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Consider the state of stress \(\sigma_{x x}=\sigma_{y y}=\) \(\sigma_{z z}=\sigma\) and \(\sigma_{x y}=\sigma_{y x}=\tau, \sigma_{x z}=\sigma_{z x}=\sigma_{y z}=\sigma_{z y}=0\) Determine the yield conditions on the basis of the Tresca and von Mises criteria. How does hydrostatic loading affect plastic yielding?

Short Answer

Expert verified
Yielding depends on \( \tau \); hydrostatic stress \( \sigma \) does not affect it.

Step by step solution

01

Understanding the Stress Conditions

The stress state is given as \( \sigma_{xx} = \sigma_{yy} = \sigma_{zz} = \sigma \) and \( \sigma_{xy} = \sigma_{yx} = \tau \), with all other shear stresses being zero. This represents a hydrostatic stress condition with an additional shear component.
02

Applying the Tresca Criterion

The Tresca yield criterion states that yielding begins when the maximum shear stress reaches a critical value. The maximum shear stress is calculated as \( \frac{1}{2} \) of the difference between the maximum and minimum principal stresses. Here, the principal stresses are \( \sigma_1 = \sigma + \tau \), \( \sigma_2 = \sigma - \tau \), \( \sigma_3 = \sigma \). Thus, the maximum shear stress is \( \max \left( \left| \frac{(\sigma_1 - \sigma_2)}{2} \right|, \left| \frac{(\sigma_2 - \sigma_3)}{2} \right|, \left| \frac{(\sigma_3 - \sigma_1)}{2} \right| \right) = \tau \). If \( \tau \) exceeds a critical value, yielding occurs.
03

Applying the von Mises Criterion

The von Mises criterion is based on the distortion energy in the material and is given by the formula \( \sqrt{\frac{1}{2}[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2]} \). Plugging in \( \sigma_1 = \sigma + \tau \), \( \sigma_2 = \sigma - \tau \), \( \sigma_3 = \sigma \), the von Mises stress simplifies to \( \sqrt{3} \tau \), which must be less than or equal to the yield stress for no yield to occur.
04

Understanding the Impact of Hydrostatic Stress

Hydrostatic stress \( \sigma \) affects the principal stresses but does not contribute to yielding in either the Tresca or von Mises criteria because it adds equally to all principal stresses. Hence, yielding is governed by the shear component \( \tau \), and not by \( \sigma \).

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Key Concepts

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

Tresca Criterion
The Tresca Criterion is a method used to determine when a material will yield under stress. It revolves around the concept of shear stress and posits that yielding begins when the maximum shear stress in the material equals its critical shear stress limit. In simple terms, it focuses on the difference between the highest and lowest principal stresses in a material.

When assessing yield using Tresca, you calculate the maximum shear stress as half of the largest difference between any two principal stresses. For our stress state, the principal stresses are given as:
  • \( \sigma_1 = \sigma + \tau \)
  • \( \sigma_2 = \sigma - \tau \)
  • \( \sigma_3 = \sigma \)

The maximum shear stress thus evaluates to \( \ au \), since it's the greatest differential. The material will yield when \( \ au \) reaches a critical value specific to the material substance. The material does not yield due to overall stresses but purely when the shear component becomes excessive.
von Mises Criterion
The von Mises Criterion takes a more energy-focused approach to understanding yield. It analyzes the distortion energy in a material, which is directly related to the differences in the three principal stresses. This criterion suggests that yield initiates when the equivalent von Mises stress exceeds a material's yield strength.

For the stress scenario presented, first, plug the principal stresses into the von Mises formula: \[ \sigma_{vm} = \sqrt{\frac{1}{2}\left((\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2\right)} \] Substituting \( \sigma_1 = \sigma + \tau \), \( \sigma_2 = \sigma - \tau \), and \( \sigma_3 = \sigma \), the von Mises stress comes out to \( \sqrt{3} \tau \). Yielding occurs when this stress matches or exceeds the material's yield point, indicating the material is deforming plastically.
Hydrostatic Stress
Hydrostatic Stress applies uniformly in all directions, much like the average stress a diver experiences underwater. In our stress example, \( \sigma \) is the hydrostatic stress, effectively raising or lowering the principal stresses equally by \( \sigma \).

