Solution to Problem 206 Axial Deformation

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 mm²
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 mm²
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