I Page Noul 12m B 2 13mC T=5u9 17m et TZ d- loomm con m m2 cis G= 16 Gpa = 16*103 MPa = 16+10 N ** Notation:- A is free end :- . ( By Right hand S; H os tres ve thumbrule TM A B C D a TM 549TI ITZ *12 13 17 Angle of twist at point of Oat corresponding to the fixed end. Oc= 0 DC7 point to ist tre Angle of twist is the cow from Right TM.. view AB=0 -vec.co from Right view TMBC= -549'N-m T.M CScanned with CamScanner ?jloc= 0.1 Ive cw from Right hand
Page No-2 a = oor t ol Angle of twist at point is from tixed and to point i Oor = (-549+[2]*17 TootCD 678 = 0.1 16*109* TT 52 c549+T2) *17 = 15,707.96 -549 +T2 = 923.99 Tz = 1472.99 Nam Torque at point "C" A is a fixed end Tg =?: is 12- 13- 17 , + (1) o 12 | Тр ТА 4- STARTO 7 reactions due to Ti AT2 , TA = -Tx Ci3 +17) +T2*(17) (12+17+13) TB = -51*12+ Tz+(12+13) 12 +17+13 TA = -16,470 +17 T2. Tos -12117 42 42 CScanned with CamScanner
| Page No-3) o it O₂ + O₂ =0; In fixed shaft both sidel * Ti= TA ; T2 = (TA - 549) ; T3 = (TA-949+72) 8,2 tomt 2 ; 0 20 Tz*i3 ; @ from eigenes Oc from a = Oc= Our + Osc 0.1 ;! - Toca Lec OAB = - TAK LAB OBC = 1. GJ TBC = TA - 549 60g gu TRE SF1s, 4+0 + 14 727+ 12%SE 15utot ++549]+13% 16x 109 * I * 0.14 =lL 42 32 2705. 244857L-5097.85 +52672 0. 13 -197646 + 204 Tz -15,470*13)–(549*42*13)+(17*13* T2) 42*16*10% * I *0.14 32 659734.4573 = -711504 +425T, l Tz= 3226.45 Nom Torque at "C" while a fixed CSscanned with CamScanner
2 page no-u) © shear stress in BC while A is fixed - TETA, T2=TA-TI, TBC = (TA-549) ; (TBC=12} - 6.45 Ta - [hts to 10 4 31643] TA= 913.80 Nom TBC = 913.80-549 TBC = 364.80 Nm . 16 TBC e shear stress in BC Td3 16* 364.8 T* 0.13 . N m2 Sc = 1857.911.*10♡ m2 I ZBC = 1.857 MPa CSScanned with CamScanner