It is an Powerpoint on Hydraulic Scissor Lift used for material handling in industries. PPT consist of design procedure, design consideration and calculation of lift.

1. Hydraulic Scissor Lift
By: www.mechaholic.com

2. 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.

3. Design Procedure
Design a hydraulic cylinder assembly to lift a load of
500 kg
Scissor Type Mechanism
Bending, Crushing, Shearing and Buckling

4. 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.

5. Design of element

6. 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

7. 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

8. 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.

9. 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˚ .

10.  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.

11.  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

12. 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)

13. 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.

14. 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)

15. 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

16. 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)

17.  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

18. 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)

19. Detail Drawing
 3D – CAD Model

20. Production
Before
Lifting
After
Lifting

21. Hydraulic Scissor Android
Application