·/πΆπΆ2
πππΆπΆπ·π·=
570.132
800= 406.31 ππππ/ππ2
The πππ·π·/πΆπΆππ is sketched with πππ·π·/πΆπΆ
π‘π‘perpendicular and πππ·π· parallel to the x-axis.
πππ·π·||ππβππ = πππ·π·/πΆπΆππ
||π·π·πΆπΆ+ πππ·π·/πΆπΆ
π‘π‘β₯π·π·πΆπΆ
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Figure 22 Finding aD
This vector polygon can be solved for πππ·π·/πΆπΆπ‘π‘ and πππ·π·.
Figure 23 Calculating aD
Complete Vector Polygon All the vector polygons are combined to produce the overall vector polygon for the acceleration.
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Figure 24 Acceleration Vector Polygon
Graphical Acceleration Results Using the initial scale, the vector polygon can be resolved to find the rest of the components in the system. The magnitudes for each of the accelerations calculated using the graphical method can be observed in the table below.
Width (pts)
Height (pts)
Length (pts)
Acceleration (mm/s2)
Angle (deg)
Aa 252.662 437.623 505.323635 505323.635 239.999976
Ab/a 289.362 6.329 289.431206 289431.206 181.252988
Ab 545.44 426.877 692.62455 692624.55 218.047753
Ac/a 168.412 39.139 172.900153 172900.153 193.083313
Ac 421.074 398.485 579.735809 579735.809 223.421191
Ad/c 233.83 398.485 462.024636 462024.636 239.595696
Ad 187.244 0 187.244 187244 180 Table 2 Graphical Acceleration Analysis Results
Graphical Instantaneous Centre Analysis For the graphical instantaneous centre of zero velocities analysis, we will be using Kennedyβs theorem. To begin Kennedyβs theorem, the number of ICβs must be found using the following.
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πΆπΆ =ππ(ππ β 1)
2 =
6(6 β 1)2
= 15
From observation of the model, the position of the following instantaneous centres can be found.
πΌπΌπΆπΆ12, πΌπΌπΆπΆ23, πΌπΌπΆπΆ34, πΌπΌπΆπΆ14, πΌπΌπΆπΆ35, πΌπΌπΆπΆ56, πΌπΌπΆπΆ16
The 8 unknown instantaneous centres left to be found can be observed in Kennedyβs circle.
Figure 25 Kennedy's Circle with Unknown and Known Instantaneous Centre
The following instantaneous centres to be found are seen as dotted lines in the Kennedyβs circle.
πΌπΌπΆπΆ13, πΌπΌπΆπΆ15, πΌπΌπΆπΆ24, πΌπΌπΆπΆ45, πΌπΌπΆπΆ46, πΌπΌπΆπΆ36, πΌπΌπΆπΆ25, πΌπΌπΆπΆ26
The instantaneous centres are found in the following order using the lines between the respective instantaneous centres.
πΌπΌπΆπΆ13 is found using the intersection of lines produced from πΌπΌπΆπΆ12 to πΌπΌπΆπΆ23 and πΌπΌπΆπΆ14 to πΌπΌπΆπΆ34.
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Figure 26 Finding IC13
πΌπΌπΆπΆ15 is found using the lines πΌπΌπΆπΆ13 to πΌπΌπΆπΆ35 and πΌπΌπΆπΆ16 to πΌπΌπΆπΆ56.
Figure 27 Finding IC15
πΌπΌπΆπΆ24 is found using the lines πΌπΌπΆπΆ12 to πΌπΌπΆπΆ14 and πΌπΌπΆπΆ23 to πΌπΌπΆπΆ34.
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Figure 28 Finding IC24
πΌπΌπΆπΆ45 is found using the lines πΌπΌπΆπΆ14 to πΌπΌπΆπΆ15 and πΌπΌπΆπΆ35 to πΌπΌπΆπΆ34.
Figure 29 Finding IC54
πΌπΌπΆπΆ46 is found using the lines πΌπΌπΆπΆ14 to πΌπΌπΆπΆ16 and πΌπΌπΆπΆ45 to πΌπΌπΆπΆ56.
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Figure 30 Finding IC64
πΌπΌπΆπΆ36 is found using the lines πΌπΌπΆπΆ13 to πΌπΌπΆπΆ16 and πΌπΌπΆπΆ35 to πΌπΌπΆπΆ56.
Figure 31 Finding IC63
πΌπΌπΆπΆ25 is found using the lines πΌπΌπΆπΆ12 to πΌπΌπΆπΆ15 and πΌπΌπΆπΆ24 to πΌπΌπΆπΆ45.
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Figure 32 Finding IC25
πΌπΌπΆπΆ26 is found using the lines πΌπΌπΆπΆ12 to πΌπΌπΆπΆ16 and πΌπΌπΆπΆ25 to πΌπΌπΆπΆ56.
Figure 33 Finding IC26
All the instantaneous centres can be seen placed onto the 6 bar linkage mechanism as seen below.
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Comlete Instantaneous Centre Diagram
Figure 34 Graphical Inspection of All Instantaneous Centres
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πΌπΌπΆπΆ15 is off the top of the graph but can be seen below.
Figure 35 Graphical IC15
Using these graphs, the velocities can be calculated with πππ΄π΄ calibrating π·π·1 using πΌπΌπΆπΆ13. Once π·π·1 is calculated πππ΅π΅ and πππΆπΆ can be calculated. Using πππΆπΆ and πΌπΌπΆπΆ15, π·π·2 can be calibrated and used to calculated πππ·π·. From the graph, the following values were calculated
Results for Instantaneous Centre of Zero Velocity Method Height
(pts) Width (pts)
Length (pts)
Velocity (mm/s)
Graphical Velocity (mm/s)
vA 8.706 5.027 10.05311718 10053.12 10053.09
vB 1.441 1.142 1.83865304 1838.653 1838.89
vC 5.202 0.491 5.225120573 5225.121 5225.43
vD 5.308 0 5.308 5308 5492 Figure 36 Joint Velocities Calculated Using Kennedy's Theorem
Analytical Method for Position, Velocity and Acceleration Position Analysis Part A - Crank To start the analytical method, the calculations are completed for the crank (link OA). The position of the crank is first calculated.
πππ¨π¨ = ππ2 cos(ππ2) = 200 cos(60) =ππππππππππ
πππ¨π¨ = ππ2 sin(ππ2) = 200 sin(60) =ππππππ.ππππππ
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Part B β Coupler & Rocker To begin calculating the coupler and rocker, the vector diagram for the system is sketched as follows.
Figure 37 Position Vector of Crank, Rocker and Coupler
Starting by using the vector form of the components to create a vector polygon.
π π π΄π΄ + π π 3 = π π 2 + π π 4
π π 3 = π π π΄π΄π΄π΄ + π π 4
π π π΄π΄π΄π΄ = π π π΄π΄ β π π π΄π΄ = ππ1ππππππ1 β ππ2ππππππ2
π π π΄π΄π΄π΄ + ππ2ππππππ2 β ππ1ππππππ1 = 0
Using this equation, the Euler identity can be used to solve the equation.
ππππππ = cos (ππ) + ππ β sin (ππ)
π π π΄π΄π΄π΄ = ππ1(cos(ππ1) + ππ β sin(ππ1)) β ππ2(cos(ππ2) + ππ β sin(ππ2))
Splitting the equation into the real and imaginary component.
π π π΄π΄π΄π΄ = (ππ2 cos(ππ2) β ππ1 cos(ππ1)) + ππ(ππ1 sin(ππ) β ππ2 sin(ππ2))
Solving for the components.
π π ππ = π π ππππππ ππππππππππππππππππ
πΌπΌππ = πΌπΌπππππΌπΌπππππππππΌπΌ ππππππππππππππππππ
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πΉπΉπΉπΉ = ππ1 cos(ππ1) β ππ2 cos(ππ2) = 800 β cos(6) β 200 β cos(60) = ππππππ.ππππππππ
π°π°ππ = ππ1 sin(ππ1) β ππ2 sin(ππ2) = 800 β sin(6) β 200 β sin(60) = βππππ.ππππππππ
Using the vector graph of the linkages, the following equations can be used to calculate ππ3 and ππ4.
