The document describes the design and finite element analysis of a scissor jack. It includes an overview of scissor jack components and operation, as well as calculations of forces and stresses on members. A series of mesh refinement studies were performed on the carrier member, lifting arms, and shaft screw to determine maximum stresses and displacements under expected loading conditions.

1. City College of New York School of Engineering
Mechanical Engineering Department
Spring-2014
Mechanical Engineering I 6500: Computer-Aided Design
Instructor : Prof. Gary Benenson
Student : Mehmet Bariskan
Final Project : Design and FEM Analysis of
Scissor Jack

2. 2
1. Overview:
A jack is a mechanical device used as a lifting device to lift heavy loads or apply great forces.
Jacks employ a screw thread or hydraulic cylinder to apply linear forces. Car jacks use
mechanical advantage to allow us to lift a vehicle by manual force alone. More powerful
jacks use hydraulic power to provide more lift over greater distance. A scissor jack is a
device constructed with a cross-hatch mechanism, much like a scissor. A commercially
available scissor jack is shown in Figure 1.
Figure 1: Scissor Jack
A scissor jack is operated by turning a lead screw. It is commonly used as car-jacks. In the
case of a scissor jack, a small force applied in the horizontal plane is used to raise or lower
large load. A scissor jack’s compressive force is obtained through the tension force applied
by its lead screw. An acme thread is most often used, as this thread is very strong and can
resist the large loads imposed on most jacks while not being dramatically weakened by
wear over many rotations. An inherent advantage is that, if the tapered sides of the screw
wear, the mating nut automatically comes into closer engagement, instead of allowing
backlash to develop (Rajput, 2007). These types are self-locking, which makes them
intrinsically safer than other jack technologies like hydraulic actuators which require
continual pressure to remain in a locked position.
The completed solidworks design of the scissor jack and its members shown in Figure 2.
Most scissor jacks are similar in design, consists of four lifting arms, a base plate, a carrier
plate, two connection members, eight connection pins, a power screw and a crank. This
crank is usually “Z” shaped. When this crank is turned, the screw turns, and this raises the

3. 3
jack. The screw acts like a gear mechanism. It has teeth (the screw thread), which turn and
move the four arms, producing work. The four arms are all connected at the corners with a
bolt that allows the corners to swivel.
Figure 2: Solidworks design of Scissor Jack and its members
The lifting members are made from c-shapes. The web of the lifting member is cut out near
the pin connections to allow proper serviceability of the scissor jack at its maximum and
minimum heights. Members 3 and 4 have ideal connections to balance the load between
the left and right side.
The connecting pins are designed with cylindrical shapes and they are subjected to tension
instead of compression. The bending moment from the screw shaft creates tension on
these members.
Other pins are used as fasteners at the various joints of the members. The existence of the
jack will depend on the ability of the pin not to fail under sudden shear, tensional and
compressive forces.

4. 4
1.1 Force and Stress Analysis
The force analysis is based on the assumption that the scissor jack is loaded vertically
symmetrical.
Figure 3: Forces in Scissor Jack members
The maximum capacity for the scissor jack is the 600 kg.
Maximum Load = 600 kg, F= 600 kg * 9.81
𝑚
𝑠2
= 5886 N
L= 145 mm, the length of the arms (From hole center to hole center)
e = 180 mm, the length from base to top center of holes (At Minimum raising height of
the jack)
𝑐𝑜𝑠𝛼 =
𝑒
2⁄
𝐿
=
180
2⁄
145
= 0.62 𝑎𝑛𝑑,
𝛼 = 51,6°

5. 5
Assuming that I can simplify the mechanism at the joint of the top section that is shown
in Figure 4.
Figure 4: Free Body Diagram of the top section
∑ 𝐹𝑥 = 0
𝐹1 ∙ 𝑠𝑖𝑛𝛼 − 𝐹2 ∙ 𝑠𝑖𝑛𝛼 = 0 , 𝐹1 ∙ 𝑠𝑖𝑛𝛼 = 𝐹2 ∙ 𝑠𝑖𝑛𝛼
𝐹1 = 𝐹2
∑ 𝐹𝑦 = 0
𝐹1 ∙ 𝑐𝑜𝑠𝛼 + 𝐹2 ∙ 𝑐𝑜𝑠𝛼 − 𝐹 = 0
2 ∙ 𝐹1 ∙ 𝑐𝑜𝑠𝛼 = 𝐹
𝐹1 =
𝐹
2∙𝑐𝑜𝑠𝛼
𝐹1 =
5886
2∙cos(51.6)
𝐹1 = 𝐹2 = 4738 𝑁
The angle is decreasing at maximum raising height of the jack. Consequently, the
maximum force is decreased. Since the maximum loading force will act at the minimum
raising height of the jack, the design stresses will be analyzed at that point.
α α
F1
F2
F

