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A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000 kg. You can ignore the weight of the post. What are (a) the stress in the post; (b) the strain in the post; and (c) the change in the post's length when the load is applied?

Short Answer

Expert verified
(a) Stress is \(1.60 \times 10^6 \text{ N/m}^2\), (b) Strain is \(8.0 \times 10^{-6}\), and (c) Length change is \(2.0 \times 10^{-5} \text{ m}\).

Step by step solution

01

Calculate the Cross-Sectional Area of the Post

The steel post is circular with a diameter of 25 cm. The radius \( r \) is half of the diameter, so \( r = \frac{25}{2} = 12.5 \) cm or 0.125 m. The cross-sectional area \( A \) of the post is given by the formula for the area of a circle: \[ A = \pi r^2 = \pi (0.125)^2 \approx 0.0491 \text{ m}^2 \].
02

Calculate the Stress in the Post

Stress is defined as the force applied per unit area. The force \( F \) here is due to the load, which is the weight of 8000 kg. Converting this to Newtons: \[ F = 8000 \times 9.8 = 78400 \text{ N} \]. Now, calculate the stress \( \sigma \) using the formula: \[ \sigma = \frac{F}{A} = \frac{78400}{0.0491} \approx 1.60 \times 10^6 \text{ N/m}^2 \].
03

Determine the Strain in the Post

Strain \( \varepsilon \) is defined as the change in length \( \Delta L \) per unit original length \( L_0 \). It is related to stress by Young's modulus \( E \): \[ \varepsilon = \frac{\sigma}{E} \]. For steel, \( E \approx 2.0 \times 10^{11} \text{ N/m}^2 \). So, \[ \varepsilon = \frac{1.60 \times 10^6}{2.0 \times 10^{11}} = 8.0 \times 10^{-6} \].
04

Calculate the Change in Length of the Post

The change in length is given by \( \Delta L = \varepsilon \times L_0 \), where \( L_0 = 2.50 \text{ m} \). Using the strain from Step 3, \[ \Delta L = 8.0 \times 10^{-6} \times 2.50 \approx 2.0 \times 10^{-5} \text{ m} \].

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Key Concepts

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

Young's Modulus
Young's modulus, denoted as \( E \), is a fundamental property in the study of materials. It is a measure of the stiffness of a material, describing the relationship between stress and strain under elastic deformation. It is defined as the ratio of stress \( \sigma \) to strain \( \varepsilon \) in the linear portion of the stress-strain curve. In formula form, it is expressed as \( E = \frac{\sigma}{\varepsilon} \).When dealing with metals like steel, Young's modulus is particularly important because it tells us how much the material will deform under a given load. For steel, Young's modulus is approximately \( 2.0 \times 10^{11} \text{ N/m}^2 \), indicating that steel is quite resistant to stretching under force. This high modulus makes steel an ideal material for structural applications where stiffness is crucial. In our exercise, Young's modulus helps determine how much the steel post will extend when subjected to the weight of an 8000 kg load.
Strain
Strain is a measure of how much an object deforms under stress. It is a dimensionless quantity, as it represents a ratio of two lengths. To calculate strain \( \varepsilon \), we use the formula: \( \varepsilon = \frac{\Delta L}{L_0} \), where \( \Delta L \) is the change in length and \( L_0 \) is the original length.Strain can be small in the case of materials like steel, where a large amount of stress leads to only minimal deformation. In our example, the calculated strain \( \varepsilon \) is \( 8.0 \times 10^{-6} \), indicating that the steel post barely stretches under the given load. This minimal strain is due to the material's high Young's modulus.Understanding strain is crucial, particularly in construction and materials engineering, because it helps predict how structures will behave under various forces.
Cross-Sectional Area
The cross-sectional area of an object is the surface area of the slice when it is cut perpendicular to its length. For round objects like posts, this area can be computed using the formula for the area of a circle: \( A = \pi r^2 \), where \( r \) is the radius of the circle.In the given problem, the steel post has a diameter of 25 cm, giving a radius of 12.5 cm or 0.125 m. The cross-sectional area \( A \) is then calculated as approximately \( 0.0491 \text{ m}^2 \).Knowing the cross-sectional area is integral in calculating stress, as stress is defined as the force exerted on an object divided by its cross-sectional area. This relationship helps engineers in ensuring that materials can handle the stresses they will encounter in practical applications.
Steel Material Properties
Steel is a commonly used material known for its durability, high tensile strength, and relatively low cost. These characteristics make it a mainstay in construction and manufacturing. The properties of steel that are crucial for this exercise include its density, which affects weight calculations, and its Young's modulus, which indicates the material's elasticity. The elasticity of steel allows it to undergo significant stress with minimal deformation, an essential trait for materials used in load-bearing structures. By knowing these properties, engineers can predict how steel will perform under different loads, ensuring safety and efficiency in design. For instance, the exercise illustrates using steel's Young's modulus to determine how much a post will stretch under a specific weight — vital in ensuring the structural integrity of buildings and bridges.

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Most popular questions from this chapter

A uniform metal bar that is 8.00 m long and has mass 30.0 kg is attached at one end to the side of a building by a frictionless hinge. The bar is held at an angle of 64.0\(^\circ\) above the horizontal by a thin, light cable that runs from the end of the bar opposite the hinge to a point on the wall that is above the hinge. The cable makes an angle of 37.0\(^\circ\) with the bar. Your mass is 65.0 kg. You grab the bar near the hinge and hang beneath it, with your hands close together and your feet off the ground. To impress your friends, you intend to shift your hands slowly toward the top end of the bar. (a) If the cable breaks when its tension exceeds 455 N, how far from the upper end of the bar are you when the cable breaks? (b) Just before the cable breaks, what are the magnitude and direction of the resultant force that the hinge exerts on the bar?

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A 350-N, uniform, 1.50-m bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of 500.0 N without breaking, and cable \(B\) can support up to 400.0 N. You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

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A relaxed biceps muscle requires a force of 25.0 N for an elongation of 3.0 cm; the same muscle under maximum tension requires a force of 500 N for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0 cm\(^2\).

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