Page 248 - Plastics Engineering
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Mechanical Behaviour of  Composites                             23 1

                Also, as before, the average fibre stress may be obtained from
                       -     [(af)max~(~ - e,)  + [(af>max~+e,  = [(af>max~ (1  - i)
                       fff =              e
                So from (3.6)

                                                                              (3.49)

                  Note that in order to get the average fibre stress as close as possible to the
                maximum fibre stress, the fibres need to be considerably longer than the critical
                length. At the critical length the average fibre stress is only half  of  the value
                achieved in continuous fibres.
                  Experiments show that equations such as (3.49) give satisfactory agreement
                with  the  measured  values  of  strength  and  modulus for polyester  sheets re-
                inforced with  chopped  strands of  glass fibre. Of  course these  strengths and
                modulus  values are only  about  20-25%  of  those  achieved with  continuous
                fibre reinforcement. This is because with randomly oriented short fibres only
                a  small percentage of  the  fibres  are  aligned along the  line of  action of  the
                applied stress. Also the packing efficiency is low and the generally accepted
                maximum value for Vf of about 0.4 is only half of that which can be achieved
                with continuous filaments.
                  In order to get the best out of  fibre reinforcement it is not uncommon to try
                to control within close limits the fibre content which will provide maximum
                stiffness for a fixed weight of  matrix and fibres. In  flexure it has been found
                that optimum stiffness is achieved when the volume fraction is 0.2 for chopped
                strand mat (CSM) and 0.37 for continuous fibre reinforcement.
                  Example 3.18  Calculate the maximum and average fibre stresses for glass
                fibres which have a diameter of  15 pm and a length of  2.5 mm. The interfacial
                shear strength is 4 MN/m2 and L,/L  = 0.3.
                  Solution Since L  > L, then
                        (gf )max  = - -A)=                            x 0.3
                                 2tyC, - 2tyL  e,
                                                     2 x 4 x 2.5 x
                                       -
                                   d
                                                            15
                                                               x
                        (af),,,  = 400 MN/m2
                            - fff = (af)max - 2) = 400 (1  - y)
                Also                    (1

                            5f = 340 MN/m2
                  In practice it should be remembered that short fibres are more likely to be
                randomly oriented rather than aligned as illustrated in Fig. 2.35. The problem
                of  analysing and predicting the performance of  randomly oriented short fibres
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