Question

Under normal operating conditions, the electric motor exerts a torque on shaft AB. Pulleys B, C, and remove torques of 1.86 kN-m, 1.01 kN- m, and 0.67 kN-m, respectively. The bearing at E allows free rotation. Determine the maximum shear stress (in MPa) in section AB of the shaft if dAB=57.4 mm, dBC-49.3 mm, dcd=45 mm, and dDE 37.8 mm. Hint: The internal torque from Dto E is OKN-m and the supplied torque at the motor (point A) is going in the opposite direction of TB. Tc, and TD Note: Your answer should include the requested units and be to 2 decimal places. dec dice del C D

Answer 1

Bearing E allows free rotation, torque at E equals to zero.

By method of section, torque at each section is determined and finally by Twisting moment equation, maximum shear stress between AB is determined.

Solution given: TeTorque at pulley E=0 To - Torque at pulley' D = 0.67 KM-m 木 E D TO С To B TR pol KM-m A TA -1.86 KN-m TE = Tosque at pulley Ca TB = Torque at pulley B Diameter of shaft daß = 57.4 mm dBc = 49.3 mm dcD = 45 mm -e = doe = 37.8 mm

Torque at TA= To Tc + TR = 0.67+ 10 +1.86 TA = 3.54 Km By method of section: Tasque Between ED TED= o kam Toque Between DC Toc = To 0.67 KMom Torsque Between CB TCB = Totto 0.67 +1.01 = 1168 KM m Torque Between BA TBA = TotTc + TA=0,67+1.01+1.86 TBA = 3:54 KN-m

Maximum Shear Stess in Section AB of the Shaft: we know that I - I tar shear stress will be maximum when r=Rie. surface Maximum shear stress (Tmax) AB so of the shaft. (['max) AB 16 TBA T dAB 6 16 ( 3.54x106) T (574) 3 (Tmax)AB = 95:33 N/mm2 (Imax) AB 95.33 MPa Answer

T HO talt ut 4 d 2 Hinti (J= polar moment of inertia for arcular section; > T= J= t 32T ra ga It d 4 for t= Tmax ; r= R① d/ → Tmax= 32 T. d 16 T T d4 Sid3 T ЛІ 32 de