## Elongation

A titanium rod with a cross sectional area of 20 cm^2 in the below figure is placed under tension. Determine the elongation if the Modulus of Elasticity (E) = 105 x 10^9 N/m^2.

##
__
__**Hint**

**Hint**

$$$\delta =\frac{PL}{AE}$$$

where
$$\delta$$
is the elastic longitudinal deformation,
$$L$$
is the length of member,
$$P$$
is the loading,
$$A$$
is the cross sectional area, and
$$E$$
is Modulus of Elasticity.

##
__
__**Hint 2**

**Hint 2**

Remember to keep track of units when solving for
$$\delta$$
.

Uniaxial Loading and Deformation:

$$$\delta =\frac{PL}{AE}$$$

where
$$\delta$$
is the elastic longitudinal deformation,
$$L$$
is the length of member,
$$P$$
is the loading,
$$A$$
is the cross sectional area, and
$$E$$
is Modulus of Elasticity.

Plugging in the values (keep track of units):

$$$\delta =\frac{PL}{AE}=\frac{45N\cdot 1m}{0.002m^2\cdot 105\times 10^9\frac{N}{m^2}}=\frac{45m}{210,000,000}$$$

$$$\delta =2.14\times 10^{-7}\:m=0.000214\:mm$$$

0.000214 mm