# JEDR Report Conclusion

Amazing Essays
Part 1 JEDR Critique
1.1. Condition Specifications
There is a mistake in the calculation of the assumed wheel circumference. Based on the obtained data from JEDR report, the average wheel diameter is 406 mm and by using the circumference formula, 2πr, the result should be 1.275 m. However, JEDR report says the wheel circumference is 1.2 m. This number should be rounded up to 1.3 m instead of rounded down since this might give critical error in further calculations.
The previous wheel circumference mistake will lead to the miscalculation of the wheel average RPM. Based on the JEDR report, the assumed average speed of a vehicle is 60 km/h or 1000 m/min. Since the wheel circumference is involved in angular velocity calculation, thus, it provides
ω
TB = P_B/ω_B = (75,000 W)/(784.31 RPM) x (1 rev)/(2π rad) x 60 s = 913.16 Nm
TA = 913.16 x (784.31 RPM)/(261.44 RPM) = 2739.45 Nm
TMotor = 2739.45 x (261.44 RPM)/(1785 RPM) = 401.23 Nm

1.4. Pulley and Belt Specification
According to the JEDR report, the pulleys’ size ratio is reversed, instead of 3:1 for pulley A to pulley B ratio, it assumed them as 1:3 for drive pulley is smaller than driven pulley. This mistake will lead to a necessary correction for contact angles. Furthermore, the incorrect RPMs will lead to the miscalculation of the centrifugal force because the velocity is wrong. Also, this cause the values of torque, F1, and F2 need to be recalculated.

1.5. Shaft Layout and Specification
Based on JEDR report drawing, the location of bearing housing on shaft B is incorrect. It is advised to place the bearing after the pulley, instead of between pulley and drum brake since it is contradicting with the first design. Due to this non-identical design, the calculation for bending moment diagrams and shear moment diagrams are also inaccurate. Furthermore, it was observed that the diameter of the shaft has been increased, which is also increase the Factor of Safety.

1.6. Bearing
Design Pulley and Belt System
Contact angle = π + 2 x sin-1((D+d)/2C)
Θ = π + 2 x sin-1((508+191.7)/(2 x 800))
As stated in JEDR, the chosen driving pulley is polyamide A – 3 (table 17-2). t = 3.3 x 10-3 Fa = 18 x 103 γ = 11.4 x 103 N/m3 f = 0.8 efΘ = e0.8 x 4.05 = 25.53
V = (π d x ω)/60 = (π x 0.508 mm x 355.2 RPM)/60 = 9.45 m/s w = γbt = 11.4 x b x 3.3 = 37.62b N/m Fc = (w x V^2)/g = (37.62b x 〖9.45〗^2)/9.81 = 342.46b
Find torque with Ks = 1.4 and nd = 1.1
T = (H x K_s x n_d)/(2π x ω/60) = (45,000 W x 1.4 x 1.1)/(2π x (939.89 RPM)/60) = 704.09 Nm
(F1)a – F2 = 2T/d = (2 x 704.09 Nm)/(0.1917 m) = 7,345.75 N
From table 17-2, the value of Fa = 18 x 103, from table 17-4, Cp = 0.7 and Polyamide belt, Cv = 1
(F1)a = b x Fa x Cp x Cv = b x 18,000 x 0.7 x 1 = 12,600b N
F2 = 12,600b – 7,345.75
((F_1 )_a-F_c)/(F_2-F_c ) ≤ efΘ (12,600b-342.46b)/(12,600b- 7,345.75-342.46b) ≤ 25.53 12,257.54b/(12,257.54b- 7,345.75) ≤ 25.53
12,600b ≤ 312,935b – 187,537 -300,335b ≤ -187,537 b = 0.62 substitute the value of b to Fc:
Fc = 342.36 x 0.62 = 212.26 N (F1)a = 12,600 x 212.26 = 7,812

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