MACHINE DESIGN II

PROJECT ONE

10 | 21 | 15

PREPARED BY:

MATT STOFLETH

2

| TABLE OF CONTENTS

0 | ABSTRACT _______________________________________________________________ 3

1 | SOLUTION APPROACH ______________________________________________________ 5

1.1 | Link Kinematics ____________________________________________________ 5 1.2 | Critical Plane Selection ______________________________________________ 6 1.3 | Critical Point Selection ______________________________________________ 6 1.4 | Factor of Safety ___________________________________________________ 7

2 | RESULTS _________________________________________________________________ 8

2.1 | Kinematic and Force Analysis _________________________________________ 8 2.2 | Critical Plane Analysis _______________________________________________ 10 2.3 | Critical Point Analysis _______________________________________________ 14 2.4 | Material Selection _________________________________________________ 19 2.5 | Final Design and Factor of Safety ______________________________________ 20

| APPENDICIES

Appendix A | Supporting Information Appendix B | MATLAB Script

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0 | ABSTRACT

A machine resembling a slider crank mechanism is used to test an Ohlins model WCJ 22/6 shock in order to receive load and velocity characteristics about the shock. A schematic of this machine may be seen below in Figure 1 along with the important features and dimensions of each machine component.

Figure 1: Conceptual layout of shock testing machine, shown (a) top view, and (b) side view

The machine uses a motor and gearbox to rotate a crank arm, which actuates a connecting rod through a crank pin to move the sider that is fixed to the end of the shock. The design uses a fixed stroke, giving the crank arm a fixed dimension. The crank shaft, arm, and pin are assumed to be one solid component for this analysis. A numbering system was used to simplify the naming convention of the machine components. The ground is referred to as link 1, the crank shaft, pin, and rod as link 2, the connecting rod as link 3, and the slider as link 4.

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Figure 2: Geometry to be determined for (a) the crank shaft, pin, and arm, and (b) the connecting rod

The purpose of this analysis was to design the crank shaft, crank pin, crank arm, and connecting rod to determine their geometry, surface finish, and material specifications so that the machine will have an infinite life factor of safety of at least 4, and a yield factor of safety of at least 5.

1040 cold rolled steel was chosen as the material for the entire assembly for the reasons discussed in Section 2.4. This material selection produced a minimum fatigue factor of safety for the machine of 4.19 in the rod end.

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1 | SOLUTION APPROACH

1.1 | Link Kinematics

A typical force versus velocity diagram for the Ohlins Shock was provided for the study.

Figure 3: Force vs. velocity diagram for the Ohlins Shock. Shown is the force on the shock.

A curve fit was created for the diagram so that the force of the shock could be used to determine the forces acting on the links in the testing machine. The velocity of the shock is equal to the velocity of the slider, or �̇�𝑅4.

The slider crank closed-form solution was used for calculating the position, velocity, and acceleration of all links. Using this data, free body diagrams were created to identify the active forces on the links for each rotation angle of the crank. The torque on the crankshaft was also calculated through this analysis. Plots displaying these kinematics and free body diagrams are located in Appendix A, along with the previously discussed closed-form solution and free body diagrams with their respective equations.

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Finally, the significance of the weight of the connecting rod and slider was explored. Because the connecting rod does not displace enough to generate significant momentum to affect the system, it may be ignored. The slider weight was also determined to be negligible, as its modification does not significantly affect any link forces or torque.

1.2 | Critical Plane Selection

Critical planes for each machine component were found by generating shear force, bending moment, and torque diagrams where applicable using the MATLAB program provided in Appendix B. Based on the magnitudes and trends of the diagrams, a strategic selection of the plane(s) at which all stresses coincide was performed. These diagrams are found in Section 2.2 and assume a positive coordinate system as declared in Figure 1, with the positive z-axis pointing out of the page in Figure 1b.

From this section of the analysis on, all forces were put into terms of a rotating coordinate system that rotated with the link being analyzed. These equations are shown in Appendix A.

Figure 4: Sample usage of rotating coordinate system facing crank shaft at rotation angle φ

1.3 | Critical Point Selection

On the critical plane, a critical point lying on that plane was established for all links. This analysis procedure consisted of using a MATLAB script to calculate all normal and shear stresses

Y

X

y’

x’

φ

7

acting on each part at the critical plane, and then combining them to find the point at which these stresses sum to the highest total stress. This point was deemed the critical point.

