Arberi Ferraj. Wentworth Institute of Technology. Design of Machine Elements MECH 420

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1 P a g e 1 Arberi Ferraj Wentworth Institute of Technology Design of Machine Elements MECH 420

2 P a g e 2 1. Executive Summary A scissor car jack was designed and must be reverse-engineered in order to discover any possible flaws or failures that the jack may potentially have. The jack must range from mm. Since the link bars are more horizontal and will observe more moment at the lowest height, there will also be the most stress at the lowest possible height. Our main focus is the 100 mm total jack height. First, the forces acting on each part of the jack have to be analyzed before it is possible to make any conclusions. We know the device is designed to withstand 4000 pounds of force. Constructing proper Free Body Diagrams made it simple to determine how much of the 4000 pound force was distributed into each part of the jack. Manipulating an Excel spreadsheet was the most efficient way in keeping track of these forces. Also, Excel simplified any sort of changes that were necessary (in case of error). Knowing all the necessary dimensions in the jack, the minimum angle, theta, was derived. This angle allowed us to find the link force, x- component of the link force, the bolt force, and all their corresponding stresses. Utilizing the Torque to raise equations was what helped calculate the required force on the handle in order for the car jack to lift the load. Inputting these forces into SolidWorks simulation helped construct the necessary Finite Element Analysis (FEA) on the critical parts of this jack. By retrieving the maximum Von Mises stress it was then possible to determine their factors of safety. If the calculated factors of safety meet the required factor of safety of 2, then the machine is safe and operable. If it does not, however, then the scissor jack would fail and not be safe. In the end, the jack was found to not be nearly safe and had several areas of failure.

3 P a g e 3 2. Contents 1. Executive Summary List of Figures List of Tables Case Study Introduction Statement of Problem Detailed Performance Specification Detailed Section on Theoretical Analysis Loading Analysis Free Body Diagrams Description of Loading Equations Maximum Loading Position Design Check Link Bar Design Check Rivet Design Check Bolt Design Check Bolt Lever Design Check Upper Part Design Check FEA Simulation Link FEA Upper Part FEA Bolt Lever FEA Base FEA Detailed Drawings of All Parts Preliminary Drawings Assembly Preliminary Drawing Exploded Assembly Findings, Discussion, Conclusion References... 55

4 P a g e 4 3. List of Figures Figure 1 - Car Jack Height... 6 Figure 2 - Bolt Lever FBD (10001)... 7 Figure 3 - Bolt Nut FBD (10002)... 8 Figure 4 - Bolt FBD (10003)... 9 Figure 5 - Base FBD (10004) Figure 6 Link Bar (10005) Figure 7 Rivets, Front (10006) Figure 8 - Rivets, Rt. Side (10006) Figure 9 - Upper Part (10010) Figure 10 - Support, Front (10011) Figure 11 Support, Rt. Side (10011) Figure 12 - Car Jack, Free Height (50001) Figure 13 - Car Jack, Max Height (50001) Figure 14 - Intro Eq. 3 Description Figure 15 - Intro Eq. 4. Description Figure 16 - Intro Eq. 5 Description Figure 17 - Intro Eq. 7 Description Figure 18 - Link Force, y-component Figure 19 - Link Cross Section Figure 20 - Link Buckling/Moment of Inertia Figure 21 - Rivet Cross-sectional Area Table 11 - Friction Coefficients Figure 22 - Rod Details Figure 23 - Acme Threads Figure 24 - Bolt Lever Cross-section Figure 25 - Link Bar FOS Plot (6.2) Figure 26 - Rivet Shear Stress Figure 27 - Bolt Axial Stress Figure 28 - Bolt Lever FOS Plot (6.2) Figure 29 - Bolt Lever Hollowed Figure 30 - Bolt Lever Filled Figure 31 - Upper Part FOS Plot (6.2) Figure 32 - Link FEA Deformation Figure 33 - Link FEA Factor of Safety Plot Figure 34 - Upper Part FEA Deformation Figure 35 - Upper Part FEA Factor of Safety Plot Figure 36 - Bolt Lever FEA Deformation Figure 37 - Bolt Lever FEA Factor of Safety Plot Figure 38 - Base FEA Deformation Figure 39 - Base FEA Factor of Safety Figure 40 - Bolt Lever Preliminary Drawing Figure 41 - Bolt Nut Preliminary Drawing Figure 42 - Bolt Preliminary Drawing Figure 43 - Base Preliminary Drawing Figure 44 - Link Bar Preliminary Drawing... 49

