Question

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4. Figure below shows a rotating shaft simply supported in ball bearings at A and D and loaded by a nonrotating force F of 6.8 kN. The shaft is made AISI/SAE cold drawn 1050 steel, estimate the life of the part. 550 C 400 B - 6.8 EN 75 -100 - 250 --- -125 Stress amplitude (MPa) 10 10 10 10 10 Fatigue life cycles) 10

Answer 1

WS 6.8kn 10 o 15 250 250 4 z (-3) 8 C. LI0 30 A L 32(d) ( L 38(D) 35 R2 30 IRI (1=3) . All dimensions are in mm. forces - Fiert we should find the reaction EMA =0 -6.88 325 + R2 (550) = 0 R2 6.8x325 -4.018 KN 550 I Fy =0 7 608 = R17 R2 → R, = 6.8 - Red = 678 – 4.018 [R = 2.782 KN

From the diagram, we observe that point 'B' has a smaller cross section, so higher bending moment and higher stress - Concentration factor the FOR AISI/SAE cold drawn 1050 steel, Sut = 690 MAR 4 Sy = 580 MPa These values are standard taken from table. Then Endurance limit sé = 0.5 Sut = 05(690) - sé = 345 mpa lum -0.265 For cold drawn surface factor Ka = 4.51 (Sut) ka = 4:51 (69090.265 ka = 0.798 -0.10+ And , size factor Kb Ce 95 95 for 2.99 3dsum -/32 -0.107 7.62 Kn=0.858 Since kc = kd = ke = Kp = || Se = kako se = (0.798)(0-858) (345) = 236MPa

32 Now, we should find geometric stress - Concentration factor kt. For that I = 2 2 2 = 1.875} from diagram - 3 = 0.093 ) ° For these values, corresponding ,kte Then fatigue stress concentration factor , hip is 1.65 + :: where a = neuber constant it is a meterial Constant. for this material a = 0.0979 mm 1.65.1 Then keq = 1+ = 1+ 0.0129 10.0979 17.0-0929 K = 1.55 Now, we should find bending moment wité". Mg = Rix250 = 2.782 X 250 Mo = 695.5 N-m)

@ Just to the left of B, - 32 the section modulus is = d' Z = II (32 32 7 = 3.2.15X103 mm The reversing bending stress is, Tran = kq MB i = 1:55469505.810-9 (3.215X103) ser = 335.1 mpa • This is greater than se and less than Sy. * Se < Prev <Sy * 236 MPa < 335-1 MPa < 580 mPa. This means we have both finite life and no yielling on the first cycle. For finite life - The ultimate strength Sut = 690 MPa a= [tsit)? Se there f = 0.894 (Fatigue strength facto)

236 ar Co.840 (690) ) 1437 mPa. b= -f log [festat ] - - f long (0.94 inche] b = -0.1308 0.844 x 690 236 Then , nev Number of geles, n = (CT) N = (33:4) 4-1508 N = 68 x 103 cycles