22-Mec-B9 Advanced Engineering Structures · Undated paper
Question 7 of 8: Orthotropic lamina: reduced stiffness, ply transformation and stresses
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Mec-B9 Advanced Engineering Structures, national examination, 3 hours, open book, non-communicating calculators permitted. Eight questions are printed; the rubric states that any five constitute a complete paper and that all problems are of equal total value (20 marks each). All eight are worked below. The mark split inside each question is the one printed on the paper.
Reference texts. T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. (thin-walled open and closed sections, shear flow, shear centre, multi-cell torsion, fatigue and fracture, laminated composites); A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. (yield criteria, torsion of non-circular prismatic bars); N. E. Dowling, Mechanical Behavior of Materials, 4th ed. (Coffin–Manson, Miner’s rule, Paris law); R. M. Jones, Mechanics of Composite Materials, 2nd ed. (reduced stiffness matrix, ply transformation).
Find. (a) the entries of the 0° lamina stiffness matrix; (b) the transformed stiffness matrix for a 30° ply; (c) the stresses $\sigma_1$, $\sigma_2$, $\tau_{12}$ produced by the given strains.
Figure 7.1 — the 30° ply: material axes 1 (along the fibres) and 2 (transverse) rotated anticlockwise from the laminate axes x, y.
Approach. Build the plane-stress reduced stiffness matrix from the four engineering constants, rotate it with the standard fourth-order transformation for part (b), and use the unrotated matrix for part (c) because the strains are given in the material axes.
Part (a) — the minor Poisson’s ratio and the reciprocity check. Orthotropic reciprocity gives $$\nu_{21} = \nu_{12}\frac{E_{22}}{E_{11}} = 0.29 \times \frac{29}{215} = 0.03912,$$ so $1 - \nu_{12}\nu_{21} = 0.988656$. The near-unity denominator is characteristic of a stiff unidirectional composite, where the transverse modulus is an order of magnitude below the longitudinal one.
Assemble the reduced stiffness matrix. For a thin lamina in plane stress the stiffness that matters is the reduced stiffness $[Q]$, not a full three-dimensional $[C]$ — a 3-D $[C]$ would additionally require $E_{33}$, $G_{13}$ and $G_{23}$, which are not supplied. Its entries are $$Q_{11} = \frac{E_{11}}{1-\nu_{12}\nu_{21}},\quad Q_{22} = \frac{E_{22}}{1-\nu_{12}\nu_{21}},\quad Q_{12} = \frac{\nu_{12}E_{22}}{1-\nu_{12}\nu_{21}},\quad Q_{66} = G_{12}.$$ Substituting, $$\boxed{\;Q_{11} = 217.47,\; Q_{22} = 29.333,\; Q_{12} = 8.506,\; Q_{66} = 19.0 \text{ GPa}\;}$$ with $Q_{16} = Q_{26} = 0$ because the material axes are also axes of orthotropic symmetry.
Part (b) — the transformation. Writing $c = \cos 30^{\circ}$ and $s = \sin 30^{\circ}$, the transformed (laminate-axis) stiffnesses are the standard fourth-order combinations $$\bar{Q}_{11} = Q_{11}c^4 + 2(Q_{12}+2Q_{66})s^2c^2 + Q_{22}s^4,$$ $$\bar{Q}_{12} = (Q_{11}+Q_{22}-4Q_{66})s^2c^2 + Q_{12}(c^4+s^4),$$ $$\bar{Q}_{22} = Q_{11}s^4 + 2(Q_{12}+2Q_{66})s^2c^2 + Q_{22}c^4,$$ $$\bar{Q}_{16} = (Q_{11}-Q_{12}-2Q_{66})sc^3 + (Q_{12}-Q_{22}+2Q_{66})s^3c,$$ $$\bar{Q}_{26} = (Q_{11}-Q_{12}-2Q_{66})s^3c + (Q_{12}-Q_{22}+2Q_{66})sc^3,$$ $$\bar{Q}_{66} = (Q_{11}+Q_{22}-2Q_{12}-2Q_{66})s^2c^2 + Q_{66}(s^4+c^4).$$
Evaluate at 30°. With $c^2 = 0.75$ and $s^2 = 0.25$ the six expressions give, in GPa,
column 1
column 2
column 6
row 1
141.60
37.34
57.38
row 2
37.34
47.53
24.08
row 6
57.38
24.08
47.83
Verify the transformation in one line. A 30° ply is fully populated: the shear–extension coupling terms $\bar{Q}_{16}$ and $\bar{Q}_{26}$ are non-zero, so an axial strain in the laminate axes produces shear stress. The cheapest check that the six expressions were typed correctly is to evaluate them at 45°, where theory demands exactly $\bar{Q}_{16} = \bar{Q}_{26} = (Q_{11}-Q_{22})/4$; the transformation returns 47.034 GPa against $(Q_{11}-Q_{22})/4 = 47.034$ GPa, an exact match.
Part (c) — stresses from the given strains. The strains are labelled $\epsilon_1$, $\epsilon_2$, $\gamma_{12}$, i.e. they are referred to the material axes, so the unrotated $[Q]$ applies and no transformation is involved: $$\begin{aligned} \sigma_1 &= Q_{11}\epsilon_1 + Q_{12}\epsilon_2 = 217.47(0.006) + 8.506(0.004) = 1338.8 \text{ MPa},\\ \sigma_2 &= Q_{12}\epsilon_1 + Q_{22}\epsilon_2 = 8.506(0.006) + 29.333(0.004) = 168.37 \text{ MPa},\\ \tau_{12} &= Q_{66}\gamma_{12} = 19.0(-0.0005) = -9.50 \text{ MPa}. \end{aligned}$$ $$\boxed{\;\sigma_1 = 1338.8 \text{ MPa},\quad \sigma_2 = 168.37 \text{ MPa},\quad \tau_{12} = -9.50 \text{ MPa}\;}$$
Sanity-check the magnitudes. The fibre-direction stress is about eight times the transverse one even though the fibre-direction strain is only 1.5 times the transverse strain, which is exactly what an $E_{11}/E_{22}$ ratio of 7.4 should produce. Note also that $\sigma_2$ is dominated by its own $Q_{22}\epsilon_2$ term but receives a 30 % contribution from the Poisson coupling to $\epsilon_1$ — dropping $Q_{12}$ would be a visible error.
