Question

Question 5 -a: For the shaft application in Question 2, the input shaft (2) is driven at a constant speed of 150 rev/min. Obtain a basic load rating for a ball bearing at A and B in shaft 1 for a life of 10 Khr with a combined reliability of R = 0.95 . Also select a proper 02-Series Deep-Groove ball bearing according to your calculations. Assume the distribution data from manufacturer 2, in table 11-6 in the textbook. 800 mm 200 mm B y: 1 200 mm T=500 N.m F = aj-zak Lo E F 400 mm 2 D Question 5 - b: Determine the loads on your bearings in your project.

Answer 1

of dution, Gurn that speed of shaft (2) N2 = 150 upm, Lize of Bearing Lion 10 khen, Combined Reliability R=0.95, Manufacture4 - 2 diameter of Guam 'C' di = 200 mm diameter of Guan 'o' d2 = 400 mm, # speed of shaft (1) is given by formula of velocity Ratio # NI d2 di Ni 400 N2 150 200 N 300 pm 19 Assuming presswie angle geau 'e' to be 20 ie (8=20 Tangential Force on Gear ? P T x2 de * Pt - 500 x103 x2 200 Pz= 5000 Radial force on Gear 'C' Pe= Pe- tan a u = 5000 · tau 20 o Py= 1819.85 N # Calculation you reactim at beaving a g 8 Let Row to Rem be the houizentow Meaction at Beaning A' 48' aloo Rav & Rev be the vertical reaction at bearing 'A' 8'8".

# Horizontal force dagram Pe= 5000N A 800 mm B C 200 1011 RBH RAH taking summation of all fouces, we have RAH + RBH - 5000 eq 0 taking moment about point 'A' we have RBH X 1000 = 5000 x 800 RoH = 4000 N then from eqo we have RAH = 1000 N #vertical Force Diagram P= 1819.85N A 800 min B C 200 mm RAV Rov Summation of all forces, we have Rart RBV = 1819.85 N eq @ 'B' we have RBV x 1000 1819.85 X 800 RBV 1455.88 N then frem eq ③ we have RAV = 363.97 N Resultant Radial force on RA= V(RAH)² + (RAV)? Bearing A is

RAE (1000)* + (363.94? RA= 1064.177 N # Resultant Radial souce on Bearing 'B' is RB = f Ron + Rov + R$ =(4000) + (1455.88)? Ro=4256.71 → N o 4.459 # From Data design Book jou manufacturer (2) we howe x = 0.0 2 , b= 1:483, L2=10 revolution Dimensionles Design Life is given ays - 60x nyx Lion 60x 300 x 10 x103 xo La 106 x = D = 180 aldo, individual Meliability Ro - TR a Ro=10.95 Ro= 0.975 176 x= xo + (0-xo)-(4-Ro) > x = 0.02 +(4:459-0-02) · (1-0.975) 1.483 + x= 0.389 # Load Rating for bearing 'À is given as RA Ha Co [40] where, a=3 you ball bearing. X =

C10 = 1064-177 1/3 180 0.389 KN لول C20 = 8231-042 Newton C10 = 8.231 #From Data Design Book Suitable ball bearing of 02- svrier Deep Groove (un ref. Clo = 8,231 KN) Bore OD = 40 mm, widen b=12 mm fillet 4= 0.6mm, Shouldeu dia. do= 19.5 mm, dw= 34 nm , Answer # Load Rating you bearing in is calculated alle /a d = 17 mm where, a=3 Co= Rox[o bearing a Cao for ball 1/3 4256.71 180 0.389 10 = С 10 91566.268 N ( 10 = 91.566 KN Bau Bearing is ús = 91-566 KN # From Data Design Book, Wuilable 02- series Deep Goove ( in Reference Co= KN) Bore widithi b= 30 mm, fillet u= 2 mm Shoulder dia ds = 104 mm ds = 104 mm, dx= 146 mm. Anuru de 90 mm OD = 160 mm 1

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