Total Weight, W = Force + Mass of Plate + Mass of Top Beam (x2)
= 3000 + 66.643 + 75.238
= 3141.88 kg
Weight of Scissors, Wo = 325.67kg
Using Scissor Jack Equation
F
Where,
F = Force provide by Jack-screw
W = Combined weight of the payload and load platform
Wo=Combined weight of the two scissor aims themselves
= Angle between the scissor aims and the horizontal
Load Force before lifting Load Force after lifting,
431.79 k N 3.2 k N
The distance (length) of screw moved = 2265 - 1539.5535 = 725.45mm
Assume using 15 Pitches Screw, so
Pitch = 15 rev/inch= 0.5906 rev/mm
Assume 2 min (120 sec) to rise up 1.5m,
Load speed, =1088.17 mm/sec
Motor speed,
= 362.724 mm / min = 214.21 rpm
Drive Torque,
= 42.838 Nm
Horse Power = = 0.140358 HP = 960.95 Watts
From the table show that the power inverse proportional to the time. Therefore, the minimum of the power motor is 650 Watts, then lifting jack only can be achieved safely lifting a load of 3 metric tonnes to a height of 1.5 meters, and to be done within 1-3 minutes.
Electrical and Motor Component
From the reference sources we have investigated that servo motor is more useful and fulfil our design, especially need to safely lifting a load of 3 metric tonnes to a height of 1.5 metres. Beside, servomotor are powerful, reliable and can spin continuously at a fix rate, since the power requirement of the design was minimum 650W if need to be done within 1-3 minutes. Base on SER servomotor, the motor of SER 39* is around 0.3 -1 K w. So, the lifting jack can be achieved by using the SER 39* servomotor.
Visual basic 6.0 is used here to allow user to control the height and speed by enter the input value. When the Data register of the parallel port receive a signal from the input of visual basic, it will send to the motor control circuit which has a motor connected. By sending a high state input, the motor will rotate. So it will achieve the adjustment of height of the lifting jack. So the flow chart of servomotor controller is show on RHS..
Structure Analysis
The size of the support beam scissor: 2441 mm long x 20 mm wide x 328mm high
For to the Material Safety Major,
Since the design module of lifting jack is using Aluminium, so the Yield Stress of Aluminium = 9.5 x 107 Nm2
= 3.2 x 107
The Principle Stress from structure analysis = 3.4 x 107 > 3.2 x 107 [PROVE]
Beside, form the graph show that the displacement of the scissor is just 5.14mm
Material Structural Analysis Comparison
The requirement mass of the lifting jack is 750 kg. So, we assume the mass of the each scissor not more than 80kg.
So, compare with 3 of the material above, from the table show that the Principle Stress of Aluminium was achieved the range of safety major. Although the price of the steel is much cheaper than Aluminium, but it’s a big gap between principle stress and safety major. Beside, titanium is more expensive compare with other material. Therefore, material of aluminium were more stability and economic between Steel and Aluminium.
Material
Aluminium has about one-third the and of . It is , and easily , , and .
Beside, the of pure aluminium is 7–11 . However, have yield strengths ranging from 100MPa to 200 MPa. It’s much higher than pure aluminium. Therefore, our mechanical design module of lifting jack would use Al – Zn – Mg alloy (7075).
Cash Analysis
Base on the market of the aluminum price (www.metalprices.com) , 1 kg of STEEL = £ 0.50 - 0.60
Assembly Drawing
In Operation
In Storage
Technical Drawing of Part
Crossbar
SER 39+ Servomotor
