Design of Hydraulic Scissor lift.pdf

introduction, types of scissor lift, advantages and disadvantages design of different components and calculation based of selected material. generated 3d model in solidworks.

1. Hydraulic scissor lift 18bme032 ,18bme050

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MECHANICAL ENGINEERING DEPARTMENT


Machine design – 1
(ME501)


“Design of Hydraulic Scissor lift”


Submitted By:
Deep Dhrafani -- 18BME032

Meet Kalola --18BME050




Submitted To:
Prof. Dhaval Patel

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Index

1. Idea generation ………………….………………………………………………….…...3
2. Selection of idea ……….……………………………………………………..….…….. 3
3. Introduction……………….…………………………………………………………..…3
4. Working principles…………………………………………………………………...….4
5. Classification of lifts ……………………………………………………………………..5
6. Material selection……………………………………………………………...…………5
• Hydraulic cylinder………………………………………………………………..5
• Base Platform…………………………………………………………………….5
• Scissors Arms…………………………………………………………………….6
• Top Platform……………………………………………………………………...6
• Movable pin ……………………………………………………………………...6

7. Design calculation of parts……………………… ……………………………………...6
• Scissors Arms…………………………………………………………………….7
• Movable pin ……………………………………………………………………..10
• Top platform …………………………………………………………………….11
• Base………………………………………………………………………………12
• Hydraulic cylinder…………..…………………………………………………...13
8. Advantages……………………………………………………………………………..13
9. Disadvantages ………………………………………………………………………….14
10. Application……………………………………………………………………………...14
11. Conclusion………………………………………………………………………………14
12. References………………………………………………………………………………14

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Hydraulic Scissor lift
Idea generation:

Problem Statement for Concept Generation The customer is in need of a transportation system that
is capable of transferring a load of up to 500 kg of non- fragile farm and garden products from a
tall loading dock to the ground and up to a distance of 10 m across a slightly inclined (~10°) gravel
parking lot.

To store items at more height and less area in warehouses. The system needs to be able to function
in all-weather conditions and safe in terms of tip -over and structural failure, and easily operated
by a single employee. with the limitations encountered in the use of ropes, ladders, scaffold and
mechanical lifts in getting to elevated height such as the amount of load to be carried,
conformability, time consumption, much energy expended etc. the idea of a lift which will
overcome the above stated limitations is used.



Selection of idea:

The objective of this project is to develop a cart-like system that will allow the company, White’s
Mill, to move multiple bags of products (up to 500 kg) from a 1 m tall storage dock and across a
gravel driveway, to their customer’s vehicles in a less strenuous but more efficient way than simply
carrying the products by hand. There are other lifts but this lift has very unique and compact design
which can be used in small area. We have selected here hydraulic scissor lift because it generates
less noise and give high efficiency.


Introduction:

A hydraulic driven Scissor lift is a mechanical gadget utilized for different applications for lifting
of the loads and weights to a height or level. A lift table is characterized as a scissor lift used to
stack, raise or lower, pass on or potentially move material between at least two rises.
Scissor lifts are a type of aerial work platform (AWP) or mobile elevated work platform (MEWP),
used to provide temporary access at height. They are commonly used for temporary maintenance
purposes and to undertake construction work. Charles Larson, of the USA, first patented the scissor
lift in 1963, but he did not invent it. It was invented by John W Parker of California.

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The main objective of the devices used for lifting purposes is to make the table adjustable to a
desired height. Unlike some other types of elevated platform, S cissor lift platforms can only move
vertically, elevated by linked, folding supports in an ‘X’ pattern, known as a pantograph. Pressure
is applied to the external side of the lowest set of supports, which elongates the crises -crossing
supports and raises the platform.

Working principle:

Hydraulic lift operated by hydraulic actuator which works Pascal law. This application uses
principles of force multiplication and with a small applied force; one can generate a higher force.


Pressure is the applied force over a surface. It is calculated as;

P=F/A

Where of is the applied force and A is the surface area.

