Let's break down the problem step by step: ### **Step 1: Find the Forces on the Shaft** #### **Torque on Shaft** - \( T = 180.0 \, \text{N} \cdot \text{m} = 180,000 \, \text{N} \cdot \text{mm} \) #### **Pitch Diameters** - \( a = 110.0 \, \text{mm} \) (Gear F) - \( b = 183.3 \, \text{mm} \) (Gear C) - \( c = 91.7 \, \text{mm} \) (Sprocket B) #### **Pressure Angle** - \( \phi = 20^\circ \) #### **Shaft Diameter** - \( d = 30 \, \text{mm} \) --- ### **Step 2: Calculate the Force at Gear C** The contact force \( F_C \) at Gear C is given by: \[ T = F_C \cdot \frac{b}{2} \] \[ F_C = \frac{2T}{b} \] \[ F_C = \frac{2 \times 180,000}{183.3} = \frac{360,000}{183.3} \approx 1964 \, \text{N} \] --- ### **Step 3: Find the Tangential and Radial Forces** - Tangential force \( F_{Ct} = F_C \cos\phi \) - Radial force \( F_{Cr} = F_C \sin\phi \) \[ F_{Ct} = 1964 \cos 20^\circ = 1964 \times 0.9397 \approx 1844 \, \text{N} \] \[ F_{Cr} = 1964 \sin 20^\circ = 1964 \times 0.3420 \approx 672 \, \text{N} \] --- ### **Step 4: Calculate Stresses in the Shaft at Gear C** #### **A. Shear Stress from Torque** \[ \tau_{\text{max}} = \frac{T \cdot r}{J} \] Where: - \( r = \frac{d}{2} = 15 \, \text{mm} \) - \( J = \frac{\pi d^4}{32} \) \[ J = \frac{\pi (30)^4}{32} = \frac{\pi \times 810,000}{32} \approx 79,577 \, \text{mm}^4 \] \[ \tau_{\text{max}} = \frac{180,000 \times 15}{79,577} \approx \frac{2,700,000}{79,577} \approx 33.9 \, \text{MPa} \] --- #### **B. Bending Stress from Tangential and Radial Forces** The maximum bending moment at section C (due to \( F_{Ct} \) and \( F_{Cr} \)) is: - **Distance from B to C**: \( f = 250 \, \text{mm} \) - The force at C creates a moment about the x-axis (vertical bending: \( M_y = F_{Ct} \cdot 0 \)), and about the y-axis (horizontal bending: \( M_z = F_{Cr} \cdot 0 \)), but let's clarify: At C: - Bending moment due to \( F_{Ct} \) (vertical): \( M_v = F_{Ct} \cdot f = 1844 \times 250 = 461,000 \, \text{N} \cdot \text{mm} \) - Bending moment due to \( F_{Cr} \) (horizontal): \( M_h = F_{Cr} \cdot f = 672 \times 250 = 168,000 \, \text{N} \cdot \text{mm} \) **Resultant bending moment:** \[ M = \sqrt{M_v^2 + M_h^2} = \sqrt{(461,000)^2 + (168,000)^2} = \sqrt{212,521,000,000 + 28,224,000,000} = \sqrt{240,745,000,000} \approx 490,637 \, \text{N} \cdot \text{mm} \] --- **Maximum bending stress:** \[ \sigma_{\text{max}} = \frac{M \cdot r}{I} \] Where \( I = \frac{\pi d^4}{64} \): \[ I = \frac{\pi (30)^4}{64} = \frac{\pi \times 810,000}{64} \approx 39,788 \, \text{mm}^4 \] \[ \sigma_{\text{max}} = \frac{490,637 \times 15}{39,788} \approx \frac{7,359,555}{39,788} \approx 185.0 \, \text{MPa} \] --- ### **Step 5: Combine the Stresses (Principal Stresses)** - Normal (bending) stress: \( \sigma = 185.0 \, \text{MPa} \) - Shear (torsional) stress: \( \tau = 33.9 \, \text{MPa} \) **Maximum principal normal stress:** \[ \sigma_{1,2} = \frac{\sigma}{2} \pm \sqrt{\left( \frac{\sigma}{2} \right)^2 + \tau^2} \] But since one principal stress is maximum and the other is minimum, the **maximum** is: \[ \sigma_{\text{max principal}} = \frac{185.0}{2} + \sqrt{ \left( \frac{185.0}{2} \right)^2 + (33.9)^2 } \] \[ = 92.5 + \sqrt{ 8,556.25 + 1,149.21 } = 92.5 + \sqrt{9,705.46} = 92.5 + 98.5 \approx 191.0 \, \text{MPa} \] **Maximum shear stress:** \[ \tau_{\text{max}} = \sqrt{ \left( \frac{\sigma}{2} \right)^2 + \tau^2 } = \sqrt{8,556.25 + 1,149.21} = \sqrt{9,705.46} \approx 98.5 \, \text{MPa} \] --- ## **Final Answers** **The maximum principal normal stress is:** \[ \boxed{191.0 \ \text{MPa}} \] **The maximum shear stress is:** \[ \boxed{98.5 \ \text{MPa}} \] --- ### **Summary Table** | Stress Type | Value (MPa) | |-------------------------------|-------------| | Maximum principal normal | 191.0 | | Maximum shear | 98.5 | If you need more detail or clarification for any step, please let me know!