Interactive Lecture
Laminate Theory, Stiffness Matrices, and Failure Criteria
by Dr Arnaldo Delli Carri
Composite materials are engineered materials made from two or more constituent materials with significantly different physical or chemical properties. When combined, they produce a material with characteristics different from the individual components — typically a stiffness-to-weight or strength-to-weight ratio that neither constituent achieves alone.
Key characteristics of composite materials:
This interactive module explores the mechanics of fibre-reinforced composites, focusing on micromechanics, macromechanics, and laminate theory.
Micromechanics deals with predicting the effective properties of a composite lamina from the properties and volume fractions of its constituents. The Rule of Mixtures provides simple yet effective estimates.
Volume Fractions:
\[V_f + V_m = 1\]
Where \(V_f\) is the fibre volume fraction and \(V_m\) is the matrix volume fraction.
Longitudinal Modulus (Parallel to fibres):
\[E_1 = E_f V_f + E_m V_m\]
Transverse Modulus (Perpendicular to fibres, Halpin-Tsai):
\[E_2 = E_m \frac{1 + \xi \eta V_f}{1 - \eta V_f}\]
\[\eta = \frac{E_f/E_m - 1}{E_f/E_m + \xi}, \quad \xi \approx 2\]
Major Poisson's Ratio:
\[\nu_{12} = \nu_f V_f + \nu_m V_m\]
In-plane Shear Modulus:
\[G_{12} = G_m \frac{1 + \xi \eta V_f}{1 - \eta V_f}, \quad \eta = \frac{G_f/G_m - 1}{G_f/G_m + \xi}\]
The constituent shear moduli come from their isotropic elastic constants: \(G_f = \dfrac{E_f}{2(1+\nu_f)}\) and \(G_m = \dfrac{E_m}{2(1+\nu_m)}\).
A lamina is a single layer of composite material with fibres oriented in one direction. The stress-strain relationship depends on the coordinate system used.
Principal Material Axes (1-2 system):
\[\begin{Bmatrix} \sigma_1 \\ \sigma_2 \\ \tau_{12} \end{Bmatrix} = \begin{bmatrix} Q_{11} & Q_{12} & 0 \\ Q_{12} & Q_{22} & 0 \\ 0 & 0 & Q_{66} \end{bmatrix} \begin{Bmatrix} \varepsilon_1 \\ \varepsilon_2 \\ \gamma_{12} \end{Bmatrix}\]
Where the reduced stiffness matrix components are:
\[Q_{11} = \frac{E_1}{1-\nu_{12}\nu_{21}}, \quad Q_{22} = \frac{E_2}{1-\nu_{12}\nu_{21}}\]
\[Q_{12} = \frac{\nu_{12}E_2}{1-\nu_{12}\nu_{21}}, \quad Q_{66} = G_{12}\]
The minor Poisson's ratio \(\nu_{21}\) is not independent — it follows from the reciprocity relation \(\nu_{21} = \nu_{12}\,E_2/E_1\).
Off-Axis Loading (x-y system rotated by angle \(\theta\)):
\[\begin{Bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{Bmatrix} = [\bar{Q}(\theta)] \begin{Bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{Bmatrix}\]
Where \([\bar{Q}] = [T]^{-1}[Q][T]^{-T}\) with transformation matrix involving \(\cos\theta\) and \(\sin\theta\).
Real composite structures consist of multiple laminae stacked at different orientations. Classical Laminate Theory (CLT) provides a framework for analysing these multi-layered structures.
Force and Moment Resultants:
\[\begin{Bmatrix} N_x \\ N_y \\ N_{xy} \end{Bmatrix} = [A] \begin{Bmatrix} \varepsilon_x^0 \\ \varepsilon_y^0 \\ \gamma_{xy}^0 \end{Bmatrix} + [B] \begin{Bmatrix} \kappa_x \\ \kappa_y \\ \kappa_{xy} \end{Bmatrix}\]
\[\begin{Bmatrix} M_x \\ M_y \\ M_{xy} \end{Bmatrix} = [B] \begin{Bmatrix} \varepsilon_x^0 \\ \varepsilon_y^0 \\ \gamma_{xy}^0 \end{Bmatrix} + [D] \begin{Bmatrix} \kappa_x \\ \kappa_y \\ \kappa_{xy} \end{Bmatrix}\]
Where:
Stiffness Matrices:
\[A_{ij} = \sum_{k=1}^{N} (\bar{Q}_{ij})_k (z_k - z_{k-1})\]
\[B_{ij} = \frac{1}{2}\sum_{k=1}^{N} (\bar{Q}_{ij})_k (z_k^2 - z_{k-1}^2)\]
\[D_{ij} = \frac{1}{3}\sum_{k=1}^{N} (\bar{Q}_{ij})_k (z_k^3 - z_{k-1}^3)\]
Unlike isotropic materials, composites exhibit complex failure modes. Several failure criteria have been developed to predict the onset of failure.
