Physics · real student question

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.

Question

Determine the efficiency (Krenchel) factor ηθ\eta_\theta of a unidirectional glass-fibre laminate when it is loaded at an angle of 8.78.7^\circ to the fibre axis. Give your answer to three decimal places.

Step-by-step solution

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

    ηθ=cos4θ\eta_\theta=\cos^4\theta

    The fourth power is not arbitrary: one factor of cos2θ\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. Set the calculator to degrees. The angle is given as 8.78.7^\circ, not radians. (In radians, 8.7=0.151848.7^\circ=0.15184 rad — using 8.78.7 rad by mistake would give a wildly wrong cos\cos.)

  3. Evaluate the cosine.

    cos8.7=0.988494\cos 8.7^\circ=0.988494

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

    0.9884942=0.977121,0.9771212=0.9547640.988494^2=0.977121,\qquad 0.977121^2=0.954764

    so ηθ=0.954764\eta_\theta=0.954764.

  5. Round to three decimal places.

    ηθ=0.955\boxed{\eta_\theta=0.955}

  6. Interpret the number. A misalignment of only 8.78.7^\circ already discards about 4.5%4.5\% of the fibres' contribution, and because of the fourth power the loss grows fast: η\eta falls to cos415=0.870\cos^4 15^\circ=0.870 at 1515^\circ and to cos445=0.25\cos^4 45^\circ=0.25 at 4545^\circ. This steep sensitivity is why unidirectional lay-ups are aligned so carefully.

Answer

ηθ=cos4(8.7)=0.955\eta_\theta=\cos^4(8.7^\circ)=0.955

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