Compute the values for the geometrical features listed in Table 8–8 for a pair of straight bevel gears having a diametral pitch of 8, a 20° pressure angle, 16 teeth in the pinion, and 48 teeth in the gear. Specify a suitable face width. The shafts are at 90°.

Question

TABLE 8–8 Geometrical Features of Straight Bevel Gears
Given Diametral pitch = [latex]P_d = N_P/d = N_G/D[/latex] or [latex]m = d/N_P = D/N_G[/latex] where [latex]N_P[/latex] = number of teeth in pinion [latex]N_G[/latex] = number of teeth in gear
Dimension	Formula
Gear ratio	[latex]m_{G}=N_{G} / N_{P}[/latex]
Pitch diameters:
Pinion	[latex]d=N_{P}/ P_{d} ext { or } d=m N_{P}[/latex]
Gear	[latex]D=N_{G} P_{d} ext { or } D=m N_{G}[/latex]
Pitch cone angles:
Pinion	[latex]\gamma= an ^{-1}\left(N_{P} / N_{G} ight) \quad ext { (lowercase Greek gamma) }[/latex]
Gear	[latex]\Gamma= an ^{-1}\left(N_{G} / N_{p} ight) \quad ext { (uppercase Greek gamma) }[/latex]
Outer cone distance	[latex]A_{o}=0.5 D / \sin (\Gamma)[/latex]
Face width must be specified:	[latex]F=[/latex]
Nominal face width	[latex]F_{ ext {nom }}=0.30 A_{o}[/latex]
Maximum face width	[latex]F_{\max }=\mathrm{A}_{0} / 3 ext { or } F_{\max }=10 / P_{d} ext { or } \mathrm{m} / 2.54 ext { (whichever is less) }[/latex]
Mean cone distance	[latex]A_{m}=A_{o}-0.5 F[/latex]
	(Note: [latex]A_m[/latex] is defined for the gear, also called [latex]A_{mG}[/latex].)
Mean circular pitch	[latex]p_{m}=\left(\pi / P_{d} ight)\left(A_{m} / A_{0} ight) \quad ext { or } \quad \pi m\left(A_{m} / A_{o} ight)[/latex]
Mean working depth	[latex]h=\left(2.00 / P_{d} ight)\left(A_{m} / A_{o} ight) ext { or } 2.00 \mathrm{~m}\left(A_{m} / A_{o} ight)[/latex]
Clearance	[latex]c=0.125 h[/latex]
Mean whole depth	[latex]h_{m}=h+c[/latex]
Mean addendum factor	[latex]c_{1}=0.210+0.290 /\left(m_{G} ight)^{2}[/latex]
Gear mean addendum	[latex]a_{G}=c_{1} h[/latex]
Pinion mean addendum	[latex]a_{P}=h-a_{G}[/latex]
Gear mean dedendum	[latex]b_{G}=h_{m}-a_{G}[/latex]
Pinion mean dedendum	[latex]b_{p}=h_{m}-a_{p}[/latex]
Gear dedendum angle	[latex]\delta_{G}= an ^{-1}\left(b_{G} / A_{m G} ight)[/latex]
Pinion dedendum angle	[latex]\delta_{P}= an ^{-1}\left(b_{P} / A_{m G} ight)[/latex]
Gear outer addendum	[latex]a_{o G}=a_{G}+0.5 F an \delta_{P}[/latex]
Pinion outer addendum	[latex]a_{o P}=a p+0.5 F an \delta_{G}[/latex]
Gear outside diameter	[latex]D_{o}=D+2 a_{o G} \cos \Gamma[/latex]
Pinion outside diameter	[latex]d_{o}=d+2 a_{o P} \cos \gamma[/latex]

Accepted Answer

Given [latex]P_d = 8; N_p = 16; N_G = 48[/latex].
Computed Values Gear Ratio
[latex]m_G = N_G/N_P = 48/16 = 3.000[/latex]
Pitch Diameter
For the pinion,
[latex]d = N_P/P_d = 16/8 = 2.000[/latex] in
For the gear,
[latex]D = N_G/P_d = 48/8 = 6.000[/latex] in
Pitch Cone Angles
For the pinion,