ππ3ππππππ3 = π π ππ + πππΌπΌππ + ππ4ππππππ4
ππ3ππβππππ3 = π π ππ β πππΌπΌππ + ππ4ππππππ4
ππ32 = π π ππ2 + πΌπΌππ2 + π π ππππ4(cos(ππ4) + ππ sin(ππ4)) β πππΌπΌππππ4(cos(ππ4) + ππ sin(ππ4))
+ π π ππππ4(cos(ππ4) + ππ sin(ππ4)) + πππΌπΌππππ4(cos(ππ4) + ππ sin(ππ4)) + ππ42 = 0
This equation can be simplified and used for finding ππ4.
π π ππ2 + πΌπΌππ2 + 2π π ππππ4 cos(ππ4) + 2πΌπΌππππ4 sin(ππ4) + ππ42 β ππ3
2 = 0
695.622 + (β89.58)2 + 2(695.62)(600) cos(ππ4) + 2(β89.58)(600) sin(ππ4) + (600)2 β (500)2
= 0
483883.73 + 8024.99 + 834741.02 cos(ππ4) β 107498.78 sin(ππ4) + 360000 β 250000 = 0
601908.72 + 834741.02 cos(ππ4) β 107498.78 sin(ππ4) = 0
834741.02 cos(ππ4) β 107498.78 sin(ππ4) = β601908.72
Solving this equation has 2 answers for different angles of ππ4.
π½π½ππ = βππππππ.ππππππ π π πΉπΉπ π π π πΉπΉπΉπΉπ π ππππ π½π½ππ = ππππππ.ππππ π π πΉπΉπ π π π πΉπΉπΉπΉπ π
The values of ππ4 are checked using the following equation at ππ4 = β142.995 to make sure that the equations are correct.
sin(ππ4) =2 tan οΏ½ππ42 οΏ½
1 + tan2 οΏ½ππ42 οΏ½
sin(β142.995) =2 tan οΏ½β142.995
2 οΏ½
1 + tan2 οΏ½β142.9952 οΏ½
β0.6019 = β0.6019
Since these values are true, the equation using cos(ππ4) should be true as well, proving that this is the angle of ππ4.
cos(ππ4) =1 β tan2 οΏ½ππ42 οΏ½
1 + tan2 οΏ½ππ42 οΏ½
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cos(β142.995) =1 β tan2 οΏ½β142.995
2 οΏ½
1 + tan2 οΏ½β142.9952 οΏ½
β0.7986 = β0.7986
The other calculated value of ππ4 = 128.31 is checked against the same equations to validate the results.
sin(128.31) =2 tan οΏ½128.31
2 οΏ½
1 + tan2 οΏ½128.312 οΏ½
0.7847 = 0.7847
cos(128.31) =1 β tan2 οΏ½128.31
2 οΏ½
1 + tan2 οΏ½128.312 οΏ½
β0.6199 = β0.6199
The following equation is used to find C which can be also used to find ππ4 and validate the results.
πͺπͺ =π π ππ2 + πΌπΌππ2 + ππ4
2 β ππ32
2ππ4=
695.622 + (β89.58)2 + 6002 β 5002
2 β 600= ππππππ.ππππ
π½π½ππ = 2 arctan οΏ½βπΌπΌππ Β± βπ π ππ2 + πΌπΌππ2 β πΆπΆ2
πΆπΆ β π π πποΏ½
= 2 arctanοΏ½β(β89.58) Β± οΏ½695.622 + (β89.58)2 β 501.592
501.59 β 695.62οΏ½
= β2.496 ππππ 2.239 ππππππ = βππππππ.ππππππ πππ π ππππππ.ππππππ π π πΉπΉπ π
Using these equations, the following vector field is used to calculate ππ3
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Figure 38 Diagram to Find AB'B
Distance between B and Bβ real component for ππ4 = β142.995.
ππ3 + ππ4 cos(ππ4) + π π ππ = 500 + 600 cos(β142.995) + 695.62 = 716.467ππππ
Distance between B and Bβ imaginary component for ππ4 = β142.995.
πΌπΌππ + ππ4 sin(ππ4) = β89.58 + 600 sin(β142.995) = β450.713ππππ
Calculating ππ3
ππ3 = 2 arctan οΏ½πΌπΌππ Β± ππ4 sin(ππ4)
ππ3 + ππ4 cos(ππ4) + π π πποΏ½ = 2 arctanοΏ½
β89.58 Β± 600 sin(β142.995)500 + 600 cos(β142.995) + 695.62
οΏ½
π½π½ππ = βππππ.ππππππ π π πΉπΉπ π ππππ ππ3 = 41.514 πππππΌπΌ
Since the link is below the x-axis, the link will not work for ππ3 = 41.514 πππππΌπΌ so therefore ππ3 =β64.346 πππππΌπΌ.
From this the X and Y position of B can be solved using the following equations.
πΏπΏπ©π© = πππ΄π΄ + ππ4 cos(ππ4) = 800 cos(6) + 600 cos(β142.995) = ππππππ.ππππππππππ
πππ©π© = πππ΄π΄ + ππ4 sin(ππ4) = 800 sin(6) + 600 sin(β142.995) = βππππππ.ππππππππ
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Figure 39 4 Bar Linkage with the First ππ3
Distance between B and Bβ real component for ππ4 = 128.32.
ππ3 + ππ2 cos(ππ4) + π π ππ = 500 + 200 cos(128.32) + 695.62 = 823.60ππππ
Distance between B and Bβ imaginary component for ππ4 = 128.32.
πΌπΌππ + ππ4 sin(ππ4) = β89.58 + 600 sin(128.32) = 381.16ππππ
Calculating ππ3 using the above values.
ππ3 = 2 arctan οΏ½πΌπΌππ Β± ππ4 sin(ππ4)
ππ3 + ππ4 cos(ππ4) + π π πποΏ½ = 2 arctanοΏ½
β89.58 Β± 600 sin(128.32)500 + 600 cos(128.32) + 695.62
οΏ½
π½π½ππ = ππππ.ππππππ π π πΉπΉπ π ππππ ππ3 = β68.458 πππππΌπΌ
Since the link is above the x-axis, the link will not work for ππ3 = β68.458 πππππΌπΌ so therefore ππ3 =49.670 πππππΌπΌ.
From this the X and Y position of B can be solved using the following equations.
πΏπΏπ©π© = πππ΄π΄ + ππ4 cos(ππ4) = 800 cos(6) + 600 cos(49.670) = ππππππ.ππππππππ
πππ©π© = πππ΄π΄ + ππ4 sin(ππ4) = 800 sin(6) + 600 sin(49.670) = ππππππ.ππππππππ
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Figure 40 4 Bar Linkage with the Second ππ3
After visually producing each of the scenarios, it can be deduced that the angle used for the system will be whenππ4 = 128.32 πππππΌπΌ and ππ3 = 49.670 πππππΌπΌ. For calculating the πΆπΆ position, the Ξ² angle will need to be used along with the previous results.
Ξ² = β12 πππππΌπΌ
πΏπΏπͺπͺ = ππ2 cos(ππ2) + πππ΄π΄πΆπΆ cos(ππ3 + Ξ²) = 200 cos(60) + 300 cosοΏ½49.67 + (β12)οΏ½ = ππππππ.ππππ ππππ
πππͺπͺ = ππ2 sin(ππ2) + πππ΄π΄πΆπΆ sin(ππ3 + Ξ²) = 200 sin(60) + 300 sinοΏ½49.67 + (β12)οΏ½ = ππππππ.ππππ ππππ
Part C β Slider From the data calculated, the following values are known.
π»π» = β50 ππππ
πππΆπΆ = 337.46 ππππ
πππΆπΆ = 356.54 ππππ
ππ5 = πΆπΆπΆπΆ = 800 ππππ
From these given values, ππ5 can be calculated through the following.
π½π½ππ = arcsin οΏ½π»π» β πππΆπΆππ5
οΏ½ = arcsin οΏ½(β50) β 356.54
800οΏ½ = β0.533 ππππππ = βππππ.ππππ π π πΉπΉπ π
From this, the position of πππ·π· and πππ·π· can be found.
