91Ó°ÊÓ

Normal stress is a fundamental concept in mechanics of materials. It measures the internal force experienced by an object in response to an applied load. In this exercise, we determined the normal stress in a steel wire. Normal stress, typically denoted by the symbol \(\sigma\), is calculated by dividing the force \(F\) applied to an object by its cross-sectional area \(A\). For our steel wire:- The force function based on the length \(L\) is \(F(L) = 250,000 (L - 10)\).- The cross-sectional area is given as \(0.1 \text{ in}^2\).The expression for normal stress in our wire is:\[ \sigma(L) = \frac{F(L)}{A} = \frac{250,000 (L - 10)}{0.1} \]\Given that the area is already converted into appropriate units, this results in:\[ \sigma(L) = 360,230,547 (L - 10) \text{ lb/in}^2 \]Normal stress helps predict how materials behave under tension and their resistance to deformation.

The work done when elongating the wire involves calculating the energy required to stretch it. This is equivalent to the area under a force versus displacement graph, which we find using integration. Understanding this concept helps in applications that involve energy conservation and mechanical work.The force changes linearly as it stretches from 10 ft to 10.01 ft, with the force function given by:\[ F(L) = 250,000 (L - 10) \]To find the work done, we need to integrate this force from length 10 ft to 10.01 ft:\[ W = \int_{10}^{10.01} F(L) \, dL = \int_{10}^{10.01} 250,000 (L - 10) \, dL \]Carrying out this integration:\[ W = 250,000 \left[ \frac{1}{2} (L - 10)^2 \right]_{10}^{10.01} = 250,000 \times 0.5 \times (0.01)^2 = 0.125 \text{ ft.lb} \]In a real-world scenario, this value can guide engineers in determining the energy requirements for stretching materials.

Deriving the force function is crucial to understanding how force relates to material deformation. In mechanics of materials, knowing the relationship between force and deformation allows for precise control in structural engineering.For our wire, the force function is linear and of the form:\[ F(L) = m(L - L_0) \]where:- \(m\) is the slope, representing the rate of change of force with length.- \(L_0\) is the initial length of the wire.- \(L\) is the current length of the wire.Given the problem:- It starts with no force at an initial length of 10 ft.- The wire experiences a 2500 lb force once stretched to 10.01 ft.The rate of change or slope \(m\) is calculated as follows:\[ m = \frac{2500 \text{ lb} - 0 \text{ lb}}{10.01 \text{ ft} - 10 \text{ ft}} = 250,000 \text{ lb/ft} \]Plugging these into the force equation gives us:\[ F(L) = 250,000 (L - 10) \]This allows engineers and students to predict how much force will be applied as the wire continues to stretch. Understanding this function is key to ensuring materials can withstand applied forces without failing.