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Applied vertical (downward) force.

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Angle to vertical of Bar AB.

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Half of mechanics is about balancing forces. Some trusses can be treated like pin-jointed structures and decomposed into axial forces, i.e., without any bending or shear forces. The solution is therefore pure trigonometry.

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A pin-jointed truss is shown in the figure below. The pivot at A is fixed, but the pivot at C is free to move vertically. The angle at B is a right angle.

\n

\"Pin-jointed

\n

If the applied force, $F$, is $\\var{force}$ kN vertically down, and the angle of Bar AB to the vertical, $\\theta$, is $\\var{theta}^\\circ$, what is the tension in Bar AC?

\n

[[0]] [Units: kN]

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Half of mechanics is about balancing forces. Some trusses can be treated like pin-jointed structures and decomposed into axial forces, i.e., without any bending or shear forces. The solution is therefore pure trigonometry.

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A pin-jointed truss is shown in the figure below. The pivot at A is fixed, but the pivot at C is free to move horizontally. The angle at B is a right angle.

\n

\"Pin-jointed

\n

If the applied force, $F$, is $\\var{force}$ kN vertically down, and the angle of Bar AB to the horizontal, $\\theta$, is $\\var{theta}^\\circ$, what is the tension in Bar AC?

\n

[[0]] [Units: kN]

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Applied vertical (downward) force.

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Angle to vertical of Bar AB.

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Angle to vertical of Bar AB.

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Applied vertical (downward) force.

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A pin-jointed truss is shown in the figure below. The pivots at A and C are fixed.

\n

\"Pin-jointed

\n

If the applied force, $F$, is $\\var{force}$ kN vertically down, and the angle between Bar AB and Bar BC, $\\theta$, is $\\var{theta}^\\circ$, what is the tension in Bar BC?

\n

[[0]] [Units: kN]

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Half of mechanics is about balancing forces. Some trusses can be treated like pin-jointed structures and decomposed into axial forces, i.e., without any bending or shear forces. The solution is therefore pure trigonometry.

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Wall thickness

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Measured hoop strain

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Diameter

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It is possible to estimate the pressure in a drinks can by measuring the hoop strain before and after opening it. (The can is an example of a thin-walled pressure vessel, and the stress can be considered as plane stress, i.e., stress through the thickness of the wall is neglected.)

", "variable_groups": [], "preamble": {"js": "", "css": ""}, "rulesets": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "parts": [{"variableReplacementStrategy": "originalfirst", "gaps": [{"strictPrecision": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "scripts": {}, "allowFractions": false, "marks": "2", "mustBeReduced": false, "minValue": "0.209*strain/(1-0.3/2)", "precision": "3", "maxValue": "0.209*strain/(1-0.3/2)", "showFeedbackIcon": true, "showPrecisionHint": true, "correctAnswerFraction": false, "type": "numberentry", "mustBeReducedPC": 0, "variableReplacements": [], "precisionType": "sigfig", "precisionMessage": "You have not given your answer to the correct precision.", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": "50", "showCorrectAnswer": true}, {"strictPrecision": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "scripts": {}, "allowFractions": false, "marks": "2", "mustBeReduced": false, "minValue": "10*(0.209*strain/(1-0.3/2))*2*thickness/diameter", "precision": "3", "maxValue": "10*(0.209*strain/(1-0.3/2))*2*thickness/diameter", "showFeedbackIcon": true, "showPrecisionHint": true, "correctAnswerFraction": false, "type": "numberentry", "mustBeReducedPC": 0, "variableReplacements": [], "precisionType": "sigfig", "precisionMessage": "You have not given your answer to the correct precision.", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": "50", "showCorrectAnswer": true}], "showFeedbackIcon": true, "scripts": {}, "type": "gapfill", "marks": 0, "variableReplacements": [], "prompt": "

A steel drinks can ($E=209$ GPa, $\\nu=0.3$) has diameter $\\var{diameter}$ mm and wall thickness $\\var{thickness}$ mm. A strain gauge is fixed circumferentially and the difference in strain, before and after opening the can, is measured as $\\var{strain}$ μm/m (microstrain).

