Now we compute the damping matrix for the two-dimensional case. The damping matrix is related to the integral of the square of the trial function. Consider the integral over a single triangle with vertices ( x i , y i ) , ( x i + 1 , y i + 1 ) , ( x i + 2 , y i + 2 ) (x_i, y_i), (x_{i+1}, y_{i+1}), (x_{i+2}, y_{i+2})
The trial function on the triangle has the form υ ( i ) ( i + 2 ) ( x , y ) \upsilon_{(i)(i+2)}(x, y) = q i ⋅ ϕ i ( x , y ) = q_i \cdot \phi_i(x, y) + q i + 1 ⋅ ϕ i + 1 ( x , y ) + q_{i+1} \cdot \phi_{i+1}(x, y) + q i + 2 ⋅ ϕ i + 2 ( x , y ) + q_{i+2} \cdot \phi_{i+2}(x, y) 5.9 { a i + b i ⋅ x i + c i ⋅ y i = 1 a i + b i ⋅ x i + 1 + c i ⋅ y i + 1 = 0 a i + b i ⋅ x i + 2 + c i ⋅ y i + 2 = 0 { a i + 1 + b i + 1 ⋅ x i + c i + 1 ⋅ y i = 0 a i + 1 + b i + 1 ⋅ x i + 1 + c i + 1 ⋅ y i + 1 = 1 a i + 1 + b i + 1 ⋅ x i + 2 + c i + 1 ⋅ y i + 2 = 0 { a i + 2 + b i + 2 ⋅ x i + c i + 2 ⋅ y i = 0 a i + 2 + b i + 2 ⋅ x i + 1 + c i + 2 ⋅ y i + 1 = 0 a i + 2 + b i + 2 ⋅ x i + 2 + c i + 2 ⋅ y i + 2 = 1 \begin{cases} a_{i} + b_{i} \cdot x_i + c_{i} \cdot y_i = 1\\ a_{i} + b_{i} \cdot x_{i+1} + c_{i} \cdot y_{i+1} = 0\\ a_{i} + b_{i} \cdot x_{i+2} + c_{i} \cdot y_{i+2} = 0 \end{cases} \begin{cases} a_{i+1} + b_{i+1} \cdot x_i + c_{i+1} \cdot y_i = 0\\ a_{i+1} + b_{i+1} \cdot x_{i+1} + c_{i+1} \cdot y_{i+1} = 1\\ a_{i+1} + b_{i+1} \cdot x_{i+2} + c_{i+1} \cdot y_{i+2} = 0 \end{cases} \begin{cases} a_{i+2} + b_{i+2} \cdot x_i + c_{i+2} \cdot y_i = 0\\ a_{i+2} + b_{i+2} \cdot x_{i+1} + c_{i+2} \cdot y_{i+1} = 0\\ a_{i+2} + b_{i+2} \cdot x_{i+2} + c_{i+2} \cdot y_{i+2} = 1 \end{cases} 5.10 { a i = ( x i + 1 ⋅ y i + 2 − x i + 2 ⋅ y i + 1 ) / det b i = ( y i + 1 − y i + 2 ) / det c i = ( − x i + 1 + x i + 2 ) / det { a i + 1 = ( − x i ⋅ y i + 2 + x i + 2 ⋅ y i ) / det b i + 1 = ( − y i + y i + 2 ) / det c i + 1 = ( x i − x i + 2 ) / det { a i + 2 = ( x i ⋅ y i + 1 − x i + 1 ⋅ y i ) / det b i + 2 = ( y i − y i + 1 ) / det c i + 2 = ( − x i + x i + 1 ) / det \begin{cases} a_{i} = (x_{i+1} \cdot y_{i+2} - x_{i+2} \cdot y_{i+1}) / \det\\ b_{i} = (y_{i+1} - y_{i+2}) / \det\\ c_{i} = (-x_{i+1} + x_{i+2}) / \det \end{cases} \begin{cases} a_{i+1} = ( - x_i \cdot y_{i+2} + x_{i+2} \cdot y_i) / \det\\ b_{i+1} = (-y_i + y_{i+2}) / \det\\ c_{i+1} = (x_i - x_{i+2}) / \det \end{cases} \begin{cases} a_{i+2} = (x_i \cdot y_{i+1} - x_{i+1} \cdot y_i) / \det\\ b_{i+2} = (y_i - y_{i+1}) / \det\\ c_{i+2} = (-x_i + x_{i+1}) / \det \end{cases}
Substitute the trial function into (6.24 ∫ △ υ 2 d S . \int_{\triangle} \upsilon^2 \,dS.