5 P a g e 5 Figure 45 - Rivet Preliminary Drawing Figure 46 - Upper Part Preliminary Drawing Figure 47 - Car Jack Assembly Preliminary Drawing Figure 48 - Car Jack Exploded Assembly List of Tables Table 1 Intro, Applied Loading Table 2 Intro, Height Definitions Table 3 - Intro, Critical Height/Angle Table 4 - Link Dimensions Table 5 - Link Buckling Force Table 6 - Upper Part Dimensions Table 7 - Rivet Dimensions Table 8 - Bolt Dimensions Table 9 - Bolt Details Table 10 - Metric Thread Dia./Area Table 11 - Friction Coefficients Figure 22 - Rod Details Table 12 - Bolt Lever Dimensions Table 13 Torque Coefficients Table AISI Steel Yield Strength Table 15 - Von Mises Stresses Table 16 - Design Check Table 17 - Excel Spreadsheet Pt Table 18 - Excel Spreadsheet Pt

6 P a g e 6 5. Case Study Introduction 5.1 Statement of Problem The intellectual property of the prototype car jack that was purchased needs to be both functionally and structurally superior to the other products on the market. The car jack must be reverse engineered using a performance simulation in order to analyze its parts where it undergoes the highest possible stress, the minimum height. Using these values, a loading analysis must be implemented to check the current design of the jack using the appropriate failure criterion. FEA simulations must occur on typical parts. By using all of these tools, the worst case loadings of the jack can be identified to meet the required specifications. 5.2 Detailed Performance Specification Being supplied with performance specifications aided the redesign process as it allowed us to identify the dead end of the designing process. This reduces the chance of over designing the jack. The jack was to be redesigned to withstand a maximum loading of 4000 pounds with a factor of safety of 2.0, which should easily allow the jack to lift ¼ of a car, at least ½ required. The jack s maximum height was set to be 265 mm, and the height at which the jack first makes contact with the car is set to 100 mm, making a total lifting distance of 165 mm (Figure 1 below depicts the car jack height). Figure 1 - Car Jack Height

7 P a g e 7 6. Detailed Section on Theoretical Analysis 6.1. Loading Analysis Free Body Diagrams Figure 2 - Bolt Lever FBD (10001) The bolt lever is applied with a force by the user and, as a result, the bolt nut reacts with the bolt as torque is applied.

8 P a g e 8 Figure 3 - Bolt Nut FBD (10002) The bolt nut absorbs the force that is applied by the user from the handle and transferred into the bolt. The roll pin holds the bolt nut in place.

9 P a g e 9 Figure 4 - Bolt FBD (10003) The applied torque originates from the applied force on the handle. This torque is then transferred into the bolt so it can twist and force the supports outward along the threads. The 4000 pound load causes the bolt to undergo tension, resulting in the forces in the bolt shown in the front view of Figure 4. The bolt must torque hard enough to overcome the 4000 pound load.

10 P a g e 10 Figure 5 - Base FBD (10004) The base plate experiences a similar force as the upper plate, the 4000 pound distributed load.

11 P a g e 11 Figure 6 Link Bar (10005) The link undergoes compression and experiences its stress beginning at the inside of the holes. The angle, theta, was derived at the highest stress where the height of the jack was the shortest.

12 P a g e 12 Figure 7 Rivets, Front (10006) Each link applies a force, at an angle theta, on the rivets depending on the height of the jack. The link forces that are applied are radial.

13 P a g e 13 Figure 8 - Rivets, Rt. Side (10006) These Free Body Diagrams of the rivets portray the shear force that is occurring within the rivets due to the link and load.

14 P a g e 14 Figure 9 - Upper Part (10010) The top surface of the upper plate supports the load. All four links react to this 4000 pound load evenly.

15 P a g e 15 Figure 10 - Support, Front (10011) One of the supports experiences all four link forces, two on each shoulder screws. This causes a shear force on the shoulder screws.

16 P a g e 16 Figure 11 Support, Rt. Side (10011) The right side view of the support represents the radial forces acting on the support from the link forces. The force of the bolt is what pushes the support outward or inward which, in turn, raises or lowers the jack height.