\n

Using Hooke's Law:

\n

$\\epsilon_h = {1 \\over E}(\\sigma_h-\\nu \\sigma_a)$

\n

and remembering that $\\sigma_a = \\sigma_h/2$:

\n
    \n
  1. The hoop stress in the can wall prior to opening it was [[0]] [Units: MPa]
  2. \n
  3. The pressure in the can prior to opening it was [[1]] [Units: bar] 
  4. \n
", "showCorrectAnswer": true}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": ""}, "ungrouped_variables": ["diameter", "thickness", "strain"], "type": "question"}, {"name": "Michael's copy of Thin-walled pressure vessel", "extensions": [], "custom_part_types": [], "resources": [["question-resources/ThinWalledCylinder-2.png", "/srv/numbas/media/question-resources/ThinWalledCylinder-2.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}, {"name": "Francis Franklin", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1887/"}], "ungrouped_variables": ["diameter", "thickness", "sYFe", "sYAl", "factor"], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": ""}, "statement": "

A closed, cylindrical, thin-walled pressure vessel can be considered as a biaxial stress case with the hoop stress and axial stress as principal stresses.

", "parts": [{"showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "gaps": [{"strictPrecision": false, "notationStyles": ["plain", "en", "si-en"], "showCorrectAnswer": true, "marks": "2", "correctAnswerStyle": "plain", "minValue": "factor*sYFe", "showPrecisionHint": true, "scripts": {}, "precisionType": "sigfig", "mustBeReduced": false, "allowFractions": false, "showFeedbackIcon": true, "type": "numberentry", "maxValue": "factor*sYFe", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "precision": "3", "precisionMessage": "You have not given your answer to the correct precision.", "mustBeReducedPC": 0, "precisionPartialCredit": "50"}, {"strictPrecision": false, "notationStyles": ["plain", "en", "si-en"], "showCorrectAnswer": true, "marks": "2", "correctAnswerStyle": "plain", "minValue": "factor*sYAl", "showPrecisionHint": true, "scripts": {}, "precisionType": "sigfig", "mustBeReduced": false, "allowFractions": false, "showFeedbackIcon": true, "type": "numberentry", "maxValue": "factor*sYAl", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "precision": "3", "precisionMessage": "You have not given your answer to the correct precision.", "mustBeReducedPC": 0, "precisionPartialCredit": "50"}], "scripts": {}, "prompt": "

\n

A closed, cylindrical, thin-walled pressure vessel has diameter $D = \\var{diameter}$ m and wall thickness $t = \\var{thickness}$ mm. The von Mises stress is given by:

\n

$\\sigma_V^2 = \\sigma_a^2 - \\sigma_a \\sigma_h + \\sigma_h^2$

\n

where $\\sigma_a$ is the axial stress and $\\sigma_h$ is the hoop stress.

\n

What is the maximum pressure (such that $\\sigma_V < \\sigma_Y$) for:

\n
    \n
  1. a steel ($\\sigma_Y=\\var{sYFe}$ MPa) pressure vessel? [[0]] [Units: MPa]
  2. \n
  3. an aluminium ($\\sigma_Y=\\var{sYAl}$ MPa) pressure vessel? [[1]] [Units: MPa]
  4. \n
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Wall thickness of thin-walled pressure vessel.

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Diameter of thin-walled pressure vessel.

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Yield stress of steel.

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pressure / yield stress

"}, "sYAl": {"group": "Ungrouped variables", "name": "sYAl", "templateType": "anything", "definition": "random(200..280#10)", "description": "

Yield stress of aluminium.

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For components in plane stress, Mohr's circle provides a quick and easy method for determining the principal stresses and the maximum shear stress.

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Angle to principal axes, doubled, not adjusted for quadrant.

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Normal stress in $x$ direction.

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Normal stress in $y$ direction.

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Maximum shear stress.

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Shear stress in $xy$ plane.

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Mean stress.

