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CST: stiffness, nodal displacement, stresses

Short answer

Plane stress CST with E=210 GPa gives nu=0. Using [k]=tA[B]^T[D][B] and the left edge fixed, free DOFs are u3=0 mm and v3=-0.01274 mm. Centroid strains are epsilon_x=0, epsilon_y=0, gamma_xy=-3.185e-5; stresses are sigma_x=0, sigma_y=0, tau_xy=-3.344 MPa.

A steel plate of uniform thickness 10 mm is being loaded as shown in the figure below. Considering the plane stress condition for this CST element, determine (a) element stiffness matrix, (b) nodal displacements, and (c) strains and stresses at the centroid of the element. Take E=210×103E = 210\times10^3 MPa and G=105×103G = 105\times10^3 MPa. The unit weight of steel is 78.5 kN/m3^3.

[Figure: right triangular plate; node 1 at the top-left, node 2 at the bottom of the left edge (300 mm below node 1, supports along the left edge), node 3 at the top-right 400 mm from node 1; a distributed load of 25 kN/m acts downward along the top edge 1-3]

Data: plane stress, one CST element, t=10t=10 mm, E=210×103E=210\times10^3 MPa. From G=E/[2(1+ν)]=105×103G=E/[2(1+\nu)]=105\times10^3 MPa, ν=E/(2G)−1=0\nu=E/(2G)-1=0. Nodes: 1 = top left (0,0)(0,0), 2 = bottom of left edge (0,−300)(0,-300), 3 = top right (400,0)(400,0). Units: N, mm.

Step 1: Geometry

  • Node 1 (top left): (0, 0)(0,\ 0) mm
  • Node 2 (bottom left): (0, −300)(0,\ -300) mm
  • Node 3 (top right): (400, 0)(400,\ 0) mm

Area: A=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=60000 mm2A=\tfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=60000\ \text{mm}^2

b1=y2−y3=−300, b2=y3−y1=0, b3=y1−y2=300b_1=y_2-y_3=-300,\ b_2=y_3-y_1=0,\ b_3=y_1-y_2=300 mm

c1=x3−x2=400, c2=x1−x3=−400, c3=x2−x1=0c_1=x_3-x_2=400,\ c_2=x_1-x_3=-400,\ c_3=x_2-x_1=0 mm

Step 2: [B] and [D] matrices (plane stress)

[B]=12A[−300000300004000−40000400−300−40000300],2A=120000 mm2[B]=\frac{1}{2A}\begin{bmatrix}-300 & 0 & 0 & 0 & 300 & 0 \\ 0 & 400 & 0 & -400 & 0 & 0 \\ 400 & -300 & -400 & 0 & 0 & 300\end{bmatrix},\quad 2A=120000\ \text{mm}^2 [D]=E1−ν2[1ν0ν10001−ν2]=105[2.1000002.1000001.050] N/mm2[D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=10^{5}\begin{bmatrix}2.100 & 0 & 0 \\ 0 & 2.100 & 0 \\ 0 & 0 & 1.050\end{bmatrix}\ \text{N/mm}^2

Step 3: Element stiffness matrix

[k]=t A [B]T[D][B]=106[1.488−0.525−0.7000−0.7870.525−0.5251.7940.525−1.4000−0.394−0.7000.5250.70000−0.5250−1.40001.40000−0.7870000.78700.525−0.394−0.525000.394] N/mm[k]=t\,A\,[B]^T[D][B]= 10^{6}\begin{bmatrix}1.488 & -0.525 & -0.700 & 0 & -0.787 & 0.525 \\ -0.525 & 1.794 & 0.525 & -1.400 & 0 & -0.394 \\ -0.700 & 0.525 & 0.700 & 0 & 0 & -0.525 \\ 0 & -1.400 & 0 & 1.400 & 0 & 0 \\ -0.787 & 0 & 0 & 0 & 0.787 & 0 \\ 0.525 & -0.394 & -0.525 & 0 & 0 & 0.394\end{bmatrix}\ \text{N/mm}

(DOF order: u1,v1,u2,v2,u3,v3u_1,v_1,u_2,v_2,u_3,v_3.)

Step 4: Load vector

Top edge load 2525 kN/m =25=25 N/mm over 400400 mm =10 000=10\,000 N, shared: 50005000 N at node 1 (absorbed by the support) and 50005000 N at node 3. Self-weight W=78.5×10−6×10×60000=47.1W=78.5\times10^{-6}\times10\times60000=47.1 N, 15.7015.70 N downward at each node. Load on the free node 3: Fy3=−(5000+15.70)=−5015.70F_{y3}=-(5000+15.70)=-5015.70 N.

{F}={00000−5016} N\{F\}=\begin{Bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -5016\end{Bmatrix}\ \text{N}

Boundary conditions and solution

Nodes 1 and 2 lie on the supported left edge (u=v=0u=v=0). Free DOFs: u3,v3u_3,v_3 (rows/columns 5 and 6).

105[7.875003.938]{u3v3}={0−5016}10^{5}\begin{bmatrix}7.875 & 0 \\ 0 & 3.938\end{bmatrix}\begin{Bmatrix}u_3 \\ v_3\end{Bmatrix}=\begin{Bmatrix}0 \\ -5016\end{Bmatrix} u3=0 mmv3=−0.01274 mm\begin{aligned} u_3&=0\ \text{mm} \\ v_3&=-0.01274\ \text{mm} \end{aligned}

Strains and stresses (constant over the element, so equal at the centroid)

Centroid: (133.33, −100.00)(133.33,\ -100.00) mm.

{ε}=[B]{d}={00−3.185×10−5}\{\varepsilon\}=[B]\{d\}=\begin{Bmatrix}0 \\ 0 \\ -3.185\times10^{-5}\end{Bmatrix} {σ}=[D]{ε}={00−3.344} MPa\{\sigma\}=[D]\{\varepsilon\}=\begin{Bmatrix}0 \\ 0 \\ -3.344\end{Bmatrix}\ \text{MPa}

Answer: (a) element stiffness matrix as given in Step 3 (N/mm); (b) u3=0u_3=0 mm, v3=−0.0127v_3=-0.0127 mm (downward), others zero; (c) at the centroid εx=0\varepsilon_x=0, εy=0\varepsilon_y=0, γxy=−3.18×10−5\gamma_{xy}=-3.18\times10^{-5} and σx=0.000\sigma_x=0.000, σy=0.000\sigma_y=0.000, τxy=−3.344\tau_{xy}=-3.344 N/mm2^2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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