A cylindrical bar has a length 𝐿 = 5 𝑚 and cross section area 𝑆 = 10 𝑚² . The bar is made of a linear elastic material with a density ρ = 2700 kg/m³ and Young’s modulus E = 70 GPa. The bar is suspended as shown in the figure and is in a state of uniaxial tension due to its self-weight. The elastic strain energy stored in the bar equals _________ J. (Rounded off to two decimal places)

Take the acceleration due to gravity as 𝑔 = 9.8 m/s² .

The correct answer is 2.08.

Step-by-step Explanation:

Consider a vertical cylindrical bar of length $L$ suspended from its upper end, subject to its own weight. Let:

$y$ = distance measured from the free (bottom) end of the bar
$\rho$ = density of the material = $2700{\text{kg/m}}^3$
$S$ = cross-sectional area = $10{\text{m}}^2$
$E$ = Young's modulus = $70\text{GPa}=70\times {10}^9{\text{N/m}}^2$
$g$ = acceleration due to gravity = $9.8{\text{m/s}}^2$

As seen in the provided diagram, the bar is suspended from the top and extends downwards vertically under the influence of gravity $g$ over its length $L$.

At any cross-section at a distance $y$ from the free bottom end, the tension force $P\left(y\right)$ is equal to the weight of the portion of the bar below that section:

$P\left(y\right)=\rho gSy$

The elastic strain energy stored in an infinitesimal element of length $dy$ is given by:

$dU=\frac{{\left[P\right(y\left)\right]}^{2}dy}{2ES}=\frac{{\left(\rho gSy\right)}^{2}dy}{2ES}=\frac{{\rho }^2{g}^{2}S{y}^{2}dy}{2E}$

Integrating this expression from $y=0$ to $y=L$ yields the total elastic strain energy stored in the bar:

$U={\int }_0^L\frac{{\rho }^2{g}^{2}S{y}^2}{2E}dy=\frac{{\rho }^2{g}^{2}S}{2E}{[\frac{y^3}{3}}_0^L=\frac{{\rho }^2{g}^{2}S{L}^3}{6E}$

Substitute the given numerical values into the formula:

$U=\frac{{\left(2700\right)}^2\times {\left(9.8\right)}^2\times 10\times 5^3}{6\times 70\times {10}^9}$

First, calculate the numerator:

${\left(2700\right)}^2=7,290,000$

${\left(9.8\right)}^2=96.04$

$5^3=125$

$\text{Numerator}=7,290,000\times 96.04\times 10\times 125=8.751645\times {10}^{11}$

Next, calculate the denominator:

$\text{Denominator}=6\times 70\times {10}^9=4.2\times {10}^{11}$

Now, compute the total elastic strain energy:

$U=\frac{8.751645\times {10}^{11}}{4.2\times {10}^{11}}\approx 2.0837\text{J}$

Rounding to two decimal places, we get:

$U\approx 2.08\text{J}$