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Tension in simple triangular truss.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "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 obtained through trigonometry.
", "advice": "\nBalancing horizontal forces at B: $P_{AB}\\sin(\\var{theta})+P_{BC}\\sin(90-\\var{theta})=0$, or more simply: $P_{AB}\\sin(\\var{theta})+P_{BC}\\cos(\\var{theta})=0$.
\nBalancing vertical forces at B: $P_{AB}\\cos(\\var{theta})=P_{BC}\\cos(90-\\var{theta})+\\var{force}$, or more simply: $P_{AB}\\cos(\\var{theta})=P_{BC}\\sin(\\var{theta})+\\var{force}$,
\nwhere the units are kN. Multiplying the first by $\\cos(\\var{theta})$ and the second by $\\sin(\\var{theta})$, we get:
\n$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})+P_{BC}\\cos(\\var{theta})\\cos(\\var{theta})=0$
\n$P_{AB}\\cos(\\var{theta})\\sin(\\var{theta})-P_{BC}\\sin(\\var{theta})\\sin(\\var{theta})=\\var{force}\\sin(\\var{theta})$
\nand subtracting the second from the first:
\n$P_{BC}\\cos(\\var{theta})\\cos(\\var{theta})+P_{BC}\\sin(\\var{theta})\\sin(\\var{theta})=-\\var{force}\\sin(\\var{theta})$, or more simply: $P_{BC}=-\\var{force}\\sin(\\var{theta})$
\nsince $\\cos^2 + \\sin^2 = 1$. Balancing vertical forces at C:
\n$P_{AC}+P_{BC}\\cos(90-\\var{theta})=0$, or more simply: $P_{AC}+P_{BC}\\sin(\\var{theta})=0$,
\nand substituting in the expression for $P_{BC}$ from above:
\n$P_{AC}-\\var{force}\\sin(\\var{theta})\\sin(\\var{theta})=0$,
\nthe final answer simplifies to:
\n$P_{AC}=\\var{force}\\sin^2(\\var{theta})=\\var{siground(Pac,3)}$ [units: kN].
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\n\nIf 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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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "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 obtained through trigonometry.
", "advice": "\nBalancing horizontal forces at B: $P_{AB}\\cos(\\var{theta})=P_{BC}\\sin(\\var{theta})$.
\nBalancing vertical forces at B: $P_{AB}\\sin(\\var{theta})+P_{BC}\\cos(\\var{theta})+\\var{force}=0$,
\nwhere units are kN. Multiplying the first by $\\sin(\\var{theta})$ and the second by $\\cos(\\var{theta})$ gives:
\n$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})-P_{BC}\\sin^2(\\var{theta})=0$
\n$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})+P_{BC}\\cos^2(\\var{theta})=-\\var{force}\\cos(\\var{theta})$
\nand subtracting the first from the second gives:
\n$P_{BC}\\sin^2(\\var{theta})+P_{BC}\\cos^2(\\var{theta})=-\\var{force}\\cos(\\var{theta})$,
\nor, since $\\sin^2+\\cos^2=1$:
\n$P_{BC}=-\\var{force}\\cos(\\var{theta})$.
\nBalancing horizontal forces at C: $P_{AC}+P_{BC}\\sin(\\var{theta})=0$
\nSubstituting in the expression for $P_{BC}$ from above, and rearranging:
\n$P_{AC}=\\var{force}\\sin(\\var{theta})\\cos(\\var{theta}) = \\var{siground(Pac,3)}$ [units: kN].
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\n\nIf 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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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "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 obtained through trigonometry.
", "advice": "Balancing vertical forces at B: $P_{BC}\\sin(\\var{theta})+\\var{force}=0$,
\nwhere units are kN. Rearranging:
\n$P_{BC}=-\\var{force}/\\sin(\\var{theta}) = \\var{siground(Pbc,3)}$ [units: kN].
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\n\nIf 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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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "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.)
", "advice": "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).
\nUsing $\\sigma_a = \\sigma_h/2$, Hooke's Law is:
\n$\\epsilon_h = {1 \\over E}(\\sigma_h-\\nu \\sigma_a) = {\\sigma_h \\over E} \\left(1-{\\nu \\over 2}\\right)$
\nand rearranging:
\n$\\sigma_h = {E \\epsilon_h \\over\\left(1-\\nu / 2\\right)} = {209 \\times 10^9 \\times \\var{strain} \\times 10^{-6} \\over (1-0.3 / 2)} $
\nso the hoop stress in the can wall prior to opening it was $\\var{siground(SH,3)}$MPa.
\nHoop stress is related to pressure by:
\n$\\sigma_h = {p D \\over 2 t}$
\nand rearranging:
\n$p = \\sigma_h {2 t \\over D} =\\var{siground(SH,3)}$MPa $\\times {2 \\times \\var{thickness} \\over\\var{diameter}} = \\var{siground(P/10,3)}$MPa.
\nThe pressure in the can prior to opening it was, therefore, $\\var{siground(P,3)}$ bar.
