In this problem, we are examining a circular solid shaft that is fixed at one end and free at the other. A torque T is applied at the free end. We are given the shaft's span L (length) and its torsional rigidity GJ. We need to evaluate two statements regarding the shaft's behavior under this torque: the rotation at the free end and the torsional strain energy stored in the shaft.

Shaft Parameters and Given Data

Let's first list the important parameters provided in the question:

• Shaft Length (Span), \(L = 5 \, \text{m}\)
• Applied Torque, \(T = 100 \, \text{kN} \cdot \text{m}\)
• Torsional Rigidity, \(GJ = 50,000 \, \text{kN} \cdot \text{m}^2 / \text{rad}\)

Rotation at Free End Analysis (Statement i)

Statement i) claims that the rotation at the free end is \(0.01 \, \text{rad}\). For a circular shaft fixed at one end and subjected to a torque at the free end, the angle of twist or rotation (\(\theta\)) is given by the formula:

\[ \theta = \frac{T L}{G J} \]

Where:

• \(T\) is the applied torque
• \(L\) is the span (length) of the shaft
• \(G\) is the shear modulus of the material
• \(J\) is the polar moment of inertia of the cross-section
• \(GJ\) is the torsional rigidity

Now, let's substitute the given values into the formula:

\[ \theta = \frac{(100 \, \text{kN} \cdot \text{m}) \times (5 \, \text{m})}{50,000 \, \text{kN} \cdot \text{m}^2 / \text{rad}} \]

\[ \theta = \frac{500 \, \text{kN} \cdot \text{m}^2}{50,000 \, \text{kN} \cdot \text{m}^2 / \text{rad}} \]

\[ \theta = \frac{500}{50,000} \, \text{rad} \]

\[ \theta = \frac{1}{100} \, \text{rad} \]

\[ \theta = 0.01 \, \text{rad} \]

The calculated rotation at the free end is \(0.01 \, \text{rad}\). This matches the value given in Statement i).

Therefore, Statement i) is correct.

Torsional Strain Energy Analysis (Statement ii)

Statement ii) claims that the torsional strain energy is \(1.0 \, \text{kN} \cdot \text{m}\). The torsional strain energy (\(U\)) stored in a shaft subjected to torque can be calculated using the formula:

\[ U = \frac{1}{2} T \theta \]

Where:

• \(T\) is the applied torque
• \( \theta \) is the rotation (angle of twist) at the free end

Using the calculated rotation (\(\theta = 0.01 \, \text{rad}\)) from the previous step:

\[ U = \frac{1}{2} \times (100 \, \text{kN} \cdot \text{m}) \times (0.01 \, \text{rad}) \]

\[ U = 50 \, \text{kN} \cdot \text{m} \times 0.01 \, \text{rad} \]

\[ U = 0.5 \, \text{kN} \cdot \text{m} \]

Alternatively, the torsional strain energy can also be calculated using the formula that directly involves the torsional rigidity:

\[ U = \frac{T^2 L}{2 G J} \]

Let's verify our result using this formula:

\[ U = \frac{(100 \, \text{kN} \cdot \text{m})^2 \times (5 \, \text{m})}{2 \times (50,000 \, \text{kN} \cdot \text{m}^2 / \text{rad})} \]

\[ U = \frac{(10,000 \, \text{kN}^2 \cdot \text{m}^2) \times (5 \, \text{m})}{100,000 \, \text{kN} \cdot \text{m}^2 / \text{rad}} \]

\[ U = \frac{50,000 \, \text{kN}^2 \cdot \text{m}^3}{100,000 \, \text{kN} \cdot \text{m}^2 / \text{rad}} \]

\[ U = 0.5 \, \text{kN} \cdot \text{m} \]

Both methods yield the same result for the torsional strain energy, which is \(0.5 \, \text{kN} \cdot \text{m}\). This value does not match the \(1.0 \, \text{kN} \cdot \text{m}\) given in Statement ii).

Therefore, Statement ii) is wrong.

Conclusion on Statements

Based on our calculations, we can conclude the following:

• Statement i): The rotation at the free end is \(0.01 \, \text{rad}\), which is consistent with our calculation. So, Statement i) is correct.
• Statement ii): The torsional strain energy is \(0.5 \, \text{kN} \cdot \text{m}\), which contradicts the \(1.0 \, \text{kN} \cdot \text{m}\) stated. So, Statement ii) is wrong.

Thus, the correct assessment is that Statement i) is correct, but Statement ii) is wrong.