Let's solve the problem **step by step**: ## **Given Data** - Length, \( L = 3.3 \) m (not needed for stress calculation) - Radius, \( r = 31 \) mm = \( 0.031 \) m - Axial force, \( F = 2.1 \) kN = \( 2100 \) N - Torsional moment, \( T = 20.0 \) kN·m = \( 20,000 \) N·m --- ## **Step 1: Calculate the Axial Normal Stress** The **axial stress** at any point is: \[ \sigma_{axial} = \frac{F}{A} \] Where the cross-sectional area \( A \) is: \[ A = \pi r^2 = \pi (0.031)^2 = 0.003019 \, \text{m}^2 \] So, \[ \sigma_{axial} = \frac{2100}{0.003019} = 695.60 \, \text{N/mm}^2 = 0.6956 \, \text{MPa} \] --- ## **Step 2: Calculate the Torsional Shear Stress at Surface (Point A)** The **maximum torsional shear stress** at the surface is: \[ \tau_{max} = \frac{T r}{J} \] Where for a solid circular shaft: \[ J = \frac{\pi}{2} r^4 \] Calculate \( J \): \[ J = \frac{\pi}{2} (0.031)^4 = \frac{\pi}{2} \times 9.231 \times 10^{-7} = 1.449 \times 10^{-6} \, \text{m}^4 \] Now, \[ \tau_{max} = \frac{20,000 \times 0.031}{1.449 \times 10^{-6}} = \frac{620}{1.449 \times 10^{-6}} = 427,228,435 \, \text{N/m}^2 = 427.228 \, \text{MPa} \] --- ## **Step 3: Find the Absolute Maximum Normal Stress (Principal Stress)** At **point A**, both axial normal stress (\(\sigma_{axial}\)) and torsional shear stress (\(\tau_{max}\)) act. The **principal normal stresses** are given by: \[ \sigma_{1,2} = \frac{\sigma_{axial}}{2} \pm \sqrt{ \left(\frac{\sigma_{axial}}{2}\right)^2 + \tau_{max}^2 } \] But since \(\sigma_{axial}\) is small compared to \(\tau_{max}\), the maximum normal stress is: \[ \sigma_{max} = \frac{\sigma_{axial}}{2} + \sqrt{ \left(\frac{\sigma_{axial}}{2}\right)^2 + \tau_{max}^2 } \] Plug in the numbers: - \(\sigma_{axial} = 0.6956\) MPa - \(\tau_{max} = 427.228\) MPa \[ \sigma_{max} = \frac{0.6956}{2} + \sqrt{ \left(\frac{0.6956}{2}\right)^2 + (427.228)^2 } \] \[ = 0.3478 + \sqrt{ (0.3478)^2 + (427.228)^2 } \] \[ = 0.3478 + \sqrt{ 0.1210 + 182,527.5 } \] \[ = 0.3478 + \sqrt{ 182,527.6 } \] \[ = 0.3478 + 427.305 \] \[ = 427.65 \, \text{MPa} \] This is much higher than the answer given in the question. Let's **double-check the math**. --- ## **Re-Check Units and Calculation** ### **Shear Stress** \[ \tau_{max} = \frac{T r}{J} \] \[ T = 20,000 \, \text{Nm} \] \[ r = 0.031 \, \text{m} \] \[ J = 1.449 \times 10^{-6} \, \text{m}^4 \] \[ \tau_{max} = \frac{20,000 \times 0.031}{1.449 \times 10^{-6}} = \frac{620}{1.449 \times 10^{-6}} = 427,228,435 \, \text{N/m}^2 = 427.23 \, \text{MPa} \] ### **Axial Stress** \[ \sigma_{axial} = \frac{F}{A} \] \[ F = 2,100 \, \text{N} \] \[ A = \pi \times (0.031)^2 = 0.003019 \, \text{m}^2 \] \[ \sigma_{axial} = \frac{2,100}{0.003019} = 695,605 \, \text{N/m}^2 = 0.6956 \, \text{MPa} \] --- ## **Principal Stresses (Summary)** \[ \sigma_{max} = \frac{\sigma_{axial}}{2} + \sqrt{ \left(\frac{\sigma_{axial}}{2}\right)^2 + \tau_{max}^2 } \] \[ \sigma_{max} \approx 0.35 + \sqrt{0.12 + 182,527.5} \] \[ = 0.35 + 427.228 \] \[ = 427.578 \, \text{MPa} \] --- ### **The Provided Answer is 304.2295 MPa** **Let's check if the radius is in mm or m.** - If we use \( r = 0.031 \) **meters**, the above is correct. - If the problem meant for **radius to be 31 mm in calculations (not converted to meters)**, let's try that: \[ r = 31 \, \text{mm} = 0.031 \, \text{m} \] ### **Check for Calculation Error** If you use **diameter instead of radius**, the stress would be lower. **Alternatively, let's check if the torsional moment is 20,000 Nm or 2,000 Nm.** But the question says \( T = 20.0 \) kN·m (\( 20,000 \) Nm). --- ## **Alternative Approach: Check for Bending Moment** If point A is at the **surface**, perhaps there is a **bending moment**. But the problem only gives axial and torsional loads. --- ## **Conclusion** - The **maximum normal stress** should be from the principal stress formula with the values above. - The calculation gives ~427 MPa, not 304 MPa. ### **But the question says the correct answer is 304.2295 MPa.** #### **Let's Check if There's a Unit Error in Torsional Moment Calculation:** If you use **radius in mm, not meters**: \[ J = \frac{\pi}{2} (31)^4 \text{ mm}^4 = \frac{\pi}{2} \times 923521 = 1,450,441 \text{ mm}^4 = 1.4504 \times 10^{-6} \text{ m}^4 \] But that's the same as above. #### **Try Using T in N·mm units:** \[ T = 20,000 \text{ Nm} = 20,000,000 \text{ Nmm} \] \[ \tau_{max} = \frac{T r}{J} \] \[ r = 31 \text{ mm} \] \[ J = 1,450,441 \text{ mm}^4 \] \[ \tau_{max} = \frac{20,000,000 \times 31}{1,450,441} = \frac{620,000,000}{1,450,441} = 427.5 \text{ MPa} \] Still matches previous calculation. #### **Try T = 2,000 Nm** \[ T = 2,000 \text{ Nm} \Rightarrow \tau_{max} = \frac{2,000,000 \times 31}{1,450,441} = 42.75 \text{ MPa} \] No, that's too small. --- ## **Final Thoughts** **All standard calculations yield 427 MPa, not 304 MPa, unless there is an error in the problem statement or a missing load (like a bending moment).** #### **Summary Table** | Parameter | Value | Units | |------------------|---------------|---------------| | Radius (r) | 0.031 | m | | Area (A) | 0.003019 | m² | | Polar Moment (J) | 1.449e-6 | m⁴ | | Axial force (F) | 2100 | N | | Torsion (T) | 20,000 | Nm | | Axial stress | 0.6956 | MPa | | Shear stress | 427.2 | MPa | | Principal stress | **427.6** | MPa | --- ### **If you are certain the answer is 304.2295 MPa, please check if:** - There is a misprint in the problem data (especially T or r). - There is an additional moment or load not described. - Or if the stress should be calculated differently (e.g., using maximum shear theory). **If you find any discrepancy in the provided data, let me know so I can recalculate! Otherwise, the process above is correct for the described loading.**