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These practice questions cover:

resolving axial forces in a truss structure
stress & strain in a closed thin-walled pressure vessel (an example of 2D plane stress)
use of Mohr's circle for determining principal stresses for a 2D plane stress case
calculation of invariants for general 3D stress case, leading to von Mises stress and principal stresses

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

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Tension in simple triangular truss.

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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 obtained through trigonometry.

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Balancing horizontal forces at B: $P_{AB}\\sin(\\var{theta})+P_{BC}\\sin(90-\\var{theta})=0$, or more simply: $P_{AB}\\sin(\\var{theta})+P_{BC}\\cos(\\var{theta})=0$.

Balancing vertical forces at B: $P_{AB}\\cos(\\var{theta})=P_{BC}\\cos(90-\\var{theta})+\\var{force}$, or more simply: $P_{AB}\\cos(\\var{theta})=P_{BC}\\sin(\\var{theta})+\\var{force}$,

where the units are kN. Multiplying the first by $\\cos(\\var{theta})$ and the second by $\\sin(\\var{theta})$, we get:

$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})+P_{BC}\\cos(\\var{theta})\\cos(\\var{theta})=0$

$P_{AB}\\cos(\\var{theta})\\sin(\\var{theta})-P_{BC}\\sin(\\var{theta})\\sin(\\var{theta})=\\var{force}\\sin(\\var{theta})$

and subtracting the second from the first:

$P_{BC}\\cos(\\var{theta})\\cos(\\var{theta})+P_{BC}\\sin(\\var{theta})\\sin(\\var{theta})=-\\var{force}\\sin(\\var{theta})$, or more simply: $P_{BC}=-\\var{force}\\sin(\\var{theta})$

since $\\cos^2 + \\sin^2 = 1$. Balancing vertical forces at C:

$P_{AC}+P_{BC}\\cos(90-\\var{theta})=0$, or more simply: $P_{AC}+P_{BC}\\sin(\\var{theta})=0$,

and substituting in the expression for $P_{BC}$ from above:

$P_{AC}-\\var{force}\\sin(\\var{theta})\\sin(\\var{theta})=0$,

the final answer simplifies to:

$P_{AC}=\\var{force}\\sin^2(\\var{theta})=\\var{siground(Pac,3)}$ [units: kN].

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Angle to vertical of Bar AB.

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Applied vertical (downward) force.

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Tension in Bar AC.

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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.

$\"Pin-jointed$

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?

[[0]] [Units: kN]

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Tension in a simple triangular truss.

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Balancing horizontal forces at B: $P_{AB}\\cos(\\var{theta})=P_{BC}\\sin(\\var{theta})$.

Balancing vertical forces at B: $P_{AB}\\sin(\\var{theta})+P_{BC}\\cos(\\var{theta})+\\var{force}=0$,

where units are kN. Multiplying the first by $\\sin(\\var{theta})$ and the second by $\\cos(\\var{theta})$ gives:

$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})-P_{BC}\\sin^2(\\var{theta})=0$

$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})+P_{BC}\\cos^2(\\var{theta})=-\\var{force}\\cos(\\var{theta})$

and subtracting the first from the second gives:

$P_{BC}\\sin^2(\\var{theta})+P_{BC}\\cos^2(\\var{theta})=-\\var{force}\\cos(\\var{theta})$,

or, since $\\sin^2+\\cos^2=1$:

$P_{BC}=-\\var{force}\\cos(\\var{theta})$.

Balancing horizontal forces at C: $P_{AC}+P_{BC}\\sin(\\var{theta})=0$

Substituting in the expression for $P_{BC}$ from above, and rearranging:

$P_{AC}=\\var{force}\\sin(\\var{theta})\\cos(\\var{theta}) = \\var{siground(Pac,3)}$ [units: kN].

Angle to vertical of Bar AB.

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Applied vertical (downward) force.

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Tension in Bar AC.

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.

$\"Pin-jointed$

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?

[[0]] [Units: kN]

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Tension in bracket-truss.

