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A solid cylindrical steel column is \(4.0 \mathrm{~m}\) long and \(9.0 \mathrm{~cm}\) in diameter. What will be its decrease in length when carrying a load of \(80000 \mathrm{~kg}\) ? \(\mathrm{Y}=1.9 \times 10^{11} \mathrm{~Pa}\) First find the $$ \begin{array}{l} \text { Cross-sectional area of column }=\pi r^{2}=\pi(0.045 \mathrm{~m})^{2}=6.36 \times 10^{-3} \mathrm{~m}^{2} \\ \text { Then, from } \mathrm{Y}=(\mathrm{F} / \mathrm{A}) /\left(\Delta L / L^{0}\right), \\ \text { Then. from } Y=(F / A) /\left(\triangle L / L_{0}\right), \\ \qquad \Delta L=\frac{F L_{0}}{A Y}=\frac{\left[\left(8.00 \times 10^{4}\right)(9.81) \mathrm{N}\right](4.0 \mathrm{~m})}{\left(6.36 \times 10^{-3} \mathrm{~m}^{2}\right)\left(1.9 \times 10^{11} \mathrm{~Pa}\right)}=2.6 \times 10^{-3} \mathrm{~m}=2.6 \mathrm{~mm} \end{array} $$

Step by step solution

01

Calculate Radius

The diameter of the cylinder is given as \(9.0 \text{ cm}\). The radius \(r\) is half of the diameter, so it is \(0.045 \text{ m}\) or \(4.5 \text{ cm}\).

02

Calculate Cross-Sectional Area

Use the formula for the area of a circle: \(A = \pi r^2\). Plug in the radius to get \(A = \pi (0.045 \text{ m})^2 = 6.36 \times 10^{-3} \text{ m}^2\).

03

Calculate Force

To find the force \(F\), multiply the mass by the acceleration due to gravity: \(F = 80000 \text{ kg} \times 9.81 \text{ m/s}^2 = 784800 \text{ N}\).

04

Use Young's Modulus Formula

Apply the Young's Modulus formula to find the change in length: \(Y = \frac{F/A}{\Delta L/L_0}\). Rearrange for \(\Delta L\): \(\Delta L = \frac{F L_0}{A Y}\).

05

Calculate Decrease in Length

Substitute the known values: \(\Delta L = \frac{(784800 \text{ N})(4.0 \text{ m})}{(6.36 \times 10^{-3} \text{ m}^2)(1.9 \times 10^{11} \text{ Pa})} = 2.6 \times 10^{-3} \text{ m}\). Convert \(2.6 \times 10^{-3} \text{ m}\) to millimeters to get **2.6 mm**.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Column

A cylindrical column is a common shape used in engineering and architecture because of its strength and ability to support weight. In this context, the column is solid and made of steel, which is known for its excellent structural properties. To better understand why a cylindrical shape is advantageous, it helps to visualize how force is distributed across its surface. Because the column is circular, any applied force is evenly spread, which prevents weak points from forming, a common issue with square or rectangular shapes.
This uniform distribution helps manage stress effectively, which makes cylindrical columns a preferred choice for supporting substantial loads in buildings and bridges. Understanding the nature of a cylindrical column helps when calculating other properties such as cross-sectional area and force.

Cross-Sectional Area

The cross-sectional area of a cylindrical column is critical because it determines how much force the column can resist. For a circle, the area can be calculated using the formula \(A = \pi r^2 \), where \(r\) is the radius of the circle.

In our given problem, the column has a diameter of 9.0 cm. The radius, which is half of the diameter, is 4.5 cm or 0.045 m when converted to meters. So, when we plug this into the formula, the cross-sectional area is approximately 6.36 x 10^{-3} m².

• This area plays a role in the column's ability to bear loads.
• Knowing the cross-sectional area is essential for further calculations that deal with force and stress applied to the column.

By ensuring the correct calculation of the cross-sectional area, you can accurately determine how the column will respond to various forces in real-world applications.

Force Calculation

Force is a vector that puts physical influence on an object, often measured in newtons (N). In engineering, it's important to calculate the forces accurately to ensure the safety and stability of structures.

In this scenario, the force acting on the cylindrical column arises from the weight of an 80000 kg load it is supporting. The force can be calculated using the equation \( F = m \, g \), where \(m\) is mass and \(g\) is the acceleration due to gravity \(9.81 \text{ m/s}^2\). So, the force \(F\) is \( 80000 \times 9.81 = 784800 \, N\).

• Accurate force calculation is crucial to ensure the material can withstand it without failing.
• This force value is further used to calculate the deformation of the column.

Being meticulous with these calculations can prevent errors in engineering and architectural design.

Deformation in Solids

Deformation refers to the change in shape of an object when a force is applied to it. In solid materials, this often involves a change in length when subjected to forces like tension or compression.

For this cylindrical column, the deformation is expressed through a decrease in length, and its calculation is guided by Young's Modulus formula, given as \( Y = \frac{F/A}{\Delta L/L_0} \). By rearranging the formula to solve for the change in length \( \Delta L \), the equation becomes \( \Delta L = \frac{F L_0}{A Y} \). Plugging in the previously calculated values, you find that \( \Delta L = 2.6 \times 10^{-3} \text{ m} \), which converts to 2.6 mm.

• Knowing the extent of deformation helps engineers to ensure that the solids do not exceed their limits and risk structural integrity.
• The amount of allowable deformation is determined by the material's properties and the design's safety requirements.

Mechanical Properties of Materials

Mechanical properties refer to how materials respond to various forces, such as tension, compression, and shear. These properties determine the suitability of a material for specific engineering applications.

Young's Modulus is one key mechanical property, representing a material's ability to withstand changes in length when under tension or compression. For steel, with a Young's Modulus of \(1.9 \times 10^{11} \, ext{Pa}\), it indicates a high resistance to deformation, meaning steel is very stiff.

• Understanding mechanical properties helps to choose appropriate materials for different parts of a structure depending on stress and load requirements.
• Other relevant properties include tensile strength, ductility, and toughness.

Proper material selection ensures that structures are not only functional but also safe and durable.