HW6_soln - -4-—5 [44777271 part/044 ofa ZLBms/om fexz‘

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Unformatted text preview: -4-—5 [44777271 part/044 ofa ZLBms/om fexz‘ oM’ZWE—fifiWAL ¢=90zz¢=5aswm (a) p _ 4P ;_ 4 (722x103 N) "A: "MHZ 7r (907ww)z = HZ MPCL ,5 = —-— — -_-_- (9.00/55 L' 5&8ww . P kN AL, mm ,MP '0 0 6 o a 3 OHAer values 7.22 0.0839 112 0.00165 57444714 My 4 14.34 0.1636 222 0.00322 21.06 0.241 326 0.00474 26.8 0.308 415 0.00606 (b) 040 g Vapl/W Cd 31.7 0.380 491 0.00748 _ , 34.1 0.484 528 0.00953 5100 ' 0' 002 ' 35.0 0.614 542 0.01209 36.0 0.924 557 0.01819 U0 '3 5 3 o M Pa» ‘ 36.5 1.279 565 0.02518 36.9 1.622 571 0.03193 37.2 1.994 576 0.03925 700 [ 600 1 01 O O J}. O O 6, Stress, MPa N 00 O O r O O / r / ' I . 0.002 I O V I _.L _.J J_ 1 J— I 0.000 0.008 0.016 0.024 0.032 0.040 8, Strain _\ C) O fl Ems/077 +€$ZL074 7075—7157 Al/ = £202 0’); = 778 W777, Flwd 510;) O7“ /&0 £4) Z/ffi. 700 600 g 50° FH— Hrs-L 5 g 400 daTa. .Pts/ by 3%: 300 I / . g “leasTsczuareS. 5 200 0168,6756 *J—E 5: 63,7 GPO. 100 7075-T651 Al ‘ 0 2 0.000 (‘ 0.004 0.008 0.012 0.016 0.020 0.002 6, Strain 700 600 At {rac‘l‘ure E 500 fracture; 100 a = 15: 3} 2 ‘21 M1262 - 400 . g 1 § A‘d-cr (kac. 4— 300 < 6 200 5 [’2‘ H57" 1 4 100 0 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 s,Strain ., A-€+er {rac'f’ul’et EH“: £4 “(Zr/E SPF = 0.1533—571/6z,675 = 0. #150 , ‘z_ 2 I 2"7 2 %/§A=/00—-50T2—L=/00—7—2g—W—7§ =26,4'Z ‘ L l 5-2—0 For-+ems/‘0M 1251 o’M Mam-T644 S+eeh wi+h 515032441044; (4) Fin/Id E) CFO, 01¢, 70/69. (’19) Defermar'm e, amd plot 3, E) 01150 @3£1‘M4 613}; (a) calculate a, amal 4H- Efg = H 5;“. 600 Pl OT beg [4444 {m9 500 a = 211,3768 . Man-Ten Steel 8 Of E 400 g 300 l,\ Lgast sggares 3? 200 00:317MPa4 {4+ of {\rst 4— b 100 Ptgfigives' 0 E = 211,900 MPc. 0.000 0.003 0.006 0.009 0.012 E = 2 “H G POL 0, Strain ‘ Tat/2e Mfg/vest G M4 “table +0 be 0"“: 57¢; MPa‘ ,7. L 1_ 7. 7,) RA 2100M 7.100%: 651.3% all 6,32 4 (b) Calcula‘l'e '6‘", 2/ as beiow; values [m fable. ’0‘1 = 0- for 2 <220 = 0.007 g: 0—( H‘E) for E>2£a where, d umkmowrm ’OV‘:O‘5_L : G‘ where d IQanOW/n I A d EV: l’n +5) where d umkrnoww‘ (I! “C. AL ._ d5 2 E- [71 T -177 puke/fa d know/Va (No‘fe. +ch fi:o*(l+£) aw; §=lw0+é§ used <03; 7.1 For the given state of stress, calculate principal normal stresses. Then use the maximum normal stress fracture criterion to obtain the effective stress and safety factor. y 2 [sz+1xy=c¢3it3, 63=62 Gmaxl > 16min| ) ‘ 6NT=MAX(GI,62,03), >0, X = Gut/6NT Stresses in MPa Si3N4 (Table 3.10) (5X 6y Oz Txy 01:3 TS 125.00 15.00 0.00 —25.00 70.00 60.42 01 02 03 Cut 6N7” X 130.42 9.58 0.00 450 130.42 3.45 ...
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HW6_soln - -4-—5 [44777271 part/044 ofa ZLBms/om fexz‘

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