Chapter 14, Problem 29P

To determine

The power rating for new gear-set.

Answer to Problem 29P

The power rating for new gear-set is $\text{8}\text{.42}\text{hp}$.

Explanation of Solution

Write the expression for diameter of pinion.

$d_p=\frac{N_p}{2p}$ (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.

$d_G=\frac{N_G}{2p}$ (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.

$V=\pi d_p{n}_p$ (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 gear ratio.

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

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

Write the expression for stress cycle factor for bending for pinion.

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

Here, the number of cycles is $N$.

Write the expression for stress cycle factor for bending for gear.

${\left(Y_N\right)}_G=1.6831{\left(\frac{N}{m_G}\right)}^{-0.0323}$ . (X)

Write the expression for allowable stress for pinion.

${\left({\sigma }_{\text{all}}\right)}_P=\frac{S_t{\left(Y_N\right)}_P}{S_F{K}_T{K}_R}$ (XI)

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 allowable stress for gear.

${\left({\sigma }_{\text{all}}\right)}_G=\frac{S_t{\left(Y_N\right)}_G}{S_F{K}_T{K}_R}$ (XII

Write the expression for load correction factor for uncrowned teeth.

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

Write the expression for pinion proportion factor.

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

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

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

$C_{pm}=1$ (XV)

Write the expression for mesh alignment factor.

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

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

Write the expression for mesh alignment correction factor.

$C_e=1$ (XVII)

Write the expression for load distribution factor $K_m$.

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

Write the expression for overload factor for pinion.

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

Here, the Lewis form factor for pinion is $Y_P$.

Write the expression for overload factor for gear.

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

Here, the lewis form factor for gear is $Y_G$.

Write the expression for transmitted load in pinion.

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

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

Write the expression for power for pinion.

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

Write the expression for transmitted load for gear.

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

Here the spur gear geometry factor for gear is $J_G$.

Write the expression for power for gear.

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

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

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

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

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

Write the expression for geometry factor.

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

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

Write the expression for hardness ratio factor $C_H$.

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

Write the expression for contact fatigue strength for pinion.

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

Write the expression for contact fatigue strength for gear.

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

Write the expression for pinion contact endurance strength.

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

Write the expression for transmitted load.

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

Here, the elastic coefficient is $C_P$.

Write the expression for power of pinion.

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

Write the expression for gear contact strength.

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

Write the expression for transmitted load.

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

Write the expression for power of gear.

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

Write the expression for rated power.

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

Conclusion:

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

$\begin{array}{c}{d}_p=\frac{22\text{teeth}}{2\times 4\text{teeth}/\text{in}}\\ =2.75\text{in}\end{array}$

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

$\begin{array}{c}{d}_G=\frac{60\text{teeth}}{2\times 4\text{teeth}/\text{in}}\\ =7.5\text{in}\end{array}$

Substitute $2.75\text{in}$ for $d_p$ and $1145\text{rpm}/\text{min}$ for $n_p$ in Equation (III).

$\begin{array}{c}V=\pi \left(2.75\text{in}\right)\left(1145\text{rev}/\text{min}\right)\\ =\frac{\pi \left(2.75\text{in}\right)\left(1145\text{rev}/\text{min}\right)}{12\text{in}/\text{ft}}\\ =\frac{9892.08}{12}\text{ft}/\text{min}\\ =824.34\text{ft}/\min \end{array}$

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 $A$, $824.34\text{ft}/\text{min}$ for $V$ and $0.8255$ for $B$ in Equation (VI).

$\begin{array}{c}{K}_v={\left(\frac{59.772+\sqrt{824.34}}{59.772}\right)}^{0.8255}\\ ={\left(\frac{59.772+28.71}{59.772}\right)}^{0.8255}\\ ={\left(\frac{88.482}{59.772}\right)}^{0.8255}\\ ={\left(1.480\right)}^{0.8255}\end{array}$

$K_v=1.38$

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

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

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

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

Substitute $3\left({10}^9\right)$ for $N$ and $2.727$ for $m_G$ in Equation (VIII).

$\begin{array}{c}{\left(Y_N\right)}_G=1.6831{\left(\frac{3\left({10}^9\right)}{2.727}\right)}^{-0.0323}\\ =1.6831\left(0.51045\right)\\ =0.8591\end{array}$

Substitute $32125\text{psi}$ for $S_t$, $0.8316$ for ${\left(Y_N\right)}_P$, $1$ 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 $32125\text{psi}$ for $S_t$, $0.8591$ for ${\left(Y_N\right)}_G$, $1$ for $K_R$, $1$ for $K_T$, $1$ for $S_F$ in Equation (IX).