17 P a g e 17 Figure 12 - Car Jack, Free Height (50001) The 4000 pound (or N) load is experienced at both the top of the upper plate and the bottom of the base plate. This is what causes the symmetry in the analyzed car jack. The free height of the jack is 100 mm, the height until contact is made with the car and the load is applied.

18 P a g e 18 Figure 13 - Car Jack, Max Height (50001) The car jack experiences the least stress at its maximum height of 265 mm.

19 P a g e Description of Loading Equations Intro: List of Intro Equations: (1) Load of Car = F load = 4000 lbs = N (2) h a = 0.1 m (3) h t = m (4) h b = m (5) h c = h a h t h b (6) θ = sin 1 ( h c ), where l = length of link bar 2l Relevant Tables: Load, F load Load, F load 4000 lbs N Table 1 Intro, Applied Loading Top height, h t Bottom height, h b Free height, h a Max height, h max 25.5 mm 14 mm 100 mm 265 mm Link Bar length, L 120 mm Table 2 Intro, Height Definitions Height (mm) Link angle, θ (radians) Link angle, θ (degrees) Table 3 - Intro, Critical Height/Angle Description of Intro Equations: (1) Applied Load: Load that the jack must overcome to lift the car. (2) Actual Jack Height: Actual height of the jack from the bottom of the base plate to the top of the upper part. In this case it is where the analyzing will take place, at 100 mm (0.1 m) (3) Top Jack Height: Vertical distance from the top of the upper part to the center of the holes of the upper part, as shown in Figure 14.

20 P a g e 20 h t Figure 14 - Intro Eq. 3 Description (4) Bottom Jack Height: Vertical distance from the bottom of the base to the center of the holes of the base, as shown in Figure 15. h b Figure 15 - Intro Eq. 4. Description (5) Effective Jack Height: Height of the jack that is used to calculate the angle theta. Also known as the height between the holes in the upper part to the holes of the base plate, as shown in Figure 16. h c Figure 16 - Intro Eq. 5 Description (6) Link Angle, Theta: The angle theta, as shown in Figure 17, is defined as the angle of the link with respect to the horizontal bolt (or ground). It is found by the law of sine. θ θ θ θ Figure 17 - Intro Eq. 7 Description

21 P a g e 21 Summary of Intro Equations: The effective jack height is not the same as the height from the very bottom to the very top of the whole jack. This is because the analysis is taking place from the center of the holes in the top and bottom parts of the jack and links where the rivets hinge. Hence subtracting the height of each. This allows us to find the angle theta, by law of sine, which is used to calculate the reaction forces at every height (1 millimeter increments). Link Bar: (1) F link = F load 4 sin θ (2) F link x = F link cos θ (3) A link = w link t link (4) σ link = F link A link (5) F buckling = π2 EI l 2 (6) I = bh3 12 List of Link Equations: Relevant Tables: Link Length 0.12 m Width m Thickness m Area m 2 Table 4 - Link Dimensions

22 P a g e 22 Link Buckling Force Steel Elastic Modulus, Es 2E+11 Pa Base, b m Height, h m Moment of Inertia, I 4.95E-11 m 4 Buckling Force, Fbuckling N Table 5 - Link Buckling Force Description of Link Equations: (1) Link Force: Compressive force on the link due to the 4000 pound load. Because of symmetry, one half of the jack is being analyzed. Therefore, it is assumed that four links are involved in the compression of the car jack. The link force is derived from the applied load. It is known that F load is equal to the sum of the y-components of the links hinged onto the upper part, as shown in Figure 18, since the applied load is in the vertical y-direction. So, the equation was originally defined as: F load = 4F link sin θ. Solving for F link defines the equation of the force in the link due to the 4000 pound compressive load in the jack. Flink-y Flink-y Figure 18 - Link Force, y-component

