Calculating Stress experienced by fibre composite.

AI Thread Summary
To determine the stress in the fibres of a composite material under a 25 kN load, the user calculated the force distribution between the matrix and fibres based on their respective moduli and volume fractions. The calculated force on the fibres was approximately 24,999.55 N, leading to a stress of 198.91 MPa when divided by the fibre's cross-sectional area. The method used is valid, as it considers the principles of load sharing and strain compatibility in composite materials. It is also suggested to verify the result against the average stress calculation. The approach effectively illustrates the mechanics of stress distribution in composite structures.
olski1
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Hi,

I need to find the stress taken up by the fibres in the longitudinal direction, when a load of 25kN is applied to a continuous aligned composite with the diameter of 2cm. 60% matirx with E=2.8Gpa and 40% glass fibre E=73Gpa.

Attempt at solution
Known values

V_m = 0.6, E_m = 2.8 Gpa ,
V_f = 0.4 , E_f = 73 Gpa
Cross sectional area A_C=πr^2= π〖(0.01)〗^2=3.142 ×〖10〗^(-4) 〖 m〗^2
Longitudinal Loading F_c= 25 ×〖10〗^3 N

i first found,
F(fibre)/F(composite) = ((73×10^9×0.4)) / (73×10^9×0.4)+(2.8×10^9 )(0.6×3.142 ×10^(-4)) => F(fibre) = 24999.548 N

Then using stress= F/A, Where A(fibre) = V(fibre)*(A(comp))
therefore, stress on fibre is 198.91 Mpa

Is that the right method and answer?
 
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One can check the answer against the average stress (load/area), and realize that assuming the two materials are of the same length, and that continuity applies, they must exhibit the same strain.
 

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