Question

Torque M is applied to the rigid disc. Determine the absolute value of the maximum shear stress in the composite rod in MPa. M-886 Nm, 1-7.7 cm, d.-2.3 cm d =7.3 cm, d =2.3 cm, 0,-11.5 cm, D =4.8 cm, G. =84 GPS, Gc=47 GPa, GA,=28 GPa. AL rigid disc St los Ips Opg Cu. М. pd24 Ø D2 - Answer:

Answer 1

Giveno AL rigid disc steel md @ube cu 01 Dz Tc C Ak da Given Torque M6=886 n-M length of each part l=707 cm, z 77mm diameters do- -2.3 CM.:23mm Called uropa 28 zl let. di = 7:3cm = 73mm, dlz=2-3 cm 323 mm, Diri.5cm = 115mm, D2 = 4.8cm 3 u8mm Modules of Rogidity of enery material is Steel Gst subipa copper Gaut Aliminium Gala 28 Gipa. Copper is solid shaft, Aluminiumes & steel are Hallow shaft. TA & tc are are Reaction Torques 27 OKT for shaft section Total applied torque TAM-M. TA TITT2 Torque on Aluminium = TI NIM Torque on TALLAL Teulla GAL JAL ㅠ 28XL0 3 x A [us" uz u78003 x T 234 on coppr = TZ Ni-M 2)' PAL= Pcci =) Gai cu لاد 2)

2) 7) 22 2) 311-884T2 - 27 TA = 312-884T; MAM e) for both sides fisced shafts Angle of twist in iz Angle of twistina Ⓡi 02 Tuxlku Tst xxst Gau Jeu Gst & Jst Tzx (TAMA-TA) uzx108x I (23) 84x10²x * [zas"-234 886 -312-884*72) 848 [a8"-23" 320115s T2.2 886-312.8848T2 zoti TA 312.8847 2.568 2) TA = 803-523M-M -> TA = 803-5231-my 886-803.523 >> Ter82.476M-M B2 2) T 2 unX23 2) 2) Tz - 2.568 N-M = 800.918N-M 2) P 7. Tc= >) shear stressin boer composite bari shear stress T = TALL Tcy RI JAL Teve 2) Ti 200918 2.568 + (15/2). ( 154 734 (154 734] 7 x 234 abslete har forquein => / 8.576 mpa Tz 820476X603 2) 长 Ist (48/2) 429 32 172 = umpa - abslite shear 2) = stressino >> Total max shearstreym- maxfrid7} =8-576mport e) Inbarz T 2 = Tst us 2