07-Str-B3 · May 2015

Question 3 of 3: Constant-Strain Triangle — Stiffness Block, Stress Field and Local Equilibrium

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

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada national examination 07-Str-B3 — Applications of the Finite Element Method, May 2015. Three hours; four pages. Closed book, with two 8½ × 11 in pages of handwritten notes permitted and one approved non-communicating calculator. Three problems, all of equal value; candidates are instructed to attempt all three. Problem 1 is a ten-part concept paper, Problems 2 and 3 are calculations.

07-Str-B3 is a finite element methods paper rather than a member-design paper: bar and beam elements, isoparametric quadrilaterals, the constant-strain triangle, numerical integration and structural dynamics. The reference list below is therefore the finite-element literature.

Reference texts.

Logan, D. L., A First Course in the Finite Element Method, 6th ed. — Ch. 3 (bar element), Ch. 4 and 5 (beam element, work-equivalent loads), Ch. 6 (constant-strain triangle, plane stress), Ch. 10 (isoparametric elements, Gauss quadrature), Ch. 16 (structural dynamics, mass matrices).
Cook, R. D., Malkus, D. S., Plesha, M. E. and Witt, R. J., Concepts and Applications of Finite Element Analysis, 4th ed. — Ch. 3–4 (formulation and convergence), Ch. 6 (isoparametric elements and numerical integration), Ch. 7 (element quality, the patch test, mesh transitions), Ch. 11 (dynamics), Ch. 15 (mixed and hybrid formulations).
Bathe, K.-J., Finite Element Procedures, 2nd ed. — Ch. 4 (variational basis and convergence), Ch. 5 (numerical integration, reduced and selective integration), Ch. 9 (time integration).
Chandrupatla, T. R. and Belegundu, A. D., Introduction to Finite Elements in Engineering, 4th ed. — Ch. 5 (CST, plane stress and plane strain), Ch. 6 (beams and frames).
Zienkiewicz, O. C., Taylor, R. L. and Zhu, J. Z., The Finite Element Method: Its Basis and Fundamentals, 7th ed. — Ch. 6 (the patch test), Ch. 11 (mixed and hybrid forms).
Hughes, T. J. R., The Finite Element Method: Linear Static and Dynamic Finite Element Analysis — Ch. 4 (mixed methods and reduced integration), Ch. 5 (beam and plate elements), Ch. 7 (consistent and lumped mass matrices).
CSA A23.3:19 Design of Concrete Structures and CSA S806:12 Design and Construction of Building Structures with Fibre-Reinforced Polymers — the Canadian code frame behind the strengthened-beam modelling question (1.9).

Question 3: Constant-Strain Triangle — Stiffness Block, Stress Field and Local Equilibrium (33.3 marks — one of three problems of 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.

Quantity	Symbol	Value
Base of the plate (along $x$)	$b$	30 mm
Height of the plate (along $y$)	$h$	20 mm
Thickness	$t$	10 mm
Modulus of elasticity	$E$	70 GPa = $70\,000$ MPa
Poisson ratio	$\nu$	0.3
Pressure on the hypotenuse	$p$	100 MPa, normal, directed inward
Support	—	the vertical edge $x=0$ is fully fixed; the base $y=0$ is free
Idealisation	—	one constant-strain triangle, plane stress

Find. (3.1) a proof of the printed closed form for the $2\times2$ stiffness block $[k_{33}]$ under plane stress; (3.2) the constant stresses $\sigma_{x},\sigma_{y},\tau_{xy}$ delivered by a single linear triangle; and (3.3) an assessment of whether that stress field satisfies equilibrium locally.

[Figure not reproduced: Figure 3-b (redrawn) — the triangular plate, modelled by a single constant-strain triangle. The vertical edge $x = 0$ is fully fixed; the pressure $p$ acts normal to the hypotenuse; the base $y = 0$ is traction-free. Nodes are numbered counter-clockwise so that the only free node is node 3. See the official exam paper.]

Approach. Partition $[B]$ by node, evaluate $[k_{33}]=t\,A_{T}\,[B_{3}]^{T}[D][B_{3}]$ with the plane-stress constitutive matrix; then apply the same block to the plate, whose fixed edge leaves node 3 as the only free node, form the consistent nodal load from the pressure on the hypotenuse, solve two equations, and recover the constant strain and stress; finally test the resulting constant field against the differential equilibrium equations and against the traction boundary conditions.

