In continuum mechanics , the Cauchy stress tensor (symbol σ {\displaystyle {\boldsymbol {\sigma }}} , named after Augustin-Louis Cauchy ), also called true stress tensor [1] or simply stress tensor , completely defines the state of stress at a point inside a material in the deformed state, placement, or configuration. The second order tensor consists of nine components σ i j {\displaystyle \sigma _{ij}} and relates a unit-length direction vector e to the traction vector T (e ) across an imaginary surface perpendicular to e :
T ( e ) = e ⋅ σ or T j ( e ) = ∑ i σ i j e i . {\displaystyle \mathbf {T} ^{(\mathbf {e} )}=\mathbf {e} \cdot {\boldsymbol {\sigma }}\quad {\text{or}}\quad T_{j}^{(e)}=\sum _{i}\sigma _{ij}e_{i}.} [a] The SI base units of both stress tensor and traction vector are newton per square metre (N/m2 ) or pascal (Pa), corresponding to the stress scalar. The unit vector is dimensionless .
The Cauchy stress tensor obeys the tensor transformation law under a change in the system of coordinates. A graphical representation of this transformation law is the Mohr's circle for stress.
The Cauchy stress tensor is used for stress analysis of material bodies experiencing small deformations : it is a central concept in the linear theory of elasticity . For large deformations, also called finite deformations , other measures of stress are required, such as the Piola–Kirchhoff stress tensor , the Biot stress tensor , and the Kirchhoff stress tensor .
According to the principle of conservation of linear momentum , if the continuum body is in static equilibrium it can be demonstrated that the components of the Cauchy stress tensor in every material point in the body satisfy the equilibrium equations (Cauchy's equations of motion for zero acceleration). At the same time, according to the principle of conservation of angular momentum , equilibrium requires that the summation of moments with respect to an arbitrary point is zero, which leads to the conclusion that the stress tensor is symmetric , thus having only six independent stress components, instead of the original nine. However, in the presence of couple-stresses, i.e. moments per unit volume, the stress tensor is non-symmetric. This also is the case when the Knudsen number is close to one, K n → 1 {\displaystyle K_{n}\rightarrow 1} , or the continuum is a non-Newtonian fluid, which can lead to rotationally non-invariant fluids, such as polymers .
There are certain invariants associated with the stress tensor, whose values do not depend upon the coordinate system chosen, or the area element upon which the stress tensor operates. These are the three eigenvalues of the stress tensor, which are called the principal stresses .
Euler–Cauchy stress principle – stress vector edit Figure 2.1a Internal distribution of contact forces and couple stresses on a differential d S {\displaystyle dS} of the internal surface S {\displaystyle S} in a continuum, as a result of the interaction between the two portions of the continuum separated by the surface Figure 2.1b Internal distribution of contact forces and couple stresses on a differential d S {\displaystyle dS} of the internal surface S {\displaystyle S} in a continuum, as a result of the interaction between the two portions of the continuum separated by the surface Figure 2.1c Stress vector on an internal surface S with normal vector n. Depending on the orientation of the plane under consideration, the stress vector may not necessarily be perpendicular to that plane, i.e. parallel to n {\displaystyle \mathbf {n} } , and can be resolved into two components: one component normal to the plane, called normal stress σ n {\displaystyle \sigma _{\mathrm {n} }} , and another component parallel to this plane, called the shearing stress τ {\displaystyle \tau } . The Euler–Cauchy stress principle states that upon any surface (real or imaginary) that divides the body, the action of one part of the body on the other is equivalent (equipollent) to the system of distributed forces and couples on the surface dividing the body ,[2] and it is represented by a field T ( n ) {\displaystyle \mathbf {T} ^{(\mathbf {n} )}} , called the traction vector , defined on the surface S {\displaystyle S} and assumed to depend continuously on the surface's unit vector n {\displaystyle \mathbf {n} } .[3] [4] : p.66–96
To formulate the Euler–Cauchy stress principle, consider an imaginary surface S {\displaystyle S} passing through an internal material point P {\displaystyle P} dividing the continuous body into two segments, as seen in Figure 2.1a or 2.1b (one may use either the cutting plane diagram or the diagram with the arbitrary volume inside the continuum enclosed by the surface S {\displaystyle S} ).
Following the classical dynamics of Newton and Euler , the motion of a material body is produced by the action of externally applied forces which are assumed to be of two kinds: surface forces F {\displaystyle \mathbf {F} } and body forces b {\displaystyle \mathbf {b} } .[5] Thus, the total force F {\displaystyle {\mathcal {F}}} applied to a body or to a portion of the body can be expressed as:
F = b + F {\displaystyle {\mathcal {F}}=\mathbf {b} +\mathbf {F} } Only surface forces will be discussed in this article as they are relevant to the Cauchy stress tensor.
