Orthotropic lamina stiffness and its transformation: 22-Mec-B9 Advanced Engineering Structures, May 2017, Q3

Question 3: Orthotropic lamina stiffness and its transformation (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 single orthotropic lamina characterised by four in-plane engineering constants, to be used at four different fibre orientations.

Given data — lamina engineering constants
Property	Symbol	Value
Longitudinal (fibre-direction) modulus	$E_{11}$	205 GPa
Transverse modulus	$E_{22}$	25 GPa
In-plane shear modulus	$G_{12}$	15 GPa
Major Poisson ratio	$\nu_{12}$	0.28
Applied strains, part (d)	$\epsilon_x,\ \epsilon_y,\ \gamma_{xy}$	0.006, 0.004, $-0.0005$

Find. (a) the on-axis reduced stiffness matrix; (b) and (c) the transformed stiffness matrices for plies at 30° and 75°; (d) the laminate-axis stresses in a 90° ply under the stated strains.

Ply nomenclature. Axes 1–2 are the material (fibre and transverse) directions; axes $x$–$y$ are the fixed laminate axes. $\theta$ is measured from $x$ to the fibre direction, shown here at 30°.

Approach. Build the plane-stress reduced stiffness from the four engineering constants, then rotate it into the laminate axes with the standard fourth-order transformation, and finally multiply the 90° transformed matrix by the applied strain vector.

Part (a) — recognise which stiffness matrix the data can support. Only $E_{11}$, $E_{22}$, $G_{12}$ and $\nu_{12}$ are given, which are the four independent in-plane constants of an orthotropic lamina. A full three-dimensional $[C]$ would also need $E_{33}$, $G_{13}$, $G_{23}$ and two more Poisson ratios, none of which is supplied, so the matrix the question is asking for is the plane-stress reduced stiffness $[Q]$ of the 0° ply.
Fix the minor Poisson ratio by reciprocity. Orthotropic symmetry requires $\nu_{12}/E_{11} = \nu_{21}/E_{22}$, so $$\nu_{21} = \nu_{12}\frac{E_{22}}{E_{11}} = 0.28 \times \frac{25}{205} = 0.03415.$$ The denominator common to every stiffness term follows as $1 - \nu_{12}\nu_{21} = 1 - 0.009561 = 0.990439$.
Assemble the on-axis reduced stiffness. With $Q_{11} = E_{11}/(1-\nu_{12}\nu_{21})$, $Q_{22} = E_{22}/(1-\nu_{12}\nu_{21})$, $Q_{12} = \nu_{12}E_{22}/(1-\nu_{12}\nu_{21})$ and $Q_{66} = G_{12}$, $$\boxed{[Q]_{0^\circ} = \begin{bmatrix} 206.98 & 7.068 & 0 \\ 7.068 & 25.241 & 0 \\ 0 & 0 & 15.00 \end{bmatrix} \ \text{GPa}}.$$ The zeros in the 16 and 26 positions are the signature of an on-axis orthotropic ply: stretching it along a material axis produces no shear.
Part (b) — set up the transformation. Rotating the reduced stiffness through an angle $\theta$ from the material axes to the laminate axes gives, with $c = \cos\theta$ and $s = \sin\theta$, $$\bar{Q}_{11} = Q_{11}c^4 + 2(Q_{12}+2Q_{66})s^2c^2 + Q_{22}s^4,$$ $$\bar{Q}_{12} = (Q_{11}+Q_{22}-4Q_{66})s^2c^2 + Q_{12}(s^4+c^4),$$ $$\bar{Q}_{22} = Q_{11}s^4 + 2(Q_{12}+2Q_{66})s^2c^2 + Q_{22}c^4,$$ $$\bar{Q}_{16} = (Q_{11}-Q_{12}-2Q_{66})sc^3 + (Q_{12}-Q_{22}+2Q_{66})s^3c,$$ $$\bar{Q}_{26} = (Q_{11}-Q_{12}-2Q_{66})s^3c + (Q_{12}-Q_{22}+2Q_{66})sc^3,$$ $$\bar{Q}_{66} = (Q_{11}+Q_{22}-2Q_{12}-2Q_{66})s^2c^2 + Q_{66}(s^4+c^4).$$ Note that these are fourth-order in $\theta$, not second order like a stress transformation.
Evaluate the 30° ply. With $c = 0.8660$ and $s = 0.5000$, $$\boxed{[\bar{Q}]_{30^\circ} = \begin{bmatrix} 131.90 & 36.71 & 56.46 \\ 36.71 & 41.03 & 22.23 \\ 56.46 & 22.23 & 44.64 \end{bmatrix} \ \text{GPa}}.$$ The off-axis terms $\bar{Q}_{16}$ and $\bar{Q}_{26}$ are no longer zero, which is the shear–extension coupling that makes a single off-axis ply twist when it is pulled.
Part (c) — evaluate the 75° ply. With $c = 0.2588$ and $s = 0.9659$ the same six expressions give $$\boxed{[\bar{Q}]_{75^\circ} = \begin{bmatrix} 27.54 & 16.95 & 5.604 \\ 16.95 & 184.92 & 39.83 \\ 5.604 & 39.83 & 24.88 \end{bmatrix} \ \text{GPa}}.$$ At 75° the ply is already close to transverse: $\bar{Q}_{22}$ has climbed to 89 % of $Q_{11}$ while $\bar{Q}_{11}$ has fallen to 13 % of it. A one-line check that the transformation has been typed correctly is available at 45°, where every term collapses to a quarter-weighted average and $\bar{Q}_{16} = \bar{Q}_{26} = (Q_{11}-Q_{22})/4 = 45.43$ GPa exactly.
Part (d) — write down the 90° matrix by inspection. At $\theta = 90^\circ$, $c = 0$ and $s = 1$, so the transformation reduces to a simple interchange of the 1 and 2 directions with the coupling terms vanishing again: $\bar{Q}_{11} = Q_{22} = 25.24$, $\bar{Q}_{22} = Q_{11} = 206.98$, $\bar{Q}_{12} = Q_{12} = 7.068$, $\bar{Q}_{66} = 15.00$ GPa, and $\bar{Q}_{16} = \bar{Q}_{26} = 0$.
Multiply through to get the ply stresses. With $\{\sigma\} = [\bar{Q}]\{\epsilon\}$ and the strains 0.006, 0.004 and $-0.0005$, $$\sigma_x = (25.24)(0.006) + (7.068)(0.004) = 0.1797\ \text{GPa},$$ $$\sigma_y = (7.068)(0.006) + (206.98)(0.004) = 0.8703\ \text{GPa},$$ $$\tau_{xy} = (15.00)(-0.0005) = -0.0075\ \text{GPa},$$ that is $$\boxed{\sigma_x = 179.7\ \text{MPa}, \quad \sigma_y = 870.3\ \text{MPa}, \quad \tau_{xy} = -7.50\ \text{MPa}}.$$