[latex]\gamma= an ^{-1}\left(N_{P} / N_{G} ight)= an ^{-1}(16 / 48)=18.43^{\circ}[/latex]
For the gear,
[latex]\Gamma= an ^{-1}\left(N_{G} / N_{P} ight)= an ^{-1}(48 / 16)=71.57^{\circ}[/latex]
Outer Cone Distance
[latex]A_{o}=0.5 \mathrm{D} / \sin (\Gamma)=0.5(6.00 \mathrm{in}) / \sin \left(71.57^{\circ} ight)=3.162 \mathrm{in}[/latex]
Face Width
The face width must be specified based on the following guidelines:
Nominal face width:
[latex]F_{ ext {nom }}=0.30 A_{o}=0.30(3.162 ext { in })=0.949 ext { in }[/latex]
Maximum face width:
[latex]F_{\max }=A_{0}/ 3=(3.162 \mathrm{in}) / 3=1.054 ext { in }[/latex]
or
[latex]F_{\max }=10 / P_{d}=10 / 8=1.25 ext { in }[/latex]
Therefore the face width should be in the range from 0.949 in to 1.054 in. Let's specify [latex]F=1.000[/latex] in.

Mean Cone Distance
[latex]A_m = A_{mG} = A_o - 0.5F = 3.162 in - 0.5(1.00 in) = 2.662[/latex] in
Ratio [latex]A_m/A_o = 2.662/3.162 = 0.842[/latex] (This ratio occurs in several following calculations.)
Mean Circular Pitch
[latex]p_m = (π/P_d)(A_m/A_o) = (π/8)(0.842) = 0.331[/latex] in
Mean Working Depth
[latex]h = (2.00/P_d)(A_m/A_o) = (2.00/8)(0.842) = 0.210[/latex] in
Clearance
[latex]c = 0.125h = 0.125(0.210 in) = 0.026[/latex] in
Mean Whole Depth
[latex]h_m = h + c =[/latex] 0.210 in + 0.026 in = 0.236 in
Mean Addendum Factor
[latex]c_1 = 0.210 + 0.290/(m_G)^2 = 0.210 + 0.290/(3.00)2 = 0.242[/latex]
Gear Mean Addendum
[latex]a_G = c_1h[/latex] = (0.242)(0.210 in) = 0.051 in
Pinion Mean Addendum
[latex]a_p = h - a_G[/latex] = 0.210 in - 0.051 in = 0.159 in

Gear Mean Dedendum
[latex]b_G = h_m - a_G[/latex] = 0.236 in - 0.051 in = 0.185 in
Pinion Mean Dedendum
[latex]b_P = h_m - a_P[/latex] = 0.236 in - 0.159 in = 0.077 in
Gear Dedendum Angle

[latex]δ _G = tan^{-1}(b_G/A_{mG}) = tan^{-1}(0.185/2.662) = 3.975°[/latex]

Pinion Dedendum Angle
[latex]δ_P = tan^{-1}(b_P/A_{mG}) = tan^{-1}(0.077/2.662) = 1.657°[/latex]
Gear Outer Addendum
[latex]a_oG = a_G + 0.5F tan δ_P[/latex]
[latex]a_{oG}[/latex] = (0.051 in) + (0.5)(1.00 in) tan(1.657°) = 0.0655 in
Pinion Outer Addendum

[latex]a_oP = a_P + 0.5F tan δ_G[/latex]
[latex]a_oP[/latex] = (0.159 in) + (0.5)(1.00 in) tan(3.975°) = 0.1937 in
Gear Outside Diameter
[latex]D_o = D + 2a_{oG} cosΓ[/latex]
[latex]D_o[/latex] = 6.000 in + 2(0.0655 in) cos (71.57°) = 6.041 in

Pinion Outside Diameter
[latex]d_o = d + 2a_{oP} cos γ[/latex]
[latex]d_o[/latex] = 2.000 in + 2(0.1937 in) cos (18.43°) = 2.368 in

Question 8.3: Compute the values for the geometrical features listed in Ta...

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