πΏπΏπ«π« = πΊπΊ = πππΆπΆ + ππ5 cos(ππ3) = 337.46 + 800 cos(β30.54) = ππππππππ.ππππππππ
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πππ«π« = π―π― = βππππππππ
A summary of all the calculated values can be seen in the table below.
Joint X Component (mm)
Y Component (mm)
A 100 173.2051
B 423.6679 554.4237
C 337.4643 356.5373
D 1026.469 -50
E 795.6175 83.62277 Table 3 Analytical Position Results
Angle Value (degrees)
ΞΈ1 6 ΞΈ2 60 ΞΈ3 49.66962 ΞΈ4 128.31 ΞΈ5 -30.5421
Table 4 Analytical Angles
Velocity Analysis Part A - Crank To begin the velocity analysis, the velocity of the crank is to be calculated.
ππ2Μ = ππ2 = 50.26 ππππππ/ππ
οΏ½ΜοΏ½π₯π΄π΄ = βππ2ππ2Μ cos(ππ2) = β200 β 50.26 cos(60) = β8706.24
οΏ½ΜοΏ½πΌπ΄π΄ = βππ2 ππ2Μsin(ππ2) = β200 β 50.26sin(60) = β5026.55
π½π½π¨π¨ = οΏ½οΏ½ΜοΏ½π₯π΄π΄2 + οΏ½ΜοΏ½πΌπ΄π΄2 = οΏ½(β8706.24)2 + (β5026.55)2 = ππππππππππ.ππππππ/π π
πΉπΉπ½π½π¨π¨ = ππ2 +ππ2
=ππ3
+ππ2
=5ππ6ππππππ = ππππππ π π πΉπΉπ π π π πΉπΉπΉπΉπ π
Part B β Coupler & Rocker To begin finding the velocities at each point, first the rotational velocity for link 3 and 4 must be found (ππ3 and ππ4). Using the closed vector loop method, these values can be found.
π π 2 + π π 3 β π π 1 β π π 4 = 0
Using the Euler identity:
ππ2ππππππ2 + ππ3ππππππ3 β ππ1ππππππ1 β ππ4ππππππ4 = 0
200 β ππππππ2 + 500 β ππππππ3 β 800 β ππππππ1 β 600 β ππππππ4 = 0
Differentiate with respect to time to find the velocity.
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ππππππ
(200ππππππ2 + 500ππππππ3 β 800ππππππ1 β 600ππππππ4) =ππππππ
(0)
200ππππππππ2ππππ2ππππ
+ 500ππππππππ3ππππ3ππππ
β 800ππππππππ1ππππ1ππππ
β 600ππππππππ4ππππ4ππππ
= 0
200ππππππππ2ππ2 + 500ππππππππ3ππ3 β 800ππππππππ1ππ1 β 600ππππππππ4ππ4 = 0
Using Eulerβs formula, the equation can be solved.
ππππππ = cos (ππ) + ππsin (ππ)
200ππππ2(cos (ππ2) + ππsin (ππ2)) + 500ππππ3(cos (ππ3) + ππsin (ππ3)) β 800ππππ1(cos(ππ1) + ππ sin(ππ1))β 600ππππ4(cos (ππ4) + ππsin (ππ4)) = 0
This equation can be divided into the real and imaginary components to solve for ππ4.
200ππ2 sin(ππ2) + 500ππ3 sin(ππ3) β 800ππ1 sin(ππ1) β 600ππ4sin (ππ4) = 0
500ππ3 =800ππ1 sin(ππ1) + 600ππ4 sin(ππ4) β 200ππ2 sin(ππ2)
sin(ππ3)
200ππ2 cos(ππ2) + 500ππ3 cos(ππ3) β 800ππ1 cos(ππ1) β 600ππ4ππππππ(ππ4) = 0
500ππ3 =800ππ1 cos(ππ1) + 600ππ4 cos(ππ4) β 200ππ2 cos(ππ2)
cos(ππ3)
These 2 equations are equal, so simultaneous equations can be used to find ππ4.
800ππ1 sin(ππ1) + 600ππ4 sin(ππ4) β 200ππ2 sin(ππ2)sin(ππ3)
=800ππ1 cos(ππ1) + 600ππ4 cos(ππ4) β 200ππ2 cos(ππ2)
cos(ππ3)
Since ππ1 = 0, the equation can be further simplified.
600ππ4 sin(ππ4) β 200ππ2 sin(ππ2)sin(ππ3) =
600ππ4 cos(ππ4) β 200ππ2 cos(ππ2)cos(ππ3)
600ππ4(cos(ππ4) sin(ππ3) β sin(ππ4) cos(ππ3)) = 200ππ2(cos(ππ2) sin(ππ3) β sin(ππ2) cos(ππ3))
ππ4 =200ππ2
600β
cos(ππ2) sin(ππ3) β sin(ππ2) cos(ππ3)cos(ππ4) sin(ππ3) β sin(ππ4) cos(ππ3)
ππ4 =200(50.26)
600β
cos(60) sin(49.67) β sin(60) cos(49.67)cos(128.31) sin(49.67) β sin(128.31) cos(49.67)
ππππ = ππ.ππππππ π π πππ π /π π
Using this value of ππ4, it can be substituted back into one of the original equations to find ππ3.
500ππ3 =600 β 3.064 sin(128.31) β 200 β 50.26 sin(60)
sin(49.67)
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ππππ = βππππ.ππππ π π πππ π /π π
Now that the angular velocities have been calculated, the velocities of the links can be calculated.
πππ΅π΅/π΄π΄ = ππππ3ππ3(cos(ππ3) + ππ sin(ππ3)) = ππ3ππ3(ππ cos(ππ3) β sin(ππ3))
πππ΅π΅/π΄π΄ = 500 β (β19.05)(ππ cos(49.67) β sin(49.67)) = (β6166.44) + ππ(β7263.40)
π½π½π©π©/π¨π¨ = οΏ½(β6166.44)2 + (β7263.40)2 = ππππππππ.ππππππππ/π π
Using the same principles, πππ΅π΅ can be found.
πππ΅π΅ = ππππ4ππ4(cos(ππ4) + ππ sin(ππ4)) = ππ4ππ4(ππ cos(ππ4) β sin(ππ4))
πππ΅π΅ = 600 β (3.064)(ππ cos(128.31) β sin(128.31)) = (β1442.83) + ππ(β1139.89)
π½π½π©π© = οΏ½(β1442.83)2 + (β1139.89)2 = ππππππππ.ππππππππ/π π
The direction of πππ΅π΅ can be found using the following equation.
πΏπΏπππ΅π΅ = arctan οΏ½πΌπΌπππ π πποΏ½ + ππ = arctan οΏ½
β1139.89β1442.83
οΏ½ + ππ = 3.810 ππππππ
πΉπΉπ½π½π©π© = ππππππ.ππππ π π πΉπΉπ π
Part C - Slider Using the currently calculated values, ππ5 can be found.
ππ = ππ2 cos(ππ2) + πππ΄π΄πΆπΆ cos(ππ3 β π½π½) + ππ5 cos(ππ5)
ππ2ππππππ2 + πππ΄π΄πΆπΆππππ(ππ3βπ½π½) β ππ5ππππππ5 β ππππππππ3 β π»π»ππππππ4 = 0
ππππππ
(ππ2ππππππ2 + πππ΄π΄πΆπΆππππ(ππ3βπ½π½) β ππ5ππππππ5 β ππππππππ3 β π»π»ππππππ4) =ππππππ
(0)
ππππππ
(200ππππππ2 + 300ππππ(ππ3βπ½π½) β 800ππππππ5 β 1026.47ππππππ3 β 50ππππππ4) = 0
200ππππ2(cos(ππ2) + ππ sin(ππ2)) + 300ππππ3(cos(ππ3 β π½π½) + ππ sin(ππ3 β π½π½)) β 800ππππ5(cos(ππ5)+ ππ sin(ππ5)) = 0
This equation can be split into its real and imaginary components to solve for ππ5.
