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Hydraulic Scissor Lift
By: www.mechaholic.com
Introduction
Hydraulic lift is a material handling device
Hydraulic lift is a type of machine that uses a
hydraulic cylinder to lift and lower objects by
applying relatively small force compared to the
weight of the object to be lifted.
It’s working is based on Pascal’s Law.
Design Procedure
Design a hydraulic cylinder assembly to lift a load of
500 kg
Scissor Type Mechanism
Bending, Crushing, Shearing and Buckling
Material Selection
Scissors Arms: Stainless steel.
Subjected to buckling load and bending load tending to break or cause bending of the components.
Hydraulic Cylinder: Mild steel
Subjected to direct compressive force, bending stress, internal compressive pressure.
Top Platform: Mild steel and the base is wood
Subjected to the weight.
Base Platform: Mild steel
Subjected to the weight of the top plat form and the scissors arms.
Design of element
Design Theory of Calculation
In this section all design concepts developed
are discussed and based on evaluation criteria
and process developed, and a final here
modified to further enhance the functionality of
the design. Considerations made during the
design and fabrication of a single acting
cylinder is as follows:
a. Functionality of the design
b. Manufacturability
c. Economic availability. i.e. General cost of
material and fabrication techniques employed
Total load calculation
The hydraulic cylinder is mounted in horizontal
position. The total load acting on the cylinder consists
of:
Mass to be put on lift : 500 kg ;Taking FOS = 1.5 for
= 750 kg rounding the mass to 800kg
Mass of top frame= 22.5 kg
Mass of each link = 40kg
Mass of links of cylinder mounting=4kg
Mass of cylinder=8.150kg
Total Mass = 874.65 kg
Total load = 8580.316N
Scissors lift calculations
For a scissor lift Force required to lift the load is dependent on, Angle of link with horizontal Mounting of cylinder on the links Length of link.
Formula used
S= a2 + L2 -2aL*cos α
Where ,
S = Distance between end points of cylinder
L= length of link = 0.6 m
α = angle of cylinder with horizontal.
Load Calculation of Link
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 (L) = 720mm.
Length of the link considered as the beam for the
calculation purpose = 360mm.
The load pattern on the top platform 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
horizontal is 20˚ .
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
Hyi=8580.316N ;
Hyi/4= (8580.316/4) = 2145.079N
2145.079 cos(20) = 2015.714N
2145.079 sin (20) = 733.66N
Now, 2015.714=(1/2)*360*W
W=11.918N/mm
DESIGN OF LINK WITH BUCKLING
As indicated before ,buckling of link with one
end fixed and other hinged :-
b=4h
E=2*105 N/m2 l=0.72m
So we get :
b=1.2m h=30mm
As
:
(Stainless
steel)
Checking in bending
Taking moment about point A,
VB * 360 – [(2015.714X*360 *(2/3))]
Therefore, VB = 1343.089N
Therefore RA = 715.026N
So, RA cos (20◦) = 671.904 N
RA sin (20◦) = 244.55N
= 62.5 Mpa,
Thus it is safe for bending.
Design of Pin
= 0.5*505/FOS
= 126.25 Mpa
126.25 = 4*F/3.14*D2 *2 = 13.15mm
D = 14 mm____________Selecting standard
value
Sut = 505 Mpa(Mild Steel)
Design of intermediate Link
Force acting on intermediate link :-
Thus =11512.48N
Now by taking moment at one end and
considering length=1m,thus
M= 11512.48Nmm
Again = 62.5Mpa
Therefore d=12cm
Design Of Hydraulic Cylinder
F = 8580.316N (Previous Calculation)
One end fixed and other free :-
L=720mm
Le=2L
Buckling
Formula
Diameter of Cylinder:
d=17.4mm → D=2d=34.8mm
E = 250 GPa
(Mild
steel)
Area on Cylinder side will be =
951.14mm2
Area on rod side will be= 237.78mm2
Now considering bigger area
Thus P= 12 N/mm2 = 120 bar
Designing of Cylinder
Taking Sut= 440 Mpa
Considering Factor of Safety =3, thus
σ = 146.66 Mpa
Now
t= 1.42 mm (for cylindrical part)
(Mild
steel)
Detail Drawing
3D – CAD Model
Production
Before
Lifting
After
Lifting
Hydraulic Scissor Android
Application