Chapter 15, Problem 1P

An uncrowned straight-bevel pinion has 20 teeth, a diametral pitch of 6 teeth/in, and a transmission accuracy number of 6. Both the pinion and gear are made of through-hardened steel with a Brinell hardness of 300. The driven gear has 60 teeth. The gearset has a life goal of 10⁹ revolutions of the pinion with a reliability of 0.999. The shaft angle is 90°, and the pinion speed is 900 rev/min. The face width is 1.25 in, and the normal pressure angle is 20°. The pinion is mounted outboard of its bearings, and the gear is straddle-mounted. Based on the AGMA bending strength, what is the power rating of the gearset? Use K₀ = 1 and S_F = S_H = 1.

To determine

The power rating of gear-set for bending.

Answer to Problem 1P

The power rating for gear-set for bending is $14.06\text{hp}$.

Explanation of Solution

Write the expression for pitch diameter of gear.

$d_G=\frac{N}{P_d}$ (I)

Here, the pitch diameter of gear is $d_G$, the number of teeth on gear is $N$ and the diametral pitch is $P_d$.

Write the expression for pitch diameter of pinion.

$d_P=\frac{n}{P_d}$ (II)

Here, the pitch diameter of pinion is $d_P$, the number of teeth on pinion is $n$ and the diametral pitch is $P_d$.

Write the expression for speed ratio of gear set.

$m=\frac{N}{n}$ (III)

Here, the speed ratio is $m$.

Write the expression for pitch line velocity.

$v_t=\pi d_P{n}_P$ (IV)

Here, the pitch line velocity is $v_t$ and the rotation of pinion per unit time is $n_P$.

Write the expression for dynamic factor.

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

Here, dynamic factor is $K_v$ and constants are $A$ and $B$.

Write the expression for constant $B$.

$B=0.25{\left(12-Q_v\right)}^{2/3}$ (VI)

Here, the transmission accuracy number is $Q_v$.

Write the expression for constant $A$.

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

Write the expression for maximum pitch line velocity.

$v_{t,\max }={\left[A+\left(Q_v-3\right)\right]}^2$ (VIII)

Here, maximum pitch line velocity is $v_{t,\max }$.

Write the expression for size factor for $0.5\le P_d\le 16\text{teeth}/\text{in}$.

$K_S=0.4867+0.2132/P_d$ (IX)

Here, size factor is $K_S$.

Write the expression for load distribution factor.

$K_m=K_{mb}+0.0036F^2$ (X)

Here, load distribution factor is $K_m$, the face width is $F$ and the constant is $K_{mb}$.

Write the critical expression for stress cycle factor for pinion.

$K_L^P=1.683{\left(N_L^P\right)}^{-0.0323}$ for $3\times {10}^6\le N_L^P\le {10}^{10}$ (XI)

Here, the stress cycle factor for pinion is $K_L^P$ and the life goal in $\text{rotation}$ for pinion is $N_L^P$.

Write the critical expression for stress cycle factor for gear.

$K_L^G=1.683{\left(\frac{N_L^P}{m}\right)}^{-0.0323}$ for $3\times {10}^6\le N_L^P\le {10}^{10}$ (XII)

Here, the stress cycle factor for gear is $K_L^G$.

Write the expression for reliability factor for bending strength.

$K_R=0.5-0.25\log \left(1-R\right)$ for $0.99\le R\le 0.999$ (XIII)

Here, the reliability factor is $K_R$ and the reliability is $R$.

Write the expression for allowable bending stress number.

$S_{at}=44H_B+2100$ (XIV)

Here, the allowable bending stress number is $S_{at}$ and the Brinell hardness is $H_B$.

Write the expression for permissible bending stress for pinion.

$S_{wt}^P=\frac{S_{at}{K}_L^P}{S_F{K}_T{K}_R}$ (XV)

Here, the permissible bending stress for pinion is $S_{wt}^P$, the bending safety factor is $S_F$ and the temperature factor is $K_T$.

Write the expression for permissible bending stress for gear.

$S_{wt}^G=\frac{S_{at}{K}_L^G}{S_F{K}_T{K}_R}$ (XVI)

Here, the permissible bending stress for gear is $S_{wt}^G$.

Write the expression for bending stress for pinion.

