FINITE ELEMENT METHOD 315so that εx
u
x lll
= = –^1
21
2∂ (^2) = – (^11) =
∂ [][]
uuBu (11.7h)
Herel is the length of the member and B is termed as the strain displacement matrix that relates the
strain at a point to the nodal displacements of an element. The stress σx in the element is
σx = Eεx = EBu (11.7i)
At this stage, an alternative weak form of the equilibrium equations in Eq. (11.2b) is introduced.
A scalar termed as the work potential is defined as the difference between the strain energy stored in
an element and work done be external loads. From Eq. (11.2c), the strain energy in a spring (or truss
element) is^1
2
uTku while the work done by the external loads is fTu, where f =
f
f
i
j
⎛
⎝
⎜
⎞
⎠
⎟. The work
potential WP is then
WP =^1
2
uTku−fTu (11.7j)
minimizing which with respect to uyields
∂
∂
WP
= – =
u
ku f 0
which is the strong form of the equilibrium condition. The strain energy stored in a truss element is
SE =^1
2
=^1
2
=^1
V x^2
T
x
V
TT T
V
dV E dV TE d V
∫∫ ∫
⎛
⎝
⎜
⎞
⎠
σε uB Bu u B B ⎟u (11.7k)
Comparing with^1
2
uTku, we realize that k =
V
TEdV
∫
BB or
k =
–1
1
–1 1
1 –1
–1 1
- –1 1
2222
VllE
ll
dV E llllAl AE
∫ l⎛⎝⎜
⎜⎞⎠⎟
⎟⎛
⎝⎞
⎠⎛⎝⎜
⎜
⎜⎞⎠⎟
⎟
⎟⎛
⎝
⎜⎞
⎠
⎟11
(11.7l)which agrees with Eq. (11.6). Note that k above is termed as the local stiffness matrix as the
displacementsu are along the axis of the truss element.
11.3.1 Transformations and Truss Element
Quite often, a truss element may be oriented arbitrarily in the x-y plane and may have different
stiffnesses for external loads along the two directions. Consider a truss element oriented at an angle
θ (Figure 11.4) where the displacements ue = [uix,uiy,ujx,ujy]T are to be determined along the x and
y axes for external loads fe = [fix,fiy,fjx,fjy]T.
Relating the displacements ue with u along ξ, we have in the matrix form
uueix
iy
jx
jyi
jTu
u
u
uu
u= =cos 0
sin 0
0 cos
0 sin=⎛⎝⎜
⎜
⎜
⎜⎜⎞⎠⎟
⎟
⎟
⎟⎟⎛⎝⎜
⎜
⎜
⎜⎞⎠⎟
⎟
⎟
⎟⎛
⎝
⎜⎞
⎠
⎟θ
θ
θ
θ (11.7m)