**Elongation due to its own weight:**
$\delta_1 = \dfrac{PL}{AE}$

Where:

P = W = 7850(1/1000)^{3}(9.81)[300(150)(1000)]

P = 3465.3825 N

L = 75(1000) = 75 000 mm

A = 300 mm^{2}

E = 200 000 MPa

Thus,

$\delta_1 = \dfrac{3\,465.3825 (75\,000)}{300 (200\,000)}$

$\delta_1 = 4.33 \, \text{ mm}$

**Elongation due to applied load:**

$\delta_2 = \dfrac{PL}{AE}$

Where:

P = 20 kN = 20 000 N

L = 150 m = 150 000 mm

A = 300 mm^{2}

E = 200 000 MPa

Thus,

$\delta_2 = \dfrac{20\,000(150\,000)}{300(200\,000)}$

$\delta_2 = 50 \, \text{ mm}$

**Total elongation:**

$\delta = \delta_1 + \delta_2$

$\delta = 4.33 + 50 = 54.33 \, \text{ mm}$ *answer*

## Comments

## In solving for P in the first

In solving for P in the first part, why did you multiply 150 to it?