Materials for engineering, 3rd Edition - (Malestrom)

Composite materials 205
σ f
*
σ = σ f + σ (1 – f )
c * f * f m ’ f
Stress
σ m
*
*
σ = σ (1 – f )
m *
f
0 f min f crit f f 1.0
6.14 Variation of fracture stress of composite with volume fraction of
fibres when ε > ε .
m * f *
shown in the model in Fig. 6.15, in which a thin fibre is embedded to a depth
x in a medium which adheres to it. If the fibre is pulled, the adhesion between
the fibre and the matrix produces a shear stress τ parallel to the surface of the
fibre. The total force on the fibre is 2πrxτ due to this shear stress. Suppose
the maximum value of shear stress which this interface can withstand is τ max ,
and the breaking stress of the fibre is σ f * : if we require that as we increase
the pull on the fibre, the fibre breaks before it pulls out of the matrix, we
must have
i.e.
πr
2
σ < 2πrxτ [6.9]
f *
σ
4τ f* max
< x d
max
The ratio x/d is called the aspect ratio of the embedded part of the fibre and,
in order to fulfil this condition, it is clear that the fibre has to be sufficiently
long and thin.
Let us now consider a ‘unit cell’ (Fig. 6.16) of matrix containing a
representative short fibre of length l. If a tensile load P is applied to each end
of the unit cell, the load is transferred into the fibre from both ends by shear
forces at the fibre/matrix interface. Consider an element of fibre at a distance
x from one end, Fig. 6.17, balancing the loads on the element we have:
P – dP + 2πr dx τ = P