Question 5 of 7: Stresses and principal stresses in a plane-strain triangular element [20 marks]

Question 5: Stresses and principal stresses in a plane-strain triangular 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 three-node constant-strain triangle in plane strain, with nodal coordinates read from the figure and the nodal displacements listed in the question:

Find. (a) the three in-plane element stresses $\sigma_x$, $\sigma_y$ and $\tau_{xy}$; (b) the in-plane principal stresses $\sigma_1$ and $\sigma_2$ and the principal angle $\theta_p$.

Approach. Compute the element area and the constant strain–displacement matrix from the nodal coordinates, multiply by the nodal displacement vector to obtain the three constant strains, apply the plane-strain constitutive matrix to obtain the stresses, then transform to principal axes with the standard two-dimensional stress transformation.

1. Compute twice the element area from the nodal coordinates. For a three-node triangle, $$2A=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$ $$2A=6(12-32)+30(32-12)+18(12-12)=-120+600+0=480\ \text{mm}^{2}$$ so $A=240\ \text{mm}^{2}$. The result is positive, confirming that nodes 1, 2, 3 are numbered counter-clockwise as required.
2. Form the geometric coefficients of the strain–displacement matrix. With the cyclic definitions $\beta_i=y_j-y_k$ and $\gamma_i=x_k-x_j$,

   Strain–displacement coefficients (mm)

   $i$	$\beta_i=y_j-y_k$	$\gamma_i=x_k-x_j$
   1	$y_2-y_3=12-32=-20$	$x_3-x_2=18-30=-12$
   2	$y_3-y_1=32-12=+20$	$x_1-x_3=6-18=-12$
   3	$y_1-y_2=12-12=0$	$x_2-x_1=30-6=+24$

   As an arithmetic check, $\sum\beta_i=0$ and $\sum\gamma_i=0$, which must hold because a rigid-body translation produces no strain.
3. Assemble the strain–displacement matrix. The constant-strain triangle gives $$[B]=\frac{1}{2A}\begin{bmatrix}\beta_1 & 0 & \beta_2 & 0 & \beta_3 & 0\\ 0 & \gamma_1 & 0 & \gamma_2 & 0 & \gamma_3\\ \gamma_1 & \beta_1 & \gamma_2 & \beta_2 & \gamma_3 & \beta_3\end{bmatrix}=\frac{1}{480}\begin{bmatrix}-20 & 0 & 20 & 0 & 0 & 0\\ 0 & -12 & 0 & -12 & 0 & 24\\ -12 & -20 & -12 & 20 & 24 & 0\end{bmatrix}$$
4. Compute the element strains. With the nodal displacement vector $\{d\}^{T}=\{0.004,\ 0.002,\ 0,\ 0,\ 0.005,\ 0\}$ mm, the strains $\{\varepsilon\}=[B]\{d\}$ evaluate term by term as $$\varepsilon_x=\frac{(-20)(0.004)+(20)(0)+(0)(0.005)}{480}=\frac{-0.080}{480}=-1.6667\times10^{-4}$$ $$\varepsilon_y=\frac{(-12)(0.002)+(-12)(0)+(24)(0)}{480}=\frac{-0.024}{480}=-5.0000\times10^{-5}$$ $$\gamma_{xy}=\frac{(-12)(0.004)+(-12)(0)+(24)(0.005)+(-20)(0.002)+(20)(0)+(0)(0)}{480}=\frac{0.032}{480}=6.6667\times10^{-5}$$ All three strains are constant over the element, which is the defining characteristic of this element.
5. Assemble the plane-strain constitutive matrix. For plane strain, $$[D]=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu & \nu & 0\\ \nu & 1-\nu & 0\\ 0 & 0 & \tfrac{1-2\nu}{2}\end{bmatrix}$$ With $E=210$ MPa and $\nu=0.3$ the leading factor is $\dfrac{210}{(1.3)(0.4)}=403.846$ MPa, so $$D_{11}=D_{22}=403.846(0.7)=282.692\ \text{MPa},\quad D_{12}=403.846(0.3)=121.154\ \text{MPa},\quad D_{33}=403.846(0.2)=80.769\ \text{MPa}$$
6. Compute the element stresses. Applying $\{\sigma\}=[D]\{\varepsilon\}$ to the strains found above, $$\sigma_x=D_{11}\varepsilon_x+D_{12}\varepsilon_y=282.692(-1.6667\times10^{-4})+121.154(-5.0\times10^{-5})=-0.047115-0.006058$$ $$\sigma_y=D_{12}\varepsilon_x+D_{11}\varepsilon_y=121.154(-1.6667\times10^{-4})+282.692(-5.0\times10^{-5})=-0.020192-0.014135$$ $$\tau_{xy}=D_{33}\gamma_{xy}=80.769\,(6.6667\times10^{-5})$$ which gives the three element stresses $$\boxed{\ \sigma_x=-0.05317\ \text{MPa},\qquad \sigma_y=-0.03433\ \text{MPa},\qquad \tau_{xy}=+0.005385\ \text{MPa}\ }$$ Because the element is in plane strain there is also an out-of-plane normal stress $\sigma_z=\nu(\sigma_x+\sigma_y)=0.3(-0.08750)=-0.02625$ MPa, which is not asked for but is needed for any yield check.
7. Locate the centre and radius of Mohr’s circle. For part (b), the average normal stress and the radius are $$\sigma_{\text{avg}}=\frac{\sigma_x+\sigma_y}{2}=\frac{-0.05317-0.03433}{2}=-0.043750\ \text{MPa}$$ $$R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}=\sqrt{(-0.0094231)^{2}+(0.0053846)^{2}}=\sqrt{1.17789\times10^{-4}}=0.010853\ \text{MPa}$$
8. Compute the principal stresses and the principal angle. The principal stresses are $\sigma_{1,2}=\sigma_{\text{avg}}\pm R$, so $$\boxed{\ \sigma_1=-0.032897\ \text{MPa},\qquad \sigma_2=-0.054603\ \text{MPa}\ }$$ and the principal direction follows from $$\tan 2\theta_p=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}=\frac{2(0.0053846)}{-0.0094231\times 2}=\frac{0.0107692}{-0.0188462}=-0.571429$$ Because the denominator $\sigma_x-\sigma_y$ is negative and the numerator $2\tau_{xy}$ is positive, $2\theta_p$ lies in the second quadrant: $2\theta_p=180^\circ-29.745^\circ=150.255^\circ$, hence $$\boxed{\ \theta_p=75.13^\circ\ \text{(measured counter-clockwise from the }x\text{ axis to the }\sigma_1\text{ direction)}}$$ The perpendicular direction, $\theta_p-90^\circ=-14.87^\circ$, carries $\sigma_2$.
9. Check the results against the stress invariants. The sum of the principal stresses must equal the sum of the in-plane normal stresses: $\sigma_1+\sigma_2=-0.032897-0.054603=-0.087500$ MPa, and $\sigma_x+\sigma_y=-0.053173-0.034327=-0.087500$ MPa. They agree. The maximum in-plane shear stress is $\tau_{\max}=(\sigma_1-\sigma_2)/2=R=0.010853$ MPa, consistent with the radius computed in step 7. Both principal stresses are compressive, which is consistent with the imposed displacement pattern: node 2 is pinned while nodes 1 and 3 move toward it in the $x$ direction, shortening the element.