Maximum Stress Criterion:
Failure occurs when any stress component exceeds its allowable value:
\[\sigma_1 < X_T, \quad \sigma_1 > -X_C, \quad \sigma_2 < Y_T, \quad \sigma_2 > -Y_C, \quad |\tau_{12}| < S\]
Tsai-Hill Criterion:
\[\left(\frac{\sigma_1}{X}\right)^2 - \frac{\sigma_1\sigma_2}{X^2} + \left(\frac{\sigma_2}{Y}\right)^2 + \left(\frac{\tau_{12}}{S}\right)^2 = 1\]
Tsai-Wu Criterion:
\[F_1\sigma_1 + F_2\sigma_2 + F_{11}\sigma_1^2 + F_{22}\sigma_2^2 + F_{66}\tau_{12}^2 + 2F_{12}\sigma_1\sigma_2 = 1\]
Where \(X_T, X_C\) are longitudinal tensile/compressive strengths, \(Y_T, Y_C\) are transverse strengths, and \(S\) is shear strength.
Composite materials are extensively used across engineering disciplines for their exceptional strength-to-weight ratio and design flexibility:
| Concept | Equation | Description |
|---|---|---|
| Longitudinal Modulus | \(E_1 = E_f V_f + E_m V_m\) | Rule of mixtures for parallel loading |
| Transverse Modulus | \(E_2 = E_m \frac{1 + \xi \eta V_f}{1 - \eta V_f}\) | Halpin-Tsai equation |
| Reduced Stiffness | \(Q_{11} = \frac{E_1}{1-\nu_{12}\nu_{21}}\) | Lamina stiffness in material axes |
| Extensional Stiffness | \(A_{ij} = \sum_{k=1}^{N} (\bar{Q}_{ij})_k (z_k - z_{k-1})\) | Laminate in-plane stiffness |
| Tsai-Hill Criterion | \(\left(\frac{\sigma_1}{X}\right)^2 - \frac{\sigma_1\sigma_2}{X^2} + \left(\frac{\sigma_2}{Y}\right)^2 + \left(\frac{\tau_{12}}{S}\right)^2 = 1\) | Interactive failure criterion |
A carbon/epoxy composite has a fibre volume fraction of 0.65. The carbon fibres have \(E_f = 230\) GPa and the epoxy matrix has \(E_m = 3.5\) GPa. Calculate the longitudinal modulus \(E_1\) of the lamina.
Show Solution
Solution:
Using the Rule of Mixtures for longitudinal modulus:
\(E_1 = E_f V_f + E_m V_m\)
where \(V_m = 1 - V_f = 1 - 0.65 = 0.35\)
\(E_1 = (230)(0.65) + (3.5)(0.35)\)
\(E_1 = 149.5 + 1.225 = 150.725\) GPa
Answer: \(E_1 \approx 150.7\) GPa
A unidirectional lamina with fibres at 30° to the loading axis experiences an applied stress of \(\sigma_x = 200\) MPa. Calculate the stress components in the material coordinate system (\(\sigma_1, \sigma_2, \tau_{12}\)).
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Solution:
Using the stress transformation equations with \(\theta = 30°\):
\(\sigma_1 = \sigma_x \cos^2\theta = 200 \times \cos^2(30°) = 200 \times 0.75 = 150\) MPa
\(\sigma_2 = \sigma_x \sin^2\theta = 200 \times \sin^2(30°) = 200 \times 0.25 = 50\) MPa
\(\tau_{12} = -\sigma_x \sin\theta\cos\theta = -200 \times \sin(30°)\cos(30°) = -200 \times 0.433 = -86.6\) MPa
The minus sign is part of the standard transformation \(\tau_{12} = -(\sigma_x-\sigma_y)\sin\theta\cos\theta + \tau_{xy}(\cos^2\theta-\sin^2\theta)\); with \(\sigma_y = \tau_{xy} = 0\) it reduces to \(-\sigma_x\sin\theta\cos\theta\).
Answer: \(\sigma_1 = 150\) MPa, \(\sigma_2 = 50\) MPa, \(\tau_{12} = -86.6\) MPa
A lamina with strengths \(X_T = 1500\) MPa, \(Y_T = 50\) MPa, and \(S = 70\) MPa experiences stresses \(\sigma_1 = 800\) MPa, \(\sigma_2 = 30\) MPa, and \(\tau_{12} = 40\) MPa. Determine if failure occurs using the Tsai-Hill criterion.
Show Solution
Solution:
Tsai-Hill criterion: \(\left(\frac{\sigma_1}{X}\right)^2 - \frac{\sigma_1\sigma_2}{X^2} + \left(\frac{\sigma_2}{Y}\right)^2 + \left(\frac{\tau_{12}}{S}\right)^2 \leq 1\)
Substituting values:
\(\left(\frac{800}{1500}\right)^2 - \frac{(800)(30)}{1500^2} + \left(\frac{30}{50}\right)^2 + \left(\frac{40}{70}\right)^2\)
\(= 0.28444 - 0.01067 + 0.36000 + 0.32653\)
\(= 0.9603\)
Since \(0.9603 < 1\), the lamina is safe with a safety factor of \(1/\sqrt{0.9603} = 1.020\)
Answer: No failure, SF = 1.02