Key Length Calculation for Power Transmission

The problem requires finding the minimum length of a key used to fix a gear on a shaft, considering the power transmitted and material properties.

Torque Calculation

First, calculate the torque ($T$) transmitted by the shaft.

Given:

• Power ($P$) = 10 kW = $10 \times 10^3$ W
• Speed ($N$) = 600 rpm

The formula relating power, torque, and speed is:

$ P = \frac{2 \pi N T}{60} $

Rearranging to find torque:

$ T = \frac{60 P}{2 \pi N} $

Substituting the values:

$ T = \frac{60 \times (10 \times 10^3 \text{ W})}{2 \pi \times (600 \text{ rpm})} = \frac{600000}{1200 \pi} = \frac{500}{\pi} \text{ N-m} $

Converting torque to N-mm:

$ T \approx \frac{500}{\pi} \times 1000 \text{ N-mm} \approx 159154.94 \text{ N-mm} $

Tangential Force Calculation

Next, calculate the tangential force ($F$) acting on the shaft surface.

Given:

• Shaft diameter ($d$) = 20 mm

The formula relating torque, tangential force, and shaft radius is:

$ T = F \times \frac{d}{2} $

Rearranging to find the force:

$ F = \frac{2 T}{d} $

Substituting the values:

$ F = \frac{2 \times 159154.94 \text{ N-mm}}{20 \text{ mm}} = 15915.49 \text{ N} $

Key Length Calculation

Finally, calculate the minimum length ($l$) of the key using the permissible shear stress.

Given:

• Key width ($w$) = 6 mm
• Permissible shear stress ($\tau$) = 80 N/mm$^2$
• Tangential force ($F$) = 15915.49 N

The shear stress is calculated as:

$ \tau = \frac{F}{\text{Shear Area}} = \frac{F}{l \times w} $

Rearranging to find the length ($l$):

$ l = \frac{F}{\tau \times w} $

Substituting the values:

$ l = \frac{15915.49 \text{ N}}{(80 \text{ N/mm}^2) \times (6 \text{ mm})} = \frac{15915.49}{480} \text{ mm} $

$ l \approx 33.157 \text{ mm} $

Conclusion

Rounding off to two decimal places, the minimum length of the key is 33.16 mm. This value lies between 32 and 34 mm, as indicated by the correct answer range.