Chapter 14, Problem 26P

To determine

The rating of speed reducer for power

The rating of speed reducer for nitrided power

Answer to Problem 26P

The rating of speed reducer for power is $58.33\text{hp}$.

The rating of speed reducer for nitrided power is $116.2\text{hp}$.

Explanation of Solution

Write the expression for diameter of pinion.

$d_p=\frac{N_p}{p}$ (I)

Here, the number of teeth on pinion is $N_P$ and the diametral pitch is $p$.

Write the expression for diameter of the gear.

$d_G=\frac{N_G}{p}$ (II)

Here, the number of teeth on gear is $N_G$ and the diametral pitch is $p$.

Write the expression for velocity of the pinion.

$V=\frac{\pi d_p{n}_p}{12}$ (III)

Here, the number of rotation made by pinion is $n_p$.

Write the expression for constant of transmission accuracy level number.

$B=0.25{\left(12-Q_v\right)}^{\frac{2}{3}}$ . (IV)

Here, the transmission accuracy level number is $Q_v$.

Write the expression for constant $A$.

$A=50+56\left(1-B\right)$ (V)

Write the expression for dynamic factor.

$K_v={\left(\frac{A+\sqrt{V}}{A}\right)}^B$ (VI)

Write the expression for allowable bending stress number through hardened steels.

$S_t=77.3HB+12800$ (VII)

Here, the brinel hardness number is $HB$.

Write the expression for stress cycle factor for bending.

$Y_N=1.6831{\left(N\right)}^{-0.0323}$ (VIII)

Here, the number of cycles is $N$.

Write the expression for allowable stress.

${\left({\sigma }_{\text{all}}\right)}_{}=\frac{S_t{Y}_N}{S_F{K}_T{K}_R}$ (IX)

Here, the reliability factor is $K_R$, the temperature factor is is $K_T$ and the safety factor against failure is $S_F$.

Write the expression for load correction factor for uncrowned teeth.

$C_{mc}=1$ . (X)

Write the expression for pinion proportion factor.

$C_{pf}=\frac{F}{10d}-0.0375+0.0125F$ (XI)

Here, the face width is $F$ and the diameter is $d$.

Write the expression for pinion proportion modifier for straddle mounted pinion.

$C_{pm}=1$ (XII)

Write the expression for mesh alignment factor.

$C_{ma}=A+BF+C{F}^2$ (XIII)

Here, the empirical constant is $A,B$ and $C$.

Write the expression for mesh alignment correction factor.

$C_e=1$ (XIV)

Write the expression for load distribution factor $K_m$.

$K_m=1+C_{mc}\left(C_{pf}{C}_{pm}+C_{ma}{C}_e\right)$ . (XV)

Write the expression for overload factor for pinion.

${\left(K_o\right)}_p=1.192{\left(\frac{F\sqrt{Y_p}}{P}\right)}^{0.0535}$ (XVI)

Write the expression for overload factor for gear.

${\left(K_o\right)}_G=1.192{\left(\frac{F\sqrt{Y_G}}{P}\right)}^{0.0535}$                                                    (XVII)

Write the expression for transmitted load in pinion.

$W_p^t=\frac{F{J}_P{\sigma }_{\text{all}}}{{\left(K_o\right)}_p{K}_v{K}_{s}p{K}_m{K}_B}$                                                      (XVIII)

Here, the spur gear geometry factor is $J_P$ and the rim thickness factor is $K_B$.

Write the expression for power for pinion.

$H_p=\frac{W_p^{t}V}{33000}$ (XIX)

Write the expression for transmitted load for gear.

$W_G^t=\frac{F{J}_P{\sigma }_{\text{all}}}{{\left(K_o\right)}_G{K}_v{K}_{s}p{K}_m{K}_B}$                                                      (XX)

Write the expression for power for gear.

$H_G=\frac{W_G^{t}V}{33000}$ (XXI)

Write the expression for gear ratio.

$m_G=\frac{N_G}{N_P}$ (XXII)

Here, the number of teeth on gear is $N_G$ and the number of teeth on pinion is $N_P$

Write the expression for pitting resistance stress cycle factor for pinion.