$\begin{array}{c}{\left({\sigma }_{all}\right)}_G=\frac{\left(32125\text{psi}\right)\times 0.8591}{1\times 1\times 1}\\ =27598.58\text{psi}\end{array}$

Substitute $1.625\text{in}$ for $F$ and $2.75\text{in}$ for $d$ in Equation (XI).

$\begin{array}{c}{C}_{pf}=\frac{1.625}{10\left(2.75\right)}-0.0375+0.0125\left(1.625\right)\\ =\frac{1.625}{27.5}-0.0375+0.0203\\ \text{=0}\text{.059}-0.0375+0.0203\\ =0.0418\end{array}$

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 $B$, $1.625\text{in}$ for $F$ and $0.930\times {10}^{-4}$ for $C$ in Equation (XIII).

$\begin{array}{c}{C}_{ma}=0.127+0.0158\left(1.625\right)-\left(0.930\times {10}^{-4}\right){\left(1.625\right)}^2\\ =0.127+0.02567-\left(0.930\times {10}^{-4}\right)\left(2.640\right)\\ =0.17835-2.4552\times {10}^{-4}\\ =0.178\end{array}$

Substitute $1$ for $C_e$, $1$ for $C_{mc}$, $1$ for $C_{pm}$, $0.0418$ for $C_{Pf}$ and $0.178$ for $C_{ma}$.

$\begin{array}{c}{K}_m=1+1\left[0.0418\times 1+0.178\times 1\right]\\ =1+0.2198\\ =1.2198\end{array}$

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

Rim thickness $K_B=1$.

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 $1.625\text{in}$ for $F$, $0.331$ for $Y_p$ and $4\text{teeth}/\text{in}$ for $P$ in Equation (XVI).

$\begin{array}{c}{\left(K_s\right)}_P=1.192{\left(\frac{1.625\sqrt{0.331}}{4}\right)}^{0.0535}\\ =1.192{\left(\frac{0.9348}{4}\right)}^{0.0535}\\ =1.192{\left(0.2337\right)}^{0.0535}\\ =1.192\times 0.925\end{array}$

${\left(K_s\right)}_P=1.1026$

Substitute $1.625\text{in}$ for $F$, $0.422$ for $Y_G$ and $4\text{teeth}/\text{in}$ for $P$ in Equation (XVI).

$\begin{array}{c}{\left(K_s\right)}_G=1.192{\left(\frac{1.625\sqrt{0.422}}{4}\right)}^{0.0535}\\ =1.192{\left(\frac{1.055}{4}\right)}^{0.0535}\\ =1.192{\left(0.26375\right)}^{0.0535}\\ =1.192\times 0.9311\end{array}$

${\left(K_s\right)}_G=1.1098$

Substitute $1.625\text{in}$ for $F$, $26715.15\text{psi}$ for ${\left({\sigma }_{all}\right)}_p$, $0.345$ for $J_P$, $1.1026$ for ${\left(K_s\right)}_P$, $1$ for $K_O$, $4\text{teeth}/\text{in}$ for $P$, $1.38$ for $K_v$, $1.2198$ for $K_m$ and $1$ for $K_B$ in Equation (XIX).

$\begin{array}{c}{W}_p^t=\frac{\left(1.625\text{in}\right)\left(0.345\right)\left(26715.15\right)}{1\times 1.38\times 1.1026\times 4\times 1.2198\times 1}\\ =\frac{14977.18}{7.42}\\ =2018.48\text{lbf}\end{array}$

Substitute $2018.48\text{lbf}$ for $W_p^t$ and $824.34\text{ft}/\min$ for $V$ un Equation (XX).

$\begin{array}{c}{H}_1=\frac{\left(2018.48\right)\left(824.34\right)}{33000}\\ =\frac{1663913.803}{33000}\\ =50.42\text{hp}\end{array}$

Substitute $1.625\text{in}$ for $F$, $27598.58\text{psi}$ for ${\left({\sigma }_{all}\right)}_G$, $0.41$ for $J_P$, $1.1098$ for ${\left(K_s\right)}_G$, $1$ for $K_O$, $4\text{teeth}/\text{in}$ for $P$, $1.38$ for $K_v$, $1.2198$ for $K_m$ and $1$ for $K_B$ in Equation (XIX).