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\nUsing Hooke's Law:
\n$\\epsilon_h = {1 \\over E}(\\sigma_h-\\nu \\sigma_a)$
\nand remembering that $\\sigma_a = \\sigma_h/2$:
\nDetermine maximum pressure in a closed thin-walled cylindrical pressure vessel before yield.
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", "advice": "A closed, cylindrical, thin-walled pressure vessel has diameter $D = \\var{diameter}$ m and wall thickness $t = \\var{thickness}$ mm.
\nUsing $\\sigma_h = {p D \\over 2 t}$ and $\\sigma_a = {p D \\over 4 t}$, the von Mises stress is given by:
\n$\\sigma_V^2 = \\sigma_a^2 - \\sigma_a \\sigma_h + \\sigma_h^2 = \\left({p D \\over 4 t}\\right)^2 - \\left({p D \\over 4 t}\\right)\\left({p D \\over 2 t}\\right) +\\left({p D \\over 2 t}\\right)^2 = 3\\left({p D \\over 4 t}\\right)^2$
\ni.e.:
\n$\\sigma_V = \\sqrt{3}\\left({p D \\over 4 t}\\right)$
\nwhich can be rearranged to give pressure:
\n$p = \\sigma_V {4 t \\over D \\sqrt{3}} = \\sigma_V {4 \\times \\var{thickness} \\times 10^{-3} \\over \\var{diameter} \\times \\sqrt{3}} = \\var{siground(factor,3)}\\sigma_V$
\nwhere $\\sigma_a$ is the axial stress and $\\sigma_h$ is the hoop stress.
\nThe maximum pressure (such that $\\sigma_V < \\sigma_Y$) for:
\nYield stress of steel.
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", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["diameter", "thickness", "sYFe", "sYAl", "factor"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "\nA 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$\\sigma_V^2 = \\sigma_a^2 - \\sigma_a \\sigma_h + \\sigma_h^2$
\nwhere $\\sigma_a$ is the axial stress and $\\sigma_h$ is the hoop stress.
\nWhat is the maximum pressure (such that $\\sigma_V < \\sigma_Y$) for:
\nUsing Mohr's Circle to calculate principal stresses in plane stress 2D case.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "For components in plane stress, Mohr's circle provides a quick and easy method for determining the principal stresses and the maximum shear stress.
", "advice": "The mean stress determines the centre of Mohr's circle, and the radius can be found by specifying a coordinate on the circle, i.e.: ($\\sigma_x,\\tau_{xy}$) = ($\\var{sigmax},\\var{tauxy}$), and then using Pythagoras to determine the length of the radius from the centre of the circle at ($\\var{siground(sigmamean,3)},0$):
\nradius = $\\sqrt{(\\sigma_x-\\sigma_m)^2+\\tau_{xy}^2} =\\sqrt{(\\var{sigmax}-(\\var{siground(sigmamean,3)}))^2+(\\var{tauxy})^2} = \\var{siground(taumax,3)}$MPa
\nThe angle, $\\theta$, between the principal axes and the $xy-$axes is given by $\\tan(2\\theta)={\\tau_{xy} \\over \\sigma_x-\\sigma_m}$:
\n$\\theta={1 \\over 2} \\tan^{-1}\\left({\\var{tauxy} \\over \\var{siground(sigmax-sigmamean,3)}}\\right) = \\var{siground(theta,3)}^\\circ$.
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\nDetermine:
\nWhat is the angle, $\\theta$, between the principal axes and the $xy-$axes? [[4]] [Units: degrees, $0\\le\\theta<180$]
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Basic Calculation", "questions": [{"name": "3D Stress - General case and von Mises calculation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Francis Franklin", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1887/"}], "tags": [], "metadata": {"description": "Calculate invariants and von Mises stress for a general 3D stress state.
", "licence": "Creative Commons Attribution 4.0 International"}, "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}\\]
\nthen the three invariants are:
\nFrom these we can easily calculate:
\nThe 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\\]
\nNote A. If the three principal stresses [Units: Pa] are different (which is usually but not always the case):
\nNote B: 1 MPa = $10^6$ Pa = $10^6$ N/m$^2$ = 1 N/mm$^2$
\nNote C: 1 MPa$^2$ = $10^{12}$ Pa$^2$, etc.
", "advice": "Calculate the invariants:
\nand thus calculate:
\nShear stress in $yz$ plane.
", "templateType": "anything"}, "tauzx": {"name": "tauzx", "group": "Ungrouped variables", "definition": "random(5..15)", "description": "Shear stress in $zx$ plane.
", "templateType": "anything"}, "I1": {"name": "I1", "group": "Ungrouped variables", "definition": "sigmax+sigmay+sigmaz", "description": "First invariant.
", "templateType": "anything"}, "sigmaz": {"name": "sigmaz", "group": "Ungrouped variables", "definition": "-random(-16..17#3)", "description": "Normal stress in $z$ direction
", "templateType": "anything"}, "sigmay": {"name": "sigmay", "group": "Ungrouped variables", "definition": "random(-18..15#3)", "description": "Normal stress in $y$ direction.
", "templateType": "anything"}, "sigmamean": {"name": "sigmamean", "group": "Ungrouped variables", "definition": "I1/3", "description": "Mean stress.
