Chapter 48, Problem 61A

To determine

(a)

The pitch diameter of the spur gear.

Answer to Problem 61A

The pitch diameter is 117 mm.

Explanation of Solution

Given:

Module (m) = 6.5mm

Number of teeth = 18

Concept Used:

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

Calculation:

  $\begin{array}{l}\mathrm{mod}ule(m)=\frac{pitchdiameter(D)}{numberofteeth(N)}\\ pitchdiameter=mN\\ pitchdiameter=18\times 6.5mm\\ pitchdiameter=117mm\end{array}$

Conclusion:

The pitch diameter is 117 mm.

To determine

(b)

The circular pitch of the spur gear.

Answer to Problem 61A

The circular pitch is 20.4204 mm.

Explanation of Solution

Given:

Module (m) = 6.5mm

Number of teeth = 18

Concept Used:

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

Calculation:

  $\begin{array}{l}Forspurgear,\\ Circularpitch(P_C)=\frac{3.1416D}{N}\\ P_C=\frac{3.1416\times 117}{18}\\ P_C=20.4204mm\end{array}$

Conclusion:

The circular pitch is 20.4204 mm.

To determine

(c)

The outside diameter of the spur gear.

Answer to Problem 61A

The outside diameter is 130 mm.

Explanation of Solution

Given:

Module (m) = 6.5mm

Number of teeth = 18

Concept Used:

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

Calculation:

  $\begin{array}{l}Pitch(P)=\frac{Numberofteeth(N)}{Pitchdiameter(D)}\\ P=\frac{18}{117}\\ P=0.1538\\ Outsidediameter(D_o)=\frac{N+2}{P}\\ D_o=\frac{18+2}{0.1538}\\ D_o=130mm\end{array}$

Conclusion:

The outside diameter is 130 mm.

To determine

(d)

The addendum of the spur gear.

Answer to Problem 61A

The addendum is 6.5 mm.

Explanation of Solution

Given:

Module (m) = 6.5mm

Number of teeth = 18

Concept Used:

  $addendum=0.3183\times Circularpitch(P_C)$

Calculation:

  $\begin{array}{l}addendum(a)=0.3183P_C\\ a=0.3183\times 20.4204mm\\ a=6.5mm\end{array}$

Conclusion:

The addendum is 6.5 mm.

To determine

(e)

The working depth of the spur gear.

Answer to Problem 61A

The working depth is 13 mm.

Explanation of Solution

Given:

Module (m) = 6.5mm

Number of teeth = 18

Concept Used:

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

Calculation:

  $\begin{array}{l}Foraspurgear,\\ Workingdepth(WD)=\frac{2.000}{Pitchofthegear(P)}\\ WD=\frac{2}{0.1538}\\ WD=13mm\end{array}$

Conclusion:

The working depth is 13 mm.

To determine

(f)

The tooth thickness of the spur gear.

Answer to Problem 61A

The thickness of the tooth is 10.21 mm.

Explanation of Solution

Given:

Module (m) = 6.5mm

Number of teeth = 18

Concept Used:

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

Calculation:

  $\begin{array}{l}Foraspurgear,\\ Thicknessofatooth(T)=\frac{1.5708}{Pitchofthegear(P)}\\ T=\frac{1.5708}{0.1538}\\ T=10.21mm\\ \end{array}$

Conclusion:

The thickness of the tooth is 10.21 mm.