3.1 Derivation of the stiffness block $[k_{33}]$

Partition the strain-displacement matrix by node. The $3\times6$ matrix printed on the paper splits into three $3\times2$ blocks, one per node. The block belonging to node 3 is $$[B_{3}]=\frac{1}{2A_{T}}\begin{bmatrix}\beta_{3}&0\\0&\gamma_{3}\\\gamma_{3}&\beta_{3}\end{bmatrix}$$ Because $[B]$ is constant over the triangle, so is every one of these blocks.
Write the element stiffness integral and use the constancy of $[B]$. In general $[k]=\int_{V}[B]^{T}[D][B]\,dV$. With $[B]$ and $[D]$ both constant and the volume equal to $A_{T}t$, the integral collapses to a product, and the block occupying rows and columns 5–6 is $$[k_{33}]=t\,A_{T}\,[B_{3}]^{T}[D][B_{3}]$$
Insert the plane-stress constitutive matrix. For plane stress $$[D]=\frac{E}{1-\nu^{2}}\begin{bmatrix}1&\nu&0\\\nu&1&0\\0&0&\dfrac{1-\nu}{2}\end{bmatrix}$$ Carrying out $[D][B_{3}]$ first: $$[D][B_{3}]=\frac{E}{2A_{T}(1-\nu^{2})}\begin{bmatrix}\beta_{3}&\nu\gamma_{3}\\\nu\beta_{3}&\gamma_{3}\\\dfrac{1-\nu}{2}\gamma_{3}&\dfrac{1-\nu}{2}\beta_{3}\end{bmatrix}$$
Pre-multiply by $[B_{3}]^{T}$ and collect the four entries. With $[B_{3}]^{T}=\dfrac{1}{2A_{T}}\begin{bmatrix}\beta_{3}&0&\gamma_{3}\\0&\gamma_{3}&\beta_{3}\end{bmatrix}$ the leading scalar becomes $t\,A_{T}\cdot\dfrac{E}{4A_{T}^{2}(1-\nu^{2})}=\dfrac{Et}{4A_{T}(1-\nu^{2})}$, and the four products are $$(1,1):\ \beta_{3}\beta_{3}+\gamma_{3}\cdot\frac{1-\nu}{2}\gamma_{3}=\beta_{3}^{2}+\frac{1-\nu}{2}\gamma_{3}^{2}$$ $$(1,2):\ \beta_{3}\,\nu\gamma_{3}+\gamma_{3}\cdot\frac{1-\nu}{2}\beta_{3}=\gamma_{3}\beta_{3}\Bigl(\nu+\frac{1-\nu}{2}\Bigr)=\Bigl(\frac{1+\nu}{2}\Bigr)\gamma_{3}\beta_{3}$$ $$(2,2):\ \gamma_{3}\gamma_{3}+\beta_{3}\cdot\frac{1-\nu}{2}\beta_{3}=\gamma_{3}^{2}+\frac{1-\nu}{2}\beta_{3}^{2}$$ with $(2,1)$ equal to $(1,2)$ by symmetry, since the middle term $\nu+\tfrac{1-\nu}{2}=\tfrac{1+\nu}{2}$ arises identically from either order.
Assemble the result. Collecting the scalar and the four entries, $$\boxed{[k_{33}]=\frac{Et}{4A_{T}(1-\nu^{2})}\begin{bmatrix}\beta_{3}^{2}+\dfrac{1-\nu}{2}\gamma_{3}^{2}&\Bigl(\dfrac{1+\nu}{2}\Bigr)\gamma_{3}\beta_{3}\\[6pt]\Bigl(\dfrac{1+\nu}{2}\Bigr)\gamma_{3}\beta_{3}&\gamma_{3}^{2}+\dfrac{1-\nu}{2}\beta_{3}^{2}\end{bmatrix}}$$ which is the expression printed on the question paper. The derivation used nothing specific to node 3, so the same form with the subscript changed gives $[k_{11}]$ and $[k_{22}]$, and replacing one subscript gives the off-diagonal blocks $[k_{ij}]$.

