Question

Please also explain, thank you!

In the system below, the shaft is solid from A to C, hollow from C to D and is made of a material with shear modulus G = 52 GPa. The solid shaft AB has diameter di = 29 mm, the solid shaft BC has diameter d2 = 51 mm and the tubular shaft CD has outer diameter d2 and inner diameter dį. A torque TB = 1421 N. m is applied at B and a torque Tc = 1386 N. m is applied at C as shown. Ends A and D are fixed to rigid walls. Consider Lj = 299 mm, L2 = 740 mm and L3 = 230 mm. B TB 4 TC LT Matlab Mathematica Python L1 = 299; % mm L2 = 740; % mm L3 = 230; % mm d1 = 29; % mm d2 = 51; % mm G = 52; % GPa Tb = 1421; % N.m Tc = 1386; % N.m copy this text

Answer 1

oli G= 52 GPa = 52 x109 Pa, d = 29mm = 0.029m, da = 51mm = 0.05ım Tc = 1386 Nm, 4y = 299 mm = 0.299 m, 4 = 740 mm = 0.7490 Lg = 280 mm = 0.23m, TB = 1421 Nm Free body diagrams: To ind, I TO JE * .

- (*) ZMx=0 + TA-TB + Tc- Tp=0 . TA - 1421 + Tc-To=0 TA-1421 + 1986 -To=0 S TA=To.* 35 Consider the free body diagram, find the torque in each segment of the shaft. TAB = TA TBC = TA-TB = TA-1421 TCD = TA-Tg+ Tc = TA-1434 41386 - TA-35 Since both the ends are fixed, Bals=0 Balb + Oblc to do=o. (honet (anal) + Cha). -- al costat 6 het (1 m) - JBC

TALI * Toch 13 t - -0 10,4 TI44 Teola (44-4,4) 32 32 . TBcl2 X + TCD²3 to. d4 do day du TO 0.0294 TA X0.299 (TA-1421)x0.74 + (TA-35) x 0,23 -0. 0.0514 0.0514-0.0294 422746.698 TA +(TA-1421 DX 109383.2484 + , A-35)(31966.8269) +0. 570095.827= 156762506 - TA = 274.9751 Nm Maximum shear stress of shaft AB, TAB = Tabri = TA (4) GAB JAB Tidu da TAB 1TA - 16X 274.9157 IT (0.029)3 TAB = 57.42 X166 Pa | TAB = 57,42 MPa]