6. 6
Figure 5: Free Body Diagram of the joint of the shaft and arms
Focusing the point of the components at the joint of the screw shaft and arms. We can say at
this point;
∑ 𝐹𝑦 = 0
𝐹1 ∙ 𝑐𝑜𝑠𝛼 − 𝐹3 ∙ 𝑐𝑜𝑠𝛼 = 0
𝐹1 ∙ 𝑐𝑜𝑠𝛼 = 𝐹3 ∙ 𝑐𝑜𝑠𝛼
𝐹1 = 𝐹3
∑ 𝐹𝑥 = 0
𝐹1 ∙ 𝑠𝑖𝑛𝛼 + 𝐹3 ∙ 𝑠𝑖𝑛𝛼 − 𝐹𝑆 = 0
𝐹𝑆 = 2 ∙ 𝐹1 ∙ 𝑠𝑖𝑛𝛼
𝐹𝑆 = 2 ∙ 4738 ∙ sin(51.6)
𝐹𝑆 = 7426 𝑁
Because of the symmetry we can write the following equation
| 𝐹1| = | 𝐹2| = | 𝐹3| = | 𝐹4| = | 𝐹1
′
| = | 𝐹2
′
| = | 𝐹3
′
| = | 𝐹4
′
|
α
α
Fs F1
F3

7. 7
2. Procedure
The first target is to predict the maximum displacement and maximum stress of the
scissor jack. Since the force is applied to the carrier member, I can predict that the
maximum displacement will happen at the top side of this member which is shown in
Figure 2 (labeled with number 6). Since the maximum force occurs on the screw shaft
which is calculated in the force and stress analysis. The maximum stress should be
occur joint of the screw and connecting pins.
2.1 Design and Analysis for the Individual Parts
Solidworks had been used to create and analyze the geometry under various boundary
conditions (restarints) and loading condition (force). The scissor jack had been analyzed
for stress and displacement.
I have followed the following steps:
I- I have started to design with the carrier member which is shown in Figure 6
and its technical drawing is shown in the drawing section.
Figure 6: Carrier Member
The top surface of the model was loaded with a compressive force of 5886 N, and the
holes were fixed as shown in Figure 6.

8. 8
The material is carbon steel (Ck 45) according to manufacturer website. If we don’t
know the material we can measure the hardness to estimate it’s tensile and yield
strength. The Ck45 graded steel has the following properties.
Applied Material Ck45, AISI 1045 Steel
Elastic Modulus 201 GPa
Tensile Strength 625 MPa
Yield Strength 530 MPa
Table 1: Properties of Ck45 Steel
Several studies were performed for creating the mesh. Then, I ran the model for stress
and displacement analysis.
Mesh Density
& Quality
(Standard
Mesh)
Maximum
Stress
(MPa)
Maximum
Displacement
(µm)
Total
# of
Element
Total
# of
Nodes
Total
# of
DOF
Global
Element
Size
(mm)
Location(s)
Local
Refinemen
t
Running
Time
(second)
Study-1
166.68 51 1699 3614 10194 5 N/A 1
Study-2
176.85 51 2897 5786 16710 4 N/A 1
Study-3
174.76 53 5941 10936 32160 3 N/A 1
Study-4
175.11 54 15994 27605 81375 2 N/A 2
Study-5
179.41 54 35041 57818 171294 1.5 N/A 3
Study-6 187.47 54 101542 159455 474117 1 N/A 9
Study-7 202.43 54 243096 366597 1092015 0.75 N/A 23
Study-8 233.07 54 786449 1141860 3410388 0.5 N/A 85
Table 2: The Results of the Mesh Studies for the carrier member
The study shows us the maximum displacement occurs at the middle section of the
member’s longitudinal axis with approximately 54 µm. The maximum stress occurs at
the same place in the first four studies. The study 5 and 6 show that the maximum
stress occurs at the upside of the holes. The study 7 and 8 show that more refinement
is giving the rising stress at the upside of the holes that is showing also the boundary
conditions are not really pairing with the real life conditions or using the solidworks is

