This preview shows
page 1. Sign up
to
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— ...
View
Full Document
- Spring '08
- Fosdick
-
Click to edit the document details