200ππ2 sin(ππ2) + 300ππ3 sin(ππ3 β π½π½) β 800ππ5 sin(ππ5) = 0
200ππππ2 cos(ππ2) + 300ππππ3 cos(ππ3 β π½π½) β 800ππππ5 cos(ππ5) = 0
One of these equations can be solved to find ππ5 since the rest of the values are known.
200 β 50.26 cos(60) + 300 β β19.05 cos(49.67 β 12) β 800ππ5 cos(β30.54) = 0
ππππ = βππ.ππππππ π π πππ π /π π
πππΆπΆ/π΄π΄ is now calculated using the same principles as previous velocity calculations.
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πππΆπΆ/π΄π΄ = πππ΄π΄πΆπΆππ3(ππ cos(ππ3 β π½π½) β sin(ππ3 β π½π½))
πππΆπΆ/π΄π΄ = 300 β (β19.06)(ππ cos(49.67 β 12) β sin(49.67 β 12))
πππΆπΆ/π΄π΄ = (3493.6) + ππ(β4525.1)
π½π½πͺπͺ/π¨π¨ = οΏ½(3493.6)2 + (β4525.1)2 = ππππππππ.ππππππππ/π π
πππΆπΆ is calculated using velocity addition.
πππΆπΆ = πππΆπΆ/π΄π΄ + πππ΄π΄
πππΆπΆ = οΏ½3493.6 + (β8706.24)οΏ½ + ππ(β4525.1 + (β5026.55))
πππΆπΆ = (β5212.67) + ππ(501.45)
π½π½πͺπͺ = οΏ½(β5212.67)2 + (501.45)2 = ππππππππ.ππππππππ/π π
πΏπΏπππΆπΆ = arctan οΏ½πΌπΌπππ π πποΏ½ + ππ = arctan οΏ½
501.45β5212.67
οΏ½ + ππ = 3.046 ππππππ
πΉπΉπ½π½πͺπͺ = ππππππ.ππππ π π πΉπΉπ π
πππ·π·/πΆπΆ is calculated using the same technique as shown in πππΆπΆ/π΄π΄.
πππ·π·/πΆπΆ = ππ5ππ5(ππ cos(ππ5) β sin(ππ5))
πππ·π·/πΆπΆ = 800 β (β0.73)(ππ cos(β30.54) β sin(β30.54))
πππ·π·/πΆπΆ = (β295.87) + ππ(β501.45)
π½π½π«π«/πͺπͺ = οΏ½(β295.87)2 + (β501.45)2 = ππππππ.ππππππππ/π π
πππ·π· is calculated using velocity addition, the same way as πππΆπΆ .
πππ·π· = πππ·π·/πΆπΆ + πππΆπΆ
πππ·π· = οΏ½β295.87 + (β5212.67)οΏ½ + ππ(β501.45 + 501.45) = (β5508.55) + ππ(0)
π½π½π«π«/πͺπͺ = οΏ½(β5508.55)2 + (0)2 = ππππππππ.ππππππππ/π π
πΏπΏπππ·π· = arctan οΏ½πΌπΌπππ π πποΏ½ + ππ = arctan οΏ½
0β5508.55
οΏ½ + ππ = 3.142 ππππππ
πΉπΉπ½π½π«π« = ππππππ π π πΉπΉπ π
Now that all the values are calculated, their values are compared to the velocities calculated in the graphical method to evaluate their differences.
Graphical
(mm/s) Analytical
(mm/s) Difference
(%) vA 10053.09 10053.10 -3.497E-05
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vB/A 9530.43 9527.96 0.02597
vC/A 5731.25 5716.78 0.2525
vC 5225.43 5236.74 -0.2164
vB 1838.89 1838.78 0.005762
vD 5492 5508.55 -0.3013
vD/C 570.13 582.23 -2.122 Table 5 Analytical Velocity Results
Observing these values obtained from the graphical and analytical methods shows that the numbers obtained from the calculations are within 3% of each other at every calculation step. This agreement between both methods proves that both the methods are effective in evaluating the numbers.
Acceleration Analysis Part A β Crank To begin the acceleration analysis, the crank acceleration is found using the following equations.
ππ2Μ = πΌπΌ2 = 0 ππππππ/ππ2
π₯π₯οΏ½ΜοΏ½π΄ = βππ2ππ2Μ sin(ππ2)βππ2ππ2Μ cos(ππ2) = β200 β 0 sin(60) β 200 β 50.26 cos(60)= β252661.87ππππ ππ2β
πΌπΌοΏ½ΜοΏ½π΄ = ππ2ππ2Μ cos(ππ2)βππ2ππ2Μ sin(ππ2) = 200 β 0 cos(60) β 200 β 50.26 sin(60) = β437623.20ππππ ππ2β
πππ¨π¨ = οΏ½π₯π₯οΏ½ΜοΏ½π΄2 + πΌπΌοΏ½ΜοΏ½π΄2 = οΏ½(β252661.87)2 + (β437623.20)2 = ππππππππππππ.ππππππππ π π ππβ
Part B β Coupler & Rocker Differentiating the velocity expression is used to obtain the acceleration.
ππ2ππππ2ππππππ2 + ππ3ππππ3ππππππ3 β ππ4ππππ4ππππππ4 β ππ1ππππ1ππππππ1 = 0
πππππποΏ½ππ2ππππ2ππππππ2 + ππ3ππππ3ππππππ3 β ππ4ππππ4ππππππ4 β 0οΏ½ =
ππππππ
(0)
200ππππ2ππππππ2ππππ22
ππππ+ 500ππππ3ππππππ3
ππππ32
ππππβ 600ππππ4ππππππ4
ππππ42
ππππ= 0
Therefore, the acceleration is equal to the following equation.
οΏ½200πππΌπΌ2ππππππ2 β 200ππ22ππππππ2οΏ½ + οΏ½500πππΌπΌ3ππππππ3 β 500ππ3
2ππππππ3οΏ½ + (β600πππΌπΌ4ππππππ4 + 600ππ42ππππππ4) = 0
πππ΄π΄ = 200πππΌπΌ2ππππππ2 β 200ππππ22ππππππ2
πππ΅π΅/π΄π΄ = 500πππΌπΌ3ππππππ3 β 500ππππ32ππππππ3
πππ΅π΅/π΄π΄ = β600πππΌπΌ4ππππππ4 + 600ππππ42ππππππ4
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Using Eulerβs formula, the equation can split into the real and imaginary component and solved for πΌπΌ3 and πΌπΌ4.
ππππππ = cos (ππ) + ππsin (ππ)
πΌπΌ2 = 0
οΏ½β200ππ22(cos(ππ2) + ππ sin(ππ2))οΏ½ + οΏ½500πππΌπΌ3(cos(ππ3) + ππ sin(ππ3))οΏ½
β οΏ½500ππ32(cos(ππ3) + ππ sin(ππ3))οΏ½
+ οΏ½β600πππΌπΌ4(cos(ππ4) + ππ sin(ππ4)) + 600ππ42(cos(ππ4) + ππ sin(ππ4))οΏ½ = 0
The real component is extracted from the equation.
(β200ππ22 cos(ππ2)) β (500πΌπΌ3 sin(ππ3)) β (500ππ3
2 cos(ππ3)) + (600πΌπΌ4 sin(ππ4))+ (600ππ42 cos(ππ4)) = 0
(β200(50.26)2 cos(60)) β (500πΌπΌ3 sin(49.67)) β (500(β19.06)2 cos(49.67))+ (600πΌπΌ4 sin(128.31)) + (600(3.06)2 cos(128.31)) = 0
The imaginary component is extracted from the equation.
(β200ππ22ππ sin(ππ2)) + (500πππΌπΌ3 cos(ππ3)) β (500ππ3
2ππ sin(ππ3)) + (β600πππΌπΌ4 cos(ππ4))+ (600ππ42ππ sin(ππ4)) = 0
(β200(50.26)2ππ sin(60)) + (500πππΌπΌ3 cos(49.67)) β (500(β19.06)2ππ sin(49.67))+ (β600πππΌπΌ4 cos(128.31)) + (600(3.06)2ππ sin(128.31)) = 0
Solving both the real and imaginary component simultaneously will solve for πΌπΌ3 and πΌπΌ4.
