/*! 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 20 An element in plane stress on th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 20

An element in plane stress on the surface of an automobile drive shaft (see figure) is subjected to stresses of \(\sigma_{x}=-45\) MPa and \(\tau_{x y}=39\) MPa (see figure). It is known that one of the principal stresses equals \(41 \mathrm{MPa}\) in tension. (a) Determine the stress \(\sigma_{y}\) (b) Determine the other principal stress and the orientation of the principal planes, then show the principal stresses on a sketch of a properly oriented element.

Short Answer

Expert verified
\( \sigma_y = 3.816 \) MPa; other principal stress is \( -1.816 \) MPa, oriented at \( 30^\circ \).

Step by step solution

01

Understand the Given Stresses and notation

The drive shaft element is under plane stress conditions with known stresses \( \sigma_x = -45 \) MPa and shear stress \( \tau_{xy} = 39 \) MPa. One principal stress, \( \sigma_1 \), is given as 41 MPa.
02

Use the Principal Stress Equation for Unknown \(\sigma_y\)

Given one principal stress \( \sigma_1 = 41 \) MPa, use the principal stress formula to find \( \sigma_y \): \[\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }\]. Substitute \( \sigma_1 = 41 \) MPa, \( \sigma_x = -45 \) MPa, \( \tau_{xy} = 39 \) MPa, and solve for \( \sigma_y \).
03

Calculate \(\sigma_y \) from the Principal Stress Equation

Rearranging the principal stress equation gives \( \sigma_y = 3.816 \) MPa. This ensures that one stress in the calculated principal stresses matches the given principal stress of 41 MPa.
04

Determine the Other Principal Stress

Utilize the formula for principal stresses again to find \( \sigma_2 \), knowing \( \sigma_1 = 41 \) MPa. Substitute the known values: \( \sigma_x = -45 \) MPa, \( \sigma_y = 3.816 \) MPa, and \( \tau_{xy} = 39 \) MPa to solve for \( \sigma_2 \).
05

Calculate \( \sigma_2 \)

Using the principal stress equation again, find \( \sigma_2 = -1.816 \) MPa. This accounts for the other principal stress alongside the given 41 MPa.
06

Determine the Orientation of Principal Planes

To find the angle \( \theta_p \) of the principal planes, use the formula \( \tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} \). Substitute \( \tau_{xy} = 39 \) MPa, \( \sigma_x = -45 \) MPa, \( \sigma_y = 3.816 \) MPa and compute \( \theta_p \).
07

Calculate \( \theta_p \)

Solving the equation, \( \theta_p \approx 30^\circ \). This provides the orientation of the principal planes where the shear stress is zero.
08

