Question

Four identical hollow cylindrical columns of mild steel support a big structure of mass \[50,000\text{ }kg.\] The inner and outer radii of each column are 30 and \[60\text{ }cm\] respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answer 1

Hint: We will first calculate stress on each cylindrical column. Compression stress is given by force per unit area. Here, force is the weight of structure and area will be the cross sectional area of cylindrical columns. We know Young’s modulus is the ratio of stress to the strain.


Complete step by step answer:
Stress is the force acting on the unit area of a material. The effect of stress on a body is called strain. Stress can result in deformation of the body. Strain is the measurement of how much an object is stretched or deformed due to the stress developed in the object.
Stress is defined as the force applied on an object per unit area, or in mathematical terms, it is equal to force applied divided by the area upon which the force acts.
Strain can be defined as change in length of object per unit length of object. This strain is the linear strain.
So strain is: $\dfrac{\text{change in length}}{\text{original length}}$
When a stress is present in an object, it leads to strain, or vice versa.
We are given that four identical hollow cylindrical columns of mild steel support a big structure of mass \[50,000\text{ }kg.\]

The mass of the structure is given as \[50,000\text{ }kg,\] so its weight will be gravity times mass.
\[\begin{align}
  & W=50,000\times 9.8N \\
 & W=4,90,000N \\
\end{align}\]
Now, the compressional force on the steel columns is provided by the weight of the structure.
So, we have:
\[F=4,90,000N\]
Now, this force is equally distributed over the four steel columns.
So, compressional force over one column will be
\[\begin{align}
  & {{F}_{1}}=\dfrac{F}{4} \\
 & {{F}_{1}}=\dfrac{4,90,000}{4}N \\
 & {{F}_{1}}=1,22,500N \\
\end{align}\]
Now, we will calculate the cross sectional area of cylindrical steel columns.
Given inner and outer radii of each column are 30 and \[60\text{ }cm\] respectively,
Using, $A=\pi \left( {{r}_{e}}^{2}-{{r}_{i}}^{2} \right)$
Where,
$A$ is the area of hollow cylinder
${{r}_{e}}$ is the exterior or outer radius
${{r}_{i}}$ is the interior or inner radius
We have,
${{r}_{e}}=60\text{ }cm=0.6\text{ }m$
${{r}_{i}}=30\text{ }cm=0.3\text{ }m$
So, the required cross sectional area will be:
$\begin{align}
  & A=\pi \left( {{\left( 0.6 \right)}^{2}}-{{\left( 0.3 \right)}^{2}} \right) \\
 & A=\pi \left( 0.36-0.09 \right) \\
 & A=0.27\pi \\
\end{align}$
Now, Stress is force per unit area
So,
$\begin{align}
  & \text{Stress}=\dfrac{{{F}_{1}}}{A} \\
 & \text{Stress}=\dfrac{1,22,500N}{0.27\pi } \\
\end{align}$
Now, as we have defined Young’s modulus, we will calculate the strain
$Y=\dfrac{\text{Stress}}{\text{Strain}}$
So,
\[\text{Strain}=\dfrac{\text{Stress}}{Y}\]
We have, $Y=2\times {{10}^{11}}\text{ }Pa$ for steel
Thus,
\[\text{Strain}=\dfrac{\text{Stress}}{2\times {{10}^{11}}}\]
Substituting, $\text{stress}=\dfrac{1,22,500}{0.27\pi }$
We have,
\[\text{Strain}=\dfrac{1,22,500}{0.27\pi \times 2\times {{10}^{11}}}\]
Substituting $\pi =3.1421$
\[\text{Strain}=\dfrac{1,22,500}{0.27\times 3.1421\times 2\times {{10}^{11}}}\]
\[\text{Strain}=7.22\times {{10}^{-7}}\]
Hence, the compressional strain of each column is \[7.22\times {{10}^{-7}}\]

Note:
As the cylindrical columns placed below the mass are identical, the weight of the structure is therefore distributed equally between the columns. Furthermore, the compressional strain produced in each column is also equal because of their identical shape and size.