When the body is subjected to external surface forces or contact forces F {\displaystyle \mathbf {F} } , following Euler's equations of motion , internal contact forces and moments are transmitted from point to point in the body, and from one segment to the other through the dividing surface S {\displaystyle S} , due to the mechanical contact of one portion of the continuum onto the other (Figure 2.1a and 2.1b). On an element of area Δ S {\displaystyle \Delta S} containing P {\displaystyle P} , with normal vector n {\displaystyle \mathbf {n} } , the force distribution is equipollent to a contact force Δ F {\displaystyle \Delta \mathbf {F} } exerted at point P and surface moment Δ M {\displaystyle \Delta \mathbf {M} } . In particular, the contact force is given by
Δ F = T ( n ) Δ S {\displaystyle \Delta \mathbf {F} =\mathbf {T} ^{(\mathbf {n} )}\,\Delta S} where T ( n ) {\displaystyle \mathbf {T} ^{(\mathbf {n} )}} is the mean surface traction .
Cauchy's stress principle asserts[6] : p.47–102 that as Δ S {\displaystyle \Delta S} becomes very small and tends to zero the ratio Δ F / Δ S {\displaystyle \Delta \mathbf {F} /\Delta S} becomes d F / d S {\displaystyle d\mathbf {F} /dS} and the couple stress vector Δ M {\displaystyle \Delta \mathbf {M} } vanishes. In specific fields of continuum mechanics the couple stress is assumed not to vanish; however, classical branches of continuum mechanics address non-polar materials which do not consider couple stresses and body moments.
The resultant vector d F / d S {\displaystyle d\mathbf {F} /dS} is defined as the surface traction ,[7] also called stress vector ,[8] traction ,[4] or traction vector .[6] given by T ( n ) = T i ( n ) e i {\displaystyle \mathbf {T} ^{(\mathbf {n} )}=T_{i}^{(\mathbf {n} )}\mathbf {e} _{i}} at the point P {\displaystyle P} associated with a plane with a normal vector n {\displaystyle \mathbf {n} } :
T i ( n ) = lim Δ S → 0 Δ F i Δ S = d F i d S . {\displaystyle T_{i}^{(\mathbf {n} )}=\lim _{\Delta S\to 0}{\frac {\Delta F_{i}}{\Delta S}}={dF_{i} \over dS}.} This equation means that the stress vector depends on its location in the body and the orientation of the plane on which it is acting.
This implies that the balancing action of internal contact forces generates a contact force density or Cauchy traction field [5] T ( n , x , t ) {\displaystyle \mathbf {T} (\mathbf {n} ,\mathbf {x} ,t)} that represents a distribution of internal contact forces throughout the volume of the body in a particular configuration of the body at a given time t {\displaystyle t} . It is not a vector field because it depends not only on the position x {\displaystyle \mathbf {x} } of a particular material point, but also on the local orientation of the surface element as defined by its normal vector n {\displaystyle \mathbf {n} } .[9]
Depending on the orientation of the plane under consideration, the stress vector may not necessarily be perpendicular to that plane, i.e. parallel to n {\displaystyle \mathbf {n} } , and can be resolved into two components (Figure 2.1c):
one normal to the plane, called normal stress σ n = lim Δ S → 0 Δ F n Δ S = d F n d S , {\displaystyle \mathbf {\sigma _{\mathrm {n} }} =\lim _{\Delta S\to 0}{\frac {\Delta F_{\mathrm {n} }}{\Delta S}}={\frac {dF_{\mathrm {n} }}{dS}},} where d F n {\displaystyle dF_{\mathrm {n} }} is the normal component of the force d F {\displaystyle d\mathbf {F} } to the differential area d S {\displaystyle dS} and the other parallel to this plane, called the shear stress τ = lim Δ S → 0 Δ F s Δ S = d F s d S , {\displaystyle \mathbf {\tau } =\lim _{\Delta S\to 0}{\frac {\Delta F_{\mathrm {s} }}{\Delta S}}={\frac {dF_{\mathrm {s} }}{dS}},} where d F s {\displaystyle dF_{\mathrm {s} }} is the tangential component of the force d F {\displaystyle d\mathbf {F} } to the differential surface area d S {\displaystyle dS} . The shear stress can be further decomposed into two mutually perpendicular vectors. Cauchy's postulate edit According to the Cauchy Postulate , the stress vector T ( n ) {\displaystyle \mathbf {T} ^{(\mathbf {n} )}} remains unchanged for all surfaces passing through the point P {\displaystyle P} and having the same normal vector n {\displaystyle \mathbf {n} } at P {\displaystyle P} ,[7] [10] i.e., having a common tangent at P {\displaystyle P} . This means that the stress vector is a function of the normal vector n {\displaystyle \mathbf {n} } only, and is not influenced by the curvature of the internal surfaces.
Cauchy's fundamental lemma edit A consequence of Cauchy's postulate is Cauchy's Fundamental Lemma ,[1] [7] [11] also called the Cauchy reciprocal theorem ,[12] : p.103–130 which states that the stress vectors acting on opposite sides of the same surface are equal in magnitude and opposite in direction. Cauchy's fundamental lemma is equivalent to Newton's third law of motion of action and reaction, and is expressed as
− T ( n ) = T ( − n ) . {\displaystyle -\mathbf {T} ^{(\mathbf {n} )}=\mathbf {T} ^{(-\mathbf {n} )}.} Cauchy's stress theorem—stress tensor edit The state of stress at a point in the body is then defined by all the stress vectors T (n ) associated with all planes (infinite in number) that pass through that point.[13] However, according to Cauchy's fundamental theorem ,[11] also called Cauchy's stress theorem ,[1] merely by knowing the stress vectors on three mutually perpendicular planes, the stress vector on any other plane passing through that point can be found through coordinate transformation equations.
