Question 14.2: Consider the data given in Example 14.1. If the axial elonga...

Using Equation 14.54,

the total area of cross section of the tension member required in each material system is given by
Carbon fiber reinforcement:

\frac{250\times 600}{0.6\times 230,000\times 0.01}=108.7 \ mm^2

Kevlar fiber reinforcement:

\frac{250\times 600}{0.6\times 125,000\times 0.01}=200.0 \ mm^2

Glass fiber reinforcement:

\frac{250\times 600}{0.6\times 75,000\times 0.01}=333.3 \ mm^2

Let us take the ply thickness as 0.5 mm and consider the following cross sections:
Carbon fiber reinforcement:

11×10.5 (area of c/s =115.5 mm²)

Kevlar fiber reinforcement:

14.5×4 (area of c/s=203 mm²)

Glass fiber reinforcement:

19×18 (area of c/s =342 mm²)

Then, using the densities estimated in Example 14.1, we can compute the corresponding mass of the tension member as follows:
Carbon/epoxy:

\frac{11\times 10.5\times 600}{1000}\times 1.528=105.9 \ g

Kevlar/epoxy:

\frac{14.5\times 14\times 600}{1000}\times 1.318=160.5 \ g

Glass/epoxy:

\frac{19\times 18\times 600}{1000}\times 1.996=409.6 \ g

Axial elongation in each case is obtained as follows:
Carbon/epoxy:

\frac{250\times 600}{0.6\times 230,000\times \left(11\times 10.5\right) } =9.41\times 10^{-3} \ mm

Kevlar/epoxy:

\frac{250\times 600}{0.6\times 125,000\times \left(14.5\times 14\right) } =9.85\times 10^{-3} \ mm

Glass/epoxy:

\frac{250\times 600}{0.6\times 75,000\times \left(19\times 18\right) } =9.75\times 10^{-3} \ mm

As we can see, all the three alternatives are acceptable from the point of view of axial elongation; however, carbon/epoxy gives us the minimum mass. Thus, we choose carbon/epoxy as the material system. The number of plies in this case is 21 and the processing method is similar to the one described in Example 14.1.