0
Before we start, some basic mathematic theorem.
Divergence
∇·v = v_{i,i}
∇·T = T_{ij,i} e_j
Gradient (dim+1)
∇\phi = \phi_{,i} e_i
∇v = v_{i,j} e_i⊗e_j
∇T = T_{ij,k} e_i⊗e_j⊗e_k
Curl (dim=)
∇ × v =ε_{ijk} v_{j,i} e_k
∇ × T =ε_{ijk} T_{mj,i} e_k⊗e_m
The product rule of differentiation
\because \bm{σ}·\bm{\nu} = \bm{σ}_{ij}\bm{\nu}_j e_i
\therefore ∇·(\bm{σ}·\bm{\nu}) = (\bm{σ}_{ij}\bm{\nu}_j)_{,i} = \bm{σ}_{ij,i}\bm{\nu}_j + \bm{σ}_{ij}\bm{\nu}_{j,i}
also
(∇·\bm{σ})·\bm{\nu} = \bm{σ}_{ij,i} \bm{\nu}_j
∇\bm{\nu} : \bm{σ} = \bm{\nu}_{i,j}\bm{σ}_{ji}
or as Prof. Zohdi suggested
∇\bm{\nu} : \bm{σ} = \bm{\nu}_{i,j}\bm{σ}_{ij}
but since \bm{σ} is symmetric
\bm{σ}_{ij} = \bm{σ}_{ji}
therefore
\tag{1} ∇·(\bm{σ}·\bm{\nu}) = (∇·\bm{σ})·\bm{\nu} + ∇\bm{\nu} : \bm{σ}
Divergence Theorem
∫_{∂R} \phi n dA = ∫_{R} ∇·\phi dV
\tag{2} ∫_{∂R} \bm{v}·n dA = ∫_{R} ∇·\bm{v} dV
∫_{∂R} \bm{T} n dA = ∫_{R} ∇·\bm{T} dV
or
∫_{∂R} \phi n_i dA = ∫_{R} \phi_{,i} dV
∫_{∂R} v_i n_i dA = ∫_{R} v_{i,i} dV
∫_{∂R} T_{ij}n_j dA = ∫_{R} T_{ij,j} dV
Stokes theorem
∫_{C} \phi d\bm{x} = ∫_{S} n × ∇\phi dA
∫_{C} v·d\bm{x} = ∫_{S} n·∇×v dA
∫_{C} T d\bm{x} = ∫_{S} (∇×T)^T n dA
or
∫_{C} \phi dx_i = ∫_{S} ε_{ijk} n_j \phi_{,k} dA
∫_{C} v_i dx_i = ∫_{S} n_i ε_{ijk} v_{k,j} dA
∫_{C} T_{ij} dx_j = ∫_{S} ε_{jpq} T_{iq,p} n_j dA
1
The SMALL deformation of the body is governed by (strong form):
∇⋅(\bm{IE}:∇\bm{u})+ρ\bm{g}=\bm{0}
since
\bm{σ} = \bm{IE}:∇\bm{u}
let
\bm{f} = ρ\bm{g}
thus
∇·\bm{σ} + \bm{f} = 0
so we can write
∫_Ω (∇·\bm{σ} + \bm{f} ) · \bm{\nu} dΩ = 0
using (1)
∫_Ω (∇·(\bm{σ}·\bm{\nu})-∇\bm{\nu}:\bm{σ}) dΩ + ∫_Ω \bm{f}·\bm{\nu} dΩ = 0
using (2)
∫_Ω ∇\bm{\nu}:\bm{σ} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{∂Ω} \bm{σ}·\bm{\nu}·n dA
because
\bm{t} = \bm{σ}·\bm{n}
therefore
∫_Ω ∇\bm{\nu}:\bm{σ} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{Γ_t} \bm{t}·\bm{\nu} dA
so we have the weak form
\begin{array}{lcr} \text{Find }\bm{u}, \bm{u}|_{Γ_u} = \bm{u}^* \text{, such that }∀\bm{\nu}, \bm{\nu}|_{Γ_u}= 0 \\ \displaystyle∫_Ω ∇\bm{\nu}:\bm{IE}:∇\bm{u} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{Γ_t} \bm{t}·\bm{\nu} dA \end{array}
where u^ is the applied boundary displacement on Γ_u, t = t^ on Γ_t
since the integrals must be finite, we improve the weak form as below
\begin{array}{lcr} \text{Find }\bm{u} ∈ \bm{H}^1(Ω), \bm{u}|_{Γ_u} = u^* \text{, such that }∀\bm{\nu} ∈ \bm{H}^1(Ω), \bm{\nu}|_{Γ_u}= 0 \\ \displaystyle∫_Ω ∇\bm{\nu}:\bm{IE}:∇\bm{u} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{Γ_t} \bm{t}·\bm{\nu} dA \end{array}
where
\bm{H}^1(Ω) \vcentcolon= [H^1(Ω)]^3
explained as below
similar to the definition of u ∈ H^1(Ω) which states
u ∈ H^1(Ω) \text{ if } ||u||_{H^1(Ω)}^2 \vcentcolon= ∫_Ω u_{,j}u_{,j} dΩ + ∫_Ω uudΩ