1.4 | Factor of Safety

To determine how the geometry, material selection, and testing environment affect how the mechanism fails, the fatigue and yield factors-of-safety were generated. The design needed to achieve the specified safety limits with an effort to avoid over-engineering the machine. This process began by determining the respective mean and alternating stresses present in the links and distinguishing them between bending, axial, and torsion stress. The endurance limit, 𝑆𝑆𝑒𝑒, was then established by taking surface finish, size, loading, temperature, reliability, and notch sensitivity factors into account. The equation for the endurance limit then becomes

𝑆𝑆𝑒𝑒 = 𝑘𝑘𝑎𝑎𝑘𝑘𝑏𝑏𝑘𝑘𝑐𝑐𝑘𝑘𝑑𝑑𝑘𝑘𝑒𝑒𝑘𝑘𝑓𝑓𝑆𝑆𝑒𝑒′

where

𝑘𝑘𝑎𝑎 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑠𝑠

𝑘𝑘𝑏𝑏 = 𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑠𝑠

𝑘𝑘𝑐𝑐 = 𝑙𝑙𝑚𝑚𝑠𝑠𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑠𝑠

𝑘𝑘𝑑𝑑 = 𝑚𝑚𝑠𝑠𝑚𝑚𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑠𝑠

𝑘𝑘𝑒𝑒 = 𝑠𝑠𝑠𝑠𝑙𝑙𝑚𝑚𝑠𝑠𝑟𝑟𝑚𝑚𝑙𝑙𝑚𝑚𝑚𝑚𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑠𝑠

𝑘𝑘𝑓𝑓 = 𝑚𝑚𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑙𝑙𝑙𝑙𝑠𝑠𝑚𝑚𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑚𝑚𝑚𝑚𝑠𝑠

In this solution, 𝑘𝑘𝑓𝑓 becomes representative of the notch sensitivity. For the material used in this analysis, 𝑆𝑆𝑒𝑒′ = 0.5 ∗ 𝑆𝑆𝑢𝑢𝑢𝑢.

Stress concentration at the select corners and holes were taken into account with notch sensitivity equations, and varied depending on corner fillet radii and hole diameter.

𝑘𝑘𝑓𝑓 = 1 +𝑘𝑘𝑢𝑢 − 1

1 + √𝑠𝑠√𝑠𝑠

In the above equation, 𝑘𝑘𝑢𝑢 refers to the theoretical stress-concentration factor, √𝑠𝑠 is the Neuber constant, and √𝑠𝑠 is simply the square root of the fillet radius or hole radius.

(1.2)

(1.1)

8

The yield factor of safety was first calculated using the comparison of the material yield strength to von Mises maximum stress:

𝑚𝑚𝑦𝑦 =𝑆𝑆𝑦𝑦

𝜎𝜎′𝑚𝑚𝑎𝑎𝑚𝑚

𝜎𝜎′𝑚𝑚𝑎𝑎𝑚𝑚 = [(𝜎𝜎𝑚𝑚 + 𝜎𝜎𝑎𝑎)2 + 3(𝜏𝜏𝑚𝑚 + 𝜏𝜏𝑎𝑎)2]1/2

This approach ignored stress concentration effects, so to determine if they were in fact insignificant the Langer first-cycle yield equation, Equation 1.5, was used. This equation ignores stress concentration effects.

𝜎𝜎𝑚𝑚′ + 𝜎𝜎𝑎𝑎′ = 𝑆𝑆𝑦𝑦𝑚𝑚

𝜎𝜎𝑚𝑚′ and 𝜎𝜎𝑎𝑎′ represent the von Mises stresses. Using MATLAB to calculate the results, a significant difference in factor of safety was seen between the two methods, with the Langer approach being a factor less than the von Mises maximum stress approach. This outcome suggests that stress factors are indeed significant in first-cycle yielding, therefore the Langer approach will be used in the following analysis for yield factor of safety.

2 | RESULTS

2.1 | Kinematic and Force Analysis

Understanding the motion of the slider attached to the shock is vital in the process of determining the forces acting on each link in the testing machine. The crank angle versus slider position, velocity, and acceleration were determined using the slider crank closed-form solution mentioned in Section 1.1 and are displayed in Appendix A.

