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A local principal stress model associated with Ernest Masson Anderson in which one principal stress is vertical, as suggested by negligible shear stress on a flat free surface. With compression positive and ${\sigma }_1\ge {\sigma }_2\ge {\sigma }_3$, vertical ${\sigma }_1$ gives normal faulting, vertical ${\sigma }_2$ gives strike-slip faulting, and vertical ${\sigma }_3$ gives reverse faulting. The other two principal stresses are horizontal. Topography, mechanical layering and local fault geometry can violate the idealization.

For a smooth objective isotropic hyperelastic material, the strict Baker-Ericksen inequalities require larger distinct principal stretches to have larger principal stresses. For $W=\Phi (v_1,v_2,v_3)$ and $J=v_1{v}_2{v}_3$, ${\sigma }_i=J^{-1}{v}_i{\Phi }_i$. The strict convention implies every stress-free state has equal stretches: if $v_i\ne v_j$ but both stresses are zero, the displayed product would vanish. Thus a natural state of a hyperelastic material has $F=vQ$, $v>0$, $Q\in SO(3)$. A weak non-strict inequality alone does not imply this rigidity.

A borehole breakout forms when the concentrated compressive stress around a drilled hole fractures and removes part of its wall. Opposed damaged sectors widen the originally circular hole. A borehole televiewer images the wall so that the enlargement and its orientation can be distinguished from an isolated natural fracture.

For an approximately vertical hole aligned with a principal stress direction, the long axis of the breakout follows the minimum horizontal compressive stress direction. The strongest tangential compression occurs there even though the distant horizontal compression in that direction is smaller. Inclined holes, pore pressure, drilling pressure and rock strength require a fuller model before orientations or magnitudes are inferred.^[ref]

Christian Otto Mohr was an engineer whose work connected graphical methods with the analysis of stressed materials. Mohr's circle represents the normal stress and shear stress on differently oriented planes by points on a circle, turning a change of coordinates into a geometric construction. TU Dresden's historical account relates this construction to his 1882 work on the representation of stress and deformation.^[ref]

For a two-dimensional stress tensor, the circle has centre $({\sigma }_x+{\sigma }_y)/2$ and radiusIts horizontal intercepts give the principal stresses. This makes Mohr's circle useful for seeing both extremal stresses and the relation between plane orientation and the components of stress.

One of a pair of fault orientations that can form under a common principal stress field with opposite senses of resolved shear stress. For newly formed, optimally oriented faults obeying the same Mohr-Coulomb failure criterion, their acute intersection angle is $90^{\circ }-\varphi$, where $\varphi$ is the friction angle. The maximum compressive principal stress bisects that acute angle. This permits an estimate $\mu =\tan \varphi$ only under the common-formation and optimal-orientation assumptions, rather than for arbitrary inherited faults.

The differential stress is the difference between the greatest and least principal stresses, $\Delta \sigma ={\sigma }_{\max }-{\sigma }_{\min }\ge 0$. It measures the spread of the principal loading, rather than the isotropic part of the stress. Adding the same pressure contribution to every principal stress leaves it unchanged.

For a symmetric three-dimensional stress tensor, the largest possible resolved shear stress is $\Delta \sigma /2$. In a Mohr circle representation this is the radius of the outer circle. A uniaxial load with principal values $(S,0,0)$ therefore has differential stress $|S|$, while hydrostatic loading has zero differential stress even when the pressure is large. This distinction is useful in yield criterion: a threshold based on maximum shear stress depends on principal-stress differences, rather than simply on the magnitude of the largest compression.

On $x^2/a^2+y^2/b^2<1$, the torsion stress function for a simply connected bar is $\varphi =\mu \Omega a^2{b}^2(1-x^2/a^2-y^2/b^2)/(a^2+b^2)$. Its Laplacian is $-2\mu \Omega$ and its boundary value is zero, proving the formula by uniqueness of the Poisson equation with Dirichlet data. Integration gives $M=\pi \mu \Omega a^3{b}^3/(a^2+b^2)$, hence the displayed stiffness $R=M/\Omega$.

The three principal stresses are $0,\pm |\nabla \varphi |$. With ${\tau }_Y$ defined as the shear yield stress, the Tresca yield criterion is $|\nabla \varphi |\le {\tau }_Y$. For $a>b$ its maximum is $2|C|/b$ at $(0,\pm b)$, where $C=\mu \Omega a^2{b}^2/(a^2+b^2)$. Equating that maximum to ${\tau }_Y$ gives the displayed first-yield twist. The ends of the minor axis yield first; geometric distance from the bar axis alone does not determine stress in a noncircular section.

