A plain weave (0/90) boron fibre fabric has a Krenchel efficiency factor of 0.5 when loaded parallel to its warp fibres. Calculate its efficiency factor, to three decimal places, when loaded at 31 degrees to the warp.

Step-by-step solution

1. 1

   Read the given 0° value as a check on the fibre split. The Krenchel factor for a set of fibre directions is

   $\eta_\theta=\sum_n a_n\cos^{4}\theta_n$

   where $a_n$ is the fraction of fibres in direction $n$. At $0^\circ$ this gives $a_0\cos^4 0^\circ+a_{90}\cos^4 90^\circ=a_0$. Since that is stated to be $0.5$, the fabric is a balanced weave: $a_0=a_{90}=0.5$.

2. 2

   Find the angle from the load to each fibre set. Loading at $31^\circ$ to the warp means the warp fibres are at $\theta_0=31^\circ$, and the weft fibres, being perpendicular to the warp, are at

   $\theta_{90}=90^\circ-31^\circ=59^\circ$

3. 3

   Write the sum.

   $\eta_\theta=0.5\cos^{4}(31^\circ)+0.5\cos^{4}(59^\circ)$

4. 4

   Evaluate each term.

   $\cos 31^\circ=0.857167\Rightarrow\cos^{4}31^\circ=0.539837$

   $\cos 59^\circ=0.515038\Rightarrow\cos^{4}59^\circ=0.070365$

5. 5

   Combine and round.

   $\eta_\theta=0.5(0.539837+0.070365)=0.5(0.610202)=0.305101$

   $\boxed{\eta_\theta=0.305}$

6. 6

   Sanity-check against the two reference angles. At $0^\circ$ (or $90^\circ$) the factor is $0.5$, and at $45^\circ$ it is $0.5\cos^4 45^\circ\times 2=0.5\times 0.25\times 2=0.25$, the minimum. The answer $0.305$ falls between those limits and, being much nearer $45^\circ$ than $0^\circ$ in effect, is correctly close to the lower end.