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HW30 - PROBLEM 7.134 SOLUTION =‘ UGO,u 278.6,4 gm...

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Unformatted text preview: PROBLEM 7.134 SOLUTION =‘ UGO ,u 278.6 ,4 gm = -5’?8.6 ,u 71,”. = Ew — Em - 273.94 + 5%.efl €x=+l€0fi Eur—480,11 Ygr‘GOO/u (GD FoV‘ Mokr‘s <20"ch O'Vsfiau'n) Pfio‘l’ Poi/9’3 X! (IéOA) 3054-!) Y: (— 980/9 400/43 C‘. (- 160;“ O) 7‘ — 300 Q .- x5 ____.- _ , (OJ +4.4 2 p arse] 310 0 9375 y 29,, = 4345‘ 9?: —Z/.52° a“; -2l.58+ <10 .~ €8.42° 9Q} -2L58° ‘ 9],: 68.142" .‘ — Isak-938.€,u r —$‘!8.é fl “ (b) é’nml,“n~rfi)a.«d\ : R XWIEn-MA) : 2R '. 877/! (c) 8‘ = 3%);- (ECJEB) “1-31;- (E, + 83,) = -2|—::—(lé0,u —480,¢¢) 1' 877/! <1 <8 7.134 through 7.137 The following state of strain has been measured on the surface PROBLEM 7.137 of a thin plate. Knowing that the surface of the plate is unstressed, determine (a) the direction and magnitude of the principal strains, (b) the maximum in-plane shearing strain, (c) the maximum shearing strain. (Use v = %) EX:—300/4 Sir—200,0: Yy:+i75’u SOLUTION Pjoi'ieai Pofw‘is X: (-300/1) 47.5,“) Y: (- ZOO/J ,427.$,u) 5"" (M3 C: (— 250,14, 0 7 y a 87.5 SD +an 29‘, 2 - 29F: -go.1c. X Qb =‘-30. 13° “ 30.:3' Ga 7 57.87” “ b R = #5014? +(87. 5,"? 2‘ 100.8 (a) at: EM+R = —2,50,u.+ 100.8/4 .- —/47,2/4 A 55’ E....‘Q= -?SO/4-loo.8,a = ~35I/4 -e (In) YW(;.-,,.,,\ : RR = 2014,44 .— 2) 1/3 : + 250 ,u 4 ’2} ac : '1—v(E“-+Eb\ ? _l-2/ (Ewifl t ‘ 2/3 Shawl. : 250% Emu}. : —' 35l/L' (c3 T...“ , 8..“ — E...“ = 250/4 + 35/}: = 6m,“ —-- 7.138 through 7.141 The lbr given state of plane strain, use Mohr’s circle to determine (a) the orientation and magnitude of the principal strains, (b) the maximum in-plane strain, (c) the maximum shearing strain. 5K:+4oo/4 sjr+200,u YxJ-‘+375fl PROBLEM 7.140 SOLUTION Ppofi'ecl Poin+5 X: (H0014 , - [87.5A‘) y: (“zoo/1, +1829.) c: (4. 300’“) 0) 29,. = Gms’ e“: 3mg") 9.: 120.2‘ R={(loo;4)‘+ (197-5,4)‘ - 212.5,“ " (a) £‘= EM+R= Boo/“+2115,“ = 5125)“ .4 Ev E...- I? = 300’u42l2.5/4 :- 87.5‘,q 4 (b) Y1.“ (in-plan) : 2R- : 425/“- s°-“‘ (c\ 8. = o 8......- 512.5;4 a... = o 7 141 7.138 through 7.141 The lbr given state of plane strain, use Mohr’s circle to PROBLEM ~ determine (a) the orientation and magnitude ofthe principal strains, (b) the maximum in-plane strain, (c) the maximum shearing strain. E, = +60/( Ejr+240fl Y‘U:—50,u SOLUTION PJO'HCA Pol-n+5 X: (CO/4’ 25/4.) Y: (2405:2514) C: 050,“, o) - 7:13] ~50 hm zap-Ere: - 60410 = 0.277773 26, = 15.52‘ 95- 174" e“: c77.74“ V \ P=~flfibydz+a$fiy = ‘73}! ,q (M EL= 8...”? = lSO/A+‘?3.H',u ’ 243.4,“ 4 £..= e...-R= I50y-‘73‘I/4? $6.6,“ 4-: a.) ngwflmp 2R = I86.8,u 4 (C) EC: 0 3...“? 293.9,“ E.;. =0 eng‘ Ens-”n = 273.4 4 ...
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