∫ △ υ ( i ) ( i + 2 ) 2 d S \displaystyle \int_{\triangle} \upsilon_{(i)(i+2)}^2 \,dS = ∫ △ [ q i ⋅ ( a i + b i ⋅ x + c i ⋅ y ) + q i + 1 ⋅ ( a i + 1 + b i + 1 ⋅ x + c i + 1 ⋅ y ) + q i + 2 ⋅ ( a i + 2 + b i + 2 ⋅ x + c i + 2 ⋅ y ) ] 2 d S \displaystyle = \int_{\triangle} \Big[ q_i \cdot (a_i + b_i \cdot x + c_i \cdot y) + q_{i+1} \cdot (a_{i+1} + b_{i+1} \cdot x + c_{i+1} \cdot y) + q_{i+2} \cdot (a_{i+2} + b_{i+2} \cdot x + c_{i+2} \cdot y) \Big]^2 \,dS Expand the square. To simplify the computations, we use the fact that for linear hat functions on the triangle the following relations hold
where S △ S_{\triangle} 6.9 S △ \displaystyle S_{\triangle} = ∣ d ∣ 2 , \displaystyle = \frac{\displaystyle |d|}{\displaystyle 2},
Using these relations and expanding the square of the trial function, we obtain
∫ △ υ ( i ) ( i + 2 ) 2 d S \displaystyle \int_{\triangle} \upsilon_{(i)(i+2)}^2 \,dS = ∫ △ [ q i 2 ⋅ ϕ i 2 + q i + 1 2 ⋅ ϕ i + 1 2 + q i + 2 2 ⋅ ϕ i + 2 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 + 1 ⋅ q i + 2 ⋅ ϕ i + 1 ⋅ ϕ i + 2 ] d S \displaystyle = \int_{\triangle} \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 + 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+1} \cdot q_{i+2} \cdot \phi_{i+1} \cdot \phi_{i+2} \Big] \,dS Apply the formulas (6.26 ∫ △ ϕ m ⋅ ϕ n d S \displaystyle \int_{\triangle} \phi_m \cdot \phi_n \,dS = { S △ 6 , m = n S △ 12 , m ≠ n \displaystyle = \begin{cases} \frac{\displaystyle S_{\triangle}}{\displaystyle 6}, & m = n\\ \frac{\displaystyle S_{\triangle}}{\displaystyle 12}, & m \neq n \end{cases}
∫ △ υ ( i ) ( i + 2 ) 2 d S \displaystyle \int_{\triangle} \upsilon_{(i)(i+2)}^2 \,dS = q i 2 ⋅ S △ 6 \displaystyle = q_i^2 \cdot \frac{\displaystyle S_{\triangle}}{\displaystyle 6} + q i + 1 2 ⋅ S △ 6 \displaystyle + q_{i+1}^2 \cdot \frac{\displaystyle S_{\triangle}}{\displaystyle 6} + q i + 2 2 ⋅ S △ 6 \displaystyle + q_{i+2}^2 \cdot \frac{\displaystyle S_{\triangle}}{\displaystyle 6} + 2 ⋅ q i ⋅ q i + 1 ⋅ S △ 12 \displaystyle + 2 \cdot q_i \cdot q_{i+1} \cdot \frac{\displaystyle S_{\triangle}}{\displaystyle 12} + 2 ⋅ q i ⋅ q i + 2 ⋅ S △ 12 \displaystyle + 2 \cdot q_i \cdot q_{i+2} \cdot \frac{\displaystyle S_{\triangle}}{\displaystyle 12} + 2 ⋅ q i + 1 ⋅ q i + 2 ⋅ S △ 12 \displaystyle + 2 \cdot q_{i+1} \cdot q_{i+2} \cdot \frac{\displaystyle S_{\triangle}}{\displaystyle 12} Simplify the expression
∫ △ υ ( i ) ( i + 2 ) 2 d S \displaystyle \int_{\triangle} \upsilon_{(i)(i+2)}^2 \,dS = S △ 12 [ 2 q i 2 + 2 q i + 1 2 + 2 q i + 2 2 + 2 q i q i + 1 + 2 q i q i + 2 + 2 q i + 1 q i + 2 ] . \displaystyle = \frac{S_{\triangle}}{12} \left[ 2 q_i^2 + 2 q_{i+1}^2 + 2 q_{i+2}^2 + 2 q_i q_{i+1} + 2 q_i q_{i+2} + 2 q_{i+1} q_{i+2} \right]. We introduce the notation for the elements of the local damping matrix of the triangle
Thus, the local damping matrix for the triangular element has the form
The global damping matrix C \mathbf{C} N × N N \times N N N
Damping matrix 1D Damping matrix 3D