Question

3. A 33-KW motor, using a set of gears, drives a shaft at a speed n, as illustredted in Fig. P3, with a safety factor = 2.2 what will be the total angle of twist between A and B. Given: G = 82 GPa, Typ = 210 MPa, n = 1500 rpm? (a) 16.64", (b) 18.64", (c) 14.64°9, (d) 10.64", (e) none of above -2m- 5 m с Motor 25 kW 8 kW off off 33 kW Fig. P3

Answer 1

SOI?: FBD! TA To E B TÅ C Tc = 6o pc 60 x(33 x103) = 210.08 Nom ZIN 2 ITXI 500 TAS 60 PA 2 IN 60x68x103) = 50.93 Nom 2 TX1500 Te = 60% (25x103) 60 PB 2 159.15 dom 2TX 1500 AC and CB : FBOOM TAT [ TB E TA Ti For Equili librium condition; i++) ΣΤ =O TA - T, -0 → Ti=TA - TAC = 50.93 xlor-mom -TB Again, TA-Tc =-TB TB = - ( TA-TC) 3 (0) (TA -Tc) TB C = (50.93- 210.08) 10, TBC -159.15 x103 N-mm TAC LAC O AB = O Act OCR + TCB Lus GI o GJ inertia Where TALLAC I= polar moment f dy 32 16 Tmax Since CB section experienced highest torque; Typ 16 / Tool FS CS Scanned with CamScanner 3 ہے Tid 3

210 16 x 159.15 x 103 2.2 3 Td do 20.401 mm into tq o, putting OAB (50.93 x 10%)2000 (-159.15X10%) *5000 + 82 x103 x TT 32 *(20.401) 82x10x (20.400) 32 OAB a - 0.49758 rad XIBO DAB -28.50q° 28.50 99 so option ® done of above is correct. (Thanuyou) CS Scanned with CamScanner