Gear Tooth and Gear Mesh Parameters Two parallel shafts A and B with center distance c are to be connected by 2 teeth/in. diametral pitch, 20° pressure angle, and spur gears l and 2 providing a velocity ratio of rs (Figure 11.9). Determine, for each gear, a. The number of teeth N b. The radius of

Question

Gear Tooth and Gear Mesh Parameters

Two parallel shafts [latex]A[/latex] and [latex]B[/latex] with center distance [latex]c[/latex] are to be connected by 2 teeth/in. diametral pitch, 20° pressure angle, and spur gears l and 2 providing a velocity ratio of [latex]r_{s}[/latex] (Figure 11.9). Determine, for each gear,

a. The number of teeth [latex]N[/latex]

b. The radius of the base circle [latex]r_{b}[/latex] and outside diameter [latex]d_{o}[/latex]

c. Clearance [latex]f[/latex]

d. The pitch-line velocity [latex]V[/latex], if gear 2 rotates at speed [latex]n_{2}[/latex]

Given: [latex] n_2=50 rpm , r_s=1 / 3, c=14 ext { in., } P=2 ext { in. }^{-1}, \phi=20^{\circ} [/latex]

Design Decision: Common stock gear sizes are considered.

Accepted Answer

a. Using Equation 11.6, we have [latex]r_{1} + r_{2} = c = 14 in., r_{1}/r_{2}[/latex] = 1/3. Hence, [latex]r_{1} = 3.5 in., r_{2} = 10.5 in.,[/latex] or [latex]d_{1} = 7 in., d_{2} = 21 in.[/latex] Equation 11.2 leads to

[latex] c=r_1+r_2=\frac{N_1+N_2}{2 P} [/latex] (11.6)

[latex]P = \frac{N}{d}[/latex] (11.2)

[latex]N_{1}[/latex] = 7(2) = 14, [latex]N_{2}[/latex] = 21(2) = 42

b. Base circle radii, applying Equation (11.9), are

[latex] r_b=r \cos \phi [/latex] (11.9)

[latex] r_{b 1}=3.5 \cos 20^{\circ}=3.289 in . [/latex]

[latex] r_{b 2}=10.5 \cos 20^{\circ}=9.867 in . [/latex]

From Table 11.1, the addendum [latex]a[/latex] = 1/2 = 0.5 in. Then

[latex] \begin{array}{c} d_{o 1}=7+2(0.5)=8 in . \ d_{o 2}=21+2(0.5)=22 in . \end{array} [/latex]

c. We have [latex]f = b_{d} − a.[/latex] Table 11.1 gives the dedendum [latex]b_{d} [/latex] = 1.25/2 = 0.625 in., and hence,

[latex] f=0.625-0.5=0.125 in [/latex]

for the pinion and gear. Note as a check that from Table 11.1, [latex]f[/latex] = 0.25/2 = 0.125 in.

d. Substituting the given data, Equation 11.7 results in

[latex] V=r_2 \omega_2=\frac{10.5}{12}\left(500 imes \frac{2 \pi}{60} ight)=45.81 fps [/latex]

Table 11.1 Commonly Used Standard Tooth Systems for Spur Gears
Item	20° Full Depth	20° Stub	25° Full Depth
Addendum [latex]a[/latex]	[latex]1/P[/latex]	[latex]0.8/P[/latex]	[latex]1/P[/latex]
Dedendum [latex]b_{d}[/latex]	[latex]1.25/P[/latex]	[latex]l/P[/latex]	[latex]1.25/P[/latex]
Clearance [latex]f[/latex]	[latex]0.25/P[/latex]	[latex]0.2/P[/latex]	[latex]0.25/P[/latex]
Working depth [latex]h_{k}[/latex]	[latex]2/P[/latex]	[latex]1.6/P[/latex]	[latex]2/P[/latex]
Whole depth [latex]h[/latex]	[latex]2.25/P[/latex]	[latex]1.8/P[/latex]	[latex]2.25/P[/latex]
Tooth thickness [latex]t[/latex]	[latex]1.571/P[/latex]	[latex]1.571/P[/latex]	[latex]1.571/P[/latex]

Question 11.2: Gear Tooth and Gear Mesh Parameters Two parallel shafts A an...

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