$\begin{array}{c}{W}_G^t=\frac{\left(1.625\text{in}\right)\left(0.41\right)\left(27598.58\right)}{1\times 1.38\times 1.2198\times 4\times 1.1098\times 1}\\ =\frac{18387.55}{7.47}\\ =2461.51\text{lbf}\end{array}$

Substitute $2461.51\text{lbf}$ for $W_G^t$ and $824.34\text{ft}/\min$ for $V$ un Equation (XX).

$\begin{array}{c}{H}_2=\frac{\left(2461.51\right)\left(824.34\right)}{33000}\\ =\frac{2029121.153}{33000}\\ =61.48\text{hp}\end{array}$

Substitute $3\left({10}^9\right)$ for $N$ and $2.727$ for $m_G$ in Equation (XXIV)

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

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

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

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

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

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

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

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 (XXX).

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

Substitute $79613.44\text{psi}$ for ${\left({\sigma }_{\text{all}}\right)}_P$, $2300$ for $C_P$, $1.625\text{in}$ for $F$, $2.75$ for $d_P$, $0.1170$ for $I$, $1.2198$ for $K_m$, $1$ for $C_f$, $1.38$ for $K_v$, $1$ for $K_o$, $1.1026$ for ${\left(K_s\right)}_p$ in Equation (XXXI).

$\begin{array}{c}{W}_3^t={\left(\frac{79613.44\text{psi}}{2300}\right)}^2\left[\frac{\left(1.625\text{in}\right)\left(2.75\right)\left(0.1170\right)}{1\times 1.38\times 1.1026\times 1.2198\times 1}\right]\\ ={\left(34.61\right)}^2\left[\frac{0.5228}{1.8560}\right]\\ =\left(1197.85\right)\left(0.2816\right)\\ =337.31\text{lbf}\end{array}$

Substitute $337.31\text{lbf}$ for $W_3^t$ and $824.34\text{ft}/\min$ for $V$ in Equation (XXXII).

$\begin{array}{c}{H}_3=\frac{\left(337.31\right)\left(824.34\text{ft}/\min \right)}{33000}\\ =\frac{278058.1254}{33000}\\ =8.42\text{hp}\end{array}$

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

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

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

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

Substitute $84172.8\text{psi}$ for ${\left({\sigma }_{\text{all}}\right)}_G$, $2300$ for $C_P$, $1.625\text{in}$ for $F$, $7.5$ for $d_G$, $0.1170$ for $I$, $1.2198$ for $K_m$, $1$ for $C_f$, $1.38$ for $K_v$, $1$ for $K_o$, $1.1098$ for ${\left(K_s\right)}_G$ in Equation (XXXV).

$\begin{array}{c}{W}_4^t={\left(\frac{84172.8\text{psi}}{2300}\right)}^2\left[\frac{\left(1.625\text{in}\right)\left(7.5\right)\left(0.1170\right)}{1\times 1.38\times 1.2198\times 1.1098\times 1}\right]\\ ={\left(36.59\right)}^2\left[\frac{1.4259}{1.8681}\right]\\ =\left(1338.82\right)\left(0.7632\right)\\ =1021.78\text{lbf}\end{array}$

Substitute $1021.78\text{lbf}$ for $W_3^t$ and $824.34\text{ft}/\min$ for $V$ in Equation (XXXVI).

$\begin{array}{c}{H}_4=\frac{\left(1021.78\right)\left(824.34\text{ft}/\min \right)}{33000}\\ =\frac{842294.1252}{33000}\\ =25.52\text{hp}\end{array}$

Substitute $50.42\text{hp}$ for $H_1$, $61.48\text{hp}$ for $H_2$, $8.42\text{hp}$ for $H_3$ and $25.52\text{hp}$ for $H_4$ in Equation (XXXVII).

$\begin{array}{c}{H}_{\text{rated}}=\text{min}\left(50.42\text{hp,61}\text{.48}\text{hp,8}\text{.42}\text{hp,25}\text{.52}\text{hp}\right)\\ =8.42\text{hp}\end{array}$

Thus, the rating of speed reducer for power is $\text{8}\text{.42}\text{hp}$.