23 P a g e 23 (2) Link Force, x-component: Force of the link in the x-direction. This is a very critical portion of the analysis because the sum of these forces help analyze the bolt. (3) Link Cross-sectional Area: Effective cross-section of specimen. Because the link is in compression, we must analyze the cross-section with respect to its width and thickness, as shown in Figure 19. t link w link Figure 19 - Link Cross Section (4) Link Stress: The stress in the link bar is compressive. It is expressed as the force in the link divided by its cross-sectional area. (5) Link Buckling Force: Force that may cause the link to buckle due to extensive compressive forces. The buckling force was calculated assuming worst case scenario. E is the elastic modulus, I is the moment of inertia, and l is the length. (6) Moment of Inertia: Direction in which the link bar will buckle. After calculating all three different combinations of moment of inertia, it was found that the smallest value came from the most vulnerable buckling scenario, as shown below in Figure 20. That is, the link buckling about its thinnest surface. Buckling b h Figure 20 - Link Buckling/Moment of Inertia

24 P a g e 24 Summary of Link Equations: The link bars experience a substantial amount of the applied load. Since the links are more horizontal than vertical at 100 mm, the highest point of stress, they will have high x- component forces. Another important analytical observation is the buckling of the link bar. The worst case scenario for the link to buckle was examined. This means the resultant of the moment of inertia of the link must be as small as possible. Upper Part: (1) σ upper = F link A upper (2) A upper = thickness width List of Upper Part Equations: Relevant Table: Upper Part Thickness m Witdth 0.01 m Area m 2 Table 6 - Upper Part Dimensions Description of Upper Part Equations: (1) Upper Part Stress: The stress in the upper part comes from the force in the links. (2) Upper Part Cross-sectional Area: The cross sectional we are concerned with is where the force is applied in the plate. In this case, it is the plate thickness multiplied by the width. Summary of Upper Part Equations: The sides of the upper part are compressed by the force of the links. The stress developed is derived from dividing the link force by the affected cross sectional area of the upper part.

25 P a g e 25 Rivet: List of Rivet Equations: (1) τ rivet = F link A rivet (2) A rivet = 1 4 πd2 Relevant Table: Rivet Diameter 0.01 m Area 7.854E-05 m 2 Table 7 - Rivet Dimensions Description of Rivet Equations: (1) Rivet Shear Stress: Shear stress in the rivet. This stress comes from the force that the link applies to the rivet since that is what the rivet is hinging. It is a shear stress because of the distance between the forces of the plate and link. (2) Rivet Cross-sectional Area: The area of the rivet in which the shear stress is experienced, as shown in Figure 21. d Figure 21 - Rivet Cross-sectional Area Summary of Rivet Equations: The rivet experiences a shear force/stress due to the compressive forces of the links.

26 P a g e 26 Bolt: List of Bolt Equations: (1) F bolt = 4F link x (2) σ bolt = F bolt A bolt F bolt [( l ) + f] πd (3) P R = m 1 ( fl ) πd m F bolt [f ( l )] πd (4) P L = m 1 + ( fl ) πd m (5) T R = F bolt 2 (l + πfd m sec α πd m fl sec α ) (6) T L = F boltd m 2 (7) πfd m sec α > l ( πfd m sec α l πd m + fl sec α ) Relevant Tables: Bolt Minor dia m Minor Area m 2 Tensile Area m 2 Table 8 - Bolt Dimensions

27 P a g e 27 Bolt Details Torque Coefficient, k 0.3 Dry Coeff of Friction, f d 0.25 Oiled Coeff of Friction, f o 0.11 Major Diameter, d m Mean Diameter, d m m Thread Pitch, l m Thread Angle, α radians Table 9 - Bolt Details Description of Bolt Equations (1) Bolt Force: Axial force in the bolt. The bolt force is derived from the summation of the x-component forces of the link bars. Since the links are compressed, they force the supports outward in the x-direction along the bolt. (2) Bolt Stress: Axial stress in the bolt. As any other stress analysis, this is derived from dividing the force by the area. However, since the bolt is in tension, the area in this case is the tensile-stress area in the bolt, as shown below in Table 10, for a 12mm major diameter coarse-pitch series. Table 10 - Metric Thread Dia./Area (3) Force Required to Raise in Screw Body: Force in threaded rod that must be applied to overcome friction in the threads and raise the load. Identified Variables: l Thread Pitch: Distance between adjacent threads (Figure 22 below). d m Mean Diameter (or Pitch Diameter): Theoretical diameter between major and minor diameters (Figure 22 below): d m = d 0.5l f Coefficient of Friction: Takes thread friction into consideration. Table 11 below shows the different friction coefficient values for each scenario.

