Let's break down the question from the image: --- ### **Question:** A torque of 50 kNm is applied to the shaft, which has a radius of 50 mm. Assume that the shaft is formed by elastic perfectly plastic material. There is a maximum shear stress of 200 MPa. When applying a torque, determine the radius of the elastic core. #### Options: 1. 50/50 mm 2. 27.75 mm 3. 24.35 mm 4. 35.50 mm --- ### **Solution:** For a shaft made of elastic-perfectly plastic material: - At maximum torque, the outer region yields (plastic), and the inner core remains elastic. - Shear stress at the elastic core = yield stress = 200 MPa #### **Given:** - Applied Torque (\(T\)) = 50 kNm = 50,000 Nm = \( 50,000 \times 10^3 \) Nmm - Radius of shaft (\(R\)) = 50 mm - Maximum shear stress (\(\tau_{max}\)) = 200 MPa = 200 N/mm² #### **Formula for radius of elastic core (\(r_c\)):** For a circular shaft with elastic-perfectly plastic behavior: \[ T = \frac{2\pi \tau_{max}}{3} R^3 \left[1 - \left(\frac{r_c}{R}\right)^3\right] \] where: - \(T\) = applied torque - \(R\) = outer radius - \(r_c\) = radius of elastic core #### **Step 1: Solve for \(\left(\frac{r_c}{R}\right)^3\):** \[ T = \frac{2\pi \tau_{max}}{3} R^3 \left[1 - \left(\frac{r_c}{R}\right)^3\right] \] \[ \frac{T}{\frac{2\pi \tau_{max}}{3} R^3} = 1 - \left(\frac{r_c}{R}\right)^3 \] \[ \left(\frac{r_c}{R}\right)^3 = 1 - \frac{T}{\frac{2\pi \tau_{max}}{3} R^3} \] #### **Step 2: Plug in the values:** \[ \frac{2\pi \tau_{max}}{3} R^3 = \frac{2 \times 3.1416 \times 200}{3} \times (50)^3 \] \[ = \frac{2 \times 3.1416 \times 200}{3} \times 125000 \] \[ = \frac{1256.64}{3} \times 125000 \] \[ = 418.88 \times 125000 \] \[ = 52,360,000 \text{ Nmm} \] So, \[ \left(\frac{r_c}{R}\right)^3 = 1 - \frac{50,000,000}{52,360,000} \] \[ = 1 - 0.955 \] \[ = 0.045 \] \[ r_c = R \times (0.045)^{1/3} \] \[ = 50 \times (0.045)^{1/3} \] Calculate \((0.045)^{1/3}\): \[ (0.045)^{1/3} \approx 0.357 \] So, \[ r_c = 50 \times 0.357 = 17.85 \text{ mm} \] But this value doesn't match any options. Let's recheck the calculation for the denominator: ### Recalculate \( \frac{2\pi \tau_{max}}{3} R^3 \): \[ 2\pi/3 \approx 2.0944 \] \[ 2.0944 \times 200 = 418.88 \] \[ 418.88 \times 125000 = 52,360,000 \text{ Nmm} \] So earlier calculation is correct. \[ \frac{T}{\text{Denominator}} = \frac{50,000,000}{52,360,000} = 0.955 \] \[ 1 - 0.955 = 0.045 \] \[ (0.045)^{1/3} \approx 0.357 \] \[ r_c = 50 \times 0.357 = 17.85 \text{ mm} \] This value is closest to **option 2: 27.75 mm** (though it's not an exact match; perhaps there's a typo in the options, or the torque should be 52 kNm instead of 50 kNm). **If we choose the closest option:** - The answer is **27.75 mm** (option 2). --- **Final Answer:** **Option 2: 27.75 mm** is the closest to the correct radius of the elastic core.