Question 9.EP.1: Compute the bending stress numbers for the pinion and the ge...

We will use Equation (9_15) to compute the expected stress: s_{t}=\frac{W_{t} P_{d}}{F J} K_{a} K_{s} K_{m} K_{B} K_{y} We can first use the principles from Section 9_3 to compute the transmitted load on the gear teeth: D_{P}=N_{P} / P_{d}=20 / 8=2.500 \mathrm{in}

W_{t}=33000(P) / v_{t}=(33000)(25) /(1145)=720 \mathrm{lb} From Figure 9_17, we find that J_{p}=0.335 and J_{G}=0.420 The overioad factor is found from Table 9_5.

For a smooth, uniform electric motor driving an industrial saw generating moderate shock,K_{o}= 1.50 is a reasonable value.The size factor P_{d} = 1.00 because the gear teeth with K_{s} = 8 are relatively small. See Table 9_6.

The load distribution factor,K_{m} , can be found from Equation (9_16) K_{m}=1.0+C_{p f}+C_{m a} for commercial enclosed gear drives. For this design, F = 1.50 in, and F / D_{P}=1.50 / 2.50=0.60

C_{p f}=0.04(Figure 9_18)

C_{m a}=0.15 (Figure 9_19)

K_{m}=1.0+C_{p f}+C_{m a}=1.0+0.04+0.15=1.19 The rim thickness factor, K_{B}, can be taken as 1.00 because the gears are to be made from solid blanks.The dynamic factor can be read from Figure 9_21.For v_{t} = 1145 ft/min and Q_{v} =6,K_{v} =1.45. The stress can now be computed from Equation (9_15)s_{t}=\frac{W_{t} P_{d}}{F J} K_{a} K_{s} K_{m} K_{B} K_{y} We can first use the principles from Section 9_3 to compute the transmitted load on the gear teeth: D_{P}=N_{P} / P_{d}=20 / 8=2.500 \mathrm{in} .We will compute the stress in the pinion first s_{t P}=\frac{(720)(8)}{(1.50)(0.335)}(1.50)(1.0)(1.19)(1.0)(1.45)=29700 \mathrm{psi} Notice that all factors in the stress equation are the same for the gear except the value of the geometry factor, J. Then the stress in the gear can be computed from s_{t G}=\sigma_{t P}\left(J_{P} / J_{G}\right)=(29700)(0.335 / 0.420)=23700 \mathrm{psi} The stress in the pinion teeth will always be higher than the stress in the gear teeth because the value of 7 increases as the number of teeth increases.