solution-hw6 - (a) For an AISI 1018 {ID—machined steel....

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Unformatted text preview: (a) For an AISI 1018 {ID—machined steel. the strengths are Eq. (3-17): S", = 440MB: 2.: H3 = % =129 a, = 370 MPa 3,“. = 067(440) = 295 MPa r 2.5 D 25 Fig.A-IS-15: E = E =o.|25, F = 55 =1.25. Kr: =14 Fig. 7-21: 95 -_—- 034 Eq. (1-31}: It}, =1+o.94{1.4-1)=1.375 For a laurel],r reversing torque of 200 N v m T __ 19,16? _ 1.376(16)(200x 103N¢mm} m _ ltd-q _ :TfZO mmF' rm“ = 175.2 MP3 1‘ Ta s; = 0.504(440) = 222 Mpa Thean factorsare k8 = 4.51:44or'1-2“ = 0.399 20 —(llfl'1 kb = = 0.902 kr=o.59, kg: 1, kfl=1 Eq. (7-17): S, = 0.899(0.902)(0-59)[222) = 106.2 MPa [[l.‘£l|{2‘£l5)]2 . - I = —— = 664 Eq (7 13} a 106.2 (7 l4)‘ b — llo HERE‘S) — 013265 Eq' ‘ — 3 3 105.2 " ' lf—fl.13265 Eq. (7-15). N — N = 23000 cycles Am. 0)) For an operating temperature of 450°C, the ternperature modification factor. fr: Table 7-6. is rd = 0.343 Thus {:71 = 0.899(0.902}(0.59)(0l343]{222) = 39.5 We _ {0.9095}? H, a — 8915 a "1'88 _ 1 0.9-:295) _ b _ —§ log 39-5 — 0.15111 1152 leLlS‘Hl ~ = H “I83 N = 14100cycles Ans. 7-11 A priori design decisions: The design decision will be: if Material and condition: 1095 HR and from Table A-EU S,” = 120. SJ, = 66 kpsi. Design factor. flf = 1.6 per problem statement. Life: [1150}(3) = 3450 cycles Function: carry 10 000 lbf load Preliminaries to iterative solution: 5;, = 0.504(120} = 60.5 kpsi kn = 2.70: 120) ‘03” = 0.759 I irrd'3 — = —— = 0.09317? 6 32 M(c1‘it.)= (2%) (1000mm) = 3000mm. in The critical location is in the middle of the shaft at the shoulder. From Fig. A-15~9: D /d = 1.5. r/d = 0.10. and K, = 1.68. With no dinect infomation consenting I. use I = 0.9. For an initial trial. seld = 2.00 in it (ACCEPT OTHER In“ M. tssumi’m M9) 100 4110? kg, = = 0.816 E S, = O.759(0.8]6)(60.5) 2 37.5 kpsi _ [0.9102(1)]2 _ a _ 37.5 _ 31m 1 0.9(120) _ = --E log 37.5 _ 0.153] SJf = 311.0(3450r'1153' = 39.3 M _ 30 _ 305.6 5" ‘ m ‘ 0.09817d3 ‘ :13 305.6 23 d 2 r:— w=fi= = 33.2 kpsi 0.2 "68 — 1 534 K; = _———_.——————-—" _ _ 1 + (NM) [(1.68 —l)/1.68](4f120l an = 3:315 = 33.2 kpsi a. = mom =tss+ {33.2) = 50.5 kpsi (Sf)34§fl _ 39.3 ‘4: ....—-—-_— or!I _ 60-5 The design is satisfactory. Reducing the diameter will reduce it, but the resulting pmferretl sizewillbed =2.Dfl in. Hf: ...
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solution-hw6 - (a) For an AISI 1018 {ID—machined steel....

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