3D stress -strain relationship

Click For Summary

Discussion Overview

The discussion centers on the 3D stress-strain relationship, focusing on the mathematical representation of stress and strain tensors in a matrix form. Participants explore the derivation of coefficients such as E/(1-v²) and the relationships between stress and strain, including the implications of off-diagonal terms in the stress tensor.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how the coefficient E/(1-v²) emerges when certain stress components are set to zero, indicating a need for clarification on the derivation.
  • Another participant points out that the off-diagonal relationships in the stress tensor are missing from the initial formulation.
  • A mathematical representation of the relationships between stress and strain is provided, including a matrix form that relates strain to stress through a compliance matrix C.
  • One participant expresses an intention to derive stress as a function of strain, suggesting the inversion of the matrix equation ε=Cσ to obtain stress components in terms of strain.
  • Additional equations are presented to define the relationships between shear modulus G, Young's modulus E, and Poisson's ratio ν, along with constants A and B used in the stress-strain relationships.
  • A later post mentions a missing expression for stress in terms of strain, proposing a matrix D that relates stress to strain, with A and B retaining their previous definitions.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the derivation of the coefficient E/(1-v²) or the completeness of the stress-strain relationships. Multiple viewpoints and approaches are presented, indicating ongoing debate and exploration of the topic.

Contextual Notes

Limitations include potential missing assumptions regarding the definitions of stress and strain, as well as unresolved mathematical steps in deriving the relationships presented. The discussion reflects a complex interplay of theoretical and mathematical considerations.

Ronankeating
Messages
62
Reaction score
0
εHi,

Based on that hoary schematic cube representation in 3D for stress-strain relationship. Stress tensor {σxx, σyy, σzz, σxy, σyz, σzx} and strain tensors {εxx, εyy, εzz, εxy, εyz, εzx} can be written interchangably. Let's suppose that σzz=σzx=σzy=0 then the reamining terms are written in matrix form then how that E/(1-v2) coefficient is emerging.

At first sight, is seems that its very simple to express it by multipliers but couldn't figured out .

Any help please,
 
Engineering news on Phys.org
You are missing the off diagonal relationships in your tensor.
 
The relationships betwenn stress and strain are

\begin{array}{l}<br /> {\varepsilon _x} = \frac{1}{E}\left[ {{\sigma _x} - \nu ({\sigma _y} + {\sigma _z}} \right] \\ <br /> {\varepsilon _y} = \frac{1}{E}\left[ {{\sigma _y} - \nu ({\sigma _x} + {\sigma _z}} \right] \\ <br /> {\varepsilon _z} = \frac{1}{E}\left[ {{\sigma _z} - \nu ({\sigma _x} + {\sigma _y}} \right] \\ <br /> {\gamma _{xy}} = \frac{1}{G}{\tau _{xy}} \\ <br /> {\gamma _{yz}} = \frac{1}{G}{\tau _{yz}} \\ <br /> {\gamma _{xz}} = \frac{1}{G}{\tau _{xz}} \\ <br /> \end{array}

Written in matrix form this becomes

\varepsilon = C\sigma

where C is the matrix

C = \frac{1}{E}\left[ {\begin{array}{*{20}{c}}<br /> 1 &amp; { - v} &amp; { - v} &amp; 0 &amp; 0 &amp; 0 \\<br /> { - v} &amp; 1 &amp; { - v} &amp; 0 &amp; 0 &amp; 0 \\<br /> { - v} &amp; { - v} &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; {\frac{E}{G}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; {\frac{E}{G}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; {\frac{E}{G}} \\<br /> \end{array}} \right]
 
(But I'm intending to write stress as function of strain)

you do this by solving (inverting) my matrix equation

ε=Cσ

this is possible and leads to

\begin{array}{l}<br /> {\sigma _x} = A\left[ {{\varepsilon _x} + B\left( {{\varepsilon _y} + {\varepsilon _z}} \right)} \right] \\ <br /> {\sigma _y} = A\left[ {{\varepsilon _y} + B\left( {{\varepsilon _x} + {\varepsilon _z}} \right)} \right] \\ <br /> {\sigma _z} = A\left[ {{\varepsilon _z} + B\left( {{\varepsilon _x} + {\varepsilon _y}} \right)} \right] \\ <br /> {\tau _{xy}} = G{\gamma _{xy}} \\ <br /> {\tau _{yz}} = G{\gamma _{yz}} \\ <br /> {\tau _{xz}} = G{\gamma _{xz}} \\ <br /> \end{array}

Also

\begin{array}{l}<br /> G = \frac{E}{{2\left( {1 + \nu } \right)}} \\ <br /> A = E\frac{{\left( {1 - \nu } \right)}}{{\left( {1 + \nu } \right)\left( {1 - 2\nu } \right)}} \\ <br /> B = \frac{\nu }{{\left( {1 - v} \right)}} \\ <br /> \end{array}

As regards to the LaTex this site uses for formulae, I am seriously deficient in latex lore so I use MathType and copy/paste.
I wouldn't recommend MT, however as it is too expensive for what it is and does not allow the inclusion of images.
 
There seems to be a missing post, however to complete a matrix expression for stress in terms of strain is

\sigma = D\varepsilon

where D is the matrix

D = A\left[ {\begin{array}{*{20}{c}}<br /> 1 &amp; B &amp; B &amp; 0 &amp; 0 &amp; 0 \\<br /> B &amp; 1 &amp; B &amp; 0 &amp; 0 &amp; 0 \\<br /> B &amp; B &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; {\frac{G}{A}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; {\frac{G}{A}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; {\frac{G}{A}} \\<br /> \end{array}} \right]

A and B have the same meaning as before
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
7K