Let us proceed to computing the damping matrix for the three-dimensional case. The damping matrix is related to the integral of the square of the trial function. Consider the integral for a single tetrahedron with vertices
( x i , y i , z i ) , \displaystyle (x_i, y_i, z_i), ( x i + 1 , y i + 1 , z i + 1 ) , \displaystyle (x_{i+1}, y_{i+1}, z_{i+1}), ( x i + 2 , y i + 2 , z i + 2 ) , \displaystyle (x_{i+2}, y_{i+2}, z_{i+2}), ( x i + 3 , y i + 3 , z i + 3 ) \displaystyle (x_{i+3}, y_{i+3}, z_{i+3}) The trial function on the tetrahedron has the form υ ( i ) ( i + 3 ) ( x , y , z ) \upsilon_{(i)(i+3)}(x, y, z) = q i ⋅ ϕ i {} = q_i \cdot \phi_i + q i + 1 ⋅ ϕ i + 1 {} + q_{i+1} \cdot \phi_{i+1} + q i + 2 ⋅ ϕ i + 2 {} + q_{i+2} \cdot \phi_{i+2} + q i + 3 ⋅ ϕ i + 3 {} + q_{i+3} \cdot \phi_{i+3} 6.13 ϕ i ( x , y , z ) \displaystyle \phi_i(x, y, z) = a i \displaystyle {} = a_i + b i ⋅ x \displaystyle {} + b_i \cdot x + c i ⋅ y \displaystyle {} + c_i \cdot y + d i ⋅ z \displaystyle {} + d_i \cdot z ϕ i + 1 ( x , y , z ) \displaystyle \phi_{i+1}(x, y, z) = a i + 1 \displaystyle {} = a_{i+1} + b i + 1 ⋅ x \displaystyle {} + b_{i+1} \cdot x + c i + 1 ⋅ y \displaystyle {} + c_{i+1} \cdot y + d i + 1 ⋅ z \displaystyle {} + d_{i+1} \cdot z ϕ i + 2 ( x , y , z ) \displaystyle \phi_{i+2}(x, y, z) = a i + 2 \displaystyle {} = a_{i+2} + b i + 2 ⋅ x \displaystyle {} + b_{i+2} \cdot x + c i + 2 ⋅ y \displaystyle {} + c_{i+2} \cdot y + d i + 2 ⋅ z \displaystyle {} + d_{i+2} \cdot z ϕ i + 3 ( x , y , z ) \displaystyle \phi_{i+3}(x, y, z) = a i + 3 \displaystyle {} = a_{i+3} + b i + 3 ⋅ x \displaystyle {} + b_{i+3} \cdot x + c i + 3 ⋅ y \displaystyle {} + c_{i+3} \cdot y + d i + 3 ⋅ z \displaystyle {} + d_{i+3} \cdot z
Substitute the trial function into (6.29 ∫ тет υ 2 d V . \int_{\text{тет}} \upsilon^2 \,dV.
∫ tet υ ( i ) ( i + 3 ) 2 d V \displaystyle \int_{\text{tet}} \upsilon_{(i)(i+3)}^2 \,dV = ∫ tet [ q i ⋅ ϕ i ( x , y , z ) + q i + 1 ⋅ ϕ i + 1 ( x , y , z ) + q i + 2 ⋅ ϕ i + 2 ( x , y , z ) + q i + 3 ⋅ ϕ i + 3 ( x , y , z ) ] 2 d V \displaystyle {} = \int_{\text{tet}} \Big[ q_i \cdot \phi_i(x, y, z) + q_{i+1} \cdot \phi_{i+1}(x, y, z) + q_{i+2} \cdot \phi_{i+2}(x, y, z) + q_{i+3} \cdot \phi_{i+3}(x, y, z) \Big]^2 \,dV For linear hat functions on the tetrahedron the following relations hold
where V tet V_{\text{tet}} 6.15 V тет \displaystyle V_{\text{тет}} = ∣ Δ ∣ 6 , \displaystyle {} = \frac{\displaystyle |\Delta|}{\displaystyle 6},
Expand the square of the trial function
∫ tet υ ( i ) ( i + 3 ) 2 d V \displaystyle \int_{\text{tet}} \upsilon_{(i)(i+3)}^2 \,dV = ∫ tet [ q i 2 ⋅ ϕ i 2 + q i + 1 2 ⋅ ϕ i + 1 2 + q i + 2 2 ⋅ ϕ i + 2 2 + q i + 3 2 ⋅ ϕ i + 3 2 + 2 ⋅ q i ⋅ q i + 1 ⋅ ϕ i ⋅ ϕ i + 1 + 2 ⋅ q i ⋅ q i + 2 ⋅ ϕ i ⋅ ϕ i + 2 + 2 ⋅ q i ⋅ q i + 3 ⋅ ϕ i ⋅ ϕ i + 3 + 2 ⋅ q i + 1 ⋅ q i + 2 ⋅ ϕ i + 1 ⋅ ϕ i + 2 + 2 ⋅ q i + 1 ⋅ q i + 3 ⋅ ϕ i + 1 ⋅ ϕ i + 3 + 2 ⋅ q i + 2 ⋅ q i + 3 ⋅ ϕ i + 2 ⋅ ϕ i + 3 ] d V \displaystyle {} = \int_{\text{tet}} \Big[ q_i^2 \cdot \phi_i^2 + q_{i+1}^2 \cdot \phi_{i+1}^2 + q_{i+2}^2 \cdot \phi_{i+2}^2 + q_{i+3}^2 \cdot \phi_{i+3}^2 + 2 \cdot q_i \cdot q_{i+1} \cdot \phi_i \cdot \phi_{i+1} + 2 \cdot q_i \cdot q_{i+2} \cdot \phi_i \cdot \phi_{i+2} + 2 \cdot q_i \cdot q_{i+3} \cdot \phi_i \cdot \phi_{i+3} + 2 \cdot q_{i+1} \cdot q_{i+2} \cdot \phi_{i+1} \cdot \phi_{i+2} + 2 \cdot q_{i+1} \cdot q_{i+3} \cdot \phi_{i+1} \cdot \phi_{i+3} + 2 \cdot q_{i+2} \cdot q_{i+3} \cdot \phi_{i+2} \cdot \phi_{i+3} \Big] \,dV Apply the formulas (6.31 ∫ тет ϕ m ⋅ ϕ n d V \displaystyle \int_{\text{тет}} \phi_m \cdot \phi_n \,dV = { V тет 10 , m = n V тет 20 , m ≠ n \displaystyle {} = \begin{cases} \frac{\displaystyle V_{\text{тет}}}{\displaystyle 10}, & m = n\\ \frac{\displaystyle V_{\text{тет}}}{\displaystyle 20}, & m \neq n \end{cases}
∫ tet υ ( i ) ( i + 3 ) 2 d V \displaystyle \int_{\text{tet}} \upsilon_{(i)(i+3)}^2 \,dV = q i 2 ⋅ V tet 10 \displaystyle {} = q_i^2 \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 10} + q i + 1 2 ⋅ V tet 10 \displaystyle {} + q_{i+1}^2 \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 10} + q i + 2 2 ⋅ V tet 10 \displaystyle {} + q_{i+2}^2 \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 10} + q i + 3 2 ⋅ V tet 10 \displaystyle {} + q_{i+3}^2 \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 10} + 2 ⋅ q i ⋅ q i + 1 ⋅ V tet 20 \displaystyle {} + 2 \cdot q_i \cdot q_{i+1} \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 20} + 2 ⋅ q i ⋅ q i + 2 ⋅ V tet 20 \displaystyle {} + 2 \cdot q_i \cdot q_{i+2} \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 20} + 2 ⋅ q i ⋅ q i + 3 ⋅ V tet 20 \displaystyle {} + 2 \cdot q_i \cdot q_{i+3} \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 20} + 2 ⋅ q i + 1 ⋅ q i + 2 ⋅ V tet 20 \displaystyle {} + 2 \cdot q_{i+1} \cdot q_{i+2} \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 20} + 2 ⋅ q i + 1 ⋅ q i + 3 ⋅ V tet 20 \displaystyle {} + 2 \cdot q_{i+1} \cdot q_{i+3} \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 20} + 2 ⋅ q i + 2 ⋅ q i + 3 ⋅ V tet 20 \displaystyle {} + 2 \cdot q_{i+2} \cdot q_{i+3} \cdot \frac{\displaystyle V_{\text{tet}}}{\displaystyle 20} Simplify the expression
∫ tet υ ( i ) ( i + 3 ) 2 d V \displaystyle \int_{\text{tet}} \upsilon_{(i)(i+3)}^2 \,dV = V tet 20 \displaystyle {} = \frac{\displaystyle V_{\text{tet}}}{\displaystyle 20} ⋅ [ 2 ⋅ q i 2 + 2 ⋅ q i + 1 2 + 2 ⋅ q i + 2 2 + 2 ⋅ q i + 3 2 + 2 ⋅ q i ⋅ q i + 1 + 2 ⋅ q i ⋅ q i + 2 + 2 ⋅ q i ⋅ q i + 3 + 2 ⋅ q i + 1 ⋅ q i + 2 + 2 ⋅ q i + 1 ⋅ q i + 3 + 2 ⋅ q i + 2 ⋅ q i + 3 ] \displaystyle {} \cdot \Big[ 2 \cdot q_i^2 + 2 \cdot q_{i+1}^2 + 2 \cdot q_{i+2}^2 + 2 \cdot q_{i+3}^2 + 2 \cdot q_i \cdot q_{i+1} + 2 \cdot q_i \cdot q_{i+2} + 2 \cdot q_i \cdot q_{i+3} + 2 \cdot q_{i+1} \cdot q_{i+2} + 2 \cdot q_{i+1} \cdot q_{i+3} + 2 \cdot q_{i+2} \cdot q_{i+3} \Big] We introduce the notation for the elements of the local damping matrix of the tetrahedron
Thus, the local damping matrix for the tetrahedral element has the form
The local damping matrix is symmetric. The global damping matrix C \mathbf{C} N × N N \times N N N
Damping matrix 2D Load vector 1D