07-Str-B3 · May 2013

Question 3 of 4: Constant-strain triangle in plane strain — element stresses and principal stresses

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations — May 2013 — 07-Str-B3 Applications of the Finite Element Method. Three hours, closed book; one of the two approved calculators (any Casio or Sharp model) and one 8.5 in by 11 in aid sheet written on both sides are permitted. The paper prints four problems, all of equal value, and instructs the candidate to answer only three (3) problems out of the four (4) proposed, the first three appearing in the answer book being the ones marked. Candidates are urged to submit a clear statement of any assumption made where a question admits more than one reading. All four problems are worked below, because this set is intended as a study resource rather than as a single exam sitting.

Reference texts: Logan, D.L., A First Course in the Finite Element Method (6th ed., Cengage) — bar and beam elements, the constant-strain triangle, and the isoparametric quadrilateral, in the same notation this paper uses; Cook, R.D., Malkus, D.S., Plesha, M.E. & Witt, R.J., Concepts and Applications of Finite Element Analysis (4th ed., Wiley) — quadrature, spurious zero-energy (hourglass) modes and element quality; Chandrupatla, T.R. & Belegundu, A.D., Introduction to Finite Elements in Engineering (4th ed., Pearson) — the CST gradient matrix in the beta/alpha form printed on page 4 of this paper; Bathe, K.-J., Finite Element Procedures (2nd ed., Prentice Hall) — isoparametric formulation and numerical integration; Zienkiewicz, O.C. & Taylor, R.L., The Finite Element Method: Its Basis and Fundamentals (7th ed., Butterworth-Heinemann) — general theory; Przemieniecki, J.S., Theory of Matrix Structural Analysis (Dover) — the plane-frame element stiffness matrix printed on page 3; Hibbeler, R.C., Mechanics of Materials (10th ed., Pearson) — statically indeterminate axially loaded members with an initial clearance; Ghali, A., Neville, A.M. & Brown, T.G., Structural Analysis: A Unified Classical and Matrix Approach (7th ed., CRC) — symmetry conditions in closed frames.

Check — the two lengths in Figure 1. Problem 1 dimensions the assembly once, as 10 in, and separately states that the copper rod is 0.005 in longer than the aluminium sleeve. The dimension line runs from the rigid support to the face of the rigid bearing plate, i.e. to the end of the rod, so the rod is taken as 10.000 in long and the sleeve as 9.995 in. Reading it the other way (sleeve 10.000 in, rod 10.005 in) changes every stress below by less than 0.1 per cent, which is far inside the precision of the data; the choice is therefore not load-bearing. Both members are treated as prismatic two-node bar elements sharing one axial degree of freedom at the loaded end.

Problem 3: Constant-strain triangle in plane strain — element stresses and principal stresses (equal value)

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 its node coordinates in millimetres, its six nodal displacements in millimetres, and isotropic elastic constants.

Given data
Node	$x$ (mm)	$y$ (mm)	$u$ (mm)	$v$ (mm)
1	5	5	0.005	0.002
2	25	5	0.000	0.000
3	15	15	0.005	0.000
Young's modulus $E = 70$ GPa; Poisson's ratio $\nu = 0.3$; plane strain

Find. The three in-plane element stresses $\sigma_x$, $\sigma_y$ and $\tau_{xy}$, and the principal stresses together with the direction in which they act.

[Figure not reproduced: Figure 3 (redrawn) — the constant-strain triangle, drawn to scale from Table 1. All dimensions in millimetres. See the official exam paper.]

Approach. Evaluate the printed gradient matrix from the node coordinates, multiply by the nodal displacement vector to obtain the constant element strains, apply the plane-strain constitutive matrix to get the stresses, and finish with the standard principal-stress transformation.

