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Q2) The solid shaft having a diameter d is fixed at end D. For a given rotation angle at C (clockwise when viewed from righta=12m b=13m c=17m T1=549(N.m) e=0.10(rad) d=100mm G=16Gpa

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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 :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| 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*i32 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

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a=12m b=13m c=17m T1=549(N.m) e=0.10(rad) d=100mm G=16Gpa Q2) The solid shaft having a diameter d is...
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