the definition of \bm{u} ∈ \bm{H}^1(Ω) states as below
\bm{u} ∈ \bm{H}^1(Ω) \text{ if } ||\bm{u}||_{\bm{H}^1(Ω)}^2 \vcentcolon= ∫_Ω u_{i,j}u_{i,j} dΩ + ∫_Ω u_iu_idΩ
2
should \bm{u}^* be different from the applied boundary displacement on Γ_u, weak form can be stated as below
\tag{3} \begin{array}{lcr} \text{Find }\bm{u} ∈ \bm{H}^1(Ω) \text{, such that }∀\bm{\nu} ∈ \bm{H}^1(Ω) \\ \displaystyle∫_Ω ∇\bm{\nu}:\bm{IE}:∇\bm{u} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{Γ_t} \bm{t}·\bm{\nu} dA + P^\star∫_{Γ_u} (\bm{u}^*-\bm{u})·\bm{\nu} dA \end{array}
or
\begin{array}{lcr} \text{Find }\bm{u} ∈ \bm{H}^1(Ω) \text{, such that }∀\bm{\nu} ∈ \bm{H}^1(Ω) \\ \displaystyle∫_Ω \nu_{i,j} IE_{ijkl} u_{k,l} dΩ = ∫_Ω f_i \nu_i dΩ + ∫_{Γ_t} t_i \nu_i dA + P^\star∫_{Γ_u} u^*_i \nu_i - u_i \nu_i dA \end{array}
where the (penalty) parameter P^\star is a large positive number.
3
To have
\begin{cases} \hat{\phi}_i(\zeta_1,\zeta_2,\zeta_3) = 1 &\text{ , } \zeta_1=\zeta_{i1},\zeta_2=\zeta_{i2},\zeta_3=\zeta_{i3} \\ \hat{\phi}_i(\zeta_1,\zeta_2,\zeta_3) = 0 &\text{ , } \text{others} \end{cases}
for each \hat{\phi}_i 8 functions (for each of the 8 points) with 8 unknowns, aka. \zeta_1^m \zeta_2^n \zeta_3^k where m,n,k=0,1 can be derived, leading to the only solution as following
\begin{cases} \hat{\phi}_1 = \frac{1}{8} (1-\zeta_1) (1-\zeta_2) (1-\zeta_3) \\ \hat{\phi}_2 = \frac{1}{8} (1+\zeta_1) (1-\zeta_2) (1-\zeta_3) \\ \hat{\phi}_3 = \frac{1}{8} (1+\zeta_1) (1+\zeta_2) (1-\zeta_3) \\ \hat{\phi}_4 = \frac{1}{8} (1-\zeta_1) (1+\zeta_2) (1-\zeta_3) \\ \hat{\phi}_5 = \frac{1}{8} (1-\zeta_1) (1-\zeta_2) (1+\zeta_3) \\ \hat{\phi}_6 = \frac{1}{8} (1+\zeta_1) (1-\zeta_2) (1+\zeta_3) \\ \hat{\phi}_7 = \frac{1}{8} (1+\zeta_1) (1+\zeta_2) (1+\zeta_3) \\ \hat{\phi}_8 = \frac{1}{8} (1-\zeta_1) (1+\zeta_2) (1+\zeta_3) \end{cases}
4
\bm{IE} has the following symmetries
IE_{ijkl} = IE_{klij} = IE_{jikl} = IE_{ijlk}
allowing us to rewrite (3) in a more compact matrix form
\tag{4} \begin{aligned} ∫_Ω ([\bm{D}]\{\bm{\nu}\})^T [\bm{IE}] ([\bm{D}]\{\bm{u}\}) dΩ = &∫_Ω \{\bm{\nu}\}^T\{\bm{f}\} dΩ \\ &+ ∫_{Γ_t} \{\bm{\nu}\}^T\{\bm{t}^*\} dA \\ &+ P^\star∫_{Γ_u} \{\bm{\nu}\}^T\{\bm{u}^*-\bm{u}\} dA \end{aligned}
where
\tag{5} [\bm{D}] \vcentcolon = \begin{bmatrix} \cfrac{∂}{∂x_1} & 0 & 0 \\ 0 & \cfrac{∂}{∂x_2} & 0 \\ 0 & 0 & \cfrac{∂}{∂x_3} \\ \cfrac{∂}{∂x_2} & \cfrac{∂}{∂x_1} & 0 \\ 0 & \cfrac{∂}{∂x_3} & \cfrac{∂}{∂x_2} \\ \cfrac{∂}{∂x_3} & 0 & \cfrac{∂}{∂x_1} \\ \end{bmatrix}, \{\bm{u}\} \vcentcolon = \begin{Bmatrix} u_1 \\ u_2 \\ u_3 \\ \end{Bmatrix}, \{\bm{f}\} \vcentcolon = \begin{Bmatrix} f_1 \\ f_2 \\ f_3 \\ \end{Bmatrix}, \{\bm{t}^*\} \vcentcolon = \begin{Bmatrix} t^*_1 \\ t^*_2 \\ t^*_3 \\ \end{Bmatrix},
and [\bm{IE}] is the elastic stiffness matrix of the material.