Cauchy's stress theorem states that there exists a second-order tensor field σ (x , t), called the Cauchy stress tensor, independent of n , such that T is a linear function of n :
T ( n ) = n ⋅ σ or T j ( n ) = σ i j n i . {\displaystyle \mathbf {T} ^{(\mathbf {n} )}=\mathbf {n} \cdot {\boldsymbol {\sigma }}\quad {\text{or}}\quad T_{j}^{(n)}=\sigma _{ij}n_{i}.} This equation implies that the stress vector T (n ) at any point P in a continuum associated with a plane with normal unit vector n can be expressed as a function of the stress vectors on the planes perpendicular to the coordinate axes, i.e. in terms of the components σij of the stress tensor σ .
To prove this expression, consider a tetrahedron with three faces oriented in the coordinate planes, and with an infinitesimal area dA oriented in an arbitrary direction specified by a normal unit vector n (Figure 2.2). The tetrahedron is formed by slicing the infinitesimal element along an arbitrary plane with unit normal n . The stress vector on this plane is denoted by T (n ) . The stress vectors acting on the faces of the tetrahedron are denoted as T (e 1 ) , T (e 2 ) , and T (e 3 ) , and are by definition the components σij of the stress tensor σ . This tetrahedron is sometimes called the Cauchy tetrahedron . The equilibrium of forces, i.e. Euler's first law of motion (Newton's second law of motion), gives:
T ( n ) d A − T ( e 1 ) d A 1 − T ( e 2 ) d A 2 − T ( e 3 ) d A 3 = ρ ( h 3 d A ) a , {\displaystyle \mathbf {T} ^{(\mathbf {n} )}\,dA-\mathbf {T} ^{(\mathbf {e} _{1})}\,dA_{1}-\mathbf {T} ^{(\mathbf {e} _{2})}\,dA_{2}-\mathbf {T} ^{(\mathbf {e} _{3})}\,dA_{3}=\rho \left({\frac {h}{3}}dA\right)\mathbf {a} ,} Figure 2.2. Stress vector acting on a plane with normal unit vector n . A note on the sign convention: The tetrahedron is formed by slicing a parallelepiped along an arbitrary plane n . So, the force acting on the plane n is the reaction exerted by the other half of the parallelepiped and has an opposite sign. where the right-hand-side represents the product of the mass enclosed by the tetrahedron and its acceleration: ρ is the density, a is the acceleration, and h is the height of the tetrahedron, considering the plane n as the base. The area of the faces of the tetrahedron perpendicular to the axes can be found by projecting dA into each face (using the dot product):
d A 1 = ( n ⋅ e 1 ) d A = n 1 d A , {\displaystyle dA_{1}=\left(\mathbf {n} \cdot \mathbf {e} _{1}\right)dA=n_{1}\;dA,} d A 2 = ( n ⋅ e 2 ) d A = n 2 d A , {\displaystyle dA_{2}=\left(\mathbf {n} \cdot \mathbf {e} _{2}\right)dA=n_{2}\;dA,} d A 3 = ( n ⋅ e 3 ) d A = n 3 d A , {\displaystyle dA_{3}=\left(\mathbf {n} \cdot \mathbf {e} _{3}\right)dA=n_{3}\;dA,} and then substituting into the equation to cancel out dA :
T ( n ) − T ( e 1 ) n 1 − T ( e 2 ) n 2 − T ( e 3 ) n 3 = ρ ( h 3 ) a . {\displaystyle \mathbf {T} ^{(\mathbf {n} )}-\mathbf {T} ^{(\mathbf {e} _{1})}n_{1}-\mathbf {T} ^{(\mathbf {e} _{2})}n_{2}-\mathbf {T} ^{(\mathbf {e} _{3})}n_{3}=\rho \left({\frac {h}{3}}\right)\mathbf {a} .} To consider the limiting case as the tetrahedron shrinks to a point, h must go to 0 (intuitively, the plane n is translated along n toward O ). As a result, the right-hand-side of the equation approaches 0, so
