Question p.6.9: It is required to form the stiffness matrix of a triangular ...
It is required to form the stiffness matrix of a triangular element 123 for use in stress analysis problems. The coordinates of the element are (1, 1), (2, 1), and (2, 2), respectively.
(a) Assume a suitable displacement field explaining the reasons for your choice.
(b) Form the [B] matrix.
(c) Form the matrix which gives, when multiplied by the element nodal displacements, the stresses in the element. Assume a general [D] matrix.
(a) There are six degrees of freedom so that the displacement field must include six coefficients. Thus
\begin{aligned}&u=\alpha_1+\alpha_7 x+\alpha_3 y (i) \\&v=\alpha_4+\alpha_5 x+\alpha_6 y (ii)\end{aligned}(b) From Eqs (i) and (ii) and referring to Fig. S.6.9
\begin{aligned}&u_1=\alpha_1+\alpha_2+\alpha_3 \quad v_1=\alpha_4+\alpha_5+\alpha_6 \\&u_2=\alpha_1+2 \alpha_2+\alpha_3 \quad v_2=\alpha_4+2 \alpha_5+\alpha_6 \\&u_3=\alpha_1+2 \alpha_2+2 \alpha_3 \quad v_3=\alpha_4+2 \alpha_5+2 \alpha_6\end{aligned}Thus
\begin{array}{lll}\alpha_2=u_2-u_1 & \alpha_3=u_3-u_2 & \alpha_1=2 u_1-u_3 \\\alpha_5=v_2-v_1 & \alpha_6=v_3-v_2 & \alpha_4=2 v_1-v_3\end{array}Therefore
\left\{\begin{array}{l}\alpha_1 \\\alpha_2 \\\alpha_3 \\\alpha_4 \\\alpha_5 \\\alpha_6\end{array}\right\}=\left[\begin{array}{cccccc}2 & 0 & 0 & 0 & -1 & 0 \\-1 & 0 & 1 & 0 & 0 & 0 \\0 & 0 & -1 & 0 & 1 & 0 \\0 & 2 & 0 & 0 & 0 & -1 \\0 & -1 & 0 & 1 & 0 & 0 \\0 & 2 & 0 & 0 & 0 & -1\end{array}\right]\left\{\begin{array}{l}u_1 \\v_1 \\u_2 \\v_2 \\u_3 \\v_3\end{array}\right\} (iii)
which is of the form
\{\alpha\}=\left[A^{-1}\right]\left\{\delta^e\right\}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 & 0 & 1 & 0 & 0 & 0 \\0 & 2 & 0 & 0 & 0 & -1 \\0 & -1 & -1 & 1 & 1 & 0\end{array}\right](c) From Eq. (6.69)
\{\sigma\}=[D][C]\left[A^{-1}\right]\left\{\delta^{\mathrm{e}}\right\} (6.69)
\{\sigma\}=[D][B]\left\{\delta^e\right\}Thus, for plane stress problems (see Eq. (6.92))
\{\sigma\}=\left\{\begin{array}{l}\sigma_x \\\sigma_y \\\tau_{x y}\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]\left\{\begin{array}{l}\varepsilon_x \\\varepsilon_y \\\gamma_{x y}\end{array}\right\} (6.92)
[D][B]=\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]\left[\begin{array}{cccccc}-1 & 0 & 1 & 0 & 0 & 0 \\0 & 2 & 0 & 0 & 0 & -1 \\0 & -1 & -1 & 1 & 1 & 0\end{array}\right]i.e.
[D][B]=\frac{E}{1-v^2}\left[\begin{array}{cccccc}-1 & 2 v & 1 & 0 & 0 & -v \\-v & 2 & v & 0 & 0 & -1 \\0 & -\frac{1}{2}(1-v) & -\frac{1}{2}(1-v) & \frac{1}{2}(1-v) & \frac{1}{2}(1-v) & 0\end{array}\right]For plain strain problems (see Eq. (6.93))
\{\sigma\}=\left\{\begin{array}{c}\sigma_x \\\sigma_y \\\tau_{x y}\end{array}\right\}=\frac{E(1-v)}{(1+v)(1-2 v)}\left[\begin{array}{ccc}1 & \frac{v}{1-v} & 0 \\\frac{v}{1-v} & 1 & 0 \\0 & 0 & \frac{(1-2 v)}{2(1-v)}\end{array}\right]\left\{\begin{array}{l}\varepsilon_x \\\varepsilon_y \\\gamma_{x y}\end{array}\right\} (6.93)
\begin{aligned}&[D][B]=\frac{F(1-v)}{(1+v)(1-2 v)}\left[\begin{array}{ccc}1 & \frac{v}{(1-v)} & 0 \\\frac{v}{(1-v)} & 1 & 0 \\0 & 0 & \frac{(1-2 v)}{2(1-v)}\end{array}\right]\\&\times\left[\begin{array}{cccccc}-1 & 0 & 1 & 0 & 0 & 0 \\0 & 2 & 0 & 0 & 0 & -1 \\0 & -1 & -1 & 1 & 1 & 0\end{array}\right]\end{aligned}\begin{aligned}{[D][B]=} & \frac{E(1-v)}{(1+v)(1-2 v)} \\& \times\left[\begin{array}{cccccc}-1 & \frac{2 v}{1-v} & 1 & 0 & 0 & -\frac{v}{1-v} \\-\frac{v}{1-v} & 2 & \frac{v}{1-v} & 0 & 0 & -1 \\0 & -\frac{1-2 v}{2(1-v)} & -\frac{1-2 v}{2(1-v)} & \frac{1-2 v}{2(1-v)} &\frac{1-2 v}{2(1-v)} & 0\end{array}\right]\end{aligned}
