302. Two solid shafts of different materials are rigidly fastened together and attached to rigid supports as shown in Fig. 3-4. The aluminum segment is 75 mm in diameter, and G, = 28 x 10° N/m?. The steel segment has a diameter of 50 mm and G, 83 x 10° N/m². The torque, T 1000 N m, is applied at the junction of the two segments. Compute the maximum shearing stress developed in the assembly. %3D %3D %3D

302. Two solid shafts of different materials are rigidly fastened together and attached to rigid supports as shown in Fig. 3-4. The aluminum segment is 75 mm in diameter, and G, = 28 x 10° N/m?. The steel segment has a diameter of 50 mm and G, 83 x 10° N/m². The torque, T 1000 N m, is applied at the junction of the two segments. Compute the maximum shearing stress developed in the assembly. %3D %3D %3D

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial forces, torsion, and so on, then such body is said to be under combined loading. A very practical example of combined loading is the crankshaft of an internal combustion engine (IC engine). The crankshaft of the IC engine is under high rpm, which is caused due to the torsional forces. Due to the presence of piston, counterbalances, and internal imbalances, the crankshaft experiences lateral forces and due to the rise in temperature, the crankshaft undergoes thermal expansion, which pulls the crankshaft along the axial direction and lateral direction. The presence of unbalanced loads along the periphery of the crankshaft also induces a bending moment.

Question

302. Two solid shafts of different materials are rigidly fastened
together and attached to rigid supports as shown in Fig. 3-4. The
aluminum segment is 75 mm in diameter, and G, 28 X 10° N/m2.
The steel segment has a diameter of 50 mm and G, 83 x 10° N/m².
The torque, T 1000 N. m, is applied at the junction of the two
segments. Compute the maximum shearing stress developed in the
assembly.
%3D
%3D
3 m
1.5 m
Ta
Aluminum
Steel
T=1.0 kN m
75 mm dia.
Ga 28 GN/m2
50 mm dia
G, = 83 GN/m2
न

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