Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts
Question Statement:
A roller chain drive connects an engine shaft running at $1200\text{ rpm}$ to an intermediate shaft at $400\text{ rpm}$. The driving sprocket has $15$ teeth. The pitch of the chain is $19.05\text{ mm}$. Calculate the pitch diameter of the driven sprocket in $\text{mm}$ (round off to 1 decimal place).
Answer Key:
273
Step-by-Step Solution & Explanation
Speed ratio is $i = \frac{N_1}{N_2} = \frac{1200}{400} = 3$.
Number of teeth on driven sprocket:
$T_2 = i \times T_1 = 3 \times 15 = 45$
Pitch diameter of driven sprocket:
$D_2 = \frac{p}{\sin(180^\circ / T_2)} = \frac{19.05}{\sin(180^\circ / 45)} = \frac{19.05}{\sin(4^\circ)} = \frac{19.05}{0.069756} \approx 273.09\text{ mm}$
Acceptable range: $271.0 - 275.0\text{ mm}$.