Applications in Solid Mechanics - Fundamentals of Finite Element Analysis - page 380

or

+

− 625

625 0 00 0

k (1) =

1
1250

1
1250

− −

625 625 0 0 625 625

0 0 0 0 − 625 625

0 . 5

− 0 . 50

= − . − .

0 51 0 5

0 − 0 . 50 . 5

The element nodal forces are readily shown to be given by

1
1
1

f (1) =

2 G A
3

which we leave in this general form for the time being.
Element 2

1
2 A (625 − 25 z )

∂ N 1
∂ y = 0

∂ N 1
∂ z

25
2 A

N 1 = =−

1
2 A (25 y )

∂ N 2
∂ y =

25
2 A

∂ N 2
∂ z

N 2 = = 0

1
2 A (25 z − 25 y )

∂ N 3

∂ y =−

25
2 A

∂ N 3

∂ z

25
2 A

N 3 = =

− 25
0
25

0
25
−

k (2) =

1
4 A

1
4 A

−

25 ] + [ − [ 025 25 0 25 ]

25

+

625 0 − 625
000

00 0

1
1250

1
1250

[ k (2) ] = −

0 625 625

0 − 625 625 − 625 0 625

0 . 5

− 0 . 5

0

= . − .

0 0 5 0 5

− 0 . 5 − 0 . 51 . 0

For element 2, the nodal forces are also given by

1
1
1

f (2) =

2 G A
3

Noting the element to global nodal correspondences, the assembled system equa-
tions are

− 0 . 50 − 0 . 5 1

2
3
4

1 2

1
2
1

− . − . 2 G A

3

0 51 0 50

=

0 − 0 . 51 − 0 . 5

− 0 . 50 − 0 . 51