Determine the Krenchel efficiency factor for a unidirectional glass-fibre laminate loaded at an angle of 8.7 degrees to the fibre direction. Give the answer to three decimal places.

Step-by-step solution

1. 1

   Recall the Krenchel orientation factor. For a bundle of fibres all lying at the same angle $\theta$ to the load, the fraction of the fibres' axial stiffness that is actually mobilised is

   $\eta_\theta=\cos^4\theta$

   The fourth power is not arbitrary: one factor of $\cos^2\theta$ comes from resolving the applied strain onto the fibre axis, and a second from resolving the resulting fibre force back onto the loading direction.

2. 2

   Set the calculator to degrees. The angle is given as $8.7^\circ$, not radians. (In radians, $8.7^\circ=0.15184$ rad — using $8.7$ rad by mistake would give a wildly wrong $\cos$.)

3. 3

   Evaluate the cosine.

   $\cos 8.7^\circ=0.988494$

4. 4

   Raise it to the fourth power. Square, then square again:

   $0.988494^2=0.977120,\qquad 0.977120^2=0.954764$

   so $\eta_\theta=0.954764$.

5. 5

   Round to three decimal places.

   $\boxed{\eta_\theta=0.955}$

6. 6

   Interpret the number. A misalignment of only $8.7^\circ$ already discards about $4.5\%$ of the fibres' contribution, and because of the fourth power the loss grows fast: $\eta$ falls to $\cos^4 15^\circ=0.870$ at $15^\circ$ and to $\cos^4 45^\circ=0.25$ at $45^\circ$. This steep sensitivity is why unidirectional lay-ups are aligned so carefully.