Page 259 - Mechanics of Asphalt Microstructure and Micromechanics

F inite Element Method and Boundar y Element Method 251

The strains can be represented as:
⎧ ∂ ⎫
⎪ ,0 ⎪
⎪ x ∂ ⎪
⎪ ∂ ⎪
L = ⎨ ,0 ⎬
⎪ y ∂ ⎪
⎪ ∂ ∂ ⎪
⎪ , ⎪
⎩ y ∂ x ∂ ⎭
⎧ β ,, β ,, β ,0 ⎫
0
0
⎪ i j k ⎪ ⎪
⎪
⎨
0
0
B = 0 γ ,, , γ , , γ k ⎬
i
j
⎪ ⎪
,,
,
,
⎩ ⎪ γβ γ β γ ,β k ⎭ ⎪
j
i
k k
i
j
Obviously, for a fixed triangle, B is a constant matrix (only related to the coordinates
of the three nodes). In other words, the strains are constant in the element if the linear
T
v
,
interpolation functions are used (e = B (, , , ,u v ) ). Typically, for higher order
u
uv
i i j j k k
polynomial interpolation functions, the strains in the element are not constant.
Step 3: This is actually the process to obtain the element stiffness matrix and the
equivalent node forces due to the boundary forces and the body forces.
K = e ∫ B CBdΩ
T
e
Ω
⎧ ⎫
⎪1 ν ,, 0 ⎪
C = E ⎪ ⎨ 10 ⎪
⎬
ν,,
1 −ν 2 ⎪ −ν ⎪
⎪00 ,, 1 ⎪
⎩ 2 ⎭
e
Apparently B CB is a constant matrix and K is a 6 6 symmetric matrix.
T
S ∫
T
F Γ e N d τ Γ: equivalent forces due to the surface tractions
e
B ∫
T
e
F Ω e N bdΩ: equivalent forces due to distributed body forces
e ∫ NbdΩ
T
Ω
⎡ N ,, N ,, N ,0⎤
0
0
⎢ i j k ⎥
0
0
⎣ , 0 ⎢ N , , N , , N ⎥
k ⎦
i
j
⎧0 ⎫ ⎪
b = ⎨ ⎬
⎩ −ρ g ⎭
Nb = ( , −ρ gN ,, −ρ gN ,, −ρ gN ) T
0
0
0
i j k
e
3
0
0
F = (, −ρ gAh/ , , −ρ gAh/ , , −ρ gAh/ ) T
0
3
3
B
τ
F = N dΓ
T
e
S ∫ Γ e
In this special case, the only non-zero component is –P at Node 24.