πΆπΆππ = ππππππ.ππππ π π πππ π /π π ππ
πΆπΆππ = ππππππππ.ππππ π π πππ π /π π ππ
Now that the alpha values are obtained, πππ΅π΅/π΄π΄, πππ΅π΅ and πππ΄π΄ can be calculated.
πππ΅π΅/π΄π΄ = ππππ3πΌπΌ3ππππππ3 β ππ3ππ32ππππππ3
πππ΅π΅/π΄π΄ = ππππ3πΌπΌ3οΏ½cos(ππ3) + ππππππππ(ππ3)οΏ½ β (ππ3ππ32(cos(ππ3) + ππππππππ(ππ3))
πππ΅π΅/π΄π΄ = ππ(ππ3πΌπΌ3cos(ππ3)βππ3ππ32ππππππ(ππ3)) β ππ3ππ3
2 cos(ππ3) β ππ3πΌπΌ3ππππππ(ππ3)
πππ΅π΅/π΄π΄ = ππ(500(442.44) cos(49.67) β 500(β19.06)2ππππππ(49.67)) β 500(β19.06)2 cos(49.67)β 500(442.44)ππππππ(49.67)
πππ΅π΅/π΄π΄ = (β286148.97) + ππ(4761.69)
π¨π¨π©π©/π¨π¨ = οΏ½(β286148.97)2 + (4761.69)2 = ππππππππππππ.ππππππππ π π ππβ
Calculating πππ΄π΄
πππ΄π΄ = ππππ2πΌπΌ2ππππππ2 β ππ2ππ22ππππππ2
36 | P a g e
πππ΄π΄ = ππππ2πΌπΌ2οΏ½cos(ππ2) + ππππππππ(ππ2)οΏ½ β (ππ2ππ22(cos(ππ2) + ππππππππ(ππ2))
πππ΄π΄ = ππ(ππ2πΌπΌ2cos(ππ2)βππ2ππ22ππππππ(ππ2)) β ππ2ππ2
2 cos(ππ2) β ππ2πΌπΌ2ππππππ(ππ2)
πΆπΆππ = ππ
πππ΄π΄ = πποΏ½βππ2ππ22ππππππ(ππ2)οΏ½ β ππ2ππ2
2 cos(ππ2)
πππ΄π΄ = πποΏ½β200(50.26)2ππππππ(60)οΏ½ β 200(50.26)2 cos(60)
πππ΄π΄ = (β252661.87) + ππ(β437623.20)
π¨π¨π¨π¨ = οΏ½(β252661.87)2 + (β437623.20)2 = ππππππππππππ.ππππππππ π π ππβ
πΏπΏπ΄π΄π΄π΄ = arctan οΏ½πΌπΌπππ π πποΏ½ + ππ = arctan οΏ½
β437623.20β252661.87
οΏ½ + ππ = 4.189 ππππππ
πΉπΉπ¨π¨π¨π¨ = ππππππ π π πΉπΉπ π
Calculating πππ΅π΅
πππ΅π΅ = ππππ4πΌπΌ4ππππππ4 β ππ4ππ42ππππππ4
πππ΅π΅ = ππππ4πΌπΌ4οΏ½cos(ππ4) + ππππππππ(ππ4)οΏ½ β (ππ4ππ42(cos(ππ4) + ππππππππ(ππ4))
πππ΅π΅ = ππ(ππ4πΌπΌ4cos(ππ4)βππ4ππ42ππππππ(ππ4)) β ππ4ππ42 cos(ππ4) β ππ4πΌπΌ4ππππππ(ππ4)
πππ΅π΅ = ππ(600 (1151.88)cos(128.31)β600(3.06)2ππππππ(128.31)) β 600(3.06)2 cos(128.31)β 600(1151.88)ππππππ(128.31)
πππ΅π΅ = (β545797.55) + ππ(β432861.51)
π¨π¨π©π© = οΏ½(β545797.55)2 + (β432861.51)2 = ππππππππππππ.ππππππππππππ π π ππβ
πΏπΏπ΄π΄π΅π΅ = arctan οΏ½πΌπΌπππ π πποΏ½ + ππ = arctan οΏ½
β432861.51β545797.55
οΏ½ + ππ = 3.812 ππππππ
πΉπΉπ¨π¨π©π© = ππππππ.ππππ π π πΉπΉπ π
Calculating πππΆπΆ/π΄π΄
πππΆπΆ/π΄π΄ = ππππ3πΌπΌ3ππππππ3 β ππ3ππ32ππππππ3
πππΆπΆ/π΄π΄ = πππππ΄π΄πΆπΆπΌπΌ3οΏ½cos(πππ΄π΄πΆπΆ) + ππππππππ(πππ΄π΄πΆπΆ)οΏ½ β (πππ΄π΄πΆπΆππ32(cos(πππ΄π΄πΆπΆ) + ππππππππ(πππ΄π΄πΆπΆ))
πππΆπΆ/π΄π΄ = ππ(πππ΄π΄πΆπΆπΌπΌ3cos(πππ΄π΄πΆπΆ)βπππ΄π΄πΆπΆππ32ππππππ(πππ΄π΄πΆπΆ)) β πππ΄π΄πΆπΆππ3
2 cos(πππ΄π΄πΆπΆ) β πππ΄π΄πΆπΆπΌπΌ3ππππππ(πππ΄π΄πΆπΆ)
πππΆπΆ/π΄π΄ = ππ(300(442.44)cos(43.67)β300(β19.06)2ππππππ(43.67)) β 300(β19.06)2 cos(43.67)β 300(442.44)ππππππ(43.67)
πππΆπΆ/π΄π΄ = (β170450.21) + ππ(20787.79)
π¨π¨πͺπͺ/π¨π¨ = οΏ½(β170450.21)2 + (20787.79)2 = ππππππππππππ.ππππππππ π π ππβ
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Calculating πππΆπΆ
πππΆπΆ = πππΆπΆ/π΄π΄ + πππ΄π΄
πππΆπΆ = (β170450.21 + (β252661.87)) + ππ(20787.79 + (β437623.20))
πππΆπΆ = (β423112.08) + ππ(β416835.41)
π¨π¨πͺπͺ = οΏ½(β423112.08)2 + (β423112.08)2 = ππππππππππππ.ππππππππ π π ππβ
πΏπΏπ΄π΄πΆπΆ = arctan οΏ½πΌπΌπππ π πποΏ½ + ππ = arctan οΏ½
β416835.41β423112.08
οΏ½ + ππ = 3.920 ππππππ
πΉπΉπ¨π¨πͺπͺ = ππππππ.πππππ π πΉπΉπ π
Part C β Slider In order to calculate πππ·π·, firstly πΌπΌ5 must be calculated as follows. Firstly, the velocity equation is differentiated.
ππππ2ππ2ππππππ2ππππ2ππππ
+ πππππ΄π΄πΆπΆππ3πππππππ΄π΄π΄π΄ππππ3ππππ
β ππππ5ππ5ππππππ5ππππ5ππππ
= 0
This produces the following acceleration equation.
οΏ½ππππ2πΌπΌ2ππππππ2βππ2ππ22ππππππ2οΏ½ + οΏ½πππππ΄π΄πΆπΆπΌπΌ3πππππππ΄π΄π΄π΄βππ3ππ3
2πππππππ΄π΄π΄π΄οΏ½ β (ππππ5πΌπΌ5ππππππ5βππ5ππ52ππππππ5) = 0
Eulerβs formula can be applied to this equation and divided into the imaginary and real components.
ππππππ = cos (ππ) + ππsin (ππ)
πΌπΌ2 = 0
β(ππ2ππ22(cos(ππ2) + ππ sin(ππ2))) + (πππππ΄π΄πΆπΆπΌπΌ3(cos(πππ΄π΄πΆπΆ) + ππ sin(πππ΄π΄πΆπΆ)))β(πππ΄π΄πΆπΆππ3
2(cos(πππ΄π΄πΆπΆ)+ ππ sin(πππ΄π΄πΆπΆ))) β (ππππ5πΌπΌ5(cos(ππ5) + ππ sin(ππ5)))+(ππ5ππ5
2(cos(ππ5) + ππ sin(ππ5))) = 0
Using the imaginary component, πΌπΌ5 can be found.
