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Chapter 12: Problem 12

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 \(80000 \mathrm{~kg}\) ? \(Y=1.9 \times 10^{11} \mathrm{~Pa}\). First find the Cross-sectional area of column \(=\pi r^{2}=\pi(0.045 \mathrm{~m})^{2}=6.36 \times 10^{-3} \mathrm{~m}^{2}\) Then, from \(Y=(F / A) /\left(\Delta L / L_{0}\right)\), $$ \Delta L=\frac{F L_{0}}{A Y}=\frac{\left[\left(8.00 \times 10^{4}\right)(9.81) \mathrm{N}\right](4.0 \mathrm{~m})}{\left(6.36 \times 10^{-3} \mathrm{~m}^{2}\right)\left(1.9 \times 10^{11} \mathrm{~Pa}\right)}=2.6 \times 10^{-3} \mathrm{~m}=2.6 \mathrm{~mm} $$

Short Answer

Expert verified
The column decreases in length by 2.6 mm.

Step by step solution

01

Understand Given Information

We have a steel column with a length of \( L_0 = 4.0 \) m and a diameter of \( 9.0 \) cm. The load applied is \( 80000 \) kg. The Young's modulus of steel, \( Y \), is given as \( 1.9 \times 10^{11} \) Pa.
02

Calculate the Radius and Cross-sectional Area

Convert the diameter to meters, \( 9.0 \) cm = \( 0.09 \) m. The radius \( r \) is half of that: \( r = 0.045 \) m. The cross-sectional area \( A \) is \( \pi r^2 \). Substitute in the radius:\[ A = \pi (0.045)^2 = 6.36 \times 10^{-3} \text{ m}^2 \].
03

Determine the Force Applied

The weight of the load is the force \( F \). This is calculated using \( F = m \cdot g \), where \( m = 80000 \) kg and \( g = 9.81 \) m/s\(^2\). Thus: \[ F = 80000 \times 9.81 = 784800 \text{ N} \].
04

Use Young's Modulus to Find Change in Length

The formula for Young's modulus is \( Y = \frac{F / A}{\Delta L / L_0} \). Rearrange to find \( \Delta L \): \[ \Delta L = \frac{F L_0}{A Y} \]. Substitute the values: \[ \Delta L = \frac{784800 \times 4.0}{6.36 \times 10^{-3} \times 1.9 \times 10^{11}} \approx 2.6 \times 10^{-3} \text{ m} = 2.6 \text{ mm} \].
05

Final Answer

The decrease in the length of the cylindrical steel column under the load is approximately \( 2.6 \) mm.

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Key Concepts

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

Cylindrical Steel Column
A cylindrical steel column is essentially a long rod shaped like a cylinder, commonly used in construction. When understanding the physics of the column, it's important to note the shape because it directly influences how the column responds to various forces. These columns maintain structural integrity while supporting heavy loads.

In this exercise, we have a cylindrical steel column with a known length and diameter. The length of the column, given as 4.0 meters, helps determine how much it might compress under a specific load. The column's cylindrical shape ensures that when a force is applied, it distributes evenly around the axis, making its mechanical behavior predictable. This shape is a key feature in architectural and engineering designs for load-bearing structures.
Cross-sectional Area
To understand how a column deforms under a load, we need to calculate its cross-sectional area. Imagine slicing the column perpendicular to its length鈥攚hat you see is the cross-section. For cylindrical shapes, the formula to find the area of this circular cross-section involves the radius.

The cross-sectional area, denoted as \(A\), can be determined using the formula \(A = \pi r^2\), where \(r\) is the radius of the column. In our case, the diameter is 9.0 cm, or 0.09 meters, so the radius is half of that, 0.045 meters. Substituting into the formula gives us the area:
  • \(A = \pi (0.045)^2 = 6.36 \times 10^{-3} \text{ m}^2\)
This value is crucial for assessing how the column will compress under applied load using Young's modulus.
Force Calculation
Calculating the force exerted on the column is a crucial step for determining its deformation. The column supports an object, adding pressure due to gravity. The given problem specifies an object weighing 80,000 kg, and we use the force formula \(F = m \cdot g\), where \(m\) represents mass and \(g\) is the acceleration due to gravity (approximately 9.81 m/s\(^2\)).

We calculate the force as follows:
  • \(F = 80000 \times 9.81 = 784800 \text{ N}\)
This force is important to understand how much stress is applied to the column. Once the force is known, combining this with the cross-sectional area and Young's modulus allows us to compute the decrease in length of the column, giving critical insights into how it behaves under stress.

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Most popular questions from this chapter

A \(60-\mathrm{kg}\) motor sits on four cylindrical rubber blocks. Each cylinder has a height of \(3.0 \mathrm{~cm}\) and a crosssectional area of \(15 \mathrm{~cm}^{2}\). The shear modulus for this rubber is \(2.0 \mathrm{MPa} .(a)\) If a sideways force of \(300 \mathrm{~N}\) is applied to the motor, how far will it move sideways? (b) With what frequency will the motor vibrate back and forth sideways if disturbed?

What is the mass of 1 liter \(\left(1000 \mathrm{~cm}^{3}\right)\) of cottonseed oil of density \(926 \mathrm{~kg} / \mathrm{m}^{3} ?\) How much does it weigh? $$ \begin{aligned} m &=\rho V=\left(926 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(1000 \times 10^{-6} \mathrm{~m}^{3}\right)=0.926 \mathrm{~kg} \\ \text { Weight } &=m g=(0.926 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=9.08 \mathrm{~N} \end{aligned} $$

What volume does \(300 \mathrm{~g}\) of mercury occupy? The density of mercury is \(13600 \mathrm{~kg} / \mathrm{m}^{3}\). From \(\rho=m / V\), $$ V=\frac{m}{\rho}=\frac{0.300 \mathrm{~kg}}{13600 \mathrm{~kg} / \mathrm{m}^{3}}=2.21 \times 10^{-5} \mathrm{~m}^{3}=22.1 \mathrm{~cm}^{3} $$

Battery acid has a specific gravity of \(1.285\) and is \(38.0\) percent sulfuric acid by weight. What mass of sulfuric acid is contained in a liter of battery acid?

A solid cube of aluminum is \(2.00 \mathrm{~cm}\) on each edge. The density of aluminum is \(2700 \mathrm{~kg} / \mathrm{m}^{3}\). Find the mass of the cube. $$ \text { Mass of cube }=\rho V=\left(2700 \mathrm{~kg} / \mathrm{m}^{3}\right)(0.0200 \mathrm{~m})^{3}=0.0216 \mathrm{~kg}=21.6 \mathrm{~g} $$
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