Longitudinal properties of a continuous aligned glass-fibre composite: 22-Mec-B8 Engineering Materials, December 2019, Q7

Question 7: Longitudinal properties of a continuous aligned glass-fibre composite (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 unidirectional composite loaded along the fibre axis, so the isostrain (Voigt) condition applies: the fibres and the matrix are bonded and of equal length, so they must undergo the same longitudinal strain.

Given data for Question 7
Quantity	Symbol	Value
Fibre volume fraction	$V_{f}$	0.50
Matrix volume fraction	$V_{m}$	0.50
Modulus of the glass fibre	$E_{f}$	69 GPa
Modulus of the hardened resin	$E_{m}$	3.4 GPa
Cross-sectional area of the composite	$A_{c}$	200 mm²
Applied longitudinal stress	$\sigma_{c}$	45 MPa

Find. (a) the longitudinal modulus of the composite; (b) the load carried by the fibres and by the matrix under the applied stress; and (c) the strain in each constituent.

Equal strain in both constituents forces the load to divide in proportion to $E V$, so half the volume of glass carries 95% of the load.

Approach. Impose equal strain on the two constituents, which converts the modulus into a volume-weighted average (the rule of mixtures). The total load follows from the applied stress and the gross area; it divides between fibre and matrix in the ratio of their $E_{i}V_{i}$ products; and the strain is then recovered independently from either constituent as a check.

Part (a) — apply the rule of mixtures in the longitudinal direction. With equal strain in both phases the composite stress is the volume-weighted average of the constituent stresses, and dividing by that common strain gives$$E_{c} = E_{f}V_{f} + E_{m}V_{m}$$Substituting the two moduli and the equal volume fractions,$$E_{c} = (69)(0.50) + (3.4)(0.50) = 34.5 + 1.7 = \boxed{36.2\ \text{GPa}}$$Half the volume of glass has lifted the stiffness of the resin by a factor of more than ten, and the matrix contributes only 1.7 of the 36.2 GPa.
Part (b) — total load on the section. The applied stress acts on the gross cross-section, so$$F_{c} = \sigma_{c}A_{c} = (45\ \text{MPa})(200\ \text{mm}^{2}) = \boxed{9000\ \text{N}} = 9.00\ \text{kN}$$
Part (b) — how that load divides. Because the strains are equal, the ratio of the loads is the ratio of the axial stiffnesses $E_{i}A_{i} = E_{i}V_{i}A_{c}$:$$\frac{F_{f}}{F_{m}} = \frac{E_{f}V_{f}}{E_{m}V_{m}} = \frac{(69)(0.50)}{(3.4)(0.50)} = \frac{34.5}{1.7} = 20.29$$Combining that ratio with $F_{f} + F_{m} = 9000$ N,$$F_{f} = 9000\left(\frac{20.29}{21.29}\right) = \boxed{8577\ \text{N}}, \qquad F_{m} = 9000 - 8577 = \boxed{423\ \text{N}}$$so the fibres take 95.3% of the load and the resin 4.7%. The two sum to 9000 N, as they must.
Part (c) — the common strain. The whole section strains together, so the strain follows from the composite stress and the composite modulus:$$\varepsilon_{c} = \frac{\sigma_{c}}{E_{c}} = \frac{45\ \text{MPa}}{36\,200\ \text{MPa}} = \boxed{1.243\times 10^{-3}}$$that is, 0.124%.
Part (c) — recover the same strain from each constituent separately. Each phase occupies half of the 200 mm², so $A_{f} = A_{m} = 100$ mm², and$$\sigma_{f} = \frac{8577}{100} = 85.8\ \text{MPa} \;\Rightarrow\; \varepsilon_{f} = \frac{85.8}{69\,000} = 1.243\times 10^{-3}$$$$\sigma_{m} = \frac{423}{100} = 4.23\ \text{MPa} \;\Rightarrow\; \varepsilon_{m} = \frac{4.23}{3400} = 1.243\times 10^{-3}$$The two agree with each other and with Step 4, which confirms the isostrain assumption has been applied consistently: $\varepsilon_{f} = \varepsilon_{m} = \varepsilon_{c}$.

