1.4.4.5 Example: Finite-element calculations for a square plate subjected to uniaxial uniform tension

Procedures 5–7 described in Section 1.4.1 will be explained by taking an example of the finite element calculations for a square plate subjected to uniaxial uniform tension, as illustrated in Fig. 1.9. The square plate model has a side of unit length 1 and a thickness of unit length 1, and consists of two constant-strain triangular elements, i.e., the model has four nodes and thus eight degrees of freedom.

Let us determine the element stiffness matrix for Element 1. From Eqs (1.73), (1.69a)–(1.69c), the [B] matrix of Element 1 is calculated as

$\begin{array}{c}\left[\mathbf{B}\right]=\left[\begin{array}{cccccc}\hfill b_{1_1}\hfill & \hfill 0\hfill & \hfill b_{2_1}\hfill & \hfill 0\hfill & \hfill b_{3_1}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill c_{1_1}\hfill & \hfill 0\hfill & \hfill c_{2_1}\hfill & \hfill 0\hfill & \hfill c_{3_1}\hfill \\ \hfill c_{1_1}\hfill & \hfill b_{1_1}\hfill & \hfill c_{2_1}\hfill & \hfill b_{2_1}\hfill & \hfill c_{3_1}\hfill & \hfill b_{3_1}\hfill \end{array}\right]\\ =\frac{1}{2{\Delta }^{\left(1\right)}}\left[\begin{array}{cccccc}\hfill y_{2_1}-y_{3_1}\hfill & \hfill 0\hfill & \hfill y_{3_1}-y_{1_1}\hfill & \hfill 0\hfill & \hfill y_{1_1}-y_{2_1}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill x_{3_1}-x_{2_1}\hfill & \hfill 0\hfill & \hfill x_{1_1}-x_{3_1}\hfill & \hfill 0\hfill & \hfill x_{2_1}-x_{1_1}\hfill \\ \hfill x_{3_1}-x_{2_1}\hfill & \hfill y_{2_1}-y_{3_1}\hfill & \hfill x_{1_1}-x_{3_1}\hfill & \hfill y_{3_1}-y_{1_1}\hfill & \hfill x_{2_1}-x_{1_1}\hfill & \hfill y_{1_1}-y_{2_1}\hfill \end{array}\right]\end{array}$

Since the area of Element 1 Δ^(e) in the above equation can be easily obtained as 1/2 without using Eq. (1.69d),

$\left[\mathbf{B}\right]=\left[\begin{array}{cccccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

Hence, from Eq. (1.85), the element stiffness matrix for Element 1 [k^(e)] is calculated as

$\left[{\mathbf{k}}^{\left(1\right)}\right]=\frac{t}{2}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill {\nu }^{\prime }\hfill & \hfill 0\hfill \\ \hfill {\nu }^{\prime }\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \alpha \hfill \end{array}\right]\left[\begin{array}{cccccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

where α = (1 − ν′)/2. After multiplications of the matrices in Eq. (1.104), the element stiffness equation is obtained from Eq. (1.84) as

$\left\{\begin{array}{c}\hfill X_{1_1}^{\left(1\right)}\hfill \\ \hfill Y_{1_1}^{\left(1\right)}\hfill \\ \hfill X_{2_1}^{\left(1\right)}\hfill \\ \hfill Y_{2_1}^{\left(1\right)}\hfill \\ \hfill X_{3_1}^{\left(1\right)}\hfill \\ \hfill Y_{3_1}^{\left(1\right)}\hfill \end{array}\right\}=\frac{t}{2}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{cccccc}\hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -1\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -{\nu }^{\prime }\hfill \\ \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill \alpha \hfill & \hfill \alpha \hfill & \hfill 0\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill \alpha \hfill & \hfill \alpha \hfill & \hfill 0\hfill \\ \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill & \hfill {\nu }^{\prime }\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left\{\begin{array}{c}\hfill u_{1_1}\hfill \\ \hfill v_{1_1}\hfill \\ \hfill u_{2_1}\hfill \\ \hfill v_{2_1}\hfill \\ \hfill u_{3_1}\hfill \\ \hfill v_{3_1}\hfill \end{array}\right\}$

