Chapter 14, Problem 1P

A steel spur pinion has a pitch of 6 teeth/in, 22 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of 1200 rev/min and transmits 15 hp to a 60-tooth gear. If the face width is 2 in, estimate the bending stress.

To determine

The bending stress on teeth of gear.

Answer to Problem 1P

The bending stress on teeth of gear is $7.63\text{ kpsi}$.

Explanation of Solution

Write the expression for pitch diameter of gear.

$d=\frac{N}{P}$ (I)

Here, the pitch diameter is $d$, the number of teeth is $N$ and the pitch of gear is $P$.

Write the expression for linear velocity of gear.

$V=\pi dn$ (II)

Here, the linear velocity is $V$ and the rotation per unit time is $n$.

Write the expression for dynamic factor of cut or milled gear profile.

$K_v=\frac{1200+V}{1200}$ (III)

Here, the dynamic factor is $K_v$ and the linear velocity in $\text{ft}/\text{min}$ is $V$.

Write the expression for transmitted load.

$W_t=33000\frac{H}{V}$ (IV)

Here, the transmitted load in $\text{lbf}$ is $W_t$, the power to be transmitted in $\text{hp}$ is $H$ and the linear velocity in $\text{ft}/\text{min}$ is $V$.

Write the expression for bending stress.

$\sigma =\frac{K_v{W}_{t}P}{FY}$ (V)

Here, the bending stress in $\text{psi}$ is $\sigma$, the face width in $\text{in}$ is $F$ and the Lewis form factor is $Y$.

Conclusion:

Substitute $22\text{teeth}$ for $N$ and $6\text{teeth}/\text{in}$ for $P$ in Equation (I).

$\begin{array}{c}d=\frac{22\text{teeth}}{6\text{teeth}/\text{in}}\\ =3.667\text{ in}\end{array}$

Substitute $3.667\text{ in}$ for $d$ $\text{1200}\text{rotation}/\text{minute}$ for $n$ in Equation (II).

$\begin{array}{c}V=\pi \left(3.667\text{in}\right)\left(1200\text{rotation}/\text{min}\right)\\ =\left(13824.26\text{in}/\text{min}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\\ \approx 1152\text{ft}/\text{min}\end{array}$

Substitute $1152\text{ft}/\text{min}$ for $V$ in Equation (III).

$\begin{array}{c}{K}_v=\frac{1200+1152\text{ft}/\text{min}}{1200}\\ =1.96\end{array}$

Substitute $1152\text{ft}/\text{min}$ for $V$ and $15\text{hp}$ for $H$ in Equation (IV).

$\begin{array}{c}{W}_t=\left(33000\right)\left(\frac{15\text{hp}}{1152\text{ft}/\text{min}}\right)\\ =429.7\text{lbf}\end{array}$

Refer to Table 14-2 “Values of the Lewis Form factor Y(These Values Are for a Normal Pressure Angle of $20°$, Full-Depth Teeth and a Diametral Pitch of Unity in the Plane of Rotation)” to obtain value of Lewis form factor as $0.331$ at 22 number of teeth.

Substitute $429.7\text{lbf}$ for $W_t$, $1.96$ for $K_v$, $6\text{teeth}/\text{in}$ for $P$, $2\text{in}$ for $F$ and $0.331$ for $Y$ in Equation (V).

$\begin{array}{c}\sigma =\frac{1.96\left(429.7\text{lbf}\right)\left(6\text{teeth}/\text{in}\right)}{\left(2\text{in}\right)\left(0.331\right)}\\ =7633\text{psi}\left(\frac{1\text{kpsi}}{1000\text{psi}}\right)\\ \approx 7.63\text{kpsi}\end{array}$

Thus the bending stress on teeth of gear is $7.63\text{kpsi}$.