3.2 Stress field from a single linear triangle

Number the nodes counter-clockwise and compute the geometric coefficients. Place node 1 at $(0,20)$, node 2 at $(0,0)$ and node 3 at $(30,0)$, all in millimetres; this ordering is counter-clockwise, which is what makes the area positive. Then $$2A_{T}=x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})=0+0+30(20-0)=600\ \text{mm}^{2},\qquad A_{T}=300\ \text{mm}^{2}$$ $$\beta_{3}=y_{1}-y_{2}=20-0=20\ \text{mm},\qquad\gamma_{3}=x_{2}-x_{1}=0-0=0$$ The vanishing of $\gamma_{3}$ is the geometric reason this problem is tractable by hand: it uncouples the two equations at node 3.
Apply the boundary conditions and identify the free degrees of freedom. The edge $x=0$ is built in, and both nodes 1 and 2 lie on it, so $u_{1}=v_{1}=u_{2}=v_{2}=0$. The single free node is node 3, and the governing system is exactly the $2\times2$ block derived in part 3.1: $$[k_{33}]\begin{Bmatrix}u_{3}\\v_{3}\end{Bmatrix}=\begin{Bmatrix}F_{x3}\\F_{y3}\end{Bmatrix}$$ This is why the question asked for $[k_{33}]$ first.
Evaluate the stiffness block numerically. With $Et/[4A_{T}(1-\nu^{2})]=(70\,000)(10)/[4(300)(1-0.09)]=700\,000/1092=641.026\ \text{N/mm}^{3}$ and $\gamma_{3}=0$, $$[k_{33}]=641.026\begin{bmatrix}20^{2}&0\\0&\dfrac{1-0.3}{2}(20)^{2}\end{bmatrix}=\begin{bmatrix}256\,410&0\\0&89\,744\end{bmatrix}\ \text{N/mm}$$
Convert the pressure into consistent nodal forces. The loaded edge runs from node 1 at $(0,20)$ to node 3 at $(30,0)$; its length and outward unit normal are $$L_{e}=\sqrt{30^{2}+20^{2}}=36.056\ \text{mm},\qquad\mathbf{n}=\frac{1}{L_{e}}\begin{Bmatrix}20\\30\end{Bmatrix}$$ The traction is $\bar{\mathbf{t}}=-p\,\mathbf{n}$, and its resultant over the edge is $$\mathbf{F}_{\text{edge}}=-p\,t\,L_{e}\,\mathbf{n}=-(100)(10)\begin{Bmatrix}20\\30\end{Bmatrix}=\begin{Bmatrix}-20\,000\\-30\,000\end{Bmatrix}\ \text{N}$$ Because the element edge is straight and interpolates linearly, the consistent load vector splits this resultant equally between the two nodes of the edge, so node 3 receives half: $$F_{x3}=-10\,000\ \text{N},\qquad F_{y3}=-15\,000\ \text{N}$$ The other half goes to node 1, where it is absorbed directly by the support.
Solve for the nodal displacements. The uncoupled system gives $$u_{3}=\frac{-10\,000}{256\,410}=-0.03900\ \text{mm},\qquad v_{3}=\frac{-15\,000}{89\,744}=-0.16714\ \text{mm}$$ $$\boxed{u_{3}=-0.0390\ \text{mm},\qquad v_{3}=-0.1671\ \text{mm}}$$ Both are negative, as expected: the pressure pushes the free corner down and to the left.
Recover the constant strains. With all of node 1 and node 2 fixed, only the node-3 block of $[B]$ contributes: $$\varepsilon_{x}=\frac{\beta_{3}u_{3}}{2A_{T}}=\frac{20(-0.03900)}{600}=-1.300\times10^{-3}$$ $$\varepsilon_{y}=\frac{\gamma_{3}v_{3}}{2A_{T}}=0$$ $$\gamma_{xy}=\frac{\gamma_{3}u_{3}+\beta_{3}v_{3}}{2A_{T}}=\frac{20(-0.16714)}{600}=-5.5714\times10^{-3}$$ The vanishing of $\varepsilon_{y}$ is a direct consequence of the element shape: with two nodes fixed on the line $x=0$, the only admissible displacement field is $u=u_{3}\,x/b$, $v=v_{3}\,x/b$, which cannot vary with $y$ at all.
Convert to stresses through the plane-stress law. $$\sigma_{x}=\frac{E}{1-\nu^{2}}(\varepsilon_{x}+\nu\varepsilon_{y})=76\,923(-1.300\times10^{-3})=-100.0\ \text{MPa}$$ $$\sigma_{y}=\frac{E}{1-\nu^{2}}(\nu\varepsilon_{x}+\varepsilon_{y})=76\,923(0.3)(-1.300\times10^{-3})=-30.0\ \text{MPa}$$ $$\tau_{xy}=G\gamma_{xy}=\frac{E}{2(1+\nu)}\gamma_{xy}=26\,923(-5.5714\times10^{-3})=-150.0\ \text{MPa}$$ $$\boxed{\sigma_{x}=-100\ \text{MPa},\qquad\sigma_{y}=-30\ \text{MPa},\qquad\tau_{xy}=-150\ \text{MPa}}$$ These three numbers are the entire "stress distribution": a constant-strain triangle delivers one constant stress state over the whole element, and with one element that means over the whole plate.
Note the closed forms hiding in the arithmetic. Carrying the algebra symbolically rather than numerically shows that the answers do not depend on $E$ at all: $$\sigma_{x}=-p,\qquad\sigma_{y}=-\nu p,\qquad\tau_{xy}=-p\,\frac{b}{h}$$ Substituting $p=100$ MPa, $\nu=0.3$, $b=30$ mm and $h=20$ mm reproduces $-100$, $-30$ and $-150$ MPa exactly. The last of these is the striking one: the shear stress the single element reports is $b/h = 1.5$ times the applied pressure, which already signals that the answer is a crude approximation rather than a faithful stress field — the subject of part 3.3.

