Stresses in a unidirectional composite skin panel: 22-Mec-B9 Advanced Engineering Structures, December 2018, Q8

Question 8: Stresses in a unidirectional composite skin panel (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.

Property or strain	Value
Longitudinal modulus $E_1$	175 GPa
Transverse modulus $E_2$	15 GPa
Shear modulus $G_{12}$	10 GPa
Major Poisson ratio $\nu_{12}$	0.31
$\varepsilon_x$	$310\times10^{-6}$
$\varepsilon_y$	$120\times10^{-6}$
$\gamma_{xy}$	$85\times10^{-6}$

Find. The in-plane stresses $\sigma_x$, $\sigma_y$ and $\tau_{xy}$ for fibres (a) along the load direction and (b) at $+30^{\circ}$ to it, under the same applied strains.

Question 8 — the same applied strain state acting on a unidirectional lamina with fibres at $0^{\circ}$ and at $+30^{\circ}$ to the $x$ load direction. Rotating the fibres leaves the strains untouched but changes the stiffness the panel presents, and introduces shear–extension coupling.

Approach. Build the plane-stress reduced stiffness matrix in the material axes, use it directly for the $0^{\circ}$ case, then transform it to the load axes for the $30^{\circ}$ case.

Get the minor Poisson ratio. The reciprocal relation fixes it:$$\nu_{21} = \nu_{12}\frac{E_2}{E_1} = 0.31\frac{15}{175} = 0.026571$$so $1-\nu_{12}\nu_{21} = 0.991763$. The near-unity value tells us the Poisson coupling barely stiffens this material, which is typical of a highly anisotropic carbon lamina.
Build the reduced stiffness matrix. Note that $[Q]$ is the plane-stress reduced stiffness, not a three-dimensional stiffness — the latter would need $E_3$, $G_{13}$ and $G_{23}$, which are not given:$$Q_{11} = \frac{E_1}{1-\nu_{12}\nu_{21}},\quad Q_{22} = \frac{E_2}{1-\nu_{12}\nu_{21}},\quad Q_{12} = \frac{\nu_{12}E_2}{1-\nu_{12}\nu_{21}},\quad Q_{66} = G_{12}$$giving $Q_{11} = 176.45$, $Q_{22} = 15.125$, $Q_{12} = 4.6886$ and $Q_{66} = 10.0$ GPa.
Compute the stresses for fibres at 0° (part a). With the fibres along $x$ the material and load axes coincide, so $[\bar Q] = [Q]$ and there is no shear–extension coupling:$$\sigma_x = Q_{11}\varepsilon_x + Q_{12}\varepsilon_y,\qquad\sigma_y = Q_{12}\varepsilon_x + Q_{22}\varepsilon_y,\qquad\tau_{xy} = Q_{66}\gamma_{xy}$$Evaluating in MPa,$$\boxed{\sigma_x = 55.26\ \text{MPa},\quad\sigma_y = 3.27\ \text{MPa},\quad\tau_{xy} = 0.85\ \text{MPa}}$$The fibre-direction stress dominates by a factor of 17, even though $\varepsilon_x$ is only 2.6 times $\varepsilon_y$ — the anisotropy of the stiffness, not the strain state, is doing the work.
Transform the stiffness to +30° (part b). With $c = \cos30^{\circ}$ and $s = \sin30^{\circ}$ the transformed stiffnesses are$$\bar Q_{11} = Q_{11}c^4 + 2(Q_{12}+2Q_{66})s^2c^2 + Q_{22}s^4$$with the companion expressions for $\bar Q_{12}$, $\bar Q_{22}$, $\bar Q_{16}$, $\bar Q_{26}$ and $\bar Q_{66}$. Evaluating all six at $30^{\circ}$ gives, in GPa,$$\bar Q_{11} = 109.46,\ \bar Q_{12} = 31.351,\ \bar Q_{22} = 28.794,\ \bar Q_{16} = 50.322,\ \bar Q_{26} = 19.535,\ \bar Q_{66} = 36.663$$A cheap check on all six fourth-order expressions at once: at $45^{\circ}$ the identity $\bar Q_{16} = \bar Q_{26} = (Q_{11}-Q_{22})/4$ must hold exactly, and it does.
Compute the stresses at +30°. Now the $\bar Q_{16}$ and $\bar Q_{26}$ terms couple shear to extension, so all three strains feed every stress:$$\sigma_x = \bar Q_{11}\varepsilon_x+\bar Q_{12}\varepsilon_y+\bar Q_{16}\gamma_{xy}$$and similarly for $\sigma_y$ and $\tau_{xy}$. The results are$$\boxed{\sigma_x = 41.97\ \text{MPa},\quad\sigma_y = 14.83\ \text{MPa},\quad\tau_{xy} = 21.06\ \text{MPa}}$$Rotating the fibres has cut $\sigma_x$ by a quarter, quadrupled $\sigma_y$, and multiplied the shear stress by 25.
Resolve onto the fibre axes as a check. Rotating the part (b) stresses through $+30^{\circ}$ gives the stresses the ply actually experiences:$$\sigma_1 = 53.43\ \text{MPa},\qquad\sigma_2 = 3.38\ \text{MPa},\qquad\tau_{12} = -1.22\ \text{MPa}$$The same three numbers come out of rotating the strains into the material axes first and then applying $[Q]$ — two independent routes to one answer — and the first stress invariant $\sigma_x+\sigma_y = \sigma_1+\sigma_2$ is preserved.

