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A machine component is made of a grade of cast iron for which \(\sigma_{U T}=8\) ksi and \(\sigma_{U C}=20\) ksi. For each of the states of stress shown and using Mohr's criterion, determine the normal stress \(\sigma_{0}\) at which rupture of the component should be expected.

Short Answer

Expert verified
The rupture stress \(\sigma_0\) is 5.71 ksi.

Step by step solution

01

Understand Mohr's Criterion

Mohr's criterion is used to predict the failure of materials under complex stress conditions. It states that failure occurs when the state of stress at a point reaches a critical combination of normal and shear stresses. For cast iron (as in this problem), Mohr's criterion is often simplified due to its brittle nature.
02

Identify Ultimate Stresses

Given are the ultimate tensile stress (\(\sigma_{UT} = 8\) ksi) and the ultimate compressive stress (\(\sigma_{UC} = 20\) ksi) for the cast iron component.
03

Mohr's Failure Criterion Equation

To apply Mohr’s failure criterion, use the equation that relates normal stresses for a brittle material:\[ \sigma_0 = \frac{\sigma_{UT} \cdot \sigma_{UC}}{\sigma_{UC} + \sigma_{UT}} \]
04

Substitute Values into the Criterion

Substitute \(\sigma_{UT} = 8\) ksi and \(\sigma_{UC} = 20\) ksi into the failure criterion equation:\[ \sigma_0 = \frac{8 \times 20}{8 + 20} \]
05

Calculate Normal Stress

Calculate \(\sigma_0\) by solving the equation:\[ \sigma_0 = \frac{160}{28} = 5.71 \text{ ksi} \]
06

Interpret the Result

The normal stress \(\sigma_0\) calculated is the stress at which rupture of the component is expected under the given material properties based on Mohr's criterion.

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Key Concepts

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

Normal Stress
Normal stress is a fundamental concept in material mechanics, representing the internal forces acting on an object per unit area. It results when forces are applied perpendicular (normal) to the surface of a material. Normal stress has a significant role in determining a material's behavior under load.

Mathematically, normal stress \( \sigma \) is given by:
  • \( \sigma = \frac{F}{A} \)
where \( F \) is the force acting normal to the surface, and \( A \) is the cross-sectional area.

In practical terms, understanding normal stress helps in evaluating whether a material or component can withstand the applied loads without failure. The evaluation of normal stress is crucial, especially in designing structures and machines to ensure safety and performance under operational conditions.
Ultimate Tensile Stress
Ultimate tensile stress (UTS), also known as tensile strength, is the maximum amount of tensile stress that a material can support before experiencing failure by breaking. When studying machine components or structural elements, knowing the UTS tells us how much stress they can endure before being pulled apart.

This property is essential in determining how durable and strong an engineering material is. UTS is commonly measured in kilopounds per square inch (ksi) or megapascals (MPa).

In the exercise, the ultimate tensile stress of cast iron is given as \( \sigma_{UT} = 8 \text{ ksi} \). This value means that, before breaking, the material can withstand normal stresses up to 8 ksi. It is important for engineers to consider UTS when designing to prevent tensile failure, especially for materials like cast iron, which are brittle and prone to cracking.
Ultimate Compressive Stress
Ultimate compressive stress (UCS) refers to the maximum compressive force that a material can take before it fails. Unlike tensile stress, compressive stress pushes together the particles of a material.

Understanding UCS helps engineers and designers ensure that a material like cast iron can withstand loading without a risk of compressive failure. Compressive stress is also measured in units like ksi or MPa.

In your exercise related to the Mohr's Criterion, the ultimate compressive stress is \( \sigma_{UC} = 20 \text{ ksi} \). This indicates that the cast iron component can withstand a stress level of up to 20 ksi in compression before failure. For this reason, this property is especially important when the material is going to face significant compressive forces during its service life.

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Most popular questions from this chapter

For the state of plane stress shown, determine the maximum shearing stress when \((a) \sigma_{x}=20 \mathrm{ksi}\) and \(\sigma_{y}=10 \mathrm{ksi},(b) \sigma_{x}=12 \mathrm{ksi}\) and \(\sigma_{y}=5\) ksi. (Hint: Consider both in-plane and out-of-plane shearing stresses.

The compressed-air tank \(A B\) has a 250 -mm outside diameter and an 8 -mm wall thickness. It is fitted with a collar by which a \(40-\mathrm{kN}\) force \(\mathbf{P}\) is applied at \(B\) in the horizontal direction. Knowing that the gage pressure inside the tank is 5 MPa, determine the maximum normal stress and the maximum shearing stress at point \(K\)

Two members of uniform cross section \(50 \times 80 \mathrm{mm}\) are glued together along plane \(a-a\) that forms an angle of \(25^{\circ}\) with the horizontal. Knowing that the allowable stresses for the glued joint are \(\sigma=800 \mathrm{kPa}\) and \(\tau=600 \mathrm{kPa},\) determine the largest centric load \(\mathbf{P}\) that can be applied.

A steel pipe of 12 -in. outer diameter is fabricated from \(\frac{1}{4}\) -in.-thick plate by welding along a helix that forms an angle of \(22.5^{\circ}\) with a plane perpendicular to the axis of the pipe. Knowing that a 40 -kip axial force \(\mathbf{P}\) and an 80 -kip \(\cdot\) in. torque T, each directed as shown, are applied to the pipe, determine the normal and in- plane shearing stresses in directions, respectively, normal and tangential to the weld.

A standard-weight steel pipe of 12 -in. nominal diameter carries water under a pressure of 400 psi. ( \(a\) ) Knowing that the outside diameter is 12.75 in. and the wall thickness is 0.375 in., determine the maximum tensile stress in the pipe. ( \(b\) ) Solve part \(a\), assuming an extra-strong pipe is used, of 12.75 -in. outside diameter and 0.5 -in. wall thickness.

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