Chapter 49, Problem 22AR

To determine

(a)

The cutting speed.

Answer to Problem 22AR

  $C=151ft/\min$

Explanation of Solution

Given information:

Diameter of shaft: $D=2.300inches$

Rotational speed of lathe: $N=250rpm$

Calculations:

The cutting speed is calculated using the given formula:

  $\begin{array}{l}C=\frac{3.1416DN}{12}\\ \text{Substituting values of }D\text{ and }N\text{:}\\ C=\frac{3.1416\left(2.300\right)\left(250\right)}{12}\\ C=150.535ft/\min \\ \text{Rounding the value to the nearer whole number:}\\ C=151ft/\min \end{array}$

Conclusion:

The cutting velocity is $C=151ft/\min$.

To determine

(b)

The RPM of the aluminum cylinder.

Answer to Problem 22AR

  $N=1273\text{ revolutions per minute (RPM)}\text{.}$

Explanation of Solution

Given information:

Diameter of the cylinder: $D=40.00mm$

Cutting speed: $C=160m/\min$

Calculations:

The RPM of the cylinder is calculated using the given formula:

  $\begin{array}{l}N=\frac{1000C}{3.1416D}\\ N=\frac{1000\left(160\right)}{3.1416\left(40.0\right)}\\ or,\\ N=1273.23\\ \text{Rounding to the nearest whole number:}\\ N=1273\text{ revolutions per minute (RPM)}\end{array}$

Conclusion:

The rotational speed of the cylinder is $N=1273\text{ RPM}\text{.}$

To determine

(c)

The total cutting time.

Answer to Problem 22AR

  $T_{total}=3.74\min$

Explanation of Solution

Given information:

Number of holes $n=20$

Diameter of each holes $D=5.50mm$

Depth of holes $d=58.00mm$

Rotational speed of drill $N=6500RPM$

Feed $F=0.05mm/rev$

Calculations:

The time taken to drill one hole is given by the relation:

  $\begin{array}{l}T=\frac{L}{FN}\\ where,\\ L=d+AP1\\ or,\\ L=d+0.5D=58.00+0.5\left(5.50\right)=60.75mm\\ Substitutingvalues:\\ T=\frac{60.75mm}{\left(6500rev/\min \right)\left(0.05mm/rev\right)}\\ T=0.1869\min \\ \text{Hence, total time taken to drill 20 holes}:\\ T_{total}=nT=20\left(0.1869\right)\\ \Rightarrow T_{total}=3.738\min =3.74\text{ min}\end{array}$

Conclusion:

The total cutting time is $T_{total}=3.74\min$

To determine

(d)

The pitch of the gear.

Answer to Problem 22AR

  $P=3.50teeth/inch$

Explanation of Solution

Given information:

Number of teeth on the gear: $N=44$

Pitch diameter: $D=12.5714inches$

Calculations:

The pitch of the gear is defined by the relation:

  $\begin{array}{l}P=\frac{N}{D}\\ substitutingvalues:\\ P=\frac{44}{12.5714}\\ \Rightarrow P=3.50teeth/inch\end{array}$

Conclusion:

The pitch of the gear is $P=3.50teeth/inch$.

To determine

(e)

The whole depth of the gear.

Answer to Problem 22AR

  $WD=0.5393inch$

Explanation of Solution

Given information:

Number of teeth on the gear: $N=20$

Pitch diameter: $D=5.0000inches$

Calculations:

The whole depth of the gear is defined by the relation:

  $\begin{array}{l}WD=\frac{2.157}{P}\\ where,P\to \text{Pitch of the gear}\\ \therefore P=\frac{N}{D}=\frac{20}{5}=4teeth/inch\\ substituting:\\ WD=\frac{2.157}{4}=0.53925inch\\ or,\\ WD=0.5393inch\end{array}$

Conclusion:

The whole depth of the gear is $WD=0.5393inch$.

To determine

(f)

The center distance between the gear and the pinion.

Answer to Problem 22AR

  $CD=8.5989inches$

Explanation of Solution

Given information:

Pitch diameter of pinion: $D_p=6.4445inches$

Pitch diameter of gear: $D_g=10.7533inches$

Calculations:

The center distance between the gear and the pinion is defined by the relation:

  $\begin{array}{l}CD=\frac{D_p+D_g}{2}\\ substituting:\\ CD=\frac{6.4445+10.7533}{2}inches\\ or,\\ \Rightarrow CD=8.5989inches\end{array}$

  $\begin{array}{l}P=\frac{N}{D}\\ substitutingvalues:\\ P=\frac{44}{12.5714}\\ \Rightarrow P=3.50teeth/inch\end{array}$

Conclusion:

The center distance between the gear and the pinion is $CD=8.5989inches$.