β(ππ2ππ22ππ sin(ππ2)) + (πππππ΄π΄πΆπΆπΌπΌ3 cos(πππ΄π΄πΆπΆ))β(πππ΄π΄πΆπΆππ3
2ππ sin(πππ΄π΄πΆπΆ)) β (ππππ5πΌπΌ5 cos(ππ5))+(ππ5ππ52ππ sin(ππ5))
= 0
β(200(50.26)2ππ sin(60)) + (300ππ(442.44) cos(43.67)) β (300(β19.06)2ππ sin(43.67))β (800πππΌπΌ5 cos(β30.54)) + (800(β0.73)2ππ sin(β30.54)) = 0
πΆπΆππ = ππππππ.πππππ π πππ π π π ππβ
Using πΌπΌ5, πππ·π·/πΆπΆ and πππ·π· can now be calculated.
πππ·π·/πΆπΆ = ππ5πΌπΌ5(cos(ππ5) + ππ sin(ππ5)) + ππ5ππ52(cos(ππ5) + ππ sin(ππ5))
πππ·π·/πΆπΆ = 800(605.29)(cos(β30.54) + ππ sin(β30.54)) + 800(β0.73)2(cos(β30.54)+ ππ sin(β30.54))
πππ·π·/πΆπΆ = (245709.84) + ππ(416835.41)
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π¨π¨πͺπͺ = οΏ½(245709.84)2 + (416835.41)2 = ππππππππππππ.ππππππππ π π ππβ
Using vector addition, πππ·π· can be found.
πππ·π· = πππ·π·/πΆπΆ + πππΆπΆ
πππ·π· = (245709.84 β 423112.08) + ππ(416835.41 + (β416835.41))
πππ·π· = (β177402.24) + ππ(0)
π¨π¨π«π« = ππππππππππππ.ππππππππ π π ππβ
πΏπΏπ΄π΄π·π· = arctan οΏ½πΌπΌπππ π πποΏ½ + ππ = arctan οΏ½
0β177402.24
οΏ½ + ππ = 3.142 ππππππ
πΉπΉπ¨π¨π«π« = πππππππ π πΉπΉπ π
Comparing the graphical method for calculating acceleration to the analytical method, similar to the velocities, both the methods have very similar values.
Graphical (mm/s2)
Analytical (mm/s2)
Difference (%)
Aa 505323.6 505323.7 2.178E-05
Ab/a 289431.2 286188.6 -1.133
Ab 692624.5 696609 0.5720
Ac/a 172900.2 171713.2 -0.6913
Ac 579735.8 593949.2 2.393
Ad/c 462024.6 483864.7 4.514
Ad 187244 177402.2 -5.548 Table 6 Analytical Acceleration Results Compared to Graphical
The variation in the methods can be attributed to the compounding small errors between calculations and difference in velocity values to calculate acceleration.
Summary of Analytical Values
Joint X Component (mm)
Y Component (mm)
A 100 173.21
B 423.67 554.42
C 337.46 356.54
D 1026.47 -50
E 795.62 83.62 Table 7 Analytical Position Results
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Angle Value (degrees)
ΞΈ1 6 ΞΈ2 60 ΞΈ3 49.66962 ΞΈ4 128.31 ΞΈ5 -30.5421
Table 8 Analytical Angles
Joint Linear Velocity (mm/s)
Linear Acceleration (mm/s2)
A 10053.1 505323.7
B 1838.78 696609
C 5236.74 593949.2
D 5508.55 177402.2 Table 9 Analytical Linear Velocity and Acceleration
Link Angular Velocity Ο (rad/s)
Angular Acceleration Ξ± (rad/s2)
2 50.26548 0
3 -19.0559 442.4406
4 3.064637 1151.876
5 -0.72779 605.2945 Table 10 Analytical Angular Velocity and Acceleration
Dynamic Forces Equations of Motion For link 2:
Figure 41 Link 2 Free Body Diagram
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First the variables used are found.
πππΊπΊ2 = 252.66ππ/ππ2
πΌπΌ2 = 0
π π 12 = π π 32 = 0.1ππ
πππΊπΊ2 = 240 πππππΌπΌ
πππΊπΊ2ππ = πππΊπΊ2cos(ππ) = 252.66cos(240) = β126.22 ππ/ππ2
πππΊπΊ2ππ = πππΊπΊ2sin(ππ) = 252.66sin(240) = β218.88 ππ/ππ2
The x and y components of each of the radii are found.
π π 12ππ = π π 12 cos(ππ12) = 0.1 cos(240) = β0.05ππ
π π 12ππ = π π 12 sin(ππ12) = 0.1 sin(240) = β0.086ππ
π π 32ππ = π π 32 cos(ππ32) = 0.1 cos(60) = 0.05ππ
π π 32ππ = π π 32 sin(ππ32) = 0.1 sin(60) = 0.086ππ
The forces and moment equation are found for link 2.
π΄π΄π΄π΄ππ = ππ2πππΊπΊ2π₯π₯ = 1 β (β126.22) = β126.22ππ
π΄π΄π΄π΄ππ = π΄π΄12ππ + π΄π΄32ππ β¦ 1
π΄π΄π΄π΄ππ = ππ2πππΊπΊ2ππ = 1 β (β218.88) = β218.88ππ
π΄π΄π΄π΄ππ = π΄π΄12ππ + π΄π΄32ππ β¦ 2
π΄π΄π΄π΄ = ππ12 + (π π 12πππ΄π΄12ππ β π π 12πππ΄π΄12ππ) + (π π 32πππ΄π΄32ππ β π π 32πππ΄π΄32ππ) = πΌπΌ2πΌπΌ2 β¦ 3
πΌπΌ2πΌπΌ2 = 0.002 β 0 = 0
For link 3:
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Figure 42 Link 3 Free Body Diagram
First the variables used are found.
πππΊπΊ3 = 584.89ππ/ππ2
πΌπΌ3 = 442.44 ππππππ/ππ2
π π 23 = 0.2731ππ
π π 43 = 0.2299ππ
π π 43 = 0.03811ππ
πππΊπΊ3 = 226.36 πππππΌπΌ
πππΊπΊ3ππ = πππΊπΊ3cos(ππ) = 584.89cos(226.36) = β403.62 ππ/ππ2
πππΊπΊ3ππ = πππΊπΊ3sin(ππ) = 252.66sin(226.36) = β423.30 ππ/ππ2
The x and y components of each of the radii are found.
π π 23ππ = π π 23 cos(ππ23) = 0.2731 cos(226.90) = β0.1994ππ
π π 23ππ = π π 23 sin(ππ23) = 0.2731 sin(226.90) = β0.1866ππ
π π 43ππ = π π 43 cos(ππ43) = 0.229 cos(57.48) = 0.1236ππ
π π 43ππ = π π 43 sin(ππ43) = 0.229 sin(57.48) = 0.1938ππ
π π 53ππ = π π 53 cos(ππ53) = 0.03811 cos(β5.026) = 0.03796ππ
π π 53ππ = π π 53 sin(ππ53) = 0.03811 sin(β5.026) = β0.003339ππ
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The forces and moment equation are found for link 3.
π΄π΄π΄π΄ππ = ππ3πππΊπΊ3ππ = 2.5 β (β403.62) = β1009.05ππ
π΄π΄π΄π΄ππ = π΄π΄23ππ + π΄π΄43ππ + π΄π΄53ππ β¦ 4
π΄π΄π΄π΄ππ = ππ3πππΊπΊ3ππ = 2.5 β (β423.30) = β1058.25ππ
π΄π΄π΄π΄ππ = π΄π΄23ππ + π΄π΄43ππ + π΄π΄53ππ β¦ 5
π΄π΄π΄π΄ = β(π π 32πππ΄π΄32ππ β π π 32πππ΄π΄32ππ) + (π π 53πππ΄π΄53ππ β π π 53πππ΄π΄53ππ) + (π π 43πππ΄π΄43ππ β π π 43πππ΄π΄43ππ) = πΌπΌ3πΌπΌ3 β¦ 6
πΌπΌ3πΌπΌ3 = 0.008 β 442.44 = 3.5395
For link 4:
Figure 43 Link 4 Free Body Diagram
First the variables used are found.
