Constant strain triangle in plane strain — element and principal stresses: 16-Civ-B9 The Finite Element Method, May 2013, Q3

Problem 3: Constant strain triangle in plane strain — element 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. One three-noded constant strain triangle in plane strain.

Quantity	Symbol	Value
Node 1 coordinates	$(x_1,y_1)$	(5, 5) mm
Node 2 coordinates	$(x_2,y_2)$	(25, 5) mm
Node 3 coordinates	$(x_3,y_3)$	(15, 15) mm
Nodal displacements, node 1	$u_1,v_1$	0.005 mm, 0.002 mm
Nodal displacements, node 2	$u_2,v_2$	0, 0
Nodal displacements, node 3	$u_3,v_3$	0.005 mm, 0
Young's modulus	$E$	70 GPa
Poisson's ratio	$\nu$	0.3
State	—	plane strain

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

Figure 3.1 — the constant strain triangle of Figure 3, with the nodal displacement vectors drawn to a greatly exaggerated scale. Node 2 does not move; nodes 1 and 3 both move 0.005 mm in x, and node 1 also rises 0.002 mm.

Approach. Evaluate the printed gradient matrix from the nodal coordinates, multiply it by the nodal displacement vector to get the (constant) strains, apply the plane strain constitutive matrix, and then rotate to the principal axes with the standard Mohr construction.

Twice the element area, and the gradient coefficients. For a triangle numbered anticlockwise, $$2A=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$ $$2A=5(5-15)+25(15-5)+15(5-5)=-50+250+0=200\ \text{mm}^{2}$$ so $A=100\ \text{mm}^{2}$, and the sign being positive confirms the anticlockwise numbering that the printed $[B]$ assumes. The coefficients are then read straight off the definitions $\beta_{ij}=y_i-y_j$ and $\alpha_{ij}=x_i-x_j$: $$\beta_{23}=5-15=-10,\quad \beta_{31}=15-5=10,\quad \beta_{12}=5-5=0$$ $$\alpha_{32}=15-25=-10,\quad \alpha_{13}=5-15=-10,\quad \alpha_{21}=25-5=20$$ all in millimetres.
Assemble the gradient matrix. Substituting into the printed form, $$[B]=\frac{1}{200}\begin{bmatrix} -10 & 0 & 10 & 0 & 0 & 0\\ 0 & -10 & 0 & -10 & 0 & 20\\ -10 & -10 & -10 & 10 & 20 & 0 \end{bmatrix}\ \text{mm}^{-1}$$ Every entry is a constant, which is the defining property of the element: the strain it can represent does not vary with position.
Element strains. With $\{q\}^{T}=\{0.005,\ 0.002,\ 0,\ 0,\ 0.005,\ 0\}$ mm, the product $\{\varepsilon\}=[B]\{q\}$ gives, row by row, $$\varepsilon_x=\frac{(-10)(0.005)+(10)(0)+(0)(0.005)}{200}=\frac{-0.050}{200} =-2.50\times10^{-4}$$ $$\varepsilon_y=\frac{(-10)(0.002)+(-10)(0)+(20)(0)}{200}=\frac{-0.020}{200} =-1.00\times10^{-4}$$ $$\gamma_{xy}=\frac{(-10)(0.005)+(-10)(0.002)+(20)(0.005)}{200}=\frac{0.030}{200} =+1.50\times10^{-4}$$ Both direct strains are compressive; the element is being squeezed and sheared at once.
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 $E=70\,000$ MPa and $\nu=0.3$ the leading coefficient is $$\frac{70\,000}{(1.3)(0.4)}=134\,615.4\ \text{MPa}$$ so $[D]$ has diagonal terms $0.7$, off-diagonal $0.3$ and shear term $0.2$ times that coefficient. Note that the plane strain matrix, unlike plane stress, has no $1-\nu^{2}$ denominator and stiffens sharply as $\nu$ approaches 0.5.
Element stresses. Carrying out $\{\sigma\}=[D]\{\varepsilon\}$, $$\sigma_x=134\,615.4\left[0.7(-2.50\times10^{-4})+0.3(-1.00\times10^{-4})\right]$$ $$\sigma_y=134\,615.4\left[0.3(-2.50\times10^{-4})+0.7(-1.00\times10^{-4})\right]$$ $$\tau_{xy}=134\,615.4\,(0.2)(1.50\times10^{-4})$$ which evaluate to $$\boxed{\sigma_x=-27.596\ \text{MPa},\quad \sigma_y=-19.519\ \text{MPa},\quad \tau_{xy}=+4.038\ \text{MPa}}$$
The out-of-plane stress that plane strain generates. Plane strain sets $\varepsilon_z=0$, which is not the same as $\sigma_z=0$; the restrained direction picks up $$\sigma_z=\nu(\sigma_x+\sigma_y)=0.3(-27.596-19.519)=-14.135\ \text{MPa}$$ This is a genuine third principal stress and it must be carried into any yield check.
In-plane principal stresses. The Mohr circle has centre and radius $$\begin{aligned}\sigma_{\text{avg}}&=\frac{\sigma_x+\sigma_y}{2}=-23.558\ \text{MPa} \\ R&=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}\end{aligned}$$ $$R=\sqrt{(-4.0385)^{2}+(4.0385)^{2}}=5.711\ \text{MPa}$$ so that $$\boxed{\begin{aligned}\sigma_1&=-17.846\ \text{MPa} \\ \sigma_2&=-29.269\ \text{MPa}\end{aligned}}$$ As a check, $\sigma_1+\sigma_2=-47.115$ MPa reproduces $\sigma_x+\sigma_y$, the first stress invariant.
Orientation and maximum shear. The principal directions follow from $$\tan 2\theta_p=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}=\frac{8.077}{-8.077}=-1 \;\Longrightarrow\; 2\theta_p=135^{\circ},\quad \theta_p=67.5^{\circ}$$ measured anticlockwise from the $x$ axis to the axis of $\sigma_1$. The maximum in-plane shear equals the circle radius, $5.711$ MPa, but because the third principal stress is $-14.135$ MPa the absolute maximum shear is larger: $$\tau_{\max}=\frac{\sigma_{\max}-\sigma_{\min}}{2} =\frac{-14.135-(-29.269)}{2}=7.567\ \text{MPa}$$ Ordering all three, $\sigma_z=-14.135 > \sigma_1=-17.846 > \sigma_2=-29.269$ MPa.