$s_t^P=\frac{W_t^P}{F}{P}_d{K}_o{K}_v\frac{K_S{K}_m}{K_x{J}^P}$ (XVII)

Here, the bending stress in pinion is $S_t^P$, the transmitted load by pinion is $W_t^P$, the overload factor is $K_o$, length wise curvature factor is $K_x$ and geometry factor for bending in pinion is $J^P$.

Write the expression for bending stress for gear.

$s_t^G=\frac{W_t^G}{F}{P}_d{K}_o{K}_v\frac{K_S{K}_m}{K_x{J}^G}$ (XVIII)

Here, the bending stress in gear is $S_t^G$, the transmitted load on gear is $W_t^G$ and geometry factor for bending in gear is $J^G$.

Write the expression for transmitted load by pinion.

$W_t^P=33000\frac{H^P}{v_t}$ (XIX)

Here, the transmitted load by pinion is $W_t^P$ and the power in $\text{hp}$ for pinion is $H^P$.

Write the expression for transmitted load on gear.

$W_t^G=33000\frac{H^G}{v_t}$ (XX)

Here, the transmitted load on gear is $W_t^G$ and the power in $\text{hp}$ for gear is $H^G$.

Conclusion:

Substitute $60\text{teeth}$ for $N$ and $6\text{teeth}/\text{in}$ for $P_d$ in Equation (I).

$\begin{array}{c}{d}_G=\frac{60\text{teeth}}{6\text{teeth}/\text{in}}\\ =10\text{in}\end{array}$

Substitute $20\text{teeth}$ for $n$ and $6\text{teeth}/\text{in}$ for $P_d$ in Equation (II).

$\begin{array}{c}{d}_P=\frac{20\text{teeth}}{6\text{teeth}/\text{in}}\\ =3.333\text{in}\end{array}$

Substitute $60\text{teeth}$ for $N$ and $20\text{teeth}$ for $n$ in Equation (III).

$\begin{array}{c}m=\frac{60\text{teeth}}{20\text{teeth}}\\ =3\end{array}$

Substitute $3.333\text{in}$ for $d_P$ and $900\text{rotation}/\text{min}$ for $n_P$ in Equation (IV).

$\begin{array}{c}{v}_t=\pi \left(3.333\text{in}\right)900\text{rotation}/\text{min}\\ =9423.835\text{in}/\text{min}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =785.3\text{ft}/\text{min}\end{array}$

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

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

Substitute $0.8255$ for $B$ in Equation (VII)

$\begin{array}{c}A=50+56\left(1-0.8255\right)\\ =59.77\end{array}$

Substitute $59.77$ for $A$, $0.8255$ for $B$ and $785.3\text{ft}/\text{min}$ for $v_t$ in Equation (V).

$\begin{array}{c}{K}_v={\left(\frac{59.77+\sqrt{785.3\text{ft}/\text{min}}}{59.77}\right)}^{0.8255}\\ ={\left(1.4688\right)}^{0.8255}\\ =1.374\end{array}$

Substitute $59.77$ for $A$ and $6$ for $Q_v$ in Equation (VIII).

$\begin{array}{c}{v}_{t,\max }={\left[59.77+\left(6-3\right)\right]}^2\\ ={\left[59.77+\left(3\right)\right]}^2\\ =3940\text{ft}/\text{min}\end{array}$

Substitute $6\text{teeth}/\text{in}$ for $P_d$ in Equation (IX).

$\begin{array}{c}{K}_S=0.4867+0.2132/6\text{teeth}/\text{in}\\ =0.5222\end{array}$

Substitute $1.1$ for $K_{mb}$ for one gear straddle-mounted and $1.25\text{in}$ for $F$ in Equation (X).

$\begin{array}{c}{K}_m=1.1+0.0036{\left(1.25\right)}^2\\ =1.106\end{array}$

Substitute ${10}^9\text{rotation}$ for $N_L^P$ in Equation (XI).

$\begin{array}{c}{K}_L^P=1.683{\left({10}^9\text{rotation}\right)}^{-0.0323}\\ =0.862\end{array}$

Substitute $3$ for $m$ and ${10}^9\text{rotation}$ for $N_L^P$ in Equation (XII).

$\begin{array}{c}{K}_L^G=1.683{\left(\frac{{10}^9}{3}\right)}^{-0.0323}\\ =0.893\end{array}$

Substitute $0.999$ for $R$ in Equation (XIII).

