6.8 Finite Element Method for Continuum Structures 209
Now,substitutingforα 1 ,α 2 ,...,α 8 inEqs.(6.96),
ui=0.00025−0.000125x−0.00175y−0.000625xy
and
vi=−0.001+0.00025x+0.002y−0.00025xy
Then,fromEqs.(6.88),
εx=
∂u
∂x
=−0.000125−0.000625y
εy=
∂v
∂y
=0.002−0.00025x
γxy=
∂u
∂y
+
∂v
∂x
=−0.0015−0.000625x−0.00025y
Therefore,atthecenteroftheelement(x=0,y= 0 ),
εx=−0.000125
εy=0.002
γxy=−0.0015
sothatfromEqs.(6.92),
σx=
E
1 −ν^2
(εx+νεy)=
200000
1 −0.3^2
(−0.000125+(0.3×0.002))
thatis,
σx=104.4N/mm^2
σy=
E
1 −ν^2
(εy+νεx)=
200000
1 −0.3^2
(0.002+(0.3×0.000125))
thatis,
σy=431.3N/mm^2
and
τxy=
E
1 −ν^2
×
1
2
( 1 −ν)γxy=
E
2 ( 1 +ν)
γxy
Thus,
τxy=