Question 14.1: Design, adopting a micromechanics-based approach, a tension ...
Design, adopting a micromechanics-based approach, a tension member of length 600 mm for carrying a static uniaxial tensile load of 200 kN. Consider the following as available material systems:
Carbon fiber reinforcement:
Kevlar fiber reinforcement:
E_{1f}=125 \ GPa,\left(\sigma ^T_{1f}\right)_{ult} =3600 \ MPa,\rho_f=1.45 \ g/cm^3Glass fiber reinforcement:
E_{f}=75 \ GPa,\left(\sigma ^T_{f}\right)_{ult} =3400 \ MPa,\rho_f=2.58 \ g/cm^3Cast epoxy resin:
E_{m}=3.6 \ GPa,\left(\sigma ^T_{m}\right)_{ult} =72 \ MPa,\rho_m=1.12 \ g/cm^3Take lower mass as a design criterion.
Let us first choose the manufacturing method for making the tension members.
For the production of limited numbers, matched-die-molding can be conveniently employed for making laminates of suitable size from which the tension members of rectangular cross section can be obtained by parting appropriately.
For matched-die-molded unidirectional composites, a high fiber volume fraction of 0.6 can be achieved. Next, taking a factor of safety as 1.25, we can readily obtain the design tensile force as 250 kN. Then, for each of the three available fiber reinforcements, the total area of cross section of the tension member required is given by
Carbon fiber reinforcement:
\frac{250\times 1000}{0.6\times 4900}=85.0 \ mm^2
Kevlar fiber reinforcement:
\frac{250\times 1000}{0.6\times 3600} =115.7 \ mm^2
Glass fiber reinforcement:
\frac{250\times 1000}{0.6\times 3400}=122.5 \ mm^2
Let us take the ply thickness as 0.5 mm and consider the following cross sections:
Carbon fiber reinforcement:
10×9.5 (area of c/s=95 mm²)
Kevlar fiber reinforcement:
11×11 (area of c/s=121 mm²)
Glass fiber reinforcement:
12×10.5 (area of c/s=126 mm²)
Note that the thickness of the tension member is a multiple of 0.5 mm. Now, we can estimate the density of each of the three material systems as follows:
Carbon/epoxy:
1.8×0.6+1.12×0.4=1.528 g/cm³
Kevlar/epoxy:
1.45× 0.6 +1.12 ×0.4=1.318 g/cm³
Glass/epoxy:
2.58×0.6+1.12×0.4=1.996 g/cm³
Then, we can compute the corresponding mass of the tension member as follows:
Carbon/epoxy:
\frac{10\times 9.5\times 600}{1000}\times 1.528=87.1 \ g
Kevlar/epoxy:
\frac{11\times 11\times 600}{1000}\times 1.318=95.7 \ g
Glass/epoxy:
\frac{12\times 10.5\times 600}{1000}\times 1.996=150.9 \ g
As we can see, carbon/epoxy gives us the lightest tension member. Thus, for weight-sensitive design, we choose carbon/epoxy as the material system.
Filament winding is a convenient technique for making unidirectional laminate.
Here, we give a possible methodology with hypothetical process parameters. Toward this, a mandrel with flat surfaces can be used. Hoop winding is to be carried out for which first we need to estimate the required bandwidth and number of spools. Let us take the filament diameter and yield of the carbon fibers as 7 µm and 12 k, respectively. Then, the total cross-sectional area \left(A_f\right) of the filaments in one tow is given by
A_f=12,000\times \frac{\pi\times \left(7\times 10^{-3}\right)^2 }{4}=0.4618 \ mm^2
For fiber volume fraction \left(A_f=0.6\right) , the total cross-sectional area \left(A_c\right) of the composite per tow is given by
A_c=\frac{0.4618}{0.6}=0.7697 \ mm^2
Then, the bandwidth \left(B_w\right) for one tow for ply thickness of 0.5 mm is given by
B_w=\frac{0.7697}{0.5}=1.54 \ mm
Note that for more number of tows, the corresponding bandwidth is obtained by multiplying the above figure by the number of tows; that is, bandwidth is given by B_w = 1.54n, n being the number of tows.
The number of spools is chosen primarily based on the size of the product being wound and the desired finish and accuracy. In general, the lower the number of spools, the better the finish and accuracy. Note further that for large number of spools, the bandwidth is high and actual hoop winding angle, which should be 90° in a strict sense, actually drifts away from 90°. On the other hand, too few spools may result in very high winding time and the choice of the number of spools turns out to be a compromise decision.
For small laminates, such as the one in this example, we can conveniently carry out hoop winding with two spools, that is, two tows. Let us then carry out hoop winding with two spools at a bandwidth of 3.1 mm. After the completion of 19 plies, the mandrel-cum-matched-die mold is closed on the flat surfaces and curing is done at high temperature in an oven. Then, the tension members are obtained from the laminates by carrying out parting (see Figure 14.12).
Note: Refer to Chapter 10 for a more detailed discussion on filament winding parameters.
