316 COMPUTER AIDED ENGINEERING DESIGN
whereλλλλλT =
cos 0
sin 0
0 cos
0 sin
θ
θ
θ
θ
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
. We can realize that λλλλλue = λλλλλλλλλλTu = I 2 × 2 u = u. Substituting u = λλλλλue in
Eq. (11.7l) gives
SE =^1
2
() =^1
2
=^1
2
uBBuue T BBuuku
V
T
ee
TT
V
T
ee
T
∫∫EdV EdV ee
⎛
⎝
⎜
⎞
⎠
⎟
⎧
⎨
⎩
⎫
⎬
⎭
⎛
⎝
⎜
⎞
⎠
⎟ (11.7n)
whereke = λλλλλT
V
TEdV
∫
⎧
⎨
⎩
⎫
⎬
⎭
BB = λλλλλTkλλλλλ is the transformed stiffness matrix of the truss element.
Using Eqs. (11.7l) and (11.7m), we have
ke AE
l
=
cos 0
sin 0
0 cos
0 sin
1–1
–1 1
cos sin 0 0
0 0 cos sin
θ
θ
θ
θ
θθ
θθ
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎜
⎞
⎠
⎟
=
cos cos sin – cos – cos sin
cos sin sin – cos sin –sin
- cos – cos sin cos cos sin
- cos sin – sin cos sin sin
22
22
22
22
AE
l
θθθ θθθ
θθ θ θθ θ
θθθθθθ
θθ θ θθ θ
⋅⋅
⋅⋅
⋅⋅
⋅⋅
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
(11.7o)
which satisfies the equilibrium condition keue = fe, where fe = λλλλλTf analogous to the transformation for
displacements. The reader may verify that the rank of ke as given in Eq. (11.7o) is one even though
it is of 4 × 4 size. This is because a truss element in two dimensions has three degrees of rigid-body
motion, i.e., translation along x and y axes, and rotation in the xy plane.
ξ
ujy,fjy
ujx,fjx
uj
θ
y
x
ui
uiy,fiy
uix,fix
Figure 11.4 A truss element oriented at an angle θθθθθ