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## 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 $d_{o}=d+2 a_{o P} \cos \gamma$

## 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
$c_1 = 0.210 + 0.290/(m_G)^2 = 0.210 + 0.290/(3.00)2 = 0.242$
$a_G = c_1h$ = (0.242)(0.210 in) = 0.051 in
$a_p = h – a_G$ = 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
$δ_P = tan^{-1}(b_P/A_{mG}) = tan^{-1}(0.077/2.662) = 1.657°$
$a_oG = a_G + 0.5F tan δ_P$
$a_{oG}$ = (0.051 in) + (0.5)(1.00 in) tan(1.657°) = 0.0655 in
$a_oP = a_P + 0.5F tan δ_G$
$a_oP$ = (0.159 in) + (0.5)(1.00 in) tan(3.975°) = 0.1937 in
$D_o = D + 2a_{oG} cosΓ$
$D_o$ = 6.000 in + 2(0.0655 in) cos (71.57°) = 6.041 in
$d_o = d + 2a_{oP} cos γ$
$d_o$ = 2.000 in + 2(0.1937 in) cos (18.43°) = 2.368 in