Figure 3.2 — Mohr circle for the in-plane stress state. Both in-plane principal stresses are compressive; the circle lies entirely to the left of the shear axis.

Result	Symbol	Value
Element area	$A$	100 mm$^{2}$
Direct strain, x	$\varepsilon_x$	$-2.50\times10^{-4}$
Direct strain, y	$\varepsilon_y$	$-1.00\times10^{-4}$
Shear strain	$\gamma_{xy}$	$+1.50\times10^{-4}$
Direct stress, x	$\sigma_x$	−27.596 MPa
Direct stress, y	$\sigma_y$	−19.519 MPa
Shear stress	$\tau_{xy}$	+4.038 MPa
Out-of-plane stress	$\sigma_z$	−14.135 MPa
Major in-plane principal stress	$\sigma_1$	−17.846 MPa
Minor in-plane principal stress	$\sigma_2$	−29.269 MPa
Principal direction	$\theta_p$	67.5$^{\circ}$ anticlockwise from x
Maximum in-plane shear	$R$	5.711 MPa
Absolute maximum shear	$\tau_{\max}$	7.567 MPa

Topic. The constant strain triangle in plane strain — from nodal displacements to principal stresses in one matrix chain.

Key relations. $\{\varepsilon\}=[B]\{q\}$ with $[B]$ as printed and $2A$ the signed twice-area; $\{\sigma\}=[D]\{\varepsilon\}$ with the plane strain $[D]$; $\sigma_z=\nu(\sigma_x+\sigma_y)$; Mohr circle $\sigma_{1,2}=\sigma_{\text{avg}}\pm R$.

Why this works. The CST assumes a linear displacement field over the element, so its derivatives — the strains — are constant, and $[B]$ contains nothing but geometry. That makes the whole calculation a pair of matrix products with no integration anywhere. Plane strain is the right idealisation when the body is long in $z$ and restrained there (a dam, a retaining wall, a long tunnel lining), and the price of that restraint is the induced $\sigma_z$, which is why the third principal stress matters even though the question is posed in two dimensions.

Common pitfall. Numbering the nodes clockwise makes $2A$ negative and silently flips the sign of every stress; always check that $2A>0$ before going on. The other frequent slip is to use the plane stress matrix $E/(1-\nu^{2})$ by habit, which here would give stresses about 12 per cent low and would lose $\sigma_z$ altogether. Finally, the shear row of $[B]$ interleaves the $\alpha$ and $\beta$ terms — copying it as two separate rows is a common transcription error.

Source. Logan, A First Course in the Finite Element Method, 6th ed., Ch. 6 (CST formulation, plane stress and plane strain matrices); Cook, Malkus, Plesha and Witt, Concepts and Applications of Finite Element Analysis, 4th ed., Ch. 3.

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

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