202 CHAPTER 6 Matrix Methods
Thus,inmatrixform,
{ε}=⎧
⎨
⎩
εx
εy
γxy⎫
⎬
⎭
=
1
E
⎡
⎣
1 −ν 0
−ν 10
002 ( 1 +ν)⎤
⎦
⎧
⎨
⎩
σx
σy
τxy⎫
⎬
⎭
(6.91)
Itmaybeshownthat(seeChapter1)
{σ}=⎧
⎨
⎩
σx
σy
τxy⎫
⎬
⎭
=
E
1 −ν^2⎡
⎣
1 ν 0
ν 10
0012 ( 1 −ν)⎤
⎦
⎧
⎨
⎩
εx
εy
γxy⎫
⎬
⎭
(6.92)
whichhastheformofEq.(6.68),thatis,
{σ}=[D]{ε}Substitutingfor{ε}intermsofthenodaldisplacements{δe},weobtain
{σ}=[D][B]{δe} (seeEq.(6.69))Inthecaseofplanestrain,theelasticitymatrix[D]takesadifferentformtothatdefinedinEq.(6.92).
Forthistypeofproblem,
εx=σx
E−
νσy
E−
νσz
E
εy=σy
E−
νσx
E−
νσz
E
εz=σz
E−
νσx
E−
νσy
E= 0
γxy=τxy
G=
2 ( 1 +ν)
EτxyEliminatingσzandsolvingforσx,σy,andτxygive
{σ}=⎧
⎨
⎩
σx
σy
τxy⎫
⎬
⎭
=
E( 1 −ν)
( 1 +ν)( 1 − 2 ν)⎡
⎢⎢
⎢
⎢
⎢
⎣
1
ν
1 −ν0
ν
1 −ν10
00
( 1 − 2 ν)
2 ( 1 −ν)⎤
⎥⎥
⎥
⎥
⎥
⎦
⎧
⎨
⎩
εx
εy
γxy⎫
⎬
⎭
(6.93)
whichagaintakestheform
{σ}=[D]{ε}Step six, in which the internal stresses{σ}are replaced by the statically equivalent nodal forces
{Fe},proceedsinanidenticalmannertothatdescribedforthebeamelement.Thus,
{Fe}=⎡
⎣
∫
vol[B]T[D][B]d(vol)⎤
⎦{δe}