πππΊπΊ4 = 345.48ππ/ππ2
πΌπΌ4 = 1151.88ππππππ/ππ2
π π 14 = π π 34 = 0.3ππ
πππΊπΊ4 = 218.80πππππΌπΌ
πππΊπΊ4ππ = πππΊπΊ4cos(ππ) = 345.48cos(218.80) = β269.27ππ/ππ2
πππΊπΊ4ππ = πππΊπΊ3sin(ππ) = 345.48sin(218.80) = β216.44ππ/ππ2
The x and y components of each of the radii are found.
π π 14ππ = π π 14 cos(ππ14) = 0.3 cos(β51.61) = 0.1863ππ
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π π 14ππ = π π 14 sin(ππ14) = 0.3 sin(β51.61) = β0.2351ππ
π π 34ππ = π π 34 cos(ππ34) = 0.3 cos(128.39) = β0.1863ππ
π π 34ππ = π π 34 sin(ππ34) = 0.3 sin(128.39) = 0.2351ππ
The forces and moment equation are found for link 4.
π΄π΄π΄π΄ππ = ππ4πππΊπΊ4ππ = 1.5 β (β269.27) = β403.91ππ
π΄π΄π΄π΄ππ = π΄π΄34ππ + π΄π΄14ππ β¦ 7
π΄π΄π΄π΄ππ = ππ4πππΊπΊ4ππ = 1.5 β (β216.44) = β324.67ππ
π΄π΄π΄π΄ππ = π΄π΄34ππ + π΄π΄14ππ β¦ 8
π΄π΄π΄π΄ = (π π 34πππ΄π΄34ππ β π π 34πππ΄π΄34ππ) + (π π 14πππ΄π΄14ππ β π π 14πππ΄π΄14ππ) = πΌπΌ4πΌπΌ4 β¦ 9
πΌπΌ4πΌπΌ4 = 0.005 β 1151.88 = 5.7594
For link 5:
Figure 44 Link 5 Free Body Diagram
First the variables used are found.
πππΊπΊ5 = 362.21 ππ/ππ2
πΌπΌ5 = 605.29 ππππππ/ππ2
π π 35 = π π 65 = 0.4ππ
πππΊπΊ5 = 213.44πππππΌπΌ
πππΊπΊ5ππ = πππΊπΊ5cos(ππ) = 362.21cos(213.44) = β302.24ππ/ππ2
πππΊπΊ5ππ = πππΊπΊ5sin(ππ) = 362.21sin(213.44) = β199.63ππ/ππ2
44 | P a g e
The x and y components of each of the radii are found.
π π 35ππ = π π 35 cos(ππ14) = 0.4 cos(149.45) = β0.344ππ
π π 35ππ = π π 35 sin(ππ23) = 0.4 sin(149.45) = 0.203ππ
π π 65ππ = π π 65 cos(ππ34) = 0.4 cos(β30.55) = 0.344ππ
π π 65ππ = π π 65 sin(ππ34) = 0.4 sin(β30.55) = β0.203ππ
The forces and moment equation are found for link 5.
π΄π΄π΄π΄ππ = ππ5πππΊπΊ5ππ = 1.8 β (β302.23) = β544.03ππ
π΄π΄π΄π΄ππ = π΄π΄35ππ + π΄π΄65ππ β¦ 10
π΄π΄π΄π΄ππ = ππ5πππΊπΊ5ππ = 1.8 β (β199.63) = β359.33ππ
π΄π΄π΄π΄ππ = π΄π΄35ππ + π΄π΄65ππ β¦ 11
π΄π΄π΄π΄ = (π π 35πππ΄π΄35ππ β π π 35πππ΄π΄35ππ) + (π π 65πππ΄π΄65ππ β π π 65πππ΄π΄65ππ) = πΌπΌ5πΌπΌ5 β¦ 12
πΌπΌ5πΌπΌ5 = 0.006 β 605.29 = 3.6318
For link 6:
Figure 45 Link 6 Free Body Diagram
First the variables used are found.
πππΊπΊ6 = 177.40ππ/ππ2
πΌπΌ6 = 0 ππππππ/ππ2
πππΊπΊ6 = 180 πππππΌπΌ
πππΊπΊ6ππ = πππΊπΊ6cos(ππ) = 177.40cos(180) = β177.40ππ/ππ2
πππΊπΊ6ππ = πππΊπΊ6sin(ππ) = 177.40sin(180) = 0ππ/ππ2
45 | P a g e
The x and y components of each of the radii are found.
π΄π΄π΄π΄ππ = ππ6πππΊπΊ6ππ = 0.9 β (177.40) = β159.66ππ
π΄π΄π΄π΄ππ = βπ΄π΄65ππ + π΄π΄16ππ β¦ 13
π΄π΄π΄π΄ππ = βπ΄π΄65ππ + πππ΄π΄16ππ β¦ 14
ππ = 0.3
The forces and moment equation are found for link 5.
π΄π΄π΄π΄ππ = ππ6πππΊπΊ6ππ = 0.9 β (0) = 0ππ
π΄π΄π΄π΄ππ = βπ΄π΄65ππ + π΄π΄16ππ
πΌπΌ6πΌπΌ6 = πΌπΌ6 β 0 = 0
Equation Matrix Using the 14 equations found from the dynamic forces equations, the following matrix can be used to represent each of the equations.
β£β’β’β’β’β’β’β’β’β’β’β’β’β‘
1 0 1 0 0 0 0 0 0 0 0 0 0 00 1 0 1 0 0 0 0 0 0 0 0 0 0
0.086628 β0.04996 β0.08663 0.04996 0 0 0 0 0 0 0 0 0 10 0 β1 0 1 0 1 0 0 0 0 0 0 00 0 0 β1 0 1 0 1 0 0 0 0 0 00 0 β0.18666 0.199477 β0.19389 0.123602 0.003339 0.037966 0 0 0 0 0 00 0 0 0 β1 0 0 0 1 0 0 0 0 00 0 0 0 0 β1 0 0 0 1 0 0 0 00 0 0 0 0.235141 0.186303 0 0 0.235141 0.186303 0 0 0 00 0 0 0 0 0 β1 0 0 0 1 0 0 00 0 0 0 0 0 0 β1 0 0 0 1 0 00 0 0 0 0 0 0.203346 0.344456 0 0 0.20335 0.344456 0 00 0 0 0 0 0 0 0 0 0 β1 0 0.3 00 0 0 0 0 0 0 0 0 0 0 β1 1 0β¦
β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β€
β
β£β’β’β’β’β’β’β’β’β’β’β’β’β’β‘π΄π΄12πππ΄π΄12πππ΄π΄32πππ΄π΄32πππ΄π΄43πππ΄π΄43πππ΄π΄53πππ΄π΄53πππ΄π΄14πππ΄π΄14πππ΄π΄65πππ΄π΄65πππ΄π΄16ππππ12 β¦
β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β€
=
β£β’β’β’β’β’β’β’β’β’β’β’β’β‘β126.221β218.875
0β1009.05β1058.253.539525β403.912β324.6655.759379β544.025β359.3313.631767β159.662
0 β¦β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β€
This matrix is solved to produce the following results for each of the unknown variables.
β£β’β’β’β’β’β’β’β’β’β’β’β’β’β‘π΄π΄12πππ΄π΄12πππ΄π΄32πππ΄π΄32πππ΄π΄43πππ΄π΄43πππ΄π΄53πππ΄π΄53πππ΄π΄14πππ΄π΄14πππ΄π΄65πππ΄π΄65πππ΄π΄16ππππ12 β¦
β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β€
=
β£β’β’β’β’β’β’β’β’β’β’β’β’β‘β1791.45β1626.381665.231407.50
61.89354.57594.29β5.32β342.02
29.9150.27
β364.65β364.65147.89 β¦
β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β₯β€
Working Model Simulation
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Analysis of the positions of each of the joints.