${\left(Z_N\right)}_p=2.466{\left(N\right)}^{-0.056}$ (XXIII)

Write the expression for pitting resistance stress cycle factor for gear.

${\left(Z_N\right)}_G=2.466{\left(\frac{N}{G}\right)}^{-0.056}$ (XXIV)

Write the expression for geometry factor.

$I=\frac{\cos {\varphi }_t\sin {\varphi }_t}{2m_N}\left(\frac{m_G}{m_G+1}\right)$ . (XXV)

Here, the pressure angle is ${\varphi }_t$.

Write the expression for hardness ratio factor $C_H$.

$C_H=\frac{H_{BP}}{H_{BG}}=1$ . (XXVI)

Write the expression for contact fatigue strength for pinion.

${}_{0.99}{\left(S_{cP}\right)}_{{10}^7}=322\left(H_B\right)+29100$ . (XXVII)

Write the expression for contact fatigue strength for gear.

${}_{0.99}{\left(S_{cG}\right)}_{{10}^7}=322\left(H_B\right)+29100$ . (XXVIII)

Write the expression for pinion contact endurance strength.

${\left({\sigma }_{\text{all}}\right)}_P=\frac{{}_{0.99}{\left(S_c\right)}_{{10}^7}\left(Z_N\right)\left(C_H\right)}{S_H\left(K_T\right)\left(K_R\right)}$ . (XXIX)

Write the expression for transmitted load.

$W_3^t={\left(\frac{{\left({\sigma }_c\right)}_p}{C_P}\right)}^2\frac{F{d}_{p}I}{K_o{K}_v{\left(K_s\right)}_p{K}_m{C}_f}$ . (XXX)

Write the expression for power of pinion.

$H_3=\frac{W_3^{t}V}{33000}$ (XXXI)

Write the expression for gear contact strength.

${\left({\sigma }_{\text{all}}\right)}_G=\frac{S_c\left(Z_N\right)\left(C_H\right)}{S_H\left(K_T\right)\left(K_R\right)}$ . (XXXII)

Write the expression for transmitted load.

$W_4^t={\left(\frac{{\left({\sigma }_c\right)}_G}{C_P}\right)}^2\frac{F{d}_{p}I}{K_o{K}_v{\left(K_s\right)}_G{K}_m{C}_f}$ (XXXIII)

Her, the elastic coefficient is $C_P$.

Write the expression for power of gear.

$H_4=\frac{W_4^{t}V}{33000}$ (XXXIV)

Write the expression for contact fatigue strength for pinion for nitride.

${}_{0.99}{\left(S_{cPN}\right)}_{{10}^7}=322\left(H_B\right)+29100$ .(XXXV)

Write the expression for contact fatigue strength for gear.

${}_{0.99}{\left(S_{cGN}\right)}_{{10}^7}=322\left(H_B\right)+29100$ . (XXXVI)

Write the expression for pinion contact endurance strength.

${\left({\sigma }_{\text{all}}\right)}_{PN}=\frac{{}_{0.99}{\left(S_{cN}\right)}_{{10}^7}\left(Z_N\right)\left(C_H\right)}{S_H\left(K_T\right)\left(K_R\right)}$ . (XXXVII)

Write the expression for transmitted load.

$W_{3N}^t={\left(\frac{{\left({\sigma }_c\right)}_p}{C_P}\right)}^2\frac{F{d}_{p}I}{K_o{K}_v{\left(K_s\right)}_p{K}_m{C}_f}$ . (XXXVIII)

Write the expression for power of pinion.

$H_{3N}=\frac{W_{3N}^{t}V}{33000}$ (XXXIX)

Write the expression for gear contact strength.

${\left({\sigma }_{\text{all}}\right)}_{GN}=\frac{S_c\left(Z_N\right)\left(C_H\right)}{S_H\left(K_T\right)\left(K_R\right)}$ . (XXXX)

Write the expression for transmitted load.

$W_{4N}^t={\left(\frac{{\left({\sigma }_c\right)}_G}{C_P}\right)}^2\frac{F{d}_{p}I}{K_o{K}_v{\left(K_s\right)}_G{K}_m{C}_f}$ (XXXXI)

Her, the elastic coefficient is $C_P$.