For an objective material isotropy law, permutation of the principal stretches permutes the labelled principal stresses. This follows by conjugating the left stretch tensor with signed permutation rotation matrices. Each principal stress is symmetric in the other two stretches, but need not be symmetric in all three. The law $\sigma =V^2$ provides the example ${\sigma }_i={\lambda }_i^2$. In an invariant representation $\sigma ={\beta }_{0}I+{\beta }_{1}V+{\beta }_2{V}^2$, the scalar coefficients are symmetric functions of the unordered stretches.

Ernest Masson Anderson related faulting to the orientations and relative magnitudes of principal stresses. Near an approximately horizontal, shear-free surface, the vertical direction is a principal direction, leading to the idealized Andersonian faulting regime.^[ref]

With compression taken positive and ${\sigma }_1\ge {\sigma }_2\ge {\sigma }_3$, vertical ${\sigma }_1$ corresponds to normal faulting, vertical ${\sigma }_3$ to reverse faulting, and vertical ${\sigma }_2$ to strike-slip faulting. Pre-existing weaknesses and rotated stress axes can produce departures from this simple classification.

An opening fracture in rock with little or no shear displacement across its faces. A geological joint is approximately normal to the least compressive principal stress when it forms in a homogeneous medium. It can remain open or become a mineral vein. Unlike a fault, a geological joint need not contain crushed fault rock.

Günter Zumpe was a civil engineer and professor of mechanics and surface structures at TU Dresden. His institutional obituary records his long association with that university.^[ref]

His historical article on Mohr's circle connects Christian Otto Mohr's work to the graphical representation of stress. The construction turns components of a two-dimensional stress tensor into a circle whose intercepts give the principal stresses. Zumpe's account places this useful engineering calculation in the history of mechanics, including Mohr's 1882 article.^[ref]

A fluid-filled fracture can open perpendicular to the least compressive principal stress. In a compression-positive contact approximation, effective stress is ${\sigma }_n^{\prime }={\sigma }_n-P_f$. Opening an initially intact plane of tensile strength $T_0$ requiresA sufficiently weak pre-existing crack can open near zero effective compression. Increasing isotropic pore fluid pressure translates all effective principal stresses without rotating their directions, so an opening fracture and a frictional normal fault can record compatible stress axes at different pressure episodes. Subsequent mineral precipitation creates a mineral vein; white appearance alone does not identify its filling mineral.

A geometric representation of resolved normal stress and shear stress. For two principal stresses, its centre is $C=({\sigma }_1+{\sigma }_3)/2$ and radius $R=({\sigma }_1-{\sigma }_3)/2$. A physical plane-normal rotation by $\theta$ corresponds to $2\theta$ on the circle, with the sense depending on the shear-sign convention. Named for Christian Otto Mohr.

For a cohesionless, optimally oriented normal fault with vertical maximum compression and horizontal minimum compression,where $\delta$ is fault dip and $\phi$ the internal friction angle. This is a geometric inference under the model, not a measurement of friction from dip alone.

For a compression-positive Mohr circle, put $C=({\sigma }_1^{\prime }+{\sigma }_3^{\prime })/2$ and $R=({\sigma }_1^{\prime }-{\sigma }_3^{\prime })/2$. Tangency to the cohesionless Mohr-Coulomb failure criterion $\tau ={\mu }_f{\sigma }_n^{\prime }$ gives $R/C=\sin \phi$. The contact point is $({\sigma }_n^{\prime },\tau )=(C{\cos }^2\phi ,C\sin \phi \cos \phi )$. The physical plane-normal angle to ${\sigma }_1^{\prime }$ is half the corresponding central angle, giving the stated dip. Equivalently the plane itself makes $45^{\circ }-\phi /2$ with ${\sigma }_1^{\prime }$.

The associated stress ratio is ${\sigma }_1^{\prime }/{\sigma }_3^{\prime }=(1+\sin \phi )/(1-\sin \phi )$. A reactivated nonoptimal plane, nonzero cohesion or unknown principal stress directions defeats the simple dip-only inversion.

