Question 6 of 7: Glass-Fibre Load Sharing in Nylon; Minimum Radius Ratio for CN=4
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2014. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Any five questions constitute
a complete paper; only the first five questions as they appear in the answer book are marked. All
seven questions are solved below for completeness.
Find. (a) Fraction of applied load carried by the glass fibres. (b) Derive the
minimum cation/anion radius ratio for 4-fold coordination.
Fig. Q6b — four anions at alternating corners of a cube (a regular
tetrahedron), with the cation at the cube centre.
Approach
Part (a) uses the isostrain (parallel, rule-of-mixtures) load-sharing assumption appropriate for
fibres and matrix loaded together along the fibre direction. Part (b) places the four anions at
alternating corners of a cube (forming a regular tetrahedron) with the cation at the cube centre,
then uses two geometric contact conditions — anion–anion touching along the cube's face
diagonal, and cation–anion touching along half the body diagonal — to solve for the
limiting radius ratio.
(a) Isostrain load fraction. Under equal strain, each constituent carries load
in proportion to $E_iV_i$:
$$\frac{F_{glass}}{F_{total}}=\frac{E_{glass}V_f}{E_{glass}V_f+E_{nylon}V_m}
=\frac{(10\times10^6)(0.10)}{(10\times10^6)(0.10)+(0.5\times10^6)(0.90)}
=\frac{1.0\times10^6}{1.45\times10^6}=\boxed{69.0\%}.$$
Despite being only 10% of the volume, the glass fibres carry the large majority of the load because
their modulus is 20× that of the nylon matrix.
(b) Setting up the geometry. Place four anions (radius $r_A$) at alternating
corners of a cube of edge $a$ (a regular tetrahedron), with the cation (radius $r_C$) at the cube
centre. At the minimum stable ratio, the anions just touch each other and the
cation touches all four anions simultaneously.
Anion–anion contact (tetrahedron edge = cube face diagonal). The
tetrahedron formed by alternating cube corners has edge length equal to the cube's face diagonal,
$a\sqrt2$; setting this equal to $2r_A$ (anions touching),
$$a\sqrt2=2r_A\ \Rightarrow\ a=r_A\sqrt2.$$
Cation–anion contact (half the body diagonal). The distance from the
cube centre to any corner is half the body diagonal, $\frac{a\sqrt3}{2}$; setting this equal to
$r_C+r_A$ (cation touching each anion),
$$\frac{a\sqrt3}{2}=r_C+r_A.$$
Solve for the ratio. Substituting $a=r_A\sqrt2$,
$$\frac{r_A\sqrt2\cdot\sqrt3}{2}=r_C+r_A\ \Rightarrow\ r_C=r_A\left(\frac{\sqrt6}{2}-1\right)
\ \Rightarrow\ \frac{r_C}{r_A}=\boxed{0.225}.$$
Below this ratio the cation would "rattle" in the tetrahedral hole without touching all four
anions (unstable); at or above it, 4-fold coordination is geometrically stable.
