Constant Strain Triangle (CST) — Solved Questions

Figure Q1 — Equilateral CST (100 mm side), pin-supported at nodes 1 & 2, with 10 kN/m² traction on edge 2–3

Step 1 — Node Coordinates & Given Data

Assign coordinates (convert mm to metres for consistent units):

Step 2 — Area of the Triangular Element

Using the determinant formula:

Step 3 — Coefficients $b_i$ and $c_i$

These are computed from cyclic differences of node coordinates:

Step 4 — Strain-Displacement Matrix $[B]$

Using the standard CST formula:

Substituting the values:

Step 5 — Constitutive (Material) Matrix $[D]$

For plane stress:

Step 6 — Element Stiffness Matrix $[K]$

Using the formula $[K] = t \cdot A \cdot [B]^T [D] [B]$:

(Units: force/m — stiffness values in the consistent unit system)

Step 7 — Load Vectors

(a) Body force (self-weight in $-y$ direction):

(b) Traction on edge 2–3: The edge length is $l_{23} = 0.1$ m. The outward normal to edge 2–3 has direction $(0.87, 0.5)$, so the traction components are:

(c) Combined load vector:

Step 8 — Apply Boundary Conditions

Nodes 1 and 2 are restrained (pinned): $u_1 = v_1 = u_2 = v_2 = 0$.

Free DOFs: $u_3$ (DOF 5) and $v_3$ (DOF 6). Extract the $2 \times 2$ sub-system.

Step 9 — Solve for Unknown Displacements

Extract the reduced $2 \times 2$ system:

Solving $[K]_{ff}\{u\}_f = \{F\}_f$:

Step 10 — Compute Element Strains and Stresses

Strain (constant throughout the element):

Stress (constant throughout the element — and equals centroid stress for CST):

Figure Q5 — Right-triangle CST (free element) with 5 kN/m² traction on edge 1–2 (left vertical edge)

Step 1 — Node Coordinates & Given Data

Assign coordinates (convert mm to metres for consistent units):

Step 2 — Area of the Triangular Element

Using the determinant formula:

Step 3 — Coefficients $b_i$ and $c_i$

These are computed from cyclic differences of node coordinates:

Step 4 — Strain-Displacement Matrix $[B]$

Using the standard CST formula:

Substituting the values:

Step 5 — Constitutive (Material) Matrix $[D]$

For plane stress:

Step 6 — Element Stiffness Matrix $[K]$

Using the formula $[K] = t \cdot A \cdot [B]^T [D] [B]$:

(Units: force/m — stiffness values in the consistent unit system)

Step 7 — Load Vector

Traction of $5$ kN/m² acts in the $+x$ direction on edge 1–2 (left vertical edge). Edge length $l_{12} = 0.5$ m.

Step 8 — Apply Boundary Conditions

No support conditions are given — this is a free (unsupported) element.

Step 9 — Solve for Unknown Displacements

Since no boundary conditions are imposed, the stiffness matrix is singular (rank 3). A unique displacement solution cannot be obtained from the free element alone.

Step 10 — Compute Element Strains and Stresses

Without support conditions, strains and stresses cannot be computed for the free element.

Figure Q6 — Right-triangle CST with nodes 1 & 2 pin-supported, horizontal traction 5 kN/m² on edge 1–3

Step 1 — Node Coordinates & Given Data

Assign coordinates (convert mm to metres for consistent units):

Step 2 — Area of the Triangular Element

Using the determinant formula:

Step 3 — Coefficients $b_i$ and $c_i$

These are computed from cyclic differences of node coordinates:

Step 4 — Strain-Displacement Matrix $[B]$

Using the standard CST formula:

Substituting the values:

Step 5 — Constitutive (Material) Matrix $[D]$

For plane stress:

Step 6 — Element Stiffness Matrix $[K]$

Using the formula $[K] = t \cdot A \cdot [B]^T [D] [B]$:

(Units: force/m — stiffness values in the consistent unit system)

Step 7 — Load Vectors

Poisson's ratio from given data: $G = E/[2(1+\nu)]$, so $\nu = E/(2G) - 1 = 200000/(2 \times 105000) - 1 = -0.05$.

(a) Body force:

(b) Traction on edge 1–3 ($l_{13} = 0.6$ m, acting in $+x$):

Step 8 — Apply Boundary Conditions

Nodes 1 and 2 are restrained: $u_1 = v_1 = u_2 = v_2 = 0$. Free DOFs: $u_3, v_3$.