since the equivalent nodal forces due to initial strains ɛ₀ and body forces F_x and F_y, {F_ɛ0}⁽¹⁾ and {F_F}⁽¹⁾ are zero. The components of the nodal displacement and force vectors are written by the element nodal numbers. By rewriting these components by the global nodal numbers as shown in Fig. 1.9, Eq. (1.105) is rewritten as

$\left\{\begin{array}{c}\hfill X_1\hfill \\ \hfill Y_1\hfill \\ \hfill X_2\hfill \\ \hfill Y_2\hfill \\ \hfill X_3\hfill \\ \hfill Y_3\hfill \end{array}\right\}=\frac{t}{2}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{cccccc}\hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -1\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -{\nu }^{\prime }\hfill \\ \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill \alpha \hfill & \hfill \alpha \hfill & \hfill 0\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill \alpha \hfill & \hfill \alpha \hfill & \hfill 0\hfill \\ \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill & \hfill {\nu }^{\prime }\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left\{\begin{array}{c}\hfill u_1\hfill \\ \hfill v_1\hfill \\ \hfill u_2\hfill \\ \hfill v_2\hfill \\ \hfill u_3\hfill \\ \hfill v_3\hfill \end{array}\right\}$

In a similar way, the element stiffness equation for Element 2 is obtained as

$\left\{\begin{array}{c}\hfill X_{1_2}^{\left(2\right)}\hfill \\ \hfill Y_{1_2}^{\left(2\right)}\hfill \\ \hfill X_{2_2}^{\left(2\right)}\hfill \\ \hfill Y_{2_2}^{\left(2\right)}\hfill \\ \hfill X_{3_2}^{\left(2\right)}\hfill \\ \hfill Y_{3_2}^{\left(2\right)}\hfill \end{array}\right\}=\frac{t}{2}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{cccccc}\hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -1\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -{\nu }^{\prime }\hfill \\ \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill \alpha \hfill & \hfill \alpha \hfill & \hfill 0\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill \alpha \hfill & \hfill \alpha \hfill & \hfill 0\hfill \\ \hfill -{\nu }^{\text{'}}\hfill & \hfill -1\hfill & \hfill {\nu }^{\prime }\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left\{\begin{array}{c}\hfill u_{1_2}\hfill \\ \hfill v_{1_2}\hfill \\ \hfill u_{2_2}\hfill \\ \hfill v_{2_2}\hfill \\ \hfill u_{3_2}\hfill \\ \hfill v_{3_2}\hfill \end{array}\right\}$

and is rewritten by using the global nodal numbers as

$\left\{\begin{array}{c}\hfill X_4\hfill \\ \hfill Y_4\hfill \\ \hfill X_3\hfill \\ \hfill Y_3\hfill \\ \hfill X_2\hfill \\ \hfill Y_2\hfill \end{array}\right\}=\frac{t}{2}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{cccccc}\hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -1\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -{\nu }^{\prime }\hfill \\ \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill \alpha \hfill & \hfill \alpha \hfill & \hfill 0\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill \alpha \hfill & \hfill \alpha \hfill & \hfill 0\hfill \\ \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill & \hfill {\nu }^{\prime }\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left\{\begin{array}{c}\hfill u_4\hfill \\ \hfill v_4\hfill \\ \hfill u_3\hfill \\ \hfill v_3\hfill \\ \hfill u_2\hfill \\ \hfill v_2\hfill \end{array}\right\}$

By assembling the two element stiffness matrices, the following global stiffness equation for the square plate subjected to uniform tension is obtained (Procedure 4):