9. 9
useless if you cannot define the appropriate boundary conditions. I am going to focus
that problem in the assembly study.
Figure 7: The maximum stress figure (left) and, the maximum displacement figure (right)
for the carrier member
II- I have completed the second analysis with the lifting arms which are labeled
in the figure 2. I have labeled them in the figure 2 with the number of 3 and 4.
The forces change with respect to its angular position. Firstly, I have applied the
compressive force that is calculated in force and stress analysis section. The arm must
be able to withstand an axial force of 4738 N for compression that is shown in Figure 8.
Figure 8: Lifting Arm

10. 10
Several studies were performed to creating the mesh. Then, I ran the model for stress
and displacement analysis.
Mesh Density
& Quality
(Standard
Mesh)
Maximum
Stress
(MPa)
Maximum
Displacement
(µm)
Total
# of
Element
Total
# of
Nodes
Total
# of
DOF
Global
Element
Size
(mm)
Location(s)
Local
Refinemen
t
Running
Time
(second)
Study-9
93.50 82 5611 11389 33843 5 N/A 3
Study-10
92.70 83 7276 14689 43743 4 N/A 4
Study-11
95.73 82 10754 21644 64500 3 N/A 5
Study-12
101.37 82 29066 53163 158769 2 N/A 12
Study-13
101.38 82 64975 110234 329514 1.5 N/A 29
Study-14
103.32 82 195905 312378 935082 1 N/A 88
Study-15
104.59 82 418984 647022 1937466 0.75 N/A 231
Study-16
103.54 82 1352434 1999109 5989299 0.5 N/A 889
Table 3: The Results of the Mesh Studies for the Lifting Arm
The study shows us the maximum displacement occurs at the top section of the lifting
arm with approximately 83 µm. Displacement occurs to the latitudinal direction because
of the arm shape causing the bending effect which is shown in Figure 9. The maximum
stress occurs in the pin holes that is shown in Figure 9. In this study, I only did
compressive study to lifting arm’s longitudinal axis. When I change the fixed place to top
holes and loaded place to bottom holes, I have the similar results for the stress and
displacement causing by the action and reaction forces.

11. 11
Figure 9: The maximum stress figure (left) and, the maximum displacement figure (right)
for the lifting arms.
III- I have completed the third analysis with the shaft screw which is labeled in
the figure 2. I have labeled it in the figure 2 with the number of 16.
I have applied a tensile force that is calculated in force and stress analysis section.
The shaft must be able to withstand an axial force of 7426 N for tension that is
shown in Figure 7.
Figure 10: Shaft Screw

12. 12
Mesh Density
& Quality
(Standard
Mesh)
Maximum
Stress
(MPa)
Maximum
Displacement
(µm)
Total
# of
Element
Total
# of
Nodes
Total
# of
DOF
Global
Element
Size
(mm)
Location(s)
Local
Refinemen
t
Running
Time
(second)
Study-17
93.81 89 1864 3842 11286 5 N/A 1
Study-18
93.69 90 4583 8344 24645 4 N/A 1
Study-19
94.01 90 7015 12348 36504 3 N/A 2
Study-20
94.27 90 23755 38472 114186 2 N/A 4
Study-21
94.35 90 53650 82739 246444 1.5 N/A 13
Study-22
94.57 90 248545 364679 1088640 0.9 N/A 55
Study-23
94.59 90 389636 565694 1689549 0.75 N/A 81
Study-24
95.35 90 1102266 1572182 4699632 0.5 N/A 532
Table 4: The Results of the Mesh Studies for the shaft screw (tensile)
The study shows us the maximum displacement occurs at the place where the force
applied of the shaft screw with approximately 90 µm. Displacement occurs to the
longitudinal direction. The maximum stress occurs at the same face with the
displacement and its magnitude approximately 95 MPa. The both pattern are shown in
Figure 11. The technical drawing of the shaft is attached in the drawing section.
Figure 11: The maximum stress figure (left) and, the maximum displacement figure
(right) for the shaft screw.

