The gear box system for the motor can be designed as-
Gear calculations
Calculations for gear A-
(A) = 2 /
= 1845.6 N
= 20 ? normal pressure angle
Ψ = 15 ? helix angle
= 20.65 ? transverse pitch angle
(A) = (A) tan
= 1842 tan 20.65
= 695 N
(A) = (A) tan Ψ
= 1842 tan15
= 494 N
Forces on gear B-
(B) = 2 /
=
= 2244 N
(B) = (B) tan
= 2244 tan 20.65
= 846 N
(B) = (B) tan Ψ
= 2244 tan15
= 602 N
Forces on gear C-
(C) = 2 /
=
= 5608 N
(C) = (C) tan
= 5608 tan 20.65
= 2114 N
(C) = (C) tan Ψ
= 5608 tan15
= 1503 N
Forces on gear D-
(D) = 2 /
= 4589 N
(D) = (D) tan
= 4589 tan 20.65
= 1730 N
(D) = (D) tan Ψ
= 4589 tan15
= 1230 N
Calculations regarding the speed and number of teeth on the gears
U =
U = 9.34
Where:
Reduction ratio for second or low speed stage
This is also given as:
So high speed stage reduction ratio becomes
Given the gear ratio the number of teeth for gear 1 and 2 can be determined. The pinion, gear 1, should have its number of teeth minimized to reduce the overall size of the gearbox. The number of teeth is assumed to be 18 as it is the minimum number required to avoid undercutting for spur gears. If helical gears are selected then with further calculations the minimum number of teeth may be reduced.
Number of teeth in gear 1
Number of teeth in gear 2
Altered high speed reduction ratio
High Speed = = = 3.95
Low Speed = 0.88 = 0.88 = 2.68
Input Shaft Speed-
= = 1460 rpm
Intermediate Shaft Speed-
= = = = 369.62 rpm
Output Shaft Speed-
= = = = 137.91 rpm
Input Shaft Speed-
= = 152.81 rad/s
Intermediate Shaft Speed-
= = 38.68 rad/s
Output Shaft Speed-
= = 14.43 rad/s
Input Shaft Torque-
= = = 49.08 Nm
Intermediate Shaft Torque-
= = = 152.81 Nm
Output Shaft Torque-
= = = 519.75 Nm
Calculations for the bending stress on the gear
- = πD / (60 x 1000)
= π x 54 x 1460 / 60000
= 4.125 m/s
= 50 / [50 + ]
= 50/ [50 + ]
= 0.677
= - 0.038 + 0.0005d where F = face width
= - 0.038 + 0.0005(52.20) d = diametrical pitch
= 0.070
= 1 + +
= 1 + 0.070 + 0.164
= 1.32
= (. / b. J).....
where = 1845 N
= 0.35
b = 40 mm
J = 0.5
= 1.5
= 1.0
= 1.25
= 1.0
= 0.65
= (. / b. J).....
= ((1843×0.35)/(40×0.5))×1.5×1×1.25×1×0.677
= 40.94 MPa
References
Kalogirou, S. A. (1996). Design and construction of a one-axis sun-tracking system. Solar Energy, 57(6), 465-469.
Ponniran, A., Hashim, A., & Joret, A. (2011). A design of low power single axis solar tracking system regardless of motor speed. International Journal of Integrated Engineering, 3(2).
Khan, M. T. A., Tanzil, S. S., Rahman, R., & Alam, S. S. (2010, December). Design and construction of an automatic solar tracking system. In Electrical and Computer Engineering (ICECE), 2010 International Conference on (pp. 326-329). IEEE.
Tudorache, T., & Kreindler, L. (2010). Design of a solar tracker system for PV power plants. Acta Polytechnica Hungarica, 7(1), 23-39.
