/*! 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 93 A bar with cross-sectional area ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
(a) \( \sigma = \frac{F}{A} \cos^2\theta \); (b) \( \tau = \frac{F}{A} \sin\theta \cos\theta \); (c) \(\theta = 0^\circ\); (d) \(\theta = 45^\circ\).

Step by step solution

01

Understanding the Problem

We are given a bar subjected to tensile forces and need to find the tensile and shear stresses on a plane inclined at an angle \(\theta\) to the perpendicular to the bar.
02

Determine Normal Stress

The tensile (normal) stress \(\sigma\) on an inclined plane at an angle \(\theta\) is given by \( \sigma = \frac{F}{A} \cos^2\theta \), where \( F \) is the force and \( A \) is the cross-sectional area.
03

Determine Shear Stress

The shear (tangential) stress \(\tau\) on the inclined plane is given by \( \tau = \frac{F}{A} \sin\theta \cos\theta \).
04

Maximize Tensile Stress

The tensile stress is maximized when \( \cos^2\theta = 1 \). This occurs at \(\theta = 0^\circ\).
05

Maximize Shear Stress

The shear stress is maximized when \( \sin\theta \cos\theta = \frac{1}{2} \sin 2\theta \) is maximum, which occurs at \(\theta = 45^\circ\).

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 91Ó°ÊÓ!

Key Concepts

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

Tensile Stress
When we talk about tensile stress, we refer to the normal stress on a plane within a material. To make this simpler, imagine a bar that is being pulled from both ends. This pulling force results in stress over the cross-sectional area perpendicular to the force. Now, if we consider a plane inside the bar that is at an angle, labeled \( \theta \), to the perpendicular, you'll observe something interesting. The tensile stress experienced by this plane isn’t just the straightforward force divided by area, \( \frac{F}{A} \), but rather \( \frac{F}{A} \cos^2\theta \). Why cosine squared? Well, it comes down to the directional component of the force over this angle. The stress decreases as the plane's angle increases, approaching zero when the plane itself is parallel to the force's direction.
Understanding tensile stress is essential because it helps in predicting materials' behavior under loads. You can never see stress, but knowing about how it acts in different scenarios ensures safety and efficiency in design. Remember:
  • The greater the angle \( \theta \), the smaller the tensile stress on that plane.
  • Tensile stress is highest when \( \theta \) is 0°, meaning the force is directly aligned with the plane.
  • This is crucial in engineering to ensure structures can bear loads without failure.
Shear Stress
Shear stress can initially seem more complex, but it's actually another layer of insight into how forces interact within materials. Picture again the bar under force, but now focus on what happens along that angular plane. Shear stress describes how the material slips along the plane, instead of stretching. This is expressed by the formula \( \tau = \frac{F}{A} \sin\theta \cos\theta \). The product of sine and cosine reflects how both the perpendicular and parallel components of the force impact the plane.
At \( \theta = 0° \) or \( 90° \), the sine function causes the shear stress to be zero, because those are points where the plane isn't cut across the exerted force. However:
  • Shear stress reaches its peak at \( \theta = 45° \), where both sine and cosine components are perfectly balanced, maximizing the tangential effect on the plane.
  • This stress impacts how layers of the material slide over each other, a crucial point in understanding potential points of failure like shear strain or cutting.
  • Knowing the shear stress can aid in designing materials that are less prone to breaking under such tangential forces.
Stress Maximization
Maximizing stress in engineering and physics is less about pushing materials to their limits and more about understanding the conditions under which these limits occur. When analyzing both tensile and shear stress, it's crucial to recognize those scenarios that lead to their maximum values.
For tensile stress, it's straightforward: the maximum occurs when the angle \( \theta \) of the plane is 0°, so the force is directly applied along the same direction. This understanding ensures that when building or testing structures, the materials chosen are best suited for the forces applied.
Shear stress, on the other hand, can't reach its maximum without a precise condition: \( \theta = 45° \). Shear strength is vital, particularly in structures that experiences torsion or when bolts and rivets are at play.
Understanding stress maximization doesn't just apply to theoretical conditions but to practical applications like:
  • Detecting potential failure points in designs.
  • Ensuring that materials used in construction have the necessary tolerance levels.
  • Fine-tuning designs to take advantage of angles where stress distribution improves performance longevity.
The analysis of angles for maximum stress is essential in avoiding over-strain and ensuring the safety and durability of engineering marvels.

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

In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 \(\mathrm{kg}\) from a vertical steel wire 0.50 \(\mathrm{m}\) long and \(2.5 \times 10^{-3} \mathrm{cm}^{2}\) in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 \(\mathrm{kg}\) . For each wire, compute (a) the tensile strain and \((\mathrm{b})\) the elongation.

Bulk Modulus of an Ideal Gas. The equation of state (the equation relating pressure, volume, and temperature) for an ideal gas is \(p V=n R T\) , where \(n\) and \(R\) are constants. (a) Show that if the gas is compressed while the temperature \(T\) is held constant, the bulk modulus is equal to the pressure. (b) When an ideal gas is compressed without the transfer of any heat into or out of it, the pressure and volume are related by \(p V^{\gamma}=\) constant, where \(\gamma\) is a constant having different values for different gases. Show that, in this case, the bulk modulus is given by \(B=\gamma p\) .

You take your dog Clea to the vet, and the doctor decides he must locate the little beast's center of gravity. It would be awkward to hang the pooch from the ceiling, so the vet must devise another method. He places Clea's front feet on one scale and ber hind feet on another. The front scale reads \(157 \mathrm{N},\) while the rear scale reads 89 \(\mathrm{N}\) . The vet next measures Clea and finds that her rear feet are 0.95 \(\mathrm{m}\) behind her front feet. How much does Clea weigh, and where is her center of gravity?

A door 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high weighs 280 \(\mathrm{N}\) and is supported by two hinges, one 0.50 \(\mathrm{m}\) from the top and the other 0.50 \(\mathrm{m}\) m from the bottom.Each hinge supports half the total weight of the door. Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

See all solutions

Recommended explanations on Physics Textbooks

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.