problem5_sol

# problem5_sol - 22.314 Problem V Solution Fall 2006 Known E...

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Unformatted text preview: 22.314 Problem V Solution Fall 2006 Known E =195000 v =0.3 A = 150 oB :=260 sB =0.54-10- 2 Derived properties EmA =0 on rsmB = sB - E smY :=0.002 smY - smA ry : mA .(oB - aA) . aAd Linear interpolation between A anb B to get yield stress arY emB - smA aY = 204.098 The 0.2% offset yield stress oY is 204.098 MPa From the uniaxial stress-strain curve, when op<=CA, se=seA=0; when aA<cp<=~B, se is linear on ap; when cp=aB, te=eeB=B-oB/E seA aB sB = B - - i =0.. 100 •e = seB.- so - 8eA Opi = te- .(oB - A) - oA I/I ap as a function of so seB sA too 200 200 0 51 o 0.002 0.004 0.006 Loading sequences and questions 1317.5 -12.5 75 a:= - 12.5 137.5 75 75 75 50 1) akk:=o 0,0 +l Oa, I + aa 2 , 2 ukk =325 i:=0..2 j =0..2 ac :=0 Sal 1 :=oa, --- S(i- j,0)- 129.167 -12.5 75 Sa =12.5 29.167 75 75 75 -58.333 une = i Sa) (Sao, -)22+ (Sao,2-2± (Sa 1 ,)2 1(Sa,2) 2. 2 (Sa2,)] oae = 204.634 Now calculate VonMises stress for comparision eigena:= eigenvals(aa) aVM .[(150 - 200) 2 (150 +25)2(200+25 cVM =204.634 We can see cae=aVM 2) Strain tensor for...
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problem5_sol - 22.314 Problem V Solution Fall 2006 Known E...

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