Figure 46 Working Model Position Analysis
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Analysis of all the velocities at each joint.
Figure 47 Working Model Velocity Analysis
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Analysis of all the accelerations at all of the joints.
Figure 48 Working Model Acceleration of Joints Analysis
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Analysis of the acceleration at the mass centre of each of the links.
Figure 49 Working Model Acceleration at Centre of Mass Analysis
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Analysis of each of the forces at each of the joints and torque moment at the motor.
Figure 50 Working Model Force Analysis
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Balancing Strategy Development To find the shaking forces, the following equation is used.
π΄π΄ππ = π΄π΄21 + π΄π΄41 + π΄π΄6
π΄π΄ππ = π΄π΄21ππ + π΄π΄21ππ + π΄π΄41ππ + π΄π΄41ππ + π΄π΄61ππ + π΄π΄61ππ
Where the values are found from the dynamic forcesβ matrix.
β£β’β’β’β’β‘π΄π΄21πππ΄π΄21πππ΄π΄41πππ΄π΄41πππ΄π΄61πππ΄π΄61ππβ¦
β₯β₯β₯β₯β€
=
β£β’β’β’β’β‘1791.451626.38342.02β29.91109.395364.65 β¦
β₯β₯β₯β₯β€
These forces can be split into their x and y components of the resultant force respectively.
Firstly, the x axis resultant force is found.
π΄π΄π΄π΄π π ππ = π΄π΄21ππ + π΄π΄41ππ + π΄π΄61ππ
π΄π΄π΄π΄π π ππ = 1791.45 + 342.02 + 109.40 = 2242.873ππ
Following with the y axis resultant force.
π΄π΄π΄π΄π π ππ = π΄π΄21ππ + π΄π΄41ππ + π΄π΄61ππ
π΄π΄π΄π΄π π ππ = 1626.38 β 29.91 + 364.65 = 1961.118ππ
π΄π΄ππ = οΏ½(π΄π΄π΄π΄π π ππ)2 + (π΄π΄π΄π΄π π ππ)2 = 2979.34ππ
The shaking moment is found for the system.
ππππ = ππ21 = βππ12 = β147.89 ππππ
Figure 51 Shaking Moment Diagram
Finally, the shaking moment is developed from the diagram above.
π΄π΄ππ = ππ21 + (π π 1π΄π΄41) + (π π 2π΄π΄61)
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π΄π΄ππ = ππ21 + (π π 1πππ΄π΄41ππ) β (π π 1πππ΄π΄41ππ) β (π π 2πππ΄π΄61ππ) β (π π 2πππ΄π΄61ππ)
π΄π΄ππ = β147.89 + οΏ½0.7956 β (β29.91)οΏ½ β (0.0836 β 342.02) β (1.0264 β 364.65)β (0.05 β 109.40) = β580.02 ππππ
In order to reduce the shaking moment transferred to the ground, some actions could be made.
Firstly, if the shaking moment isnβt too great, a damper can be used to absorb the energy being transferred to the ground to help reduce the overall effect on the ground.
If the shaking moment is too great or a damper is unable to be installed, another solution is to remove or add mass to the system at points which can reduces the overall moments at each of the ground points.
Results Graphical Method Analytical Method Computational
Method Linear Velocity Vector
(ππππ/π π ) IC
(ππππ/π π ) ΞΈ
(π π πΉπΉπ π ) Vector
(ππππ/π π ) ΞΈ
(π π πΉπΉπ π ) Vector
(ππππ/π π ) π½π½π¨π¨ 10053.09298 10053.1171
8 150.00 10053.09649 150 10053.1
π½π½π©π© 1838.888167 1838.65304 218.39 1838.782206 218.31 2520.1
π½π½πͺπͺ 5225.428643 5225.120573
174.61 5236.737183 174.5051768
5617.9
π½π½π«π« 5492 5308 180 5508.545817 180 5676.2
Angular Velocity Vector (π π πππ π /π π ) Vector (π π πππ π /π π ) Vector (π π πππ π /π π )
ππππ 50.26548246 50.26548246
ππππ -19.0608557 -19.05591673
ππππ 3.064816259 3.064637009
ππππ -0.712661395 -0.727786137
Linear Acceleration
Vector (ππππ/π π ππ)
ΞΈ (π π πΉπΉπ π ) Vector (ππππ/π π ππ)
ΞΈ (π π πΉπΉπ π ) Vector (ππππ/π π ππ)
πππ¨π¨ 505323.6353 239.9999761 505323.7453 240 505323.7
πππ©π© 692624.5496 218.0477535 696608.9655 218.4172561
671303.7
πππͺπͺ 579735.8094 223.4211914 593949.1532 224.571854 563438.9
πππ«π« 187244 180 177402.2398 180 160192.2
Angular Acceleration
Vector (ππππ/π π ππ) Vector (ππππ/π π ππ) Vector (ππππ/π π ππ)
πΆπΆππ 0 0
πΆπΆππ 445.6566542 442.4406331
πΆπΆππ 1154.375247 1151.875838
πΆπΆππ 577.5306514 605.2945123
Table 11 Velocity and Acceleration Results for ππ2 = 60 degrees
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Analytical Method Computational Method
Raw Data (N) Joint Forces (N) Software Generated (N)
πππππππΏπΏ -1791.454367 π΄π΄12 2419.5905
ππππππππ -1626.379177
πππππππΏπΏ 1665.233433 π΄π΄32 2180.383 π΄π΄π¨π¨ 2195.65
ππππππππ 1407.504103
πππππππΏπΏ 61.88841423 π΄π΄ππππ 359.93553 π΄π΄π©π© 438.99
ππππππππ 354.5749709
πππππππΏπΏ 594.2925285 π΄π΄ππππ 594.31632 π΄π΄πͺπͺ 709.01
ππππππππ -5.317468002
πππππππΏπΏ -342.0237588 π΄π΄14 343.32907 π΄π΄π¬π¬ 433.53
ππππππππ 29.90989642
πππππππΏπΏ 50.26751607 π΄π΄ππππ 368.09677
ππππππππ -364.6483324
ππππππππ -364.6483324 π΄π΄16 380.70404 π΄π΄πΊπΊ 2642.40
πππππππΏπΏ -109.3944997
π»π»ππππ (π΅π΅ππ) 147.8827066
ππ12 (ππππ) 157.32
Table 12 Force Results for ππ2 = 60 degrees
The stroke distance was determined as seen below:
Stroke Distance = 1026.4mm
The shaking force, shaking torque and shaking moment at ππ2 = 60 degrees were found as seen below:
π΄π΄π π = 2979.34ππ
πππ π = β147.883ππππ
π΄π΄π π = β580.02ππππ
Discussion The results at every stage of the process produced values which were within an acceptable tolerance of other values found using different methods. Most values were within 5% of other calculated values for other methods, however, the values which were outside this calculation were still relatively close to the other methods calculations.
This discrepancy between some of the values can be attributed to measurement inaccuracy for the different methods. Some of the errors in the graphical method can be attributed to rounding of numbers and measurement tools resolution being unable to 100% accurately determine values. For the Working Model results, the snapshot which the results were taken may not have been at exactly when ππ2 = 60 degrees causing a slightly different situation to be analysed. Across all methods, compounding differences and rounding errors in the methods will attribute to some of the larger differences in results obtained towards the end of each section. Since all the results have relatively
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similar magnitude and results, it can be assumed that these values are correct for if this mechanism was to be created in the real world.
Conclusion Overall the values that were produced from all the methods may not be completely accurate but most of the values are close enough that any of them can be used as an approximate used for a system and thus confirms that the methods are relatively accurate in determining the values for the position, velocity, acceleration and forces on the system.
References 1. Matthew West 2015, Four-Bar Linkages, Dynamics, Viewed September 12 2020,
<http://dynref.engr.illinois.edu/aml.html> 2. Course material of MIET1077: Mechanics of Machines provided by Prof. Firoz Alam
Attachments Attached to this document is a zip folder containing all the Working Model files used.