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## Q. 9.3

Compute the bending stress numbers for the pinion and gear of the pair of gears similar to the pair shown in Figure 9–2(a) and (b). The pinion rotates at 1750 rpm, driven directly by an electrical motor. The driven machine is an industrial saw requiring 25 hp. The gear unit is enclosed and is made to commercial standards. Gears are straddle mounted between bearings. The following gear data apply:
$N_p = 20$        $N_g = 70$           $Av = 10$
The gear teeth are 20°, full depth, involute teeth and the gear blanks are solid. The gears will be made from steel, so Figure 9–11 can be used to select an initial value for the diametral pitch.  ## Verified Solution

The equation for the design power is:

$P_{design} = K_o . P$
The required power for the industrial saw is known to be 25 hp. The overload factor is found from Table 9–1. For a smooth, uniform electric motor driving an industrial saw generating moderate shock, a reasonable value would be:
$K_o = 1.5$
$P_{design}$= 1.5 . (25 hp) = 37.5 hp
Since the gears will be made of steel, Figure 9–11 can be used to find an initial diametral pitch based on the design power and the pinion angular velocity, $n_p$ = 1750 rpm.

The diametral pitch selected is then $P_d = 6$. Recall that this value has the unit of teeth/in or in${}^{-1}$. We will use this form later to ensure proper units in the calculation for stresses in the pinion and the gear.
Data that are useful to visualize the overall size of the gear pair and that serve as input to later decisions are now computed:
Pitch diameters of gear set:

$D_P=\frac{N_P}{P_d}=\frac{20}{6}=3.333$ in

$D_G=\frac{N_G}{P_d}=\frac{70}{6}=11.667$ in

Center distance of gear set

$C=\frac{D_P+D_G}{2}=\frac{3.333 in + 11.667 in}{2}=7.500$ in

Velocity ratio of gear set:

$VR=m_G=\frac{n_P}{n_G}=\frac{N_G}{N_P}=\frac{70}{20}=3.50$

The speed of the gear can be found from by rewriting the equation for the velocity ratio:

$VR=\frac{n_P}{n_G}$

$n_G=\frac{n_P}{VR}=\frac{1750 rpm}{3.50}= 500$ rpm

The pitch line speed (in ft/min or fpm) can be calculated using the pitch diameter and the angular velocity of the pinion:

$v_{t}=\frac{D_{p}}{2} \cdot n_{p}=\frac{3.333 \mathrm{in}}{2} \cdot 1750 \mathrm{rev} / \mathrm{min} \cdot \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \cdot \frac{1 \mathrm{ft}}{12 \mathrm{in}}=1527 \mathrm{fpm}$

We can also use the gear pitch diameter and angular velocity to calculate the pitch line speed:

$v_{t}=\frac{D_{G}}{2} \cdot n_{G}=\frac{11.667 \mathrm{in}}{2} \cdot 500 \mathrm{rev} / \mathrm{min} \cdot \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \cdot \frac{1 \mathrm{ft}}{12 \mathrm{in}}=1527 \mathrm{fpm}$

We can then use the principles from Section 9–3 to compute the transmitted load on the gear teeth. First find the torque on the pinion and gear:

\begin{aligned}&T_{P}=\frac{\text { Power }}{n_{P}}=\frac{25 \mathrm{hp}}{1750 \mathrm{rev} / \mathrm{min}} \cdot\frac{33000 \frac{\mathrm{lb} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2\pi \mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=900.4 \mathrm{lb} \cdot \mathrm{in} \\&T_{G}=\frac{\text { Power }}{n_{G}}=\frac{25 \mathrm{hp}}{500 \mathrm{rev} / \mathrm{min}} \cdot \frac{33000\frac{\mathrm{b} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2 \pi\mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=3151.3 \mathrm{lb} \cdot \mathrm{in}\end{aligned}

We can see that a smaller torque on the pinion shaft will produce a larger torque on the gear shaft.

