316 COMPUTER AIDED ENGINEERING DESIGN
whereλλλλλT =
cos 0
sin 0
0 cos
0 sinθ
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
VT
ee
TT
VT
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 sin1–1
–1 1cos sin 0 0
0 0 cos sinθ
θ
θ
θθθ
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
22AE
lθθθ 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.
ξujy,fjyujx,fjxujθyxuiuiy,fiyuix,fixFigure 11.4 A truss element oriented at an angle θθθθθ