"}}, "parts": [{"showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "gaps": [{"notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "showCorrectAnswer": true, "marks": "2", "minValue": "sigmamean", "precision": "3", "mustBeReduced": false, "allowFractions": false, "precisionMessage": "You have not given your answer to the correct precision.", "precisionType": "sigfig", "showFeedbackIcon": true, "type": "numberentry", "maxValue": "sigmamean", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "scripts": {}, "correctAnswerStyle": "plain", "showPrecisionHint": true, "mustBeReducedPC": 0, "precisionPartialCredit": "50"}, {"notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "showCorrectAnswer": true, "marks": "2", "minValue": "sigmamean+taumax", "precision": "3", "mustBeReduced": false, "allowFractions": false, "precisionMessage": "You have not given your answer to the correct precision.", "precisionType": "sigfig", "showFeedbackIcon": true, "type": "numberentry", "maxValue": "sigmamean+taumax", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "scripts": {}, "correctAnswerStyle": "plain", "showPrecisionHint": true, "mustBeReducedPC": 0, "precisionPartialCredit": "50"}, {"notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "showCorrectAnswer": true, "marks": "2", "minValue": "sigmamean-taumax", "precision": "3", "mustBeReduced": false, "allowFractions": false, "precisionMessage": "You have not given your answer to the correct precision.", "precisionType": "sigfig", "showFeedbackIcon": true, "type": "numberentry", "maxValue": "sigmamean-taumax", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "scripts": {}, "correctAnswerStyle": "plain", "showPrecisionHint": true, "mustBeReducedPC": 0, "precisionPartialCredit": "50"}, {"notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "showCorrectAnswer": true, "marks": "2", "minValue": "taumax", "precision": "3", "mustBeReduced": false, "allowFractions": false, "precisionMessage": "You have not given your answer to the correct precision.", "precisionType": "sigfig", "showFeedbackIcon": true, "type": "numberentry", "maxValue": "taumax", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "scripts": {}, "correctAnswerStyle": "plain", "showPrecisionHint": true, "mustBeReducedPC": 0, "precisionPartialCredit": "50"}, {"notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "showCorrectAnswer": true, "marks": "2", "minValue": "theta", "precision": "3", "mustBeReduced": false, "allowFractions": false, "precisionMessage": "You have not given your answer to the correct precision.", "precisionType": "sigfig", "showFeedbackIcon": true, "type": "numberentry", "maxValue": "theta", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "scripts": {}, "correctAnswerStyle": "plain", "showPrecisionHint": true, "mustBeReducedPC": 0, "precisionPartialCredit": "50"}], "scripts": {}, "prompt": "

A sheet steel component is subject to stresses $\\sigma_x=\\var{sigmax}$ MPa, $\\sigma_y=\\var{sigmay}$ MPa and $\\tau_{xy}=\\var{tauxy}$ MPa.

\n

Determine:

\n
    \n
  1. The mean stress, $\\sigma_m=$[[0]] [Units: MPa]
  2. \n
  3. The maximum principal stress, $\\sigma_1=$[[1]] [Units: MPa]
  4. \n
  5. The minimum principal stress, $\\sigma_2=$[[2]] [Units: MPa]
  6. \n
  7. The maximum shear stress, $\\tau_\\text{max}=$[[3]] [Units: MPa]
  8. \n
\n

What is the angle, $\\theta$, between the principal axes and the $xy-$axes? [[4]] [Units: degrees, $0\\le\\theta<180$]

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Second invariant.

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Normal stress in $y$ direction.

", "definition": "random(-18..15#3)", "name": "sigmay"}, "tauzx": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Shear stress in $zx$ plane.

", "definition": "random(5..15)", "name": "tauzx"}, "sigmamean": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Mean stress.

", "definition": "I1/3", "name": "sigmamean"}, "J2": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Second deviatoric invariant.

", "definition": "I2-I1^2/3", "name": "J2"}, "sigmav": {"templateType": "anything", "group": "Ungrouped variables", "description": "

von Mises stress.

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Normal stress in $z$ direction

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

Normal stress in $x$ direction.

", "definition": "random(-17..16#3)", "name": "sigmax"}, "I3": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Third invariant.

