# Step-by-Step Solution to Chain Drive Design Problem Let's systematically solve each part of the problem as follows: --- ## **Given Data** - **Chain Type:** 60 (ANSI No. 60) - **Number of Strands:** 3 - **Input Speed (\(N_1\)):** 600 rpm - **Input Power (\(P_{in}\)):** 50 hp - **Output Speed (\(N_2\)):** 210 rpm --- ## (a) **Design Power** The **design power** considers service factors for load type, operating hours, and potential overloads. ### **Step 1: Select Service Factor (SF)** - Assume medium shock, 10+ hours/day: **SF ≈ 1.3** (from typical chain drive tables). ### **Step 2: Calculate Design Power** \[ \text{Design Power} = P_{in} \times SF = 50 \text{ hp} \times 1.3 = \boxed{65\ \text{hp}} \] --- ## (b) **Required Power per Strand** \[ \text{Power per strand} = \frac{\text{Design Power}}{\text{Number of Strands}} = \frac{65}{3} = \boxed{21.67\ \text{hp}} \] --- ## (c) **Chain Pitch** For ANSI No. 60 chain: - **Pitch (\(p\)) = .75 inch** (from standard chain tables) \[ \boxed{p = .75\ \text{inch}} \] --- ## (d) **Number of Teeth on Sprockets** Let's select a practical number: - **Driver Sprocket (\(T_1\)):** Typically 17–23 teeth for drives; let’s use \(T_1 = 20\). - **Speed Ratio:** \(\frac{N_1}{N_2} = \frac{600}{210} \approx 2.857\) - **Driven Sprocket (\(T_2\)):** \[ T_2 = T_1 \times \frac{N_1}{N_2} = 20 \times 2.857 = 57.14 \approx 57 \text{ teeth} \] \[ \boxed{T_1 = 20,\ T_2 = 57} \] --- ## (e) **Pitch Diameter of Sprockets** \[ D = \frac{p}{\sin\left(\frac{180^\circ}{T}\right)} \] where: - \(D\) = pitch diameter (inches) - \(p\) = pitch (inches) - \(T\) = number of teeth ### **Driver Sprocket (\(T_1 = 20\)):** \[ D_1 = \frac{.75}{\sin(180/20)} = \frac{.75}{\sin(9^\circ)} \approx \frac{.75}{.1564} = 4.8\ \text{in} \] ### **Driven Sprocket (\(T_2 = 57\)):** \[ D_2 = \frac{.75}{\sin(180/57)} \] \[ \sin(3.16^\circ) \approx .0552 \] \[ D_2 = \frac{.75}{.0552} = 13.6\ \text{in} \] \[ \boxed{D_1 = 4.8\ \text{in},\ D_2 = 13.6\ \text{in}} \] --- ## (f) **Actual Chain Length in Inches** Chain length in pitches (\(L_p\)): \[ L_p = \frac{2C}{p} + \frac{T_1 + T_2}{2} + \frac{(T_2 - T_1)^2}{4\pi^2} \cdot \frac{p}{C} \] where \(C\) = center distance (inches), \(p\) = pitch. Let’s **assume a center distance** (\(C\)) of 24 inches (good practice is 30–50 times the pitch; \(30 \times .75 = 22.5\) in). \[ L_p = \frac{2 \times 24}{.75} + \frac{20+57}{2} + \frac{(57-20)^2}{4\pi^2} \cdot \frac{.75}{24} \] \[ = 64 + 38.5 + \frac{37^2}{39.48} \cdot .03125 \] \[ = 64 + 38.5 + \frac{1369}{39.48} \cdot .03125 \] \[ = 64 + 38.5 + 34.68 \cdot .03125 \] \[ = 64 + 38.5 + 1.08 = 103.58\ \text{pitches} \] Chain length in inches: \[ L = L_p \times p = 103.58 \times .75 = \boxed{77.7\ \text{inches}} \] --- ## (g) **Acceptable Angle of Contact on the Driver Sprocket** \[ \theta = 180^\circ - \left( \frac{T_2 - T_1}{T_2 + T_1} \right) \times 60^\circ \] \[ = 180^\circ - \left( \frac{57-20}{57+20} \right) \times 60^\circ \] \[ = 180^\circ - \left( \frac{37}{77} \right) \times 60^\circ \] \[ = 180^\circ - (.4805 \times 60) \] \[ = 180^\circ - 28.83^\circ = \boxed{151.2^\circ} \] --- ## (h) **Actual Center Distance in Inches** As assumed earlier for calculations, \[ \boxed{C = 24\ \text{inches}} \] --- ## (i) **Lubrication Type** For a No. 60 chain, 600 rpm, and high power: - **Type:** Oil Bath or Drip Lubrication \[ \boxed{\text{Oil bath or drip lubrication}} \] --- ## **Summary Table** | Parameter | Value | |-----------------------------------|-------------------| | (a) Design Power | 65 hp | | (b) Power per Strand | 21.67 hp | | (c) Chain Pitch | .75 in | | (d) Teeth on Sprockets | 20 (driver), 57 (driven) | | (e) Pitch Diameter | 4.8 in (driver), 13.6 in (driven) | | (f) Chain Length | 77.7 in | | (g) Angle of Contact | 151.2° | | (h) Center Distance | 24 in | | (i) Lubrication | Oil bath/drip | --- ### **Let me know if you need clarification on any step!**