A 6.35 mm module, straight bevel pinion of 14 teeth drives a gear of 20 teeth. The shaft angle is 90°. Calculate the addendum and dedendum, circular tooth thickness for each gear, and the pitch and base radii of the equivalent spur gear.

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Accepted Answer

Given [latex] m=6.35 mm , z_{1}=14, z_{2}=20, \Sigma=90^{\circ}, h_{a}=h_{f}=? [/latex].

[latex] i=z_{2} / z_{1}=20 / 14=1.4286 [/latex].

[latex] d_{1}=m z_{1}=6.35 imes 14=88.9 mm , d_{2}=6.25 imes 20=127 mm [/latex].

[latex] L=0.5\left[d_{1}^{2}+d_{2}^{2} ight]^{0.5}=0.5\left[(88.9)^{2}+(127)^{2} ight]^{0.5}=77.51 mm [/latex].

[latex] \sin \delta_{1}=d_{1} /(2 L)=48 /(2 imes 53.66)=0.4472 [/latex].

[latex] \delta_{1}=26.56^{\circ} [/latex].

[latex] an \delta_{1}=\sin \Sigma /(\cos \Sigma+i) [/latex].

[latex] =\sin 90^{\circ} /\left(\cos 90^{\circ}+1.4286 ight)=0.7 [/latex].

[latex] \delta_{1}=35^{\circ}, \delta_{2}=90-35=55^{\circ} [/latex].

[latex] h_{a}=m=6.35 mm . h_{f}=1.25 m =1.25 imes 6.35=7.94 mm [/latex].

[latex] ext { For a bevel gear, } z_{y}=z / \cos \delta [/latex].

[latex] z_{v 1}=z_{1} / \cos \delta_{1}=14 / \cos 35^{\circ}=17, z_{v 2}=20 / \cos 55^{\circ}=35 [/latex].

[latex] ext { For the equivalent spur gear, } d=m z_{v} ext {, The pitch diameters are: } [/latex]

[latex] d_{1}=6.35 imes 17=108 mm , d_{2}=6.35 imes 35=222.25 mm [/latex].

[latex] ext { Circular tooth thickness }=\pi imes 108 / 17=111.96 mm [/latex].

Question 14.37: A 6.35 mm module, straight bevel pinion of 14 teeth drives a...

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