Step 9 — Solve for Unknown Displacements

Extract the reduced $2 \times 2$ system:

Solving $[K]_{ff}\{u\}_f = \{F\}_f$:

Step 10 — Compute Element Strains and Stresses

Strain (constant throughout the element):

Stress (constant throughout the element — and equals centroid stress for CST):

Figure Q7 — Triangular plate with body force and 15 kN/m² horizontal traction on edge 2–3

Step 1 — Node Coordinates & Given Data

Assign coordinates (convert mm to metres for consistent units):

Step 2 — Area of the Triangular Element

Using the determinant formula:

Step 3 — Coefficients $b_i$ and $c_i$

These are computed from cyclic differences of node coordinates:

Step 4 — Strain-Displacement Matrix $[B]$

Using the standard CST formula:

Substituting the values:

Step 5 — Constitutive (Material) Matrix $[D]$

For plane stress:

Step 6 — Element Stiffness Matrix $[K]$

Using the formula $[K] = t \cdot A \cdot [B]^T [D] [B]$:

(Units: force/m — stiffness values in the consistent unit system)

Step 7 — Load Vectors

(a) Body force ($\gamma = 80$ kN/m³):

(b) Traction on edge 2–3 (right vertical edge, $l_{23} = 0.5$ m). The traction acts in the $-x$ direction with magnitude $15$ MPa $= 15000$ kN/m².

(c) Combined:

Step 8 — Apply Boundary Conditions

Nodes 1 and 2 are restrained: $u_1 = v_1 = u_2 = v_2 = 0$. Free DOFs: $u_3, v_3$.

Step 9 — Solve for Unknown Displacements

Extract the reduced $2 \times 2$ system:

Solving $[K]_{ff}\{u\}_f = \{F\}_f$:

Step 10 — Compute Element Strains and Stresses

Strain (constant throughout the element):

Stress (constant throughout the element — and equals centroid stress for CST):

Figure Q10 — Fixed left-edge triangular plate with 5 kN/m² downward traction on edge 1–2

Step 1 — Node Coordinates & Given Data

Assign coordinates (convert mm to metres for consistent units):

Step 2 — Area of the Triangular Element

Using the determinant formula:

Step 3 — Coefficients $b_i$ and $c_i$

These are computed from cyclic differences of node coordinates:

Step 4 — Strain-Displacement Matrix $[B]$

Using the standard CST formula:

Substituting the values:

Step 5 — Constitutive (Material) Matrix $[D]$

For plane stress:

Step 6 — Element Stiffness Matrix $[K]$

Using the formula $[K] = t \cdot A \cdot [B]^T [D] [B]$:

(Units: force/m — stiffness values in the consistent unit system)

Step 7 — Load Vectors

(a) Body force:

(b) Traction on edge 1–2 (bottom edge, $l_{12} = 0.8$ m). Uniform pressure $5$ MPa $= 5000$ kN/m² acting downward ($-y$).

(c) Combined:

Step 8 — Apply Boundary Conditions

Left edge is fixed (wall), meaning nodes 1 and 3 (both at $x = 0$) are fully restrained: $u_1 = v_1 = u_3 = v_3 = 0$.

Free DOFs: $u_2, v_2$ (node 2 only).

Step 9 — Solve for Unknown Displacements

Extract the reduced $2 \times 2$ system:

Solving $[K]_{ff}\{u\}_f = \{F\}_f$:

Step 10 — Compute Element Strains and Stresses

Strain (constant throughout the element):

Stress (constant throughout the element — and equals centroid stress for CST):

Figure Q11 — Right-triangle CST with 25 kN/m² horizontal traction on edge 2–3 and self-weight

Step 1 — Node Coordinates & Given Data

Assign coordinates (convert mm to metres for consistent units):

Step 2 — Area of the Triangular Element

Using the determinant formula:

Step 3 — Coefficients $b_i$ and $c_i$

These are computed from cyclic differences of node coordinates:

Step 4 — Strain-Displacement Matrix $[B]$

Using the standard CST formula:

Substituting the values:

Step 5 — Constitutive (Material) Matrix $[D]$

For plane stress:

Step 6 — Element Stiffness Matrix $[K]$

Using the formula $[K] = t \cdot A \cdot [B]^T [D] [B]$:

(Units: force/m — stiffness values in the consistent unit system)

Step 7 — Load Vectors

(a) Body force:

(b) Traction on edge 2–3 (right vertical edge, $l_{23} = 0.4$ m). Uniform $25$ MPa acting in the $-x$ direction:

(c) Combined:

Step 8 — Apply Boundary Conditions

Nodes 1 and 2 are restrained (pinned): $u_1 = v_1 = u_2 = v_2 = 0$. Free DOFs: $u_3, v_3$.

