Question

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Problem 1 (5 Points) Bearing reactions R, and Rz are exerted on the shaft shown in the figure below, which rotates at 950 rev/min and supports an 8-kip bending force. Use a 1020 HR steel. Specify a diameter d using a design factor of ng = 1.6 for a life of 10 hr. The surfaces are machined. F = 8 kip — 10 in— +5 in 5 in d/10 R. 1.4

Answer 1

E-skip loin sin/sini ) 9 inkR2 N = 950 rev / min 1020 HR steel na = 1.6 he lour Sut = 55 kpsi Su 30 kpsi Foto N=.950 x 10x60 eyles, b ) N= 570000 luces. Listy - A lsin 18 kip ksin & Sthy Taking moment about Ri 8x15 - R2 x 200 8x 15 = R2X20 _R2 = SX15 hor. 20 R2 = 6. Kips similarly taking moment about R2 Ri 20 8X5=0 Ri=2 &ips

Bending moment diagram BMR = OUT B Mc = R2x5=30 BMR = 6R2 x 10-885= 20 BMA =0 12.00 E It at 20 kipin ! zo kip.in critical location is not obvious, evaluate any point, so, Eraluating at point B 82 M M = 20 kipin Grer = Me M (d/2) I ad/64 - :02 32720190 ) T d. Gover = 203.7 Kapsi 2 03, Now, we will get notch sensitirity and stress concentration factor - Notch sensitivity depends an unknown Dameter errore Now estimate value of q = 0.85

charts of factor are for stress concentration factor theoretical stress concentration considered from charts, Da 1.4d - 1.4 N oild tool ke = 1.65 - Kfiz Itackt - 1 annos = 1+0.85 (1651) = 1.55 2 Now, evaluate for fatigue strength - Se = 0.5 Gut = 0.5455=27.5kpsi 1790 .265 Ka=2.7055) = LLEL. 0.90 Ka=0.98 since diameter is not known, assume ralue of Kb =0.85 -80, Ses Kaike. Selon = 0.93*0.85 X 27.5 Se = 21.73 k psi Determine desired fatigue strength from en dia. fatigue strength fraction of sog to 2 a of Suta (6.985) UB 2. a = 112:9572

b = I log (f Sut) and og 10.9x55 3: 24.73 23 = -0.1191 ! b fatigue strength, Sf9. N. Fotter _ = 112.751570000). S = 23.25 kpsi. compare strength to stress and solve Sf hd 'ha6 Kf Oren 23.251 -1.55 X 203.7 d² = 21.728 ad d = 2.79 m. since , size factor& notch sensitivity were guessed, check now... - Kb=0.91 -0.154 -0.157_ 0.91(2.179) in 2 Kb=.01771. Istori so, from graph it appears. I value is low, so we choose q=0.91, and again eraluate problem with new values ef=lt ogill:65-1) = ling -

na = St k, Gey 22. 28 | у9 У 202.) 3 . 22.28 а - 2.81 ip.