Thermal stresses in a thin triangular plate modelled by one CST: 16-Civ-B9 The Finite Element Method, December 2016, Q3

Problem 3: Thermal stresses in a thin triangular plate modelled by one CST (one of three equal-value problems)

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. One constant-strain triangle, clamped along its base, heated uniformly, with no mechanical load at all — the entire load vector comes from the restrained thermal expansion.

Given data (page 4 and Appendix A)
Quantity	Symbol	Value
Node 1 (bottom-left)	$(x_{1},y_{1})$	$(0,\,0)$ mm
Node 2 (bottom-right)	$(x_{2},y_{2})$	$(25,\,0)$ mm
Node 3 (apex)	$(x_{3},y_{3})$	$(0,\,45)$ mm
Thickness	$t$	$1$ mm
Elastic modulus	$E$	$30{,}000$ $\text{N/mm}^{2}$
Poisson ratio	$\nu$	$0.25$
Thermal expansion coefficient	$\alpha$	$11\times 10^{-6}$ per °C
Temperature rise	$\Delta T$	$30^\circ\text{C}$
Boundary condition	—	edge 1–2 fully clamped, edges 1–3 and 3–2 free

Find. The state of stress $\{\sigma_{x},\sigma_{y},\tau_{xy}\}$ in the element, and a comment on how well that stress field satisfies equilibrium locally.

Figure 3.1 — The plate of Figure 3 as a single constant-strain triangle. Numbering 1, 2, 3 counter-clockwise makes the signed area positive, which Appendix A requires; the clamped base leaves node 3 as the only free node.

Approach. Build $[B]$ and $[D]$ from Appendix A, convert the temperature rise into an equivalent nodal load vector, solve the two free equations for the apex displacement, then recover the total strain, subtract the free thermal strain and multiply by $[D]$.

