Question 5 of 7: Stresses in a constant-strain triangular plane-strain element

Question 5: Stresses in a constant-strain triangular plane-strain element (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single three-node triangular element in plane strain with all six nodal displacements prescribed.

Find. The three in-plane stress components, the two in-plane principal stresses with the principal angle, and a discussion of the constant-strain triangle in bending.

Approach. Compute twice the element area from the nodal coordinates, build the constant strain-displacement matrix $[\mathbf{B}]$, multiply by the nodal displacement vector to get the strains, apply the plane-strain constitutive matrix, and finish with the standard principal-stress transformation.

1. Compute twice the element area. $$2A = x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) = 5(8-18) + 15(18-6) + 10(6-8) = -50 + 180 - 20 = 110\ \text{mm}^{2}$$ so $A = 55\ \text{mm}^{2}$. The value is positive, confirming the counter-clockwise node numbering.
2. Form the geometric coefficients. With $\beta_i = y_j - y_k$ and $\gamma_i = x_k - x_j$ taken in cyclic order, $$\beta_1 = 8-18 = -10,\quad \beta_2 = 18-6 = 12,\quad \beta_3 = 6-8 = -2$$$$\gamma_1 = 10-15 = -5,\quad \gamma_2 = 5-10 = -5,\quad \gamma_3 = 15-5 = 10$$ A useful arithmetic check is that both sets must sum to zero: $-10+12-2 = 0$ and $-5-5+10 = 0$.
3. Assemble the strain-displacement matrix. For a three-node triangle the displacement field is linear, so $[\mathbf{B}]$ is constant over the element — this is what gives the element its name: $$[\mathbf{B}] = \frac{1}{2A}\begin{bmatrix}\beta_1 & 0 & \beta_2 & 0 & \beta_3 & 0\\ 0 & \gamma_1 & 0 & \gamma_2 & 0 & \gamma_3\\ \gamma_1 & \beta_1 & \gamma_2 & \beta_2 & \gamma_3 & \beta_3\end{bmatrix} = \frac{1}{110}\begin{bmatrix}-10&0&12&0&-2&0\\ 0&-5&0&-5&0&10\\ -5&-10&-5&12&10&-2\end{bmatrix}$$
4. Compute the element strains. With $\{d\}^{T} = \{0.003\ \ 0\ \ 0\ \ 0\ \ 0.005\ \ 0.003\}\ \text{mm}$, $$\varepsilon_x = \frac{(-10)(0.003)+(12)(0)+(-2)(0.005)}{110} = \frac{-0.040}{110} = -3.6364\times10^{-4}$$ $$\varepsilon_y = \frac{(-5)(0)+(-5)(0)+(10)(0.003)}{110} = \frac{0.030}{110} = +2.7273\times10^{-4}$$ $$\gamma_{xy} = \frac{(-5)(0.003)+(-10)(0)+(-5)(0)+(12)(0)+(10)(0.005)+(-2)(0.003)}{110} = \frac{0.029}{110} = +2.6364\times10^{-4}$$ The element is being compressed in $x$ while stretching in $y$ and shearing positively.
5. Build the plane-strain constitutive matrix. Plane strain (not plane stress) is specified, so $$[\mathbf{D}] = \frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu&\nu&0\\ \nu&1-\nu&0\\ 0&0&\tfrac{1-2\nu}{2}\end{bmatrix} = \frac{210}{(1.3)(0.4)}\begin{bmatrix}0.7&0.3&0\\0.3&0.7&0\\0&0&0.2\end{bmatrix}$$ with the leading factor $210/0.52 = 403.846\ \text{MPa}$, giving $D_{11} = 282.692$, $D_{12} = 121.154$ and $D_{33} = 80.769\ \text{MPa}$.
6. Part (a) — multiply out the stresses. $\{\sigma\} = [\mathbf{D}]\{\varepsilon\}$ gives $$\sigma_x = (282.692)(-3.6364\times10^{-4}) + (121.154)(2.7273\times10^{-4}) = -0.10280 + 0.03304$$ $$\boxed{\;\sigma_x = -0.06976\ \text{MPa},\quad \sigma_y = +0.03304\ \text{MPa},\quad \tau_{xy} = +0.02129\ \text{MPa}\;}$$ For completeness, plane strain also carries an out-of-plane normal stress $\sigma_z = \nu\left(\sigma_x+\sigma_y\right) = -0.01101\ \text{MPa}$, which is what the constraint $\varepsilon_z = 0$ requires.
7. Part (b) — transform to principal axes. The centre and radius of Mohr's circle are $$\sigma_{\text{avg}} = \frac{\sigma_x+\sigma_y}{2} = -0.018357\ \text{MPa},\qquad R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2} + \tau_{xy}^{2}} = \sqrt{(-0.051398)^{2}+(0.021294)^{2}} = 0.055635\ \text{MPa}$$ so that $$\boxed{\;\sigma_1 = +0.03728\ \text{MPa},\qquad \sigma_2 = -0.07399\ \text{MPa}\;}$$ and the principal direction follows from $\tan 2\theta_p = 2\tau_{xy}/(\sigma_x-\sigma_y)$, which with the negative denominator places $2\theta_p$ in the second quadrant: $$\boxed{\;\theta_p = 78.75^\circ\ \text{measured counter-clockwise from the }x\text{-axis to }\sigma_1\;}$$ The invariant checks hold: $\sigma_1+\sigma_2 = \sigma_x+\sigma_y = -0.03671$ and $\sigma_1\sigma_2 = \sigma_x\sigma_y-\tau_{xy}^{2}$. The maximum in-plane shear is $R = 0.05563\ \text{MPa}$ on planes $45^\circ$ from the principal directions.
8. Part (c) — the constant-strain triangle in bending. Because the assumed displacement field is linear, the strain and hence the stress are constant throughout the element. Pure bending, however, requires the axial strain to vary linearly through the depth, from tension on one face to compression on the other. A single layer of constant-strain triangles simply cannot represent that gradient: the element responds by developing a spurious shear strain instead, absorbing energy in shear that should have gone into bending. The element is therefore far too stiff in bending — this is shear locking, or the parasitic-shear problem — and computed deflections can be several times too small on a coarse mesh.

Four remedies are standard, in rough order of preference. Use higher-order elements — the linear-strain triangle (six-node LST) or an eight-node quadrilateral reproduces the linear strain variation of bending directly and removes the problem at source. Otherwise refine the mesh strongly through the depth, using at least four to six constant-strain triangles across the thickness and arranging them in crossed (union-jack) patterns rather than a single diagonal, so the directional bias of the mesh cancels; convergence is guaranteed but slow. Third, use a quadrilateral with incompatible bending modes, in which internal degrees of freedom are added specifically to reproduce the bending shape and then statically condensed out. Fourth, where the structural action really is bending, use a beam, plate or s​hell element whose formulation is built on bending kinematics rather than forcing a membrane element to imitate them.