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Plate stiffness matrix with two elements

Short answer

The plate is discretized by two CST elements (each 900 mm²); their individual stiffness matrices are assembled into an 8×8 global matrix (N/mm). The full matrix is provided in the solution. With left edge fixed (u1=v1=u4=v4=0), the reduced system can be solved for displacements.

Find the stiffness matrix using two elements for the following plate loaded as shown in the figure. Take γ=78.5\gamma = 78.5 kN/m3^3.

[Figure: rectangular plate 30 mm high and 60 mm long, fixed along the left edge, E=70E = 70 GPa, ν=0.33\nu = 0.33, thickness =10= 10 mm; a 10 kN downward load acts at the free end]

Idealisation: plane stress CST. The 60 mm x 30 mm plate (thickness 10 mm, E=70 000E=70\,000 MPa, ν=0.33\nu=0.33) is divided into two triangles by the diagonal 1-3. Nodes: 1 (0,0)(0,0), 2 (60,0)(60,0), 3 (60,30)(60,30), 4 (0,30)(0,30) with the fixed edge 1-4. Element 1 = (1, 2, 3), element 2 = (1, 3, 4), both counter-clockwise. Units: N, mm.

[D]=E1−ν2[1ν0ν10001−ν2],[k]=t A [B]T[D][B],[B]=12A[b10b20b300c10c20c3c1b1c2b2c3b3][D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix},\qquad [k]=t\,A\,[B]^T[D][B],\quad [B]=\frac{1}{2A}\begin{bmatrix}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{bmatrix}

with b1=y2−y3b_1=y_2-y_3, b2=y3−y1b_2=y_3-y_1, b3=y1−y2b_3=y_1-y_2, c1=x3−x2c_1=x_3-x_2, c2=x1−x3c_2=x_1-x_3, c3=x2−x1c_3=x_2-x_1. Each element has A=900A=900 mm2^2.

Element stiffness matrices

Element 1 (nodes 1-2-3): A=900A=900 mm2^2, b=(−30,30,0)b=(-30,30,0), c=(0,−60,60)c=(0,-60,60)

[k](1)=105[1.9640−1.9641.2960−1.29600.6581.316−0.658−1.3160−1.9641.3164.595−2.612−2.6321.2961.296−0.658−2.6128.5131.316−7.8550−1.316−2.6321.3162.6320−1.29601.296−7.85507.855] N/mm[k]^{(1)}=10^{5}\begin{bmatrix}1.964 & 0 & -1.964 & 1.296 & 0 & -1.296 \\ 0 & 0.658 & 1.316 & -0.658 & -1.316 & 0 \\ -1.964 & 1.316 & 4.595 & -2.612 & -2.632 & 1.296 \\ 1.296 & -0.658 & -2.612 & 8.513 & 1.316 & -7.855 \\ 0 & -1.316 & -2.632 & 1.316 & 2.632 & 0 \\ -1.296 & 0 & 1.296 & -7.855 & 0 & 7.855\end{bmatrix}\ \text{N/mm}

Element 2 (nodes 1-3-4): A=900A=900 mm2^2, b=(0,30,−30)b=(0,30,-30), c=(−60,0,60)c=(-60,0,60)

[k](2)=105[2.63200−1.316−2.6321.31607.855−1.29601.296−7.8550−1.2961.9640−1.9641.296−1.316000.6581.316−0.658−2.6321.296−1.9641.3164.595−2.6121.316−7.8551.296−0.658−2.6128.513] N/mm[k]^{(2)}=10^{5}\begin{bmatrix}2.632 & 0 & 0 & -1.316 & -2.632 & 1.316 \\ 0 & 7.855 & -1.296 & 0 & 1.296 & -7.855 \\ 0 & -1.296 & 1.964 & 0 & -1.964 & 1.296 \\ -1.316 & 0 & 0 & 0.658 & 1.316 & -0.658 \\ -2.632 & 1.296 & -1.964 & 1.316 & 4.595 & -2.612 \\ 1.316 & -7.855 & 1.296 & -0.658 & -2.612 & 8.513\end{bmatrix}\ \text{N/mm}

DOF order for each element is the order of its nodes (u,vu,v for each node).

Assembly

Adding the element matrices to the global DOFs (u1,v1,u2,v2,u3,v3,u4,v4)(u_1,v_1,u_2,v_2,u_3,v_3,u_4,v_4) gives the 8×88\times8 global stiffness matrix:

[K]=105[4.5950−1.9641.2960−2.612−2.6321.31608.5131.316−0.658−2.61201.296−7.855−1.9641.3164.595−2.612−2.6321.296001.296−0.658−2.6128.5131.316−7.855000−2.612−2.6321.3164.5950−1.9641.296−2.61201.296−7.85508.5131.316−0.658−2.6321.29600−1.9641.3164.595−2.6121.316−7.855001.296−0.658−2.6128.513] N/mm[K]=10^{5}\begin{bmatrix}4.595 & 0 & -1.964 & 1.296 & 0 & -2.612 & -2.632 & 1.316 \\ 0 & 8.513 & 1.316 & -0.658 & -2.612 & 0 & 1.296 & -7.855 \\ -1.964 & 1.316 & 4.595 & -2.612 & -2.632 & 1.296 & 0 & 0 \\ 1.296 & -0.658 & -2.612 & 8.513 & 1.316 & -7.855 & 0 & 0 \\ 0 & -2.612 & -2.632 & 1.316 & 4.595 & 0 & -1.964 & 1.296 \\ -2.612 & 0 & 1.296 & -7.855 & 0 & 8.513 & 1.316 & -0.658 \\ -2.632 & 1.296 & 0 & 0 & -1.964 & 1.316 & 4.595 & -2.612 \\ 1.316 & -7.855 & 0 & 0 & 1.296 & -0.658 & -2.612 & 8.513\end{bmatrix}\ \text{N/mm}

Nodes 1 and 3 are shared by both elements, so their terms are sums of both elements.

Loads and displacements (for completeness)

Self-weight of each element is 78.5×10−6×10×900=0.70778.5\times10^{-6}\times10\times900=0.707 N, i.e. 0.236 N downward at each of its nodes. The 10 kN end load is shared equally by the free-end nodes 2 and 3 (5000 N each, downward). With u1=v1=u4=v4=0u_1=v_1=u_4=v_4=0, solving the reduced 4×44\times4 system gives

u2=−0.0267 mm,v2=−0.128 mmu3=0.0214 mm,v3=−0.12 mm\begin{aligned}u_2&=-0.0267\ \text{mm},& v_2&=-0.128\ \text{mm}\\ u_3&=0.0214\ \text{mm},& v_3&=-0.12\ \text{mm}\end{aligned}

Answer: the 8×88\times8 global stiffness matrix above (N/mm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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