--- title: "CST plate: deflection and stresses at centroid" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:cst-plane-stress-jestha" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-plane-stress-jestha site: BCT Engineering --- # CST plate: deflection and stresses at centroid Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2079 Jestha - 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/cst-plane-stress-jestha - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question A plate of thickness 8 mm is being loaded as shown in the figure. Determine the deflection at the point of load application. Also calculate the stresses at the centroid of the plate. Take $E = 69900$ MPa and $\nu = 0.3$. [Figure: triangular plate with node 1 at (35, 25) mm, node 2 at (0, 25) mm and node 3 at (0, 0) mm; thickness $t = 8$ mm; the left edge is fixed; a 90 N vertical load and a 45 N horizontal load act at node 1] ## Short answer Using a constant strain triangle, the stiffness matrix is assembled and solved for the free displacements. Deflection at load point: u1=2.05×10^-4 mm, v1=-1.17×10^-3 mm. Stresses at centroid: σx=0.450, σy=0.135, τxy=-0.900 N/mm^2. ## Answer A constant strain triangle (CST) is used with plane stress ($E=69900$ MPa, $\nu=0.3$, $t=8$ mm). Units: N, mm. Node 1 is the loaded node; the nodes 2 and 3 on the left edge are fixed. #### Step 1: Geometry - Node 1: $(35,\ 25)$ mm - Node 2: $(0,\ 25)$ mm - Node 3: $(0,\ 0)$ mm Area: $A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=437.5\ \text{mm}^2$ $b_1=y_2-y_3=25,\ b_2=y_3-y_1=-25,\ b_3=y_1-y_2=0$ mm $c_1=x_3-x_2=0,\ c_2=x_1-x_3=35,\ c_3=x_2-x_1=-35$ mm #### Step 2: [B] and [D] matrices (plane stress) $$ [B]=\frac{1}{2A}\begin{bmatrix}25 & 0 & -25 & 0 & 0 & 0 \\ 0 & 0 & 0 & 35 & 0 & -35 \\ 0 & 25 & 35 & -25 & -35 & 0\end{bmatrix},\quad 2A=875\ \text{mm}^2 $$ $$ [D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{4}\begin{bmatrix}7.681 & 2.304 & 0 \\ 2.304 & 7.681 & 0 \\ 0 & 0 & 2.688\end{bmatrix}\ \text{N/mm}^2 $$ #### Step 3: Element stiffness matrix $$ [k]=t\,A\,[B]^T[D][B]= 10^{5}\begin{bmatrix}2.195 & 0 & -2.195 & 0.922 & 0 & -0.922 \\ 0 & 0.768 & 1.075 & -0.768 & -1.075 & 0 \\ -2.195 & 1.075 & 3.700 & -1.997 & -1.506 & 0.922 \\ 0.922 & -0.768 & -1.997 & 5.070 & 1.075 & -4.302 \\ 0 & -1.075 & -1.506 & 1.075 & 1.506 & 0 \\ -0.922 & 0 & 0.922 & -4.302 & 0 & 4.302\end{bmatrix}\ \text{N/mm} $$ (DOF order: $u_1,v_1,u_2,v_2,u_3,v_3$.) #### Step 4: Load vector Loads act at node 1 only. Assumed directions: 45 N horizontal toward $+x$ and 90 N vertical downward ($-y$). Self-weight is neglected (not given). $$ \{F\}=\begin{Bmatrix}45 \\ -90 \\ 0 \\ 0 \\ 0 \\ 0\end{Bmatrix}\ \text{N} $$ #### Boundary conditions and solution The left edge is fixed, so nodes 2 and 3 have $u=v=0$. Only $u_1$ and $v_1$ are unknown, so use rows and columns 1 and 2 of $[k]$. $$ 10^{5}\begin{bmatrix}2.195 & 0 \\ 0 & 0.768\end{bmatrix}\begin{Bmatrix}u_1 \\ v_1\end{Bmatrix}=\begin{Bmatrix}45 \\ -90\end{Bmatrix} $$ $$ \begin{aligned} u_1&=2.05\times10^{-4}\ \text{mm} \\ v_1&=-0.001172\ \text{mm} \end{aligned} $$ #### Strains and stresses (constant over the element, so equal at the centroid) Centroid: $(11.67,\ 16.67)$ mm. $$ \{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}5.858\times10^{-6} \\ 0 \\ -3.348\times10^{-5}\end{Bmatrix} $$ $$ \{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}0.45 \\ 0.135 \\ -0.9\end{Bmatrix}\ \text{MPa} $$ **Answer:** deflection at the load point (node 1): $u_1=2.05\times10^{-4}$ mm (horizontal), $v_1=-0.00117$ mm (vertical, downward). Stresses at the centroid: $\sigma_x=0.450$, $\sigma_y=0.135$, $\tau_{xy}=-0.900$ N/mm$^2$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are the stresses at the centroid equal to the element stresses everywhere? A. The CST has constant strain, hence constant stress, over the element B. The centroid is the only point where stress is defined C. Stress varies linearly and the centroid is the average D. Because the load acts at the centroid Answer: A. [B] contains only constants, so strain and stress are constant over a CST. (Key point: Stress at centroid) 2. Why are only rows and columns 1 and 2 of [k] used? A. Nodes 2 and 3 on the fixed left edge have u = v = 0, so only u1, v1 are unknown B. The load acts at nodes 2 and 3 C. The stiffness matrix is singular otherwise D. Only node 1 has stiffness Answer: A. Fixed nodes have zero displacements, so their rows and columns drop out. (Key point: Reduction by boundary conditions) 3. What is the area of the triangle with nodes (35,25), (0,25), (0,0)? A. 875 mm^2 B. 437.5 mm^2 C. 35 mm^2 D. 1750 mm^2 Answer: B. A = 1/2 x 35 x 25 = 437.5 mm^2 (2A = 875). (Key point: Area of the element) 4. The plane stress [D] is E/(1-nu^2) times A. [1 nu 0; nu 1 0; 0 0 (1-nu)/2] B. [1 0 0; 0 1 0; 0 0 1] C. [1-nu nu 0; nu 1-nu 0; 0 0 1] D. [1 nu 0; nu 1 0; 0 0 1-nu^2] Answer: A. This is the standard plane-stress constitutive matrix with shear term (1-nu)/2. 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