Free body diagrams of each machine link and their respective forces were created to build equations relating the forces between the links from the shock and may be found in Appendix A. Figure 6 shows the relationship between the individual forces in the connecting rod and crank angle, and Figure 7 shows the relationship between the magnitude of those forces and the angle of the crank.

(1.3)

(1.5)

(1.4)

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Figure 6: Connecting rod forces vs. crank angle

Figure 7: Connecting rod magnitude force vs. crank angle

As the crank arm rotates from 0 to 180 degrees, the machine is pulling on the rod. The shock produces more resistance when extended in tension, as described in Figure 3, thus, the magnitude force acting on the connecting rod is higher than when the shock is compressed.

The torque on the shaft was then determined based on the forces acting on the crank arm.

0

50

100

150

200

250

300

350

400

450

500

0 50 100 150 200 250 300 350

Forc

e (lb

)

Crank Position (degrees)

Connecting Rod Magnitude Force

TENSION COMPRESSION

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Figure 8: Output shaft torque vs. crank angle

As expected, the torque in the shaft shows a similar trend to the forces acting on it. Therefore, the torque in the shaft increases when the shock is being pulled in tension.

2.2 | Critical Plane Analysis

Figures 9 and 10 display the shear force and bending moment diagrams for the crank shaft, pin, and arm. The critical plane is the section at which the max shear force and bending moment magnitudes coincide. The shape of the plots will remain constant for any rotation angle, so the diagrams were all gathered for an arbitrary rotation angle of 45 degrees. These diagrams represent the machine with its final chosen dimensions and material.

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(a) (b)

Figure 9: Shear force and bending moment diagrams for (a) the shaft and (b) the pin

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(a) (b)

Figure 10: Shear force, bending moment, and torque diagrams analyzed from (a) the Z direction and (b) the X direction

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In all of these cases, the shear force will be constant at any arbitrary crank angle, so the bending moment diagrams become the key to identifying the critical planes.

In Figure 9a, the maximum moment occurs at Z = 0 inches, but since these diagrams imply nothing about stress concentrations it was not necessarily chosen as the critical plane immediately. In a complete analysis of the shaft, stress concentration effects would be added and the maximum stresses between the two planes would be compared. In this case the plane with stress concentrations was assumed to be the critical plane, which is the plane at Z = length of the shaft. Because the length of the shaft is so small, the stress concentration effects are likely to create failure at this location so this assumption is a good one.

The maximum moments in Figure 9b point to Z = 0 inches as the critical plane for the pin. Stress concentration effects are also present at this plane, so this choice makes sense.

Figures 10a and 10b show that all of the moments acting on the crank arm are greatest at the plane X = 0 inches and Z = 0 inches. The length of the arm was assumed to be the distance between the right edge of the crank shaft in the x’ direction, up to the natural end of the arm.

Figure 11: Critical plane analysis for crank arm

Z

x’

𝐿𝐿𝑎𝑎𝑎𝑎𝑚𝑚

SHAFT

PIN

ARM

Critical Plane in Z

Critical Plane in x’

x’ = 0

Z = 0

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The critical plane on the connecting rod was determined from understanding that because the rod only experiences a normal axial stress along its length, the plane on the rod with the smallest area will see the largest amount of stress. The plane in the end of the rod that is perpendicular with the length of the arm and passes through the hole centerline has the smallest area and is subject to stress concentration effects from the hole. This plane was determined to be the critical plane for the rod.

2.3 | Critical Point Analysis

A MATLAB script was set to output the maximum normal and shear stresses acting at all angles, φ, on any critical plane. The script was set to analyze all stresses at a particular angle of φ for 360 degrees of 𝜃𝜃2 and store the maximum normal and shear stress values into vectors that updated for each angle of φ. Once all 360 degrees of φ were iterated through, the maximum value of the vectors were output for use in future analysis. These stress values are the stress values at the critical point on the critical plane for any particular link. To find what angle of φ these maximum values occurred at, the maximum stress vectors were plotted against φ.

Figure 12: Shaft critical point determination

CRITICAL POINT

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The φ at which these maximum stress coincide is the critical point of that element on the critical plane.

Figure 13: Pin critical point determination

For both the shaft and the pin, the angle φ at which the maximum normal and shear stresses coincide is 90 degrees. Therefore, the critical point on the critical planes for the crank shaft and pin is at φ = 90 degrees.

Because the crank arm is not of circular geometry, a different method of finding the critical point was used.

