HO CHI MINHUNIVERSITY OF TECHNOLOGYDESIGN PROJECT Lecturer: Le Thanh LongStudent: Phan Cuong Quoc KhanhMSSV: 2153445... StudentPhan Cuong Quoc KhanhStudent: Phan Cuong Quoc Khanh - 21534
SELECTING ELECTRIC MOTOR AND SPEED RATIO DISTRIBUTION
Transmission Efficiency
In which: η gear : efficiency of gear drive η V−belt : efficiency of V - belt drive η bearing : efficiency of bearing η coupling : efficiency of coupling η chain drive : efficiency of chain
Select desired efficiencies: η gear : 0.97 η V−belt :0.96 η bearing :0.99 η coupling :1
Load calculation
Required power: P ct =P td η=3.890.85=4.57kW
Preliminary rotation speed
Working shaft rotation speed: n lv `×1000×0.87
Whole system transmission ratio: u t =u 2 gear × u V−belt =4 2 ×2 32 Rotation speed of the motor shaft: n preliminary =n lv × u t @×32 1280= (rpm)
Motor selection
According to the calculation, select motor with P ≥ P ct and n dc ≈ n preliminary
Therefore, select motor 4A112M4Y3 with 1425 rpm and 5.5 kW of power.
Transmission ratio
Overall transmission ratio: u sys =n dc n w
40 5.6 u gear box , we have transmission ratio of V-drive as: u V−belt = u sys u gear box
Load on each shaft
Load on belt conveyor (shaft 4): P 5= P η coupling ×η bearing
Load on speed reducer’s driven gear shaft (shaft 3): P 4= P 5 η bearing
=4.34 0.99=4.38kW Load on speed reducer’s driving gear shaft (shaft 2):
Load on each speed reducer’s driven gear shaft (shaft 2): P 2=P 3
2 =2.28kW Load on each speed reducer’s driving gear shaft (shaft 1):
Load on motor shaft: P motor = 2P 1 η bearing ×η V−belt
Torque and rotation speed on each shaft
Rotation speed on shaft 1: n 1= n dc u V−belt 25
Rotation speed on shaft 2: n 2= n 1 u gear 1 q2.5 4.915rpm
Rotation speed on shaft 3: n 3= n 2 u gear 2
5 3.26Erpm Torque of motor shaft:
Characteristic table
Shaft Motor Shaft 1 Shaft 2 Shaft 3 End effector
DESIGNING V-BELT DRIVE
Specification
Material
Select rubber for belt material.
Diameter d 1 and d 2 of small pulley
Rotation speed with respect to chosen d 1 : v 1=π d 1 n
Choose ε=0.01, we have diameter d 2 as: d 2 = u d 1
Center distance calculation
2 +(200 100− ) 2 4×240 1.66mm Select the standard length as 1600 mm.
Number of rotations in each second
According to table (4.15) we check the rotation per second:, i=v l=7.45
Calculating again the distance due to the standard l
Now, we calculate again the distance a following the standard l00mm
Contact angle α of belt and selection of belt thickness
Number of belts
-With v=7.45m/s, d 10mm, choose [ P 0 ] =1.85kW (Table 4.19)
The width of belt, due to 4.17 and table 4.21:
- The diameter of belt: d a =d+2h 00+2×3.36.6mm
Determine the initial tension and the force acting on the shaft
According to table (4.21), the force acting on the shaft is:
Calculating the belt durability
Maximum stress in 1 belt: σ max =σ 1+σ v +σ ul =σ 0+0.5σ t +σ v +σ ul =F 0
DESIGNING HELICAL GEAR TRANSMISSION
Specification
One-directional rotations, two shifts, low level load (279 days of working per year, 8 hours per shift)
Material
We choose the material for the pair of helical gear:
Small pulley: Steel 45, annealing at HB'0, having [ σ b 1 ] 0( MPa);
Big pulley: Steel 45, annealing at HB$0, having [ σ b 2 ] u0( MPa); [ σ CH 2 ] E0 (MPa)
Allowable stress
According to the formula 6.1a and 6.2b
In basic, we choose: { Z Y R R × Z ×Y V S × K × K xF xH =1 =1
Where: ¿: respectively are bending stress and contact stress allowing base cycle, their values are founded at table 6.2: σ ¿¿
S F , S H : respectively are safety coefficient when calculating bending and contacting searched in table 6.2:
K FC : coefficient affects the load, take K Fc =1 (Load is put on one side)
K HL , K FL : Age coefficient, determined by formula 6.3 and 6.4:
K HL = m H √ N N HO HE ; K FL = m F √ N N FO FE
Where: m H , m F : the degree of the curve in each time contacting and bending with HB