2. A shaft is designed to carry torque as shown below. The shaft consists of a steel rod which is fixed against rotation at both ends. In addition, the left half of the rod is well bonded to an aluminum sleeve. The assembly is loaded by a torque of 2000 N-m as shown. - The steel rod has shear modulus G = 80 GPa and radius r= 1.5 cm (hence polar moment of inertia Ip = 7.95 x 108 m². The aluminum sleeve has shear modulus G = 26 GPa, outer radius 2.6 cm and inner radius 1.5 cm (hence polar moment of inertia I, = 7.95 x 108 m²). 2000 N H rod sleeve -0.517 a. Calculate the total reaction torques at the two walls (points A and C). b. Calculate the maximum shear stresses in the steel rod and the aluminum sleeve. c. Calculate the angle of twist at point B.

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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2. A shaft is designed to carry torque as shown below. The shaft consists of a steel
rod which is fixed against rotation at both ends. In addition, the left half of the rod
is well bonded to an aluminum sleeve. The assembly is loaded by a torque of 2000
N-m as shown.
-
The steel rod has shear modulus G = 80 GPa and radius r= 1.5 cm (hence polar
moment of inertia Ip = 7.95 x 108 m². The aluminum sleeve has shear modulus G
= 26 GPa, outer radius 2.6 cm and inner radius 1.5 cm (hence polar moment of
inertia I, = 7.95 x 108 m²).
2000 N
sleeve
-0.57
rod
a. Calculate the total reaction torques at the two walls (points A and C).
b. Calculate the maximum shear stresses in the steel rod and the aluminum sleeve.
c. Calculate the angle of twist at point B.
Transcribed Image Text:2. A shaft is designed to carry torque as shown below. The shaft consists of a steel rod which is fixed against rotation at both ends. In addition, the left half of the rod is well bonded to an aluminum sleeve. The assembly is loaded by a torque of 2000 N-m as shown. - The steel rod has shear modulus G = 80 GPa and radius r= 1.5 cm (hence polar moment of inertia Ip = 7.95 x 108 m². The aluminum sleeve has shear modulus G = 26 GPa, outer radius 2.6 cm and inner radius 1.5 cm (hence polar moment of inertia I, = 7.95 x 108 m²). 2000 N sleeve -0.57 rod a. Calculate the total reaction torques at the two walls (points A and C). b. Calculate the maximum shear stresses in the steel rod and the aluminum sleeve. c. Calculate the angle of twist at point B.
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