Interpolation Functions for General Element Formulation - Fundamentals of Finite Element Analysis - page 173

■ Solution
Using s = x / L , we have s 1 = 0, s 2 = 1 / 3, s 3 = 2 / 3 , and s 4 = 1 . The monomial terms of
interest are s , s − 1 / 3, s − 2 / 3 , and s − 1 . The monomial products
N 1 ( s ) = C 1 s −

s − ( s − 1)

1
3

2
3

N 2 ( s ) = C 2 s s −

( s − 1)

2
3

N 3 ( s ) = C 3 s s −

( s − 1)

1
3

C 4 s s

s

1
3

2
3

N 4 ( s ) = − −

automatically satisfy the required zero-value conditions for each interpolation function.
Hence, we need evaluate only the constants C i such that N i ( s = s i ) = 1, i

= 1, 4 . Apply-

ing each of the four unity-value conditions, we obtain
N 1 (0) = 1 = C 1

( − 1)

1
3

2
3

− −

N 2 1
3

= 1 = C 2 1
3

1
3

2
3

− −

N 3 2
3

= 1 = C 3 2
3

1
3

1
3

−

C 4 (1) 2
3

1
3

N 4 (1) = 1 =

9
2 , C 2 =

27
2

27
2

9
2 .

from which C 1 =− , C 3 =− , C 4 =

The interpolation functions are then given as

s − s − ( s − 1)

9
2

1
3

2
3

N 1 ( s ) =−

s s −

( s − 1)

27
2

2
3

N 2 ( s ) =

s s −

( s − 1)

27
2

1
3

N 3 ( s ) =−

2 s s

s

9 1

3

2
3

N 4 ( s ) = − −