HW6 - P K 6W 2 1C(Pi‘Po K ” T(200 ~[4.7 bb 24 m 2"...

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Unformatted text preview: P- K 6W : 2 1C _ (Pi‘Po) K ” T (200% ~ [4.7 bb/.'.. ). 24 m _ 2" E: .‘n : I42. 3 k“ we WnIUz/r Wfaq F2 WWW abth Mr 55%” P21 + 6mm ‘ (MK I: I 1 HM __ (3X105Fu - wasp-600mm “‘ W : 75”“? 6k‘fl : 2-6—ZISMPCK W (5—38; 24:» data/(M GIMME W W W66- __6: -— MFR :: 3‘75MFK 7- ‘ 2 1. ‘5 + 0‘3)IL1P0\ 16+; : L:—“ : icy/MFR ‘6th Pi : "5+0‘3‘Mra:7\66MP‘1 :1 :— _________fi_fl mm W)‘ 91mm WK 9W Shim 7,25 7.55MPA P _; {E350 g/Jclrndb fmf If: 1g“ a}. 250 ' .A ‘ r\ ‘ 5W- a M m F.“ :5 m (a) 1:543 estate?» 23130 -. bio 7tth 5x,“ «)5 w: (,st _ PK __ [80x 6 I ‘ 2t; “3????” :: Wfllé "2 4320 W m h3fi9¢imzm git/(44.5 mtms in 22m;- 7910? “I? I 6+ i O 80 I firm: 1h '2? I : { :, 2250 PM (E) 2:271» UN. C7 LLIJM ‘W’t W. -e_%*«t9i5® 5 r “5 o jet fl ’ w M m O 2: 6 Go .41 E) m t m 2 x g“ . . ?0 x 4 3 f7 {LSQ ~21? mt!” 5 u 3 4‘ 3 6h 1. we; 1 B, 719W r 6% Sufi,” face rwzf ‘I 6y 2/ 6a 7/ a: @fla’ / [X I 6F— 51300 136099350" :: 424233 F3: 7:?“ {Eco-SM“ [378-88 w 6Q :: 5460'! {89002555 :2 6515102. in; 9° W fibre/we); Fewmw f» m may WM. 3x]: 6? 1 4241.33 pg; pmflxfi f» w mm} M 67f”: 6Q 1"— 6557\92" “€367” : $7568?! f Ex 1:2; M“;- 1‘ *_'§”’fi523 Jr :2! 'SMRQ —— "0.09241‘9‘9011 “@9524» 64mg); c __0. gm? " Z + W 2” C8550 1' “tin-- «Sf/1,1500 : #0‘00222 egg“ 6— 8 iv EV i :2. @9924 ? 033:2. ' “- 0£¢24~~ 030i ‘T mm N #:2158125: .2. 7 M1. H m (/9550 n‘ : f} 00 [012- me figme 6" (i i I WK Ex W EV . T #33: .: - l SW39 4* 2' {.9526 $3024" {it-Om: K _ a mums: ---~»~w---—-~;--~~ ‘svnSo + mmmmfifw gay-.550 Aw / B 6-3-2 A“ affilz [ff-15¢ 0n 1%}; ffilnmd L18 am 2Q Ifng M'h‘l”r$ (Lari; , 5%)»? (In; SAC/IQ ' C K C I f (_,< 1 066?? ()1: ~G.OOZA Tra _,_,_W __._ f! as” ; 5‘3"}; 6—3.9 W W; C2 : 9.968 «0.594. 2 —— : was - {a _: JWMB" : 0‘0067/ $0 ’Ffimk?h Sega;fl cm W 5:1“: CE“; R 3 “:2 {100875 E; : Cg" R: 0002-" O‘OOé‘fl : ~- @0043??? __n_‘ .{W :ZR : {10057in :: (1084- W HW6 3-D Mohr’s Circle. (Handout) 5, = 0.0030, s, = 0.0060, g, = 0.0095, ya, 2 00010, y], = yz : 0 Set up the coordinate as shown. (Let the shear stram on the Z plane be zero.) So a, = -0.0095 should be one principle strain, thus we only need to calculate the other two principle strains based on the strains on the X-y plane. First draw the 2D Mohr’s circle to get the two unknown principle strains, and then draw the 3D Mohr’s circle as above. r ,u 4" _, ‘ a .. @2957? r."- ffii 4",?" < Via/é ...—HM'— f g v“ n , A - “ "" )fl—mmum ‘ "A :11 , «3 0,525+ 3' (atria: -- We?” 7‘ "5035 fr—w. I - __ r .r .- .. , W 1,, , _.’ r15 : «filer u-- g" a??? 59-15? 3? 1" ' :- ;I,Jr " 1 if! v? x -v - ‘ ’ J‘ I 5‘ .‘ W A" “J m. :k m“qu O f, V . , __ ~ r 0.006? t ,. ,. “5;! r: MILK t Watt: yéijfi‘? 7,? “Y x ; -awwflMMw )( ‘1: 7; f‘ 3 {3 .Z, mmmm w. w- :- ; 2' f5 "3 “HQ WWJ‘J.-fztl.e y tat 155, ,7 u. 0‘93“ ‘9" "*1 , t r ‘ " w 6m” Wfl":—_“ x :3- H N 7 Ham/mi 1. For the stress state given by: 0')r = ~—40 MPa cry = 20 MPa 6: = 100 MPa TX), = —10 MPa TX: 2 0 MPa 7 = 0 MPa Lt Wfiatt-s I From Equation 5—26: II = crx +03, +02 =80 2 2 2 12 =GX*O'y+O'Z*O",+G'X*O_z—T .{v‘hr sz—T z.\‘ 2-2900 13 = ax * a"), *crz —oi\. "‘rgfi7 —cry *rzn — 0': * rzxy + 21' xy *1“ y; * r 3: -90,000 From Equation 5-25: (I3 4102 +120 —13 = 0 0'3 —80c72 - 29000 +90,000 = 0 Solve and get three roots (principle stresses) are: 100MPa, 21.6MPa, -41.6MPa. C?) n (00 fl (_ '6 Tmr -— ‘————:-)/L1Fq :7. Z 7O\8MFQ (’3) W 73412 M Princisz W is MM fig ffimflz 90min; CAM: E 2205641: , 722-027 , : l;gqffi[2f‘6yfq— 0.27({00MF0\ '4l'6M/7a )J : 2.28” x10"; “f E5 : T‘ELQ — v’mmafi : 23W“ ['4{.6MR # 0.270620% +2Ir6M/é >1 _ —4 ~— ~ 3‘7Sx :0 VM : me _ 70.8 MP4. CT fi Squq : 885x“)? “b 60] ...
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