[\bm{IE}] \vcentcolon = \begin{bmatrix} IE_{1111} & IE_{1122} & IE_{1133} & IE_{1112} & IE_{1123} & IE_{1113} \\ IE_{2211} & IE_{2222} & IE_{2233} & IE_{2212} & IE_{2223} & IE_{2213} \\ IE_{3311} & IE_{3322} & IE_{3333} & IE_{3312} & IE_{3323} & IE_{3313} \\ IE_{1211} & IE_{1222} & IE_{1233} & IE_{1212} & IE_{1223} & IE_{1213} \\ IE_{2311} & IE_{2322} & IE_{2333} & IE_{2312} & IE_{2323} & IE_{2313} \\ IE_{1311} & IE_{1322} & IE_{1333} & IE_{1312} & IE_{1323} & IE_{1313} \\ \end{bmatrix}
Further rewrite {\bm{u}^h}
\tag{6} \{\bm{u}^h\} = [\phi]\{\bm{a}\}
where
\{\bm{a}\} \vcentcolon = \begin{Bmatrix} a_1 \\ a_2 \\ a_3 \\ .\\ .\\ .\\ a_{3N}\\ \end{Bmatrix}, [\phi] \vcentcolon = \begin{bmatrix} \phi_1 & \phi_2 & ... & \phi_N & & & & & & & & \\ & & & & \phi_1 & \phi_2 & ... & \phi_N & & & & \\ & & & & & & & & \phi_1 & \phi_2 & ... & \phi_N \\ \end{bmatrix},
choose \bm{\nu} with the same basis, but a different linear combination
\tag{7} \{\bm{\nu}^h\} = [\phi]\{\bm{b}\}
with (6),(7) we can rewrite (4)
\tag{8} \begin{aligned} ∫_Ω ([\bm{D}][\phi]\{\bm{b}\})^T [\bm{IE}] ([\bm{D}][\phi]\{\bm{a}\}) dΩ = &∫_Ω ([\phi]\{\bm{b}\})^T\{\bm{f}\} dΩ \\ & + ∫_{Γ_t} ([\phi]\{\bm{b}\})^T\{\bm{t}^*\} dA \\ & + P^\star∫_{Γ_u} ([\phi]\{\bm{b}\})^T\{\bm{u}^*-[\phi]\{\bm{a}\}\} dA \end{aligned}
We can rewrite (8)
\{\bm{b}\}^T \{ [\bm{K}] \{\bm{a}\} - \{\bm{R}\} \} = 0
where
\tag{9} [\bm{K}] \vcentcolon = ∫_Ω ([\bm{D}][\phi])^T [\bm{IE}] ([\bm{D}][\phi]) dΩ + P^\star∫_{Γ_u} [\phi]^T[\phi] dA
\tag{10} \{\bm{R}\} \vcentcolon = ∫_Ω [\phi]^T\{\bm{f}\} dΩ + ∫_{Γ_t} [\phi]^T\{\bm{t}^*\} dA + P^\star∫_{Γ_u} [\phi]^T\{\bm{u}^*\} dA
Consider
[\bm{K}] = \sum_{e=1}^{N_e}{[\bm{K}]^e}
we can rewrite (9),(10) for e=1,2..,N_e
\tag{11} [\bm{K}]^e \vcentcolon = ∫_{Ω_e} ([\bm{D}][\phi])^T [\bm{IE}] ([\bm{D}][\phi]) dΩ_e + P^\star∫_{Γ_{u,e}} [\phi]^T[\phi] dA_e
\tag{12} \{\bm{R}\}^e \vcentcolon = ∫_{Ω_e} [\phi]^T\{\bm{f}\} dΩ_e + ∫_{Γ_{t,e}} [\phi]^T\{\bm{t}^*\} dA_e + P^\star∫_{Γ_{u,e}} [\phi]^T\{\bm{u}^*\} dA_e
where
Γ_{u,e} = Γ_u ∩ ∂Ω_e
Γ_{t,e} = Γ_t ∩ ∂Ω_e
5
In 3D,
[\hat\phi] \vcentcolon = \begin{bmatrix} \hat\phi_1 & \hat\phi_2 & ... & \hat\phi_8 & & & & & & & & \\ & & & & \hat\phi_1 & \hat\phi_2 & ... & \hat\phi_8 & & & & \\ & & & & & & & & \hat\phi_1 & \hat\phi_2 & ... & \hat\phi_8 \\ \end{bmatrix}
express [\bm{D}] in terms ζ_1, ζ_2, ζ_3
[D(\phi(x1, x2, x3))] = [\hat D \phi(M_{x_1}(ζ1, ζ2, ζ3),M_{x_2}(ζ1, ζ2, ζ3),M_{x_3}(ζ1, ζ2, ζ3))]
so that