", "templateType": "anything"}, "sigmax": {"name": "sigmax", "group": "Ungrouped variables", "definition": "random(-17..16#3)", "description": "Normal stress in $x$ direction.
", "templateType": "anything"}, "tauxy": {"name": "tauxy", "group": "Ungrouped variables", "definition": "random(-15..-5)", "description": "Shear stress in $xy$ plane.
", "templateType": "anything"}, "I2": {"name": "I2", "group": "Ungrouped variables", "definition": "sigmax*sigmay+sigmay*sigmaz+sigmaz*sigmax-tauzx^2-tauxy^2-tauyz^2", "description": "Second invariant.
", "templateType": "anything"}, "J2": {"name": "J2", "group": "Ungrouped variables", "definition": "I2-I1^2/3", "description": "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.
", "templateType": "anything"}, "sigmav": {"name": "sigmav", "group": "Ungrouped variables", "definition": "sqrt(-3*J2)", "description": "von Mises stress.
", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["sigmax", "sigmay", "sigmaz", "tauxy", "tauyz", "tauzx", "I1", "I2", "I3", "J2", "sigmav", "sigmamean"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "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]}\\]
\nCalculate the invariants:
\nand thus calculate:
\nDetermine the principal stresses for a 3D stress state (with null third invariant).
", "licence": "Creative Commons Attribution 4.0 International"}, "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}\\]
\nthen the three invariants are:
\nFrom these we can easily calculate:
\nThe 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\\]
\nNote A. If the three principal stresses [Units: Pa] are different (which is usually but not always the case):
\nNote B: 1 MPa = $10^6$ Pa = $10^6$ N/m$^2$ = 1 N/mm$^2$
\nNote C: 1 MPa$^2$ = $10^{12}$ Pa$^2$, etc.
", "advice": "Calculate the invariants:
\nTo calculate the principal stresses, solve the cubic equation:
\n\\[\\lambda^3 - I_1 \\lambda^2 + I_2 \\lambda - I_3 = 0\\]
\nwhich, since $I_3 \\approx 0$, simplifies to:
\n\\[\\lambda^3 - I_1 \\lambda^2 + I_2 \\lambda =\\lambda \\left( \\lambda^2 - I_1 \\lambda + I_2 \\right) = 0\\]
\nwhich has a root at $\\lambda = 0$ and the quadratic formula can be used to find the other two roots:
\n$\\lambda = {I_1 \\pm \\sqrt{I_1^2 - 4 I_2} \\over 2} = {\\var{siground(I1,3)} \\pm \\sqrt{(\\var{siground(I1,3)})^2 - 4 \\times (\\var{siground(I2,3)})} \\over 2} = \\var{siground(lambda1,3)}$MPa or $\\var{siground(lambda2,3)}$MPa.
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", "templateType": "anything", "can_override": false}, "sigmamax": {"name": "sigmamax", "group": "Ungrouped variables", "definition": "if(lambda1<0,0,lambda1)", "description": "Maximum principal stress.
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", "templateType": "anything", "can_override": false}, "sigmamin": {"name": "sigmamin", "group": "Ungrouped variables", "definition": "if(lambda2>0,0,lambda2)", "description": "Minimum principal stress.
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", "templateType": "anything", "can_override": false}, "tauzx": {"name": "tauzx", "group": "Ungrouped variables", "definition": "random(5..15)/10", "description": "Shear stress in $zx$ plane.
", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["sigmax", "sigmay", "sigmaz", "tauxy", "tauyz", "tauzx", "I1", "I2", "I3", "delta", "lambda1", "lambda2", "sigmamax", "sigmamin", "sigmamiddle"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "The stress at a particular point in a component has been calculated as:
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\nCalculate the invariants:
\nAssuming $I_3 \\approx 0$ and can be neglected, determine:
\n(The answer here should be close to zero.)
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", "licence": "Creative Commons Attribution 4.0 International"}, "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}\\]
\nthen the three invariants are:
\nFrom these we can easily calculate:
\nThe 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\\]
\nNote A. If the three principal stresses [Units: Pa] are different (which is usually but not always the case):
\nNote B: 1 MPa = $10^6$ Pa = $10^6$ N/m$^2$ = 1 N/mm$^2$
\nNote C: 1 MPa$^2$ = $10^{12}$ Pa$^2$, etc.
", "advice": "Calculate the invariants:
\nThe von Mises stress is $\\sigma_V=\\sqrt{-3J_2}=\\sqrt{I_1^2 - 3 I_2}=\\sqrt{(\\var{sigmaz})^2 - 3 \\times (\\var{I2})} = \\var{siground(sigmav,3)}$MPa.
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", "templateType": "anything"}, "sigmaz": {"name": "sigmaz", "group": "Ungrouped variables", "definition": "-random(50..150)", "description": "Maximum compressive axial stress.
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\nCalculate the invariants:
\nAnd thus the von Mises stress is $\\sigma_V=$[[2]] [Units: MPa].
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", "showactualmark": true, "feedbackmessages": [], "advicethreshold": 0, "showtotalmark": true, "showanswerstate": true}, "name": "Keith's copy of Introduction to Stresses", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "These practice questions cover:
\n