28 P a g e 28 Table 11 - Friction Coefficients Figure 22 - Rod Details (4) Force Required to Lower in Screw Body: Force in threaded rod that must be applied to lower the load. (5) Torque Required to Raise in Screw Body: Required torque in threaded rod that must overcome friction in the screw body to begin raising the load. Since the bolt in the machine has acme threads, the angle, α, in each thread must be considered. In this case, α is equal to 30 degrees. Figures 23 below draws the proper illustration. Figure 23 - Acme Threads (6) Torque Required to Lower in Screw Body: Required torque in threaded rod that must overcome friction in the screw body to begin lowering the load. (7) Self-locking Condition: If the product of pi, the friction coefficient, mean diameter, and secant of the threaded angle are greater than the pitch, then the jack will self-lock, resulting in no negative lowering forces.

29 P a g e 29 Bolt Lever: List of Bolt Lever Equations: (1) T total = kdf bolt (2) F applied = T total l lever (3) σ lever = F applied A lever (4) A lever = π 4 OD lever 2 π 4 ID lever 2 Relevant Tables: Bolt Lever Length 0.3 m Outside dia m Inside dia m Inertia E-10 m 4 Area E-05 m 2 Table 12 - Bolt Lever Dimensions Description of Bolt Lever Equations: (1) Total Torque Required to Raise: Actual torque required to lift the load. This torque differs from the torque to raise in the bolt because it accounts for the bolt collar friction. This is where the connection between the bolt nut and the bolt lever is. The torque coefficient, k, was chosen to be 0.3 (Table 13 below) because that is the highest possible coefficient, assuming the most friction at the worst case scenario. Table 13 Torque Coefficients

30 P a g e 30 (2) Force on Lever Required to Raise: The force applied to the handle by the user in order to raise the load. Derived from dividing the distance of the total required torque by the length of the handle (assuming user applies force at the very end of the handle). (3) Stress on Lever: The stress the lever experiences as the user applies the necessary force to raise the load. This is expressed as the force per unit area. (4) Cross-sectional Area: The area of the lever that is affected by the force to produce the stress. Since the lever is hollowed out, the cross-section is between the outer and inner diameters, displayed in Figure 24 below. ID OD Figure 24 - Bolt Lever Cross-section Maximum Loading Position Microsoft Excel allowed me to analyze this car jack from the free height, 100 mm, to the maximum height, 265 mm, in increments of 1 mm. After doing so, conclusions were very simple to make. However, in terms of examining areas of failure only one height is necessary: the free height. The reasoning is because that is where the most stress occurs in every part of this car jack. If the jack is low, its x-component forces must be extremely high in order to fight the opposing, vertical load. The force acting on the links are more perpendicular to the applied load than it is parallel. It is very difficult for an almost horizontal member to resist a vertical force. In a sense, the links are weakest at the free height. Therefore, it can be concluded that the most possible stress in the car jack will occur at the lowest height of 100 mm. This is the focal point of the reverse engineering process.

31 P a g e Design Check Design Check Introduction: Yield Strength 1020 Steel, Sy-steel Pa Table AISI Steel Yield Strength Stress Analysis Bolt Link Bolt Lever Rivet (τ max) Upper Part Von Mises Stress, σ (Pa) Design Check Table 15 - Von Mises Stresses Part FOS - Steel Safe? Link No Rivet No Bolt No Bolt Lever No Upper Part No Table 16 - Design Check Yield Strength: The stress at which a specific amount of plastic deformation is produced. The yield strength of 1020 AISI steel is low in comparison to other metals. This means the material will more easily experience yielding and deform quicker. Von Mises Stress: The concept of Von Mises stress originates from the distortion energy theory. The distortion energy theory is used for ductile materials, such as the 1020 AISI steel that the jack is made out of. This theory was applied to all parts listed above in Table 15 except for the Rivet. Calculations were made using the equations below to arrive to the rivet s max shear stress and the bolt s Von Mises stress. The FEA simulation in SolidWorks helped find the Von Mises for the other three analyzed parts.

