Constant-strain triangle in plane strain — stresses and principal values: 22-Mec-B10 Finite Element Analysis, May 2017, Q6

Question 6: Constant-strain triangle in plane strain — stresses and principal values (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 with nodes at $1(6,12)$, $2(30,12)$ and $3(18,32)$ mm, plate thickness $t = 1$ mm, plane-strain conditions, and the nodal displacement vector below.

Question 6 — given data
Quantity	Symbol	Value
Node 1 coordinates	$(x_1,y_1)$	$(6,\ 12)$ mm
Node 2 coordinates	$(x_2,y_2)$	$(30,\ 12)$ mm
Node 3 coordinates	$(x_3,y_3)$	$(18,\ 32)$ mm
Node 1 displacements	$(u_1,v_1)$	$(0.002,\ 0.004)$ mm
Node 2 displacements	$(u_2,v_2)$	$(0,\ 0)$ mm
Node 3 displacements	$(u_3,v_3)$	$(0,\ 0.006)$ mm
Thickness	$t$	$1$ mm
Elastic constants	$E,\ \nu$	$210$ MPa, $0.3$

Find. The three in-plane element stresses $\sigma_x$, $\sigma_y$, $\tau_{xy}$, then the principal stresses $\sigma_1$, $\sigma_2$ and the principal angle $\theta_p$.

Question 6 — the constant-strain triangle with its nodal coordinates and the imposed nodal displacements (arrows indicate direction only).

Check: the paper prints $E = 210$ MPa, three orders of magnitude below the $210$ GPa expected for a steel plate. The solution below uses the value exactly as printed. If $210$ GPa was intended, every stress in this answer simply scales by $10^{3}$ (so $\sigma_y = 46.44$ MPa, $\sigma_1 = 52.37$ MPa and so on); the strains, the principal directions and the principal angle $\theta_p$ are entirely unaffected, because the constitutive matrix scales uniformly.

Approach. Compute twice the element area from the nodal coordinates, form the constant strain–displacement matrix $[B]$, extract the three strains, multiply by the plane-strain constitutive matrix $[D]$, then take the principal values from Mohr's circle.

Compute twice the element area. $$2A = x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) = 6(12-32) + 30(32-12) + 18(12-12) = -120 + 600 + 0 = 480\ \text{mm}^{2}$$ so $A = 240\ \text{mm}^{2}$. The result is positive, which confirms that nodes 1, 2, 3 are numbered counter-clockwise — a negative value would mean the element is inverted and every subsequent sign would be wrong.
Form the geometric coefficients. With the usual definitions $\beta_i = y_j - y_m$ and $\gamma_i = x_m - x_j$ taken cyclically, $$\beta_1 = y_2-y_3 = -20,\quad \beta_2 = y_3-y_1 = 20,\quad \beta_3 = y_1-y_2 = 0$$ $$\gamma_1 = x_3-x_2 = -12,\quad \gamma_2 = x_1-x_3 = -12,\quad \gamma_3 = x_2-x_1 = 24$$ As a check, $\sum\beta_i = 0$ and $\sum\gamma_i = 0$, which is what guarantees zero strain under rigid-body translation.
Assemble the strain–displacement matrix. $$[B] = \frac{1}{2A}\begin{bmatrix} \beta_1 & 0 & \beta_2 & 0 & \beta_3 & 0\\ 0 & \gamma_1 & 0 & \gamma_2 & 0 & \gamma_3\\ \gamma_1 & \beta_1 & \gamma_2 & \beta_2 & \gamma_3 & \beta_3\end{bmatrix} = \frac{1}{480}\begin{bmatrix} -20 & 0 & 20 & 0 & 0 & 0\\ 0 & -12 & 0 & -12 & 0 & 24\\ -12 & -20 & -12 & 20 & 24 & 0\end{bmatrix}$$ Every entry is a constant, which is the defining property of this element: strain, and therefore stress, is uniform throughout the triangle.
Extract the strains. With $\{d\} = \{0.002,\ 0.004,\ 0,\ 0,\ 0,\ 0.006\}^{T}$ mm, $$\varepsilon_x = \frac{(-20)(0.002)}{480} = -8.3333\times10^{-5},\qquad \varepsilon_y = \frac{(-12)(0.004)+(24)(0.006)}{480} = \frac{0.096}{480} = 2.0000\times10^{-4}$$ $$\gamma_{xy} = \frac{(-12)(0.002)+(-20)(0.004)}{480} = \frac{-0.104}{480} = -2.16667\times10^{-4}$$ The element is being stretched in $y$, compressed slightly in $x$ and sheared — all three strains are of order $10^{-4}$, consistent with micron-scale displacements over a 20 mm element.
Form the plane-strain constitutive matrix. For plane strain (not plane stress), $$[D] = \frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix} 1-\nu & \nu & 0\\ \nu & 1-\nu & 0\\ 0 & 0 & \tfrac{1-2\nu}{2}\end{bmatrix} = \frac{210}{(1.3)(0.4)}\begin{bmatrix} 0.7 & 0.3 & 0\\ 0.3 & 0.7 & 0\\ 0 & 0 & 0.2\end{bmatrix}$$ with the leading factor $E/[(1+\nu)(1-2\nu)] = 403.846$ MPa.
Compute the stresses. Multiplying $\{\sigma\} = [D]\{\varepsilon\}$, $$\sigma_x = 403.846\big[0.7(-8.3333\times10^{-5}) + 0.3(2.0\times10^{-4})\big] = 403.846\,(1.6667\times10^{-6})$$ $$\sigma_y = 403.846\big[0.3(-8.3333\times10^{-5}) + 0.7(2.0\times10^{-4})\big] = 403.846\,(1.15\times10^{-4}),\qquad \tau_{xy} = 403.846\,(0.2)(-2.16667\times10^{-4})$$ which evaluates to $$\boxed{\,\sigma_x = 6.731\times10^{-4}\ \text{MPa},\qquad \sigma_y = 4.6442\times10^{-2}\ \text{MPa},\qquad \tau_{xy} = -1.7500\times10^{-2}\ \text{MPa}\,}$$ The very small $\sigma_x$ is genuine and instructive: the compressive $\varepsilon_x$ and the Poisson coupling from the tensile $\varepsilon_y$ very nearly cancel, so this element sits within a whisker of a uniaxial-$y$ stress state. The plane-strain condition also generates an out-of-plane normal stress $\sigma_z = \nu(\sigma_x+\sigma_y) = 1.4135\times10^{-2}$ MPa, which the question does not ask for but which must exist for $\varepsilon_z$ to remain zero.