13. 13
IV- I have completed the fourth analysis with the shaft screw again. In this study,
I will focus the torsional stress in the shaft screw.
Firstly, I need to find out that what is the maximum torque that a human can apply to the
crank shaft? The crank shaft is labeled with number 9 in the figure 2.
The maximum force we could produce with a down push is (m * g) where m is our
mass. Considering the turning the shaft is not just a down push, I have made several
searches. I went to a gym and there are different type of people that they can pull and
push between 30 kg to 50 kg. I am going to apply the maximum force with 50 Kg, since
the load could apply from the stronger person to the jack.
F= 50 kg. * 9.81
𝑚
𝑠2
= 490.5 N
When I consider the head of the shaft screw that has a diameter of 23 mm. The torque
that we may create:
490.5 N *
0.023
2
m = 5.64 Nm
We could apply the torque to the head of the outer face of the shaft directly that is
shown in Figure 12. The fixed surface has chosen in the joint of the shaft screw and the
middle pin. Also, the fixed hinge restraint has applied to the between the middle pin and
the head of the shaft which is shown in the Figure 12.
Figure 12: Boundary conditions of the shaft screw #1 (torque)

14. 14
Mesh Density
& Quality
(Standard
Mesh)
Maximum
Stress
(MPa)
Maximum
Displacement
(µm)
Total
# of
Element
Total
# of
Nodes
Total
# of
DOF
Global
Element
Size
(mm)
Location(s)
Local
Refinemen
t
Running
Time
(second)
Study-25
54.95 251 2030 4109 11949 5 N/A 2
Study-26
56.36 253 4655 8435 24783 4 N/A 3
Study-27
59.28 253 6679 11946 35052 3 N/A 5
Study-28
65.28 253 24270 39109 115383 2 N/A 16
Study-29
70.78 253 67293 102722 302505 1.5 N/A 46
Study-30
80.67 253 277011 404403 1199013 0.9 N/A 279
Study-31
86.20 253 388592 564211 1679241 0.75 N/A 386
Study-32
110 253 1118962 1594211 4752015 0.5 N/A 1410
Table 5: The Results of the Mesh Studies #1 for the shaft screw (torque)
The results are not converging. I have decided to change the boundary conditions.
Since there is a movement between the teeth’s, I have decided to change the fixed
place to the shaft face that is shown in Figure 13.
Figure 13: Boundary conditions of the shaft screw #2 (torque)

15. 15
Mesh Density
& Quality
(Standard
Mesh)
Maximum
Stress
(MPa)
Maximum
Displacement
(µm)
Total
# of
Element
Total
# of
Nodes
Total
# of
DOF
Global
Element
Size
(mm)
Location(s)
Local
Refinemen
t
Running
Time
(second)
Study-33
54.95 303 2030 4109 12243 5 N/A 1
Study-34
56.35 305 4655 8435 25221 4 N/A 3
Study-35
56.16 305 6679 11946 35673 3 N/A 5
Study-36
57.85 305 24270 39109 116958 2 N/A 19
Study-37
57.13 305 67293 102722 305691 1.5 N/A 56
Study-38
57.10 305 189352 282335 834324 1 N/A 418
Study-39
57.45 305 388592 564211 1690752 0.75 N/A 514
Study-40
58.11 305 1118962 1594211 4778511 0.5 N/A 2024
Table 6: The Results of the Mesh Studies #2 for the shaft screw (torque)
Figure 14: The maximum Stress of the shaft screw (torque)

16. 16
Figure 15: The maximum Displacement of the shaft screw (torque)
The study shows us the maximum displacement occurs at the head section of the shaft
screw with approximately 305 µm. Displacement occurs to the circular direction. The
maximum stress occurs between the shaft and head of the screw which is sown in
Figure 15 with approximately 58 MPa.
V- I have completed the fifth analysis with the base plate that is labeled with
number 1 in the Figure 2. In this study, I will focus the compressive load by
causing of the lower lifting arms that is 4738 N. I have created the new planes
with 51.6 degree towards to lifting arms that is shown in Figure 16. The fixed
surface has chosen the bottom surface of the base plate. The left arms and
the right arms cause the force of 4738 N separately that were applied to the
base channel. After the studies, I have found the maximum displacement
approximately 3 µm around at the four holes and the maximum stress were
found approximately 114 MPa in the four holes. Both patterns are showing in
Figure 17 and Figure 18 respectively.