Fix the geometry and check the node ordering. With the coordinates above the signed area is positive, so nodes 1, 2, 3 run counter-clockwise as Appendix A assumes:
$$2A=\left(x_{2}-x_{1}\right)\left(y_{3}-y_{1}\right) -\left(x_{3}-x_{1}\right)\left(y_{2}-y_{1}\right)=25\times 45-0=1125\ \text{mm}^{2}, \qquad A=562.5\ \text{mm}^{2}$$ Form the coordinate differences Appendix A needs. Using $x_{ij}=x_{i}-x_{j}$ and $y_{ij}=y_{i}-y_{j}$:
$$y_{23}=-45,\quad y_{31}=45,\quad y_{12}=0,\qquad x_{32}=-25,\quad x_{13}=0,\quad x_{21}=25\ \ \text{(mm)}$$ Write the strain-displacement matrix. Substituting into the printed form of $[B]$ with $2A=1125$ $\text{mm}^{2}$:
$$[B]=\frac{1}{1125}\begin{bmatrix} -45&0&45&0&0&0\\ 0&-25&0&0&0&25\\ -25&-45&0&45&25&0\end{bmatrix}\ \text{mm}^{-1}$$ Build the elastic matrix from $H_{1}$, $H_{2}$ and $G$. The plate is thin, so this is plane stress and $a=0$ in the Appendix A definitions:
$$G=\frac{E}{2(1+\nu)}=12{,}000,\qquad H_{1}=\frac{2G}{1-\nu}=\frac{E}{1-\nu^{2}}=32{,}000,\qquad H_{2}=\nu H_{1}=8{,}000\ \ \text{N/mm}^{2}$$ $$[D]=\begin{bmatrix}32{,}000&8{,}000&0\\ 8{,}000&32{,}000&0\\ 0&0&12{,}000\end{bmatrix}\ \text{N/mm}^{2}$$ Convert the temperature rise into nodal forces. A uniform rise wants to produce the free strain $\{\varepsilon_{0}\}$, and the work-equivalent load vector is $\{f\}=tA[B]^{T}[D]\{\varepsilon_{0}\}$:
$$\{\varepsilon_{0}\}=\alpha\Delta T\{1,1,0\}^{T} =\left\{3.3\times 10^{-4},\ 3.3\times 10^{-4},\ 0\right\}^{T}$$ $$\{f\}=\left\{-297,\ -165,\ +297,\ 0,\ 0,\ +165\right\}^{T}\ \text{N}$$ Apply the boundary conditions. Both base nodes are clamped, so only $u_{3}$ and $v_{3}$ survive. The corresponding block of $[k]=tA[B]^{T}[D][B]$ happens to be diagonal, because $x_{13}=0$ for a vertical clamped edge:
$$\begin{bmatrix}3333.33&0\\ 0&8888.89\end{bmatrix} \begin{Bmatrix}u_{3}\\ v_{3}\end{Bmatrix} =\begin{Bmatrix}0\\ 165\end{Bmatrix}\ \text{N}$$ Solve for the apex displacement. The horizontal component of the thermal load at node 3 is zero because $y_{12}=0$ and $[D]\{\varepsilon_{0}\}$ has no shear term, so the apex rises straight up:
$$\boxed{u_{3}=0,\qquad v_{3}=\frac{165}{8888.89}=0.0185625\ \text{mm}}$$ Recover the total strain. Multiplying $[B]$ by $\{0,0,0,0,0,v_{3}\}^{T}$ leaves only the term $x_{21}v_{3}/2A=v_{3}/45$:
$$\{\varepsilon\}=[B]\{d\}=\left\{0,\ \frac{0.0185625}{45},\ 0\right\}^{T} =\left\{0,\ 4.125\times 10^{-4},\ 0\right\}^{T}$$ Note that $\varepsilon_{y}$ is larger than the free thermal strain $3.3\times 10^{-4}$: because expansion in $x$ is completely prevented, the Poisson coupling lets the plate grow further in $y$ than it would if it were free.
Compute the stresses. Only the difference between the total strain and the free thermal strain produces stress:
$$\{\sigma\}=[D]\left(\{\varepsilon\}-\{\varepsilon_{0}\}\right) =\begin{Bmatrix}32000(-3.3)+8000(0.825)\\ 32000(0.825)+8000(-3.3)\\ 0\end{Bmatrix} \times 10^{-4}$$ $$\boxed{\sigma_{x}=-9.9\ \text{N/mm}^{2},\qquad \sigma_{y}=0,\qquad \tau_{xy}=0}$$ so the principal stresses are $0$ and $-9.9$ $\text{N/mm}^{2}$ on the $x$ and $y$ axes, and the von Mises stress is $9.9$ $\text{N/mm}^{2}$.
Interpret the result. The answer is a pure uniaxial compression, and it is exactly the fully restrained one-dimensional value:
$$\sigma_{x}=-E\alpha\Delta T=-30000\times 11\times 10^{-6}\times 30 =-9.9\ \text{N/mm}^{2}$$ That is not a coincidence. A single CST clamped along a straight edge has just one free node, so it can only stretch in one direction; the mesh is far too coarse to represent the true two-dimensional field, and the model degenerates into a bar restrained in $x$ and free in $y$. For comparison, a fully restrained plate would carry the biaxial value $-E\alpha\Delta T/(1-\nu)=-13.2$ $\text{N/mm}^{2}$.
Check nodal equilibrium. The reactions follow from $\{r\}=[k]\{d\}-\{f\}$ and come out as a self-equilibrating horizontal pair on the clamped base, with no vertical reaction at all:
$$r_{x1}=+222.75\ \text{N},\qquad r_{x2}=-222.75\ \text{N},\qquad \sum F_{x}=\sum F_{y}=0$$ This is consistent with the stress field: the resultant of $\sigma_{x}$ over the 45 mm vertical edge is $9.9\times 1\times 45=445.5$ N, and the base delivers it as a couple of two $222.75$ N forces.
Comment on local equilibrium. Inside the element the stress is constant, so every derivative in $\sigma_{ij,j}+b_{i}=0$ vanishes and the interior differential equations are satisfied identically — trivially, not accurately. The traction boundary conditions are a different story. Both the vertical edge 1–3 and the hypotenuse 3–2 are free surfaces, so the traction $t_{i}=\sigma_{ij}n_{j}$ on each ought to be zero:
$$\text{edge }1\text{-}3\ \left(\mathbf{n}=(-1,0)\right):\quad \sigma_{n}=-9.9\ \text{N/mm}^{2},\ \tau=0$$ $$\text{edge }3\text{-}2\ \left(\mathbf{n}=(0.8742,\,0.4856)\right):\quad \left|\mathbf{t}\right|=8.654,\ \sigma_{n}=-7.565,\ \tau=-4.203\ \text{N/mm}^{2}$$ A residual of $9.9$ $\text{N/mm}^{2}$ on a surface that should carry nothing is the entire computed stress, and $8.65$ $\text{N/mm}^{2}$ on the hypotenuse is almost as bad. That is the honest verdict: one CST satisfies interior equilibrium for free and violates the free-surface conditions completely, and only mesh refinement drives those residuals down.