Shear stress due to torsion in a rectangular body can be described as

𝜏𝜏 =𝑇𝑇

𝛼𝛼1𝑚𝑚𝑠𝑠2 𝑠𝑠𝑚𝑚𝑠𝑠 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴1 𝑚𝑚𝑚𝑚 𝐹𝐹𝑚𝑚𝐹𝐹𝑠𝑠𝑠𝑠𝑠𝑠 14

CRITICAL POINT

16

𝜏𝜏 =𝑇𝑇

𝛼𝛼2𝑚𝑚𝑠𝑠2 𝑠𝑠𝑚𝑚𝑠𝑠 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴1 𝑚𝑚𝑚𝑚 𝐹𝐹𝑚𝑚𝐹𝐹𝑠𝑠𝑠𝑠𝑠𝑠 14

where T is the torque about x’ in the arm for this solution.

Figure 14: Torsion of a rectangular cross-section

𝛼𝛼1 and 𝛼𝛼2 refer to constants related to the ratio of 𝑑𝑑𝑐𝑐 based on the geometry of the crank arm.

A ratio of 3 was selected for this solution, which corresponds to 𝛼𝛼1 = 0.267 and 𝛼𝛼2 = 0.355. This ratio was selected after discovering that too small of a ratio limits the size of the crank arm thickness, making it difficult to achieve the desired factors of safety.

Based on the above equations, it follows that only the four points in the middle of each face of the arm could be analyzed for the critical point determination.

Figure 15: Cross-section of crank arm critical plane in x’

Z

y’

1

4

3

2

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These points were labeled on the cross-section of the crank arm for ease of script writing in MATLAB. For each point, the script determined which equation to use for the shear stress due to torsion and then combined it with other shear stresses in that part of the arm. The normal stresses were also calculated at each point and then both were put into vectors. The maximum values of these vectors were extracted for future use.

Selection of a critical point resulted from comparing the maximum normal and shear stresses against each point, similar to the method used for the shaft and pin.

Figure 16: Arm critical point determination

CRITICAL POINT

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The overlap of maximum normal stress and shear stress occurs at point 3. Thus, the critical point of the crank arm is at point 3 on the critical plane in x’, denoted in Figure 11 above. This point also lies on the arm critical plane in Z, which was expected.

Because there is only normal stress due to axial force in the connecting rod, which is constant throughout the entire critical plane cross-section, there is no distinguishable critical point for the link. The rod will fail uniformly at the critical plane.

The stress concentration results for all links are displayed in Table 1.

Table 1: Stress concentration results

2.4 | Material Selection

The material used for each component in the machine was first narrowed down by yield and ultimate tensile strengths. This filtered down the selection to three main contenders: 1035, 1040, and 1045 cold drawn steel. Steel was selected instead of Aluminum, Titanium, or some other special material because weight isn’t a limiting factor in this design, assuming the testing fixture will not need to be moved frequently. Steel costs less than Aluminum and significantly less than Titanium, so it was a reasonable choice.

1045 steel is more resistant to corrosion but assuming that the lab environment is not a corrosive one, this option appeared unnecessary. 1035 steel is typically used for levers, bolts, studs, and other similar parts that are headed or extruded. This doesn’t particularly fit this machine application, so other options were investigated. 1040 steel is a medium carbon,

ka (Cold-drawn Surf. Fin.)

kb kc (Von Mises) kd (room temp) ke (90%)

Shaft 0.832 0.862 1.000 1.000 0.897 Arm 0.832 1.000 1.000 1.000 0.897 Pin 0.832 0.879 1.000 1.000 0.897 Rod 0.832 1.000 1.000 1.000 0.897 Rod End

0.832 1.000 1.000 1.000 0.897

kt kts kf kfs Se' (psi)

Se (psi)

Shaft 1.950 1.700 1.862 1.864 42500 50964 Arm - - 1.000 - 42500 31714 Pin 1.800 - 1.722 - 42500 47991 Rod 2.700 - 2.700 - 42500 79889 Rod End

2.500 - 2.500 - 42500 79889

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medium tensile strength steel that shows good strength, toughness, and wear resistance. This testing machine is likely to be run several times, so good wear resistance is an attractive characteristic to have. Applications for this cold rolled steel include forged crankshafts and couplings, so this material fits the application quite well. With an ultimate tensile strength of 85 kpsi and a yield strength of 71 kpsi, 1040 CD steel became the material selected for all components in the machine.

