Chapter 14, Problem 25P

The speed reducer of Prob. 14-24 is to be used for an application requiring 40 hp at 1145 rev/min. For the gear and the pinion, estimate the AGMA factors of safety for bending and wear, that is, (S_F)_P, (S_F)_G, (S_H)_P, and (S_H)_G. By examining the factors of safety, identify the threat to each gear and to the mesh.

14-24 A speed-reducer has 20° full-depth teeth, and the single-reduction spur-gear gearset has 22 and 60 teeth. The diametral pitch is 4 teeth/in and the face width is $3\frac{1}{4}$ in. The pinion shaft speed is 1145 rev/min. The life goal of 5-year 24-hour-per-day service is about 3(10⁹) pinion revolutions. The absolute value of the pitch variation is such that the quality number is 6. The materials are 4340 through-hardened grade 1 steels, heat-treated to 250 Brinell, core and case, both gears. The load is moderate shock and the power is smooth. For a reliability of 0.99, rate the speed reducer for power.

To determine

The factor of safety for bending for pinion.

The factor of safety for bending for gear.

The factor of safety for wear for pinion.

The factor of safety for wear for gear.

Answer to Problem 25P

The factor of safety for bending for pinion is $3.47$.

The factor of safety for bending for gear is $4.13$.

The factor of safety for wear for pinion is $1.07$.

The factor of safety for wear for gear is $1.87$.

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 transmitted load for gear.

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

Write the expression for gear ratio.

$m_G=\frac{N_G}{N_p}$ (XX)

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.

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

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

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

Write the expression for geometry factor.

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

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

Write the expression for hardness ratio factor $C_H$.

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

Write the expression for contact fatigue strength for pinion.

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

Write the expression for contact fatigue strength for gear.

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

Write the expression for pinion contact endurance strength.

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

Write the expression for transmitted load.

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

Write the expression for gear contact strength.

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

Write the expression for transmitted load.

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

Her, the elastic coefficient is $C_P$.

Write the expression for transmitted load.

$W^t=\frac{33000K_{O}H}{V}$ (XXXI)

Write the expression for factor of safety based on load and stress for pinion bending.

${\left(S_F\right)}_P=\frac{W_{{}_1}^t}{W^t}$ (XXXII)

Write the expression for factor of safety based on load and stress for gear bending.

${\left(S_F\right)}_G=\frac{W_2^t}{W^t}$ (XXXIII)

Write the expression for factor of safety based on load.

$n_3=\frac{W_3^t}{W^t}$ . (XXXIV)

Write the expression for based on stress.

${\left(S_H\right)}_P=\sqrt{n_3}$ (XXXV)

Write the expression for factor of safety on load.

$n_4=\frac{W_4^t}{W^t}$                                                                (XXXVI)

Write the expression for factor of safety based on stress.

${\left(S_H\right)}_P=\sqrt{n_4}$ (XXXVII)

Conclusion:

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

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

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

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 $3.25\text{in}$ 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 $3.25\text{in}$ 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 $60$ 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 $N$ 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 (XXVI).

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

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 (XXVII).

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

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

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

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 (XXIX).

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

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 (XXX).

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

Substitute $1.25$ for $k_o$, $40\text{hp}$ for $H$ and $1648.68\text{ft}/\text{min}$ for $V$ in Equation (XXXI)

$\begin{array}{c}{W}^t=\frac{33000\left(1.25\right)\left(40\text{hp}\right)}{1648.68}\\ =\frac{1650000}{1648.68}\\ =1000.80\text{lbf}\end{array}$

Substitute $3477.80\text{lbf}$ for $W_1^t$ and $1000\text{lbf}$ for $W^t$ in Equation (XXXII).

$\begin{array}{c}{\left(S_F\right)}_P=\frac{3477.80\text{lbf}}{1000\text{lbf}}\\ =3.47\end{array}$

Thus, the factor of safety for bending for pinion is $3.47$.

Substitute $4133.046\text{lbf}$ for $W_2^t$ and $1000\text{lbf}$ for $W^t$ in Equation (XXXIII).

$\begin{array}{c}{\left(S_F\right)}_G=\frac{4133.046\text{lbf}}{1000\text{lbf}}\\ =4.13\end{array}$

Thus, the factor of safety for bending for gear is $4.13$.

Substitute $1167.66\text{lbf}$ for $W_3^t$ and $1000\text{lbf}$ for $W^t$ in Equation (XXXIV).

$\begin{array}{c}{n}_3=\frac{1167.66\text{lbf}}{1000\text{lbf}}\\ =1.16\end{array}$

Substitute $1.16$ for $n_3$ in Equation (XXXV).

$\begin{array}{c}{\left(S_H\right)}_P=\sqrt{1.16}\\ =1.07\end{array}$

Thus, the factor of safety for wear for pinion is $1.07$.

Substitute $3528.59\text{lbf}$ for $W_4^t$ and $1000\text{lbf}$ for $W^t$ in Equation (XXXVI).

$\begin{array}{c}{n}_4=\frac{3528.59\text{lbf}}{1000\text{lbf}}\\ =3.5\end{array}$

Substitute $3.5$ for $n_4$ in Equation (XXXVII).

$\begin{array}{c}{\left(S_H\right)}_P=\sqrt{3.5}\\ =1.87\end{array}$

Thus, the factor of safety for wear for gear is $1.87$.

The factor of safety for wear for pinion is $1.07$ which least all of them so wear is the most threatening mode of failure.