Let's solve the problem step by step. --- ### **Given Data** - Diameter of shaft, \( d = 50 \) mm \( = 0.05 \) m - From \( A \) to \( C \) (0 to 0.5 m): distributed torque \( w_1 = 500 \) N·m/m - From \( C \) to \( B \) (0.5 to 1.0 m): distributed torque \( w_2 = 1000 \) N·m/m - Material: A992 steel (but modulus of rigidity \( G \) not given, we'll use \( G \) symbolically) --- ## **(a) Determine the internal torque as a function of \( x \).** We'll use the **method of sections**. Cut the shaft at location \( x \) from the fixed end \( A \). ### **For \( 0 \leq x < 0.5 \) m ("Region 1")** - The distributed torque is \( w_1 = 500 \) N·m/m. - The internal torque at section \( x \): Let \( T(x) \) be the internal torque. Sum of distributed torque to the left of section at \( x \): \[ T(x) = w_1 \cdot x = 500x \quad \text{(N·m)} \] --- ### **For \( 0.5 \leq x \leq 1.0 \) m ("Region 2")** - For \( x > 0.5 \), left of \( x \) includes two regions: - From 0 to 0.5 m: \( w_1 = 500 \) N·m/m - From 0.5 to \( x \): \( w_2 = 1000 \) N·m/m So, \[ T(x) = (w_1 \cdot 0.5) + [w_2 \cdot (x - 0.5)] = 500 \cdot 0.5 + 1000(x - 0.5) \] \[ T(x) = 250 + 1000x - 500 \] \[ T(x) = 1000x - 250 \quad \text{(N·m)} \] --- ### **Summary: Internal Torque Function** \[ \boxed{ T(x) = \begin{cases} 500x & 0 \leq x < 0.5 \\ 1000x - 250 & 0.5 \leq x \leq 1.0 \end{cases} } \] --- ## **(b) Angle of Twist as a Function of \( x \) (\( \theta(x) \))** The angle of twist for a shaft subject to torque: \[ \frac{d\theta}{dx} = \frac{T(x)}{GJ} \] where \( J \) is the polar moment of inertia for a solid shaft: \[ J = \frac{\pi}{32} d^4 = \frac{\pi}{32} (0.05)^4 = 6.136 \times 10^{-6} \text{ m}^4 \] Let \( G \) be the modulus of rigidity (use symbolically). ### **Region 1: \( 0 \leq x < 0.5 \)** \[ \frac{d\theta}{dx} = \frac{500x}{GJ} \] \[ \theta(x) = \int_0^x \frac{500\xi}{GJ} d\xi = \frac{500}{GJ} \int_0^x \xi d\xi = \frac{500}{GJ} \left[ \frac{\xi^2}{2} \right]_0^x = \frac{250x^2}{GJ} \] --- ### **Region 2: \( 0.5 \leq x \leq 1.0 \)** We must account for both regions to get the total angle from \( 0 \) to \( x \): \[ \theta(x) = \theta(0.5) + \int_{0.5}^{x} \frac{1000\xi - 250}{GJ} d\xi \] From Region 1: \[ \theta(0.5) = \frac{250 \cdot (0.5)^2}{GJ} = \frac{250 \cdot 0.25}{GJ} = \frac{62.5}{GJ} \] Now, \[ \int_{0.5}^{x} (1000\xi - 250) d\xi = 1000 \int_{0.5}^{x} \xi d\xi - 250 \int_{0.5}^{x} d\xi \] \[ = 1000 \left[ \frac{\xi^2}{2} \right]_{0.5}^{x} - 250[x]_{0.5}^{x} \] \[ = 1000 \left( \frac{x^2}{2} - \frac{0.5^2}{2} \right) - 250(x - 0.5) \] \[ = 500x^2 - 500 \cdot 0.25 - 250x + 125 \] \[ = 500x^2 - 125 - 250x + 125 \] \[ = 500x^2 - 250x \] So, \[ \theta(x) = \frac{62.5}{GJ} + \frac{1}{GJ}(500x^2 - 250x) \] \[ = \frac{62.5 + 500x^2 - 250x}{GJ} \] --- ### **Summary: Angle of Twist Function** \[ \boxed{ \theta(x) = \begin{cases} \displaystyle \frac{250x^2}{GJ} & 0 \leq x < 0.5 \\ \displaystyle \frac{500x^2 - 250x + 62.5}{GJ} & 0.5 \leq x \leq 1.0 \end{cases} } \] Where: - \( G \) = modulus of rigidity for A992 steel - \( J = 6.136 \times 10^{-6} \text{ m}^4 \) --- **If you need a plot or want to substitute a value for \( G \), let me know!**