Chapter 48, Problem 63A

To determine

(a)

The pitch diameter of the spur gear.

Answer to Problem 63A

The pitch diameter is 25 mm.

Explanation of Solution

Given:

Module (m) = 2.5mm

Number of teeth = 10

Concept Used:

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

Calculation:

  $\begin{array}{l}\mathrm{mod}ule(m)=\frac{pitchdiameter(D)}{numberofteeth(N)}\\ pitchdiameter=mN\\ pitchdiameter=2.5\times 10mm\\ pitchdiameter=25mm\end{array}$

Conclusion:

The pitch diameter is 25 mm.

To determine

(b)

The circular pitch of the spur gear.

Answer to Problem 63A

The circular pitch is 7.854 mm.

Explanation of Solution

Given:

Module (m) = 2.5 mm

Number of teeth = 10

Concept Used:

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

Calculation:

  $\begin{array}{l}Forspurgear,\\ Circularpitch(P_C)=\frac{3.1416D}{N}\\ P_C=\frac{3.1416\times 25}{10}\\ P_C=7.854mm\end{array}$

Conclusion:

The circular pitch is 7.854 mm.

To determine

(c)

The outside diameter of the spur gear.

Answer to Problem 63A

The outside diameter is 30 mm.

Explanation of Solution

Given:

Module (m) = 2.5 mm

Number of teeth = 10

Concept Used:

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

Calculation:

  $\begin{array}{l}Pitch(P)=\frac{Numberofteeth(N)}{Pitchdiameter(D)}\\ P=\frac{10}{25}\\ P=0.4\\ Outsidediameter(D_o)=\frac{N+2}{P}\\ D_o=\frac{10+2}{0.4}\\ D_o=30mm\end{array}$

Conclusion:

The outside diameter is 30 mm.

To determine

(d)

The addendum of the spur gear.

Answer to Problem 63A

The addendum is 2.5 mm.

Explanation of Solution

Given:

Module (m) = 2.5 mm

Number of teeth = 10

Concept Used:

  $addendum=0.3183\times Circularpitch(P_C)$

Calculation:

  $\begin{array}{l}addendum(a)=0.3183P_C\\ a=0.3183\times 7.854mm\\ a=2.5mm\end{array}$

Conclusion:

The addendum is 2.5 mm.

To determine

(e)

The working depth of the spur gear.

Answer to Problem 63A

The working depth is 5 mm.

Explanation of Solution

Given:

Module (m) = 2.5 mm

Number of teeth = 10

Concept Used:

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

Calculation:

  $\begin{array}{l}Foraspurgear,\\ Workingdepth(WD)=\frac{2.000}{Pitchofthegear(P)}\\ WD=\frac{2}{0.4}\\ WD=5mm\end{array}$

Conclusion:

The working depth is 5 mm.

To determine

(f)

The tooth thickness of the spur gear.

Answer to Problem 63A

The thickness of the tooth is 3.927 mm.

Explanation of Solution

Given:

Module (m) = 2.5 mm

Number of teeth = 10

Concept Used:

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

Calculation:

  $\begin{array}{l}Foraspurgear,\\ Thicknessofatooth(T)=\frac{1.5708}{Pitchofthegear(P)}\\ T=\frac{1.5708}{0.4}\\ T=3.927mm\\ \end{array}$

Conclusion:

The thickness of the tooth is 3.927 mm.