$\begin{array}{c}{K}_R=0.5-0.25\log \left(1-0.999\right)\\ =1.25\end{array}$

Substitute $300$ for $H_B$ in Equation (XIV).

$\begin{array}{c}{S}_{at}=44\left(300\right)+2100\\ =15300\text{psi}\end{array}$

Substitute $15300\text{psi}$ for $S_{at}$, $0.862$ for $K_L^P$, $1$ for $S_F$, $1$ for $K_T$ and $1.25$ for $K_R$ in Equation (XV).

$\begin{array}{c}{S}_{wt}^P=\frac{\left(15300\text{psi}\right)\text{0}\text{.862}}{1\times 1\times 1.25}\\ =10550.88\text{psi}\end{array}$

Refer to Figure 15-7 “Bending factor $J\left(Y_J\right)$ for coniflex straight-bevel gears with a $20°$ normal pressure angle and $90°$ shaft angle” to obtain value of $J^P$ as $0.249$.

Substitute $10550.88\text{psi}$ for $S_t^P$, $1.25\text{in}$ for $F$, $6\text{teeth}/\text{in}$ for $P_d$, $1$ for $K_o$, $1.374$ for $K_v$, $0.5222$ for $K_S$, $0.249$ for $J^P$ and $1.106$ for $K_m$ in Equation (XVII).

$\begin{array}{l}10550.88\text{psi}=\frac{W_t^P}{1.25\text{in}}\left(6\text{teeth}/\text{in}\right)\left(1\right)\left(1.374\right)\left(\frac{0.5222\times 1.106}{1\times 0.249}\right)\\ W_t^P=\frac{10550.88\times 1.25\times 0.249}{6\times 1.374\times 0.5222\times 1.106}\text{lbf}\\ W_t^P=689.71\text{lbf}\end{array}$

Substitute $689.71\text{lbf}$ for $W_t^P$ and $785.3\text{ft}/\text{min}$ for $v_t$ in Equation (XIX).

$\begin{array}{l}689.71\text{lbf}=33000\frac{H^P}{785.3\text{ft}/\text{min}}\\ H^P=\frac{689.71\times 785.3}{33000}\text{hp}\\ H^P=16.41\text{hp}\end{array}$

Substitute $15300\text{psi}$ for $S_{at}$, $0.893$ for $K_L^P$, $1$ for $S_F$, $1$ for $K_T$ and $1.25$ for $K_R$ in Equation (XVI).

$\begin{array}{c}{S}_{wt}^G=\frac{\left(15300\text{psi}\right)\times 0.893}{1\times 1\times 1.25}\\ =10930.32\text{psi}\end{array}$

Refer to Figure 15-7 “Bending factor $J\left(Y_J\right)$ for coniflex straight-bevel gears with a $20°$ normal pressure angle and $90°$ shaft angle” to obtain value of $J^G$ as $0.206$.

Substitute $10930.32\text{psi}$ for $S_t^G$, $1.25\text{in}$ for $F$, $6\text{teeth}/\text{in}$ for $P_d$, $1$ for $K_o$, $1.374$ for $K_v$, $0.5222$ for $K_S$, $0.206$ for $J^G$ and $1.106$ for $K_m$ in Equation (XVIII).

$\begin{array}{l}10930.32\text{psi}=\frac{W_t^G}{1.25\text{in}}\times 6\text{teeth}/\text{in}\times 1\times 1.374\times \frac{0.5222\times 1.106}{1\times 0.206}\\ W_t^G=\frac{10930.32\times 1.25\times 0.206}{6\times 1.374\times 0.5222\times 1.106}\text{lbf}\\ W_t^G=591.13\text{lbf}\end{array}$

Substitute $591.13\text{lbf}$ for $W_t^G$ and $785.3\text{ft}/\text{min}$ for $v_t$ in Equation (XX).

$\begin{array}{l}591.13\text{lbf}=33000\frac{H^G}{785.3\text{ft}/\text{min}}\\ H^G=\frac{785.3\times 591.13}{33000}\\ H^G=14.06\text{hp}\end{array}$

The power rating for a gear-set for bending is defined as minimum of power rating in bending for pinion and gear, so that the bending of teeth on both gear and pinion can be saved.

Thus, the power rating for gear-set for bending is $14.06\text{hp}$.