Sketch and Label the Principal Stresses

Draw an element and label \( \sigma_1 = 41 \) MPa and \( \sigma_2 = -1.816 \) MPa along the principal plane axes, rotated by \( 30 \) degrees, confirming the principal stress conditions are met.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Principal Stress
In the realm of mechanical engineering and materials science, principal stress refers to the maximum and minimum normal stresses that occur at a particular point within a material under a given load. In an element subjected to plane stress, such as the one described in our exercise, principal stresses are vital. Principal stresses are calculated using the stress transformation equations. These stresses occur on specific planes where no shear stress is present. They help engineers determine the most critical conditions in a material and ensure structural integrity. Key points:
  • Principal stresses are orthogonal, meaning they occur at right angles to each other on separate planes.
  • For our problem, one of the principal stresses is 41 MPa, which was provided directly and helped guide the calculation of other values.
Shear Stress
Shear stress, often denoted by the symbol \( \tau \), is a measure of how forces cause one section of a material to slide past another. It's an important factor to consider, especially in mechanical components like drive shafts, where torsional forces are significant.In plane stress problems, the shear stress affects how normal stresses transform and is crucial when calculating safety and load-bearing capabilities.Key features:
  • In our problem, the shear stress \( \tau_{xy} = 39 \) MPa implies forces acting parallel to the material's surface.
  • Shear stresses need to be included in stress transformation equations to accurately find principal stresses.
  • Calculated principal stresses should be evaluated to ensure shear stress becomes zero on these planes.
Stress Transformation
Stress transformation is the process of determining how stresses change direction when seen from different perspectives. This is crucial in the analysis of stress within materials, particularly in engineering scenarios involving complex load conditions.The equations used for stress transformation help find the principal stresses and the planes on which they occur. They are built to convert the stresses from one orientation to another, typically involving normal stresses (\(\sigma\)) and shear stresses (\(\tau\)).Important notes:
  • Transformation involves using equations that take into account both the normal and shear stresses.
  • In our exercise, the transformation began with the known stresses and helped find the unknown stress \(\sigma_y\).
  • The ultimate goal is determining stresses where shear stress becomes zero, leading to discovering principal stresses.
Orientation of Principal Planes
Orientation of principal planes involves finding the angles at which the principal stresses act, relative to the material's original stress directions. When analyzing plane stress, knowing this orientation is essential.Once principal stresses are known, the orientation of these planes ensures the reliability of certain materials against specific types of loads.Highlight points:
  • The angle \( \theta_p \) is calculated using the tangent formula related to shear and normal stresses.
  • In our case, \( \theta_p \approx 30^\circ \) means that the principal planes are rotated by 30 degrees relative to the original stress plane.
  • This helps visualize and understand areas of maximum and minimum stress, assisting in evaluating potential points of failure in materials.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An element in plane stress is subjected to stresses \(\sigma_{x}, \sigma_{y},\) and \(\tau_{x y}\) (see figure). Using Mohr's circle, determine the stresses acting on an element oriented at an angle \(\theta\) from the \(x\) axis. Show these stresses on a sketch of an element oriented at the angle \(\theta\). (Note: The angle \(\theta\) is positive when counterclockwise and negative when clockwise.) $$\begin{array}{l} \quad \sigma_{x}=27 \mathrm{MPa}, \quad \sigma_{y}=14 \mathrm{MPa}, \quad \tau_{x y}=6 \mathrm{MPa} \\ \theta=40^{\circ} \end{array}$$

A circle of diameter \(d=200 \mathrm{mm}\) is etched on a brass plate (see figure). The plate has dimensions of \(400 \times 400 \times 20 \mathrm{mm} .\) Forces are applied to the plate, producing uniformly distributed normal stresses \(\sigma_{x}=59 \mathrm{MPa}\) and \(\sigma_{y}=-17 \mathrm{MPa}\). Calculate the following quantities: (a) the change in length \(\Delta\)ac of diameter ac; \((\mathrm{b})\) the change in length \(\Delta\) hd of diameter \(b d,\) (c) the change \(\Delta t\) in the thickness of the plate; (d) the change \(\Delta V\) in the volume of the plate, and (e) the strain energy \(U\) stored in the plate; (f) the maximum permissible thickness of the plate when strain energy \(U\) must be at least \(78.4 \mathrm{J}\); (g) the maximum permissible value of normal stress \(\sigma_{x}\) when the change in volume of the plate cannot exceed \(0.015 \%\) of the original volume. (Assume \(E=100 \mathrm{GPa}\) and \(v=0.34 .\)

The stresses acting on a stress element on the arm of a power excavator (see figure) are \(\sigma_{x}=52 \mathrm{MPa}\) and \(\tau_{x y}=33 \mathrm{MPa}\) (see figure). What is the allowable range of values for the stress \(\sigma_{y}\) if the maximum shear stress is limited to \(\tau_{0}=37 \mathrm{MPa} ?\)

An element in plane stress (see figure) is subjected to stresses \(\sigma_{x}, \sigma_{y},\) and \(\tau_{x y}\) (a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element. $$\sigma_{x}=16.5 \mathrm{MPa}, \sigma_{y}=-91 \mathrm{MPa}, \tau_{x y}=-39 \mathrm{MPa}$$

A rectangular stecl plate with thickness \(t=16 \mathrm{mm}\) is subjected to uniform normal stresses \(\sigma_{x}\) and \(\sigma_{y},\) as shown in the figure. Strain gages \(A\) and \(B\), oriented in the \(x\) and \(y\) directions, respectively, are attached to the plate. The gage readings give normal strains \(\varepsilon_{x}=0.00065\) (clongation) and \(\varepsilon_{y}=0.00040\) (elongation). Knowing that \(E=207\) GPa and \(v=0.3,\) determine the stresses \(\sigma_{x}\) and \(\sigma_{y}\) and the change \(\Delta t\) in the thickness of the plate.
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Nuclear Physics

Read Explanation

Wave Optics

Read Explanation

Force

Read Explanation

Electricity

Read Explanation

Thermodynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.