Answer:
Δ L = 2.57 x 10⁻⁵ m
Explanation:
given,
cross sectional area = 1.6 m²
Mass of column = 26600 Kg
Elastic modulus, E = 5 x 10¹⁰ N/m²
height = 7.9 m
Weight of the column = 26600 x 9.8
= 260680 N
we know,
Young's modulus=[tex]\dfrac{stress}{strain}[/tex]
stress = [tex]\dfrac{P}{A}[/tex]
= [tex]\dfrac{260680}{1.6}[/tex]
= 162925
strain = [tex]\dfrac{\Delta L}{L}[/tex]
now,
[tex]Y = \dfrac{stress}{strain}[/tex]
[tex]\Delta L = \dfrac{162925}{Y}\times L[/tex]
[tex]\Delta L = \dfrac{162925}{5 \times 10^10}\times 7.9[/tex]
Δ L = 2.57 x 10⁻⁵ m
The column is shortened by Δ L = 2.57 x 10⁻⁵ m
