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QB_FM_ADV_009
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NAT • 2 Marks
Section 2: Farm Machinery
Machine Design
Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts
Question Statement:
A pair of spur gears in a tractor transmission gearbox has a module $m = 4\text{ mm}$ and a pressure angle $\phi = 20^\circ$. The pinion has $z_1 = 20$ teeth and the gear wheel has $z_2 = 40$ teeth. Both gears have a standard addendum of $a = 1.0\,m = 4\text{ mm}$. The length of the path of contact in $\text{mm}$ is ________ (round off to 2 decimal places).
Step-by-Step Solution & Explanation
```
Path of Contact Schematic:
Pitch Point (P)
K <------ P ------> L
[Path of Approach] [Path of Recess]
```
**Step-by-step Calculation:**
1. Pitch radii:
$r = \frac{m z_1}{2} = \frac{4 \times 20}{2} = 40\text{ mm}$
$R = \frac{m z_2}{2} = \frac{4 \times 40}{2} = 80\text{ mm}$
2. Base circle radii:
$r_b = r \cos\phi = 40 \cos 20^\circ = 40 \times 0.93969 = 37.588\text{ mm}$
$R_b = R \cos\phi = 80 \cos 20^\circ = 80 \times 0.93969 = 75.175\text{ mm}$
3. Addendum radii:
$r_a = r + a = 40 + 4 = 44\text{ mm}$
$R_a = R + a = 80 + 4 = 84\text{ mm}$
4. Path of approach ($KP$):
$KP = \sqrt{R_a^2 - R_b^2} - R \sin\phi = \sqrt{84^2 - 75.175^2} - 80 \sin 20^\circ$
$\sqrt{7056 - 5651.34} = \sqrt{1404.66} = 37.479\text{ mm}$
$80 \sin 20^\circ = 80 \times 0.34202 = 27.362\text{ mm}$
$KP = 37.479 - 27.362 = 10.117\text{ mm}$
5. Path of recess ($PL$):
$PL = \sqrt{r_a^2 - r_b^2} - r \sin\phi = \sqrt{44^2 - 37.588^2} - 40 \sin 20^\circ$
$\sqrt{1936 - 1412.86} = \sqrt{523.14} = 22.872\text{ mm}$
$40 \sin 20^\circ = 13.681\text{ mm}$
$PL = 22.872 - 13.681 = 9.191\text{ mm}$
6. Total length of path of contact ($KL$):
$KL = KP + PL = 10.117 + 9.191 = 19.308\text{ mm} \approx 19.31\text{ mm}$
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