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22-Mec-B8 Engineering Materials · December 2013

Question 4 of 8: E-glass/PVC composite — modulus, load sharing and strain

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2013 — 07-Mec-B8 Engineering Materials. Three hours, open book; any non-communicating calculator permitted. Eight questions, all of equal value; any FIVE constitute a complete paper, so each question is worth 20 marks. Candidates are urged to state any assumptions made. All eight questions are solved below, because the set as a whole is the study resource.

Reference texts (22-Mec-B8 Engineering Materials).

  • Askeland & Wright, The Science and Engineering of Materials, 7th ed. — the primary syllabus text.
  • Callister & Rethwisch, Materials Science and Engineering: An Introduction, 10th ed.
  • Shackelford, Introduction to Materials Science for Engineers, 8th ed. — ceramics, glasses and glass-ceramics.
  • Dieter, Mechanical Metallurgy, 3rd ed. — true stress–strain and the necking instability.
  • Fontana, Corrosion Engineering, 3rd ed. — galvanic series, cathodic protection and Faraday’s law.
  • Groover, Fundamentals of Modern Manufacturing, 7th ed. — composite shaping processes.

Question 4: E-glass/PVC 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 continuous, aligned E-glass/PVC composite loaded along the fibre direction:

Given data
QuantitySymbolValue
Modulus of E-glass fibreEf72 GPa
Modulus of hardened PVC matrixEm2.5 GPa
Volume fraction of PVC matrixVm0.62
Volume fraction of glass fibreVf0.38
Cross-sectional areaA500 mm2
Longitudinal loadP75 000 N

Find. (a) the longitudinal modulus of the composite, (b) the percentage of the applied load carried by the fibres, and (c) the strain under the stated load.

unidirectional laminate loaded along the fibres — equal strain in both phasesE-glass Vₜ = 0.38 | PVC Vₘ = 0.62load share94.6 %5.4 % matrixE_c = 28.91 GPa (rule of mixtures)
Continuous aligned composite loaded along the fibres. Both phases are bonded and stretch together, so the strain is common; the stiffer phase therefore carries the higher stress, and the load divides in proportion to EiVi.

Approach. Loading along continuous aligned fibres is the isostrain (Voigt) case: fibre and matrix suffer the same strain, so the composite modulus is the volume-weighted average of the two moduli and the load divides in proportion to the product of modulus and volume fraction. The strain then follows from the composite stress and the composite modulus.

  1. Fix the volume fractions. The matrix occupies 62 % of the volume, so the reinforcement occupies the rest: $$V_f = 1 - V_m = 1 - 0.62 = 0.38$$ It is worth pausing on this line, because the question quotes the matrix fraction while every formula that follows is written in terms of the fibre fraction.
  2. Apply the rule of mixtures for the longitudinal modulus. In the isostrain condition the composite modulus is $$E_c = E_fV_f + E_mV_m = 72(0.38) + 2.5(0.62) = 27.36 + 1.55$$ $$\boxed{\ E_c = 28.91\ \text{GPa}\ }$$ Almost 95 % of that stiffness comes from the 38 % of the volume that is glass, which is the entire point of reinforcing at all.
  3. Partition the load between the phases. Since both phases carry the same strain ε, the stress in each is σi = Eiε and the force in each is that stress times its share of the area, which for aligned fibres equals its volume fraction. Hence $$\frac{P_f}{P_c} = \frac{E_fV_f\varepsilon}{(E_fV_f + E_mV_m)\varepsilon} = \frac{E_fV_f}{E_c} = \frac{27.36}{28.91}$$ $$\boxed{\ \frac{P_f}{P_c} = 0.9464\ \text{, i.e. } 94.64\ \% \text{ carried by the fibres}\ }$$ leaving only 5.36 % to the PVC. The strain cancels, which is why the load split is a property of the material and not of the load applied.
  4. Find the composite stress under the stated load. Referring the load to the whole section, $$\sigma_c = \frac{P}{A} = \frac{75\,000\ \text{N}}{500\ \text{mm}^2} = 150.0\ \text{MPa}$$
  5. Obtain the strain from Hooke’s law for the composite. With Ec = 28.91 GPa = 28 910 MPa, $$\varepsilon = \frac{\sigma_c}{E_c} = \frac{150.0}{28\,910}$$ $$\boxed{\ \varepsilon = 5.19\times 10^{-3} = 0.519\ \%\ }$$
  6. Check the answer against the phase stresses. A useful closing check: at this common strain the fibre stress is σf = 72 000(0.005189) = 373.6 MPa and the matrix stress is σm = 2500(0.005189) = 12.97 MPa. Recombining them over their areas returns 373.6(0.38)(500) + 12.97(0.62)(500) = 70 980 + 4020 = 75 000 N, the applied load, and the fibre share 70 980/75 000 = 94.6 % confirms part (b).
Results for the E-glass/PVC composite
QuantitySymbolValue
Fibre volume fractionVf0.38
(a) Longitudinal modulus of the compositeEc28.91 GPa
(b) Share of the load carried by the glassPf/Pc94.64 %
Share carried by the PVC matrixPm/Pc5.36 %
Composite stress under 75 kNσc150.0 MPa
(c) Strain under 75 kNε5.19 × 10−3 (0.519 %)

Check: the solution assumes continuous, perfectly aligned fibres with a perfect fibre–matrix bond, loading along the fibre axis, and both phases still elastic. Discontinuous or misaligned fibres require a length-efficiency factor, and a transverse load would call for the isostress (Reuss) rule of mixtures, which would give a composite modulus of only about 3.95 GPa — less than a seventh of the longitudinal value. This anisotropy is the defining feature of an aligned composite, not a defect in the calculation.