Part (a)
From the given geometry we have
\begin{array}{ll}a_{1}=x_{2} y_{3}-x_{3} y_{2}=12 \quad a_{2}=x_{3} y_{1}-x_{1} y_{3}=0 & a_{3}=x_{1} y_{2}-x_{2} y_{1}=0 \\b_{1}=y_{2}-y_{3}=-4 \quad b_{2}=y_{3}-y_{1}=4 & b_{3}=y_{1}-y_{2}=0 \\c_{1}=x_{3}-x_{2}=0 \quad c_{2}=x_{1}-x_{3}=-3 & c_{3}=x_{2}-x_{1}=3 \\G=a_{1}+a_{2}+a_{3}=12 &\end{array} \qquad (a)
Matrix \mathbf{B} is now obtained from Equation 18.72
\begin{aligned}\varepsilon &=\mathbf{L}(\mathbf{N}) \mathbf{q}^e=\mathbf{B q}^e \\&=\frac{1}{G}\left[\begin{array}{cccccc}b_1 & 0 & b_2 & 0 & b_3 & 0 \\0 & c_1 & 0 & c_2 & 0 & c_3 \\c_1 & b_1 & c_2 & b_2 & c_3 & b_3\end{array}\right]\left\{\begin{array}{l}u_1 \\v_1 \\u_2 \\v_2 \\u_3 \\v_3\end{array}\right\}\end{aligned}
(18.72)
\mathbf{B}=\frac{1}{12}\left[\begin{array}{rrrrrr}-4 & 0 & 4 & 0 & 0 & 0 \\0 & 0 & 0 & -3 & 0 & 3 \\0 & -4 & -3 & 4 & 3 & 0\end{array}\right] \qquad (b)
Substituting for \mathrm{B}, t=1, A=G / 2=6
, and matrix D for plane strain condition we get
\mathbf{K}=1.389 \times 10^{6}\left[\begin{array}{cccccc}12.8 & 0 & -12.8 & 2.4 & 0 & -2.4 \\0 & 4.8 & 3.6 & -4.8 & -3.6 & 0 \\-12.8 & 3.6 & 15.5 & -6.0 & -2.7 & 2.4 \\2.4 & -4.8 & -6.0 & 12.0 & 3.6 & -7.2 \\0 & -3.6 & -2.7 & 3.6 & 2.7 & 0 \\-2.4 & 0 & 2.4 & -7.2 & 0 & 7.2\end{array}\right] \mathrm{kN} / \mathrm{m} \qquad (c)
The interpolation functions are given by
\begin{aligned}&N_{1}=\frac{1}{G}\left(a_{1}+b_{1} x+c_{1} y\right)=\frac{1}{G}(12-4 x) \\&N_{2}=\frac{1}{G}\left(a_{2}+b_{2} x+c_{2} y\right)=\frac{1}{G}(4 x-3 y) \\&N_{3}=\frac{1}{G}\left(a_{3}+b_{3} x+c_{3} y\right)=\frac{1}{G}(3 y)\end{aligned} \qquad (d)
The mass matrix is obtained from
\begin{aligned}\mathbf{M} &=\rho t \int_{A}\left[\begin{array}{cccccc}N_{1}^{2} & 0 & N_{1} N_{2} & 0 & N_{1} N_{3} & 0 \\0 & N_{1}^{2} & 0 & N_{1} N_{2} & 0 & N_{1} N_{3} \\N_{1} N_{2} & 0 & N_{2}^{2} & 0 & N_{2} N_{3} & 0 \\0 & N_{1} N_{2} & 0 & N_{2}^{2} & 0 & N_{2} N_{3} \\N_{1} N_{3} & 0 & N_{2} N_{3} & 0 & N_{3}^{2} & 0 \\0 & N_{1} N_{3} & 0 & N_{2} N_{3} & 0 & N_{3}^{2}\end{array}\right] d A \\&=1.2\left[\begin{array}{llllll}2 & 0 & 1 & 0 & 1 & 0 \\0 & 2 & 0 & 1 & 0 & 1 \\1 & 0 & 2 & 0 & 1 & 0 \\0 & 1 & 0 & 2 & 0 & 1 \\1 & 0 & 1 & 0 & 2 & 0 \\0 & 1 & 0 & 1 & 0 & 2\end{array}\right] \text { tonne }\end{aligned} \qquad (e)
As an example, the integral \int_{A} N_{1}^{2} d A is obtained from
\begin{aligned}\int_{A} N_{1}^{2} d A &=\frac{1}{G^{2}} \int_{0}^{3} \int_{0}^{4 x / 3}(12-4 x)^{2} d y d x \\&=\frac{1}{G^{2}} \frac{64}{3} \int_{0}^{3}\left(9 x+x^{3}-6 x^{2}\right) d x=1\end{aligned} \qquad (f)
\operatorname{Part}(\mathrm{b})
Loads equivalent to the gravity loads are obtained from
\begin{aligned}\mathrm{p}_{b} &=\int_{A} \mathbf{N}^{T}\left\{\begin{array}{c}0 \\-\rho g\end{array}\right\} t d A \\&=\int_{0}^{3} \int_{0}^{4 x / 3}\left[\begin{array}{cc}N_{1} & 0 \\0 & N_{1} \\N_{2} & 0 \\0 & N_{2} \\N_{3} & 0 \\0 & N_{3}\end{array}\right]\left\{\begin{array}{c}0 \\-\rho g\end{array}\right\} d x d y=-23.544\left\{\begin{array}{l}0 \\2 \\0 \\2 \\0 \\2\end{array}\right\} \mathrm{kN}\end{aligned} \qquad (g)
As an example, the integral \int_{A} N_{1} d A is obtained from
\begin{aligned}\int_{A} N_{1} d A &=\frac{1}{G} \int_{0}^{3} \int_{0}^{4 x / 3} a_{1}+b_{1} x+c_{1} y d x d y \\&=\frac{1}{G}\left(6 a_{1}+12 b_{1}+8 c_{1}\right)=2\end{aligned} \qquad (h)
It will be noted that the equivalent load vector is comprised of a load of \rho g t A / 3, or 1 / 3 of the total weight, at each node directed along the negative y axis, where A is the area of the triangle.
To obtain the nodal forces equivalent to the hydrostatic pressure on face BC, we define a local \xi axis with origin at \mathrm{B} and directed along BC. The interpolation functions that correspond to the \xi axis are
\mathrm{L}=\left[\begin{array}{ll}\left(1-\frac{\xi}{l}\right) & \frac{\xi}{l}\end{array}\right] \qquad (i)
where l=4 is the length of side BC. The nodal forces are now given by
\begin{aligned}\mathrm{p}_{S} &=-\int_{0}^{l}\left\{\begin{array}{r}1-\frac{\xi}{l} \\\frac{\xi}{l}\end{array}\right\}\left(90-\frac{40 \xi}{l}\right) t d \xi (j)\\&=-\left\{\begin{array}{l}153.3 \\126.7\end{array}\right\} \quad \mathrm{kN} \qquad (k)\end{aligned}
The nodal forces are shown in Figure E18.7b. They are directed along the negative x direction and are, in fact, statically equivalent to the distributed force.
