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Chapter 8

Q. 8.3

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°.

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

Step-by-Step

Verified Solution

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

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

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

Gear Mean Dedendum
b_G = h_m – a_G = 0.236 in – 0.051 in = 0.185 in
Pinion Mean Dedendum
b_P = h_m – a_P = 0.236 in – 0.159 in = 0.077 in
Gear Dedendum Angle

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

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

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

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