--- title: "Plate stiffness matrix with two elements" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:stiffness-two-element-plate" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/stiffness-two-element-plate site: BCT Engineering --- # Plate stiffness matrix with two elements Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2077 Chaitra, 8 marks - Video: [Finite element modelling of 2D two-dimensional stress analysis with constant strain triangles (CST)](https://www.youtube.com/watch?v=lOJ9rNW_6mo) — Mahesh Gadwantikar - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/stiffness-two-element-plate - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Find the stiffness matrix using two elements for the following plate loaded as shown in the figure. Take $\gamma = 78.5$ kN/m$^3$. [Figure: rectangular plate 30 mm high and 60 mm long, fixed along the left edge, $E = 70$ GPa, $\nu = 0.33$, thickness $= 10$ mm; a 10 kN downward load acts at the free end] ## 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. ## Answer **Idealisation:** plane stress CST. The 60 mm x 30 mm plate (thickness 10 mm, $E=70\,000$ MPa, $\nu=0.33$) is divided into two triangles by the diagonal 1-3. Nodes: 1 $(0,0)$, 2 $(60,0)$, 3 $(60,30)$, 4 $(0,30)$ with the fixed edge 1-4. Element 1 = (1, 2, 3), element 2 = (1, 3, 4), both counter-clockwise. Units: N, mm. $$ [D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix},\qquad [k]=t\,A\,[B]^T[D][B],\quad [B]=\frac{1}{2A}\begin{bmatrix}b_1&0&b_2&0&b_3&0\\0&c_1&0&c_2&0&c_3\\c_1&b_1&c_2&b_2&c_3&b_3\end{bmatrix} $$ with $b_1=y_2-y_3$, $b_2=y_3-y_1$, $b_3=y_1-y_2$, $c_1=x_3-x_2$, $c_2=x_1-x_3$, $c_3=x_2-x_1$. Each element has $A=900$ mm$^2$. #### Element stiffness matrices **Element 1** (nodes 1-2-3): $A=900$ mm$^2$, $b=(-30,30,0)$, $c=(0,-60,60)$ $$ [k]^{(1)}=10^{5}\begin{bmatrix}1.964 & 0 & -1.964 & 1.296 & 0 & -1.296 \\ 0 & 0.658 & 1.316 & -0.658 & -1.316 & 0 \\ -1.964 & 1.316 & 4.595 & -2.612 & -2.632 & 1.296 \\ 1.296 & -0.658 & -2.612 & 8.513 & 1.316 & -7.855 \\ 0 & -1.316 & -2.632 & 1.316 & 2.632 & 0 \\ -1.296 & 0 & 1.296 & -7.855 & 0 & 7.855\end{bmatrix}\ \text{N/mm} $$ **Element 2** (nodes 1-3-4): $A=900$ mm$^2$, $b=(0,30,-30)$, $c=(-60,0,60)$ $$ [k]^{(2)}=10^{5}\begin{bmatrix}2.632 & 0 & 0 & -1.316 & -2.632 & 1.316 \\ 0 & 7.855 & -1.296 & 0 & 1.296 & -7.855 \\ 0 & -1.296 & 1.964 & 0 & -1.964 & 1.296 \\ -1.316 & 0 & 0 & 0.658 & 1.316 & -0.658 \\ -2.632 & 1.296 & -1.964 & 1.316 & 4.595 & -2.612 \\ 1.316 & -7.855 & 1.296 & -0.658 & -2.612 & 8.513\end{bmatrix}\ \text{N/mm} $$ DOF order for each element is the order of its nodes ($u,v$ for each node). #### Assembly Adding the element matrices to the global DOFs $(u_1,v_1,u_2,v_2,u_3,v_3,u_4,v_4)$ gives the $8\times8$ global stiffness matrix: $$ [K]=10^{5}\begin{bmatrix}4.595 & 0 & -1.964 & 1.296 & 0 & -2.612 & -2.632 & 1.316 \\ 0 & 8.513 & 1.316 & -0.658 & -2.612 & 0 & 1.296 & -7.855 \\ -1.964 & 1.316 & 4.595 & -2.612 & -2.632 & 1.296 & 0 & 0 \\ 1.296 & -0.658 & -2.612 & 8.513 & 1.316 & -7.855 & 0 & 0 \\ 0 & -2.612 & -2.632 & 1.316 & 4.595 & 0 & -1.964 & 1.296 \\ -2.612 & 0 & 1.296 & -7.855 & 0 & 8.513 & 1.316 & -0.658 \\ -2.632 & 1.296 & 0 & 0 & -1.964 & 1.316 & 4.595 & -2.612 \\ 1.316 & -7.855 & 0 & 0 & 1.296 & -0.658 & -2.612 & 8.513\end{bmatrix}\ \text{N/mm} $$ 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 $78.5\times10^{-6}\times10\times900=0.707$ 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 $u_1=v_1=u_4=v_4=0$, solving the reduced $4\times4$ system gives $$ \begin{aligned}u_2&=-0.0267\ \text{mm},& v_2&=-0.128\ \text{mm}\\ u_3&=0.0214\ \text{mm},& v_3&=-0.12\ \text{mm}\end{aligned} $$ **Answer:** the $8\times8$ global stiffness matrix above (N/mm). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 60 mm x 30 mm plate divided into two CST elements? A. Two rectangles of 30 x 30 mm B. Two triangles formed by the diagonal 1-3 C. Two triangles formed by a vertical line at mid-length D. One triangle and one rectangle Answer: B. The diagonal 1-3 splits the rectangle into elements (1,2,3) and (1,3,4), both numbered counter-clockwise. (Key point: Idealisation) 2. Which expression gives the stiffness matrix of a CST element in plane stress? A. [k] = A [B][D][B]^T B. [k] = t A [B]^T [D] [B] C. [k] = (AE/L)[1 -1; -1 1] D. [k] = t A [D]^T [B] [D] Answer: B. Because B and D are constant, the integral of B^T D B over the volume reduces to t A B^T D B. (Key point: Element stiffness formula) 3. Why are the entries for nodes 1 and 3 in the global matrix sums of two element terms? A. They have the largest displacements B. They are the supported nodes C. Both elements contain nodes 1 and 3, so both contribute stiffness to those DOFs D. They carry the 10 kN load Answer: C. In direct assembly the stiffness terms of every element that shares a DOF are added. (Key point: Assembly of shared nodes) 4. Why is the global stiffness matrix 8 x 8? A. There are 2 elements with 4 terms each B. There are 4 nodes with 2 DOFs (u, v) each C. There are 3 strains per element and 2 elements, plus 2 extra D. There are 8 elements Answer: B. 4 nodes x 2 displacement DOFs per node = 8 DOFs, so K is 8 x 8. 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