A composite assembly of length, L = 300 mm, consisting of a steel [G = 79 GPa] core (2) connected by rigid plates at the ends of an aluminum [G 33 GPa] tube (1) is shown. The aluminum tube has outside diameter D₁ = 50 mm and inside diameter d₁ = 35 mm. The solid steel core has diameter d₂ = 25 mm. The allowable shear stress of aluminum tube (1) is (Tallow) = 85 MPa, and the allowable shear stress of steel core (2) is (Tallow)2 = 140 MPa. Determine (a) the allowable torque T that can be applied to the composite shaft. (b) the corresponding torques produced in tube (1) and core (2). (c) the angle of twist produced by the allowable torque T. T A (1) (2) T B d₂ (1) d₁ D₁

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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A composite assembly of length, L = 300 mm, consisting of a steel [G = 79 GPa] core (2) connected by rigid plates at the ends of an
aluminum [G = 33 GPa] tube (1) is shown. The aluminum tube has outside diameter D₁ = 50 mm and inside diameter d₁ = 35 mm.
The solid steel core has diameter d₂ = 25 mm. The allowable shear stress of aluminum tube (1) is (Tallow)1 = 85 MPa, and the
allowable shear stress of steel core (2) is (Tallow)2 = 140 MPa. Determine
(a) the allowable torque T that can be applied to the composite shaft.
(b) the corresponding torques produced in tube (1) and core (2).
(c) the angle of twist produced by the allowable torque T.
T
A
(1)
Answer: J₁ =
2
Calculate the polar moment of inertia, J₁, of the aluminum tube.
.4662688
L
(10%) mm 4.
T
d₂
(1)
d₁
D₁
Transcribed Image Text:A composite assembly of length, L = 300 mm, consisting of a steel [G = 79 GPa] core (2) connected by rigid plates at the ends of an aluminum [G = 33 GPa] tube (1) is shown. The aluminum tube has outside diameter D₁ = 50 mm and inside diameter d₁ = 35 mm. The solid steel core has diameter d₂ = 25 mm. The allowable shear stress of aluminum tube (1) is (Tallow)1 = 85 MPa, and the allowable shear stress of steel core (2) is (Tallow)2 = 140 MPa. Determine (a) the allowable torque T that can be applied to the composite shaft. (b) the corresponding torques produced in tube (1) and core (2). (c) the angle of twist produced by the allowable torque T. T A (1) Answer: J₁ = 2 Calculate the polar moment of inertia, J₁, of the aluminum tube. .4662688 L (10%) mm 4. T d₂ (1) d₁ D₁
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