Mechanics – Mark Ma's Blog

Core Concepts in FEA

0

Before we start, some basic mathematic theorem¹².

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

∫_{∂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

∫_{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 $u \in H^1(Ω)$ 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}

\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 ²³

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 density⁴ $q$

\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

Tensor derivative (continuum mechanics) Cartesian coordinates, https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)#Cartesian_coordinates_2 ↩
Lallit Anand and Sanjay Govindjee, 2019, Introduction to Continuum Mechanics of Solid ↩ ↩
Allan F. Bower, 2012, Applied Mechanics of Solids, Chapter 3 ↩
Thermal conduction differential form, https://en.wikipedia.org/wiki/Thermal_conduction#Differential_form ↩