LetAt maximum pressure, the stress is the field from part (c) with $P=P_m$ and $s=s_m$. Under the assumed elastic unloading, previously accumulated plastic strain is frozen. Therefore it is the stress increment, rather than the total stress, that satisfies homogeneous linear elasticity and radial elastic compatibility. Applying the superposition principle, a pressure change $-\Delta$ produces the elastic incrementsThey have zero far-field stress and change the wall traction from $-P_m$ to $-P$. These increments are the same in the formerly plastic and formerly elastic regions because their elastic Lamé parameters agree. Adding them to the maximum-load field givesandThese formulae retain $s_m$: it records the plastic loading history and is not the instantaneous loading-front formula with $P$ substituted for $P_m$. If the entire return is elastic, setting $\Delta =P_m$ gives the residual stress after removal of the applied pressure.

To check admissibility, examine the signed difference of the principal stresses in the formerly plastic shell:It immediately moves into the interior of the Tresca yield criterion when unloading starts. It can next reach the opposite limit $D=+2{\tau }_Y$. For fixed $\Delta >0$, $D$ is largest at $r=a$ and stays above $-2{\tau }_Y$. At the wall,There is no earlier violation in the outer region. Indeed,so its magnitude is no larger than $|D(s_m)|$, which has already been controlled by the shell calculation. Hence the first reverse yielding occurs at the cavity wall, withIf $P_c<P_m<2P_c$, the stress stays within the yield surface throughout unloading to zero. If $P_m=2P_c$, it reaches reverse yield exactly at $P=0$; further unloading would produce reverse plastic deformation. If $P_m>2P_c$, reverse plastic deformation starts while the applied pressure is still positive. ThusIf “yield again while unloaded” includes first touching the yield surface at the final endpoint, the corresponding condition is $P_m\ge 8{\tau }_Y/3$. The displayed elastic unloading fields are physically valid only for $0\le \Delta \le \min (P_m,2P_c)$; beyond that interval a new reverse-plastic region must be solved for. Continuing the elastic formula to $P=0$ when $P_m>2P_c$ would violate the given Tresca yield criterion.

Let $B=A^{\mathsf{T}}A$. Since $\det A>0$, $B$ is a positive-definite matrix. The spectral theorem gives a unique principal square root of a positive semidefinite matrixThen $R^{\mathsf{T}}R=U^{-1}B{U}^{-1}=I$ and $\det R=\det A/\det U=1$, so $R$ is a rotation matrix. Define $V=RU{R}^{\mathsf{T}}$. It is a symmetric matrix, is positive definite, and satisfies $V^2=A{A}^{\mathsf{T}}$. Consequently the polar decomposition in continuum mechanics isHere $U$ and $V$ are the right stretch tensor and left stretch tensor.

For the Cauchy stress tensor, use both material isotropy and material frame indifference. Their respective transformation laws areThe first law and $A=VR$ give $\sigma (A)=\sigma (V)$. Combining the two laws gives $\sigma (QV{Q}^{\mathsf{T}})=Q\sigma (V)Q^{\mathsf{T}}$. In a principal stretch basis, $V=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$. Each half-turn about a coordinate axis leaves $V$ unchanged; its conjugation changes the signs of the corresponding off-diagonal stress tensor entries. Those entries must therefore vanish. If two principal stretches coincide, rotations within their eigenspace additionally force the stress tensor to be scalar on that eigenspace. Thus the Cauchy stress tensor and left stretch tensor have common principal axes, including at repeated principal stretches. This proves coaxiality of isotropic elastic stress.

The precise symmetry conclusion is equivariance of isotropic principal stresses. Every permutation of the principal stretches can be realized by a signed permutation rotation matrix, sowhen the indexing is permuted consistently. Equivalently, permuting the arguments permutes the labelled principal stresses in the same way. In particular, ${\sigma }_1$ is symmetric in ${\lambda }_2,{\lambda }_3$, and analogous statements hold for ${\sigma }_2,{\sigma }_3$.

For distinct principal stretches, interpolate the three values ${\sigma }_i$ at the three points ${\lambda }_i$:Uniqueness of this interpolation implies that the scalar coefficients ${\beta }_j$ are symmetric functions of the unordered principal stretches; they may be expressed through the elementary symmetric invariants. Smooth isotropic constitutive laws have the corresponding invariant representation across repeated stretches as well. The wording about symmetric functions must be understood in this collective sense: each labelled principal stress need not be symmetric in all three arguments. For example, the objective isotropic law $\sigma =V^2$ has ${\sigma }_i={\lambda }_i^2$, which disproves that stronger interpretation.

