18 Budynas-Nisbett:Shigley's Ill.Design of Mechanical 14.Spur and Helical Gears T©The McGraw-Hill Mechanical Engineering Elements Companies,2008 Design,Eighth Edition 720 Mechanical Engineering Design estimating the capacity of gear drives when life and reliability are not important con- siderations.The equations can be useful in obtaining a preliminary estimate of gear sizes needed for various applications. EXAMPLE 14-1 A stock spur gear is available having a diametral pitch of 8 teeth/in,a 1-in face,16 teeth,and a pressure angle of 20 with full-depth teeth.The material is AISI 1020 steel in as-rolled condition.Use a design factor of nd=3 to rate the horsepower output of the gear corresponding to a speed of 1200 rev/m and moderate applications. Solution 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 S=55 kpsi and Sy=30 kpsi.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 V=Td=π(21200 12= =628 ft/min 12 The velocity factor from Eq.(14-4b)is found to be K,=1200+V=1200+628 =1.52 1200 1200 Table 14-2 gives the form factor as Y=0.296 for 16 teeth.We now arrange and sub- stitute in Eq.(14-7)as follows: W= FYoa_1.5(0.296)10000 =3651bf KuP 1.52(8) The horsepower that can be transmitted is W1V365(628) Answer hp=33000=33000 =6.95hp 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 under- stand some of the fundamentals that will be involved in the AGMA approach. EXAMPLE 14-2 Estimate the horsepower rating of the gear in the previous example based on obtaining an infinite life in bending. Solution The rotating-beam endurance limit is estimated from Eq.(6-8) S%=0.5Sr=0.5(55)=27.5kpsi To obtain the surface finish Marin factor k we refer to Table 6-3 for machined surface. finding a =2.70 and b=-0.265.Then Eq.(6-19)gives the surface finish Marin factor ka as ka=a50=2.70(55)-0265=0.934Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears 718 © The McGraw−Hill Companies, 2008 720 Mechanical Engineering Design EXAMPLE 14–1 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 as-rolled condition. Use a design factor of nd = 3 to rate the horsepower output of the gear corresponding to a speed of 1200 rev/m and moderate applications. Solution 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 Sut = 55 kpsi and Sy = 30 kpsi. 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 V = πdn 12 = π(2)1200 12 = 628 ft/min The velocity factor from Eq. (14–4b) is found to be Kv = 1200 + V 1200 = 1200 + 628 1200 = 1.52 Table 14–2 gives the form factor as Y = 0.296 for 16 teeth. We now arrange and sub￾stitute in Eq. (14–7) as follows: Wt = FYσall Kv P = 1.5(0.296)10 000 1.52(8) = 365 lbf The horsepower that can be transmitted is Answer hp = Wt V 33 000 = 365(628) 33 000 = 6.95 hp 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 under￾stand some of the fundamentals that will be involved in the AGMA approach. estimating the capacity of gear drives when life and reliability are not important con￾siderations. The equations can be useful in obtaining a preliminary estimate of gear sizes needed for various applications. EXAMPLE 14–2 Estimate the horsepower rating of the gear in the previous example based on obtaining an infinite life in bending. Solution The rotating-beam endurance limit is estimated from Eq. (6–8) S e = 0.5Sut = 0.5(55) = 27.5 kpsi To obtain the surface finish Marin factor ka we refer to Table 6–3 for machined surface, finding a = 2.70 and b = −0.265. Then Eq. (6–19) gives the surface finish Marin factor ka as ka = aSb ut = 2.70(55) −0.265 = 0.934