// Numbas version: exam_results_page_options {"name": "3D Stress - General case and von Mises calculation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "3D Stress - General case and von Mises calculation", "tags": [], "metadata": {"description": "

Calculate invariants and von Mises stress for a general 3D stress state.

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The principal stresses and maximum shear stress, and the von Mises stress, can all be determined from the 3D stress matrix and its invariants, i.e., if:

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\\[\\sigma=\\begin{pmatrix}\\sigma_x & \\tau_{xy} & \\tau_{zx} \\\\ \\tau_{xy} & \\sigma_y & \\tau_{yz} \\\\ \\tau_{zx} & \\tau_{yz} & \\sigma_z\\end{pmatrix}\\]

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then the three invariants are:

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    \n
  1. The sum of the diagonal elements (the 'trace'): $I_1 = \\sigma_x + \\sigma_y + \\sigma_z$ [Units: Pa].
  2. \n
  3. An expression that is a sum of sub-determinants: $I_2 = \\sigma_x \\sigma_y + \\sigma_y \\sigma_z + \\sigma_z \\sigma_x - \\tau_{xy}^2 - \\tau_{yz}^2 - \\tau_{zx}^2$ [Units: Pa$^2$].
  4. \n
  5. The determinant: $I_3 = \\sigma_x \\sigma_y \\sigma_z + 2 \\tau_{xy} \\tau_{yz} \\tau_{zx} - \\sigma_x \\tau_{yz}^2 -\\sigma_y \\tau_{zx}^2 - \\sigma_z \\tau_{xy}^2$ [Units: Pa$^3$].
  6. \n
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From these we can easily calculate:

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  1. The mean ('hydrostatic') stress, which is the sum of the diagonal elements divided by three: $\\sigma_\\text{mean} = I_1 / 3$ [Units: Pa].
  2. \n
  3. The second deviatoric stress invariant: $J_2 = I_2 - I_1^2 / 3$ [Units: Pa$^2$].
  4. \n
  5. The von Mises ('equivalent') stress: $\\sigma_V = \\sqrt{-3J_2}$ [Units: Pa].
  6. \n
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The principal stresses can be found by solving the eigenvalue/eigenvector matrix problem (the eigenvalues are the principal stresses), or by finding the three roots (the three roots - $\\lambda_1$, $\\lambda_2$, $\\lambda_3$ - are the principal stresses) of the equation:

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\\[\\lambda^3 - I_1 \\lambda^2 + I_2 \\lambda - I_3 = 0\\]

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Note A. If the three principal stresses [Units: Pa] are different (which is usually but not always the case):

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  1. The maximum principal stress, $\\sigma_\\text{max}$, is the most tensile (most positive or least negative) of the principal stresses.
  2. \n
  3. The minimum principal stress, $\\sigma_\\text{min}$, is the most compressive (most negative or least positive) of the principal stresses.
  4. \n
  5. The middle principal stress, $\\sigma_\\text{middle}$, is in between the other two, so that $\\sigma_\\text{min} \\le \\sigma_\\text{middle} \\le \\sigma_\\text{max}$.
  6. \n
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Note B: 1 MPa = $10^6$ Pa = $10^6$ N/m$^2$ = 1 N/mm$^2$

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Note C: 1 MPa$^2$ = $10^{12}$ Pa$^2$, etc.

", "advice": "

Calculate the invariants:

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    \n
  1. $I_1=\\sigma_x + \\sigma_y + \\sigma_z = \\var{sigmax} +\\var{sigmay} +\\var{sigmaz} = \\var{I1}$MPa.
  2. \n
  3. $I_2=\\sigma_x \\sigma_y + \\sigma_y \\sigma_z + \\sigma_z \\sigma_x - \\tau_{xy}^2 - \\tau_{yz}^2 - \\tau_{zx}^2 = (\\var{sigmax}) \\times (\\var{sigmay}) + (\\var{sigmay}) \\times (\\var{sigmaz}) + (\\var{sigmaz}) \\times (\\var{sigmax}) - (\\var{tauxy})^2 - (\\var{tauyz})^2 - (\\var{tauzx})^2 = \\var{I2}$MPa$^2$.
  4. \n
  5. $I_3=\\sigma_x \\sigma_y \\sigma_z + 2 \\tau_{xy} \\tau_{yz} \\tau_{zx} - \\sigma_x \\tau_{yz}^2 -\\sigma_y \\tau_{zx}^2 - \\sigma_z \\tau_{xy}^2$ = $(\\var{sigmax}) \\times (\\var{sigmay}) \\times (\\var{sigmaz}) + 2 \\times (\\var{tauxy}) \\times (\\var{tauyz}) \\times (\\var{tauzx}) - (\\var{sigmax}) \\times (\\var{tauyz})^2 - (\\var{sigmay}) \\times (\\var{tauzx})^2 - (\\var{sigmaz}) \\times (\\var{tauxy})^2 = \\var{I3}$MPa$^3$.
  6. \n
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and thus calculate:

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  1. The mean ('hydrostatic') stress: $\\sigma_\\text{mean} = I_1 / 3 = \\var{I1} / 3 = \\var{siground(sigmamean,3)}$MPa.
  2. \n
  3. The second deviatoric stress invariant: $J_2 = I_2 - I_1^2 / 3 = \\var{I2} - (\\var{I1})^2 / 3 = \\var{siground(J2,3)}$MPa$^2$.
  4. \n
  5. The von Mises stress: $\\sigma_V = \\sqrt{-3 J_2} = \\sqrt{-3 \\times (\\var{siground(J2,3)})} = \\var{siground(sigmav,3)}$MPa.
  6. \n
\n

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Shear stress in $yz$ plane.

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Shear stress in $zx$ plane.

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First invariant.

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Normal stress in $z$ direction

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Normal stress in $y$ direction.

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Mean stress.

", "templateType": "anything"}, "sigmax": {"name": "sigmax", "group": "Ungrouped variables", "definition": "random(-17..16#3)", "description": "

Normal stress in $x$ direction.

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Shear stress in $xy$ plane.

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Second invariant.

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Second deviatoric invariant.

", "templateType": "anything"}, "I3": {"name": "I3", "group": "Ungrouped variables", "definition": "sigmax*sigmay*sigmaz+2*tauxy*tauyz*tauzx-sigmax*tauyz^2-sigmay*tauzx^2-sigmaz*tauxy^2", "description": "

Third invariant.

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von Mises stress.

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The stress at a particular point in a component has been calculated as:

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\\[\\sigma=\\begin{pmatrix} \\var{sigmax} & \\var{tauxy} & \\var{tauzx} \\\\ \\var{tauxy} & \\var{sigmay} & \\var{tauyz} \\\\ \\var{tauzx} & \\var{tauyz} & \\var{sigmaz} \\end{pmatrix} \\text{[Units: MPa]}\\]

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Calculate the invariants:

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  1. $I_1=$[[0]] [Units: MPa].
  2. \n
  3. $I_2=$[[1]] [Units: MPa$^2$].
  4. \n
  5. $I_3=$[[2]] [Units: MPa$^3$].
  6. \n
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and thus calculate:

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  1. The mean stress: $\\sigma_\\text{mean}=$[[3]] [Units: MPa].
  2. \n
  3. The second deviatoric stress invariant: $J_2=$[[4]] [Units: MPa$^2$]
  4. \n
  5. The von Mises stress: $\\sigma_V=$[[5]] [Units: MPa].
  6. \n
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