T ( n ) = T ( e 1 ) n 1 + T ( e 2 ) n 2 + T ( e 3 ) n 3 . {\displaystyle \mathbf {T} ^{(\mathbf {n} )}=\mathbf {T} ^{(\mathbf {e} _{1})}n_{1}+\mathbf {T} ^{(\mathbf {e} _{2})}n_{2}+\mathbf {T} ^{(\mathbf {e} _{3})}n_{3}.} Assuming a material element (Figure 2.3) with planes perpendicular to the coordinate axes of a Cartesian coordinate system, the stress vectors associated with each of the element planes, i.e. T (e 1 ) , T (e 2 ) , and T (e 3 ) can be decomposed into a normal component and two shear components, i.e. components in the direction of the three coordinate axes. For the particular case of a surface with normal unit vector oriented in the direction of the x 1 -axis, denote the normal stress by σ 11 , and the two shear stresses as σ 12 and σ 13 :
T ( e 1 ) = T 1 ( e 1 ) e 1 + T 2 ( e 1 ) e 2 + T 3 ( e 1 ) e 3 = σ 11 e 1 + σ 12 e 2 + σ 13 e 3 , {\displaystyle \mathbf {T} ^{(\mathbf {e} _{1})}=T_{1}^{(\mathbf {e} _{1})}\mathbf {e} _{1}+T_{2}^{(\mathbf {e} _{1})}\mathbf {e} _{2}+T_{3}^{(\mathbf {e} _{1})}\mathbf {e} _{3}=\sigma _{11}\mathbf {e} _{1}+\sigma _{12}\mathbf {e} _{2}+\sigma _{13}\mathbf {e} _{3},} T ( e 2 ) = T 1 ( e 2 ) e 1 + T 2 ( e 2 ) e 2 + T 3 ( e 2 ) e 3 = σ 21 e 1 + σ 22 e 2 + σ 23 e 3 , {\displaystyle \mathbf {T} ^{(\mathbf {e} _{2})}=T_{1}^{(\mathbf {e} _{2})}\mathbf {e} _{1}+T_{2}^{(\mathbf {e} _{2})}\mathbf {e} _{2}+T_{3}^{(\mathbf {e} _{2})}\mathbf {e} _{3}=\sigma _{21}\mathbf {e} _{1}+\sigma _{22}\mathbf {e} _{2}+\sigma _{23}\mathbf {e} _{3},} T ( e 3 ) = T 1 ( e 3 ) e 1 + T 2 ( e 3 ) e 2 + T 3 ( e 3 ) e 3 = σ 31 e 1 + σ 32 e 2 + σ 33 e 3 , {\displaystyle \mathbf {T} ^{(\mathbf {e} _{3})}=T_{1}^{(\mathbf {e} _{3})}\mathbf {e} _{1}+T_{2}^{(\mathbf {e} _{3})}\mathbf {e} _{2}+T_{3}^{(\mathbf {e} _{3})}\mathbf {e} _{3}=\sigma _{31}\mathbf {e} _{1}+\sigma _{32}\mathbf {e} _{2}+\sigma _{33}\mathbf {e} _{3},} In index notation this is
T ( e i ) = T j ( e i ) e j = σ i j e j . {\displaystyle \mathbf {T} ^{(\mathbf {e} _{i})}=T_{j}^{(\mathbf {e} _{i})}\mathbf {e} _{j}=\sigma _{ij}\mathbf {e} _{j}.} The nine components σij of the stress vectors are the components of a second-order Cartesian tensor called the Cauchy stress tensor , which can be used to completely define the state of stress at a point and is given by
σ = σ i j = [ T ( e 1 ) T ( e 2 ) T ( e 3 ) ] = [ σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33 ] ≡ [ σ x x σ x y σ x z σ y x σ y y σ y z σ z x σ z y σ z z ] ≡ [ σ x τ x y τ x z τ y x σ y τ y z τ z x τ z y σ z ] , {\displaystyle {\boldsymbol {\sigma }}=\sigma _{ij}=\left[{\begin{matrix}\mathbf {T} ^{(\mathbf {e} _{1})}\\\mathbf {T} ^{(\mathbf {e} _{2})}\\\mathbf {T} ^{(\mathbf {e} _{3})}\\\end{matrix}}\right]=\left[{\begin{matrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\\\end{matrix}}\right]\equiv \left[{\begin{matrix}\sigma _{xx}&\sigma _{xy}&\sigma _{xz}\\\sigma _{yx}&\sigma _{yy}&\sigma _{yz}\\\sigma _{zx}&\sigma _{zy}&\sigma _{zz}\\\end{matrix}}\right]\equiv \left[{\begin{matrix}\sigma _{x}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{y}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{z}\\\end{matrix}}\right],} where σ 11 , σ 22 , and σ 33 are normal stresses, and σ 12 , σ 13 , σ 21 , σ 23 , σ 31 , and σ 32 are shear stresses. The first index i indicates that the stress acts on a plane normal to the Xi -axis, and the second index j denotes the direction in which the stress acts (For example, σ12 implies that the stress is acting on the plane that is normal to the 1st axis i.e.;X 1 and acts along the 2nd axis i.e.;X 2 ). A stress component is positive if it acts in the positive direction of the coordinate axes, and if the plane where it acts has an outward normal vector pointing in the positive coordinate direction.
