Composite Modulus

#407 / Materials Science Medium

If a composite material has the known characteristics shown, and a strain of 0.10, what is its modulus of elasticity in kPa?

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,
$$$\sigma_c=f_1 \sigma_1+f_2\sigma_2+f_3\sigma_3$$$
$$$=(0.7)(30kPa)+(0.25)(20kPa)+(0.05)(100kPa)$$$
$$$=21kPa+5kPa+5kPa=31\:kPa$$$
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,
$$$E=\frac{\sigma_c}{\varepsilon}=\frac{31kPa}{0.10}=310\:kPa$$$
310 kPa


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