Evaluate stress in fibre:
Example, the matrix material can be stressed to its full strength of 80 MPa, calculate the allowable % of broken fibres. What will be the stress in fibre in both the cases?
Solution
σ′ = 1200 MPa, σ′ = 80 MPa, V _{f}= 0.7, E _{f}/ E_{m} = 24
∴ x = (1200 - 80 + 0.7 (1 - 24) 80) /(1200 + 80 (1 - 24))
= 0.2625
26.25% fibre might be broken.
The fibre stress when σ′_{m} = 64 MPa .
σ′_{f }= σ′_{m} × E_{ f} / E_{m}
= 64 × 24 = 1536 MPa
When
σ′_{m }= 80 MPa, σ′_{f} = 80 × 24 = 1920 MPa
Note down that in both the cases the fibre stress is less than its strengths of 2400 MPa.
This is concluded that less broken fibres are allowable if the matrix material is permitted to be stressed to its capacity but in such case the fibre shall be subjected to higher stress.