Calculating Stress experienced by fibre composite.

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SUMMARY

The discussion focuses on calculating the stress experienced by glass fibres in a continuous aligned composite under a longitudinal load of 25 kN. The composite consists of 60% matrix with a modulus of elasticity (E) of 2.8 GPa and 40% glass fibre with E of 73 GPa. The calculated stress on the fibres is determined to be 198.91 MPa, using the formula F(fibre)/F(composite) and the stress formula stress = F/A. The method employed is validated by checking against the average stress based on load and area, confirming the correctness of the approach.

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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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