[\hat \bm{D}] = \begin{bmatrix} \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_1} & 0 & 0 \\ 0 & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_2} & 0 \\ 0 & 0 & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_3} \\ \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_1} & 0 \\ 0 & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_2} \\ \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_3} & 0 & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_1} \\ \end{bmatrix}
therefore
[\hat \bm{D}][\hat\phi] = \begin{bmatrix} \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_1} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_1} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_1} & & & & & & & \\ & & & & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_2} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_2} & & & \\ & & & & & & & & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_3} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_3} \\ \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_2} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_1} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_1} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_1} & & & \\ & & & & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_3} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_2} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_2} \\ \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_3} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_3} & & & & & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_1} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_1} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_1} \\ \end{bmatrix}
with Gaussian Quadrature
\int_0^L{F(x)dx} = \int_{-1}^1{F(x(\zeta))J(\zeta)d\zeta} = \sum_1^G{w_i}F(\zeta_i)J(\zeta_i)
we can rewrite (11),(12)
[\bm{K}]^e = \sum_1^G{\sum_1^G{\sum_1^G{w_q w_r w_s ([\hat \bm{D}][\hat\phi])^T[\hat{\bm{IE}}][\hat \bm{D}][\hat\phi]J}}}+ P^\star\sum_1^G{\sum_1^G{w_q w_r [\phi]^T[\phi] J_s}}
\{\bm{R}\}^e = \sum_1^G{\sum_1^G{\sum_1^G{w_q w_r w_s [\phi]^T\{\bm{f}\} J}}} + \sum_1^G{\sum_1^G{w_q w_r[\phi]^T\{\bm{t}^*\} J_s}} + P^\star\sum_1^G{\sum_1^G{w_q w_r [\phi]^T\{\bm{u}^*\} J_s}}
where
\bm{F} \vcentcolon= ∇x(ζ_1,ζ_2,ζ_3)
J \vcentcolon= |\bm{F}|
and
J_s = J\bm{F}^{-T}·\bm{N}·\bm{n}
derived as below
Nanson’s formula
\bm{n} dA_e = J\bm{F}^{-T}·\bm{N} d\hat{Ae}
where A_e is an area of a region in the current configuration, \hat{Ae} is the same area in the reference configuration, and \bm{n} is the outward normal to the area element in the current configuration while \bm{N} is the outward normal in the reference configuration.
