Torque Calculation for Aluminum Bars

This problem involves determining the torque in one aluminum bar given the torque in another, considering differences in their dimensions but the same angle of twist. The relationship between torque, material properties, dimensions, and angle of twist for a circular shaft under torsion is described by the torsion formula.

The torsion formula is given by:

\(T = \frac{GJ\theta}{L}\)

Where:

• \(T\) is the applied torque
• \(G\) is the shear modulus of the material
• \(J\) is the polar moment of inertia of the cross-section
• \(\theta\) is the angle of twist
• \(L\) is the length of the shaft

For a solid circular shaft with diameter \(d\), the polar moment of inertia \(J\) is:

\(J = \frac{\pi d^4}{32}\)

We are given two aluminum bars (let's call them bar 1 and bar 2) with the following relationships:

• Length of bar 1, \(L_1 = 2L_2\)
• Diameter of bar 1, \(d_1 = 2d_2\)
• Angle of twist is identical for both bars, \(\theta_1 = \theta_2\)
• Torque in bar 2, \(T_2 = 50 \text{ N-m}\)

Since both bars are made of aluminum, their shear modulus \(G\) is the same (\(G_1 = G_2 = G\)).

Let's write the torsion formula for both bars:

For bar 1: \(T_1 = \frac{G J_1 \theta_1}{L_1}\)

For bar 2: \(T_2 = \frac{G J_2 \theta_2}{L_2}\)

Now, let's find the relationship between the polar moments of inertia \(J_1\) and \(J_2\):

\(J_1 = \frac{\pi d_1^4}{32}\)

Substitute \(d_1 = 2d_2\):

\(J_1 = \frac{\pi (2d_2)^4}{32} = \frac{\pi (16 d_2^4)}{32} = 16 \left(\frac{\pi d_2^4}{32}\right)\)

Since \(J_2 = \frac{\pi d_2^4}{32}\), we have:

\(J_1 = 16 J_2\)

Now, let's set up a ratio of the torques \(T_1\) and \(T_2\):

\(\frac{T_1}{T_2} = \frac{\frac{G J_1 \theta_1}{L_1}}{\frac{G J_2 \theta_2}{L_2}}\)

Simplify the ratio:

\(\frac{T_1}{T_2} = \frac{G J_1 \theta_1}{L_1} \times \frac{L_2}{G J_2 \theta_2}\)

Since \(G_1 = G_2 = G\) and \(\theta_1 = \theta_2\), these terms cancel out:

\(\frac{T_1}{T_2} = \frac{J_1}{L_1} \times \frac{L_2}{J_2}\)

Substitute the given relationships \(L_1 = 2L_2\) and \(J_1 = 16J_2\):

\(\frac{T_1}{T_2} = \frac{16J_2}{2L_2} \times \frac{L_2}{J_2}\)

Cancel out the common terms \(J_2\) and \(L_2\):

\(\frac{T_1}{T_2} = \frac{16}{2} = 8\)

So, the relationship between \(T_1\) and \(T_2\) is:

\(T_1 = 8 \times T_2\)

We are given \(T_2 = 50 \text{ N-m}\). Substitute this value:

\(T_1 = 8 \times 50 \text{ N-m}\)

\(T_1 = 400 \text{ N-m}\)

Thus, the torque value for bar 1 is 400 N-m.