Question p.6.8: It is required to formulate the stiffness of a triangular el...
It is required to formulate the stiffness of a triangular element 123 with coordinates (0, 0), (a, 0), and (0, a), respectively, to be used for ‘plane stress’ problems.
(a) Form the [B] matrix.
(b) Obtain the stiffness matrix [K^e].
Why, in general, is a finite element solution not an exact solution?
(a) The element is shown in Fig. S.6.8. The displacement functions for a triangular element are given by Eqs (6.82). Thus
\left.\begin{array}{l}u(x, y)=\alpha_1+\alpha_2 x+\alpha_3 y \\v(x, y)=\alpha_4+\alpha_5 x+\alpha_6 y\end{array}\right\} (6.82)
\left.\begin{array}{l}u_1=\alpha_1, \quad v_1=\alpha_4 \\u_2=\alpha_1+a \alpha_2, \quad v_2=\alpha_4+a \alpha_5 \\u_3=\alpha_1+a \alpha_3, \quad v_3=\alpha_4+a \alpha_6\end{array}\right\} (i)
From Eq. (i)
\begin{array}{lll}\alpha_1=u_1 & \alpha_2=\left(u_2-u_1\right) / a & \alpha_3=\left(u_3-u_1\right) / a \\\alpha_4=v_1 &\alpha_5=\left(v_2-v_1\right) / a & \alpha_6=\left(v_3-v_1\right) / a\end{array}Hence in matrix form
\left\{\begin{array}{l}\alpha_1 \\\alpha_2 \\\alpha_3 \\\alpha_4 \\\alpha_5 \\\alpha_6\end{array}\right\}=\left[\begin{array}{cccccc}1 & 0 & 0 & 0 & 0 & 0 \\-1 / a & 0 & 1 / a & 0 & 0 & 0 \\-1 / a & 0 & 0 & 0 & 1 / a & 0 \\0 & 1 & 0 & 0 & 0 & 0 \\0 & -1 / a & 0 & 1 / a & 0 & 0 \\0 & -1 / a & 0 & 0 & 0 & 1 / a\end{array}\right]\left\{\begin{array}{l}u_1 \\v_1 \\u_2 \\v_2 \\u_3 \\v_3\end{array}\right\}which is of the form
\{x\}=\left[A^{-1}\right]\left\{\delta^e\right\}Also, from Eq. (6.89)
\{\varepsilon\}=\left[\begin{array}{llllll}0 & 1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 1 \\0 & 0 & 1 & 0 & 1 & 0\end{array}\right]\left\{\begin{array}{l}\alpha_1 \\\alpha_2 \\\alpha_3 \\\alpha_4 \\\alpha_5 \\\alpha_6\end{array}\right\} (6.89)
[C]=\left[\begin{array}{llllll}0 & 1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 1 \\0 & 0 & 1 & 0 & 1 & 0\end{array}\right]Hence
[B]=[C]\left[A^{-1}\right]=\left[\begin{array}{cccccc}-1 / a & 0 & 1 / a & 0 & 0 & 0 \\0 & -1 / a & 0 & 0 & 0 & 1 / a \\-1 / a & -1 / a & 0 & 1 / a & 1 / a & 0\end{array}\right](b) From Eq. (6.94)
\left[K^{\mathrm{e}}\right]=\left[[B]^{\mathrm{T}}[D][B] A t\right] (6.94)
\begin{aligned}{\left[K^e\right] } &=\left[\begin{array}{ccc}-1 / a & 0 & -1 / a \\0 & -1 / a & -1 / a \\1 / a & 0 & 0 \\0 & 0 & 1 / a \\0 & 0 & 1 / a \\0 & 1 / a & 0\end{array}\right] \frac{E}{1-v^2}\left[\begin{array}{ccc}1 & v & 0 \\v & 1 & 0 \\0 & 0 & \frac{1}{2}(1-v)\end{array}\right] \\& \times\left[\begin{array}{cccccc}-1 / a & 0 & 1 / a & 0 & 0 & 0 \\0 & -1 / a & 0 & 0 & 0 & 1 / a \\-1 / a & -1 / a & 0 & 1 / a & 1 / a & 0\end{array}\right] \frac{1}{2} a^2 t\end{aligned}which gives
\left[K^e\right]=\frac{E t}{4\left(1v^2\right)}\left[\begin{array}{cccccc}3-v & 1+v & -2 & -(1-v) & -(1-v) & -2 v \\1+v & 3-v & -2 v & -(1-v) & -(1-v) & -2 \\-2 & -2 v & 2 & 0 & 0 & 2 v \\-(1-v) & -(1-v) & 0 & 1-v & 1-v & 0 \\-(1-v) & -(1-v) & 0 & 1-v & 1-v & 0 \\-2 v & -2 & -2 v & 0 & 0 & 2\end{array}\right]Continuity of displacement is only ensured at nodes, not along their edges.
