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Single fibre pull out

**Quick proof:**

Consider a fibre with a remaining embedded length of x being pulled out an increment of distance dx. The work done is given by the product of the force acting on the fibre and the distance moved in the direction of the force

dW = ( 2 π r x ) τ_{i*} dx

The work done in pulling this fibre out completely is therefore given by

\[\Delta W = \int\limits_0^{{x_0}} {} 2\pi rx{\tau _{i*}}{\rm{d}}x = \pi rx_0^2{\tau _{i*}}\]

The number of fibres per unit cross-sectional area, N, is related to the fibre volume fraction, f, and the fibre radius, r by

\[N = \frac{f}{{\pi {r^2}}}\]

so the pull-out work of fracture per unit crack area, G_{cp}, is given by

\[{G_{{\rm{cp}}}} = N\Delta W = \frac{{f\pi rx_0^2{\tau _{i*}}}}{{\pi {r^2}}} = 4f {s^2} r {\tau _{i*}}\]

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