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Balancing vertical forces at B: $P_{BC}\\sin(\\var{theta})+\\var{force}=0$,

where units are kN. Rearranging:

$P_{BC}=-\\var{force}/\\sin(\\var{theta}) = \\var{siground(Pbc,3)}$ [units: kN].

Angle to vertical of Bar AB.

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Applied vertical (downward) force.

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Tension in Bar BC.

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

$\"Pin-jointed$

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?

[[0]] [Units: kN]

Work backwards from hoop strain measurement to calculate prior gauge pressure in a drinks can.

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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.)

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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).

Using $\\sigma_a = \\sigma_h/2$, Hooke's Law is:

$\\epsilon_h = {1 \\over E}(\\sigma_h-\\nu \\sigma_a) = {\\sigma_h \\over E} \\left(1-{\\nu \\over 2}\\right)$

and rearranging:

$\\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)} $

so the hoop stress in the can wall prior to opening it was $\\var{siground(SH,3)}$MPa.

Hoop stress is related to pressure by:

$\\sigma_h = {p D \\over 2 t}$

and rearranging:

$p = \\sigma_h {2 t \\over D} =\\var{siground(SH,3)}$MPa $\\times {2 \\times \\var{thickness} \\over\\var{diameter}} = \\var{siground(P/10,3)}$MPa.

The pressure in the can prior to opening it was, therefore, $\\var{siground(P,3)}$ bar.

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Measured hoop strain

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Diameter

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Wall thickness

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Hoop stress.

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Pressure [bar].

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Using Hooke's Law:

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

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

The hoop stress in the can wall prior to opening it was [[0]] [Units: MPa]
The pressure in the can prior to opening it was [[1]] [Units: bar]

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Determine maximum pressure in a closed thin-walled cylindrical pressure vessel before yield.

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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.

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A closed, cylindrical, thin-walled pressure vessel has diameter $D = \\var{diameter}$ m and wall thickness $t = \\var{thickness}$ mm.

Using $\\sigma_h = {p D \\over 2 t}$ and $\\sigma_a = {p D \\over 4 t}$, the von Mises stress is given by:

$\\sigma_V^2 = \\sigma_a^2 - \\sigma_a \\sigma_h + \\sigma_h^2 = \\left({p D \\over 4 t}\\right)^2 - \\left({p D \\over 4 t}\\right)\\left({p D \\over 2 t}\\right) +\\left({p D \\over 2 t}\\right)^2 = 3\\left({p D \\over 4 t}\\right)^2$

i.e.:

$\\sigma_V = \\sqrt{3}\\left({p D \\over 4 t}\\right)$

which can be rearranged to give pressure:

$p = \\sigma_V {4 t \\over D \\sqrt{3}} = \\sigma_V {4 \\times \\var{thickness} \\times 10^{-3} \\over \\var{diameter} \\times \\sqrt{3}} = \\var{siground(factor,3)}\\sigma_V$

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

The maximum pressure (such that $\\sigma_V < \\sigma_Y$) for:

a steel ($\\sigma_Y=\\var{sYFe}$ MPa) pressure vessel is $\\var{siground(factor*sYFe,3)}$MPa.
an aluminium ($\\sigma_Y=\\var{sYAl}$ MPa) pressure vessel is $\\var{siground(factor*sYAl,3)}$MPa.

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Yield stress of steel.

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Wall thickness of thin-walled pressure vessel.

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Yield stress of aluminium.

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Diameter of thin-walled pressure vessel.

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pressure / yield stress

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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:

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

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

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

a steel ($\\sigma_Y=\\var{sYFe}$ MPa) pressure vessel? [[0]] [Units: MPa]
an aluminium ($\\sigma_Y=\\var{sYAl}$ MPa) pressure vessel? [[1]] [Units: MPa]

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Using Mohr's Circle to calculate principal stresses in plane stress 2D case.

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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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The mean stress, $\\sigma_m=(\\sigma_x+\\sigma_y)/2=\\var{siground(sigmamean,3)}$MPa.

