Question 9.6: A gear pair is to be designed to transmit 15.0 kilowatts (kW...

The nominal velocity ratio is

VR = 575/275 = 2.09

Specify an overload factor of K_o = 1.50 from Table 9-5 for a uniform power source and moderate shock for the meat grinder. Then computer design power,

P_{des} = K_oP = (1.50)( 15 \ kW) = 22.5 kW

From Figure 9-27, m= 4 is a reasonable trial module. Then

N_p = 18 (design decision)
D_p = N_pm = (18)(4) = 72 mm
N_G = N_p(VR) = (18)(2.09) = 37.6 (Use 38.)
D_G = N_Gm = (38)(4) = 152 mm

Final output speed = n_G = n_p(N_p/N_G)

n_G = 575 rpm × (18/38) = 272 rpm (okay)

Center distance = C=(N_p/N_G)m/2 [Equation (8-18)]

C=\frac{1}{2} \left[\frac{N_G}{P_d} +\frac{N_P}{P_d} \right] =\frac{(N_G + N_p)}{2P_d}                     (8-18)

C = (18 + 38)(4)/2 = 112 mm (okay)

In SI units, the pitch line speed in meters per second (m/s) is

V_t = πD_pn_p/(60 000)

where D_p  is in mm and n_p  is in revolutions per minute (rpm). Then

V_t = [(π)(72)(575)]/(60000) = 2.17 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)/(2.17) = 6920 N

In the U.S. Customary Unit System, it was recommended that the face width be approximately F = 12/P_d in. The equivalent SI value is F = 12(m) mm. For this problem, F = 12(4) = 48 mm. Let’s use F = 50 mm.
Other factors are found as before.

K_s= K_B= 1.00
K_v = 1.34 (Figure 9-21)
K_m= 1.21 (Figures 9-19 and 9-19) (F/D_p = 50/72 = 0.69)
J_P = 0.315        J_G = 0.380 [Figure 9-17(a)]

Then the stress in the pinion is found from Equation (9-15), modified by letting P_d = 1/m:

S_t=\frac{W_tP_d}{FJ}K_oK_sK_mK_BK_v

S_{tp}=\frac{W_tK_oK_sK_BK_mK_v}{FmJ_p}=\frac{(6920)(1.50)(1)(1.21)(1.34)}{(50)(4)(0.315)} =269MPa

This is a reasonable stress level. The required hardness of grade 1 material is HB 360, as found in Figure 9-10. Proceed with design for pitting resistance. For two steel gears,

K_s = 1.0
C_P = 191 (Table 9-10)
I = 0.092 (Figure 9-23)
K_v = 1.34
K_o = 1.50
K_m= 1.21

The contact stress [Equation (9-25)] gives

S_{c}=C_P\sqrt{\frac{W_tK_oK_sK_BK_mK_v}{FD_pI}}

S_{c}=C_P\sqrt{\frac{W_tK_oK_sK_BK_mK_v}{FD_pI}}=191\sqrt{\frac{(6920)(1.50)(1.0)(1.21)(1.34)}{(50)(72)(0.092)} }=1367 MPa

Converting to ksi gives

s_c = 1367 MPa × 1 ksi/6.895 MPa = 198 ksi

From Table 9-3, the required surface hardness is HRC 58-64, case-carburized. Grade 2. Material selection from Appendix 5, for carburized steels, is as follows:

AISI 4320 SOQT 300; s_a = 1500 MPa ; 13% elongation. Grade 2
Case-harden by carburizing to HRC 58 minimum
Case depth: 0.6 mm minimum (Figure 9-12)
Comment: Redesigning the gears to permit using Grade 1 material is recommended