$\frac{T_G}{T_P}=\frac{3151.3 lb . in}{900.4 lb . in}= 3.50$

You will notice this is the same value as the velocity ratio. The output torque will increase in proportion to the decrease in angular velocity. This means that the motor torque required is 900.4 lb-in for an output
torque of 3151.3 lb-in.
We will use the Equation (9–16) to compute the bending stress number:

$s_t=\frac{W_tP_d}{FJ}K_oK_sK_mK_BK_v$

Let’s go through each term of this equation:
The tangential force (gear driving force) is based on the pinion torque and pinion pitch diameter:

$W_t= \frac{T_P}{(\frac{D_P}{2})}=\frac{900.4 lb . in}{(\frac{3.333in}{2})}=540.3$ lb

The tangential force can also be calculated using the gear torque and gear pitch diameter:

$W_t= \frac{T_G}{(\frac{D_G}{2})}=\frac{3151.3 lb . in}{(\frac{10.667in}{2})}=540.3$ lb

Although the radial and normal gear forces are not required for the bending stress equation, we will calculate all the forces on the spur gear teeth. The radial force (gear separating force) is:

$W_t= W_t.tan(\varphi)$ = 540.3 lb . tan(20°) = 196.6 lb

The normal force along the line of action is:

$W_n=\sqrt{W_t^2+W_t^2} = \sqrt{(540.3 lb)^2+(196.6 lb)^2}= 575 lb$

The nominal face width of the gears has been defined as $F = 12/P_d$ and we know $Pd = 6$.

$F=\frac{12}{P_d}=\frac{12}{6}=2.00$ in

The geometry factor, J, will be determined for both the pinion and the gear. The pressure angle is 20° so we use Figure 9–10(a).

$J_P = 0.335$
$J_G = 0.420$
$K_o = 1.5$
Teeth with a diametral pitch, $P_d = 6$, are relatively small so, from Table 9–2, the size factor is:
$K_S = 1$

The load-distribution factor, $K_m$, can be found from Equation (9–17) ($K_m = 1.0 + C_{pf} + C_{ma}$) for commercial enclosed gear drives. For this design, $F = 2.00$ in and

$\frac{F}{D_P}=\frac{2.00 in}{3.333in}=0.60$

From Figure 9–12 we can find the approximate value for the pinion proportion factor:
$C_{pf} ≈ 0.04$
To obtain a more precise value for the pinion proportion factor, we can use the equation that represents
the curve in Figure 9–12, when 1.0 ≤ $F$ < 15:

$C_{pf}=\frac{F}{10D_P}- 0.0375 + 0.0125F =\frac{2.00}{10 *3.333}- 0.0375 + 0.0125 * 2.00 = 0.047$

The mesh alignment factor, $C_{ma}$, can be determined from Figure 9–13 using the commercial enclosed gear unit curve:
$C_{ma}$ = 0.16
We can also use the equation that represents the commercial enclosed gear unit curve to calculate a more precise value:

$C_{ma}=0.127 + 0.0158 . F- 1.093 × 10^{-4}. F^2$ $C_{ma}=0.127 + 0.0158 *2.00- 1.093 × 10^{-4}. (2.00)^2=0.158$

Substitute these two factors into Equation (9–17) to calculate the load distribution factor:
$K_m = 1.0 + C_{PF} + C_{ma}$ = 1.0 + 0.047 + 0.158 = 1.21
The rim thickness factor, $K_B$, can be taken as 1.00 because the gears are to be made from solid blanks with no cast or machined rim.
$K_B$ = 1.00

The dynamic factor, $K_v$, can be read from Figure 9–16. For a pitch line velocity, $v_t = 1527$ fpm and a gear quality number of, $A_v = 10$, the dynamic factor is:
$K_v = 1.41$
The bending stress can now be computed from Equation (9–16). We will compute the bending stress of the pinion first:

$s_{tp}=\frac{540lb*6 in^{-1}}{2.00 in *0.335}*1.50*1.0*1.21*1.0*1.41=12376$ psi

Notice all factors in the stress equation are the same for the gear except the value of the geometry factor, $J$. The bending stress number for the gear is

$s_{tG}=\frac{540lb*6 in^{-1}}{2.00 in *0.420}*1.50*1.0*1.21*1.0*1.41=9871$ psi

The stress in the pinion teeth will always be higher than the stress in the gear teeth because the value of $J$ increases as the number of teeth increases.

 TABLE 9–1 Suggested Overload Factors, $K_o$ Driven Machine Power source Uniform Light shock Moderate shock Heavy shock Uniform 1.00 1.25 1.50 1.75 Light shock 1.20 1.40 1.75 2.25 Moderate shock 1.30 1.70 2.00 2.75

 TABLE 9–2 Suggested Size Factors, $K_s$ Diametral pitch, $P_d$ Metric module, m Size factor, $K_s$ ≥ 5 ≤ 5 1.00 4 6 1.05 3 8 1.15 2 12 1.25 1.25 20 1.40    