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 ($\\var{siground(sigmamean,3)},0$):

radius = $\\sqrt{(\\sigma_x-\\sigma_m)^2+\\tau_{xy}^2} =\\sqrt{(\\var{sigmax}-(\\var{siground(sigmamean,3)}))^2+(\\var{tauxy})^2} = \\var{siground(taumax,3)}$MPa

The maximum principal stress is the mean stress plus the radius: $\\var{siground(sigmamean+taumax,3)}$MPa.
The minimum principal stress is the mean stress minus the radius: $\\var{siground(sigmamean-taumax,3)}$MPa.
The maximum shear stress is just the radius: $\\var{siground(taumax,3)}$MPa.

The angle, $\\theta$, between the principal axes and the $xy-$axes is given by $\\tan(2\\theta)={\\tau_{xy} \\over \\sigma_x-\\sigma_m}$:

$\\theta={1 \\over 2} \\tan^{-1}\\left({\\var{tauxy} \\over \\var{siground(sigmax-sigmamean,3)}}\\right) = \\var{siground(theta,3)}^\\circ$.

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Shear stress in $xy$ plane.

", "templateType": "anything", "can_override": false}, "sigmax": {"name": "sigmax", "group": "Ungrouped variables", "definition": "random(-4..40#2)", "description": "

Normal stress in $x$ direction.

", "templateType": "anything", "can_override": false}, "theta": {"name": "theta", "group": "Ungrouped variables", "definition": "if(sigmaxAngle to principal axes, adjusted for quadrant.

", "templateType": "anything", "can_override": false}, "taumax": {"name": "taumax", "group": "Ungrouped variables", "definition": "sqrt((sigmax-sigmamean)^2+tauxy^2)", "description": "

Maximum shear stress.

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

", "templateType": "anything", "can_override": false}, "sigmay": {"name": "sigmay", "group": "Ungrouped variables", "definition": "random(-3..13#2)", "description": "

Normal stress in $y$ direction.

", "templateType": "anything", "can_override": false}, "sigmamean": {"name": "sigmamean", "group": "Ungrouped variables", "definition": "(sigmax+sigmay)/2", "description": "

Mean stress.

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

Determine:

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

What 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", "pickingStrategy": "all-ordered", "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:

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

then the three invariants are:

The sum of the diagonal elements (the 'trace'): $I_1 = \\sigma_x + \\sigma_y + \\sigma_z$ [Units: Pa].
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$^2$].
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$^3$].

From these we can easily calculate:

The mean ('hydrostatic') stress, which is the sum of the diagonal elements divided by three: $\\sigma_\\text{mean} = I_1 / 3$ [Units: Pa].
The second deviatoric stress invariant: $J_2 = I_2 - I_1^2 / 3$ [Units: Pa$^2$].
The von Mises ('equivalent') stress: $\\sigma_V = \\sqrt{-3J_2}$ [Units: Pa].

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:

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

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

The maximum principal stress, $\\sigma_\\text{max}$, is the most tensile (most positive or least negative) of the principal stresses.
The minimum principal stress, $\\sigma_\\text{min}$, is the most compressive (most negative or least positive) of the principal stresses.
The middle principal stress, $\\sigma_\\text{middle}$, is in between the other two, so that $\\sigma_\\text{min} \\le \\sigma_\\text{middle} \\le \\sigma_\\text{max}$.

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

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

", "advice": "

Calculate the invariants:

$I_1=\\sigma_x + \\sigma_y + \\sigma_z = \\var{sigmax} +\\var{sigmay} +\\var{sigmaz} = \\var{I1}$MPa.
$I_2=\\sigma_x \\sigma_y + \\sigma_y \\sigma_z + \\sigma_z \\sigma_x - \\tau_{xy}^2 - \\tau_{yz}^2 - \\tau_{zx}^2 = (\\var{sigmax}) \\times (\\var{sigmay}) + (\\var{sigmay}) \\times (\\var{sigmaz}) + (\\var{sigmaz}) \\times (\\var{sigmax}) - (\\var{tauxy})^2 - (\\var{tauyz})^2 - (\\var{tauzx})^2 = \\var{I2}$MPa$^2$.
$I_3=\\sigma_x \\sigma_y \\sigma_z + 2 \\tau_{xy} \\tau_{yz} \\tau_{zx} - \\sigma_x \\tau_{yz}^2 -\\sigma_y \\tau_{zx}^2 - \\sigma_z \\tau_{xy}^2$ = $(\\var{sigmax}) \\times (\\var{sigmay}) \\times (\\var{sigmaz}) + 2 \\times (\\var{tauxy}) \\times (\\var{tauyz}) \\times (\\var{tauzx}) - (\\var{sigmax}) \\times (\\var{tauyz})^2 - (\\var{sigmay}) \\times (\\var{tauzx})^2 - (\\var{sigmaz}) \\times (\\var{tauxy})^2 = \\var{I3}$MPa$^3$.

and thus calculate:

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

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Shear stress in $yz$ plane.

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

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.

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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.

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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:

\\[\\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]}\\]

Calculate the invariants:

$I_1=$[[0]] [Units: MPa].
$I_2=$[[1]] [Units: MPa$^2$].
$I_3=$[[2]] [Units: MPa$^3$].

and thus calculate:

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

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Determine 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:

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

then the three invariants are:

The sum of the diagonal elements (the 'trace'): $I_1 = \\sigma_x + \\sigma_y + \\sigma_z$ [Units: Pa].
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$^2$].
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$^3$].

From these we can easily calculate:

The mean ('hydrostatic') stress, which is the sum of the diagonal elements divided by three: $\\sigma_\\text{mean} = I_1 / 3$ [Units: Pa].
The second deviatoric stress invariant: $J_2 = I_2 - I_1^2 / 3$ [Units: Pa$^2$].
The von Mises ('equivalent') stress: $\\sigma_V = \\sqrt{-3J_2}$ [Units: Pa].

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

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

The maximum principal stress, $\\sigma_\\text{max}$, is the most tensile (most positive or least negative) of the principal stresses.
The minimum principal stress, $\\sigma_\\text{min}$, is the most compressive (most negative or least positive) of the principal stresses.
The middle principal stress, $\\sigma_\\text{middle}$, is in between the other two, so that $\\sigma_\\text{min} \\le \\sigma_\\text{middle} \\le \\sigma_\\text{max}$.

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

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

", "advice": "

Calculate the invariants:

$I_1=\\sigma_x + \\sigma_y + \\sigma_z = \\var{sigmax} +\\var{sigmay} +\\var{sigmaz} = \\var{siground(I1,3)}$MPa.
$I_2=\\sigma_x \\sigma_y + \\sigma_y \\sigma_z + \\sigma_z \\sigma_x - \\tau_{xy}^2 - \\tau_{yz}^2 - \\tau_{zx}^2 = (\\var{sigmax}) \\times (\\var{sigmay}) + (\\var{sigmay}) \\times (\\var{sigmaz}) + (\\var{sigmaz}) \\times (\\var{sigmax}) - (\\var{tauxy})^2 - (\\var{tauyz})^2 - (\\var{tauzx})^2 = \\var{siground(I2,3)}$MPa$^2$.
$I_3=\\sigma_x \\sigma_y \\sigma_z + 2 \\tau_{xy} \\tau_{yz} \\tau_{zx} - \\sigma_x \\tau_{yz}^2 -\\sigma_y \\tau_{zx}^2 - \\sigma_z \\tau_{xy}^2$ = $(\\var{sigmax}) \\times (\\var{sigmay}) \\times (\\var{sigmaz}) + 2 \\times (\\var{tauxy}) \\times (\\var{tauyz}) \\times (\\var{tauzx}) - (\\var{sigmax}) \\times (\\var{tauyz})^2 - (\\var{sigmay}) \\times (\\var{tauzx})^2 - (\\var{sigmaz}) \\times (\\var{tauxy})^2 = \\var{siground(I3,3)}$MPa$^3$.