The very large $\sigma_y$ is correct and is the physical point of part (d): in a 90° ply the laminate $y$-axis is the fibre direction, so the modest transverse strain of 0.004 is resisted by the full 205 GPa fibre stiffness. The same strain field applied to a 0° ply would put 1270 MPa in $\sigma_x$ and only 143 MPa in $\sigma_y$ — the ply orientation, not the strain, decides where the load goes.

Final results
Quantity	Symbol	Value
Minor Poisson ratio	$\nu_{21}$	0.03415
(a) $Q_{11}$, $Q_{22}$	—	206.98, 25.241 GPa
(a) $Q_{12}$, $Q_{66}$	—	7.068, 15.00 GPa
(b) 30° ply, $\bar{Q}_{11}/\bar{Q}_{22}/\bar{Q}_{12}$	—	131.90 / 41.03 / 36.71 GPa
(b) 30° ply, $\bar{Q}_{66}/\bar{Q}_{16}/\bar{Q}_{26}$	—	44.64 / 56.46 / 22.23 GPa
(c) 75° ply, $\bar{Q}_{11}/\bar{Q}_{22}/\bar{Q}_{12}$	—	27.54 / 184.92 / 16.95 GPa
(c) 75° ply, $\bar{Q}_{66}/\bar{Q}_{16}/\bar{Q}_{26}$	—	24.88 / 5.604 / 39.83 GPa
(d) 90° ply stresses	$\sigma_x,\ \sigma_y,\ \tau_{xy}$	179.7, 870.3, $-7.50$ MPa

Topic. Plane-stress stiffness of an orthotropic lamina and its rotation into laminate axes.

Key relations. $Q_{11} = E_{11}/(1-\nu_{12}\nu_{21})$, $Q_{22} = E_{22}/(1-\nu_{12}\nu_{21})$, $Q_{12} = \nu_{12}E_{22}/(1-\nu_{12}\nu_{21})$, $Q_{66} = G_{12}$, with $\nu_{21} = \nu_{12}E_{22}/E_{11}$; the $[\bar{Q}]$ terms are fourth-order in $\sin\theta$ and $\cos\theta$.

Why this works. A thin lamina is loaded only in its own plane, so $\sigma_3 = \tau_{13} = \tau_{23} = 0$ and the three-dimensional compliance collapses to a 3×3 system in $\sigma_1$, $\sigma_2$, $\tau_{12}$. Inverting that 3×3 compliance gives the reduced stiffness $[Q]$ — "reduced" because it is what survives the plane-stress assumption, and it is not simply the corresponding block of the full three-dimensional $[C]$. Rotating a fourth-rank tensor is what produces the fourth powers of $\sin\theta$ and $\cos\theta$ in the transformation.

Common pitfall. Quoting the reduced stiffness as if it were the three-dimensional $[C]$, and dropping the $(1-\nu_{12}\nu_{21})$ denominator so that $Q_{11}$ is reported as $E_{11}$. Also very common: transforming with the second-order (Mohr-circle) rules used for stress instead of the fourth-order rules required for stiffness.

Source. R. M. Jones, Mechanics of Composite Materials, 2nd ed., Ch. 2.5–2.6 (reduced stiffnesses and the transformed reduced stiffness matrix).

Question 3 of 8: Orthotropic lamina stiffness and its transformation

Question 3: Orthotropic lamina stiffness and its transformation (20 marks)