$\frac{t}{2}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{cccccccc}\hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -1\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 1+\alpha \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 0\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill \\ \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1+\alpha \hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill & \hfill -\alpha \hfill & \hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -\alpha \hfill & \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill \end{array}\right]\left\{\begin{array}{c}\hfill u_1\hfill \\ \hfill v_1\hfill \\ \hfill u_2\hfill \\ \hfill v_2\hfill \\ \hfill u_3\hfill \\ \hfill v_3\hfill \\ \hfill u_4\hfill \\ \hfill v_4\hfill \end{array}\right\}=\left\{\begin{array}{c}\hfill X_1\hfill \\ \hfill Y_1\hfill \\ \hfill X_2\hfill \\ \hfill Y_2\hfill \\ \hfill X_3\hfill \\ \hfill Y_3\hfill \\ \hfill X_4\hfill \\ \hfill Y_4\hfill \end{array}\right\}$

where the left-hand and right-hand sides of the equation are replaced with each other.

Let us now impose boundary conditions on the nodes. Namely, node 1 is clamped in both the x- and y-directions, node 2 is clamped only in the x-direction, and nodes 2 and 4 are subjected to equal nodal forces X₂ = X₄ = (p × 1 × 1)/2 = p/2, respectively, in the x-direction. A pair of the equal nodal forces p/2 applied to nodes 2 and 4 in the x-direction is the finite element model of a uniformly distributed tension force p per unit area exerted on the side $\overline{24}$ in the x-direction, as illustrated in Fig. 1.9. The geometrical and mechanical boundary conditions for the present case are

$u_1=v_1=v_2=u_3=0$

and

$X_2=X_4=\left(\mathit{pt}\right)/2=p/2,Y_3=Y_4=0$

respectively. Substitution of Eqs (1.110), (1.111) into Eq. (1.109) gives the global stiffness equation, i.e.,

$\frac{t}{2}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{cccccccc}\hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -1\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 1+\alpha \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 0\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill \\ \hfill -\alpha \hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill \\ \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1+\alpha \hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -\alpha \hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill & \hfill -\alpha \hfill & \hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -\alpha \hfill & \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 1+\alpha \hfill \end{array}\right]\left\{\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill u_2\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill v_3\hfill \\ \hfill u_4\hfill \\ \hfill v_4\hfill \end{array}\right\}=\left\{\begin{array}{c}\hfill X_1\hfill \\ \hfill Y_1\hfill \\ \hfill p/2\hfill \\ \hfill Y_2\hfill \\ \hfill X_3\hfill \\ \hfill 0\hfill \\ \hfill p/2\hfill \\ \hfill 0\hfill \end{array}\right\}$

Rearrangement of Eq. (1.112) by collecting unknown variables for forces and displacements in the left-hand side and known values of the forces and displacements in the right-hand side brings about the following simultaneous equations (Procedure 5):

$\begin{array}{l}\frac{t}{2}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{cccccccc}\hfill -\frac{2}{t}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -\frac{2}{t}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1+\alpha \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -\frac{2}{t}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -{\nu }^{\prime }\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -\frac{2}{t}\cdot \frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill -{\nu }^{\prime }\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {\nu }^{\prime }+\alpha \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1+\alpha \hfill & \hfill -\alpha \hfill & \hfill -\alpha \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -\alpha \hfill & \hfill 1+\alpha \hfill & \hfill {\nu }^{\prime }+\alpha \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -\alpha \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -\alpha \hfill & \hfill {\nu }^{\text{'}}+\alpha \hfill & \hfill 1+\alpha \hfill \end{array}\right]\hfill \\ \left\{\begin{array}{c}\hfill X_1\hfill \\ \hfill Y_1\hfill \\ \hfill u_2\hfill \\ \hfill Y_2\hfill \\ \hfill X_3\hfill \\ \hfill v_3\hfill \\ \hfill u_4\hfill \\ \hfill v_4\hfill \end{array}\right\}=\left\{\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill p/2\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill p/2\hfill \\ \hfill 0\hfill \end{array}\right\}\hfill \end{array}$

Eq. (1.113) can be solved numerically by, for instance, the Gauss elimination procedure. The solutions for Eq. (1.113) are

$u_2=u_4=p/E^{\prime },v_3=v_4=-{\nu }^{\prime }p/E^{\prime },X_1=X_3=-p/2,Y_1=Y_2=0\text{.}$

(Procedure 6).