To calculate the principal stresses, solve the cubic equation:

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

which, since $I_3 \\approx 0$, simplifies to:

\\[\\lambda^3 - I_1 \\lambda^2 + I_2 \\lambda =\\lambda \\left( \\lambda^2 - I_1 \\lambda + I_2 \\right) = 0\\]

which has a root at $\\lambda = 0$ and the quadratic formula can be used to find the other two roots:

$\\lambda = {I_1 \\pm \\sqrt{I_1^2 - 4 I_2} \\over 2} = {\\var{siground(I1,3)} \\pm \\sqrt{(\\var{siground(I1,3)})^2 - 4 \\times (\\var{siground(I2,3)})} \\over 2} = \\var{siground(lambda1,3)}$MPa or $\\var{siground(lambda2,3)}$MPa.

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Root of cubic poly - principal stress.

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Shear stress in $yz$ plane.

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Normal stress in $y$ direction.

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Normal stress in $z$ direction

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Third invariant.

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Part of root solution.

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Shear stress in $xy$ plane.

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Second invariant.

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Maximum principal stress.

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First invariant.

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Middle principal stress.

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Minimum principal stress.

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Root of cubic poly - principal stress.

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Normal stress in $x$ direction.

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Shear stress in $zx$ plane.

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

Calculate the invariants:

$I_1=$[[0]] [Units: MPa].
$I_2=$[[1]] [Units: MPa$^2$].
$I_3=$[[2]] [Units: MPa$^3$].

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

The maximum principal stress: $\\sigma_\\text{max}=$[[3]] [Units: MPa].
The middle principal stress: $\\sigma_\\text{middle}=$[[4]] [Units: MPa]
The minimum principal stress: $\\sigma_\\text{min}=$[[5]] [Units: MPa].

(The answer here should be close to zero.)

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Calculate the von Mises stress at the base of a post subject to axial compression, bending and torsion.

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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:

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

then the three invariants are:

The sum of the diagonal elements (the 'trace'): $I_1 = \\sigma_x + \\sigma_y + \\sigma_z$ [Units: Pa].
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$^2$].
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$^3$].

From these we can easily calculate:

The mean ('hydrostatic') stress, which is the sum of the diagonal elements divided by three: $\\sigma_\\text{mean} = I_1 / 3$ [Units: Pa].
The second deviatoric stress invariant: $J_2 = I_2 - I_1^2 / 3$ [Units: Pa$^2$].
The von Mises ('equivalent') stress: $\\sigma_V = \\sqrt{-3J_2}$ [Units: Pa].

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

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

The maximum principal stress, $\\sigma_\\text{max}$, is the most tensile (most positive or least negative) of the principal stresses.
The minimum principal stress, $\\sigma_\\text{min}$, is the most compressive (most negative or least positive) of the principal stresses.
The middle principal stress, $\\sigma_\\text{middle}$, is in between the other two, so that $\\sigma_\\text{min} \\le \\sigma_\\text{middle} \\le \\sigma_\\text{max}$.

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

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

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Calculate the invariants:

$I_1 = \\sigma_x + \\sigma_y + \\sigma_z = 0 + 0 + (\\var{sigmaz}) = \\var{sigmaz}$MPa.
$I_2 = \\sigma_x \\sigma_y + \\sigma_y \\sigma_z + \\sigma_z \\sigma_x - \\tau_{xy}^2 - \\tau_{yz}^2 - \\tau_{zx}^2 =0 \\times 0 + 0 \\times (\\var{sigmaz}) + (\\var{sigmaz}) \\times 0 - 0^2 - 0^2 - (\\var{tauzx})^2 = \\var{I2}$MPa$^2$.

The 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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Shear stress from torsion / twist.

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Second deviatoric invariant.

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First invariant.

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von Mises stress.

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Second invariant.

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Maximum compressive axial stress.

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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.

Calculate the invariants:

$I_1=$[[0]] [Units: MPa].
$I_2=$[[1]] [Units: MPa$^2$].

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