A stock spur gear is available having a diametral pitch of 8 teeth/in, a 11/2 -in face, 16 teeth, and a pressure angle of 20◦ with full-depth teeth. The material is AISI 1020 steel in asrolled condition. Use a design factor of nd = 3 to rate the horsepower output of the gear corresponding to a speed

Question

A stock spur gear is available having a diametral pitch of 8 teeth/in, a [latex]1 \frac{1}{2}[/latex] -in face, 16 teeth, and a pressure angle of 20◦ with full-depth teeth. The material is AISI 1020 steel in asrolled condition. Use a design factor of [latex]n_{d}=3[/latex] to rate the horsepower output of the gear corresponding to a speed of 1200 rev/m and moderate applications.

Accepted Answer

The term moderate applications seems to imply that the gear can be rated by using the yield strength as a criterion of failure. From Table A–20, we find [latex]S_{u t}=55 kpsi[/latex] and [latex]S_{y}=30 kpsi[/latex]. A design factor of 3 means that the allowable bending stress is 30/3 = 10 kpsi. The pitch diameter is N/P = 16/8 = 2 in, so the pitch-line velocity is

[latex]V=\frac{\pi d n}{12}=\frac{\pi(2) 1200}{12}=628 ft / \min[/latex]

The velocity factor from Eq. (14–4b) is found to be

[latex]K_{v}=\frac{1200+V}{1200} \quad( ext { cut or milled profile })[/latex] (14–4b)

[latex]K_{v}=\frac{1200+V}{1200}=\frac{1200+628}{1200}=1.52[/latex]

Table 14–2 gives the form factor as Y = 0.296 for 16 teeth. We now arrange and substitute in Eq. (14–7) as follows:

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)
Number of Teeth	Y	Number of Teeth	Y
12	0.245	28	0.353
13	0.261	30	0.359
14	0.277	34	0.371
15	0.290	38	0.384
16	0.296	43	0.397
17	0.303	50	0.409
18	0.309	60	0.422
19	0.314	75	0.435
20	0.322	100	0.447
21	0.328	150	0.460
22	0.331	300	0.472
24	0.337	400	0.480
26	0.346	Rack	0.485

[latex]\sigma_{ ext {all }}=\left\{\begin{array}{ll}\frac{S_{t}}{S_{F}} \frac{Y_{N}}{K_{T} K_{R}} & ext { (U.S. customary units) } \\frac{S_{t}}{S_{F}} \frac{Y_{N}}{Y_{ heta} Y_{Z}} & ext { (SI units) }\end{array} ight.[/latex] (14–17)

[latex]W^{t}=\frac{F Y \sigma_{ ext {all }}}{K_{v} P}=\frac{1.5(0.296) 10000}{1.52(8)}=365 lbf[/latex]

The horsepower that can be transmitted is

[latex]h p=\frac{W^{t} V}{33000}=\frac{365(628)}{33000}=6.95 hp[/latex]

It is important to emphasize that this is a rough estimate, and that this approach must not be used for important applications. The example is intended to help you understand some of the fundamentals that will be involved in the AGMA approach.

Question 14.1: A stock spur gear is available having a diametral pitch of 8...

Related Answered Questions