32 P a g e 32 Distortion Energy Theory n = S y σ Max Shear Stress Theory n = S y 2τ max Design Check: Five major parts of this jack were analyzed in the design check: the link bars, rivet pins, bolt, bolt lever, and the upper part. With the design check, conclusions on the factor of safety for each part can be made. If the derived safety factor is found to be less than 2, then the part will not be considered safe and fail. In this design, every part resulted in a failure with every part having a factor of safety of well under 2. Other types of failures such as buckling of the link bar must also be analyzed Link Bar Design Check Figure 25 - Link Bar FOS Plot (6.2)

33 P a g e 33 Figure 25 above portrays the factor of safety plot for the link bar. The blue is where the link is safe and the red is where the link is not safe. The links proved to be the second safest part of the whole car jack. Using the Distortion Energy theory, the factor of safety came out to be 0.516, as displayed in table 16 above. Von Mises stress in this part is somewhat high due to the thinness of the link. This part is dealing with a lot of stress because of the high loading it experiences with such a small cross-sectional area. With a low yield strength of 351 MPa, the Von Mises stress must be at least half of that to meet the 2.0 factor of safety requirement. In this case, it is not. The FEA simulation in SolidWorks produced a Von Mises stress of 681 MPa, almost double the yield strength. The distortion energy theory was used to determine this factor of safety because the material is ductile and does not undergo shear stress. The link is not safe in the red area shown in Figure 25 because of the buckling. The buckling force was 6785 N, assuming worst case scenario. The force on the link experiences a force well over this number which means the link will buckle. Again, this failure comes from the link being very thin Rivet Design Check The rivets were the safest parts of this machine. The factor of safety was found using max shear stress theory (MSS theory) because the rivets are in shear due to an unevenly applied compression, as shown below in Figure 26. This maximum shear stress that the rivet experiences is found by calculating the force per unit area, similar to any other stress. The force that is applied to the rivet pins comes from the link bars. To improve this part, the rivet must have a greater cross-sectional area. A much wider diameter on the rivets is needed in order for them to reach the safety requirement. Figure 26 - Rivet Shear Stress

34 P a g e Bolt Design Check As critical of a part as the bolt is, it proved to be the second least safe part of the machine with a factor of safety of For this part, the factor of safety was found using the distortion energy theory. However, simulating an FEA on this part would be rather difficult because of the threads. So, the Von Mises stress was calculated using the equation below: σ = 1 2 (σ x σ y ) 2 + (σ y σ z ) 2 + (σ z σ x ) 2 + 6(τ xy 2 + τ yz 2 + τ zx 2 ) However, the bolt only experiences horizontal forces, meaning forces in the x-direction. So its y and z components all become 0 leaving the equation as: σ = 2σ x 2 + 6τ xy 2 where σ x is the bolt s axial stress and τ xy is the bolt s shear stress. The shear stress was derived with the equation below: 2 τ xy = T Rd bolt 2J and J = π 4 2 (d bolt ) 2 where T R is the torque to raise, d bolt is the bolt diameter, and J is the polar moment Inertia ( m 4 ). The axial stress comes from the sum of the forces of each link acting on the bolt in the horizontal x-direction, as shown in Figure 27 below. This high Von Mises stress is what led to a poor factor of safety. In order to fix this, the threaded rod must have a much larger diameter in order to have a greater cross-sectional area and a lower axial stress. Lower axial stress results in lower Von Mises stress which, as a result, means a higher factor of safety. Figure 27 - Bolt Axial Stress

35 P a g e Bolt Lever Design Check Figure 28 - Bolt Lever FOS Plot (6.2) The bolt lever was the absolute least safe part of the car jack with a factor of safety of The factor of safety was calculated using the distortion energy theory. A simple FEA was performed to obtain the Von Mises stress of 7,332 MPa. As shown in the figure above, the handle is safe only at the fixed are and where the force is being applied (assuming point force at very end of handle to account for full 300 mm length). This part of the design can easily be fixed by filling the hollow part of the lever and increasing the diameter. Figures 29 and 30 below show how the issue can be resolved. Fill Figure 29 - Bolt Lever Hollowed Figure 30 - Bolt Lever Filled

36 P a g e Upper Part Design Check Figure 31 - Upper Part FOS Plot (6.2) Although the upper plate was the third safest part of the car jack with a factor of safety of 0.476, it is still considered very unsafe since it does not meet the required FOS of 2. The safest parts in the upper plate are at the middle, where the fixture is located. The upper part experiences forces from all four attached link bars. The stress in this part comes from the summation of those four link bars. The distortion energy theory equation was also utilized in this part to obtain the final factor of safety. Again, the best way to improve the upper part is to increase its overall thickness. Doing so will result in a considerably lower Von Mises stress which should result in a greater factor of safety.

