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## Q. 9.8

A gear pair is to be designed to transmit 15.0 kilowatts (kW) of power to a large meat grinder in a commercial meat processing plant. The pinion is attached to the shaft of an electric motor rotating at 575 rpm.
The gear must operate at 270 to 280 rpm. The gear unit will be enclosed and of commercial quality.

Commercially hobbed (quality number A11), 20°, full-depth, involute gears are to be used in the metric module system. The maximum center distance is to be 200 mm. Specify the design of the gears.

## Verified Solution

The nominal velocity ratio is
$VR = 575/275 = 2.09$
Specify an overload factor of $K_o = 1.50$ from Table 9–1 for a uniform power source and moderate shock for the meat grinder. Then compute design power,

$P_{des} = K_{o}P = (1.50)(15$ kW) = 22.5 kW
From Figure 9–11, m = 5 is a reasonable trial module. Then
$N_P = 18$ (design decision)
$D_P = N_Pm = (18)(5) = 90$ mm
$N_G = N_P(VR) = (18)(2.09) = 37.6$ (Use 38)
$D_G = N_Gm = (38)(5) = 190$ mm
Final output speed = $n_G = n_P(N_P/N_G)$
$n_G = 575 rpm × (18/38) = 272$ rpm (OK)

Center distance = $C = (N_P + N_G)$m/2[Table 8–1]
$C = (18 + 38)(5)/2 = 140$ mm (OK)
In SI units, the pitch line speed in meters per second (m/s) is
$v_t = πD_Pn_P/(60 000) = [(π)(90)(575)]/(60 000) = 2.71$ m/s
In SI units, the transmitted load, $W_t$, is in newtons (N). If the power, P, is in kW, and $v_t$ is in m/s,
$W_t = 1000(P)/v_t = (1000)(15)/(2.71) = 5536$ N
Face width, $F$: Let’s specify the nominal $F$ = 12m = 12(5) = 60 mm

Factors in stress analysis:

$K_o = 1.50$ (found earlier)
$K_s = 1.00$ (Table 9–2; m = 5)              $K_B = 1.00$ (Use solid gear blanks)
$K_R = 1.00$ (Table 9–11; 0.99 reliability)                  $S_F = 1.00$ (No unusual conditions)
$K_v = 1.31$ (Figure 9–16;      $A_v = 11$)
$K_m = 1.21$ (Figures 9–12 and 9–13;  $F = 60$ mm;  $F/D_P = 60/90 = 0.67$)
$J_P = 0.315; J_G = 0.380$ (Figure 9–10; $N_P = 18, N_G = 38$)
$C_p = 191$ (Table 9–7)      $I = 0.092$ (Figure 9–17)

Pinion contact stress:
(Equation 9–23)

$S_{c}=C_{P} \sqrt{\frac{W_{t} K_{o} K_{s} K_{m} K_{v}}{F D_{P} l}}=191 \sqrt{\frac{(5536)(1.50)(1.0)(1.21)(1.31)}{(60)(90)(0.092)}}=983 \mathrm{MPa}$
Adjustments for number of cycles, from Figures 9-21 and 9-22:
$Y_{N p}=0.94 \quad Z_{N P}=0.91 \quad Y_{N G}=0.96 \quad Z_{N G}=0.92$
Required $s_{a C P}=s_{C}(S F)\left(K_{R}\right) / Z_{N P}=983 \mathrm{MPa}(1.0)(1.0) / 0.91=1080 \mathrm{MPa}$
Using $S_{a c P}=1080 \mathrm{MPa}$, Figure 9-19 shows the required hardness $=\mathrm{HB} 396$ for through-hardened Grade 1 steel. This is acceptable but near the upper end of recommended range.

Material specification:

From Figure A4–5 (other possibilities exist),
SAE 4340 OQT 800; HB 415; $s_y$ = 1324 MPa; $s_u$ = 1448 MPa; 12% elongation.

Check other stresses:

The contact stress for the gear and the bending stress for the pinion and the gear are expected to require less material hardness and strength.
Required $s_{acG} = s_c(SF)(K_R)/Z_{NG} = 983$ MPa(1.0)(1.0)/0.92 = 1068 MPa
This is slightly lower than for the pinion (OK)

\begin{aligned}S_{t P} &=\frac{W_{t} K_{0} K_{S} K_{B} K_{m} K_{v}}{F m J_{P}}=\frac{(5536)(1.50)(1)(1)(1.21)(1.31)}{(60)(5)(0.315)}=139 \mathrm{MPa} \\\text { Required } S_{a t P} &=S_{t P}(S F)\left(K_{R}\right) / Y_{N P}=139 \mathrm{MPa}(1.0)(1.0) / 0.94=148 \mathrm{MPa}\end{aligned}
Referring to Figure 9-18, it is obvious that bending stress requires far lower hardness for the gear teeth, less than HB 180. The stress in the gear is always less than that in the pinion so it will obviously be safe as well.