Step 9 — Solve for Unknown Displacements

Extract the reduced $2 \times 2$ system:

Solving $[K]_{ff}\{u\}_f = \{F\}_f$:

Step 10 — Compute Element Strains and Stresses

Strain (constant throughout the element):

Stress (constant throughout the element — and equals centroid stress for CST):

Figure Q15 — Thin triangular plate fixed at nodes 1 & 2, with point loads at node 3

Step 1 — Node Coordinates & Given Data

Assign coordinates (in inches):

Step 2 — Area of the Triangular Element

Using the determinant formula:

Step 3 — Coefficients $b_i$ and $c_i$

These are computed from cyclic differences of node coordinates:

Step 4 — Strain-Displacement Matrix $[B]$

Using the standard CST formula:

Substituting the values:

Step 5 — Constitutive (Material) Matrix $[D]$

For plane stress:

Step 6 — Element Stiffness Matrix $[K]$

Using the formula $[K] = t \cdot A \cdot [B]^T [D] [B]$:

(Units: lb/in — stiffness values in the consistent unit system)

Step 7 — Load Vector

Concentrated point loads applied directly at node 3:

(No body force or traction — just point loads at node 3.)

Step 8 — Apply Boundary Conditions

Nodes 1 and 2 are fixed (left edge): $u_1 = v_1 = u_2 = v_2 = 0$. Free DOFs: $u_3, v_3$.

Step 9 — Solve for Unknown Displacements

Extract the reduced $2 \times 2$ system:

Solving $[K]_{ff}\{u\}_f = \{F\}_f$:

Step 10 — Compute Element Strains and Stresses

Strain (constant throughout the element):

Stress (constant throughout the element — and equals centroid stress for CST):

Figure Q16 — Smaller fixed-edge triangular plate with 5 kN/m² downward traction on edge 1–2

Step 1 — Node Coordinates & Given Data

Assign coordinates (convert mm to metres for consistent units):

Step 2 — Area of the Triangular Element

Using the determinant formula:

Step 3 — Coefficients $b_i$ and $c_i$

These are computed from cyclic differences of node coordinates:

Step 4 — Strain-Displacement Matrix $[B]$

Using the standard CST formula:

Substituting the values:

Step 5 — Constitutive (Material) Matrix $[D]$

For plane stress:

Step 6 — Element Stiffness Matrix $[K]$

Using the formula $[K] = t \cdot A \cdot [B]^T [D] [B]$:

(Units: force/m — stiffness values in the consistent unit system)

Step 7 — Load Vectors

(a) Body force:

(b) Traction on edge 1–2 (bottom edge, $l_{12} = 0.4$ m). Uniform $5$ MPa downward:

(c) Combined:

Step 8 — Apply Boundary Conditions

Left edge fixed: $u_1 = v_1 = u_3 = v_3 = 0$. Free DOFs: $u_2, v_2$.

Step 9 — Solve for Unknown Displacements

Extract the reduced $2 \times 2$ system:

Solving $[K]_{ff}\{u\}_f = \{F\}_f$:

Step 10 — Compute Element Strains and Stresses

Strain (constant throughout the element):

Stress (constant throughout the element — and equals centroid stress for CST):

Figure Q17 — Symbolic CST setup with usual notation, traction T on edge 2–3

Step 1 — Node Coordinates & Given Data

Let the three nodes have coordinates:

Given: $E = 30 \times 10^5$ psi, $\nu = 0.3$, $t = 0.3$ in, $\gamma = 460$ lb/in³, traction $T = 360$ psi on edge 2–3.

Step 2 — Area of the Triangular Element

Step 3 — Coefficients $b_i$ and $c_i$

Step 4 — Strain-Displacement Matrix $[B]$

Step 5 — Constitutive Matrix $[D]$ (Plane Stress)

Step 6 — Element Stiffness Matrix $[K]$

Substitute $t = 0.3$ in, $A$ from Step 2, $[B]$ from Step 4, and $[D]$ from Step 5 to get the $6 \times 6$ stiffness matrix.

Step 7 — Load Vectors

(a) Body force (self-weight in $-y$):

(b) Traction on edge 2–3 (length $l_{23}$). Resolve $T = 360$ psi into components $t_x, t_y$ perpendicular to edge 2–3:

(c) Total: $\{F\} = \{f_b\} + \{f_t\}_{23}$.

Step 8 — Apply Boundary Conditions

If nodes 1 and 2 are restrained: $u_1 = v_1 = u_2 = v_2 = 0$.

Partition the system and extract the $2 \times 2$ reduced system for the free DOFs $(u_3, v_3)$.

Step 9 — Solve for Displacements

Step 10 — Compute Strains and Stresses