Question 8 of 8: Stresses in a unidirectional carbon/epoxy laminate at 0° and +45°

Question 8: Stresses in a unidirectional carbon/epoxy laminate at 0° and +45° (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. Every ply of the laminate carries the same fibre direction, so the stack behaves as one homogeneous orthotropic sheet in plane stress and the lamina constitutive law can be applied directly to the laminate strains:

Find. The direct stresses σ_x and σ_y and the shear stress τ_xy produced by that strain state, first with the fibres along x and then with the fibres rotated to +45°.

Approach. Build the plane-stress reduced stiffness matrix of the lamina from the four engineering constants, apply it directly for the 0° case, then transform it through 45° to obtain the stiffnesses in the loading axes and apply it again.

1. Find the minor Poisson ratio from the reciprocal relation. The compliance matrix of an orthotropic lamina is symmetric, which requires $$\frac{\nu_{12}}{E_1} = \frac{\nu_{21}}{E_2} \quad\Longrightarrow\quad \nu_{21} = \nu_{12}\frac{E_2}{E_1} = 0.30\times\frac{15}{190} = 0.02368$$ The two Poisson ratios are emphatically not equal, and the small one is the physically sensible result: stretching a stiff fibre-dominated direction contracts the compliant transverse direction a lot, whereas stretching the transverse direction barely disturbs the fibres. The denominator that recurs throughout is $$1 - \nu_{12}\nu_{21} = 1 - 0.30(0.02368) = 0.99289$$
2. Assemble the reduced stiffness matrix. For plane stress in the material axes, $$Q_{11} = \frac{E_1}{1-\nu_{12}\nu_{21}},\quad Q_{22} = \frac{E_2}{1-\nu_{12}\nu_{21}},\quad Q_{12} = \frac{\nu_{12}E_2}{1-\nu_{12}\nu_{21}},\quad Q_{66} = G_{12}$$ Substituting the given constants, $$Q_{11} = \frac{190}{0.99289} = 191.36,\quad Q_{22} = \frac{15}{0.99289} = 15.107$$ $$Q_{12} = \frac{0.30(15)}{0.99289} = 4.532,\quad Q_{66} = 10.00 \qquad \text{(all GPa)}$$
3. Part (a): apply the law with the fibres along x. At θ = 0° the material axes and the loading axes coincide, so no transformation is needed and $$\begin{bmatrix}\sigma_x\\ \sigma_y\\ \tau_{xy}\end{bmatrix} = \begin{bmatrix}Q_{11} & Q_{12} & 0\\ Q_{12} & Q_{22} & 0\\ 0 & 0 & Q_{66}\end{bmatrix}\begin{bmatrix}\varepsilon_x\\ \varepsilon_y\\ \gamma_{xy}\end{bmatrix}$$ Working the three rows in turn, with the stiffnesses in GPa and the strains dimensionless so that the products come out in GPa and are reported in MPa: $$\sigma_x = 191.36(300\times10^{-6}) + 4.532(100\times10^{-6}) = 0.05741 + 0.00045\ \text{GPa}$$ $$\sigma_y = 4.532(300\times10^{-6}) + 15.107(100\times10^{-6}) = 0.00136 + 0.00151\ \text{GPa}$$ $$\tau_{xy} = 10.00(75\times10^{-6}) = 0.00075\ \text{GPa}$$ so that $$\boxed{\ \sigma_x = 57.86\ \text{MPa},\quad \sigma_y = 2.87\ \text{MPa},\quad \tau_{xy} = 0.75\ \text{MPa}\ }$$
4. Read the physical meaning of the 0° answer before moving on. The longitudinal stress is twenty times the transverse one although the strain along x is only three times that along y, because the fibre direction is thirteen times stiffer. The zeros in the third row and column matter just as much: with the fibres aligned to the load axes, a direct strain produces no shear stress and a shear strain produces no direct stress. The material is specially orthotropic in these axes.