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Types of scissor lifts:

• Vehicle Mounted Scissor Lifts
• Twin Scissors Lifts
• Pit Mounted Scissor Lifts
• Semi-Electric Mobile Scissor Lifts
• Tandem Scissors Lifts
• Self Propelled Scissor Lifts
• Floor Mounted Scissor Lifts

Design specification:

Full height of lift 1.050 m
Lift capacity 500 kg
Breath of lift 810 mm
Length of lift 780 mm
Weight of lift 874.65 kg

Material Selection:

It is necessary to evaluate the particular type of forces imposed on components with a view to
determining the exact mechanical properties and necessary material for each equipment. A very
brief analysis of each component follows thus:
1. Scissors arms
2. Hydraulic cylinder
3. Top plat form
4. Base plat form
5. Movable pin

Hydraulic Cylinder: This part is considered as a support with both ends pinned. It is subjected to
direct compressive force which imposes a bending stress which may cause buckling of the
component. It is also subjected to internal compressive pressure which generates circumferential
and longitudinal stresses all around the wall thickness. Hence necessary material property must
include strength, ductility, toughness and hardness. The recommended material is mild steel.

Pin: movable pins are used to joint link with each other and with platform and base. These pins
are generally made up of mild steel. Pins designed under shearing criteria. Movable pin is very
important part of scissor lift.

Base Platform: This component is subjected to the weight of the top plat form and the scissors
arms. It is also responsible for the stability of the whole assembly. Therefore strength, hardness
and stiffness are needed mechanical properties. Grey Cast iron is used.

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Scissors Arms: This part is subjected to buckling load and bending load tending to break or cause
bending of the components. Hence based on strength, stiffness, plasticity and hardness. A
recommended material is mild steel.
Top Platform: This component is subjected to the weight of the work person and his equipment,
hence strength is required, the frame of the plat form is 2024 ALCLAD (Aluminum Alloy) .


Design calculation of parts:
Mass to be put on top platform: 500 kg
∴By taking F.O.S = 1.5
∴(500)∙(1.5) =750 kg ≅765kg

Mass of top frame = 23 kg
Mass of each link = 5kg
∴ mass of all links = (5*8) = 40 kg

Mass of cylinder mounting link = 6.5 kg
Mass of hydraulic cylinder = 8.150 kg

Total mass of assembly = 23+800+40+4+8.150 = 874.65 kg


Weight of whole assembly = (875.15)∙(9.81)=8580.316 ????????????

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Scissor arm:


Now Let H
y0 = Mass applied on the lift = 800kg
B = Mass of the lift which the cylinder needs to lift = 22.5+40+4+8.150 =74.65 kg
H
yi=Total weight =8580.316N

Link Design:

For the link design it has been considered that, the entire load is acting on half of the link length.

Length of the entire link = 720mm.
Length of the link considered as the beam for the calculation purpose = 360mm.

The load pattern on the top platf orm is considered to be U.D.L. Hence, the load pattern on the link
is uniformly varying load (U.V.L.) due to its inclination with horizontal. The calculation is done
for the link in shut height position, i.e., when the angle made by the links with hor izontal i s 200.

The length of the pin from the intermediate pin to the bottom roller is considered as a beam. The
forces acting on the beam are:

 The reaction offered by the base to the roller, RA resolved into 2 components.
 The reactions offered by the intermediate pin, HB, VB.
 The force due to (Payload + Platform weight) resolved into two components, along the
length of the link and perpendicular to the length of the link.

W = force per unit length of the beam can be evaluated as follows,

As the load pattern of U.V.L. is a triangle, we can say,

W (total force perpendicular to the link) = (1/2) ×base×w
H
yi = 8580.316N

H
yi
4
= 2145.079N

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2145.079 cos (20°) = 2015.714N
2145.079 sin (20°)= 733.66N

Now, 2015.714 = (
1
2
) ×360×W
W = 11.918 N/mm

Taking moment about point A,

V
B×360 – [(2015.714×360 × (2/3)] = 0
V
B = 1343.089N

∑F
y = 0, gives
V
B + [????????????
1×cos (20°)] – 2015.714=0

Putting value of V from equation (1) in equation (2), we get,
1343.809+????????????
????????????cos(20°)– 2015.714 = 0

Therefore, ????????????
????????????= 715.026 N
????????????
????????????cos (20°)= 671.904 N
????????????
????????????sin (20°) = 244.55N

∑F
x = 0, gives
????????????
????????????sin (20°) +733.63 = ????????????
????????????
Therefore, ????????????
???????????? = 489.08N

Where, M = Maximum Bending moment on the link considered as beam.
Y = distance of the neutral axis from the farthest fiber = h/2.