Write the expression for power of gear.

$H_{4N}=\frac{W_{4N}^{t}V}{33000}$ (XXXXII)

Write the expression for rated power.

$H_{rated}=\text{min}\left(H_1,H_2,H_3,H_4\right)$ . (XXXXIII)

Write the expression for rated power for nitrided.

$H_{ratedN}=\text{min}\left(H_1,H_2,H_{3N},H_{4N}\right)$ (XXXXIV)

Conclusion:

Substitute $22\text{teeth}$ for $N_P$ and $4\text{teeth}/\text{in}$ for $p$ in Equation (I).

$\begin{array}{c}{d}_p=\frac{22\text{teeth}}{4\text{teeth}/\text{in}}\\ =5.5\text{in}\end{array}$

Substitute $60\text{teeth}$ for $N_P$ and $4\text{teeth}/\text{in}$ for $p$ in Equation (II)

$\begin{array}{c}{d}_G=\frac{60\text{teeth}}{4\text{teeth}/\text{in}}\\ =15\text{in}\end{array}$

Substitute $5.5\text{in}$ for $d_p$ and $1145\text{rpm}/\text{min}$ for $n_p$ in Equation (III).

$\begin{array}{c}V=\frac{\pi \left(5.5\text{in}\right)\left(1145\text{rev}/\text{min}\right)}{12\text{in}/\text{ft}}\\ =\frac{\pi \left(5.5\text{in}\right)\left(1145\text{rev}/\text{min}\right)}{12\text{in}/\text{ft}}\\ =\frac{19784.17}{12}\text{ft}/\text{min}\\ =1648.68\text{ft}/\min \end{array}$

Substitute $6$ for $Q_v$ in Equation (IV).

$\begin{array}{c}B=0.25{\left(12-6\right)}^{\frac{2}{3}}\\ =0.25{\left(6\right)}^{\frac{2}{3}}\\ =0.25\times 3.3019\\ =0.8255\end{array}$

Substitute $0.8255$ for $B$ in Equation (V).

$\begin{array}{c}A=50+56\left(1-0.8255\right)\\ =50+56\left(0.1745\right)\\ =50+9.772\\ =59.772\end{array}$

Substitute $59.772$ for $A$, $1648.68\text{ft}/\text{min}$ for $V$ and $0.8255$ for $B$ in Equation (VI).

$\begin{array}{c}{K}_v={\left(\frac{59.772+\sqrt{1648.68}}{59.772}\right)}^{0.8255}\\ ={\left(\frac{59.772+40.603}{59.772}\right)}^{0.8255}\\ ={\left(\frac{100.375}{59.772}\right)}^{0.8255}\\ ={\left(1.679\right)}^{0.8255}\end{array}$

$K_v=1.53$

Substitute $250$ for $HB$ in Equation (VII).

$\begin{array}{c}{S}_t=\left(77.3\left(250\right)+12800\right)\text{psi}\\ \text{=}\left(\text{19325+12800}\right)\text{psi}\\ =32125\text{psi}\end{array}$

Substitute $3\left({10}^9\right)$ for $N$ in Equation (VIII).

$\begin{array}{c}{Y}_N=1.6831{\left(3\left({10}^9\right)\right)}^{-0.0323}\\ =1.6831\left(0.4941\right)\\ =0.8316\end{array}$

Substitute $32125\text{psi}$ for $S_t$, $0.8316$ for $Y_N$, $1$ for $K_R$, $1$ for $K_T$, $1$ for $S_F$ in Equation (IX).

$\begin{array}{c}{\left({\sigma }_{all}\right)}_P=\frac{\left(32125\text{psi}\right)\times 0.8316}{1\times 1\times 1}\\ =26715.15\text{psi}\end{array}$

Substitute $3.25\text{in}$ for $F$ and $5.5\text{in}$ for $d$ in Equation (XI).

