Computer Aided Engineering Design

FINITE ELEMENT METHOD 327

=Ai/A,Nj(x,y) = Aj/Aand Nk(x,y) = Ak/A are all greater than or equal to zero. Further, if P is at node
j,Aj = A and so Nj(x,y) = 1 while Ni(x,y) = Nk(x,y) = 0. Following the same argument to choose the
interpolation scheme for u in Eq. (11.10a), we can use a similar expression for y displacements as
well, that is

vvvvvvv = + + = ( , ) + ( , ) + ( , )

A

A

A

A

A

A

i i j j k k Nxyiij jN xy N xykk(11.10f)

Ifu = [ui vi uj vj uk vk]T is the displacement vector for the element, then

u Nxy N xy N xy

Nxy Nxy Nxy

ijk

v iii

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

=

(, ) 0 (, ) 0 (, ) 0

0 (, ) 0 (, ) 0 (, )

uNu = (11.10g)

The strain εx along the x direction is defined as ∂
∂

u

x

while that in the ydirection,εy is given by ∂

∂

v

y

.

The shear strain γxy is expressed as ∂
∂

∂

∂

u

yx

+ v. The strain vector εεεεεmay be written as

 = =

0

0 =

0

0

ε

ε

γ

x

y

xy

x

y

yx

u

x

y

yx

⎛

⎝

⎜

⎜⎜

⎞

⎠

⎟

⎟⎟

∂ ∂ ∂ ∂ ∂ ∂

∂

∂

⎛

⎝

⎜

⎜

⎜

⎜

⎜

⎞

⎠

⎟

⎟

⎟

⎟

⎟

⎛

⎝

⎜

⎞

⎠

⎟

∂ ∂ ∂ ∂ ∂ ∂

∂

∂

⎛

⎝

⎜

⎜

⎜

⎜

⎜

⎞

⎠

⎟

⎟

⎟

⎟

⎟

v

Nu

=

(, ) 0 (, ) 0 (, ) 0

0 (, ) 0 (, ) 0 (, )

( , ) ( , ) ( , ) ( ,

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

x

Nxy

x

Nxy

x

Nxy

y

Nxy

y

Nxy

y

Nxy

y

Nxy

x

Nxy

y

Nxy

x

Nxy

ijk

ijk

ii j j)) ( , ) ( , )

∂

∂

∂

∂

⎛

⎝

⎜

⎜

⎜

⎜

⎜

⎜⎜

⎞

⎠

⎟

⎟

⎟

⎟

⎟

⎟⎟

y

Nxy

x

kkNxy

(11.10h)
Now

∂

∂

∂

∂

Nxy

x

yy

A

Nxy

y

xx

A

i(, ) = j – k i k j

2

,

(, )

=

• 2
  ∂
  ∂

∂

∂

Nxy

x

yy

A

Nxy

y

xx

A

j(, ) k i j i k

=

• 2
  ,

(, )

=

• 2

∂

∂

∂

∂

Nxy

x

yy

A

Nxy

y

xx

A

k(, ) = ij – k ji

2

,

(, )

=

• 2

so that

 =^1

2

• 0 – 0 – 0
  0 – 0 – 0 –


• – – – – –

=

A

yy yy yy

xx xx xx

xxyyxxyyxxyy

j kkiij

k jik ji

k jjk i kkij ii j

⎛

⎝

⎜

⎜⎜

⎞

⎠

⎟

⎟⎟

uBu (11.10i)