Problem 3 — results
Quantity	Symbol	Value
Element area	$A$	$562.5$ $\text{mm}^{2}$
Plane-stress constants	$H_{1}$, $H_{2}$, $G$	$32{,}000$ / $8{,}000$ / $12{,}000$ $\text{N/mm}^{2}$
Free thermal strain	$\alpha\Delta T$	$3.3\times 10^{-4}$
Thermal load vector	$\{f\}$	$\{-297,-165,297,0,0,165\}$ N
Apex displacement	$u_{3}$, $v_{3}$	$0$ and $+0.0185625$ mm
Total strains	$\varepsilon_{x},\varepsilon_{y},\gamma_{xy}$	$0$, $4.125\times 10^{-4}$, $0$
Stresses	$\sigma_{x},\sigma_{y},\tau_{xy}$	$-9.9$, $0$, $0$ $\text{N/mm}^{2}$
Principal stresses	$\sigma_{1},\sigma_{2}$	$0$ and $-9.9$ $\text{N/mm}^{2}$
von Mises stress	$\sigma_{e}$	$9.9$ $\text{N/mm}^{2}$
Base reactions	$r_{x1}$, $r_{x2}$	$+222.75$ and $-222.75$ N
Spurious traction, free edge 1–3	$\sigma_{n}$	$-9.9$ $\text{N/mm}^{2}$
Spurious traction, free hypotenuse	$\sigma_{n}$, $\tau$	$-7.565$ and $-4.203$ $\text{N/mm}^{2}$

Topic. The constant-strain triangle under an initial (thermal) strain: equivalent nodal loads from $\{\varepsilon_{0}\}$, stress recovery as $[D](\{\varepsilon\}-\{\varepsilon_{0}\})$, and what a one-element mesh can and cannot tell you.

Key relations.

$$\{f\}_{\Delta T}=t\,A\,[B]^{T}[D]\{\varepsilon_{0}\},\qquad \{\varepsilon_{0}\}=\alpha\Delta T\{1,1,0\}^{T}$$ $$\{\sigma\}=[D]\left([B]\{d\}-\{\varepsilon_{0}\}\right),\qquad t_{i}=\sigma_{ij}n_{j}$$

Why this works. Thermal expansion is an initial strain, not a force, so it enters the variational statement through $\int[B]^{T}[D]\{\varepsilon_{0}\}dV$ and is handled exactly like any other load once converted. Subtracting $\{\varepsilon_{0}\}$ before multiplying by $[D]$ is what makes an unrestrained heated body come out stress free, which is the check to run on any thermal model.

Common pitfall. Writing $\{\sigma\}=[D][B]\{d\}$ and forgetting the initial-strain subtraction, which here would return $\sigma_{y}=+13.2$ $\text{N/mm}^{2}$ instead of zero — a tensile stress in a heated free direction, which is physically impossible. A second trap is numbering the nodes clockwise, which makes the signed area negative and flips the sign of every entry of $[B]$; the third is asserting that a constant-strain element “violates equilibrium” without saying which equilibrium, since the interior equations are satisfied exactly and only the traction conditions fail.

Source. Logan, A First Course in the Finite Element Method, 6th ed., Ch. 6 (the CST, $[B]$ and $[D]$) and §15.2 (thermal stress and the initial-strain load vector); Cook, Malkus, Plesha & Witt, 4th ed., §3.5 and §4.1; Appendix A of this examination paper for the $H_{1}$, $H_{2}$, $G$ form of $[D]$.

Question 3 of 3: Thermal stresses in a thin triangular plate modelled by one CST

Problem 3: Thermal stresses in a thin triangular plate modelled by one CST (one of three equal-value problems)