17. 17
Figure 16: Base Channel
Mesh Density
& Quality
(Standard
Mesh)
Maximum
Stress
(MPa)
Maximum
Displacement
(µm)
Total
# of
Element
Total
# of
Nodes
Total
# of
DOF
Global
Element
Size
(mm)
Location(s)
Local
Refinemen
t
Running
Time
(second)
Study-41
99.34 3 4146 8758 22671 5 N/A 1
Study-42
111.59 3 5896 12272 31653 4 N/A 1
Study-43
113.85 3 10008 20328 52035 3 N/A 2
Study-44
113.59 3 33095 59249 158874 2 N/A 3
Study-45
113.30 3 72743 123194 336435 1.5 N/A 7
Study-46
115.96 3 213035 341571 950610 1 N/A 20
Study-47
113.76 3 504062 774165 219506 0.75 N/A 48
Study-48
114.57 3 1588087 2342876 6736581 0.5 N/A 347
Table 7: The Results of the Mesh Studies for the base plate

18. 18
Figure 17: The maximum stress of the base plate
Figure 18: The maximum displacement of the base plate

19. 19
2.2 Assembly Analysis
I- Assembly
After, I have created the parts for the scissor jack. I have combined them in a solidworks
assembly file. We need to use the mate command properly to attach the parts each
other’s. I have started with the carrier member, right lifting arm and left lifting arm. There
is a hinge command in the mechanical mates section. I have selected that command to
create a hinge connection between the hole of the carrier member and the hole of the
left arm, chosen concentric and coincident selections are shown in Figure 19.
Figure 19: The Hinge (Mechanical Mates) Selections for upside connections
I have completed the remaining three holes with the same method. Then, I have used
the same command for the joint of the upper and bottom lifting arms. In this step,
selection of the concentric and coincident selections are shown in Figure 20. Following
step was to connect the lifting arms to the base plate which is shown in Figure 21. The
operation was applied to the all pin connection holes as described above.

20. 20
Figure 20: The Hinge (Mechanical Mates) Selections for arm connections

21. 21
Figure 21: The Hinge Mate (Mechanical Mates) Selections for bottom side connections
The screw mate that, I have applied between the shaft screw and middle pin. The outer
face of the shaft screw and the inner face of the hole in the middle pin were chosen as
mate faces. The screw mate command with distance / revolution option was applied that
is shown in Figure 22. I have applied a concentric mate to the same faces. The
concentric mate also was applied between the remaining middle pin and outer face of
the shaft screw.
Figure 22: The Screw Mate (Mechanical Mates) Selections
The coincident mate was applied to two panel that are shown in in Figure 23. Then, the
hinge mate was applied between outer face of the middle pin’s bearing and inner face of
the holes of the arms at the connection places. A tangent mate was applied between
the one face of the shaft’s head and the outer face of the middle pin to prevent
movement between two components that is shown in Figure 24.

22. 22
Figure 23: Coincident mate selections
Figure 24: Tangent mate selections
II – FE Analysis
I have followed the following steps to perform static studies in Solidworks program.

23. 23
Firstly, I have applied the material to the all component that is described in Table 1.
Secondly, I have applied a fixture to fix the bottom face of the scissor jack. I have
applied an external load of 5886 N to upside of the carrier member. The both boundary
conditions are shown in Figure 25.
Figure 25: The boundary conditions for the assembly
The third step was to apply the connectors to the pin holes. I have chosen pin type of
connector and apply to the inner face of the holes with the connection type of the “With
retaining ring (No translation)” option. I have applied the pin connectors to the 12
connections with the 24 inner face off the holes which are shown in Figure 26.
The fourth step was to define contact sets. We have to specify the contact sets to
program otherwise all parts has considered as a single part. I have defined total of 30
contact sets between the faces of the components. We need to focus to movement of
the scissor jack and inspect the faces working towards. The outer flank faces of the
upper lifting arms and inner flank faces of the lower lifting arms can be examples of the
contact sets have to be considered. After two faces are chosen in the program, we need
to choose also the type of the contact sets. I have completed all the contact sets with
the “No Penetration” option.
The fifth step was to mesh the assembly and run the study. The results are shown in
Table 8.