Part (b) — principal stresses and principal angle. With only $\sigma_x$, $\sigma_y$ and $\tau_{xy}$ in the plane, Mohr's circle gives the centre and radius directly:

$$\sigma_{\text{avg}} = \frac{\sigma_x+\sigma_y}{2} = 2.3558\times10^{-2}\ \text{MPa},\qquad R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}} = 2.8809\times10^{-2}\ \text{MPa}$$ $$\boxed{\,\sigma_1 = 5.2367\times10^{-2}\ \text{MPa},\qquad \sigma_2 = -5.2513\times10^{-3}\ \text{MPa},\qquad \theta_p = -71.30^{\circ}\,}$$

The angle follows from $\tan 2\theta_p = 2\tau_{xy}/(\sigma_x-\sigma_y)$, evaluated with a two-argument arctangent so that the quadrant is unambiguous: both $2\tau_{xy}$ and $\sigma_x-\sigma_y$ are negative, which places $2\theta_p$ in the third quadrant and gives $\theta_p = -71.30^{\circ}$ measured from the $x$ axis. Rotating the stress state through that angle returns $\sigma_1$ on the rotated normal and zero shear, so $\theta_p$ is the $\sigma_1$ direction and $\sigma_2$ acts on the perpendicular plane at $+18.70^{\circ}$. Three invariant checks close the answer: $\sigma_1+\sigma_2 = \sigma_x+\sigma_y$, $\sigma_1\sigma_2 = \sigma_x\sigma_y - \tau_{xy}^{2}$, and $\tfrac12(\sigma_1-\sigma_2) = R = \tau_{\max}$ in the plane. Physically, the element is close to pure tension along a direction some $19^{\circ}$ off the $y$ axis, with a small compressive principal stress at right angles — the signature of a strongly sheared, lightly stretched patch of plate.

Question 6 — final results
Quantity	Result
Element area	$A = 240\ \text{mm}^{2}$ ($2A = 480$)
Strains	$\varepsilon_x = -8.333\times10^{-5}$, $\varepsilon_y = 2.000\times10^{-4}$, $\gamma_{xy} = -2.167\times10^{-4}$
$\sigma_x$ (a)	$6.731\times10^{-4}$ MPa
$\sigma_y$ (a)	$4.6442\times10^{-2}$ MPa
$\tau_{xy}$ (a)	$-1.7500\times10^{-2}$ MPa
Out-of-plane stress	$\sigma_z = 1.4135\times10^{-2}$ MPa
$\sigma_1$ (b)	$5.2367\times10^{-2}$ MPa
$\sigma_2$ (b)	$-5.2513\times10^{-3}$ MPa
$\theta_p$ (b)	$-71.30^{\circ}$ from the $x$ axis ($\sigma_1$ direction)

Topic. The constant-strain triangle in plane strain: strain–displacement matrix, constitutive law, and principal stress transformation.

Key relations. $2A = x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$; $\{\varepsilon\} = [B]\{d\}$ with $[B]$ constant; plane strain $[D] = \dfrac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu&\nu&0\\ \nu&1-\nu&0\\ 0&0&\tfrac{1-2\nu}{2}\end{bmatrix}$; $\sigma_z = \nu(\sigma_x+\sigma_y)$; $\sigma_{1,2} = \sigma_{\text{avg}} \pm R$; $\tan2\theta_p = 2\tau_{xy}/(\sigma_x-\sigma_y)$.

Why this works. Linear interpolation over a triangle makes the displacement gradients constant, so a single $[B]$ describes the entire element and no integration is needed to get the stresses — they follow from one matrix product. Plane strain, rather than plane stress, is the correct idealisation when the body is long in $z$ and restrained there; the difference is not cosmetic, since the plane-strain $[D]$ contains the factor $(1-2\nu)$ in its denominator and stiffens sharply as $\nu \to 0.5$. The principal transformation is then pure two-dimensional stress algebra applied to the uniform state, with $\sigma_z$ acting as a third principal stress that happens to lie out of plane.

Common pitfall. Using the plane-stress $[D] = \dfrac{E}{1-\nu^{2}}[\cdot]$ when the question says plane strain is the single most common error and changes every stress. Close behind is losing the quadrant of $\theta_p$ by using a single-argument arctangent: $\tan2\theta_p$ has period $180^{\circ}$ in $2\theta_p$, so the naive inverse can return the $\sigma_2$ axis instead of the $\sigma_1$ axis. Finally, a negative $2A$ signals clockwise node numbering, not a computational slip — renumber rather than take the absolute value, or the shear sign will be wrong.

Source. Logan, A First Course in the Finite Element Method, 6th ed., Ch. 6 (development of the plane stress and plane strain stiffness equations); Cook et al., 4th ed., Ch. 3.

Question 6 of 7: Constant-strain triangle in plane strain — stresses and principal values

Question 6: Constant-strain triangle in plane strain — stresses and principal values (20 marks)