3.3 Comment on local equilibrium

The differential equations of equilibrium for a plane body without body forces are

$$\frac{\partial\sigma_{x}}{\partial x}+\frac{\partial\tau_{xy}}{\partial y}=0,\qquad\frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\sigma_{y}}{\partial y}=0$$

Every stress component delivered by the constant-strain triangle is a constant, so every derivative in those two equations is identically zero and both are satisfied. Taken at face value this looks like an excellent result — but it is a vacuous one. The element satisfies interior equilibrium not because it found the right stress field but because any constant field satisfies it. The two equations impose no restriction whatsoever on a constant-strain triangle, so they cannot be used to judge the answer.

The real test is the traction boundary conditions, and there the field fails badly. On the loaded hypotenuse the traction implied by the computed stresses is

$$\boldsymbol{\sigma}\cdot\mathbf{n}=\begin{bmatrix}-100&-150\\-150&-30\end{bmatrix}\frac{1}{36.056}\begin{Bmatrix}20\\30\end{Bmatrix}=\begin{Bmatrix}-180.3\\-108.2\end{Bmatrix}\ \text{MPa}$$

whereas the applied traction is $-p\,\mathbf{n}=\{-55.5,\ -83.2\}^{T}$ MPa. The two are not close, and they do not even have the same direction. Worse, on the base $y=0$, which carries no load at all and whose outward normal is $\{0,-1\}^{T}$, the computed field produces

$$\boldsymbol{\sigma}\cdot\mathbf{n}=\{-\tau_{xy},\ -\sigma_{y}\}^{T}=\{+150,\ +30\}^{T}\ \text{MPa}$$

on a surface that ought to be traction-free. A single constant-stress element simply has no freedom left to satisfy a boundary condition: three stress components are fixed by two nodal displacements, and the requirement of a traction-free edge alone would demand $\tau_{xy}=\sigma_{y}=0$, which contradicts the loading.

What the model does get right is equilibrium in the global and weak senses. Integrating the computed tractions around the whole boundary gives exactly zero resultant, as any constant stress field must; and the traction resultant on the fixed edge, $\{20\,000,\ 30\,000\}^{T}$ N, is exactly equal and opposite to the applied load resultant $\{-20\,000,\ -30\,000\}^{T}$ N. The reactions are therefore correct even though the stresses are not. This is the general behaviour described in question 1.1, seen in its starkest form: the finite element method enforces equilibrium in an integral, work-equivalent sense at the nodes, and leaves pointwise equilibrium — particularly on the boundary — to be recovered only as the mesh is refined.

The practical conclusion for a candidate or a designer is that this one-element answer is usable as an order-of-magnitude check and nothing more. Refining the plate into a graded mesh of constant-strain triangles would drive the boundary tractions towards their correct values roughly in proportion to the element size, while a single quadratic (six-node) triangle, whose stresses vary linearly, would already reduce the violation substantially at the same node count.

Check: assumptions carried through Problem 3. The figure is read as a right triangle with the built-in edge vertical on the $y$ axis, the 30 mm dimension measured along the base and the 20 mm dimension along the fixed edge, with $p$ acting normal to and directed into the hypotenuse and the base traction-free. The consistent nodal load is computed for a uniform pressure over a straight linear edge, which splits the resultant equally between the two edge nodes.

