What is the angle of twist (in radians) of an aluminum cylindrical shaft (YS = 255 MPa, G = 186 MPa) with a diameter of 3.00 m and a length of 20.0 cm, when a torque of 8 Nm is applied?

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Answer :

yashothase37 yashothase37 · Answer 1 · 2024-01-31T17:30:11-05:00

The angle of twists in the aluminum cylindrical shaft is approximately 0.00018 radians.

To calculate the angle of twists in the aluminum cylindrical shaft, we need to use the formula:

θ = (T * L) / (G * J)

First, let's calculate the polar moment of inertia (J) for the shaft using the formula:

J = (π * D^4) / 32

Given that the diameter (D) of the shaft is 3.00m, we can substitute this value into the formula:

J = (π * (3.00m)^4) / 32

Calculating this, we find:

J ≈ 0.297m^4

Now, we can substitute the given values into the formula for the angle of twist:

θ = (8.00Nm * 0.20m) / (186MPa * 0.297m^4)

Calculating this, we find:

θ ≈ 0.00018 radians

Therefore, the angle of twists in the aluminum cylindrical shaft is approximately 0.00018 radians.

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