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Composite Materials Solutions to Composite Materials Questions Problems 1. For a polymer-matrix fibre-reinforced composite, list three functions of the matrix phase; (b) compare the desired mechanical characteristics of matrix and fibre phases; and (c) cite two reasons why there must be a strong bond between fibre and matrix at their interface. a) Three functions of the polymer-matrix phase are: 1. to bind the fibres together so that applied stress is distributed among the fibres; 2. to protect the surface of the fibres from being damaged; and 3. to separate the fibres and inhibit crack propagation. b) The matrix phase must be ductile and is usually relatively soft, whereas the fibre phase must be stiff and strong c) There must be a strong interfacial bond between fibre and matrix in order to: 1. maximise the stress transmittance between matrix and fibre phases; and 2. minimise fibre pull-out, and the probability of failure.

2. A continuous and aligned fibre-reinforced composite is to be produced consisting of 45 vol% aramid fibres and 55 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows:

Aramid fibre Polycarbonate

Modulus of Elasticity (GPa) 131 2.4

Tensile Strength (MPa) 3600 65

For this composite, compute (a) the longitudinal tensile strength, and (b) the longitudinal modulus of elasticity. (1656 MPa, 60.3 GPa). (a) The longitudinal tensile strength is determined by using the equation below; (TS)c = (TS)mVm + (TS)f Vf = (65x106 x 0.55) + (3600 x 0.45) =1656 MPa a) The longitudinal tensile modulus is computed by:

Ecl = EmVm + EfVf = (2.4x109 x 0.55) + (131x109 x 0.45) =60.3 GPa 3. Is it possible to produce a continuous and oriented aramid fibre-epoxy matrix composite having longitudinal and transverse moduli of elasticity of 57.1 GPa and 4.12 GPa respectively? Why or why not? Assume that the modulus of elasticity of the epoxy is 2.4 GPa. (Possible)

Using the data in the table in Q2 above, E for aramid fibres = 131 GPa. For the longitudinal modulus Ecl = Em(1-Vfl) + EfVfl 57.1 GPa = (2.4x109) (1-Vfl) + (131x109) Vfl Solving this expression for Vfl yields Vfl = 0.425 1

Repeating this procedure for the transverse modulus Ect

E mE f E ct  [1  Vft ] E f  Vft Em 4.12GPa 

(2.4GPa)(131GPa) [1 Vft ] 131GPa  Vft (2.4GPa)

Solving this expression for Vft leads to Vft = 0.425. Thus, since Vfl and Vft are equal, the proposed composite is possible. 4. For a continuous and oriented fibre-reinforced composite, the moduli of elasticity in the longitudinal and transverse directions are 33.1 GPa and 3.66 GPa respectively. If the volume fraction of fibres is 0.30, determine the moduli of elasticity of fibre and matrix phases. (2.6 GPa, 104 GPa). Ecl = Em(1-Vf) + EfVf 33.1x109 = Em(1-0.30) + Ef(0.30) And

E ct 

E mE f [1  Vf ] E f  Vf Em

E mE f 3.66GPa  [1  0.30] E f  0.30Em Solving these two expressions simultaneously for E m and Ef leads to: Em = 2.6 GPa Ef = 104 GPa

5. In an aligned and continuous carbon fibre-reinforced nylon 6,6 composite, the fibres are to carry 97% of a load applied in the longitudinal direction. (a) Using the data provided, determine the volume fraction of fibres that will be required. (b) What will be the tensile strength of this composite? (Vf = 0.258, TS = 495 MPa).

Carbon fibre Nylon 6,6

Modulus of Elasticity (GPa) 260 2.8

Tensile Strength (MPa) 1700 76

Ff EV Ef Vf  f f  Fm Em Vm Em (1  Vf ) Now:

Ff 0.97  32 .33 Fm 0.03 Substituting in for Ef and Em

Ff 260 GPa 32.33  Fm (2.8 GPa)(1 Vf ) 2

And solving for Vf yields: Vf = 0.258 (b) We are now asked for the tensile strength of this composite.

