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