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Chapter 11: Problem 28

A vertical, solid steel post 25 \(\mathrm{cm}\) in diameter and 2.50 \(\mathrm{m}\) long is required to support a load of 8000 \(\mathrm{kg}\) . 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: \( 1,598,776 \, N/m^2 \); (b) Strain: \( 8 \times 10^{-6} \); (c) Change in length: 0.02 mm.

Step by step solution

01

Calculate the Cross-Sectional Area

First, determine the cross-sectional area of the post, which approximately has a circular shape. The diameter of the post is given as 25 cm, so the radius is half of that, 12.5 cm, or 0.125 m. The area \( A \) of a circle is calculated by the formula \( A = \pi r^2 \). Thus, \( A = \pi (0.125)^2 \approx 0.0491 \, m^2 \).
02

Compute the Stress

Stress \( \sigma \) is defined as the force divided by the area over which the force is applied. The force \( F \) is equal to the weight of the load, which is \( 8000 \times 9.81 \, N = 78480 \, N \). Hence, the stress is \( \sigma = \frac{78480}{0.0491} \approx 1,598,776 \, N/m^2 \).
03

Calculate the Strain

Strain \( \varepsilon \) is the deformation per unit length, calculated as stress divided by the Young's modulus \( E \) for the material. Assuming steel with \( E \approx 2 \times 10^{11} \, N/m^2 \), the strain is \( \varepsilon = \frac{1,598,776}{2 \times 10^{11}} \approx 8 \times 10^{-6} \).
04

Determine the Change in Length

The change in length \( \Delta L \) can be found by multiplying the original length \( L \) by the strain: \( \Delta L = L \varepsilon = 2.50 \, m \times 8 \times 10^{-6} \approx 0.00002 \, m \). This is equivalent to a change of 0.02 mm.

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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, named after the British scientist Thomas Young, is a measure of the stiffness of a material. It is a fundamental property that quantifies how much a material will deform under a given load.Young's Modulus is defined by the ratio of stress to strain:\[ E = \frac{\sigma}{\varepsilon} \]where \( E \) is Young's Modulus, \( \sigma \) is stress, and \( \varepsilon \) is strain. For materials like steel, it has a high value, indicating that steel requires a lot of force to deform. The units of Young's Modulus are typically in pascals (Pa), or newtons per square meter (N/m虏).Understanding Young's Modulus is crucial when designing structures that need to withstand specific loads without undergoing too much deformation. For instance, in the exercise with the steel post, knowing the Young's Modulus of steel (about \( 2 \times 10^{11} \, \text{N/m}^2 \)) allows us to compute the strain and predict how much the post will elongate.
Cross-Sectional Area
The cross-sectional area of an object is the area of a particular section through the object, perpendicular to its length. It plays a crucial role in determining how much load a material can support.In our exercise, the steel post has a circular cross-section with a given diameter of 25 cm. To find the cross-sectional area \( A \), we'd use the formula for the area of a circle:\[ A = \pi r^2 \]Here, the radius \( r \) is half of the diameter, which is 12.5 cm or 0.125 m. Plugging in the values:\[ A = \pi \times (0.125)^2 \approx 0.0491 \, \text{m}^2 \]This area is crucial for calculating stress as it determines the area over which the force is applied. A larger cross-sectional area would mean that the stress for the same load would be less, making the material safer against deformation or failure.
Elastic Deformation
Elastic deformation refers to the temporary shape change that is reversible when the applied force is removed. It occurs when the material returns to its original form after the stress is released.This concept is significant in materials like steel, which undergo elastic deformation when subjected to stresses within a certain limit (elastic limit). Beyond this limit, the material may experience permanent, or plastic, deformation.Elastic deformation is quantified by calculating strain, the proportional change in the length of a material relative to its original length. Using Young's Modulus, we calculate strain as:\[ \varepsilon = \frac{\sigma}{E} \]In the steel post scenario, with a known stress and Young's Modulus, the strain is determined to be a very small value (\( 8 \times 10^{-6} \)), indicative of a small elongation under an applied load that remains in the elastic range. Thus, when the load is removed, the post returns to its original length.

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Most popular questions from this chapter

A uniform ladder 5.0 \(\mathrm{m}\) long rests against a frictionless, vertical wall with its lower end 3.0 \(\mathrm{m}\) from the wall. The ladder weighs 160 \(\mathrm{N}\) . The coefficient of static betwoen the foot of the ladder and the ground is \(0.40 .\) A man weighing 740 \(\mathrm{N}\) climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder. (a) What is the maximum frictional force that the ground can exert on the ladder at its lower end? (b) What is the actual frictional force when the man has climbed 1.0 \(\mathrm{m}\) along the ladder? (c) How far along the ladder can the man climb before the ladder starts to slip?

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In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8 \(\mathrm{N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{mm},\) what is the breaking stress of the alloy?

A bar with cross-sectional area \(A\) is subjected to equal and opposite tensile forces \(\vec{F}\) at its ends. Consider a plane through the bar making an angle through the bar making an angle \(\theta\) with a plane at right angles to the bar (Fig. 11.64\()\) . (a) What is the tensile (normal) stress at this plane in terms of \(F, A,\) and \(\theta ?(b)\) What is the shear (tangential) stress at the plane in terms of \(F, A,\) and \(\theta ?\) (c) For what value of \(\theta\) is the tensile stress a maximum? (d) For what value of \(\theta\) is the shear stress a maximum?

Three vertical forces act on an airplane when it is flying at a constant altitude und with a constant velocity. These ure the weight of the airplane, an aerodynamic force on the wing of the airplane, and an aerodynamic force on the airplane's horizontal tail. (The aerodynamic forces are exerted by the surrounding air, and are reactions to the forces that the wing and tail exert on the air as the aerodynamic forces are exerted by the surrounding air, and are reactions to the forces that the wing and tail exert on the air as the airplane flies through it) For a particular light airplane with a weight of 6700 \(\mathrm{N}\) , the center of gravity is 0.30 \(\mathrm{m}\) m front of the point where the wing's vertical aerodynamic force acts and 3.66 \(\mathrm{m}\) in front of the point where the tail's vertical aerodynamic force acts. Determine the magnitude and direction (upward or down-ward) of each of the two vertical aerodvamic forces.
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