Deviatoric Stress

3 Key excerpts on "Deviatoric Stress"

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  eBook - ePub

  Advanced Solid Mechanics

  Simplified Theory

  • Farzad Hejazi, Tan Kar Chun(Authors)
  • 2021(Publication Date)
  • CRC Press

    (Publisher)

  It can be observed that when Deviatoric Stress along an axis is high, the distortion of body along that axis is significant, as indicated in Fig. 2.18. Figure 2.18 Hydrostatic and Deviatoric Stresses. The hydrostatic stress can be written as follows: σ m = σ x + σ y + σ z 3. (2.45) Applying the relationship in Eq. (2.35) to the equation above yields the following expression: σ m = I 1 3. (2.46) Dilation due to hydrostatic stress is the uniform displacement of particles across a solid, whether it is towards or away from each other. In contrast, distortion due to Deviatoric Stress is differential displacement of particles across the solid. Therefore, distortion in this case is very likely to break the bond between particles compared to dilation. In other words, Deviatoric Stress is the kind of stress responsible for yielding. It is essential for engineers and material scientists when the plastic behaviour of solid is concerned. Let the total stress be σ. The Deviatoric Stress, σ ′, is defined as follows σ ' = σ − σ m. Substituting (2.46) into the equation above gives the following equation: σ ' = σ − I 1 3. In other words, we can express the total stress in the following form: σ = σ ' + I 1 3. Substituting the equation above into Eq. (2.34) we can get (σ ' + I 1 3) 3 − I 1 (σ ' + I 1 3) 2 + I 2 (σ ' + I 1 3) − I 3 = 0. Expanding the equation above leads to the. following: σ ′ 3 + 2 I 1 σ ′ 2 3 + I 1 2 σ ′ 9 + I 1 σ ′ 2 3 + 2 I 1 2 σ ′ 9 + I 1 3 27 − I 1 σ ′ 2 − 2 I 1 2 σ ′ 3 − I 1 3 9 + I 2 σ ′ + I 1 I 2 3 − I 3 = 0 Rearranging the equation above and factorising it with common terms yields the. following: σ ' 3 + (2 I 1 3 + I 1 3 − I 1) σ ' 2 + (I 1 2 9 + 2 I 1 2 9 − 2 I 1 2 3 + I 2) σ ' − I 1 3 9 + I 1 3 27 + I 1 I 2 3 − I 3 = 0. Simplification of the equation above produces the following. expression: σ ' 3 − (I 1 2 3 − I 2) σ ' − (2 I 1 3 27 − I 1 I 2 3 + I 3) = 0. σ ′ is the unknown that needs to be solved

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  eBook - ePub

  Mechanics of Sheet Metal Forming

  • Jack Hu, Zdzislaw Marciniak, John Duncan(Authors)
  • 2002(Publication Date)
  • Butterworth-Heinemann

    (Publisher)

  Levy–Mises Flow Rule. This states that the ratio of the strain increments will be the same as the ratio of the Deviatoric Stresses, i.e.

  (2.13a)

  or

  (2.13b)

  If a material element is deforming in a plane stress, proportional process, as described by Equation 2.6 , then Equation 2.13(b) can be integrated and expressed in terms of the natural or true strains, i.e.

  (2.13c)

  2.5.2 Relation between the stress and strain ratios

  From the above, we obtain the relation between the stress and strain ratios:

  (2.14)

  It may be seen that while the flow rule gives the relation between the stress and strain ratios, it does not indicate the magnitude of the strains. If the element deforms under a given stress state (i.e. α is known) the ratio of the strains can be found from Equation 2.13 , or 2.14 . The relationship can be illustrated for different load paths as shown in Figure 2.9 ; the small arrows show the ratio of the principal strain increments and the lines radiating from the origin indicate the loading path on an element. It may be seen that each of these strain increment vectors is perpendicular to the von Mises yield locus. (It is possible to predict this from considerations of energy or work.)

  Figure 2.9 Diagram showing the strain increment components for different stress states around the von Mises yield locus.

  2.5.3 (Worked example) stress state

  The current flow stress of a material element is 300 MPa. In a deformation process, the principal strain increments are 0.012 and 0.007 in the 1 and 2 directions respectively. Determine the principal stresses associated with this in a plane stress process.

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  eBook - ePub

  Practical Guide to the Packaging of Electronics

  Thermal and Mechanical Design and Analysis, Third Edition

  • Ali Jamnia(Author)
  • 2016(Publication Date)
  • CRC Press

    (Publisher)

  11 Mechanical and Thermomechanical Concerns

  Introduction

  An important aspect of electronics packaging is developing an understanding of the stresses that its components undergo and their relationship to the system’s failure and/or reliability. The cause of these stresses may be temperature and its variations, vibration, or physical properties such as weight. It may occur at the board and component level, enclosure levels, and up to the system itself.

  Stresses are internal distributed forces, which are caused by external applied loads. Strains are changes in the form under the same loads. Consider a rod of length L and diameter A . One may intuitively recognize that the displacement of the end of this rod depends directly on the magnitude of the applied force—very similar to the force–deflection relationship of a spring–mass system as shown in Figure 11.1 .

  Now consider what happens inside of this rod in Figure 11.2 . The concentrated load is (internally) developed over the area of the cross section. Thus, one may express this distributed force as follows:

  σ = F A

  Similarly, a distributed (average) displacement may also be calculated.

  ε = Δ L

  It turns out that σ and ε have a relationship similar to a force–deflection curve in a spring–mass system. The slope of this line (E ) is called tensile modulus, Young’s modulus, or modulus of elasticity.

  Figure 11.1 Force–deflection relationship.

  Figure 11.2 Internal forces.

  Consider another scenario. A block under a shear force will also deflect. In shear, the force–deflection relationships are defined as follows:

  F = τA

  where A is the area and τ is the shear stress. Furthermore, there is a relationship between the shear stress and shear strain (γ) similar to that of the stress–strain relationship.

  τ = G γ

  where G is shear modulus and γ is shear strain.

  In general, both normal and shear stresses develop in solids under a general loading. For example, a cantilever beam under a simple load at the free end exhibits both normal and shear stresses, as shown in Figure 11.3

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