Question

The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. If the shaft has a diameter of 40 mm. [10 marks) a-Draw the torque diagram. [2 marks] b-Determine the angle of twist of end B with respect to end A. [3 marks] C- Determine the maximum shear stress the shaft is subjected to. [2 marks] d- If the allowable shear stress in the shaft is 300 MPa, and the allowable angle of twist between B and A is 0.05 rad. The shaft is replaced by a pipe with inner diameter of 20 mm and an outer diameter of 30 mm, determine the maximum torques that can be applied to the pipe. [3 marks) 400 Nm 1200 Nm 600 Nm 600 mm 500 mm 400mm

Answer 1

Greien-A992 sted. solution Torque diagram 400 + B 200 Property of 1992 stel. Syt = 3 45MPa. Sut = 450 MPa. G=15G Pa. Torque diagram Assumption Anticho kurse Torque-negative clockwise - positive. 6. ORA From Torque diagram TAC = - 200 Nm TAD = -800 Nam ToB 400 Nm. foo

Op = OAC +00g + Ope. G5 - Tacle & Teales & Top Loe GI G-75Gpa. I = I dt = I( 40 32 32 ( 40/1900) 4= 2.512107mn 20070-4 + (-800) (-5)+4001.63 OBA 7 75x1042.51x10 .80 4.00 + 240 240 OBA: 5-3x10-5 [ ORA = -0.01212 rad OBA= 0.01272 grad (cow) Tmar where Tornare = 8000-m Tmar = 165 mase 16 X 800 x103 Td3 =63-66 N/mmi 7 (40) Tmane=63-66 MPa a. the same of assume Latz assume T be the torque proportion of Torque ocaur at each sation re) At A = 07,, B2T C-3T , D-67. 2 25 where Tmax = OTT 3T B D (solid) с 267 lo Tallowalles 300 MPa. Tallowable = 16 Tmas Tid3 T dola 400~)*-500.-I 600-5 32 Chellow) I. (d54-d14) For do 30com di= 20mm. 300 - T(3012) -) T= 1276272.01N-mm. (304-209) I 1276-27N-m TI 32

(1) ORA-0.05. 2T A C В J=1 (204-d; 4) P 32 - 6-38 x10-8 m4 GJ= 4T 786.02 4 0.05 = eac +9o+S - GI [TaclAct Toolest Toplan [TC0-4) -67)(0-5)+27(0.6)] 1 4.786-02 [-0.47-27+1025] 0.05 4786.02 -1-21 239-3 T= 199-4 N-m At C, maximum Torque of 67 is applied (e) 1196-5Nm. SO so |Tmax=1196.50-m
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