Chapter 48, Problem 65A

To determine

(a)

The pitch diameter of the spur gear.

Answer to Problem 65A

The pitch diameter is 260 mm.

Explanation of Solution

Given:

Module (m) = 10mm

Number of teeth = 26

Concept Used:

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

Calculation:

  $\begin{array}{l}\mathrm{mod}ule(m)=\frac{pitchdiameter(D)}{numberofteeth(N)}\\ pitchdiameter=mN\\ pitchdiameter=10\times 26mm\\ pitchdiameter=260mm\end{array}$

Conclusion:

The pitch diameter is 260 mm.

To determine

(b)

The circular pitch of the spur gear.

Answer to Problem 65A

The circular pitch is 31.416 mm.

Explanation of Solution

Given:

Module (m) = 10 mm

Number of teeth = 26

Concept Used:

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

Calculation:

  $\begin{array}{l}Forspurgear,\\ Circularpitch(P_C)=\frac{3.1416D}{N}\\ P_C=\frac{3.1416\times 260}{26}\\ P_C=31.416mm\end{array}$

Conclusion:

The circular pitch is 31.416 mm.

To determine

(c)

The outside diameter of the spur gear.

Answer to Problem 65A

The outside diameter is 280 mm.

Explanation of Solution

Given:

Module (m) = 10 mm

Number of teeth = 26

Concept Used:

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

Calculation:

  $\begin{array}{l}Pitch(P)=\frac{Numberofteeth(N)}{Pitchdiameter(D)}\\ P=\frac{26}{260}\\ P=0.1\\ Outsidediameter(D_o)=\frac{N+2}{P}\\ D_o=\frac{26+2}{0.1}\\ D_o=280mm\end{array}$

Conclusion:

The outside diameter is 280 mm.

To determine

(d)

The addendum of the spur gear.

Answer to Problem 65A

The addendum is 10 mm.

Explanation of Solution

Given:

Module (m) = 10 mm

Number of teeth = 26

Concept Used:

  $addendum=0.3183\times Circularpitch(P_C)$

Calculation:

  $\begin{array}{l}addendum(a)=0.3183P_C\\ a=0.3183\times 31.416mm\\ a=10mm\end{array}$

Conclusion:

The addendum is 10 mm.

To determine

(e)

The working depth of the spur gear.

Answer to Problem 65A

The working depth is 20 mm.

Explanation of Solution

Given:

Module (m) = 10 mm

Number of teeth = 26

Concept Used:

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

Calculation:

  $\begin{array}{l}Foraspurgear,\\ Workingdepth(WD)=\frac{2.000}{Pitchofthegear(P)}\\ WD=\frac{2}{0.1}\\ WD=20mm\end{array}$

Conclusion:

The working depth is 20 mm.

To determine

(f)

The tooth thickness of the spur gear.

Answer to Problem 65A

The thickness of the tooth is 15.708 mm.

Explanation of Solution

Given:

Module (m) = 10 mm

Number of teeth = 26

Concept Used:

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

Calculation:

  $\begin{array}{l}Foraspurgear,\\ Thicknessofatooth(T)=\frac{1.5708}{Pitchofthegear(P)}\\ T=\frac{1.5708}{0.1}\\ T=15.708mm\\ \end{array}$

Conclusion:

The thickness of the tooth is 15.708 mm.