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8.5-8 - M L=2.flm d=4flmm M=tiflkg G=EflGPa cram = 100...

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Unformatted text preview: M L=2.flm d=4flmm M=tiflkg G=EflGPa cram = 100 MP3 Tam = 54:: MPa 11th 2 = T = 12.5664 mm W = M3 = {ISO kg][9.31 W52} = 533.6 N w :r T W Gt 1;!) Teams.- T= ’Lm (Es. 3—15) T smasmsss: .— =I—" (m. 3—11} T = (%) (a = Grim“ = (31:: x1135 115.)qu 1' = 3D £35m“ Units: 1' = MPa trim = radians W TENSIIE STRESS (I; = I = 415.339 NIPa a" = fl 0' = a". = 46.339 MP1: 3? 1'1? = —BD film (MPa) PRWCJFAL STRESSES i— n"; + U}. 3 I [II—ti}. 2 i \“l( 2 j J”; cr1.2=23.42u i animus-100%.; (MPH) Nate flint 0'1 is positive and 0'2 is negative. Therefore, flie maximum in—plame shear stress is greater fliait the maximum Dut—uf—plane shear stress. H112: MIAMI-{UM APELII. CI'F RD'HLTICIN BASED ON TENSILE STRESS 0'1 = maximum tensile stress 0' = 100 MP3 511m 11x: MPH = 23420 + sigma}? + 5400 sm? (1m — 23.420)2 = (23.421301 + 6400 [Hm 5316 = 154m sin (pm = 0.9114 rad = 52.2° MIAMI-{UM ANGLE CI'F RD'Efl'lflN BASED ON lN-PLANE SHEAR STRESS Ill U1— ”'1. 1 2 fl mem +ij= (23.43)] +64mrbm 2 1-me = 54:: MPa 50 = “23420)? + 64130 41mg (51))2 = (23.4213)2 + 15400 {bin solving, rpm = 0.5522 rad = 31.150 Sm stxsss Gflvssxs (tam = 0.552 rad = 31.6“ d— ...
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