Composite Material

#315 / Materials Science Medium

If a composite material has the known characteristics shown, and a measured strain of 0.05, what is its modulus of elasticity?

Expand Hint
$$$\sigma_c=\sum f_i\sigma_i$$$
where $$\sigma_c$$ is the strength parallel to the fiber direction, $$f_i$$ is the volume fraction of the individual material, and $$\sigma_i$$ is the individual material’s strength.
Hint 2
Hooke’s Law:
$$$\sigma=E\varepsilon $$$
where $$\sigma$$ is the stress, $$E$$ is the elastic modulus (modulus of elasticity or Young’s modulus), and $$\varepsilon $$ is the strain.
A composite material is a material formed from two or more “base” materials. The base/basic materials have notably dissimilar physical or chemical properties, but when merged to create a composite, produce properties unlike the individual elements. To find a composite’s strength:
$$$\sigma_c=\sum f_i\sigma_i$$$
where $$\sigma_c$$ is the strength parallel to the fiber direction, $$f_i$$ is the volume fraction of the individual material, and $$\sigma_i$$ is the individual material’s strength. Thus,
$$$\sigma_c=f_1 \sigma_1+f_2\sigma_2$$$
$$$=(0.6)(50kPa)+(0.4)(25kPa)=30kPa+10kPa=40\:kPa$$$
Hooke’s Law:
$$$\sigma=E\varepsilon $$$
where $$\sigma$$ is the stress, $$E$$ is the elastic modulus (modulus of elasticity or Young’s modulus), and $$\varepsilon $$ is the strain. Therefore,
$$$E=\frac{\sigma_c}{\varepsilon}=\frac{40kPa}{0.05}=800\:kPa$$$
800 kPa


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