It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calculate (a) the critical fiber length, and (b) the volume fraction of fibers necessary if (1) the average fiber diameter is 0.030 mm, (2) the average fiber length is 2.4 mm, (3) the fiber fracture strength is 5100 MPa, (4) the fiber-matrix bond strength is 17 MPa, (5) the matrix stress at fiber failure is 17.5 MPa.

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Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength [tex](\sigma_{cd})=630\ MPa = 630*10^6\ Pa[/tex], Fracture strength

[tex](\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa[/tex]

a) The critical length ([tex]L_c[/tex]) is given by:

[tex]L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm[/tex]

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

[tex]\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456[/tex]

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