Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

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=

200000

2 ( 1 +0.3)

×(−0.0015)