Let's break down the problem step by step: ### **Given Data** - **Power transmitted by driving shaft (P):** 30 kW - **Speed of driving shaft (N₁):** 2400 rpm - **Speed of layshaft (N₂):** 1600 rpm - **Speed of driven shaft in lowest gear (N₄,low):** 400 rpm - **Distance between axes of driving shaft and layshaft:** 180 mm - **Module of all teeth (m):** 6 mm We are to **arrange a 4-speed gearbox with speeds in geometric progression** (GP). --- ## **1. Gearbox Layout Sketch** A typical 4-speed gearbox with driving shaft (input), layshaft (intermediate), and driven shaft (output): ``` [Input shaft]---(A)---(Gear 1) (Gear 2)---(B)---[Layshaft]---(C)---(Gear 3) (Gear 4)---(D)---[Output shaft] ``` Generally: - Gear A meshes with Gear B (fixed pair, input to layshaft). - Gears on layshaft mesh with gears on the output shaft through dog clutches to give different speed ratios. *(A sketch is required in the exam; please draw a typical 4-speed sliding-mesh or constant-mesh gearbox as per above description.)* --- ## **2. Calculation of Gear Ratios** ### **a) Geometric Progression of Output Speeds** Let the output speeds be: \( N_{4,1} \), \( N_{4,2} \), \( N_{4,3} \), \( N_{4,4} \) Given lowest speed \( N_{4,1} = 400\, \text{rpm} \) Highest speed (let's assume direct drive for highest gear): - In direct gear, output shaft = input shaft = 2400 rpm Thus, \( N_{4,4} = 2400\, \text{rpm} \) Let the speeds be in GP: Let the common ratio = \( r \), so, \[ N_{4,1} = 400 \\ N_{4,2} = 400r \\ N_{4,3} = 400r^2 \\ N_{4,4} = 400r^3 = 2400 \] Solving for \( r \): \[ 400r^3 = 2400 \implies r^3 = \frac{2400}{400} = 6 \implies r = \sqrt[3]{6} \approx 1.817 \] Thus, the four output speeds are: - 1st: 400 rpm - 2nd: \( 400 \times 1.817 \approx 727 \) rpm - 3rd: \( 400 \times 1.817^2 \approx 400 \times 3.303 \approx 1321 \) rpm - 4th: 2400 rpm **Summary of output speeds:** - 400 rpm (low) - 727 rpm - 1321 rpm - 2400 rpm (high) --- ### **b) Speed Ratios** Speed ratio (input/output): | Gear | Output rpm | Ratio (Input/Output) | |------|------------|----------------------| | 1st | 400 | 2400 / 400 = 6.00 | | 2nd | 727 | 2400 / 727 ≈ 3.30 | | 3rd | 1321 | 2400 / 1321 ≈ 1.82 | | 4th | 2400 | 2400 / 2400 = 1.00 | --- ## **3. Number of Teeth Calculation** ### **a) Gear Train Concept** Assume: - **Primary drive**: Input shaft gear (A) to layshaft gear (B) - **Secondary drive**: Layshaft to output shaft via gear pairs for each speed Given: - Input to layshaft: 2400 rpm to 1600 rpm ⇒ ratio = 1.5 Let: - \( Z_A \), \( Z_B \) = Teeth on input and layshaft gears; \( Z_A \) on input shaft, \( Z_B \) on layshaft; \( Z_A \) drives \( Z_B \) - \( Z_{Lx} \), \( Z_{Ox} \) = teeth on layshaft and output shaft for gear x (x = 1,2,3) For each gear, the total gear ratio is: \[ \text{Overall ratio} = \frac{N_{in}}{N_{out}} = \left(\frac{Z_B}{Z_A}\right) \times \left(\frac{Z_{Ox}}{Z_{Lx}}\right) \] ### **b) Assigning Teeth for Primary Drive** \[ \frac{N_{in}}{N_{lay}} = \frac{2400}{1600} = 1.5 = \frac{Z_B}{Z_A} \] Let’s pick suitable teeth numbers (must be integers, and avoid undercutting by keeping minimum teeth ~18 for 20° pressure angle, but since module is large and no mention of pressure angle, we can go lower). Let’s take: - \( Z_A = 24 \) - \( Z_B = 36 \) - Check: \( 36/24 = 1.5 \) ✔️ --- ### **c) Calculation for Each Gear Pair** For each gear, \[ \frac{N_{in}}{N_{out}} = \frac{Z_B}{Z_A} \times \frac{Z_{Ox}}{Z_{Lx}} \] Let \( R_x = \frac{N_{in}}{N_{out}} \), \( \frac{Z_B}{Z_A} = 1.5 \) So, \[ \frac{Z_{Ox}}{Z_{Lx}} = \frac{R_x}{1.5} \] #### **1st Gear (Lowest Speed)** \[ R_1 = 6.0 \implies \frac{Z_{O1}}{Z_{L1}} = \frac{6.0}{1.5} = 4.0 \] #### **2nd Gear** \[ R_2 = 3.30 \implies \frac{Z_{O2}}{Z_{L2}} = \frac{3.30}{1.5} = 