# Explanation of the Problem This problem is about finding the **maximum power** transmitted by a **stepped shaft** subjected to a maximum allowable shearing stress, and understanding how this changes with frequency (speed). Let's break it down step by step: --- ## **Given Data** - **Fillet radius:** \( r = C \) mm (C is a variable/constant) - **Allowable shearing stress:** \( \tau_{max} = D \) MPa (D is a variable/constant) - **Shaft dimensions:** Large diameter \( D = 60 \) mm, small diameter \( d = 30 \) mm - **Modulus of rigidity:** \( G = 77.2 \) GPa - **Rotational frequency:** \( f = 60 \) Hz (for part a), \( f = 50 \) Hz (for part b) - **Stress concentration chart:** Provided (for K) --- ## **Step 1: Find the Maximum Torque** The shaft will fail where the **maximum shear stress** is highest, usually at the fillet (change in diameter). ### **Shear Stress with Stress Concentration** \[ \tau_{max} = K_t \cdot \frac{T \cdot r_{outer}}{J} \] Where: - \( K_t \) = stress concentration factor (from the chart, depends on \( r/d \) and \( D/d \)) - \( T \) = torque - \( r_{outer} \) = outer radius at the small diameter (\( d/2 \)) - \( J \) = polar moment of inertia for the small shaft (\( J = \frac{\pi}{32} d^4 \) for a solid shaft) ### **Solving for Maximum Torque** Rearrange for \( T \): \[ \tau_{max} = K_t \cdot \frac{T \cdot (d/2)}{J} \implies T_{max} = \frac{\tau_{max} \cdot J}{K_t \cdot (d/2)} \] Substitute \( J \): \[ T_{max} = \frac{\tau_{max} \cdot \frac{\pi}{32} d^4}{K_t \cdot (d/2)} = \frac{\tau_{max} \cdot \pi d^3}{16 K_t} \] --- ## **Step 2: Find the Stress Concentration Factor \( K_t \)** From the chart: - Find \( r/d \) (with \( r = C \), \( d = 30 \) mm) - Find \( D/d \) (with \( D = 60 \), \( d = 30 \), so \( D/d = 2 \)) - Enter the chart with these values to get \( K_t \). --- ## **Step 3: Find the Maximum Power** Power transmitted by a rotating shaft: \[ P = T \cdot \omega \] Where: - \( T \) = maximum torque (from above) - \( \omega \) = angular speed \( = 2\pi f \) So, \[ P_{max} = T_{max} \cdot 2\pi f \] --- ## **Step 4: Effect of Rotational Speed** For part **b**: - If \( f \) changes from 60 Hz to 50 Hz, \( P_{max} \) will change **proportionally** (since max torque is limited by material, not by speed). \[ P_{max} \propto f \] So, **if speed decreases, max allowable power decreases**. --- ## **Summary / Answers** ### a) **Maximum Power Calculation** 1. **Find \( K_t \)** from the chart using given \( r/d \) and \( D/d \). 2. **Compute \( T_{max} \):** \[ T_{max} = \frac{\tau_{max} \cdot \pi d^3}{16 K_t} \] 3. **Compute \( P_{max} \):** \[ P_{max} = T_{max} \cdot 2\pi f \] ### b) **Effect of Frequency** - **At lower frequency (50 Hz), the maximum power is:** - **LOWER**, because power is proportional to rotational speed. --- ## **Key Steps to Solve** 1. Find \( K_t \) from the chart using \( r/d \) and \( D/d \). 2. Plug values into the torque equation. 3. Use the new torque to find power at the required frequency. --- ### **Let me know if you'd like to see a full sample calculation with numbers!**