Question

Question: "Consider the shaft in problem 10-37. Calculate the effective internal torques resulting from the applied torques. Your answer should feature a value for section AC, CD, and DB."

PoF RusselLC_Hibbeler_Statics.a anics.pdf Open with ▼ 0.5 m Prob. 10-38 Prob. 10-36 10-39. The 60-mm-diameter shaft is made of 6061-T6 aluminum having an allowable shear stress of Tlow 80 MPa Determine the maximum allowable torque T. Also, find the 10-37. The splined ends and gears attached to the A992 corresponding angle of twist of disk A relative to disk C steel shaft are subjected to the torques shown. Determine -40. The 60-mm-diameter shaft is made of 6061-T6 the angle of twist of end B with respect to end A. The shaft aluminum. If the allowable shear stress is allo-80 MPa, and the angle of twist of disk A relative to disk C is limited has a diameter of 40 mm so that it does not exceed 0.06 rad, determine the maximum allowable torque T 400 N-m 1200 N-m 600 N-m 1.20 m 600 mm 500 mm 200 N-m 400mm Prob. 10-37 Probs 10-39/40 484 CHAPTER 10 TORSION 10-41. The 50-mm-diameter A992 steel shaft is subjected 10-44. The rotating f to the torques shown. Determine the angle of twist of the end A when brought to a sudden stop at D, begins to oscillate clockwise-counter- oint A on the outer edge of the rough a 6-mm arc. Determine the eveloped in the tubular A-36 steel shaft due to this oscillation. The shaft has an inner diameter Page 501 933

Answer 1

Use the right hand rule and the established sign convention that positive torque is directed away from the second end of the shaft. Calculate the torque applied at point A T, =200 N·m AC Calculate the net torque applied at point C. TCD-200+ 600 -800 N m Calculate the net torque applied at point D. TDB-200+ 600-1200 =-400 N·m Draw the torque diagram of the shaft as follows: T(N m) 800 200 -400

Calculate the polar moment of inertia by using the following equation 32 1000 mm 32 2.51327×10-7 m4 Obtain the modules of rigidity of A-992 steel G= 75 GPa Calculate the angle of twist of end A with respect to end B by using the following equation TL JG JG JG JG 200 N mx (400 mm) 2.5133x10-7 m4 x75x109 Pa 2.5133x10 m4 x 75x10 Pa 800 N mx(500 mm) -400 N mx (600mm) 2.5133 x10-7 m*x75x109 Pa 200 N mx(400 x10-3 m 2.5133x10-7 m4 x75x109 Pa 2.5133x10 m4 x 75x10 Pa )800 N mx(500x10m) 400 N mx(600x10*m) 2.5133 x10 m4 x75x109 Pa 4.244 x103 +2.122x102-1.2732 x10-2 -0.012732 rad - 0.012732x : O. 7300 180° ( 1π= 180°)