Thus, using the components of the stress tensor
T ( n ) = T ( e 1 ) n 1 + T ( e 2 ) n 2 + T ( e 3 ) n 3 = ∑ i = 1 3 T ( e i ) n i = ( σ i j e j ) n i = σ i j n i e j {\displaystyle {\begin{aligned}\mathbf {T} ^{(\mathbf {n} )}&=\mathbf {T} ^{(\mathbf {e} _{1})}n_{1}+\mathbf {T} ^{(\mathbf {e} _{2})}n_{2}+\mathbf {T} ^{(\mathbf {e} _{3})}n_{3}\\&=\sum _{i=1}^{3}\mathbf {T} ^{(\mathbf {e} _{i})}n_{i}\\&=\left(\sigma _{ij}\mathbf {e} _{j}\right)n_{i}\\&=\sigma _{ij}n_{i}\mathbf {e} _{j}\end{aligned}}} or, equivalently,
T j ( n ) = σ i j n i . {\displaystyle T_{j}^{(\mathbf {n} )}=\sigma _{ij}n_{i}.} Alternatively, in matrix form we have
[ T 1 ( n ) T 2 ( n ) T 3 ( n ) ] = [ n 1 n 2 n 3 ] ⋅ [ σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33 ] . {\displaystyle \left[{\begin{matrix}T_{1}^{(\mathbf {n} )}&T_{2}^{(\mathbf {n} )}&T_{3}^{(\mathbf {n} )}\end{matrix}}\right]=\left[{\begin{matrix}n_{1}&n_{2}&n_{3}\end{matrix}}\right]\cdot \left[{\begin{matrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\\\end{matrix}}\right].} The Voigt notation representation of the Cauchy stress tensor takes advantage of the symmetry of the stress tensor to express the stress as a six-dimensional vector of the form:
σ = [ σ 1 σ 2 σ 3 σ 4 σ 5 σ 6 ] T ≡ [ σ 11 σ 22 σ 33 σ 23 σ 13 σ 12 ] T . {\displaystyle {\boldsymbol {\sigma }}={\begin{bmatrix}\sigma _{1}&\sigma _{2}&\sigma _{3}&\sigma _{4}&\sigma _{5}&\sigma _{6}\end{bmatrix}}^{\textsf {T}}\equiv {\begin{bmatrix}\sigma _{11}&\sigma _{22}&\sigma _{33}&\sigma _{23}&\sigma _{13}&\sigma _{12}\end{bmatrix}}^{\textsf {T}}.} The Voigt notation is used extensively in representing stress–strain relations in solid mechanics and for computational efficiency in numerical structural mechanics software.
Transformation rule of the stress tensor edit It can be shown that the stress tensor is a contravariant second order tensor, which is a statement of how it transforms under a change of the coordinate system. From an xi -system to an xi ' -system, the components σij in the initial system are transformed into the components σij ' in the new system according to the tensor transformation rule (Figure 2.4):
σ i j ′ = a i m a j n σ m n or σ ′ = A σ A T , {\displaystyle \sigma '_{ij}=a_{im}a_{jn}\sigma _{mn}\quad {\text{or}}\quad {\boldsymbol {\sigma }}'=\mathbf {A} {\boldsymbol {\sigma }}\mathbf {A} ^{\textsf {T}},} where A is a rotation matrix with components aij . In matrix form this is
[ σ 11 ′ σ 12 ′ σ 13 ′ σ 21 ′ σ 22 ′ σ 23 ′ σ 31 ′ σ 32 ′ σ 33 ′ ] = [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] [ σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33 ] [ a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ] . {\displaystyle \left[{\begin{matrix}\sigma '_{11}&\sigma '_{12}&\sigma '_{13}\\\sigma '_{21}&\sigma '_{22}&\sigma '_{23}\\\sigma '_{31}&\sigma '_{32}&\sigma '_{33}\\\end{matrix}}\right]=\left[{\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\\end{matrix}}\right]\left[{\begin{matrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\\\end{matrix}}\right]\left[{\begin{matrix}a_{11}&a_{21}&a_{31}\\a_{12}&a_{22}&a_{32}\\a_{13}&a_{23}&a_{33}\\\end{matrix}}\right].} Figure 2.4 Transformation of the stress tensor
Expanding the matrix operation , and simplifying terms using the symmetry of the stress tensor , gives