Compute twice the element area. For a triangle numbered anticlockwise, $$2A=\left(x_2-x_1\right)\left(y_3-y_1\right)-\left(x_3-x_1\right)\left(y_2-y_1\right)=(25-5)(15-5)-(15-5)(5-5)=200\ \text{mm}^{2},$$ so $A = 100\ \text{mm}^{2}$. The result is positive, confirming that nodes 1–2–3 are indeed listed anticlockwise and that no sign correction is needed.
Evaluate the beta and alpha coefficients. Using $\beta_{ij}=y_i-y_j$ and $\alpha_{ij}=x_i-x_j$ exactly as defined on the question paper, $$\beta_{23}=y_2-y_3=5-15=-10,\ \ \beta_{31}=y_3-y_1=15-5=10,\ \ \beta_{12}=y_1-y_2=5-5=0,$$ $$\alpha_{32}=x_3-x_2=15-25=-10,\ \ \alpha_{13}=x_1-x_3=5-15=-10,\ \ \alpha_{21}=x_2-x_1=25-5=20.$$ That $\beta_{12}=0$ simply records that nodes 1 and 2 lie on the same horizontal line, which is visible in Figure 3.
Assemble the gradient matrix. Substituting into the printed form with $2A = 200$ mm², $$[B]=\frac{1}{200}\begin{bmatrix}-10 & 0 & 10 & 0 & 0 & 0\\ 0 & -10 & 0 & -10 & 0 & 20\\ -10 & -10 & -10 & 10 & 20 & 0\end{bmatrix}\ \text{mm}^{-1}.$$ Each entry has units of reciprocal length, as it must, since $[B]$ maps displacements onto dimensionless strains.
Compute the constant element strains. With the nodal displacement vector $\{d\}=\{0.005,\ 0.002,\ 0,\ 0,\ 0.005,\ 0\}^{T}$ mm, the product $\{\varepsilon\}=[B]\{d\}$ gives $$\varepsilon_x=\frac{(-10)(0.005)+(10)(0)+(0)(0.005)}{200}=-2.500\times10^{-4},$$ $$\varepsilon_y=\frac{(-10)(0.002)+(-10)(0)+(20)(0)}{200}=-1.000\times10^{-4},$$ $$\gamma_{xy}=\frac{(-10)(0.005)+(-10)(0.002)+(-10)(0)+(10)(0)+(20)(0.005)+(0)(0)}{200}=+1.500\times10^{-4}.$$ These are the same everywhere inside the element, which is what the name constant-strain triangle records.
Form the plane-strain constitutive matrix. For plane strain, $$[D]=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu & \nu & 0\\ \nu & 1-\nu & 0\\ 0 & 0 & \tfrac{1-2\nu}{2}\end{bmatrix},\ \ \frac{E}{(1+\nu)(1-2\nu)}=\frac{70\,000}{(1.3)(0.4)}=134\,615.4\ \text{MPa},$$ so $D_{11}=D_{22}=94\,230.8$ MPa, $D_{12}=40\,384.6$ MPa and $D_{33}=26\,923.1$ MPa. Note that the plane-strain matrix, not the plane-stress one, must be used here; at this Poisson's ratio the plane-strain leading diagonal is about 23 per cent larger than the plane-stress value $E/(1-\nu^{2})=76\,923$ MPa.
Recover the element stresses. Multiplying, $\{\sigma\}=[D]\{\varepsilon\}$: $$\sigma_x=94\,230.8(-2.5\times10^{-4})+40\,384.6(-1.0\times10^{-4})=\boxed{-27.60\ \text{MPa}}$$ $$\sigma_y=40\,384.6(-2.5\times10^{-4})+94\,230.8(-1.0\times10^{-4})=\boxed{-19.52\ \text{MPa}}$$ $$\tau_{xy}=26\,923.1(1.5\times10^{-4})=\boxed{+4.04\ \text{MPa}}$$ Plane strain also generates an out-of-plane direct stress, $\sigma_z=\nu\left(\sigma_x+\sigma_y\right)=0.3(-47.12)=-14.13$ MPa, which is needed for any yield check even though the question does not ask for it.
Transform to principal axes. The centre and radius of the in-plane Mohr circle are $$\sigma_{\text{avg}}=\frac{\sigma_x+\sigma_y}{2}=-23.56\ \text{MPa},\ \ R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}=\sqrt{(-4.038)^{2}+(4.038)^{2}}=5.711\ \text{MPa},$$ so the in-plane principal stresses are $$\sigma_1=\sigma_{\text{avg}}+R=\boxed{-17.85\ \text{MPa}},\ \ \sigma_2=\sigma_{\text{avg}}-R=\boxed{-29.27\ \text{MPa}}$$ and the principal direction follows from $\tan 2\theta_p = 2\tau_{xy}/(\sigma_x-\sigma_y) = -1$, giving $2\theta_p = 135\,{}^{\circ}$ and $\theta_p = 67.5\,{}^{\circ}$ measured anticlockwise from the $x$ axis to the axis of $\sigma_1$. Because $\sigma_x - \sigma_y$ is negative while $\tau_{xy}$ is positive, the correct quadrant is the second one, which the two-argument arctangent selects automatically and a bare arctangent does not.
Assemble the three-dimensional stress state. Ordering all three principal values, $\sigma_z=-14.13$ MPa is the algebraically largest, then $-17.85$ MPa and then $-29.27$ MPa. The maximum in-plane shear is $R = 5.71$ MPa, but the absolute maximum shear stress in the element is $$\tau_{\max}=\frac{\sigma_{\max}-\sigma_{\min}}{2}=\frac{-14.13-(-29.27)}{2}=7.57\ \text{MPa},$$ acting on a plane inclined to the $z$ axis. The element is in triaxial compression with a modest shear, which is characteristic of plane-strain conditions.