$\begin{array}{c}{C}_{pf}=\frac{3.25}{10\left(5.5\right)}-0.0375+0.0125\left(3.25\right)\\ =\frac{3.25}{55}-0.0375+0.04062\\ \text{=0}\text{.059}-0.0375+0.04062\\ =0.06212\end{array}$

Refer to table 14-9, “Empirical constant $A$,$B$ and $C$ for Eq.(14-34),face width $F$ in inches”, obtain the empirical constant $A$,$B$ and $C$ as $0.127$,$0.0158$ and $0.930\times {10}^{-4}$.

Substitute $0.127$ for $A$, $0.0158$ for $B$, $3.25\text{in}$ for $F$ and $0.930\times {10}^{-4}$ for $C$ in Equation (XIII).

$\begin{array}{c}{C}_{ma}=0.127+0.0158\left(3.25\right)-\left(0.930\times {10}^{-4}\right){\left(3.25\right)}^2\\ =0.127+0.05135-\left(0.930\times {10}^{-4}\right)\left(10.5625\right)\\ =0.17835-9.823\times {10}^{-4}\\ =0.17736\end{array}$

Substitute $1$ for $C_e$, $1$ for $C_{mc}$, $1$ for $C_{pm}$, $0.06212$ for $C_{Pf}$ and $0.17736$ for $C_{ma}$.

$\begin{array}{c}{K}_m=1+1\left[0.06212\times 1+0.17736\times 1\right]\\ =1+0.23948\\ =1.23\end{array}$

Refer Figure 14-6, “spur gear geometry factor”, to obtain the geometry factor for number of teeth $22T$ on pinion and number of teeth $60T$ on gear as $J_P=0.345$ and $J_G=0.41$.

Since the thickness of gear is constant so $K_B=1$.

Since the loading is uniform so $K_o=1$.

Refer table 14-2, “values of the lewis form factor $Y$ for pressure angle of $20°$, full-depth teeth, and a diametral pitch of unity in the plane of rotation” to obtain the lewis form factor for number of teeth $22T$ on pinion and number of teeth $60T$ on gear as $Y_P=0.331$ and $Y_G=0.422$.

Substitute $3.25\text{in}$ for $F$, $0.331$ for $Y_p$ and $4\text{teeth}/\text{in}$ for $P$ in Equation (XVI).

$\begin{array}{c}{\left(K_s\right)}_P=1.192{\left(\frac{3.25\sqrt{0.331}}{4}\right)}^{0.0535}\\ =1.192{\left(\frac{1.8698}{4}\right)}^{0.0535}\\ =1.192{\left(0.4674\right)}^{0.0535}\\ =1.192\times 0.960\end{array}$

${\left(K_s\right)}_P=1.14432$

Substitute $3.25\text{in}$ for $F$, $0.422$ for $Y_p$ and $4\text{teeth}/\text{in}$ for $P$ in Equation (XVI).

$\begin{array}{c}{\left(K_s\right)}_G=1.192{\left(\frac{3.25\sqrt{0.422}}{4}\right)}^{0.0535}\\ =1.192{\left(\frac{2.111}{4}\right)}^{0.0535}\\ =1.192{\left(0.5277\right)}^{0.0535}\\ =1.192\times 0.966\end{array}$

${\left(K_s\right)}_G=1.1514$

Substitute $3.25\text{in}$ for $F$, $26715.15\text{psi}$ for $\left({\sigma }_{all}\right)$, $0.345$ for $J_P$, $1.14432$ for ${\left(K_s\right)}_P$, $1$ for $K_O$, $4\text{teeth}/\text{in}$ for $P$, $1.53$ for $K_v$, $1.23$ for $K_m$ and $1$ for $K_B$ in Equation (XIX).

$\begin{array}{c}{W}_p^t=\frac{\left(3.25\text{in}\right)\left(0.345\right)\left(26715.15\right)}{1\times 1.53\times 1.14432\times 4\times 1.23\times 1}\\ =\frac{29954.36}{8.613}\\ =3477.80\text{lbf}\end{array}$

Substitute $3477.80\text{lbf}$ for $W_p^t$ and $1648.68\text{ft}/\min$ for $V$ un Equation (XX).