Quantity	Symbol	Result
Triangle area	$A_{T}$	300 mm$^{2}$
Geometric coefficients at node 3	$\beta_{3},\gamma_{3}$	20 mm, 0
Stiffness block	$[k_{33}]$	$\mathrm{diag}(256\,410,\ 89\,744)$ N/mm
Consistent nodal load at node 3	$F_{x3},F_{y3}$	$-10\,000$ N, $-15\,000$ N
Displacement of the free corner	$u_{3},v_{3}$	$-0.0390$ mm, $-0.1671$ mm
Strains (constant)	$\varepsilon_{x},\varepsilon_{y},\gamma_{xy}$	$-1.300\times10^{-3}$, $0$, $-5.571\times10^{-3}$
Normal stress along $x$	$\sigma_{x}$	$-100$ MPa (equals $-p$)
Normal stress along $y$	$\sigma_{y}$	$-30$ MPa (equals $-\nu p$)
Shear stress	$\tau_{xy}$	$-150$ MPa (equals $-pb/h$)
Interior equilibrium	$\nabla\cdot\boldsymbol{\sigma}$	satisfied identically (constant field)
Traction on the loaded edge: computed / required	—	$\{-180.3,-108.2\}$ / $\{-55.5,-83.2\}$ MPa
Traction on the free base: computed / required	—	$\{+150,+30\}$ / $\{0,0\}$ MPa
Reaction resultant on the fixed edge	$R_{x},R_{y}$	$+20\,000$ N, $+30\,000$ N ✓ balances the load

Topic: The constant-strain triangle in plane stress — its stiffness blocks, its consistent edge loading, and the exact sense in which its stress field satisfies equilibrium.

Key relations: $\{\varepsilon\}=[B]\{d\}$ with $[B]=\dfrac{1}{2A_{T}}[\beta_{i}\ 0;\ 0\ \gamma_{i};\ \gamma_{i}\ \beta_{i}]$; $[k]=t\,A_{T}[B]^{T}[D][B]$; plane stress $[D]=\dfrac{E}{1-\nu^{2}}[1\ \nu\ 0;\ \nu\ 1\ 0;\ 0\ 0\ \tfrac{1-\nu}{2}]$; consistent edge load $\mathbf{F}=\int_{S}[N]^{T}\bar{\mathbf{t}}\,t\,dS=\tfrac{1}{2}\bar{\mathbf{t}}\,t\,L_{e}$ per node; equilibrium $\partial\sigma_{x}/\partial x+\partial\tau_{xy}/\partial y=0$.

Why this works: The linear triangle interpolates displacement linearly, so its strains — the first derivatives — are constant, and with them the stresses. That single fact drives everything in this question. It makes $[B]$ come outside the stiffness integral, reducing $\int[B]^{T}[D][B]\,dV$ to the algebraic product $t\,A_{T}[B]^{T}[D][B]$ and giving the neat closed form for $[k_{33}]$. It makes the whole plate a single stress state in part 3.2. And it makes the interior equilibrium equations trivially true but completely uninformative in part 3.3, so that the only meaningful test of the answer is on the boundary, where it fails. The generalisation worth remembering is that a displacement-based element satisfies equilibrium only where the virtual-work statement puts it — at the nodes, in resultant — and the coarser the element, the more visible the gap.

Common pitfall: Three traps recur. Numbering the nodes clockwise makes $2A_{T}$ come out negative and silently flips the sign of every stress; always check the ordering before computing $\beta$ and $\gamma$. Resolving the pressure into "horizontal and vertical components of $p$" by multiplying by the projected edge lengths and then applying $p$ again to the inclined length double-counts the load; the safe route is $\mathbf{F}=-p\,t\,L_{e}\,\mathbf{n}$, in which $L_{e}\mathbf{n}=\{h,\ b\}^{T}$ collapses to the projected dimensions automatically. Finally, concluding in part 3.3 that "equilibrium is satisfied" because the derivatives vanish misses the point of the question — the marks are for recognising that the test is vacuous for a constant field and for turning instead to the traction boundary conditions.

Source: Logan, A First Course in the Finite Element Method, 6th ed., §6.2–6.4 (CST derivation, plane stress and plane strain, treatment of surface tractions); Chandrupatla & Belegundu, Introduction to Finite Elements in Engineering, 4th ed., §5.2–5.5; Cook, Malkus, Plesha & Witt, 4th ed., §3.6 (the constant-strain triangle) and §4.4 (equilibrium and the meaning of a finite element stress field); Bathe, Finite Element Procedures, 2nd ed., §4.2.

← Question 2 Back to the paper →