Page 253 - Mechanics of Asphalt Microstructure and Micromechanics

F inite Element Method and Boundar y Element Method 245

While at each point it will result in some residual forces, one may release the
strong satisfaction of Equation (8-11) with a weak satisfaction in the following integral
format:
∫ Rw d Ω 0 (8-13)
Ω j j
∫ (Lu + b )w d Ω = 0 (8-14)
Ω j j j
Where w is a weighting function or test function. It can be proved that if the above
weak equations are satisfied for any chosen test function w j , one will have the strong
satisfaction. Equation (8-14) is the weak solution statement. It can be further represent-
ed in the following equation with domain decomposition. The different domains are
finite in size and sum up to occupy the entire object.
Ω n
∑ ∫ ( Lu + b w d Ω = 0 (8-15)
)
j
j
j
Ω 1 Ω
These domains can be considered as finite meshes.
The following considers the integral of the first term in Equation (8-14):
∂ 2
L = D
jk ijkl ∂∂
xx
i l
Ω ∫ Du w d Ω = 0
,
k il
j
ijkl
Since
u w = ( uw ), − uw (8-16)
,
kil j ki , j l ki , j l ,
∫ Du w d Ω = ∫ D [( u w ), −u w ] dΩ (8-17)
Ω ijkl k il , j Ω ijkl k i , j l k i , j,l l
By using the divergence theorem, one has
Ω ∫ D ( u w ), l dΩ = Γ ∫ D u w n dΓ
j l
k i
ijkl
j
,
ijkl
k i
,
∫ Du w d Ω = ∫ Duw n d Γ − ∫ Du wdΩ
Ω ijkl k il , j Γ ijkl k i , j l Ω ijkl k ki , j l , (8-18)
Since
σ = Du τ = σ n
jl ijkl k i , j jl l (8-19)
The surface integral of the first term on the right then becomes:
Γ ∫ Duw n dΓ Γ ∫ τ w dΓ
ijkl k i , j l j j (8-20)
∫ D u wd Ω = ∫ τ wd Γ − ∫ D u w d Ω (8-21)
,
Ω ijkl k il j Γ j j Ω ijkl k i , j l ,
By using Equation (8-6) or (8-11), one will arrive at:
Ω ∫ Duw dΩ Γ ∫ τ w dΓ + Ω ∫ b w dΩ (8-22)
ijkl k i , j l , j j j j