????????????
???????????? = allowable bending stress =
????????????????????????????????????
????????????.????????????.????????????
=
250
4
= 62.5 MPa
I = Moment of Inertia of the link c/s about the X-X (horizontal) axis =
????????????ℎ
3
12

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Where, b = width of the link
h = thickness of the link





h

b

Now the maximum bending moment is at the point of zero shear force. And Maximum bending
moment is given by
(????????????×????????????
2
)
(9√3)
.

????????????
???????????????????????????????????? =
(11.198×3602)
(9√3)

= 93098.423 N.mm

Substituting in
????????????
(????????????)
=
????????????
????????????
(????????????)


Assume Y=h/2 and b=4h ∴ h=13.07
Rounding th e value to available dimensions are h =15mm and b=60mm


Pin:

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FBD of pin:

????????????
???????????????????????????????????????????????????????????????????????????????????????????????????????????? =
0.5 ×????????????
????????????????????????
????????????.????????????.????????????
=
0.5 ×380
????????????.????????????.????????????
= 63.33 Mpa

63.33 = 4*[
????????????
2????????????????????????
2
]
= 10.76mm

D = 12 mm………. selecting standard value .



Top platform:

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FBD for top platform:



Design of the Top Platform:

Material is 2024 ALCLAD (Aluminum Alloy)
????????????
(????????????)
=
????????????
????????????
(????????????)
; I =
????????????????????????
3
12
; y =
????????????
2


????????????=
(????????????×????????????
2
)
8
=
(800×9.8×(810)
2
)
8 = 794610 ????????????∙????????????????????????

????????????????????????????????????
????????????.????????????.????????????.
= ????????????
????????????????????????????????????=
68.94
2
=34.47
????????????
????????????????????????
2� ;I =
????????????????????????
3
12


????????????
????????????????????????????????????????????????????????????????????????=
????????????∗????????????
(????????????)
=
????????????∗
????????????
2
????????????????????????
3
12


∴????????????
2
=176.89

∴????????????=13.30 ????????????????????????

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Also checking for deflection:
∆
(????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????) =
????????????×????????????×????????????
????????????
????????????????????????????????????×????????????×????????????


∆
(????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????) =
????????????????????????????????????
????????????????????????????????????
=????????????.???????????? ????????????????????????

∆=
????????????×(????????????????????????????????????×????????????.????????????)×(????????????????????????????????????)
????????????
????????????????????????????????????×????????????????????????,????????????????????????????????????×
????????????????????????????????????(????????????????????????.????????????????????????)
????????????
????????????????????????
=????????????.????????????????????????????????????????????????

∆<∆
(????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????)

E = 73Gpa , ????????????
????????????=68.94
????????????
????????????????????????
2�
????????????=
???????????????????????????????????????????????? ???????????????????????? ???????????????????????????????????? ????????????????????????????????????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????? ???????????????????????? ???????????????????????????????????? ???????????????????????????????????? ????????????????????????????????????????????????????????????????????????????????????????????????


Volume = 0.780×0.810×0.0133=0.008408971????????????
3


Mass = ????????????×????????????=22.704Kg

So, we take mass as 23 Kg which was our assumption.



Base:


Design of Base:

F = 8580.316 N,
D(thickness) = 25mm, b =48 mm
F.O.S. = 5


????????????
???????????????????????? = 344.735 Mpa

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????????????
????????????????????????????????????=
????????????
????????????????????????×0.5
5

????????????
????????????????????????????????????=34.47
????????????
????????????????????????
2�

????????????
????????????ℎ???????????????????????????????????? =
????????????
????????????
=
8580.316
25×48


????????????
????????????ℎ???????????????????????????????????? =7.15
????????????
????????????????????????
2�

∴ ????????????
????????????ℎ????????????????????????????????????<????????????
????????????????????????????????????