multiply \bm{n} on both sides
dA_e = J\bm{F}^{-T}·\bm{N}·\bm{n} d\hat{Ae}
therefore
J_s = J\bm{F}^{-T}·\bm{N}·\bm{n}
6
6.1
∇⋅(\bm{IK}⋅∇θ)+ρs=0
let the heat flux density q
\bm{q} = \bm{IK}⋅∇θ
and let
f = ρs
therefore
∇⋅\bm{q} + f = 0
so we can write
\tag{13} ∫_Ω (∇·\bm{q} + f ) · \nu dΩ = 0
now consider
(∇⋅\bm{q})·\nu = q_{i,i}\nu
∇⋅(\bm{q}\nu) = (q_i \nu)_{,i} = q_{i,i}\nu + q_i \nu_{,i}
(∇\nu)·\bm{q} = q_i \nu_{,i}
therefore we can write (13) as
∫_Ω (∇⋅(\bm{q}\nu) - (∇\nu)·\bm{q}) dΩ + ∫_Ω f\nu dΩ= 0
∫_Ω (∇\nu)·\bm{q} dΩ = ∫_Ω f\nu dΩ + ∫_Ω ∇⋅(\bm{q}\nu)
using (2)
∫_Ω (∇\nu)·\bm{q} dΩ = ∫_Ω f\nu dΩ + ∫_{∂Ω} \bm{q}⋅\bm{n} \nu dA
so we have the weak form
\begin{array}{lcr} \text{Find }θ ∈ H^1(Ω), θ|_{Γ_θ} = θ^* \text{, such that }∀\nu ∈ H^1(Ω), \nu|_{Γ_θ}= 0 \\ \displaystyle∫_Ω (∇\nu)·\bm{IK}⋅∇θ dΩ = ∫_Ω f \nu dΩ + ∫_{∂Ω} \bm{q}⋅\bm{n} \nu dA \end{array}
6.2
should θ^* be different from the applied boundary displacement on Γ_θ, weak form can be stated as below
\tag{14} \begin{array}{lcr} \text{Find }θ ∈ H^1(Ω), θ|_{Γ_θ} = θ^* \text{, such that }∀\nu ∈ H^1(Ω), \nu|_{Γ_θ}= 0 \\ \displaystyle∫_Ω (∇\nu)·\bm{IK}⋅∇θ dΩ = ∫_Ω f \nu dΩ + ∫_{∂Ω} \bm{q}⋅\bm{n} \nu dA + P^\star∫_{Γ_θ} (θ^*-θ) \nu dA \end{array}
where the (penalty) parameter P^\star is a large positive number.
6.3
same as 3
6.4
rewrite θ^h
\tag{15} θ^h = [\phi]\{\bm{a}\} = \phi_i a_i
where
\{\bm{a}\} \vcentcolon = \begin{Bmatrix} a_1 \\ a_2 \\ a_3 \\ .\\ .\\ .\\ a_{N}\\ \end{Bmatrix}, [\phi] \vcentcolon = \begin{bmatrix} \phi_1 & \phi_2 & ... & \phi_N \end{bmatrix},
choose \bm{\nu} with the same basis, but a different linear combination
\tag{16} \nu^h = [\phi]\{\bm{b}\} = \phi_j b_j
with (15),(16) we can rewrite (14)
\tag{17} \begin{aligned} ∫_Ω (∇[\phi]\{\bm{b}\})·\bm{IK}⋅∇[\phi]\{\bm{a}\} dΩ = & ∫_Ω f [\phi]\{\bm{b}\} dΩ \\ & + ∫_{∂Ω} \bm{q}⋅\bm{n} [\phi]\{\bm{b}\} dA \\ & + P^\star∫_{Γ_θ} (θ^*-[\phi]\{\bm{a}\})·[\phi]\{\bm{b}\} dA \end{aligned}
We can rewrite (17)
\{\bm{b}\}^T \{ K \{\bm{a}\} - \{\bm{R}\} \} = 0
where
\tag{18} K \vcentcolon = ∫_Ω (∇[\phi])·\bm{IK}⋅∇[\phi] dΩ + P^\star∫_{Γ_θ} [\phi]^T[\phi] dA
\tag{19} \{\bm{R}\} \vcentcolon = ∫_Ω [\phi]^Tf dΩ + ∫_{∂Ω} [\phi]^T\bm{q}⋅\bm{n} dA + P^\star∫_{Γ_θ} [\phi]^Tθ^* dA
Consider
K= \sum_{e=1}^{N_e}K^e
we can rewrite (18),(19) for e=1,2..,N_e
K^e \vcentcolon = ∫_{Ω_e} (∇[\phi])·\bm{IK}⋅∇[\phi] dΩ_e + P^\star∫_{Γ_{u,e}} [\phi]^T[\phi] dA_e
\{\bm{R}\}^e \vcentcolon = ∫_{Ω_e} [\phi]^Tf dΩ_e + ∫_{∂Ω,e} [\phi]^T\bm{q}⋅\bm{n} dA_e + P^\star∫_{Γ_{u,e}} [\phi]^Tθ^* dA_e
where
Γ_{θ,e} = Γ_θ ∩ ∂Ω_e
7
Key difference is that the diemsion of the equations derived for thermodynamics are generally lower than those for linear elasticity.
This is because θ and s are scalers rather than vectors as \bm{u} and \bm{g}, and \bm{IK} is a second order tensor rather than a forth order tensor as \bm{IE}.
Reference
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Changsha Culture and Art Center, by Zaha Hadid.
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