", "definition": "sigmax*sigmay*sigmaz+2*tauxy*tauyz*tauzx-sigmax*tauyz^2-sigmay*tauzx^2-sigmaz*tauxy^2", "name": "I3"}, "I1": {"templateType": "anything", "group": "Ungrouped variables", "description": "

First invariant.

", "definition": "sigmax+sigmay+sigmaz", "name": "I1"}, "tauyz": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Shear stress in $yz$ plane.

", "definition": "random(-5..5)", "name": "tauyz"}, "tauxy": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Shear stress in $xy$ plane.

", "definition": "random(-15..-5)", "name": "tauxy"}}, "ungrouped_variables": ["sigmax", "sigmay", "sigmaz", "tauxy", "tauyz", "tauzx", "I1", "I2", "I3", "J2", "sigmav", "sigmamean"], "statement": "

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:

\n

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

\n

then the three invariants are:

\n
    \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
\n

From these we can easily calculate:

\n
    \n
  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
\n

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:

\n

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

\n

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

\n
    \n
  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
\n

Note B: 1 MPa = $10^6$ Pa = $10^6$ N/m$^2$ = 1 N/mm$^2$

\n

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

", "advice": "", "parts": [{"variableReplacements": [], "gaps": [{"allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReduced": false, "precisionType": "sigfig", "scripts": {}, "maxValue": "I1", "precisionPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "minValue": "I1", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "mustBeReducedPC": 0, "showCorrectAnswer": true, "precision": "3", "variableReplacements": [], "correctAnswerFraction": false, "showPrecisionHint": true, "marks": "2", "type": "numberentry"}, {"allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReduced": false, "precisionType": "sigfig", "scripts": {}, "maxValue": "I2", "precisionPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "minValue": "I2", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "mustBeReducedPC": 0, "showCorrectAnswer": true, "precision": "3", "variableReplacements": [], "correctAnswerFraction": false, "showPrecisionHint": true, "marks": "2", "type": "numberentry"}, {"allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReduced": false, "precisionType": "sigfig", "scripts": {}, "maxValue": "I3", "precisionPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "minValue": "I3", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "mustBeReducedPC": 0, "showCorrectAnswer": true, "precision": "3", "variableReplacements": [], "correctAnswerFraction": false, "showPrecisionHint": true, "marks": "2", "type": "numberentry"}, {"allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReduced": false, "precisionType": "sigfig", "scripts": {}, "maxValue": "sigmamean", "precisionPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "minValue": "sigmamean", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "mustBeReducedPC": 0, "showCorrectAnswer": true, "precision": "3", "variableReplacements": [], "correctAnswerFraction": false, "showPrecisionHint": true, "marks": "2", "type": "numberentry"}, {"allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReduced": false, "precisionType": "sigfig", "scripts": {}, "maxValue": "J2", "precisionPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "minValue": "J2", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "mustBeReducedPC": 0, "showCorrectAnswer": true, "precision": "3", "variableReplacements": [], "correctAnswerFraction": false, "showPrecisionHint": true, "marks": "2", "type": "numberentry"}, {"allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReduced": false, "precisionType": "sigfig", "scripts": {}, "maxValue": "sigmav", "precisionPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "minValue": "sigmav", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "mustBeReducedPC": 0, "showCorrectAnswer": true, "precision": "3", "variableReplacements": [], "correctAnswerFraction": false, "showPrecisionHint": true, "marks": "2", "type": "numberentry"}], "showCorrectAnswer": true, "prompt": "

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

\n

\\[\\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]}\\]

\n

Calculate the invariants:

\n
    \n
  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
\n

and thus calculate:

\n
    \n
  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
", "scripts": {}, "marks": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "type": "gapfill"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"js": "", "css": ""}, "functions": {}, "tags": [], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": ""}, "type": "question"}, {"name": "Michael's copy of 3D Stress - Principal stresses calculation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}, {"name": "Francis Franklin", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1887/"}], "variable_groups": [], "ungrouped_variables": ["sigmax", "sigmay", "sigmaz", "tauxy", "tauyz", "tauzx", "I1", "I2", "I3", "delta", "lambda1", "lambda2", "sigmamax", "sigmamin", "sigmamiddle"], "statement": "

