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 rs (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=500 rpm , r_s=1 / 3, c=14 ext { in., } P=2 ext { in. }^{-1}, \phi=20^{\circ} ext {. } [/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 ext { in., } r_1 / r_2=1 / 3 ext {. } [/latex] Hence, [latex] r_1=3.5 ext { in., } r_2=10.5 ext { in. } [/latex] ,or [latex] d_1=7 ext { 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=7(2)=14, \quad N_2=21(2)=42 [/latex]

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

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

[latex] \begin{aligned} & r_{b 1}=3.5 \cos 20^{\circ}=3.289 in . \ & r_{b 2}=10.5 \cos 20^{\circ}=9.867 in . \end{aligned} [/latex]

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

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

c. We have [latex] f=b_d-a [/latex]. Table 11.1 gives the dedendum [latex] b_d=1.25 / 2=0.625 in . [/latex], 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_1 \omega_1=r_2 \omega_2 [/latex] (11.7)

[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
$\text { Addendum } a$	$1 / P$	$0.8 / P$	$1 / P$
Dedendum $b_d$	$1.25 / P$	$1 / P$	$1.25 / P$
Clearance $f$	$0.25 / P$	$0.2 / P$	$0.25 / P$
Working depth $h_k$	$2 / P$	$1.6 / P$	$2 / P$
Whole depth $h$	$2.25 / P$	$1.8 / P$	$2.25 / P$
Tooth thickness $t$	$1.571 / P$	$1.571 / P$	$1.571 / P$

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

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