The strains and stresses in the square palate can be calculated by substituting the solutions (1.114) into Eqs (1.73), (1.74), respectively (Procedure 7). The resultant strains and stresses are given by the following equations:

$\left\{\begin{array}{c}\hfill {\varepsilon }_x\hfill \\ \hfill {\varepsilon }_y\hfill \\ \hfill {\gamma }_{\mathit{xy}}\hfill \end{array}\right\}=\left[\mathbf{B}\right]\left\{\begin{array}{c}\hfill u_4\hfill \\ \hfill v_4\hfill \\ \hfill u_3\hfill \\ \hfill v_3\hfill \\ \hfill u_2\hfill \\ \hfill v_2\hfill \end{array}\right\}=\frac{p}{E^{\prime }}\left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\left\{\begin{array}{c}\hfill 1\hfill \\ \hfill -{\nu }^{\prime }\hfill \\ \hfill 0\hfill \\ \hfill {\nu }^{\prime }\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right\}=\frac{p}{E^{\prime }}\left\{\begin{array}{c}\hfill 1\hfill \\ \hfill -{\nu }^{\prime }\hfill \\ \hfill 1-{\nu }^{\prime }+{\nu }^{\prime }-1\hfill \end{array}\right\}=\frac{p}{E^{\prime }}\left\{\begin{array}{c}\hfill 1\hfill \\ \hfill -{\nu }^{\prime }\hfill \\ \hfill 0\hfill \end{array}\right\}$

and

$\begin{array}{l}\left\{\begin{array}{c}\hfill {\sigma }_x\hfill \\ \hfill {\sigma }_y\hfill \\ \hfill {\tau }_{\mathit{xy}}\hfill \end{array}\right\}=\frac{E^{\prime }}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill {\nu }^{\prime }\hfill & \hfill 0\hfill \\ \hfill {\nu }^{\prime }\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1-{\nu }^{\prime }}{2}\hfill \end{array}\right]\left\{\begin{array}{c}\hfill {\varepsilon }_x\hfill \\ \hfill {\varepsilon }_y\hfill \\ \hfill {\gamma }_{\mathit{xy}}\hfill \end{array}\right\}\\ =\frac{p}{1-{{\nu }^{\prime }}^2}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill {\nu }^{\prime }\hfill & \hfill 0\hfill \\ \hfill {\nu }^{\prime }\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1-{\nu }^{\prime }}{2}\hfill \end{array}\right]\left\{\begin{array}{c}\hfill 1\hfill \\ \hfill -{\nu }^{\prime }\hfill \\ \hfill 0\hfill \end{array}\right\}=\frac{p}{1-{{\nu }^{\prime }}^2}\left\{\begin{array}{c}\hfill 1-{{\nu }^{\prime }}^2\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right\}\\ =p\left\{\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right\}\end{array}$

The results obtained by the present finite element calculations imply that a square plate subjected to uniaxial uniform tension in the x-direction is elongated by a uniform strain of p/E′ in the loading direction, whereas it is contacted by a uniform strain of –ν′p/E′ in the direction perpendicular to the loading direction and that only a uniform normal stress of σ_x = p is induced in the plate. The result that the nodal reaction forces at nodes 1 and 3 are equal to –p/2, i.e. X₁ = X₃ = − p/2 implies that a uniform reaction force of –p is produced along the side $\overline{13}$. It is concluded that the above results obtained by the finite element method agree well with the physical interpretations of the present problem.