2.20 \] #### **3rd Gear** \[ R_3 = 1.82 \implies \frac{Z_{O3}}{Z_{L3}} = \frac{1.82}{1.5} = 1.21 \] #### **4th Gear (Top speed, direct drive)** For direct drive, there is no gear pair; input and output shafts are coupled directly. So, skip gear calculation for top gear. --- ### **d) Determining Number of Teeth** Given **center distance between driving shaft and layshaft = 180 mm**, **module = 6 mm**. The **pitch circle diameter** for a gear is \( d = m \times Z \). For a pair of gears in mesh: \[ \frac{d_1 + d_2}{2} = \text{center distance} \] Or, \[ d_1 + d_2 = 2 \times 180 = 360\, \text{mm} \] So, for each pair: \[ m(Z_1 + Z_2) = 360 \implies (Z_1 + Z_2) = \frac{360}{6} = 60 \] Now, for each pair: #### **Primary Pair (A and B):** \[ Z_A + Z_B = 60 \\ Z_A = 24,\, Z_B = 36 \quad (\text{already assumed, matches sum}) \] #### **1st Gear Pair:** \[ Z_{L1} + Z_{O1} = 60 \\ \frac{Z_{O1}}{Z_{L1}} = 4.0 \implies Z_{O1} = 4 Z_{L1} \] So, \[ Z_{L1} + 4Z_{L1} = 60 \implies 5Z_{L1} = 60 \implies Z_{L1} = 12,\, Z_{O1} = 48 \] #### **2nd Gear Pair:** \[ Z_{L2} + Z_{O2} = 60 \\ \frac{Z_{O2}}{Z_{L2}} = 2.20 \implies Z_{O2} = 2.2 Z_{L2} \] So, \[ Z_{L2} + 2.2 Z_{L2} = 60 \implies 3.2 Z_{L2} = 60 \implies Z_{L2} = 18.75,\, Z_{O2} = 41.25 \] We need integer teeth. Closest integer values, keeping sum = 60 and ratio ≈ 2.2: Try \( Z_{L2} = 19 \), \( Z_{O2} = 41 \): sum = 60, ratio = 41/19 ≈ 2.158 Or \( Z_{L2} = 20 \), \( Z_{O2} = 40 \): sum = 60, ratio = 2.0 Or \( Z_{L2} = 18 \), \( Z_{O2} = 42 \): sum = 60, ratio = 2.333 \( Z_{L2} = 19, Z_{O2} = 41 \) gives closest ratio to 2.2. So, select: - \( Z_{L2} = 19 \), \( Z_{O2} = 41 \) #### **3rd Gear Pair:** \[ Z_{L3} + Z_{O3} = 60 \\ \frac{Z_{O3}}{Z_{L3}} = 1.21 \implies Z_{O3} = 1.21 Z_{L3} \] So, \[ Z_{L3} + 1.21Z_{L3} = 60 \implies 2.21Z_{L3} = 60 \implies Z_{L3} = 27.15,\, Z_{O3} = 32.85 \] Try integer values: - \( Z_{L3} = 27 \), \( Z_{O3} = 33 \): sum = 60, ratio = 33/27 = 1.222 - \( Z_{L3} = 28 \), \( Z_{O3} = 32 \): sum = 60, ratio = 32/28 = 1.143 \( Z_{L3} = 27 \), \( Z_{O3} = 33 \) is closer to the requirement. --- ### **e) **Summary Table of Teeth Numbers** | Gear Pair | Layshaft Teeth (\( Z_L \)) | Output Shaft Teeth (\( Z_O \)) | Ratio (\( Z_O/Z_L \)) | |-----------|----------------------------|----------------------------------|-----------------------| | Primary | 36 | 24 | 1.5 | | 1st | 12 | 48 | 4.0 | | 2nd | 19 | 41 | 2.158 | | 3rd | 27 | 33 | 1.222 | | 4th | Direct drive | - | - | --- ## **4. Torque Calculations** ### **a) Output Torque in Lowest Gear** \[ \text{Power} (P) = 2\pi N T / 60 \implies T = \frac{P \times 60}{2\pi N} \] Given: - \( P = 30 \text{ kW} = 30000 \text{ W} \) - \( N = 400 \text{ rpm} \) (lowest gear) \[ T = \frac{30000 \times 60}{2\pi \times 400} = \frac{1,800,000}{2513.27} \approx 716 \text{ Nm} \] --- ### **b) Torque on the Gearbox Frame in Lowest Gear** This is the **reaction torque** due to gear forces, i.e., the difference in torque between input and output, acting in the opposite direction to input rotation. Input shaft torque: \[ T_{in} = \frac{P \times 60}{2\pi N_{in}} = \frac{30000 \times 60}{2\pi \times 2400} = \frac{1,800,000}{15079.65} \approx 119.4 \text{ Nm} \] **Gearbox frame torque** is the reaction equal to the difference: \[ T_{frame} = T_{out} - T_{in} = 716 - 119.4 = 596.6 \text{ Nm} \] **Direction:** - The reaction torque on the frame acts opposite to the direction of the input torque (to balance the moment applied by the input shaft). --- ## **5. Summary of Answers** ### **(i) Number of Teeth in Each Pair** - **Primary (Input shaft to layshaft):** - Input gear: **24 teeth** - Layshaft gear: **36 teeth** - **1st Gear (Layshaft to output):** - Layshaft gear