Question

A 3.0 m tall, 40 cm diameter concrete column supports a 235,000 kg load. by how much is the column compressed? assume young's modulus for concrete to be 3.0 ✕ 1010 n/m2.

Answer 1

The column is compressed as much as 1.8 × 10⁻³ m

Further explanation

Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.

$\boxed {F = k \times \Delta x}$

F = Force ( N )

k = Spring Constant ( N/m )

Δx = Extension ( m )

The formula for finding Young's Modulus is as follows:

$\boxed {E = \frac{F / A}{\Delta x / x_o}}$

E = Young's Modulus ( N/m² )

F = Force ( N )

A = Cross-Sectional Area ( m² )

Δx = Extension ( m )

x = Initial Length ( m )

Let us now tackle the problem !

Given:

x₀ = 3 m

d = 40 cm = 0,4 m

m = 235000 kg

E = 3 × 10¹⁰ N/m²

Unknown:

Δx = ?

Solution:

$F = w$

$F = m \times g$

$F = 235000 \times 9,8$

$\boxed {F = 2303000 ~ Newton}$

$A = \frac{1}{4} \pi d^2$

$A = \frac{1}{4} \pi 0.4^2$

$\boxed {A = 0.04 \pi ~ m^2}$

$E = \frac{F / A}{\Delta x / x_o}$

$3 \times 10^{10} = \frac{2303000 / ( 0.04 \pi )}{\Delta x / 3}$

$\Delta x / 3 = \frac{2303000 / ( 0.04 \pi )}{3 \times 10^{10}}$

$\Delta x = 3 \times \frac{2303000 / ( 0.04 \pi )}{3 \times 10^{10}}$

$\large {\boxed {\Delta x \approx 1.8 \times 10^{-3} ~ m} }$

Learn more

• Young's modulus : brainly.com/question/6864866
• Young's modulus for aluminum : brainly.com/question/7282579
• Young's modulus of wire : brainly.com/question/9755626

Answer details

Grade: College

Subject: Physics

Chapter: Elasticity

Keywords: Elasticity , Diameter , Concrete , Column , Load , Compressed , Stretched , Modulus , Young

Answer 2

The Young modulus E is given by:
$E= \frac{F L_0}{A \Delta L}$
where 
F is the force applied
A is the cross-sectional area perpendicular to the force applied
$L_0$ is the initial length of the object
$\Delta L$ is the increase (or decrease) in length of the object.

In our problem, $L_0 = 3.0 m$ is the initial length of the column, $E=3.0 \cdot 10^{10}N/m^2$ is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
$r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m$
and the cross-sectional area is
$A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2$
The force applied to the column is the weight of the load:
$W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N$

Now we have everything to calculate the compression of the column:
$\Delta L = \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m$
So, the column compresses by 1.83 millimeters.