## Composite Modulus

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

##
__
__**Hint**

**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**

**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