Chapter 48, Problem 62A

To determine

(a)

The pitch diameter of the spur gear.

Answer to Problem 62A

The pitch diameter is 216 mm.

Explanation of Solution

Given:

Module (m) = 9 mm

Number of teeth = 24

Concept Used:

  $\mathrm{mod}ule=\frac{pitchdiameter}{numberofteeth}$

Calculation:

  $\begin{array}{l}\mathrm{mod}ule(m)=\frac{pitchdiameter(D)}{numberofteeth(N)}\\ pitchdiameter=mN\\ pitchdiameter=9\times 24mm\\ pitchdiameter=216mm\end{array}$

Conclusion:

The pitch diameter is 216 mm.

To determine

(b)

The circular pitch of the spur gear.

Answer to Problem 62A

The circular pitch is 28.2744 mm.

Explanation of Solution

Given:

Module (m) = 9 mm

Number of teeth = 24

Concept Used:

  $\begin{array}{l}Forspurgear,\\ Circularpitch(P_C)=\frac{3.1416\times Pitchdiameter(D)}{Numberofteeth(N)}\end{array}$

Calculation:

  $\begin{array}{l}Forspurgear,\\ Circularpitch(P_C)=\frac{3.1416D}{N}\\ P_C=\frac{3.1416\times 216}{24}\\ P_C=28.2744mm\end{array}$

Conclusion:

The circular pitch is 28.2744 mm.

To determine

(c)

The outside diameter of the spur gear.

Answer to Problem 62A

The outside diameter is 234 mm.

Explanation of Solution

Given:

Module (m) = 9 mm

Number of teeth = 24

Concept Used:

  $\begin{array}{l}Pitch(P)=\frac{Numberofteeth(N)}{Pitchdiameter(D)}\\ Outsidediameter(D_o)=\frac{N+2}{P}\end{array}$

Calculation:

  $\begin{array}{l}Pitch(P)=\frac{Numberofteeth(N)}{Pitchdiameter(D)}\\ P=\frac{24}{216}\\ P=0.111\\ Outsidediameter(D_o)=\frac{N+2}{P}\\ D_o=\frac{24+2}{0.111}\\ D_o=234mm\end{array}$

Conclusion:

The outside diameter is 234 mm.

To determine

(d)

The addendum of the spur gear.

Answer to Problem 62A

The addendum is 9 mm.

Explanation of Solution

Given:

Module (m) = 9 mm

Number of teeth = 24

Concept Used:

  $addendum=0.3183\times Circularpitch(P_C)$

Calculation:

  $\begin{array}{l}addendum(a)=0.3183P_C\\ a=0.3183\times 28.2744mm\\ a=9mm\end{array}$

Conclusion:

The addendum is 9 mm.

To determine

(e)

The working depth of the spur gear.

Answer to Problem 62A

The working depth is 18 mm.

Explanation of Solution

Given:

Module (m) = 9 mm

Number of teeth = 24

Concept Used:

  $\begin{array}{l}Foraspurgear,\\ Workingdepth(WD)=\frac{2.000}{Pitchofthegear(P)}\end{array}$

Calculation:

  $\begin{array}{l}Foraspurgear,\\ Workingdepth(WD)=\frac{2.000}{Pitchofthegear(P)}\\ WD=\frac{2}{0.111}\\ WD=18mm\end{array}$

Conclusion:

The working depth is 18 mm.

To determine

(f)

The tooth thickness of the spur gear.

Answer to Problem 62A

The thickness of the tooth is 14.15 mm.

Explanation of Solution

Given:

Module (m) = 9 mm

Number of teeth = 24

Concept Used:

  $\begin{array}{l}Foraspurgear,\\ Thicknessofatooth(T)=\frac{1.5708}{Pitchofthegear(P)}\\ \end{array}$

Calculation:

  $\begin{array}{l}Foraspurgear,\\ Thicknessofatooth(T)=\frac{1.5708}{Pitchofthegear(P)}\\ T=\frac{1.5708}{0.111}\\ T=14.15mm\\ \end{array}$

Conclusion:

The thickness of the tooth is 14.15 mm.