Question

1) [40 pts] Determination of minimum shaft diameter using DE-Gerber, DE-ASME Elliptic, DE-Soderberg, and DE-Goodman criteria A shaft is loaded in bending and torsion such that M. = 66 Nm, T. = 42.496 Nm, Mm = 51.858 Nm, and Tm = 33 N m. For the shaft, S, = 700 MPa and S, = 560 MPa, and a fully corrected endurance limit of S, = 210 MPa is assumed. Let K = 2.2 and Kis = 1.8. With a design factor of 2.0, determine the minimum acceptable diameter of the shaft using the methods mentioned in the subsection. a) Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Gerber criterion. The diameter of the shaft is mm. b) Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Ellipitic criterion. The diameter of the shaft is mm. c) Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Soderberg criterion. The diameter of the shaft is mm. d) Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Goodman criterion The diameter of the shaft is mm 2) (15 pts] Determine the catalog rating for a ball bearing Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10000 hours at 1800 rev/min with a load of 2 kN with a reliability of 90 percent. The catalog rating (C10) for a ball bearing is KN.

Answer 1

0 solution hiven, Ma -66 Nm = 66X103 Nmm, Ta=42:496 Wm = 42-496x103Nmm Mm=518f8Nm, tm >330m, 344=toorpa, Sy4 S60MPa Se=210 MPa, Kf = 2.2 Kes 718 designfactoo (N)=2 Let d= dicimeter of Shaft (mm) Amplitude Bending stress (5) = f ( 32 Ma 7103 1479,748103 = 2:2 ( 328 66x103 (6)m = ko le ) – 2-4 (29 Xb4x403 (Bm = m = 0,2 ( 328818583103 1162.6768103 Now, d3 +18 ( 16x42-496X103) 6b - MPa d3 Mean 疥d3 MPa Amplitude shear streep ima-kps ( Ta =18 Mean sheer stress Om = kRs ( 1657 168 421 Nd3 muallin 389.77x10 Rupa 13 im 21:8 16X33X103 302-678103 d3 MPa using von Niges distortion energy theory, Now, allennating vom Miles component (ca) 6a (6 +39 1479.74810312 d3 43 389.47X103)2 d3 69 1626.46X103 d3 MPa

meam von-Mises component () = 100)+ 3M m 6m = 1162.676X103 d3 1275 UXIO3 302.67X1032 d3 cm MPa 23 2 ) herber et criterion No 60 Nom 12 + .se Sut =) + d By You 2x 1626:46X103 2x 1275 4x103 d3 x 20 15. uaxlo 13,278x106 + 13 =) (23)2 Let dax > a ?- 15:49 x103-13.248X106-)

on + = solving, 2= 16304.38 d3=16304:38 d=25:36mm diameter of shaft (d) = 2% 36mm b) Elliptic coiterion Na 17 ( nguy 2 = 2x 162646x103)2 2x1275-4X10312 d3x210 d3x560 d=28.27 mm diameter of shaft (d) = 25.27 mm Soderberg.coiterion de + syt 1626 46 X 103 1275.48103 d3x210 d3x560 d=27.164 mm diameter of shaft (d) = 27.164 mm d) Goodman criterion, =t = -1626:468103 + 1276 4X103 d3x210 d=26.75mm diameter of shaft (d) = 26.75 mm. om zt + 6m se + Sut d3 x 100 =)