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The motor produces a torque of T = 20 N # m on gear A. If
Chapter 5, Problem 5-73(choose chapter or problem)
The motor produces a torque of \(T=20 \mathrm{~N} \cdot \mathrm{m}\) on gear A. If gear C is suddenly locked so it does not turn, yet B can freely turn, determine the angle of twist of F with respect to E and F with respect to D of the L2-steel shaft, which has an inner diameter of 30 mm and an outer diameter of 50 mm. Also, calculate the absolute maximum shear stress in the shaft. The shaft is supported on journal bearings at G at H.
Questions & Answers
QUESTION:
The motor produces a torque of \(T=20 \mathrm{~N} \cdot \mathrm{m}\) on gear A. If gear C is suddenly locked so it does not turn, yet B can freely turn, determine the angle of twist of F with respect to E and F with respect to D of the L2-steel shaft, which has an inner diameter of 30 mm and an outer diameter of 50 mm. Also, calculate the absolute maximum shear stress in the shaft. The shaft is supported on journal bearings at G at H.
ANSWER:
Problem 5-73The motor produces a torque of T = 20 N . m on gear A. If gear C is suddenly locked so it doesnot turn, yet B can freely turn, determine the angle of twist of F with respect to E and F withrespect to D of the L2-steel shaft, which has an inner diameter of 30 mm and an outer diameterof 50 mm. Also, calculate the absolute maximum shear stress in the shaft. The shaft issupported on journal bearings at G at H. Step-by-step solution Step 1 of 4 ^Calculate the polar moment of inertia of the shaft. J = d d 4 32 ( o i)Here, do is the outer diameter and d iis the inner diameter.Substitute 0.05 m for d oand 0.03 for di. 4 4 J =32 ( 0.05 0.03 ) = 5.34071 x 10 m 4Radius of gear A,rA= 0.03 mRadius of gear F, r = 0.1 m FCalculate the torque applied to EF as gear C is suddenly locked. T T rF = AHere, r A is the radius of gear A and rF is the radius of gear F.\nSubstitute 0.03 m for r ,A0.1 m for r , Fnd 20 N-m for TA. T 20 0.1 = 0.03 20 × 0.1 T = 0.03 = 66.667 N-m