Let's break the problem into parts as requested and proceed step by step with **clear explanations**. --- ### **Given Data** - **Torque at F, \( T_P \):** 150 Nm - **Allowable Shear Stress, \( \tau_{allow} \):** 75 MPa = \( 75 \times 10^6 \) N/m² - **Gear radii:** - \( r_D = 200 \) mm = 0.2 m - \( r_G = 75 \) mm = 0.075 m We are to find the **required diameter** of: - (a) **Shaft CDE** - (b) **Shaft FGH** --- ## **STEP 1: Find the Maximum Torque in Each Shaft** Let's clarify the typical gear train arrangement (common in similar problems): - **Motor applies torque at point F.** - Shaft **FGH** connects gears at **F** and **G**. - Shaft **CDE** connects gears at **C**, **D**, and **E**. - **Gears D and G** are connected by a gear mesh. Let's assume: - **Torque is applied at F:** \( T_F = 150 \) Nm (**on shaft FGH**) - **Gears D and G** mesh: **Torque is transmitted from shaft FGH via G to shaft CDE via D**. #### **A. Torque on Shaft FGH (T_FGH)** - **Shaft FGH** carries the **motor torque**: \( T_{FGH} = 150 \) Nm #### **B. Torque on Shaft CDE (T_CDE)** - **Gears G (radius \( r_G \))** and **D (radius \( r_D \))** mesh. - Torque **transmitted through gears**: - Force at gear mesh \( = \frac{T_{FGH}}{r_G} \) - Torque at D: \( T_D = \text{Force} \times r_D = \frac{T_{FGH}}{r_G} \times r_D \) \[ T_{CDE} = T_D = T_{FGH} \times \frac{r_D}{r_G} \] \[ T_{CDE} = 150 \times \frac{200}{75} = 150 \times \frac{8}{3} = 400 \text{ Nm} \] --- ## **STEP 2: Find Required Shaft Diameters** **Formula:** The maximum shear stress in a solid circular shaft: \[ \tau_{max} = \frac{T \cdot r}{J} \] where \( J = \frac{\pi}{32} d^4 \) (polar moment of inertia for a solid shaft) \[ \tau_{max} = \frac{T \cdot (d/2)}{\frac{\pi}{32} d^4} = \frac{16T}{\pi d^3} \] Set \( \tau_{max} = \tau_{allow} \): \[ \tau_{allow} = \frac{16T}{\pi d^3} \] \[ d^3 = \frac{16T}{\pi \tau_{allow}} \] \[ d = \left( \frac{16T}{\pi \tau_{allow}} \right)^{1/3} \] --- ### **(a) Shaft CDE** - **Maximum torque:** \( T_{CDE} = 400 \) Nm = **400,000 Nmm** - **Allowable shear stress:** 75 MPa = **75 N/mm²** \[ d_{CDE} = \left( \frac{16 \times 400,000}{\pi \times 75} \right)^{1/3} \] Calculate numerator: \[ 16 \times 400,000 = 6,400,000 \] Denominator: \[ \pi \times 75 \approx 235.62 \] So, \[ d_{CDE}^3 = \frac{6,400,000}{235.62} \approx 27,165.6 \] \[ d_{CDE} = (27,165.6)^{1/3} \] Take cube root: \[ d_{CDE} \approx 30.0 \text{ mm} \] --- ### **(b) Shaft FGH** - **Maximum torque:** \( T_{FGH} = 150 \) Nm = **150,000 Nmm** - **Allowable shear stress:** 75 N/mm² \[ d_{FGH} = \left( \frac{16 \times 150,000}{\pi \times 75} \right)^{1/3} \] Calculate numerator: \[ 16 \times 150,000 = 2,400,000 \] Denominator: \[ \pi \times 75 \approx 235.62 \] So, \[ d_{FGH}^3 = \frac{2,400,000}{235.62} \approx 10,188.3 \] \[ d_{FGH} = (10,188.3)^{1/3} \] Take cube root: \[ d_{FGH} \approx 21.8 \text{ mm} \] --- ## **Conclusion** ### **Required shaft diameters are:** - **(a) Shaft CDE:** **Diameter = 30.0 mm** - **(b) Shaft FGH:** **Diameter = 21.8 mm** --- ### **Summary Table** | Shaft | Maximum Torque (Nm) | Required Diameter (mm) | |----------|---------------------|-----------------------| | CDE | 400 | 30.0 | | FGH | 150 | 21.8 | **Explanation:** - Shaft CDE needs to be thicker because torque is increased by the gear ratio (\( r_D/r_G \)). - Shaft FGH only needs to transmit the smaller input torque. **Always round up to the nearest standard size for manufacturing.** --- **If you have a diagram or a specific arrangement, let me know if the gear labeling is different!**