Question 2 of 8: Stiffness matrices of an orthotropic lamina

Question 2: Stiffness matrices of an orthotropic lamina (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 unidirectional lamina with four independent in-plane elastic constants:

Find. The four independent entries of the reduced stiffness matrix in material axes, the transformed stiffness matrices for 90° and 45° plies, and the stresses generated in a 90° ply by the stated strain state.

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

1. Recover the minor Poisson ratio and the common denominator. Reciprocity fixes$$\nu_{21} = \nu_{12}\frac{E_{22}}{E_{11}} = 0.30\left(\frac{19}{190}\right) = 0.0300,$$so the denominator common to every stiffness term is $1 - \nu_{12}\nu_{21} = 1 - 0.30(0.030) = 0.991$. Only four constants are supplied, which is exactly the number a plane-stress lamina needs; a full three-dimensional $[C]$ would also require $E_{33}$, $G_{13}$ and $G_{23}$, so the matrix asked for is the reduced (plane-stress) stiffness.
2. Assemble the 0° lamina stiffness matrix (part a). With$$Q_{11}=\frac{E_{11}}{1-\nu_{12}\nu_{21}},\quad Q_{22}=\frac{E_{22}}{1-\nu_{12}\nu_{21}},\quad Q_{12}=\frac{\nu_{12}E_{22}}{1-\nu_{12}\nu_{21}},\quad Q_{66}=G_{12},$$the numbers are $Q_{11} = 190/0.991 = 191.73$, $Q_{22} = 19/0.991 = 19.173$, $Q_{12} = 0.3(19)/0.991 = 5.752$ and $Q_{66} = 12.00$, all in GPa, giving$$\boxed{[C]_{0^\circ} = \begin{bmatrix} 191.73 & 5.752 & 0\\ 5.752 & 19.173 & 0\\ 0 & 0 & 12.00\end{bmatrix}\ \text{GPa}} .$$The zeros in the third row and column are the signature of orthotropy in material axes: extension and shear do not couple.
3. Rotate to a 90° ply (part b). The transformation$$\bar{Q}_{11}=Q_{11}c^{4}+2(Q_{12}+2Q_{66})s^{2}c^{2}+Q_{22}s^{4},\qquad c=\cos\theta,\ s=\sin\theta$$(with the companion expressions for $\bar{Q}_{12},\bar{Q}_{22},\bar{Q}_{16},\bar{Q}_{26},\bar{Q}_{66}$) collapses at $\theta = 90^\circ$, where $c = 0$ and $s = 1$, to a simple interchange of the 1 and 2 directions:$$\boxed{[\bar{Q}]_{90^\circ} = \begin{bmatrix} 19.173 & 5.752 & 0\\ 5.752 & 191.73 & 0\\ 0 & 0 & 12.00\end{bmatrix}\ \text{GPa}} .$$The shear-coupling terms stay zero because 90° is still a principal material direction.
4. Rotate to a 45° ply (part c). Now $c^{2} = s^{2} = \tfrac12$, so $c^{4} = s^{4} = s^{2}c^{2} = \tfrac14$ and every term reduces to a quarter-weighted average. For instance$$\bar{Q}_{11} = \tfrac14 Q_{11} + \tfrac12 (Q_{12}+2Q_{66}) + \tfrac14 Q_{22} = \tfrac14(191.73) + \tfrac12(29.752) + \tfrac14(19.173) = 67.60\ \text{GPa},$$and the shear-coupling terms become $\bar{Q}_{16} = \bar{Q}_{26} = \tfrac14 (Q_{11}-Q_{22}) = \tfrac14(191.73-19.17) = 43.14$ GPa. Collecting all six,$$\boxed{[\bar{Q}]_{45^\circ} = \begin{bmatrix} 67.60 & 43.60 & 43.14\\ 43.60 & 67.60 & 43.14\\ 43.14 & 43.14 & 49.85\end{bmatrix}\ \text{GPa}} .$$The matrix is now fully populated: a 45° ply stretched along $x$ also wants to shear, which is precisely why single off-axis plies are never used alone.
5. Apply the strain state to the 90° ply (part d). Multiplying $[\bar{Q}]_{90^\circ}$ (in MPa) by $\{0.0007,\ 0.004,\ -0.0015\}$,$$\sigma_x = 19\,173(0.0007) + 5\,752(0.004) = 13.42 + 23.01 = 36.43\ \text{MPa},$$$$\sigma_y = 5\,752(0.0007) + 191\,726(0.004) = 4.03 + 766.90 = 770.93\ \text{MPa},$$$$\tau_{xy} = 12\,000(-0.0015) = -18.00\ \text{MPa},$$so$$\boxed{\sigma_x = 36.4\ \text{MPa},\quad \sigma_y = 770.9\ \text{MPa},\quad \tau_{xy} = -18.0\ \text{MPa}} .$$In a 90° ply the laminate $y$-axis is the fibre direction, which is why the modest 0.004 strain along $y$ generates by far the largest stress.