Despite its contribution to the principal stresses, hydrostatic stress doesn't influence yield according to either the Tresca or von Mises criteria. This is because yield dependence in these criteria focuses on shear stresses, which are unaffected by uniform hydrostatic increases. The yielding process remains governed solely by the varying shear component, \( \tau \), which can tilt or distort the body, unlike the even pressure of \( \sigma \).
Shear Stress
Shear Stress is pivotal in determining the onset of yield in materials. It occurs due to forces that act parallel to a surface, essentially trying to slide material components over each other. In the context of the problem, shear stresses are expressed as \( \sigma_{xy} = \tau \) and \( \sigma_{yx} = \tau \), directly affecting the material's integrity.

This stress type is critical in both the Tresca and von Mises yield criteria. It's the value of \( \tau \) that potentially triggers yield, as opposed to the equally distributed hydrostatic stress. Yield occurs when \( \tau \) reaches levels beyond what the material can endure, leading to material deformation through slippage or distortion. Understanding how shear stress interacts with yield criteria helps predict material behavior under different stress conditions.
Principal Stresses
Principal Stresses describe the normal stresses acting on a plane where shear stress is zero. These are the maximum and minimum normal stresses at any point in the material, simplifying complex stress systems into simpler, more manageable components.

In this case, the principal stresses are defined as:
  • \( \sigma_1 = \sigma + \tau \)
  • \( \sigma_2 = \sigma - \tau \)
  • \( \sigma_3 = \sigma \)

These represent the inherent stress states across distinct planes in the material devoid of shear effects. Principal stresses greatly aid in assessing the potential of yielding based on simplified stress states. By analyzing them, engineers can foresee whether a complex stress situation might lead to material failure, applying suitable yield criteria to make informed predictions.

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Most popular questions from this chapter

A theoretical estimate of the strength of a crystalline solid is its binding energy per unit volume. Evaluate the strength of forsterite if its binding energy is \(10^{3} \mathrm{~kJ} \mathrm{~mole}^{-1}\) and its mean atomic volume is \(6.26 \times 10^{-6} \mathrm{~m}^{3} \mathrm{~mole}^{-1}\). The presence of grain boundaries and dislocations weakens a crystalline solid considerably below its theoretical strength.

Consider the one-dimensional diffusion of radioactive tracer atoms initially absent from a crystalline solid but deposited uniformly at time \(t=0\) on the surface \(x=0\) of the semi-infinite solid. The number of radioactive atoms deposited at \(t=0\) is \(C\) per unit surface area. Show that the concentration of radioactive atoms \(n\) (number per unit volume) in the solid must satisfy the diffusion equation $$\frac{\partial n}{\partial t}=D \frac{\partial^{2} n}{\partial x^{2}} $$Equation \((7-79)\) can be obtained by first deriving the equation of conservation of tracer atoms $$\frac{\partial n}{\partial t}=-\frac{\partial J}{\partial x}$$ where we assume that tracer atoms diffuse in the \(x\) direction only. The actual decay of the tracer atoms has been ignored in formulating the mass balance. Solve Equation \((7-79)\) subject to the initial and boundary conditions $$\begin{array}{ll}n(x, t=0)=0 & (7-81) \\\\\int_{0}^{\infty} n(x, t) d x=C . & (7-82)\end{array}$$ Show that \(n(x, t)\) is given by $$n(x, t)=\frac{C}{(\pi D t)^{1 / 2}} \exp \left(\frac{-x^{2}}{4 D t}\right)$$

The average kinetic energy \(\bar{\phi}\) of an atom in a crystalline solid is given by $$\bar{\phi}=\frac{1}{n} \int_{0}^{\infty} \phi d n_{\phi}$$ Verify that the Maxwell-Boltzmann distribution gives \(\bar{\phi}=\frac{3}{2} k T\) by carrying out the integration.

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