Question 5 of 7: Plane-strain constant-strain triangle — element and principal stresses

Question 5: Plane-strain constant-strain triangle — element and principal stresses (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 three-node triangular element with nodes at (7, 8), (17, 8) and (12, 20) mm, a known set of nodal displacements, and plane-strain elastic constants as printed on the paper.

Find. The three in-plane element stresses and then the two principal stresses with the principal angle.

Approach. Compute twice the element area and the strain-displacement coefficients, form the constant strains, multiply by the plane-strain constitutive matrix, and close with the plane-stress transformation formulae for the principal values and direction.

1. Compute twice the area and the geometric coefficients. For a CST, $2A=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$:$$2A=7(8-20)+17(20-8)+12(8-8)=-84+204+0=120\ \text{mm}^2\ \Rightarrow\ A=60\ \text{mm}^2$$The coefficients follow directly from the coordinates: $\beta_1=y_2-y_3=-12$, $\beta_2=y_3-y_1=12$, $\beta_3=y_1-y_2=0$; $\gamma_1=x_3-x_2=-5$, $\gamma_2=x_1-x_3=-5$, $\gamma_3=x_2-x_1=10$ (mm). A positive $2A$ confirms the nodes are numbered counter-clockwise.
2. Form the strain-displacement matrix. The CST strains are constant over the element, $\{\varepsilon\}=[B]\{d\}$ with$$[B]=\frac{1}{2A}\begin{bmatrix}\beta_1&0&\beta_2&0&\beta_3&0\\0&\gamma_1&0&\gamma_2&0&\gamma_3\\\gamma_1&\beta_1&\gamma_2&\beta_2&\gamma_3&\beta_3\end{bmatrix}=\frac{1}{120}\begin{bmatrix}-12&0&12&0&0&0\\0&-5&0&-5&0&10\\-5&-12&-5&12&10&0\end{bmatrix}$$
3. Compute the element strains. With $\{d\}=\{0.004,\,0.002,\,0,\,0,\,0.005,\,0\}^{T}$ mm,$$\varepsilon_x=\frac{-12(0.004)+12(0)+0(0.005)}{120}=-4.000\times10^{-4}$$$$\varepsilon_y=\frac{-5(0.002)-5(0)+10(0)}{120}=-8.333\times10^{-5}$$$$\gamma_{xy}=\frac{-5(0.004)-5(0)+10(0.005)-12(0.002)+12(0)+0(0)}{120}=+5.000\times10^{-5}$$The element is compressed in both directions with a small shear — note that the millimetre units of the coordinates and the displacements cancel, so the strains are dimensionless as they must be.
4. Assemble the plane-strain constitutive matrix. For plane strain,$$[D]=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu&\nu&0\\\nu&1-\nu&0\\0&0&\tfrac{1-2\nu}{2}\end{bmatrix}=\frac{210}{(1.3)(0.4)}\begin{bmatrix}0.7&0.3&0\\0.3&0.7&0\\0&0&0.2\end{bmatrix}$$$$[D]=\begin{bmatrix}282.69&121.15&0\\121.15&282.69&0\\0&0&80.77\end{bmatrix}\ \text{MPa}$$
5. (a) Multiply out the element stresses. $\{\sigma\}=[D]\{\varepsilon\}$ gives$$\sigma_x=282.69(-4.000\times10^{-4})+121.15(-8.333\times10^{-5})=-0.11308-0.01010$$$$\sigma_y=121.15(-4.000\times10^{-4})+282.69(-8.333\times10^{-5})=-0.04846-0.02356$$$$\boxed{\sigma_x=-0.1232\ \text{MPa},\quad\sigma_y=-0.0720\ \text{MPa},\quad\tau_{xy}=80.77(5.000\times10^{-5})=+0.00404\ \text{MPa}}$$Both direct stresses are compressive, consistent with the negative strains.
6. (b) Transform to principal values. Using the standard in-plane transformation,$$\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}=-0.09760\pm\sqrt{(-0.02558)^{2}+(0.00404)^{2}}$$$$\boxed{\sigma_1=-0.0717\ \text{MPa},\qquad\sigma_2=-0.1235\ \text{MPa}}$$and the principal direction follows from$$\tan 2\theta_p=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}=\frac{2(0.004038)}{-0.05115}=-0.1579$$$$\boxed{\theta_p=85.5^\circ\ \text{(to }\sigma_1),\qquad \theta_p=-4.5^\circ\ \text{(to }\sigma_2)}$$The two roots of $\tan2\theta_p$ are $-4.49^\circ$ and $85.51^\circ$; substituting back into the transformation equation shows that the larger (algebraically) principal stress $\sigma_1$ acts on the plane at $85.5^\circ$ from the x-axis. As a check, $\sigma_1+\sigma_2=-0.1952=\sigma_x+\sigma_y$, and the out-of-plane stress $\sigma_z=\nu(\sigma_x+\sigma_y)=-0.0586$ MPa is non-zero, as plane strain requires.