Page 252 - Mechanics of Asphalt Microstructure and Micromechanics

244 Ch a p t e r E i gh t

∂σ
Equilibrium equations (static problem): ij + b = 0 (8-3)
∂x j
i
For isotropic elasticity: D = λδ δ + μ δ δ +δ δ ) (8-4)
2(
ijkl ij kl ik jl il jk

l and m are Lamé constants.
The equilibrium equations represented by displacements will take the following
format:
(λ + ) μ u + μu + b = 0 (8-5)
kk , j , j ii j

In a more general format, it can be expressed as the following operator format:
Lu + b = 0 (8-6)
j j
Where L is a linear operator.
This set of equations can be also written in the vector format.
Lu + b = 0
The following is to obtain the specific format of the L operator.
Equations (8-3) and (8-1) can be written in the following format:
∂ε
D kl + b = 0 (8-7)
ijkl x ∂ j
i
u ∂ u ∂
ε = ( k + l )/2
kl x ∂ x ∂
l k (8-8)
Therefore:
∂ 2 u ∂ 2 u
2
D ( k + l )/ + b = 0 (8-9)
ijkl ∂∂ ∂∂ j
xx
xx
i l i k
∂ 2 u ∂ 2 u u ∂ u ∂
Since D = D , one can prove that D k = D l and D k k = D l
ijkl ijlk ijkl ∂∂ ijkl ∂∂ ijkl x ∂ ijkl x ∂
xx
xx
∂ 2 i l i k l k
So D u + b = 0 (8-10)
ijkl ∂∂ k j
xx
i l
∂ 2
Denote L = D , one has
jk ijkl ∂∂
xx
i l Lu + b = 0 (8-11)
jk k j
Here, L jk is a linear operator.
For solving the equilibrium equations using the analytical approach, it is equivalent
to finding displacement functions that satisfy the above equilibrium equations at any
point and the corresponding boundary conditions. For relatively simple problems, ana-
lytical solutions can be obtained. For most of the complicated geometric and loading
conditions, it is almost impossible to find analytical solutions. Numerical methods are
often used to obtain approximate solutions. Approximate solutions may result in some
residual forces:
Lu + b = R (8-12)
jk k j j
Where R j is the residual force.