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Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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
Transcribed Image Text:Problem 3.15. A hollow aluminium tube of rectangular cross-
section as shown in Fig. 3·27 is subjected to a torque of 56-5 kN.m along
its longitudinal axis. Determine the shearing stresses and the angle of
twist. Assume G = 28 GPa.
0.5m
t2 = 0.006
0.25m
t₁ =
0.012
0.01
(a)
h
t4=0.006
h
=
ti
13
(b) Membrane surface
Fig. 3.27.
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- Strength of materials 2arrow_forwardMechanics Lesson The 6 kN force is in the vertical direction, while the P=7.5 kN force is in the z direction.Since the safe shear stress is 60 MPa, determine the smallest allowable diameter of the AD shaft.arrow_forwardA steel shaft with a length of 3.00 m. One meter of the steel is contained in a brass tube and firmly attached. The diameter d1 = 70 mm and the diameter d2 = 90 mm. Brass G = 39GPa and steel G = 77.20 GPa. Determine: The torque that can be applied if the deformation at one end is 10 degrees. Answer: T = 8.64 kN – m The torque that can be applied if the maximum shear stress in brass is 70MPa and in steel is 110MPa. Answer: T = 6.352 kN – m Recommendation: analyze by segments. The first segment is only brass, then the composite part between brass and steel, and the last part only steel. In the composite segment, use only one material. In this same segment, equalize the deformations.arrow_forward
- An aluminum thin‐walled tube of hollow rectangular cross‐section, Figure 1,is subjected to a torque of magnitude, T = 15360 N.m. If the tube lengthis, L = 1 m , and the shear modulus of aluminum is, G = 28 GPa.1. Determine the total angle of twist of the one‐meter long tube.2. Determine the shear stresses on the walls of the tube cross‐section.Where does the maximum shear stress occur?arrow_forwardAnswer number 2 Good handwritingarrow_forwardPlease help. Do not just guess. Prove your answer. I will give a good feedback. thank you :)arrow_forward
- Show free body diagramarrow_forward1. A solid cylindrical rod is made of brass (G = 39 GPa); the rod has a diameter of 42 mm and is 750 mm long. The maximum allowable shear stress is 55 MPa and the twist cannot exceed 4° degrees. Determine the maximum allowable torque that can be applied. (Hint: check both conditions.) Ans. T= 800 Nmarrow_forward8.8 (A/B). Opposing axial torques are applied at the ends of a straight bar ABCD. Each of the parts AB, BC and CD is 500 mm long and has a hollow circular cross-section, the inside and outside diameters being, respectively, AB 25 mm and 60 mm, BC 25 mm and 70 mm, CD 40 mm and 70 mm. The modulus of rigidity of the material is 80 GN/m³ throughout. Calculate: (a) the maximum torque which can be applied if the maximum shear stress is not to exceed 75 MN/m²; (b) the maximum torque if the twist of D relative to A is not to exceed 2. [E.I.E.] [3.085 kNm, 3.25 kNm]arrow_forward
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