Bearing calculations raz 1150 lbf ray 3567 lbf ra 375

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Bearing calculations: RAz = 115.0 lbf RAy = 356.7 lbf RA = 375 lbf RBz = 1776.0 lbf RBy = 725.3 lbf RB = 1918 lbf FRB = 16 400 lbf Cylindrical roller bearing at right end of shaf C = 18 658 lbf, ID = 1.181 1 in, OD = 2.834 6 in, W = 1.063 in Shoulder diameter = 1.45 in to 1.53 in, and maximum fillet radius = 0.043 in Deep-groove ball bearing at lef end of shaf C = 5058 lbf,ID = 1.000 in, OD = 2.500 in, W = 0.75 in Shoulder diameter = 1.3 in to 1.4 in, and maximum fillet radius = 0.08 in Key Design: Transmitted torque: T = 3240 lbf-in Bore diameters: d3 = d4 = 1.625 in Gear hub lengths: l3 = 1.5 in, l4 = 2.0 in for a shaf diameter of 1.625 in, choose a square key with side dimension t = 3 8 in. Choose 1020 CD material, with Sy = 57 kpsi. The force on the key at the surface of the shaf is F = T/ r = 3240 /1.625/2 = 3988 lbf Checking for failure by crushing, we find the area of one-half the face of the key is used n = Sy/ σ Solving for l gives l = 2Fn/ t Sy 2(3988)(2) /(0.375)(57000) = 0.75 in Since both gears have the same bore diameter and transmit the same torque, the same key specification can be used for both.
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We have chosen the vehicle Volvo C30 EV 2011 model. Following are its major specifications: Design constraints: For the gear reducer layout there are two gear-pairs. By considering fixed number of teeth Z i varying thickness values b i and power delivered P, each gear must satisfy two constraints mentioned The bending and compressive stresses developed in the i th gear-pair are calculated Bending stress developed in the i th gear-pair: Compressive stress developed in the i th gear-pair: The transmission ratio is defined as the ratio of the number of teeth Z wi in wheel to the number of teeth Z pi in pinion The thickness value should lie between lower and upper limit values,
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The maximum gear-ratio r max in any gear-pair must not exceed a limit Number of teeth in each gear pair should be integers and value must be greater than its lower limit. Z wi , Z pi Z ( L) for i = 1,2. Since multiple-criteria are on different scales, to reflect their actual contribution to the multiple- criterion objective function their values have to be normalized to the same scale Two Stages Gearbox details: Wheel radius: The wheel radius is half the wheel diameter, the distance from the centre of the wheel, the axle, to the outer edge. We have chosen wheel of 8inch radius Weight: 3200 lbs | 1451.496 kg Dimensions: Wheel base: 2,640 mm (103.9 in) Length: 4,266 mm (168.0 in)
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Width: 1,782 mm (70.2 in) Height: 1,447 mm (57.0 in) Gear Ratio: Calculation of Transmissibility Ratio: I = (Z2/Z1) = (30/20) = 1.5 Calculation of tangential load: Ft = (K0*103*W) / Vm K0 = 1.5 (for median life) = (1.5*750)/Vm Vm = (П d1N1)/60 = (ПmZ1N1)/ (60*1000) = (П*m*20*300)/ (60*1000) Vm = 0.314m Which implies, Ft = (1.5*750)/Vm = (3582.8)/m Calculation of initial dynamic load: Fd = Ft * Cv Cv = (6 + Vm)/ 6 = 3 (Assume Vm = 12) Fd = (3*3582.8)/m = 10748.4/m Calculation of Beam Strength: b= 10m FB= *σb+ by*Pa = 720*10*m*y*П*m Y=0.1084 FB = 2450.70 m2 Calculation of module: 2450.70m2 = (10748.4)/m Revaluation of Beam strength: FB = 98028 N Ft = 1791.4 NCalculation of Dynamic load:
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Vm = 0.628mm d1 = mZ1 = 2*20= 40mm Fd = 1796.17 N Fs > Fd (Hence design is safe) Fw = d1*Q*Kb = 20*1.2*1.1*20 = 528 N Q = 2(1.5)/ (1.5+1) = 1.2 Free Body diagram , Analysis and pseudo code: An important effect when accelerating or braking is the effect of dynamic weight transfer. The weight distribution dramatically affects the maximum traction force per wheel. This is because
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