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?

Question

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 [latex][K^e][/latex].
Why, in general, is a finite element solution not an exact solution?

Accepted Answer

(a) The element is shown in Fig. S.6.8. The displacement functions for a triangular element are given by Eqs (6.82). Thus

[latex]\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} ight\}[/latex] (6.82)

[latex]\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} ight\}[/latex] (i)

From Eq. (i)

[latex]\begin{array}{lll}\alpha_1=u_1 & \alpha_2=\left(u_2-u_1 ight) / a & \alpha_3=\left(u_3-u_1 ight) / a \\alpha_4=v_1 &\alpha_5=\left(v_2-v_1 ight) / a & \alpha_6=\left(v_3-v_1 ight) / a\end{array}[/latex]

Hence in matrix form

[latex]\left\{\begin{array}{l}\alpha_1 \\alpha_2 \\alpha_3 \\alpha_4 \\alpha_5 \\alpha_6\end{array} ight\}=\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} ight]\left\{\begin{array}{l}u_1 \v_1 \u_2 \v_2 \u_3 \v_3\end{array} ight\}[/latex]

which is of the form

[latex]\{x\}=\left[A^{-1} ight]\left\{\delta^e ight\}[/latex]

Also, from Eq. (6.89)

[latex]\{\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} ight]\left\{\begin{array}{l}\alpha_1 \\alpha_2 \\alpha_3 \\alpha_4 \\alpha_5 \\alpha_6\end{array} ight\}[/latex] (6.89)

[latex][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} ight][/latex]

Hence

[latex][B]=[C]\left[A^{-1} ight]=\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} ight][/latex]

(b) From Eq. (6.94)

[latex]\left[K^{\mathrm{e}} ight]=\left[[B]^{\mathrm{T}}[D][B] A t ight][/latex] (6.94)

[latex]\begin{aligned}{\left[K^e ight] } &=\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} ight] \frac{E}{1-v^2}\left[\begin{array}{ccc}1 & v & 0 \v & 1 & 0 \0 & 0 & \frac{1}{2}(1-v)\end{array} ight] \& imes\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} ight] \frac{1}{2} a^2 t\end{aligned}[/latex]

which gives

[latex]\left[K^e ight]=\frac{E t}{4\left(1v^2 ight)}\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} ight][/latex]

Continuity of displacement is only ensured at nodes, not along their edges.

Question p.6.8: It is required to formulate the stiffness of a triangular el...

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