$\begin{array}{c}{H}_1=\frac{\left(3477.80\right)\left(1648.68\right)}{33000}\\ =\frac{5733779.304}{33000}\\ =173.75\text{hp}\end{array}$

Substitute $3.25\text{in}$ for $F$, $26715.15\text{psi}$ for $\left({\sigma }_{all}\right)$, $0.41$ for $J_P$, $1.1514$ for ${\left(K_s\right)}_P$, $1$ for $K_O$, $4\text{teeth}/\text{in}$ for $P$, $1.53$ for $K_v$, $1.23$ for $K_m$ and $1$ for $K_B$ in Equation (XIX).

$\begin{array}{c}{W}_G^t=\frac{\left(3.25\text{in}\right)\left(0.41\right)\left(26715.15\right)}{1\times 1.53\times 1.14432\times 4\times 1.23\times 1}\\ =\frac{35597.93}{8.613}\\ =4133.046\text{lbf}\end{array}$

Substitute $4133.046\text{lbf}$ for $W_G^t$ and $1648.68\text{ft}/\min$ for $V$ un Equation (XX).

$\begin{array}{c}{H}_2=\frac{\left(4133.046\right)\left(1648.68\right)}{33000}\\ =\frac{6814070.27}{33000}\\ =206.48\text{hp}\end{array}$

Substitute $60$ for $N_G$ and $22$ for $N_P$ in Equation (XXI)

$\begin{array}{c}G=\frac{60}{22}\\ =2.727\end{array}$

Substitute $3\left({10}^9\right)$ for $N$ in Equation (XX)

$\begin{array}{c}{\left(Z_N\right)}_G=2.466{\left(3\left(\frac{{10}^9}{2.727}\right)\right)}^{-0.056}\\ =2.466\times 0.3116\\ =0.768\end{array}$

Substitute $3\left({10}^9\right)$ for $N$ in Equation (XXVI).

$\begin{array}{c}{\left(Z_N\right)}_P=2.466{\left(3\left({10}^9\right)\right)}^{-0.056}\\ =2.466\times 0.2946\\ =0.7264\end{array}$

Substitute $20°$ for ${\varphi }_t$, $2.727$ for $m_G$ and $1$ for $m_N$ in Equation (XXVII).

$\begin{array}{c}I=\frac{\cos \left(20°\right)\sin \left(20°\right)}{2\times 1}\left(\frac{2.727}{1+2.727}\right)\\ =\frac{0.32139}{2}\left(\frac{2.727}{3.727}\right)\\ =0.160\times 0.7316\\ =0.1170\end{array}$

Substitute $250$ for $H_B$ in Equation (XXVIII).

$\begin{array}{c}{}_{0.99}{\left(S_{cp}\right)}_{{10}^7}=322\left(250\right)+29100\\ =80500+29100\\ =109600\text{psi}\end{array}$

Substitute $109600\text{psi}$ for ${}_{0.99}{\left(S_{cp}\right)}_{{10}^7}$, $0.7264$ for ${\left(Z_N\right)}_P$, $1$ for $S_H$, $1$ for $C_H$, $1$ for $K_T$ and $1$ for $K_R$ in Equation (XXIX).

$\begin{array}{c}{\left({\sigma }_{\text{all}}\right)}_P=\frac{\left(109600\text{psi}\right)\left(0.7264\right)\left(1\right)}{\left(1\right)\left(1\right)}\\ =79613.44\text{psi}\end{array}$

Substitute $79613.44\text{psi}$ for ${\left({\sigma }_{\text{all}}\right)}_P$, $2300$ for $C_P$, $3.25\text{in}$ for $F$, $5.5$ for $d_P$, $0.1170$ for $I$, $1.23$ for $K_m$, $1$ for $C_f$, $1.53$ for $K_v$, $1$ for $K_o$, $1.14$ for ${\left(K_s\right)}_p$ in Equation (XXX).

$\begin{array}{c}{W}_3^t={\left(\frac{79613.44\text{psi}}{2300}\right)}^2\left[\frac{\left(3.25\text{in}\right)\left(5.5\right)\left(0.1170\right)}{1\times 1.53\times 1.14\times 1.23\times 1}\right]\\ ={\left(34.61\right)}^2\left[\frac{2.0913}{2.1453}\right]\\ =\left(1197.85\right)\left(0.9748\right)\\ =1167.66\end{array}$

Substitute $1167.66$ for $W_3^t$ and $1648.68\text{ft}/\min$ for $V$ in Equation (XXXI).