(TS)c = (TS)mVm + (TS)f Vf =76 Mpa x (1 - 0.258) + (1700 Mpa)(0.258) = 495 MPa 6. A continuous and aligned fibre composite having a cross-sectional area of 1130 mm2 is subject to an external tensile load. If the stresses sustained by the fibre and matrix phases are 156 MPa and 2.75 MPa respectively, the force sustained by the fibre phase is 74,000 N and the total longitudinal strain is 1.25 x 10 -3, determine: (a) the force sustained by the matrix phase, (b) the modulus of elasticity of the composite material in the longitudinal direction, and, the moduli of elasticity for the fibre and matrix phases. (Fm = 1802 N, Ec = 53.7 GPa, Ef = 124.8 GPa) (a) For this problem, we are asked to calculate the force sustained by the matrix phase. It is first necessary to compute the volume fraction of the matrix phase, V m. This may be accomplished by first determining V f and then Vm from Vm = 1-Vf. The value of Vf may be calculated realising that:

f 

Ff F  f A f V fA c

Or

Vf 

Ff 74000 N  0.420  f A c (156 x 106 )(1.13 x 10  3 )

Now: Vm = 1-Vf = 1-0.420 = 0.580 Therefore since:

m 

Fm F  m A m V mA c

Then: Fm = VmσmAc = (0.58)(2.75x106)(1.13x10-3) = 1802 N (b) We are now asked to calculate the modulus of Elasticity in the longitudinal direction. This is possible realising that Ec 

Fm  Ff c and that  c  , thus: Ac  Ec  =

Fm  Ff  Ac

1802 N  74000 N 53.7 GPa   (1.25 x 10 3 )(1.13 x 10 3 )

(c) Finally it is necessary to determine the modulus of elasticity for the fibre and matrix phases. This is possible as follows: 3

2.75 x 106   9 Em  m  m   2.2 x 10 Pa or 2.2 GPa 1.25 x 10  3 c m

Ef 

6 f  f 156 x 10   1.248 x 1011 Pa or 124.8 GPa  c 1.25 x 10 3 f

7. Disregard this question. 8. Describe one desirable characteristic and one less desirable characteristic discontinuous-oriented, and (2) discontinuous random fibre-reinforced composites. (a) List four reasons why glass fibres are most commonly used for reinforcement. (b) Why is the surface protection of glass fibres so important? (c) What measures are taken to protect the surface of glass fibres?

for

each

of

(1)

For discontinuous-oriented fibre-reinforced composites one desirable characteristic is that the composite is relatively strong and stiff in one direction; a less desirable characteristic is that the mechanical properties are anisotropic. For discontinuous and random fibre reinforced, one desirable characteristic is that the properties are isotropic; a less desirable characteristic is there is no single strength direction. The four reasons why glass fibres are most commonly used for reinforcement are listed below:    

It is easily drawn into high-strength fibres from the molten state. It is readily available and may be fabricated into a glass-reinforced plastic economically using a wide variety of composite-manufacturing techniques. As a fibre, it is relatively strong, and when embedded in a plastic matrix, it produces a composite having a very high specific strength. When coupled with the various plastics, it possesses a chemical inertness that renders the composite useful in a variety of corrosive environments.

The surface protection of glass fibres is important because surface flaws or cracks will act as points of stress concentration, which dramatically reduce the tensile strength of the material. Care must be taken not to rub or abrade the surface after the fibres are drawn. As a surface protection, newly drawn fibres are coated with a protective surface film called a ‘size’. 9. Briefly describe pultrusion, filament winding, and prepreg production fabrication processes; describe the advantages and disadvantages of each. For pultrusion, the advantages are: the process may be automated, production rates are relatively high, a wide variety of shapes having constant cross-sections are possible, and very long pieces can be produced. The chief disadvantage is that shapes are limited to those having a constant cross-section. For filament winding, the advantages are: the process may be automated, a variety of winding patterns are possible, and a high degree of control over winding uniformity and orientation is afforded. The chief disadvantage is that the variety of shapes is somewhat limited. For prepreg production, the advantages are: resin does not need to be added to the prepreg, the lay-up arrangement relative to the orientation of the individual piles is variable, and the lay-up process may be automated. The chief disadvantages of these techniques of this technique are that the final curing is necessary after fabrication and thermoset prepregs must be stored at sub-ambient temperatures to prevent complete curing.

10. Briefly describe laminar composites. What is the prime reason for fabricating these materials?

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Laminar composites are a series of sheets or panels, each of which has a preferred high strength direction. These sheets are stacked and then cemented together such that the orientation of the high strength direction varies from layer to layer. These composites are constructed in order to have a relatively high strength in virtually all direction within the plane of the laminate. 11. Briefly describe sandwich panels. What is the prime reason for fabricating these structural composites? What are the functions of the faces and the core? Sandwich panels consist of two outer face sheets of a high strength material which are separated by a layer of a less dense lower strength core material. The prime reason for fabricating these composites is to produce structures having high in plane strengths, high shear rigidities and low densities. The faces function so as to bear the majority of in-plane loading and also transverse bending stresses. On the other hand, the core separates the faces and resists deformations perpendicular to the faces.

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