- #1
olski1
- 15
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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?
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?