the allowable torque Te that can be applied to the compound shaft at flange B. the magnitudes of the internal torques in segments (1) and (2). the rotation angle of flange B (relative to support A) that is produced by the allowable torque TB.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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The compound shaft consists of a solid aluminum segment (1) and a hollow brass segment (2) that are connected at flange B and
securely attached to rigid supports at A and C. Aluminum segment (1) has a diameter of 0.55 in., a length of L₁ = 40 in., a shear modulus
of 3800 ksi, and an allowable shear stress of 5.8 ksi. Brass segment (2) has an outside diameter of 0.45 in., a wall thickness of 0.15 in., a
length of L₂ = 30 in., a shear modulus of 6400 ksi, and an allowable shear stress of 8.2 ksi. Determine:
(a) the allowable torque Te that can be applied to the compound shaft at flange B.
(b) the magnitudes of the internal torques in segments (1) and (2).
(c) the rotation angle of flange B (relative to support A) that is produced by the allowable torque TB.
L₁
L2
TB
(1)
Answer:
(a) TB=
(b) T₁ =
T2=
(c) OB=
B
lb-ft
lb-ft
lb-ft
rad
(2)
C
Transcribed Image Text:The compound shaft consists of a solid aluminum segment (1) and a hollow brass segment (2) that are connected at flange B and securely attached to rigid supports at A and C. Aluminum segment (1) has a diameter of 0.55 in., a length of L₁ = 40 in., a shear modulus of 3800 ksi, and an allowable shear stress of 5.8 ksi. Brass segment (2) has an outside diameter of 0.45 in., a wall thickness of 0.15 in., a length of L₂ = 30 in., a shear modulus of 6400 ksi, and an allowable shear stress of 8.2 ksi. Determine: (a) the allowable torque Te that can be applied to the compound shaft at flange B. (b) the magnitudes of the internal torques in segments (1) and (2). (c) the rotation angle of flange B (relative to support A) that is produced by the allowable torque TB. L₁ L2 TB (1) Answer: (a) TB= (b) T₁ = T2= (c) OB= B lb-ft lb-ft lb-ft rad (2) C
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