## Composite Strain

If a composite material has the known characteristics shown, and a modulus of elasticity of 500 kPa, what is its strain?

##
__
__**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,

$$$\sigma_c=f_1 \sigma_1+f_2\sigma_2+f_3\sigma_3$$$

$$$=(0.6)(50kPa)+(0.3)(10kPa)+(0.1)(70kPa)$$$

$$$=30kPa+3kPa+7kPa=40\:kPa$$$

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,

$$$\varepsilon=\frac{\sigma_c}{E}=\frac{40kPa}{500kPa}=0.08$$$

0.08