Mohr's circle for the in-plane stress state. The out-of-plane plane-strain stress is marked separately because it does not lie on this circle.

Problem 3 — final results
Quantity	Value
Element area $A$ (and $2A$)	100 mm² (200 mm²)
$\varepsilon_x$ / $\varepsilon_y$ / $\gamma_{xy}$	$-2.500\times10^{-4}$ / $-1.000\times10^{-4}$ / $+1.500\times10^{-4}$
$\sigma_x$	$-27.60$ MPa
$\sigma_y$	$-19.52$ MPa
$\tau_{xy}$	$+4.04$ MPa
$\sigma_z$ (plane strain)	$-14.13$ MPa
Principal stress $\sigma_1$	$-17.85$ MPa
Principal stress $\sigma_2$	$-29.27$ MPa
Principal direction $\theta_p$	$67.5\,{}^{\circ}$ anticlockwise from the $x$ axis
Maximum in-plane shear / absolute maximum shear	5.71 MPa / 7.57 MPa

Topic. The constant-strain triangle: linear displacement interpolation, the resulting constant gradient matrix, plane-strain constitutive behaviour, and recovery of principal stresses from an element result.

Key relations. $\{\varepsilon\}=[B]\{d\}$ with $[B]$ as printed, $2A=(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1)$, $\{\sigma\}=[D]\{\varepsilon\}$, plane strain $[D]=\dfrac{E}{(1+\nu)(1-2\nu)}\left[\begin{smallmatrix}1-\nu&\nu&0\\ \nu&1-\nu&0\\ 0&0&(1-2\nu)/2\end{smallmatrix}\right]$, $\sigma_z=\nu(\sigma_x+\sigma_y)$, and $\sigma_{1,2}=\sigma_{\text{avg}}\pm\sqrt{\left((\sigma_x-\sigma_y)/2\right)^{2}+\tau_{xy}^{2}}$.

Why this works. Assuming the displacement field to be linear in $x$ and $y$ across the triangle forces its derivatives, and hence the strains, to be constant, so $[B]$ contains only geometry and the strain recovery is a single matrix product with no integration. The coefficients $\beta_{ij}$ and $\alpha_{ij}$ are simply the coordinate differences of the two nodes opposite the one being differentiated, which is why each column of $[B]$ depends on the other two nodes and not on its own. Plane strain then imposes $\varepsilon_z = 0$, which does not eliminate $\sigma_z$ but fixes it at $\nu(\sigma_x+\sigma_y)$ — a real stress that a plane-strain analysis must report before any yield criterion is applied.

Common pitfall. Using the plane-stress constitutive matrix, which at $\nu = 0.3$ understates $\sigma_x$ by about 22 per cent and silently deletes $\sigma_z$; taking $\beta_{ij}$ as a difference of $x$ coordinates instead of $y$; numbering the nodes clockwise, which returns a negative $2A$ and flips every stress sign; and reporting only the two in-plane principal stresses when the out-of-plane value governs the absolute maximum shear, as it does here.

Source. Logan, A First Course in the Finite Element Method (6th ed.), Ch. 6 — the constant-strain triangle, its $[B]$ matrix and plane stress against plane strain; Chandrupatla & Belegundu, Introduction to Finite Elements in Engineering (4th ed.), Ch. 6, which uses the same beta and alpha notation as this paper; Cook et al., Concepts and Applications of Finite Element Analysis (4th ed.), Ch. 3, on constant-strain elements and stress recovery.

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