$\begin{array}{c}{H}_3=\frac{\left(1167.66\right)\left(1648.68\text{ft}/\min \right)}{33000}\\ =\frac{1925097.6888}{33000}\\ =58.33\text{hp}\end{array}$

Substitute $250$ for $H_B$ in Equation (XXXII).

$\begin{array}{c}{}_{0.99}{\left(S_{cG}\right)}_{{10}^7}=322\left(250\right)+29100\\ =80500+29100\\ =109600\text{psi}\end{array}$

Substitute $109600\text{psi}$ for ${}_{0.99}{\left(S_{cG}\right)}_{{10}^7}$, $0.768$ for ${\left(Z_N\right)}_P$, $1$ for $S_H$, $1$ for $C_H$, $1$ for $K_T$ and $1$ for $K_R$ in Equation (XXXIII).

$\begin{array}{c}{\left({\sigma }_{\text{all}}\right)}_G=\frac{\left(109600\text{psi}\right)\left(0.768\right)\left(1\right)}{\left(1\right)\left(1\right)}\\ =84172.8\text{psi}\end{array}$

Substitute $84172.8\text{psi}$ for ${\left({\sigma }_{\text{all}}\right)}_G$, $2300$ for $C_P$, $3.25\text{in}$ for $F$, $15$ for $d_G$, $0.1170$ for $I$, $1.23$ for $K_m$, $1$ for $C_f$, $1.53$ for $K_v$, $1$ for $K_o$, $1.1514$ for ${\left(K_s\right)}_G$ in Equation (XXXIV).

$\begin{array}{c}{W}_4^t={\left(\frac{84172.8\text{psi}}{2300}\right)}^2\left[\frac{\left(3.25\text{in}\right)\left(15\right)\left(0.1170\right)}{1\times 1.53\times 1.15\times 1.23\times 1}\right]\\ ={\left(36.59\right)}^2\left[\frac{5.70375}{2.1641}\right]\\ =\left(1338.82\right)\left(2.6356\right)\\ =3528.59\end{array}$

Substitute $3528.59$ for $W_3^t$ and $1648.68\text{ft}/\min$ for $V$ in Equation (XXXV).

$\begin{array}{c}{H}_4=\frac{\left(3528.59\right)\left(1648.68\text{ft}/\min \right)}{33000}\\ =\frac{5817515.76}{33000}\\ =176.28\text{hp}\end{array}$

Substitute $390$ for $H_B$ in Equation (XXXVI).

$\begin{array}{c}{}_{0.99}{\left(S_{cpN}\right)}_{{10}^7}=322\left(390\right)+29100\\ =80500+29100\\ =154680\text{psi}\end{array}$

Substitute $390$ for $H_B$ in Equation (XXXVII).

$\begin{array}{c}{}_{0.99}{\left(S_{cGN}\right)}_{{10}^7}=322\left(390\right)+29100\\ =125580+29100\\ =154680\text{psi}\end{array}$

Substitute $154680\text{psi}$ for ${}_{0.99}{\left(S_{cpN}\right)}_{{10}^7}$, $0.7264$ for ${\left(Z_N\right)}_P$, $1$ for $S_H$, $1$ for $C_H$, $1$ for $K_T$ and $1$ for $K_R$ in Equation (XXXVIII).

$\begin{array}{c}{\left({\sigma }_{\text{all}}\right)}_{PN}=\frac{\left(154680\text{psi}\right)\left(0.7264\right)\left(1\right)}{\left(1\right)\left(1\right)}\\ =112359.55\text{psi}\end{array}$

Substitute $112359.55\text{psi}$ for ${\left({\sigma }_{\text{all}}\right)}_{PN}$, $2300$ for $C_P$, $3.25\text{in}$ for $F$, $5.5$ for $d_P$, $0.1170$ for $I$, $1.23$ for $K_m$, $1$ for $C_f$, $1.53$ for $K_v$, $1$ for $K_o$, $1.14$ for ${\left(K_s\right)}_p$ in Equation (XXXIX).

