n05_static page 1 of 6 L = 0.75 in r = 0.05 in D = 0.75 in d = 0.5 in 9T #41 roller chain sprocket SAE 1045 cold-rolled steel 1 HP at 400 rpm P=T 33000 ft .lbf T = P / = 1 HP HP. min min rev 12 in = 157.6 in.lbf 400 rev 2 rad ft rSPR F #41 ANSI chain, pitch p = 0.5 inch, maximum working load = 314 lbf, McMaster-Carr pitch circumference = 2 rSPR ≈ (9T) p rSPR ≈ (9T) p / 2 = 0.7162 in rSPR = 360° / number of teeth = 40° p = 0.5 in sin /2 = (p/2) / rSPR EXACT rSPR = (p/2) / (sin /2) = 0.731 in T = F rSPR F = T / rSPR = 220 lbf bending moment M = F L = 165 in.lbf torsion moment T = 157.6 in.lbf I = d4 / 64 = 3.068 x 10-3 in4 c = d / 2 = 0.25 in J = d4 / 32 = 6.136 x 10-3 in4 p n05_static 165 in .lbf 0.25 in NOM = M c / I = NOM = T c / J = 3.068x10 3 in 4 157.6 in .lbf 0.25 in 6.136x10 3 in 4 page 2 of 6 = 13.45 ksi = 6.421 ksi maximum shear failure criterion (OLD) x = 13.45 ksi x / 2 MAX = y = 0 0.5 S y MAX xy = 6.421 ksi 2 2 xy 1045 cold-rolled steel N= = 4.14 = 9.30 ksi Sy = 77 ksi y -xy Table A-9 Norton Eq 5.10 and 5.11 Norton von Mises failure criterion x = 13.45 ksi ’ = y = 0 xy = 6.421 ksi 2x 2y x y 3 2xy = 17.45 ksi 1045 cold-rolled steel N= Sy ' = 4.41 Sy = 77 ksi Eq 5.7d Norton Table A-9 Norton Eq 5.8a Norton iterative solution for new diameter NOM = M c / I = x xy M d2 64 32 M d4 d3 NOM = T c / J = try NEW d = 0.625 in 3 0 .5 NEW ’ = OLD ’ = 0.512 OLD ’ 0.625 T d2 32 16 T d4 d3 n05_static page 3 of 6 static stress concentration factors shaft bending D/d = 1.5 A = 0.93836 b = -0.25759 Figure C-2 Norton Kt = A(r/d)b = 1.698 r/d = 0.1 = Kt NOM = 22.84 ksi shaft torsion D/d = 1.5 interpolated A = 0.8526 b = -0.2334 Figure C-3 Norton Kt = A(r/d)b = 1.459 r/d = 0.1 = Kt NOM = 9.368 ksi maximum shear failure criterion (OLD) x = 22.84 ksi x / 2 MAX = y = 0 0.5 S y MAX xy = 9.368 ksi 2 2 xy 1045 cold-rolled steel N= = 2.61 = 14.77 ksi Sy = 77 ksi y -xy Table A-9 Norton Eq 5.10 and 5.11 Norton von Mises failure criterion x = 22.84 ksi ’ = y = 0 xy = 9.368 ksi 2x 2y x y 3 2xy = 28.02 ksi 1045 cold-rolled steel N= Sy ' = 2.75 x xy Sy = 77 ksi Eq 5.7d Norton Table A-9 Norton Eq 5.8a Norton key length assume 1020 hot-rolled Sy = 30 ksi Table A-9 Norton n05_static d = 0.5 inch DIA shaft w = 0.125 inch square key page 4 of 6 Table 10-2 Norton assume LKEY = 0.625 inch < L overhang T = 157.6 in.lbf rSHAFT = 0.25 inch ASHEAR = w L = 0.078125 in2 ' = 3 = 13.98 ksi NYIELD = SY / ' = 2.15 FKEY = T / rSHAFT = 630.4 lbf = FKEY / ASHEAR = 8069.1 psi n05_static page 5 of 6 fatigue stress concentration (notch sensitivity) factors full reversed bending NOM = 13.45 ksi 1045 cold-rolled steel Sut = 91 ksi Neuber a = 0.0692 1 q= 1 a in = 0.7637 Kt = 1.698 Table A-9 Norton interpolated Table 6-6 Norton Eq 6.13 Norton r Kf = 1 + q ( Kt – 1 ) = 1.533 Eq 6.11b Norton = Kf NOM = 20.62 ksi alt’ = = 20.62 ksi NOM = 6.421 ksi constant torsion Kt = 1.459 no fatigue stress concentration (notch sensitivity) factor mean’ = 3 2xy = 16.23 ksi = Kt NOM = 9.368 ksi Eq 5.7d Norton modified endurance limit 1045 cold-rolled steel Sut = 91 ksi Se’ = 0.5 Sut = 45.5 ksi CLOAD = 1.0 Eq 6.5a Norton bending Eq 6.7a Norton CSIZE = 0.869 d -0.097 = 0.9294 CSURF = 0.76 machined CSURF = A(Sut)b = 0.817 Table A-9 Norton circular section Eq 6.7b Norton Figure 6-26 Norton A = 2.7 b = -0.265 machined Eq 6.7e and Table 6-3 Norton use CSURF = 0.76 CTEMP = 1.0 CRELI = 0.814 normal environmental temperature range R = 99% Table 6-4 Norton Eq 6.7f Norton n05_static page 6 of 6 Se = CLOAD CSIZE CSURF CTEMP CRELI Se’ = (1) (0.9294) (0.76) (1) (0.814) 45.5 ksi = 26.16 ksi fatigue factor of safety 400 rev 60 min min hr 24 hr = 576,000 rev/day day use infinite life N>106 alt’ = 20.62 ksi mean’ = 16.23 ksi Nalt = SN / alt’ = 1.269 SN = Se = 26.16 ksi Sut = 91 ksi Nmean = Sut / mean’ = 5.607 NGOODMAN = Nalt Nmean / (Nalt + Nmean) = 1.03 1045 cold-rolled steel Sy = 77 ksi NFIRST = Sy / (alt’ + mean’) = 2.09 Table A-9 Norton
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