24. 24
Figure 26: The pin connections
Mesh Density
& Quality
Maximum
Stress
(MPa)
Maximum
Displacement
(µm)
Total
# of
Element
Total
# of
Nodes
Total
# of
DOF
Global
Element
Size
(mm)
Running
Time
(second)
Study-49
Curvature 211.61 417 71805 139476 416247 10 2194
Study-50
218.38 409 161672 297903 889521 5 4612
Study-51
222.20 438 264319 468019 1397634 3.5 7316
Study-52
2
Study-53
1.5
Study-54
1
Study-55
0.75
Table 8: The Results of the Mesh Studies for the assembly

25. 25
Figure 27: von-Misses stress contour of the scissor jack at 5886 N load
Figure 28: Displacement contour of the scissor jack at 5886 N load

26. 26
3. Results
The convergence studies were completed for the individual parts with eight studies. The
tabulated data for von Mises stress and displacement are gathered in Table 9 and 10.
Maximum
Stress
(MPa)
Part # 6
Carrier
Member
Part #
2,3,4,5
Arms
Part # 16
Shaft Screw
Tension
Part # 16
Shaft Screw
Torque
Part # 1 Base
Study 166.68 93.50 93.81 54.95 99.34
Study 176.85 92.70 93.69 56.35 111.59
Study 174.76 95.73 94.01 56.16 113.85
Study 175.11 101.37 94.27 57.85 113.59
Study 179.41 101.38 94.35 57.13 113.30
Study 187.47 103.32 94.57 57.10 115.96
Study 202.43 104.59 94.59 57.45 113.76
Study 233.07 103.54 95.35 58.11 114.57
Table 9: von Mises Stresses for the individual parts
Maximum
Displacement
(µm)
Part # 6
Carrier
Member
Part #
2,3,4,5
Arms
Part # 16
Shaft Screw
Tension
Part # 16
Shaft Screw
Torque
Part # 1 Base
Study 51 82 251 303 3
Study 51 83 253 305 3
Study 53 82 253 305 3
Study 54 82 253 305 3
Study 54 82 253 305 3
Study 54 82 253 305 3
Study 54 82 253 305 3
Study 54 82 253 305 3
Table 10: Displacement results for the individual parts

27. 27
4. Discussion
All the studies that I have completed in this project are showing us the maximum stress
occurs at the carrier member. The study for that individual part has converged at 5 point
within 5 % error and it occurs top surface of the member. But, when I made the study
using the smaller element size. I have found the maximum stress bigger than the
acceptable window for the convergence study but the place of the maximum stress has
change its place from top surface of the part to edge of the pin holes. I have skipped
that part after I couldn’t converging the number even I have added the fillets to sharp
edges and trying the different boundary conditions for the holes. In the assembly study,
I have reached the maximum stress of 222 MPa. I am not trusting this number also
since, I couldn’t complete the studies as I have targeted. If, I can trust my first study
solution with 5 points in the safe window for the carrier member. The maximum stress is
187.47 MPa according to from study 2 to study 8. When I consider the material
mechanical properties of the tensile stress which is 625 MPa.
If (
𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑆𝑡𝑟𝑒𝑠𝑠
) ≥ 2 (2 is taken as a constant for Factor of Safety)
625 𝑀𝑃𝑎
187.47 𝑀𝑃𝑎
= 3.3
We can say that design is acceptable for the carrier member.
The lifting arm was studied under the tensile load of 4738 N from the two holes towards
to another two holes while these holes fixed. The maximum stress is approximately 104
MPa that is shown in Figure 9 and the convergence study is in the Table 3. Even that
design is acceptable for this study, there are different stresses could be cause the lifting
arm to fail. Focusing the teeth section in the real jack there is a contact between them
while jack works. If that teethes loaded in the operation, we should consider the bending
effect as well.
The shaft screw and the base member also in the acceptable stresses when we
compare with the material tensile strength.
In this project, I have learned that the procedure of the FE Analysis for the assembly
projects. I have spent most of the project time while trying to find accurate boundary
conditions for the scissor jack. I am still not sure that the applied boundary conditions
are pairing with the real object or not. Because of the project time not enough for me to
finish the assembly study. I am going to continue to work on this project to understand if
solidworks can give an accurate answer or its just targeting us to some information.

28. 28
5. Drawings
Figure 29: Technical Drawing of the Carrier Member

29. 29
Figure 4: Technical Drawing of the Lifting Arm

30. 30
Figure 4: Technical Drawing of the Screw Shaft

31. 31
Figure 4: Technical Drawing of the Base Plate

32. 32