σ 11 ′ = a 11 2 σ 11 + a 12 2 σ 22 + a 13 2 σ 33 + 2 a 11 a 12 σ 12 + 2 a 11 a 13 σ 13 + 2 a 12 a 13 σ 23 , σ 22 ′ = a 21 2 σ 11 + a 22 2 σ 22 + a 23 2 σ 33 + 2 a 21 a 22 σ 12 + 2 a 21 a 23 σ 13 + 2 a 22 a 23 σ 23 , σ 33 ′ = a 31 2 σ 11 + a 32 2 σ 22 + a 33 2 σ 33 + 2 a 31 a 32 σ 12 + 2 a 31 a 33 σ 13 + 2 a 32 a 33 σ 23 , σ 12 ′ = a 11 a 21 σ 11 + a 12 a 22 σ 22 + a 13 a 23 σ 33 + ( a 11 a 22 + a 12 a 21 ) σ 12 + ( a 12 a 23 + a 13 a 22 ) σ 23 + ( a 11 a 23 + a 13 a 21 ) σ 13 , σ 23 ′ = a 21 a 31 σ 11 + a 22 a 32 σ 22 + a 23 a 33 σ 33 + ( a 21 a 32 + a 22 a 31 ) σ 12 + ( a 22 a 33 + a 23 a 32 ) σ 23 + ( a 21 a 33 + a 23 a 31 ) σ 13 , σ 13 ′ = a 11 a 31 σ 11 + a 12 a 32 σ 22 + a 13 a 33 σ 33 + ( a 11 a 32 + a 12 a 31 ) σ 12 + ( a 12 a 33 + a 13 a 32 ) σ 23 + ( a 11 a 33 + a 13 a 31 ) σ 13 . {\displaystyle {\begin{aligned}\sigma _{11}'={}&a_{11}^{2}\sigma _{11}+a_{12}^{2}\sigma _{22}+a_{13}^{2}\sigma _{33}+2a_{11}a_{12}\sigma _{12}+2a_{11}a_{13}\sigma _{13}+2a_{12}a_{13}\sigma _{23},\\\sigma _{22}'={}&a_{21}^{2}\sigma _{11}+a_{22}^{2}\sigma _{22}+a_{23}^{2}\sigma _{33}+2a_{21}a_{22}\sigma _{12}+2a_{21}a_{23}\sigma _{13}+2a_{22}a_{23}\sigma _{23},\\\sigma _{33}'={}&a_{31}^{2}\sigma _{11}+a_{32}^{2}\sigma _{22}+a_{33}^{2}\sigma _{33}+2a_{31}a_{32}\sigma _{12}+2a_{31}a_{33}\sigma _{13}+2a_{32}a_{33}\sigma _{23},\\\sigma _{12}'={}&a_{11}a_{21}\sigma _{11}+a_{12}a_{22}\sigma _{22}+a_{13}a_{23}\sigma _{33}\\&+(a_{11}a_{22}+a_{12}a_{21})\sigma _{12}+(a_{12}a_{23}+a_{13}a_{22})\sigma _{23}+(a_{11}a_{23}+a_{13}a_{21})\sigma _{13},\\\sigma _{23}'={}&a_{21}a_{31}\sigma _{11}+a_{22}a_{32}\sigma _{22}+a_{23}a_{33}\sigma _{33}\\&+(a_{21}a_{32}+a_{22}a_{31})\sigma _{12}+(a_{22}a_{33}+a_{23}a_{32})\sigma _{23}+(a_{21}a_{33}+a_{23}a_{31})\sigma _{13},\\\sigma _{13}'={}&a_{11}a_{31}\sigma _{11}+a_{12}a_{32}\sigma _{22}+a_{13}a_{33}\sigma _{33}\\&+(a_{11}a_{32}+a_{12}a_{31})\sigma _{12}+(a_{12}a_{33}+a_{13}a_{32})\sigma _{23}+(a_{11}a_{33}+a_{13}a_{31})\sigma _{13}.\end{aligned}}} The Mohr circle for stress is a graphical representation of this transformation of stresses.
Normal and shear stresses edit The magnitude of the normal stress component σ n of any stress vector T (n ) acting on an arbitrary plane with normal unit vector n at a given point, in terms of the components σij of the stress tensor σ , is the dot product of the stress vector and the normal unit vector:
σ n = T ( n ) ⋅ n = T i ( n ) n i = σ i j n i n j . {\displaystyle {\begin{aligned}\sigma _{\mathrm {n} }&=\mathbf {T} ^{(\mathbf {n} )}\cdot \mathbf {n} \\&=T_{i}^{(\mathbf {n} )}n_{i}\\&=\sigma _{ij}n_{i}n_{j}.\end{aligned}}} The magnitude of the shear stress component τ n , acting orthogonal to the vector n , can then be found using the Pythagorean theorem :
τ n = ( T ( n ) ) 2 − σ n 2 = T i ( n ) T i ( n ) − σ n 2 , {\displaystyle {\begin{aligned}\tau _{\mathrm {n} }&={\sqrt {\left(T^{(\mathbf {n} )}\right)^{2}-\sigma _{\mathrm {n} }^{2}}}\\&={\sqrt {T_{i}^{(\mathbf {n} )}T_{i}^{(\mathbf {n} )}-\sigma _{\mathrm {n} }^{2}}},\end{aligned}}} where
( T ( n ) ) 2 = T i ( n ) T i ( n ) = ( σ i j n j ) ( σ i k n k ) = σ i j σ i k n j n k . {\displaystyle \left(T^{(\mathbf {n} )}\right)^{2}=T_{i}^{(\mathbf {n} )}T_{i}^{(\mathbf {n} )}=\left(\sigma _{ij}n_{j}\right)\left(\sigma _{ik}n_{k}\right)=\sigma _{ij}\sigma _{ik}n_{j}n_{k}.} Balance laws – Cauchy's equations of motion edit Figure 4. Continuum body in equilibrium Cauchy's first law of motion edit According to the principle of conservation of linear momentum , if the continuum body is in static equilibrium it can be demonstrated that the components of the Cauchy stress tensor in every material point in the body satisfy the equilibrium equations:
σ j i , j + F i = 0 {\displaystyle \sigma _{ji,j}+F_{i}=0} ,where σ j i , j = ∑ j ∂ j σ j i {\displaystyle \sigma _{ji,j}=\sum _{j}\partial _{j}\sigma _{ji}}
For example, for a hydrostatic fluid in equilibrium conditions, the stress tensor takes on the form:
σ i j = − p δ i j , {\displaystyle {\sigma _{ij}}=-p{\delta _{ij}},} where p {\displaystyle p} is the hydrostatic pressure, and δ i j {\displaystyle {\delta _{ij}}\ } is the kronecker delta .