$\begin{array}{c}{W}_{3N}^t={\left(\frac{112359.55\text{psi}}{2300}\right)}^2\left[\frac{\left(3.25\text{in}\right)\left(5.5\right)\left(0.1170\right)}{1\times 1.53\times 1.14\times 1.23\times 1}\right]\\ ={\left(48.85\right)}^2\left[\frac{2.0913}{2.1453}\right]\\ =\left(2386.32\right)\left(0.9748\right)\\ =2326.18\text{lbf}\end{array}$

Substitute $2326.18\text{lbf}$ for $W_{3N}^t$ and $1648.68\text{ft}/\min$ for $V$ in Equation (XXXX).

$\begin{array}{c}{H}_{3N}=\frac{\left(2326.18\right)\left(1648.68\text{ft}/\min \right)}{33000}\\ =\frac{3835126.442}{33000}\\ =116.2\text{hp}\end{array}$

Substitute $154680\text{psi}$ for ${}_{0.99}{\left(S_{cGN}\right)}_{{10}^7}$, $0.768$ for ${\left(Z_N\right)}_P$, $1$ for $S_H$, $1$ for $C_H$, $1$ for $K_T$ and $1$ for $K_R$ in Equation (XXXXI).

$\begin{array}{c}{\left({\sigma }_{\text{all}}\right)}_{GN}=\frac{\left(154680\text{psi}\right)\left(0.768\right)\left(1\right)}{\left(1\right)\left(1\right)}\\ =118794.24\text{psi}\end{array}$

Substitute $118794.24\text{psi}$ for ${\left({\sigma }_{\text{all}}\right)}_{GN}$, $2300$ for $C_P$, $3.25\text{in}$ for $F$, $15$ for $d_G$, $0.1170$ for $I$, $1.23$ for $K_m$, $1$ for $C_f$, $1.53$ for $K_v$, $1$ for $K_o$, $1.1514$ for ${\left(K_s\right)}_G$ in Equation (XXXIV).

$\begin{array}{c}{W}_{4N}^t={\left(\frac{118794.24\text{psi}}{2300}\right)}^2\left[\frac{\left(3.25\text{in}\right)\left(15\right)\left(0.1170\right)}{1\times 1.53\times 1.15\times 1.23\times 1}\right]\\ ={\left(51.64\right)}^2\left[\frac{5.70375}{2.1641}\right]\\ =\left(2666.68\right)\left(2.6356\right)\\ =7028.30\text{lbf}\end{array}$

Substitute $7028.30\text{lbf}$ for $W_{4N}^t$ and $1648.68\text{ft}/\min$ for $V$ in Equation (XXXXII).

$\begin{array}{c}{H}_{4N}=\frac{\left(7028.30\right)\left(1648.68\text{ft}/\min \right)}{33000}\\ =\frac{11587417.64}{33000}\\ =351.13\text{hp}\end{array}$

Substitute $173.75\text{hp}$ for $H_1$ , $206.48\text{hp}$ for $H_2$ , $58.33\text{hp}$ for $H_3$ and $176.28\text{hp}$ for $H_4$ in Equation (XXXXIII).

$\begin{array}{c}{H}_{\text{rated}}=\text{min}\left(173.75\text{hp,206}\text{.48}\text{hp,58}\text{.33}\text{hp,176}\text{.28}\text{hp}\right)\\ =58.33\text{hp}\end{array}$

Thus, the rating of speed reducer for power is $58.33\text{hp}$.

Substitute $173.75\text{hp}$ for $H_1$, $206.48\text{hp}$ for $H_2$ , $116.2\text{hp}$ for $H_{3N}$ and $351.13\text{hp}$ for $H_{4N}$ in Equation (XXXXIV).

$\begin{array}{c}{H}_{\text{ratedN}}=\text{min}\left(173.75\text{hp,206}\text{.48}\text{hp,116}\text{.2}\text{hp,351}\text{.13}\text{hp}\right)\\ =116.2\text{hp}\end{array}$

Thus, the rating of speed reducer for nitrided power is $116.2\text{hp}$.