37 P a g e FEA Simulation Link FEA Figure 32 - Link FEA Deformation

38 Figure 33 - Link FEA Factor of Safety Plot P a g e 38

39 P a g e Upper Part FEA Figure 34 - Upper Part FEA Deformation

40 Figure 35 - Upper Part FEA Factor of Safety Plot P a g e 40

41 P a g e Bolt Lever FEA Figure 36 - Bolt Lever FEA Deformation

42 Figure 37 - Bolt Lever FEA Factor of Safety Plot P a g e 42

43 P a g e Base FEA Figure 38 - Base FEA Deformation

44 Figure 39 - Base FEA Factor of Safety P a g e 44

45 P a g e Detailed Drawings of All Parts Preliminary Drawings Figure 40 - Bolt Lever Preliminary Drawing

46 Figure 41 - Bolt Nut Preliminary Drawing P a g e 46

47 Figure 42 - Bolt Preliminary Drawing P a g e 47

48 Figure 43 - Base Preliminary Drawing P a g e 48

49 Figure 44 - Link Bar Preliminary Drawing P a g e 49

50 Figure 45 - Rivet Preliminary Drawing P a g e 50

51 Figure 46 - Upper Part Preliminary Drawing P a g e 51

52 P a g e Assembly Preliminary Drawing Figure 47 - Car Jack Assembly Preliminary Drawing

53 P a g e Exploded Assembly Figure 48 - Car Jack Exploded Assembly

54 P a g e Findings, Discussion, Conclusion The design of the scissor car jack was thoroughly examined throughout this technical report. After properly analyzing each individual part of the machine, it became rather simple to say that this product would fail miserably in almost every aspect. Running the simulation in Excel helped me conclude that the highest stress would occur at the lowest possible height with the applied 4000 pound load. Also, SolidWorks Simulation assisted in deriving the max Von Mises stresses for some parts (handle, link bar, upper part) in order to calculate the factor of safeties. Not only were stresses in each part incredibly high, but the derived factor of safeties were also very low. Other failures within the link also made the jack very dangerous. Overall, failures for five of the main analyzed parts in this jack were easily noticed. As previously stated, the bolt experiences an axial stress. This is due to the supports pushing outward from the links as the torque of the bolt fights this force. So in terms of bolt force, we are concerned with the sum of the forces of the link bars in the horizontal x-direction. The sum of these forces (68300 N) proves to be too high for the bolt to handle since it develops an extremely high Von Mises stress which results in a low safety factor. Another failure that stood out was the link bar. Each link undergoes compressive forces of N. After calculating the buckling force of 6785 N, assuming worst case scenario of buckling, it was clear that the link would fail and buckle. This is because the links experience forces that greatly exceed the 6785 N buckling mark. Other failures occurred at the rivet pins. Again, the high stress proved to be too much for the pins to handle, resulting in a shear failure. The least safe part of the whole design was the bolt lever. The ductile steel is not nearly strong enough to handle the required applied force of 820 N. This is because of how thin and hollow the lever was designed. Lastly, the upper plate had failures about the sides of the part. This results from how thin the part was designed. Finite Element Analysis allowed me to see the deformation and conclude that the Von Mises stress was too high in order to get a satisfying factor of safety. All in all, the design of this car jack was poorly engineered. All of the failures and low safety factors come from the incredible amount of stress the jack experiences. Even after applying a 1000 pound load in the Excel simulation, there are still parts that are unsafe. The bolt itself is not safe until the load is as low as 400 pounds. The best way to enhance the car jack s performance is to thicken every part. This will ease the stress and result in higher factors of safety. With as many failures as this car jack would experience, it is simple to conclude that it is simply very unsafe.

55 P a g e References [1] Budynas, Richard G., Nisbett, J. Keith., and Shigley, Joseph Edward., 2011, Shigley s Mechanical Engineering Design, McGraw-Hill, New York, NY. [2] Duva, Le, and Talpasanu, 2014, Design of Machine Elements: MECH420 Minor Project, Boston, MA. [3] 2014, Mechanics. from Table 17 - Excel Spreadsheet Pt. 1

56 Table 18 - Excel Spreadsheet Pt. 2 P a g e 56

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