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:

\n

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

\n

then the three invariants are:

\n
    \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
\n

From these we can easily calculate:

\n
    \n
  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
\n

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:

\n

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

\n

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

\n
    \n
  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
\n

Note B: 1 MPa = $10^6$ Pa = $10^6$ N/m$^2$ = 1 N/mm$^2$

\n

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

", "variables": {"I2": {"group": "Ungrouped variables", "name": "I2", "templateType": "anything", "definition": "sigmax*sigmay+sigmay*sigmaz+sigmaz*sigmax-tauzx^2-tauxy^2-tauyz^2", "description": "

Second invariant.

"}, "sigmamin": {"group": "Ungrouped variables", "name": "sigmamin", "templateType": "anything", "definition": "if(lambda2>0,0,lambda2)", "description": "

Minimum principal stress.

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

Third invariant.

"}, "tauyz": {"group": "Ungrouped variables", "name": "tauyz", "templateType": "anything", "definition": "random(-5..5)/10", "description": "

Shear stress in $yz$ plane.

"}, "tauxy": {"group": "Ungrouped variables", "name": "tauxy", "templateType": "anything", "definition": "random(-15..-5)/10", "description": "

Shear stress in $xy$ plane.

"}, "delta": {"group": "Ungrouped variables", "name": "delta", "templateType": "anything", "definition": "sqrt(I1^2-4*I2)", "description": "

Part of root solution.

"}, "sigmamax": {"group": "Ungrouped variables", "name": "sigmamax", "templateType": "anything", "definition": "if(lambda1<0,0,lambda1)", "description": "

Maximum principal stress.

"}, "sigmaz": {"group": "Ungrouped variables", "name": "sigmaz", "templateType": "anything", "definition": "-siground((2*tauxy*tauyz*tauzx-sigmax*tauyz^2-sigmay*tauzx^2)/(sigmax*sigmay-tauxy^2),3)", "description": "

Normal stress in $z$ direction

"}, "tauzx": {"group": "Ungrouped variables", "name": "tauzx", "templateType": "anything", "definition": "random(5..15)/10", "description": "

Shear stress in $zx$ plane.

"}, "sigmax": {"group": "Ungrouped variables", "name": "sigmax", "templateType": "anything", "definition": "random(-17..16#3)/10", "description": "

Normal stress in $x$ direction.

"}, "lambda2": {"group": "Ungrouped variables", "name": "lambda2", "templateType": "anything", "definition": "(I1-delta)/2", "description": "

Root of cubic poly - principal stress.

"}, "lambda1": {"group": "Ungrouped variables", "name": "lambda1", "templateType": "anything", "definition": "(I1+delta)/2", "description": "

Root of cubic poly - principal stress.

"}, "I1": {"group": "Ungrouped variables", "name": "I1", "templateType": "anything", "definition": "sigmax+sigmay+sigmaz", "description": "

First invariant.

"}, "sigmamiddle": {"group": "Ungrouped variables", "name": "sigmamiddle", "templateType": "anything", "definition": "if(lambda1<0,lambda1,if(lambda2>0,lambda2,0))", "description": "

Middle principal stress.

"}, "sigmay": {"group": "Ungrouped variables", "name": "sigmay", "templateType": "anything", "definition": "random(-16..17#3)/10", "description": "

Normal stress in $y$ direction.