The comparison between the two parts is the point of the question. The applied strain state is identical; only the fibre orientation changes, and yet the stress state is transformed. Two features are worth naming. First, the appearance of a large $\tau_{xy}$ under a strain state that is almost pure extension is shear–extension coupling, an effect with no counterpart in isotropic materials: it exists because $\bar Q_{16}$ and $\bar Q_{26}$ are non-zero for any fibre angle other than $0^{\circ}$ or $90^{\circ}$. In a real laminate this is why plies are laid up in balanced $\pm\theta$ pairs, so that the coupling terms cancel at the laminate level. Second, the fibre-axis stresses in the last step are the ones that matter for strength, because a unidirectional ply is enormously stronger along the fibres than across them; at $30^{\circ}$ the ply carries 53.4 MPa along the fibres against only 3.4 MPa transversely, and it is the transverse and shear components that would reach their (much lower) allowables first.

Quantity	(a) fibres at 0°	(b) fibres at +30°
$\sigma_x$	55.26 MPa	41.97 MPa
$\sigma_y$	3.27 MPa	14.83 MPa
$\tau_{xy}$	0.85 MPa	21.06 MPa
$\sigma_1$ along the fibres	55.26 MPa	53.43 MPa
$\sigma_2$ across the fibres	3.27 MPa	3.38 MPa
$\tau_{12}$ in the fibre frame	0.85 MPa	−1.22 MPa

Topic. Plane-stress constitutive behaviour of an orthotropic unidirectional lamina, and stiffness transformation to an off-axis orientation.

Key relations. Reciprocity $\nu_{21} = \nu_{12}E_2/E_1$; reduced stiffnesses $Q_{11} = E_1/(1-\nu_{12}\nu_{21})$, $Q_{22} = E_2/(1-\nu_{12}\nu_{21})$, $Q_{12} = \nu_{12}E_2/(1-\nu_{12}\nu_{21})$, $Q_{66} = G_{12}$; the transformed stiffness $[\bar Q]$ from the fourth-order expressions in $\sin\theta$, $\cos\theta$; stress rotation to the material axes.

Why this works. A unidirectional lamina is orthotropic, so in its own material axes the stiffness matrix has the sparse form with no coupling between extension and shear; four independent constants describe it in plane stress, and the given data supply exactly those four. The reciprocal relation is not an extra assumption but a consequence of the strain-energy function being a proper potential, which forces the compliance matrix to be symmetric. Rotating to an off-axis orientation is a pure tensor transformation, and because stiffness is a fourth-rank tensor the transformation involves fourth powers of $\sin\theta$ and $\cos\theta$ — which is why the $[\bar Q]$ expressions look so much heavier than the familiar second-rank stress-rotation formulae.

Common pitfall. Using engineering shear strain and tensor shear strain interchangeably. The $[\bar Q]$ expressions quoted here are written for engineering shear strain $\gamma_{xy}$, which is twice the tensor component; mixing the conventions halves or doubles the shear stress. The other frequent error is transforming the stresses from the $0^{\circ}$ answer instead of transforming the stiffness — that would be correct only if the stresses were the given quantity, whereas here it is the strains that are held fixed while the material rotates beneath them.

Source. Megson, Aircraft Structures for Engineering Students, 6th ed., ch. 12 (composite laminates, stiffness transformation); Jones, Mechanics of Composite Materials, 2nd ed., ch. 2.

Question 8 of 8: Stresses in a unidirectional composite skin panel

Question 8: Stresses in a unidirectional composite skin panel (20 marks)