Question

A solid cylindrical steel column is \(4 \mathrm{~m}\) long and \(9 \mathrm{~cm}\) in diameter \(\cdot\left(\mathrm{Y}_{\text {steel }}=1.9 \times 10^{11} \mathrm{Nm}^{-2}\right)\). The decrease in length of the column, while carrying a load of \(80000 \mathrm{~kg}\) is........ (A) \(3.2 \mathrm{~mm}\) (B) \(1.8 \mathrm{~mm}\) (C) \(4.4 \mathrm{~mm}\) (D) \(2.6 \mathrm{~mm}\)

Short answer

The decrease in length of the column while carrying a load of $80000 \mathrm{~kg}$ is approximately \(1.8 \mathrm{~mm}\) (Option B).

Step-by-step solution

Calculate Area of Cross-Section

We are given the diameter of the cylindrical column as \(9 cm\). To calculate the area of the cross-section A, let's first convert diameter to radius. radius \(r = \frac{diameter}{2} = \frac{9}{2} cm\) Now, the area of the cross-section can be calculated as: A = \(\pi r^2\) A = \(\pi\left(\frac{9}{2}\right)^2 cm^2\) A \(= \pi\left(\frac{81}{4}\right) cm^2\) Now, convert area A to \(m^2\): A \(= \pi\left(\frac{81}{4} \cdot 10^{-4}\right) m^2\)

Calculate Stress

A force of \(80000 kg\) is applied on the column, we need to convert this to Newtons (N) using the gravitational acceleration (\(g = 9.81 m/s^2\)). Force F = mass × acceleration_due to gravity F = \(80000 kg × 9.81 m/s^2\) F = \(784800 N\) Now, calculate the stress (σ) applied on the column using the formula: Stress σ = \(\frac{Force (F)}{Area (A)}\) σ = \(\frac{784800 N}{\pi\left(\frac{81}{4} \cdot 10^{-4}\right) m^2}\)

Calculate Strain

To calculate the strain (ε) in the column, use Young's modulus (Y): Strain ε = \(\frac{Stress (σ)}{Young's Modulus (Y)}\) ε = \(\frac{\frac{784800 N}{\pi\left(\frac{81}{4} \cdot 10^{-4}\right) m^2}}{1.9 \times 10^{11} Nm^{-2}}\)

Find Decrease in Length

Lastly, find the decrease in length of the column by multiplying the strain (ε) and the initial length (L) Decrease in length = Strain (ε) × Initial Length (L) Decrease in length = ε × \(4 m\) Decrease in length = \(\frac{\frac{784800 N}{\pi\left(\frac{81}{4} \cdot 10^{-4}\right) m^2}}{1.9 \times 10^{11} Nm^{-2}} \times 4 m\) Plug in the values and evaluate the expression to find the decrease in length (in meters) and then convert it to millimeters. The result will be the correct answer among the options provided.

Key concepts

Young's modulus

Young's modulus, also known simply as the modulus of elasticity, is a fundamental mechanical property that describes how a material responds to stress. It measures the stiffness of a given material. Simply put, it is the ratio of stress (force per unit area) to strain (proportional deformation) in a material in the linear elasticity regime of uniaxial deformation.
For our cylindrical steel column, Young's modulus (\( Y \)) is given as \( 1.9 \times 10^{11} \mathrm{~Nm^{-2}} \). This value is characteristic of steel and helps us calculate how much the material will deform under a specific load. Young's modulus allows us to relate physical dimensions and applied forces to numerical strain results through the relationship:

• Stress \( (\sigma) = \frac{F}{A} \)
• Strain \( (\epsilon) = \frac{\sigma}{Y} \)

In our problem, the ability to connect stress and strain via Young's modulus was key to determining how much the column would shorten under a load of 80,000 kg.

Stress and strain

Understanding stress and strain is essential when examining how materials deform. Stress is the force applied to a material, generally measured as force per area. For our column, calculating the stress involves the formula:\[ Stress (\sigma) =\frac{Force (F)}{Area (A)} \]
For the current example, the force applied due to the weight of the load is calculated as \( 784800 \mathrm{~N}\) while the cross-sectional area was converted into square meters.
Strain, on the other hand, is a dimensionless quantity describing deformation representing the relative change in shape. Once stress is known, strain can be calculated using Young's modulus:\[ Strain (\epsilon) = \frac{Stress (\sigma)}{Young's Modulus (Y)} \]
Both concepts are fundamental in assessing the elasticity of materials, showing how much they stretch or compress under specific conditions. They are also key to understanding the extent to which physical properties will change when subjected to various loads.

Cylindrical geometry calculation

In physics, the geometry of objects dictates the complexity of calculations, especially for determining material responses under stress. In our exercise, we handle a solid cylindrical column whose geometry directs the primary equations: area and volume.
We begin with converting the diameter into radius (radius = \( \frac{diameter}{2} \)), then use these dimensions for calculating the cross-sectional area \( A \) using the equation:\[A = \pi r^2\]
This area is vital in calculating the stress experienced by the cylinder, as discussed earlier. The transformation into the appropriate units, such as converting \( \mathrm{cm^2} \) to \( \mathrm{m^2} \), ensures that all components remain consistent in the SI unit system.
These geometry-related calculations are a bedrock of applied physics, allowing us to translate physical properties into meaningful results, such as estimating how much a column would shorten when carrying massive loads.