Derivation of equilibrium equations Consider a continuum body (see Figure 4) occupying a volume V {\displaystyle V} , having a surface area S {\displaystyle S} , with defined traction or surface forces T i ( n ) {\displaystyle T_{i}^{(n)}} per unit area acting on every point of the body surface, and body forces F i {\displaystyle F_{i}} per unit of volume on every point within the volume V {\displaystyle V} . Thus, if the body is in equilibrium the resultant force acting on the volume is zero, thus: ∫ S T i ( n ) d S + ∫ V F i d V = 0 {\displaystyle \int _{S}T_{i}^{(n)}dS+\int _{V}F_{i}dV=0} By definition the stress vector is T i ( n ) = σ j i n j {\displaystyle T_{i}^{(n)}=\sigma _{ji}n_{j}} , then
∫ S σ j i n j d S + ∫ V F i d V = 0 {\displaystyle \int _{S}\sigma _{ji}n_{j}\,dS+\int _{V}F_{i}\,dV=0} Using the Gauss's divergence theorem to convert a surface integral to a volume integral gives
∫ V σ j i , j d V + ∫ V F i d V = 0 {\displaystyle \int _{V}\sigma _{ji,j}\,dV+\int _{V}F_{i}\,dV=0} ∫ V ( σ j i , j + F i ) d V = 0 {\displaystyle \int _{V}(\sigma _{ji,j}+F_{i}\,)dV=0} For an arbitrary volume the integral vanishes, and we have the equilibrium equations
σ j i , j + F i = 0 {\displaystyle \sigma _{ji,j}+F_{i}=0}
Cauchy's second law of motion edit According to the principle of conservation of angular momentum , equilibrium requires that the summation of moments with respect to an arbitrary point is zero, which leads to the conclusion that the stress tensor is symmetric , thus having only six independent stress components, instead of the original nine:
σ i j = σ j i {\displaystyle \sigma _{ij}=\sigma _{ji}} Derivation of symmetry of the stress tensor Summing moments about point O (Figure 4) the resultant moment is zero as the body is in equilibrium. Thus, M O = ∫ S ( r × T ) d S + ∫ V ( r × F ) d V = 0 0 = ∫ S ε i j k x j T k ( n ) d S + ∫ V ε i j k x j F k d V {\displaystyle {\begin{aligned}M_{O}&=\int _{S}(\mathbf {r} \times \mathbf {T} )dS+\int _{V}(\mathbf {r} \times \mathbf {F} )dV=0\\0&=\int _{S}\varepsilon _{ijk}x_{j}T_{k}^{(n)}dS+\int _{V}\varepsilon _{ijk}x_{j}F_{k}dV\\\end{aligned}}} where r {\displaystyle \mathbf {r} } is the position vector and is expressed as
r = x j e j {\displaystyle \mathbf {r} =x_{j}\mathbf {e} _{j}} Knowing that T k ( n ) = σ m k n m {\displaystyle T_{k}^{(n)}=\sigma _{mk}n_{m}} and using Gauss's divergence theorem to change from a surface integral to a volume integral, we have
0 = ∫ S ε i j k x j σ m k n m d S + ∫ V ε i j k x j F k d V = ∫ V ( ε i j k x j σ m k ) , m d V + ∫ V ε i j k x j F k d V = ∫ V ( ε i j k x j , m σ m k + ε i j k x j σ m k , m ) d V + ∫ V ε i j k x j F k d V = ∫ V ( ε i j k x j , m σ m k ) d V + ∫ V ε i j k x j ( σ m k , m + F k ) d V {\displaystyle {\begin{aligned}0&=\int _{S}\varepsilon _{ijk}x_{j}\sigma _{mk}n_{m}\,dS+\int _{V}\varepsilon _{ijk}x_{j}F_{k}\,dV\\&=\int _{V}(\varepsilon _{ijk}x_{j}\sigma _{mk})_{,m}dV+\int _{V}\varepsilon _{ijk}x_{j}F_{k}\,dV\\&=\int _{V}(\varepsilon _{ijk}x_{j,m}\sigma _{mk}+\varepsilon _{ijk}x_{j}\sigma _{mk,m})dV+\int _{V}\varepsilon _{ijk}x_{j}F_{k}\,dV\\&=\int _{V}(\varepsilon _{ijk}x_{j,m}\sigma _{mk})dV+\int _{V}\varepsilon _{ijk}x_{j}(\sigma _{mk,m}+F_{k})dV\\\end{aligned}}} The second integral is zero as it contains the equilibrium equations. This leaves the first integral, where x j , m = δ j m {\displaystyle x_{j,m}=\delta _{jm}} , therefore
∫ V ( ε i j k σ j k ) d V = 0 {\displaystyle \int _{V}(\varepsilon _{ijk}\sigma _{jk})dV=0} For an arbitrary volume V, we then have
ε i j k σ j k = 0 {\displaystyle \varepsilon _{ijk}\sigma _{jk}=0} which is satisfied at every point within the body. Expanding this equation we have
σ 12 = σ 21 {\displaystyle \sigma _{12}=\sigma _{21}} , σ 23 = σ 32 {\displaystyle \sigma _{23}=\sigma _{32}} , and σ 13 = σ 31 {\displaystyle \sigma _{13}=\sigma _{31}} or in general
σ i j = σ j i {\displaystyle \sigma _{ij}=\sigma _{ji}} This proves that the stress tensor is symmetric
However, in the presence of couple-stresses, i.e. moments per unit volume, the stress tensor is non-symmetric. This also is the case when the Knudsen number is close to one, K n → 1 {\displaystyle K_{n}\rightarrow 1} , or the continuum is a non-Newtonian fluid, which can lead to rotationally non-invariant fluids, such as polymers .