5. Part (b): transform the stiffness matrix through +45°. With m = cos θ and n = sin θ, the transformed reduced stiffnesses are $$\bar{Q}_{11} = Q_{11}m^4 + 2(Q_{12}+2Q_{66})m^2n^2 + Q_{22}n^4$$ $$\bar{Q}_{22} = Q_{11}n^4 + 2(Q_{12}+2Q_{66})m^2n^2 + Q_{22}m^4$$ $$\bar{Q}_{12} = (Q_{11}+Q_{22}-4Q_{66})m^2n^2 + Q_{12}(m^4+n^4)$$ $$\bar{Q}_{66} = (Q_{11}+Q_{22}-2Q_{12}-2Q_{66})m^2n^2 + Q_{66}(m^4+n^4)$$ $$\bar{Q}_{16} = (Q_{11}-Q_{12}-2Q_{66})m^3n - (Q_{22}-Q_{12}-2Q_{66})mn^3$$ $$\bar{Q}_{26} = (Q_{11}-Q_{12}-2Q_{66})mn^3 - (Q_{22}-Q_{12}-2Q_{66})m^3n$$ At θ = 45° every one of the products m⁴, n⁴, m²n², m³n and mn³ equals 0.25, which makes the arithmetic transparent: $$\bar{Q}_{11} = \bar{Q}_{22} = 0.25(191.36+15.107) + 0.5(4.532+20) = 51.617 + 12.266 = 63.883$$ $$\bar{Q}_{12} = 0.25(191.36+15.107-40) + 0.5(4.532) = 41.617 + 2.266 = 43.883$$ $$\bar{Q}_{66} = 0.25(191.36+15.107-9.064-20) + 0.5(10.00) = 44.351 + 5.000 = 49.351$$ $$\bar{Q}_{16} = \bar{Q}_{26} = 0.25\left[(191.36-4.532-20) - (15.107-4.532-20)\right] = 44.063$$ all in GPa. The two coupling terms are no longer zero, and they are as large as the diagonal terms — that is the whole content of part (b).
6. Apply the transformed law to the same strain state. In the loading axes the stiffness matrix is now full, $$\begin{bmatrix}\sigma_x\\ \sigma_y\\ \tau_{xy}\end{bmatrix} = \begin{bmatrix}\bar{Q}_{11} & \bar{Q}_{12} & \bar{Q}_{16}\\ \bar{Q}_{12} & \bar{Q}_{22} & \bar{Q}_{26}\\ \bar{Q}_{16} & \bar{Q}_{26} & \bar{Q}_{66}\end{bmatrix}\begin{bmatrix}\varepsilon_x\\ \varepsilon_y\\ \gamma_{xy}\end{bmatrix}$$ and substituting row by row, with the strains entered in units of $10^{-6}$ so that every product carries a common factor of $10^{-6}\ \text{GPa} = 10^{-3}\ \text{MPa}$, $$\sigma_x = \left[63.883(300) + 43.883(100) + 44.063(75)\right] = 19165 + 4388 + 3305 = 26858$$ $$\sigma_y = \left[43.883(300) + 63.883(100) + 44.063(75)\right] = 13165 + 6388 + 3305 = 22858$$ $$\tau_{xy} = \left[44.063(300) + 44.063(100) + 49.351(75)\right] = 13219 + 4406 + 3701 = 21326$$ Restoring the factor of $10^{-3}\ \text{MPa}$ gives $$\boxed{\ \sigma_x = 26.86\ \text{MPa},\quad \sigma_y = 22.86\ \text{MPa},\quad \tau_{xy} = 21.33\ \text{MPa}\ }$$
7. Check the 45° result by an independent route. Rotating the strains into the material axes first, with m = n = 0.7071, $$\varepsilon_1 = \varepsilon_x m^2 + \varepsilon_y n^2 + \gamma_{xy}mn = 150+50+37.5 = 237.5\times10^{-6}$$ $$\varepsilon_2 = \varepsilon_x n^2 + \varepsilon_y m^2 - \gamma_{xy}mn = 150+50-37.5 = 162.5\times10^{-6}$$ $$\gamma_{12} = 2(\varepsilon_y-\varepsilon_x)mn + \gamma_{xy}(m^2-n^2) = -200\times10^{-6}$$ Applying the untransformed law in those axes gives σ₁ = 46.18 MPa, σ₂ = 3.53 MPa and τ₁₂ = −2.00 MPa, and rotating that stress state back through −45° reproduces 26.86, 22.86 and 21.33 MPa exactly. Two independent routes agreeing is the check worth spending two minutes on in an exam.
8. Interpret the comparison. The same strain state that produced 57.86 MPa of direct stress and almost no shear at 0° now produces less than half that direct stress and a shear stress twenty-eight times larger. Nothing about the loading changed; only the material orientation did. At 45° the fibres run along neither loading axis, so the stiff direction is no longer aligned with ε_x, and the shear–extension coupling terms convert the direct strains into shear. This is why a real laminate is balanced and symmetric: a stack of +45° plies alone would twist and warp under simple tension, whereas pairing every +45° ply with a −45° ply cancels the coupling terms at the laminate level.