Question 7 — final results
Quantity	Symbol / basis	Value
(a) Longitudinal modulus	$E_{f}V_{f}+E_{m}V_{m}$	36.2 GPa
(b) Total load	$\sigma_{c}A_{c}$	9000 N (9.00 kN)
(b) Load on the glass fibres	$F_{c}\,E_{f}V_{f}/(E_{c}V_{c})$	8577 N (95.3%)
(b) Load on the resin matrix	$F_{c}-F_{f}$	423 N (4.7%)
(c) Strain in the fibres	$\sigma_{f}/E_{f}$	1.243 × 10⁻³
(c) Strain in the matrix	$\sigma_{m}/E_{m}$	1.243 × 10⁻³
(c) Strain of the composite	$\sigma_{c}/E_{c}$	1.243 × 10⁻³ (0.124%)
Stress in the fibres	$F_{f}/A_{f}$	85.8 MPa
Stress in the matrix	$F_{m}/A_{m}$	4.23 MPa

Check: the calculation assumes a perfect fibre–matrix bond with no interfacial slip, fibres that are genuinely continuous and perfectly aligned with the load, and both constituents still elastic. The last is easily checked: at 0.124% strain, E-glass (failure strain about 2.5%) and a typical hardened polyester or epoxy (failure strain 1–4%) are both far from failure, so linear elasticity is safe. Note also that the answer is direction-specific — loaded transversely, the same laminate follows the isostress (Reuss) rule and its modulus collapses to $E_{f}E_{m}/(V_{m}E_{f}+V_{f}E_{m}) = 6.48$ GPa, a factor of 5.6 lower.

Concept & Theory

Topic. Isostrain (Voigt) loading of a continuous aligned composite: the rule of mixtures, load sharing in proportion to $EV$, and the anisotropy that follows from it.

Key relations. $\varepsilon_{c} = \varepsilon_{f} = \varepsilon_{m}$ (isostrain); $E_{cl} = E_{f}V_{f} + E_{m}V_{m}$; $F_{f}/F_{m} = E_{f}V_{f}/(E_{m}V_{m})$; $V_{f}+V_{m}=1$; transverse (isostress) modulus $E_{ct} = E_{f}E_{m}/(V_{m}E_{f}+V_{f}E_{m})$.

Why this works. If the fibres and the matrix are bonded and both run the full length of the specimen, they have no choice but to extend by the same amount, so the strain is the shared quantity and the stress is what differs between them. Each phase then carries the stress its own modulus demands at that strain, $\sigma_{i} = E_{i}\varepsilon$, and each contributes to the total load in proportion to the area it occupies, which is why the load ratio is the ratio of the $EV$ products and why the composite modulus is a simple volume average. The practical consequence is the striking imbalance seen here: glass is only twenty times stiffer than the resin, yet at equal volume fractions it takes twenty times the load, so 95% of the force passes through the fibres and the matrix is left doing what it is really there for — holding the fibres in alignment, transferring load into them across the interface by shear, and protecting them from abrasion and the environment. The same reasoning also explains why the rule of mixtures is a longitudinal result only. Turn the load through ninety degrees and the constituents are now in series rather than in parallel, the shared quantity becomes the stress instead of the strain, and the compliant matrix dominates the answer.

Common pitfall. Dividing the fibre load by the full 200 mm² instead of by the fibre area of 100 mm², which halves $\sigma_{f}$ and produces two different strains for two phases that must strain equally — that inconsistency is the built-in check on the answer. A second is confusing volume fraction with weight fraction: they differ whenever the densities differ, and glass at 2.5 Mg/m³ against resin at 1.2 gives 68 wt% glass for this 50 vol% laminate. A third is quoting the longitudinal modulus as though it were the property of the material rather than of the material in one direction.

Source. Callister & Rethwisch, 10th ed., §16.5 (continuous aligned fibre composites, Eqs. 16.10–16.17); Askeland & Wright, Ch. 17; Shackelford, Ch. 12; Ashby, Materials Selection in Mechanical Design, 5th ed., Ch. 20 (hybrid materials).

Question 7 of 7: Longitudinal properties of a continuous aligned glass-fibre composite

Question 7: Longitudinal properties of a continuous aligned glass-fibre composite (20 marks)