Question 5 of 7: Fiber-Reinforced Composite Load Sharing and Minimum Radius Ratio for CN 4

Question 5: Fiber-Reinforced Composite Load Sharing and Minimum Radius Ratio for CN 4 (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. Continuous-fiber composite loaded parallel to the fibers (isostrain); $V_f=0.15$ (glass), $V_m=0.85$ (nylon); $E_f=10.2\times10^6$ psi, $E_m=0.65\times10^6$ psi.

Find. (a) Fraction of the total applied force carried by the glass fibers. (b) Derive the minimum cation:anion radius ratio for four-fold (tetrahedral) coordination.

Approach

(a) Under isostrain loading (fibers and matrix experience the same strain, the standard assumption for continuous aligned fibers loaded along their axis), the load carried by each phase is proportional to $E_iV_i$, so the fiber-to-matrix load ratio and hence the fiber load fraction follow directly from the rule of mixtures. (b) The minimum radius ratio is set by the geometric condition at which the central cation just touches all four surrounding anions while those anions are simultaneously touching each other — below this ratio the anions would overlap, which is geometrically forbidden.

1. (a) Load ratio (isostrain / rule of mixtures). Equal strain in both phases means each phase's force is $F_i=\sigma_iA_i=E_i\varepsilon A_i$, so with $A_i\propto V_i$: $$\frac{F_f}{F_m}=\frac{E_fV_f}{E_mV_m}=\frac{(10.2\times10^6)(0.15)}{(0.65\times10^6)(0.85)}$$ $$\boxed{\frac{F_f}{F_m}\approx2.769}$$
2. (a) Fraction on the fibers. $$\frac{F_f}{F_f+F_m}=\frac{F_f/F_m}{1+F_f/F_m}=\frac{2.769}{3.769}$$ $$\boxed{\approx0.735\ \ (73.5\%\ \text{of the load})}$$ Even at only $15$ vol% glass, the fibers carry nearly three-quarters of the applied load, because their modulus is roughly $16\times$ that of the nylon matrix — the essence of stiff-fiber reinforcement of a compliant polymer.
3. (b) Geometric construction. Place four anions of radius $R$ at alternating corners of a cube of edge $a$ (this reproduces the regular tetrahedral arrangement), with the cation of radius $r$ at the cube centre. Anion–anion contact occurs along a face diagonal of the cube (the tetrahedron edge), which has length $a\sqrt2$ and equals $2R$ at the minimum ratio: $$a\sqrt2=2R\quad\Rightarrow\quad a=R\sqrt2$$
4. (b) Cation–anion contact along the body diagonal. The cation at the cube centre touches each anion along the line from the centre to a cube corner, a distance $\tfrac12(a\sqrt3)$ (half the body diagonal), which equals $R+r$ at the minimum ratio: $$R+r=\frac{a\sqrt3}{2}=\frac{(R\sqrt2)\sqrt3}{2}=\frac{R\sqrt6}{2}$$ Solving for the ratio: $$\frac{r}{R}=\frac{\sqrt6}{2}-1$$ $$\boxed{\frac rR\approx0.225}$$ This is the minimum ratio because for any smaller $r/R$, the anions would need to move even closer together to keep touching the cation, which would force them to overlap each other — a geometrically impossible, unstable configuration. Below $r/R\approx0.225$ the structure switches to a lower coordination number (CN 3).