"}}, "parts": [{"showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "gaps": [{"notationStyles": ["plain", "en", "si-en"], "showPrecisionHint": true, "showCorrectAnswer": true, "marks": "2", "correctAnswerStyle": "plain", "precision": "3", "mustBeReduced": false, "allowFractions": false, "precisionType": "sigfig", "mustBeReducedPC": 0, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "maxValue": "I1", "variableReplacements": [], "strictPrecision": false, "showFeedbackIcon": true, "correctAnswerFraction": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "minValue": "I1", "precisionPartialCredit": "50"}, {"notationStyles": ["plain", "en", "si-en"], "showPrecisionHint": true, "showCorrectAnswer": true, "marks": "2", "correctAnswerStyle": "plain", "precision": "3", "mustBeReduced": false, "allowFractions": false, "precisionType": "sigfig", "mustBeReducedPC": 0, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "maxValue": "I2", "variableReplacements": [], "strictPrecision": false, "showFeedbackIcon": true, "correctAnswerFraction": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "minValue": "I2", "precisionPartialCredit": "50"}, {"notationStyles": ["plain", "en", "si-en"], "showPrecisionHint": true, "showCorrectAnswer": true, "marks": "2", "correctAnswerStyle": "plain", "precision": "3", "mustBeReduced": false, "allowFractions": false, "precisionType": "sigfig", "mustBeReducedPC": 0, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "maxValue": "I3+10^-8", "variableReplacements": [], "strictPrecision": false, "showFeedbackIcon": true, "correctAnswerFraction": false, "scripts": {}, "precisionMessage": "

(The answer here should be close to zero.)

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

\n

\\[\\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]}\\]

\n

Calculate the invariants:

\n
    \n
  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
\n

Assuming $I_3 \\approx 0$ and can be neglected, determine:

\n
    \n
  1. The maximum principal stress: $\\sigma_\\text{max}=$[[3]] [Units: MPa].
  2. \n
  3. The middle principal stress: $\\sigma_\\text{middle}=$[[4]] [Units: MPa]
  4. \n
  5. The minimum principal stress: $\\sigma_\\text{min}=$[[5]] [Units: MPa].
  6. \n
", "variableReplacements": [], "showFeedbackIcon": true, "type": "gapfill"}], "variablesTest": {"condition": "", "maxRuns": 100}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": ""}, "preamble": {"js": "", "css": ""}, "rulesets": {}, "advice": "", "tags": [], "functions": {}, "type": "question"}, {"name": "Michael's copy of Sign Post - von Mises Stress at Base", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}, {"name": "Francis Franklin", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1887/"}], "rulesets": {}, "tags": [], "variables": {"sigmav": {"templateType": "anything", "group": "Ungrouped variables", "description": "

von Mises stress.

", "definition": "sqrt(-3*J2)", "name": "sigmav"}, "I2": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Second invariant.

", "definition": "-tauzx^2", "name": "I2"}, "sigmaz": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Maximum compressive axial stress.

", "definition": "-random(50..150)", "name": "sigmaz"}, "tauzx": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Shear stress from torsion / twist.

", "definition": "random(5..25)", "name": "tauzx"}, "J2": {"templateType": "anything", "group": "Ungrouped variables", "description": "

Second deviatoric invariant.

", "definition": "I2-I1^2/3", "name": "J2"}, "I1": {"templateType": "anything", "group": "Ungrouped variables", "description": "

First invariant.

", "definition": "sigmaz", "name": "I1"}}, "ungrouped_variables": ["sigmaz", "tauzx", "I1", "I2", "J2", "sigmav"], "statement": "

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:

\n

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

\n

then the three invariants are:

\n
    \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
\n

From these we can easily calculate:

\n
    \n
  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
\n

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:

\n

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

\n

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

\n
    \n
  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
\n

Note B: 1 MPa = $10^6$ Pa = $10^6$ N/m$^2$ = 1 N/mm$^2$

\n

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

", "advice": "", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"js": "", "css": ""}, "functions": {}, "parts": [{"prompt": "

A sign post is subject to bending and torsion from the applied wind load, as well as axial compression from the weight of the sign. At the point of maximum compression, from the combined weight and bending, the axial stress is $\\sigma_z = \\var{sigmaz}$ MPa. The only other component of stress at this location is the shear stress from the torsion: $\\tau_{zx}=\\var{tauzx}$ MPa.

\n

Calculate the invariants:

\n
    \n
  1. $I_1=$[[0]] [Units: MPa].
  2. \n
  3. $I_2=$[[1]] [Units: MPa$^2$].
  4. \n
\n

And thus the von Mises stress is $\\sigma_V=$[[2]] [Units: MPa].

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A series of questions covering the Introduction to Stresses part of the Mechanics module.

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These practice questions cover:

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