2.5 | Final Design and Factor of Safety

The procedure for determining the factors of safety for each link began with calculating the mean and alternating stresses in each link.

𝜎𝜎𝑚𝑚 =𝜎𝜎𝑚𝑚𝑎𝑎𝑚𝑚 + 𝜎𝜎𝑚𝑚𝑚𝑚𝑚𝑚

2

𝜎𝜎𝑎𝑎 = �𝜎𝜎𝑚𝑚𝑎𝑎𝑚𝑚 − 𝜎𝜎𝑚𝑚𝑚𝑚𝑚𝑚

2�

𝜏𝜏𝑚𝑚 =𝜏𝜏𝑚𝑚𝑎𝑎𝑚𝑚 + 𝜏𝜏𝑚𝑚𝑚𝑚𝑚𝑚

2

𝜏𝜏𝑎𝑎 = �𝜏𝜏𝑚𝑚𝑎𝑎𝑚𝑚 − 𝜏𝜏𝑚𝑚𝑚𝑚𝑚𝑚

2�

With stress concentration features present, the von Mises stresses 𝜎𝜎𝑎𝑎′ and 𝜎𝜎𝑚𝑚′ were calculated for use in the factor of safety equations. These equations also take into account the load modification factor, 𝑘𝑘𝑐𝑐, which is why this factor is set to 1 for all links in Table 1. These equations are shown in Appendix A.

As discussed in Section 1.5, the yield factors of safety were determined using the Langer yield equation (1.5). The fatigue factors of safety were determined by using the modified Goodman equation:

(2.1)

(2.2)

(2.3)

(2.4)

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𝑚𝑚𝑓𝑓 =1

𝜎𝜎𝑎𝑎′𝑆𝑆𝑒𝑒

+ 𝜎𝜎𝑚𝑚′ 𝑆𝑆𝑢𝑢𝑢𝑢

Table 2 displays the aforementioned values for all links at the critical φ.

Table 2: Stress properties

Sigma_a (psi)

Sigma_m (psi)

Tau_a (psi)

Tau_m (psi)

Sigma_a' (psi)

Sigma_m' (psi)

Ny (Langer)

N (mod-Goodman)

Shaft 4045.2 2488.6 277 2673.2 4073.6 5256.6 7.61 4.74 Arm 5080.2 622.78 761.8 700.3 5248.8 1363.6 12.85 5.51 Pin 3945.7 2427.4 398.9 398.9 3945.7 2427.4 11.14 5.88 Rod End

4884.7 4906.9 0 0 5746.7 4906.9 6.66 4.19

The fatigue factors of safety are close to the requested values, so this combination of link geometry and material choice together create a good design for this testing machine. The rod end factor of safety governs the safety of the entire mechanism because it is the lowest.

It was discovered that modifying the fillet radii had less of an effect on the factor of safety than changing the geometry of the part. Particularly, any dimension that extended the distance from the connecting rod centerline to the shaft created a larger bending moment on the machine, thus reducing the factors of safety dramatically. With this in mind, the pin length, rod width, and arm width were kept as small as possible.

(2.5)

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Table 3: Final geometry and material selections

The final geometry of each part and their critical planes are displayed in the series of following figures.

Figure 17: Shaft critical point

Material of Links 1040 CD Steel

Ultimate Tensile Stress (kpsi) 85 Yield Strength (kpsi) 71 Heat Treatment NONE Surface Finish of Shaft, Arm, and Rod

Cold Rolled

Surface Finish of Pin Machined Pin Diameter (in) 1 Pin Fillet Radius (in) 0.08 Pin Shoulder Diameter (in) 1.16 Crank Arm Width (in) 2.5 Crank Arm Thickness (in) 0.85 Rod Width (in) 1 Rod Thickness (in) 1.5 Rod Hole Diameter (in) 1 Rod End Thickness (in) 0.6 Rod Fillet Radius (in) 0.08 Shaft Diameter (in) 1.2 Shaft Fillet Radius (in) 0.08 Shaft Shoulder Diameter (in) 2.5

y’

x’

CRITICAL POINT AT φ = 90°

φ

22

Figure 18: Pin critical point

Figure 19: Arm critical point

y’

x’

φ

CRITICAL POINT AT φ = 90°

y’

Z

23

Figure 20: Final machine design

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| APPENDIX A

Figure 5: Slider Kinematics