Use both material isotropy and material frame indifference. The polar decomposition in continuum mechanics writes $F=VR$, where $V$ is the left stretch tensor. Isotropy removes the reference rotation: $\sigma (F)=\sigma (V)$. Choose an orthonormal eigenbasis of $V$. A half-turn about any principal axis leaves $V$ fixed but reverses appropriate off-diagonal entries of $\sigma (V)$. Frame indifference therefore makes those entries zero. If a principal stretch is repeated, rotations in its eigenspace additionally force the stress to be isotropic there. Thus the principal stresses share the stretch axes: this proves coaxiality of isotropic elastic stress with $V$ and with $B=V^2=F{F}^T$.

For simple shear, the left Cauchy-Green deformation tensor isIts principal axes make an angle $\vartheta$ with the coordinate axes satisfying $2\cot (2\vartheta )=\gamma$ when $\gamma \ne 0$. If the two principal stresses are $s_1,s_2$, rotating their diagonal matrix back givesConsequently the universal simple-shear normal-stress identity isEquivalently, it follows immediately from the off-diagonal entry of $\sigma B-B\sigma =0$. At $\gamma =0$, the undeformed material has isotropic stress, so the identity holds there too. No particular elastic constitutive equation has been used.

The first Piola-Kirchhoff stress tensor $T_R$ maps a reference normal to force per reference area. The Cauchy stress tensor $T$ maps a current normal to force per current area. Nanson's formula gives $\mathbf{n}da=J{F}^{-T}\mathbf{N}dA$. Equating the actual surface force in both descriptions gives $T_R=JT{F}^{-T}$, and thereforeFor incompressibility $\boxed{J=1}$. The strain-energy derivative only fixes the constitutive stress up to a constraint reaction in hyperelastic stress: $T_R=DW-p\mathrm{cof}F$, hence $T=DW{F}^T-pI$. The scalar $p$ is a Lagrange multiplier determined by equilibrium and boundary tractions, rather than by the isochoric energy alone.

In principal stretch axes, the extra principal stress is $v_i\partial W/\partial v_i=2\alpha v_i^2-2\beta v_i^{-2}$, up to that arbitrary spherical reaction. With $B=F{F}^T$, its eigenvalues are $v_i^2$, so isotropy yieldsFor the prescribed simple shear,The $x_3$ faces retain normal $\pm {\mathbf{e}}_3$, and zero traction there gives $p=2(\alpha -\beta )$. ThusThis constant stress satisfies the interior equilibrium equation, and the remaining face forces maintain the deformation.

To compute the angle, the material face $x_1=1$ becomes the spatial plane $y_1-\gamma y_2=1$. Its outward unit normal and an in-plane unit tangent areResolve the current traction $\mathbf{t}=T\mathbf{n}$ into these directions:Since $N<0$ for nonzero $\gamma$, the angle lies above $\pi /2$, andUsing $\arctan z\sim z$ at zero givesUsing the undeformed normal ${\mathbf{e}}_1$ would give the wrong coefficient. The transformed unit normal is essential to this simple-shear traction angle for a Mooney-Rivlin solid.

For an Andersonian faulting regime appropriate to the stated normal fault, take ${\sigma }_1$ vertical, ${\sigma }_2$ north–south along strike, and ${\sigma }_3$ east–west. Thus the minimum horizontal compression is normal to the fault’s north–south strike. In an east–up cross-section, draw ${\sigma }_1$ vertically and ${\sigma }_3$ horizontally, with ${\sigma }_2$ perpendicular to the page. These are plausible regional principal stress directions, rather than measured directions uniquely recoverable from one photograph.

Use the same conditional optimal normal-fault dip relation as in question 1, with $\delta =55^{\circ }$:The second panel of the original Mohr circle figure has ${\sigma }_3^{\prime }=1$ and ${\sigma }_1^{\prime }=2.03961$; its tangent from the origin has slope $0.364$. Resolving the stresses onto a $55^{\circ }$ plane places its stress point at that tangency. The diagram assumes negligible cohesion, a vertical maximum principal stress and an optimally oriented new or equivalently oriented sliding plane. Without those assumptions, a dip and a statement that sliding began do not uniquely determine the coefficient of friction.