Computer Aided Engineering Design

314 COMPUTER AIDED ENGINEERING DESIGN

constants gives

c

uu

xx

ji

ji

2 =

• 
  
  
  
  • and c
  
  
  
  

ux u x

xx

ij ji

ji

1 =

• 
  
  
  
  • or =
  
  
  
  

+

2

and =

• 
  

(^122)
d
uu
d
⎛ ij uuji
⎝⎜
⎞
⎠⎟
Thus, Eq. (11.7a) becomes
ux
ux u x
xx
x
uu
xx
xx
xx
u
xx
xx
uNxuNx
ij ji
ji
ji
ji
j
ji
i
i
ji
() = ji i j

• 
  
  
  
  • 
    
    
    
    • 
      
      
      
      • 
        
      
      
      • 
        

        =

        
        
        
        
        • 
          
        
        
        
        
      
      
      
      
    
    
    
    
  
  
  
  


• 
  

• 
  

• 
  


• 
  

  = () + ()
  ⎛
  ⎝
  ⎜

  
  

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟ uuj

(11.7b)

which can be expressed in the matrix form as

u(x) = [Ni(x)Nj(x)]

u

u

i

j

⎛

⎝

⎞

⎠

= N(x)u (11.7c)

Alternatively,

u

uu u u

uuNuNu

ij ji

() = ijiijj

+

2

+

• 2
  =

1 –

2

+

1 +

2

ξξ = () + ()

ξξ

ξξ

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟ (11.7d)

which in matrix form is

u(ξ) = [Ni(ξ) Nj(ξ)]

u

u

i

j

⎛

⎝

⎞

⎠

= N(ξ)u (11.7e)

Here, Nx

xx

xx

Nx

xx

i xx

j

ji

j

i

ji

() =

• 
  


• 
  

  , ( ) =

  
  


• 
  


• 
  

  or NNij() =

  
  

1 –

2

, ( ) =

1 +

2

ξ

ξ

ξ

⎛ ξ

⎝

⎜

⎞

⎠

⎟ are termed as the

shape or interpolating functions. Note that Ni(xi) = 1 while Ni(xj) = 0. Similarly, Nj(xi) = 0 while
Ni(xj) = 1, that is, generically in any finite element the value of the shape function, Nj(x) (or Nj(ξ))
is one at node j and is zero at all the other nodes of that element. The functions are positive and at
any point within the element, they sum to 1. In other words, the finite element shape functions are
barycentric similar to Bernstein polynomials or B-spline basis functions discussed in Chapters 4 and
5, respectively. In fact, we can relate the local coordinates x and ξby comparing the coefficients in
Eqs. (11.7b) and (11.7d). Comparing Nj(x) with Nj(ξ) gives

xx

xx

i

ji

• 
  
  
  
  • =
    1 +
    2
  
  
  
  

ξ

⇒ xx x = ij

1 –

2

+

1 +

2

= ( )

ξξ

ξ

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟ Nx (11.7f)

We can either use Eq. (11.7c) or (11.7e) to compute the axial strain εx at location P in the element.
Using the latter we have

εξ

ξ

ξ

ξξ

x

u

xx

d

dx

d

dx

= = ( ) = ( ) = –^1

2

1

2

∂

∂

∂

∂

∂

∂ []

Nu N u u (11.7g)

From Eq. (11.7f), dx
d

x

x

i xxl

j

ji

ξ

= –^1

2

1

2

=

( – )

2

=

[] 2

⎛

⎝⎜

⎞

⎠⎟