B5.9EH2: Mechanical Engineering Science 8

Part 1: Mechanics of Materials

Topic 1: Mechanics of composite materials

Outline Tutorial Solutions on Unidirectional FRPs

(Covering Topics 1.1 and 1.2)

1. The tensile modulus of a thermoset polyester matrix is 3 GPa and that of E-glass reinforcing fibres is 76 GPa. Calculate the tensile modulus for the composite series (Vf = 0 to 1) where the fibres are continuous and a stress is applied in each of the directions parallel (E1) and transverse (E2) to the fibres.

Using the equations developed for E1 and E2 and substituting values:


E1=Vf×76+1−Vf×3=3+73Vf and E2=76×31−Vf×76+Vf×3=22876−73Vf

2. A composite is made from 50% (by volume) of carbon fibres (uniaxial tensile modulus 230 GPa and uniaxial tensile strength 3200 MPa) continuously uniaxially aligned in an epoxy matrix (uniaxial tensile modulus 2.4 GPa and uniaxial tensile strength 60 MPa).

(i) Calculate the tensile strength and tensile modulus of the composite.

For the tensile modulus

E1=Vf×Ef+1−Vf×Em=0.5×230+0.5×2.4=116GPa

Fεf=σf/Ef=1.39×10−2or the strength, we need to find out if the composite conforms to the “brittle fibre” or “brittle matrix” model, so first find the failure strains:

aεm=σm/Em=2.5×10−2nd

So model is a brittle fibre model. At 50% fibres, this is likely to be a case of high fibre volume fraction, but can check by determining V'f, the value of Vf at which s*1 for the low Vf model is the same as s*1 for the high Vf model:

1−Vf'σm=1−Vf'σm'+Vf'σf, so Vf'=σm−σm'σf−σm'+σm

Tσm'=Emεf=2.4×13.9=33.4MPahe value of s'm is found from:

sσ1=Vfσf+1−Vfσm'o V'f = 8.2 × 10-3, and so the composite strength is given by:

i.e:

σ1=0.5×3200+0.5×33.4=1620MPa

(ii) Find the minimum volume fraction of fibres required to produce strengthening of the composite and calculate the tensile modulus at that fibre loading.

Here we need to find V"f, which is when s*1 for the high Vf model equals the strength of the unreinforced matrix, s*m:

σm=1−Vf''σm'+Vf''σf

i.e.

Vf''=σm−σm'σf−σm'=60−33.43200−33.4=8.4×10−3

aE1=8.4×10−3×230+1−8.4×10−3×2.4=4.31GPand the corresponding modulus is:

which shows that, even for this small fibre addition, the modulus is almost doubled.

3. Calculate the tensile strength for the E-glass / polyester composite series in Q1 where the stress is applied parallel to the fibre direction. The tensile strengths of fibres and matrix are 1800 MPa and 55 MPa, respectively

As before, first determine the failure strains:

εf=1800/76×103=2.37×10−2

and so this is a “brittle matrix” composite, where the strength is given by:

At low volume fraction of fibres: σ1=Vfσf'+1−Vfσm

and at high volume fraction of fibres: σ1=Vfσf

To find the stress in the fibres when the matrix fails: σf'=Efεm=76×1.83×10−2=1390MPa

so the composite strength is given by the larger of:

σ1=1390Vf+551−Vf=55+1335Vf

aσ1=1800Vfnd


4. A nylon 6.6 bar (uniaxial tensile modulus 2.7 GPa, shear modulus 1.015 GPa and uniaxial tensile strength 70 MPa) is reinforced by 20% by volume of carbon fibres (uniaxial tensile modulus 230 GPa and uniaxial tensile strength 3200 MPa). The bar is produced by injection moulding in such a way that the fibres (of length 400 mm and diameter 6 mm) are aligned along the axis of the bar.

(i) Find the mean tensile stress carried by the fibres and the overall stress carried by the bar when a tensile strain of 10-4 is applied to it.

The mean fibre stress is given by σf=Efε1η1, so we need to know the aspect ratio: a=l/d=4006=66.7 and the inter-fibre spacing, R. Assuming fibres to be arranged in a square array with a fibre at each corner, the relationship between volume (or area) fraction and R is given by:

Vf=πd24/2R2=0.2 and so: 2Rd=π4×0.2=1.982 and so the parameter n can now be found:

Now, the fibre length correction factor can be found:

η1=1−tanhnana=1−tanh0.1136×66.70.1136×66.7=0.868

and the mean fibre stress is σf=230×103×10−4×0.868=20MPa

The total stress in the composite is obtained from:

σ1=Vfσf+1−Vfσm=0.2×20+0.8×2.7×103×10−4=4.22MPa

(ii) Calculate the tensile strength of the bar and the mean fibre stress at failure. Find also the tensile strengths for composites containing 30% by volume of fibres of the same size and 30% by volume of fibres of length 600 mm. The shear strength of the fibre-matrix interface can be taken to be 32 MPa.

The critical length for this type of fibre is given by:

lc=σf2τid=32002×32×6=300μm

so that it will be possible for some fibres to reach their breaking stress at the failure plane. Since the matrix is a thermoplastic, the model is a brittle fibre one, which is confirmed by calculating the matrix and fibre failure strains: εm=σm/Em=2.59×10−2 and εf=1.39×10−2

When the fibres fail, the composite strain (e1) is equal to the fibre failure strain (above) and the matrix stress at this point is:

σm'=Emεf=2700×0.0139=37.5MPa

whereas the mean fibre stress (allowing for interfacial failures) is:

σf=σf1−lc2l=32001−300800=2000MPa

and so the stress in the composite when the strain is sufficient to cause fibre breakage is given by: σ1=Vfσf+1−Vfσm'=0.2×2000+0.8×37.5=430MPa.

The composite may not fail at this point if the broken fibres slipping within the matrix are able to withstand a higher stress (low volume fraction case) and the maximum stress carried by the composite would then be given by:

σ1=lc2lVfσf+1−Vfσm3008000.2×3200+0.8×70=296MPa.

So, the composite strength in this case is 430 MPa.

More generally, the composite strength is the higher of σ1=300×32002l−70Vf+70 (low volume fraction fibres) and σ1=3162.5−300×32002lVf+37.5(high volume fraction fibres):