/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 The load P causes the open coils... [FREE SOLUTION] | 91Ó°ÊÓ

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The load P causes the open coils of the spring to make an angle \(\theta\) with the horizontal when the spring is stretched. Show that for this position this causes a torque \(T=P R \cos \theta\) and a bending moment \(M=P R \sin \theta\) at the cross section. Use these results to determine the maximum normal stress in the material.

Short Answer

Expert verified
To calculate the maximum normal stress, one would first calculate the torque and bending moment using the provided formulas, and then substitute these values into the formula for stress. Without specific values provided for the load, radius, angle, section module, and polar moment of inertia, a numerical solution cannot be provided.

Step by step solution

01

Understanding The Problem

It's mentioned that the load \(P\) causes the open coils of the spring to make an angle \(\theta\) with the horizontal when the spring is stretched. This gives two parameters - torque \(T\) and bending moment \(M\) at the cross section, represented by the equations \(T=P R \cos \theta\) and \(M=P R \sin \theta\) respectively.
02

Calculation of Torque and Bending Moment

With the given load \(P\), radius \(R\) and angle \(\theta\), you can plug these values into the equations for \(T\) and \(M\) to calculate the torque and bending moment respectively.
03

Determining the Maximum Normal Stress

The maximum normal stress \(\sigma\) in the material can be determined using the equation \(\sigma = \frac{M}{S} + \frac{T}{J}\) where \(S\) is the section module and \(J\) is the polar moment of inertia. Substituting the values obtained for \(M\) and \(T\) from the previous step, calculate for \(\sigma\).

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Key Concepts

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

Torque in Materials
When applied to a material, torque represents the rotational force causing an object to twist. In essence, it's the rotational equivalent of linear force. In materials science, it plays a critical role in understanding how objects will respond when forces aim to rotate them.

In the exercise, the angle \(\theta\) is the pivotal factor in determining the amount of torque \(T\) at the cross section of the spring, given by the formula \(T=P R \cos \theta\). It's essential to understand that torque can induce shear stress within the material, potentially causing deformation. When calculating torque, we consider the force applied \(P\), the distance from the pivot point \(R\), and the angle in relation to the force direction \(\theta\), which informs us about the effectiveness of the force in causing rotation.
Bending Moment
The bending moment \(M\) refers to the measure of the bending effect that can occur when a force is applied at a certain distance from the point of support. It's a concept that factors heavily into beam design and analysis, as it quantifies the internal moments that cause a material to bend.

In our exercise, the bending moment is given as \(M=P R \sin \theta\), highlighting how the perpendicular component of the force contributes to the bending. It is the sine component that relates to the force's distance from the neutral axis, promoting bending. Understanding the interplay of force, distance, and the resulting bending moment is essential for effectively designing and analyzing materials and structures in engineering.
Mechanics of Materials
Mechanics of materials, also known as strength of materials, is a subject that deals with the behavior of solid objects subject to stresses and strains. Its primary purpose is to determine the load capacity of a material and how it will deform under various types of forces. This includes understanding concepts like stress, strain, torque, and bending moments.

An in-depth knowledge of mechanics of materials is fundamental when approaching problems like the one in our exercise. Analyzing how a spring stretches under a load \(P\) and determining the resulting normal stress incorporates applying principles such as equilibrium, compatibility of deformation, and constitutive relations of materials—concepts central to the discipline of mechanics of materials.
Normal Stress in Materials
Normal stress in materials arises when an external force acts perpendicular to the surface area of a material. It is a measure of the intensity of the internal forces these materials experience, which can either compress or elongate them.

In the case of this exercise, the normal stress is of particular interest because it reveals how the spring's material will behave under the combined effects of torque and bending moment. The formula \(\sigma = \frac{M}{S} + \frac{T}{J}\) integrates the bending moment \(M\) and the torque \(T\) to calculate the maximum normal stress in a cross-section of the spring. Here, \(S\) is the section modulus, which concerns the cross-section's size and shape, and \(J\) is the polar moment of inertia, reflecting the distribution of the material around the axis of rotation. Understanding the maximum normal stress is crucial to prevent material failure caused by excessive load.

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Most popular questions from this chapter

The weight of 175 lb is dropped from a height of \(4 \mathrm{ft}\) from the top of the \(\mathrm{A} 992\) steel beam. Determine the maximum deflection and maximum stress in the beam if the supporting springs at \(A\) and \(B\) each have a stiffness of \(k=500 \mathrm{lb} / \mathrm{in.}\) The beam is 3 in. thick and 4 in. wide.

A steel cable having a diameter of 0.4 in. wraps over a drum and is used to lower an elevator having a weight of 800 lb. The elevator is 150 ft below the drum and is descending at the constant rate of \(2 \mathrm{ft} / \mathrm{s}\) when the drum suddenly stops. Determine the maximum stress developed in the cable when this occurs. \(E_{\mathrm{st}}=29\left(10^{3}\right) \mathrm{ksi}, \sigma_{Y}=50 \mathrm{ksi}\)

The collar has a weight of \(50 \mathrm{lb}\) and falls down the titanium bar. If the bar has a diameter of 0.5 in. determine the maximum stress developed in the bar if the weight is (a) dropped from a height of \(h=1 \mathrm{ft}\), (b) released from a height \(h=0,\) and (c) placed slowly on the flange at \(A\). \(E_{\mathrm{ti}}=16\left(10^{3}\right) \mathrm{ksi}, \sigma_{Y}=60 \mathrm{ksi}\).

The strain-energy density must be the same whether the state of stress is represented by \(\sigma_{x}, \sigma_{y},\) and \(\tau_{x y},\) or by the principal stresses \(\sigma_{1}\) and \(\sigma_{2} .\) This being the case, equate the strain-energy expressions for each of these two cases and show that \(G=E /[2(1+\nu)]\)

Determine the speed \(v\) of the 50 -Mg mass when it is just over the top of the steel post, if after impact, the maximun stress developed in the post is \(550 \mathrm{MPa}\). The post has a length of \(L=1 \mathrm{m}\) and a cross-sectional area of \(0.01 \mathrm{m}^{2}\) \\[ E_{\mathrm{st}}=200 \mathrm{GPa}, \sigma_{Y}=600 \mathrm{MPa} \\]

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