Question

1. The machine shaft in Fig. 1 is made of 1030 normalized steel, with machined surfaces. The shaft is to be designed for a lifetime of 5x105 cycles, with a design factor of 1.7. Determine an acceptable diameter d for the shouldered section of the shaft. Use a value of kp=0.85 for the first iteration calculation. In this problem, the design factor is defined as na = $70, where Sy is the fatigue strength at 500,000 cycles. F = 8 kip - 10 in— 5 in -5 in I diSR. d/IOR E Fig. 1

Answer 1

IMARO Skip 2) R2X200 8x 15 R2 28x15 Isin = 6 kip. Ri= akip We have to design for sxios cycles Sé = 5 Sut ..5x76 100 psi = 38050psi Ka I surface finish factor) 2.78 Kb = .85 ke ( relaibility factor ) - Let's design for 901 relarbility Kc = -897 kg (stress concentration factor from table Kf = 1.7 Sea Kakket I Se 78 X 85 X 897x 1 x 38050 1.7 133 11 psi 2

log 10 9 Sut = 4.8356 log, osat log 10 se Клу 10 р = 4.1242 los 10 15x105] logio set = 5.6989 "3 log on 6 By similarity of triangles los 10 9 Sut - log ost log 10N-3 log 10.9 sut - logiose 6-3 = 4.8356 - log ost 5.6989 -3 4.8356 - 4.1242 - 3 2) los lo sf = 4.1956 St = 15689 psi da a St a 15689 2 9228.9 psi °a? M.y = 32 * RIXID 32X 2x105 x 10 ad 3 Ad 3 xd3 2) 32*2*10*XL0 = 9228.9 2 d3 = 22.07 in 3 2) da 2.805 in Dus valo of Kb is whona ko must be o 75

Toz x 38050 so our New Se= Kakakek L se! • 78x:75X1897 x 11745 psi log iosf = log, o'q sut - (log, 6'9 Sut-logose) x logion - 3 = 4.8356 - (4.8356- 4.0698) x 5.6989-3 & 4.1467 Sfa 44017.18 o ao St a 8245.4 psi da = 32x2 x 103 x 10 - 8245.4 nd 3 =) d 3 = 24.7 a) da 2.91 in