Hydraulic cylinder:






Now the maximum force will act on the cylinder when the cylinder is in shut down position i.e.,
when the scissor links are closed.

For calculations we will consider α=30°
Thus, substituting α=30° in eqn (1), we get F=8580.316 N

Force = 8580.316 N
????????????
????????????=2×????????????
????????????=
????????????
64
×????????????
4

E=250 Gpa (for mild steel)
Buckling formula, F =
????????????
2
????????????????????????
????????????
????????????
2

Diameter of cylinder: d = 17.4 mm ≅ 18 mm

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A=Area of cylinder side will be = 254.46 ????????????????????????
2


p =
????????????
????????????
=
8580.316
254.46
= 33.71 N/????????????????????????
2

????????????
????????????????????????=440 ???????????????????????????????????? ???????????????????????????????????? ???????????????????????????????????????????????? ????????????????????????????????????????????????????????????
????????????
????????????=
????????????
????????????????????????
????????????????????????????????????

Considering, FOS = 3
????????????=146.66 ????????????????????????????????????
Now, t =
????????????
????????????????????????
????????????
2????????????????????????
, t = 3.44 ≅ 4 ????????????????????????

Advantages:
• because of robust it is easy to handle
• compact design,
• more flexibility and mobility
• easy to adjust at any height
• used on uneven surfaces
• most economical, dependable, and versatile method of lifting heavy loads
• because of few moving parts it provides high stability
• less noise
• high efficiency compared to other lifts

Disadvantages:
• high initial cost
• take more time for lifting heavy load
• during high load, hydraulic
• motor generate more heat

Application:

Modern scissor lifts and tilters are utilized for a wide variety of applications in numerous ventures
which incorporate manufacturing, warehousing, schools, grocery distribution, military, medical
and printing, construction engineering, to place heavy obje cts above ground for working.
Conclusion:
Thus, a scissor lift gives most monetary trustworthy and flexible techniques for lifting loads; it has
some moving parts which may just require oil for lubrication. This lift table raises load easily to
any desired height. The scissor lift can be utilized in mix with any of utilizations, for example,
pneumatic, hydraulic, mechanical, and so on. Lift tables may consolidate rotating stages (manual
or controlled); tilt stages, and so forth, as a part of design.

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References
https://www.researchgate.net/publication/329586031_Design_and_Fabrication_of_
Hydraulic_Scissor_Lift
https://www.hydraulicequipment.in/hydraulic-cylinders.html
https://www.parker.com/parkerimages/mobilecylinder/cat/english/0001q.pdf
https://www.colle.eu/en/rent/working-at-heights/aerial-lifts/scissor -lifts/s280-25 -
d4wds-ab-lift
https://www.hunker.com/12460647/how-to-calculate -scissor-lift-length
https://www.google.com/url?sa=t&source=web&rct=j&url=http://pnrsolution.org/
Datacenter/Vol3/Issue2/309.pdf&ved=2ahUKEwiFzrqbhbntAhU-
63MBHZb7AqAQFjABegQIARAF&usg=AOvVaw34ablxXdJvsS6AsD5Zj0T0&c
shid=1607247013356
https://www.google.com/url?sa=t&source=web&rct=j&url=https://pdfs.semanticsc
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UK7GDsXf8gypQBNyiF
https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.addybuild
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work_23.html%3Fm%3D1&ved=2ahUKEwign-
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9fc4ppwqGn8PUgU
https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.heavyloa
ded.com/scissor-
lift/&ved=2ahUKEwjCtdzTiLntAhVHQH0KHTjqB5YQFjAcegQIEBAB&usg=A
OvVaw0ewj5BtuCboXvCM0szefOt
https://en.wikipedia.org/wiki/Aerial_work_platform