Properties of a graphite-epoxy-reinforced plastic

Assume that a graphite-epoxy-reinforced plastic with longitudinal fibers contains 15% graphite fibers, which have a strength of 2000 MPa and elastic modulus of 250 GPa. The strength of the epoxy matrix is 100 MPa, and it has an elastic modulus of 80 GPa. Calculate (a) the elastic modulus of the composite and (b) the fraction of the load supported by the fibers.

Question

Properties of a graphite-epoxy-reinforced plastic

Assume that a graphite-epoxy-reinforced plastic with longitudinal fibers contains 15% graphite fibers, which have a strength of 2000 MPa and elastic modulus of 250 GPa. The strength of the epoxy matrix is 100 MPa, and it has an elastic modulus of 80 GPa. Calculate (a) the elastic modulus of the composite and (b) the fraction of the load supported by the fibers.

Accepted Answer

(a) The data given are [latex] x = 0.15[/latex], [latex]E_{f} = 250 [/latex] GPa, [latex]E_{m} = 80 [/latex] GPa, [latex]σ_{f} = 2000 [/latex] MPa, and [latex]σ_{m}=100 [/latex] MPa. From Eq. (10.16),

[latex]E _{c}=xE_{f} + (1 - x)E_{m} .[/latex] (10.16)

[latex]E_{c}=0.15(250)+(1-0.15)80=37.5+68=105.5[/latex] GPa.

(b) The load fraction [latex]F_f/F_m[/latex] is obtained from Eq. (10.15):

[latex]\frac{F_f}{F_m} =\frac{A_fE_f}{A_mE_m} . [/latex] (10.15)

[latex]\frac{F_f}{F_m} =\frac{(0.15)(250)}{(0.85)(80)}=0.55.[/latex]

Since

[latex]F_c = F_f + F_m[/latex]

and

[latex]F_{m}=\frac{F_{f}}{0.55},[/latex]

we obtain

[latex]F_c =F_f+ \frac{F_f}{0.55} = 2.82F_f,[/latex]

or

[latex]F_f = 0.35F_c .[/latex]

Therefore, the fibers support 35% of the load, even though they occupy only 15% of the cross-sectional area (and hence volume) of the composite.

Question 10.4: Properties of a graphite-epoxy-reinforced plastic Assume tha...

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