Summary of the Design:
$P=15.0 \mathrm{~kW}$ from an electric motor to a large meat grinder
Pinion speed: $n_{P}=575 \mathrm{rpm}$
Gear speed: $n_{G}=272 \mathrm{rpm}$
Number of teeth: $N_{P}=18 ; N_{G}=38$ Center distance: $C=140.00 \mathrm{~mm}$
Module: $m=5 \mathrm{~mm}$
Diameters: $D_{P}=90 \mathrm{~mm} ; D_{G}=190 \mathrm{~mm}$
Material: Steel-SAE 4340 OQT 800

Comment:

A redesign may be considered with several possible approaches:
1. Increase the face width, F, to lower the stresses and permit the choice of a material with more moderate required hardness and better ductility. The recommended upper limit of face width is $16m = 16(5) = 80$ mm.
2. Increase the size of pinion and its number of teeth (same module) to lower stresses.
Possible trial: Module: $m = 5$ mm Number of teeth: $N_P = 22; N_G = 46$
Center distance: $C = 170.00$ mm Diameters: $D_P = 110[latex] mm; [latex]D_G = 230$ mm
3. Consider case-hardened steel for the initial design, rather than through-hardened steel. A smaller design is possible.

 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 9–11 Reliability Factor, $K_R$ Reliability $K_R$ 0.90, one failure in 10 0.85 0.99, one failure in 100 1.00 0.999, one failure in 1000 1.25 0.9999, one failure in 10 000 1.5
 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
 TABLE 9–7 Elastic Coefficient, $C_p$ Gear material and modulus of elasticity, $E_G, lb/in^2 (MPa)$ Pinion material Modulus of elasticity, $E_P,lb/in^2$ (MPa) Steel $30×10^6$$(2×10^5)$ Malleable iron $25×10^6$$(1.7×10^5)$ Nodular iron $24×10^6$$(1.7×10^5)$ Cast iron $22×10^6$$(1.5×10^5)$ Aluminum bronze $17.5×10^6$$(1.2×10^5)$ Tin bronze $16×10^6$$(1.1×10^5)$ Steel $30×10^6$ 2300 2180 2160 2100 1950 1900 $(2×10^5)$ 191 181 179 174 162 158 Mall. Iron $25×10^6$ 2180 2090 2070 2020 1900 1850 $(1.7×10^5)$ 181 174 172 168 158 154 Nod. Iron $24×10^6$ 2160 2070 2050 2000 1880 1830 $(1.7×10^5)$ 179 172 170 166 156 152 Cast iron $22×10^6$ 2100 2020 2000 1960 1850 1800 $(1.5×10^5)$ 174 168 166 163 154 149 Al. bronze $1.75×10^6$ 1950 1900 1880 1850 1750 1700 $(1.2×10^5)$ 162 158 156 154 145 141 Tin bronze $16×10^6$ 1900 1850 1830 1800 1700 1650 $(1.1×10^5)$ 158 154 152 149 141 137

 TABLE 8–1 Gear and Tooth Features, Diameters, Center Distance for a Gear Pair Formulas U.S. Full-depth involute system Metric module system (mm) Number of teeth and Pitches Symbol Definition Typical unit General formula Coarse pitch $P_d < 20$ (in) Fine pitch $P_d ≥ 20$ (in) Number of teeth N Integer count of teeth on a gear Circular pitch P Arc distance between corresponding points on adjacent teeth in or mm $p = πD/N$ $p = π/P_d$ $p = πm$ Diametral pitch $P_d$ Number of teeth per inch of pitch diameter $in^{-1}$ $P_d = N/D$ Module m Pitch diameter divided by number of teeth mm m = D/N $m = 25.4/P_d$ Diameters Pitch diameter D Kinematic characteristic diameter for a gear; Diameter of the pitch circle in or mm $D = N/P_d$ $D = mN$ Outside diameter $D_o$ Diameter to the outside surface of the gear teeth in or mm $D_o = (N + 2)/P_d$ $D_o = m(N + 2)$ Root diameter $D_R$ Diameter to the root circle of the gear at the base of the teeth in or mm $D_R = D - 2b$ Gear Tooth Features Addendum a Radial distance from pitch circle to outside of tooth in or mm $a = 1.00/P_d$ a = 1.00m Dedendum b Radial distance from pitch circle to bottom of tooth space in or mm $b = 1.25/P_d$ $b = 1.20/P_d + 0.002$ $b = 1.25m^1$ Clearance c Radial distance from top of fully engaged tooth of mating gear to bottom of tooth space in or mm $c = 0.25/P_d$ $c = 0.20/P_d + 0.002$ $c = 0.25m^1$ Whole depth $h_t$ Radial distance from top of a tooth to bottom of tooth space in or mm $h_t = a + b$ $h_t = 2.00/P_d$ $h_t = 2.20/P_d + 0.002$ $h_t = 2.25m^1$ Working depth $h_k$ Radial distance a gear tooth projects into tooth space of mating gear in or mm $h_k = a + a = 2a$ $h_k = 2.25/P_d$ $h_k = 2.25/P_d$ $h_k = 2.00m^1$ Tooth thickness Tensile strength Theoretical arc distance equal to 1/2 of circular pitch in or mm $t = p/2$ $t = π/[2(P_d)]$ $t = πm/2$ Face width F Width of tooth parallel to axis of gear in or mm Design decision Approximately $12/P_d$ Pressure angle $\phi$ Angle between the tangent to the pitch circle and the perpendicular to the gear tooth surface degrees Design decision Most common value = 20° Others: 14 1/2°, 25° Center Distance C Distance from between centerlines of mating gears in or mm $C = (D_P + D_G)/2$ $C = (N_P + N_G)/2P_d$ $C = m(N_P + N_G)/2$