Question 3 of 8: Glass-fibre composite — modulus, load sharing and strain

Question 3: Glass-fibre composite — modulus, load sharing and strain (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 continuous-fibre composite loaded along the fibres.

Find. The longitudinal modulus of the composite, the share of the applied load carried by the fibres, and the strain produced by the stated load.

Approach. Take the isostrain (Voigt) assumption for loading parallel to continuous fibres, apply the rule of mixtures to get the composite modulus, use the same assumption to split the load between the phases, and finally divide the applied stress by the composite modulus to get the strain.

1. Fix the volume fractions. The question states the plastic content, not the fibre content, and it is the fibre fraction that the rule of mixtures needs first: $$V_{f} = 1 - V_{m} = 1 - 0.60 = 0.40$$ Reading the given 60 per cent as the reinforcement is the single most common way to lose this question.
2. Apply the rule of mixtures for the longitudinal modulus. Because the phases are bonded and loaded in parallel they share a common strain, so their stiffnesses add in proportion to volume: $$E_{c} = E_{f}V_{f} + E_{m}V_{m} = 75(0.40) + 5(0.60)$$ which gives $$\boxed{E_{c} = 30.0 + 3.0 = 33.0\ \text{GPa}}$$ The fibres supply 30 of those 33 gigapascals even though they occupy only two-fifths of the volume.
3. Split the load between the phases. With a common strain $\varepsilon$, each phase carries $\sigma_{i} = E_{i}\varepsilon$ over its own area $V_{i}A$, so the force ratio is $$\frac{P_{f}}{P_{c}} = \frac{E_{f}V_{f}}{E_{f}V_{f} + E_{m}V_{m}} = \frac{30.0}{33.0}$$ giving $$\boxed{\frac{P_{f}}{P_{c}} = 0.9091 \;\Rightarrow\; 90.91\%\ \text{carried by the glass}}$$ and 9.09 per cent left to the plastic. Notice that the ratio of the stresses in the two phases is simply the ratio of their moduli, $\sigma_{f}/\sigma_{m} = 75/5 = 15$; the load ratio differs from it only because the fibres occupy less of the section.
4. Find the mean stress on the composite. The applied load acts on the full section, so $$\sigma_{c} = \frac{P}{A} = \frac{75\,900}{500} = 151.8\ \text{MPa}$$ with the newton and the square millimetre giving megapascals directly.
5. Divide by the composite modulus to obtain the strain. Working in consistent units with $E_{c} = 33\,000\ \text{MPa}$, $$\varepsilon = \frac{\sigma_{c}}{E_{c}} = \frac{151.8}{33\,000}$$ so that $$\boxed{\varepsilon = 4.60\times10^{-3} = 0.460\%}$$
6. Check the answer by rebuilding the load from the phase stresses. At that strain the fibres carry $\sigma_{f} = 75\,000(4.60\times10^{-3}) = 345.0\ \text{MPa}$ and the matrix $\sigma_{m} = 5\,000(4.60\times10^{-3}) = 23.0\ \text{MPa}$. Multiplying by the respective areas, $$P_{f} = 345.0(0.40)(500) = 69\,000\ \text{N}, \; P_{m} = 23.0(0.60)(500) = 6\,900\ \text{N}$$ and $69\,000 + 6\,900 = 75\,900\ \text{N}$, which reproduces the applied load exactly and confirms both the modulus and the 90.91 per cent load split.

A strain of 0.46 per cent is a realistic working value for a glass-reinforced plastic: E-glass fibre fails at roughly 2 to 3 per cent strain, so the laminate is at perhaps a fifth of its fibre-limited capacity, while the matrix at 23 MPa is well inside its own elastic range. That comfortable margin is exactly what the isostrain analysis is for — it converts a load on a section into the strain that both phases must survive.