3D Mohr's Circle

CalculationInputs﻿﻿σx
:-100MPa​
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﻿﻿σy
:10MPa​
﻿
﻿﻿σz
:50MPa​
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OutputAverage Normal Stresses﻿﻿σ1
:-45.0MPa​
﻿
﻿﻿σ2
:30.0MPa​
﻿
﻿﻿σ3
:-25.0MPa​
﻿
Shear Stresses﻿﻿τxy
:55.0MPa​
﻿
﻿﻿τyz
:20.0MPa​
﻿
﻿﻿τxz
:75.0MPa​
﻿
﻿﻿τmax
:75.0MPa​
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ExplanationThe 3D stresses are usually given by six stress components, the normal stresses at the x- y- and z- directions and the shear stresses at the xy , yz, and xz planes. These stresses are then put in a 3 x 3 symmetric matrix shown below. 
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T3=[σxτxyτxzτxyσyτyzτxzτyzσz]T_3 = \begin{bmatrix}    \sigma_x & \tau_{xy} & \tau_{xz} \\    \tau_{xy} & \sigma_y & \tau_{yz} \\    \tau_{xz} & \tau_{yz} & \sigma_z \\\end{bmatrix}T3​=​σx​τxy​τxz​​τxy​σy​τyz​​τxz​τyz​σz​​​
In order to get the maximum normal and shear stresses, we get the eigenvalues of the matrix above. Say, that the cut plane has the following directional cosine:
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v=(vx, vy, vz)\mathbf{v} = (v_x,\ v_y,\ v_z)v=(vx​, vy​, vz​)
We can find the normal stress on this plane by,
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σv=σxvx2+σyvy2+σzvz2+2τxyvxvy+2τyzvyvz+2τxzvxvz\sigma_v = \sigma_x  v_{x}^{2} + \sigma_y v_{y}^{2} + \sigma_z  v_{z}^{2} + 2 \tau_{xy} v_{x}  v_{y} + 2 \tau_{yz}  v_{y}  v_{z} + 2 \tau_{xz}  v_{x}  v_{z}σv​=σx​vx2​+σy​vy2​+σz​vz2​+2τxy​vx​vy​+2τyz​vy​vz​+2τxz​vx​vz​
To find the eigenvalues and eigenvectors, we use the following characteristic polynomial equation: 
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det(σI3−T3) = σ3−Aσ3−Bσ−C=0det(\sigma I_3-T_3)\ =\ \sigma^3- A\sigma^3 - B\sigma-C=0det(σI3​−T3​) = σ3−Aσ3−Bσ−C=0
Wherein 
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A=σx+σy+σzB=σxσy+σyσz+σxσz−τxy2−τyz2−τxz2C=σxσyσz+2τxyτyzτxz−σxτyz2−σyτxz2−σzτxy2A = \sigma_x + \sigma_y + \sigma_z \\B = \sigma_x \sigma_y + \sigma_y \sigma_z + \sigma_x \sigma_z - \tau_{xy}^2 - \tau_{yz}^2 - \tau_{xz}^2 \\C = \sigma_x \sigma_y \sigma_z + 2 \tau_{xy} \tau_{yz} \tau_{xz} - \sigma_x \tau_{yz}^2 - \sigma_y \tau_{xz}^2 - \sigma_z \tau_{xy}^2A=σx​+σy​+σz​B=σx​σy​+σy​σz​+σx​σz​−τxy2​−τyz2​−τxz2​C=σx​σy​σz​+2τxy​τyz​τxz​−σx​τyz2​−σy​τxz2​−σz​τxy2​
We can then find the normal stresses using the following system of equations:
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A=σ1+σ2+σ3B=σ1σ2+σ2σ3+σ1σ3C=σ1σ2σ3A = \sigma_1 + \sigma_2 + \sigma_3 \\B = \sigma_1 \sigma_2 + \sigma_2 \sigma_3 + \sigma_1 \sigma_3 \\C = \sigma_1 \sigma_2 \sigma_3A=σ1​+σ2​+σ3​B=σ1​σ2​+σ2​σ3​+σ1​σ3​C=σ1​σ2​σ3​
For a more graphical approach, we can use the following approach:
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σave1 = σx + σy2σave2 = σy + σz2σave3 = σx + σz2\sigma_{ave1}\ =\ \frac{\sigma_x\ +\ \sigma_y}{2}\\\sigma_{ave2}\ =\ \frac{\sigma_y\ +\ \sigma_z}{2}\\\sigma_{ave3}\ =\ \frac{\sigma_x\ +\ \sigma_z}{2}σave1​ = 2σx​ + σy​​σave2​ = 2σy​ + σz​​σave3​ = 2σx​ + σz​​
The obtained values will serve as the centers of our circles. The radii of the circles can be obtained by:
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radius1 = ∣σx −σave1∣radius2 = ∣σy −σave2∣radius3 = ∣σz −σave3∣radius_1\ =\ |\sigma_x\ -\sigma_{ave1}|\\radius_2\ =\ |\sigma_y\ -\sigma_{ave2}|\\radius_3\ =\ |\sigma_z\ -\sigma_{ave3}|radius1​ = ∣σx​ −σave1​∣radius2​ = ∣σy​ −σave2​∣radius3​ = ∣σz​ −σave3​∣
ReferencesMech of Materials Ch 07 lecture 04 Transformation of 3D Principal Stress with Mohr Circles by Cheng-fu Chen 
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