Principal stresses and stress invariants edit Stress components on a 2D rotating element . Example of how stress components vary on the faces (edges) of a rectangular element as the angle of its orientation is varied. Principal stresses occur when the shear stresses simultaneously disappear from all faces. The orientation at which this occurs gives the principal directions . In this example, when the rectangle is horizontal, the stresses are given by [ σ 11 σ 12 σ 21 σ 22 ] = [ − 10 10 10 15 ] . {\displaystyle \left[{\begin{matrix}\sigma _{11}&\sigma _{12}\\\sigma _{21}&\sigma _{22}\end{matrix}}\right]=\left[{\begin{matrix}-10&10\\10&15\end{matrix}}\right].} At every point in a stressed body there are at least three planes, called principal planes , with normal vectors n {\displaystyle \mathbf {n} } , called principal directions , where the corresponding stress vector is perpendicular to the plane, i.e., parallel or in the same direction as the normal vector n {\displaystyle \mathbf {n} } , and where there are no normal shear stresses τ n {\displaystyle \tau _{\mathrm {n} }} . The three stresses normal to these principal planes are called principal stresses .
The components σ i j {\displaystyle \sigma _{ij}} of the stress tensor depend on the orientation of the coordinate system at the point under consideration. However, the stress tensor itself is a physical quantity and as such, it is independent of the coordinate system chosen to represent it. There are certain invariants associated with every tensor which are also independent of the coordinate system. For example, a vector is a simple tensor of rank one. In three dimensions, it has three components. The value of these components will depend on the coordinate system chosen to represent the vector, but the magnitude of the vector is a physical quantity (a scalar) and is independent of the Cartesian coordinate system chosen to represent the vector (so long as it is normal ). Similarly, every second rank tensor (such as the stress and the strain tensors) has three independent invariant quantities associated with it. One set of such invariants are the principal stresses of the stress tensor, which are just the eigenvalues of the stress tensor. Their direction vectors are the principal directions or eigenvectors .
A stress vector parallel to the normal unit vector n {\displaystyle \mathbf {n} } is given by:
T ( n ) = λ n = σ n n {\displaystyle \mathbf {T} ^{(\mathbf {n} )}=\lambda \mathbf {n} =\mathbf {\sigma } _{\mathrm {n} }\mathbf {n} } where λ {\displaystyle \lambda } is a constant of proportionality, and in this particular case corresponds to the magnitudes σ n {\displaystyle \sigma _{\mathrm {n} }} of the normal stress vectors or principal stresses.
Knowing that T i ( n ) = σ i j n j {\displaystyle T_{i}^{(n)}=\sigma _{ij}n_{j}} and n i = δ i j n j {\displaystyle n_{i}=\delta _{ij}n_{j}} , we have
T i ( n ) = λ n i σ i j n j = λ n i σ i j n j − λ n i = 0 ( σ i j − λ δ i j ) n j = 0 {\displaystyle {\begin{aligned}T_{i}^{(n)}&=\lambda n_{i}\\\sigma _{ij}n_{j}&=\lambda n_{i}\\\sigma _{ij}n_{j}-\lambda n_{i}&=0\\\left(\sigma _{ij}-\lambda \delta _{ij}\right)n_{j}&=0\\\end{aligned}}} This is a homogeneous system , i.e. equal to zero, of three linear equations where n j {\displaystyle n_{j}} are the unknowns. To obtain a nontrivial (non-zero) solution for n j {\displaystyle n_{j}} , the determinant matrix of the coefficients must be equal to zero, i.e. the system is singular. Thus,
| σ i j − λ δ i j | = | σ 11 − λ σ 12 σ 13 σ 21 σ 22 − λ σ 23 σ 31 σ 32 σ 33 − λ | = 0 {\displaystyle \left|\sigma _{ij}-\lambda \delta _{ij}\right|={\begin{vmatrix}\sigma _{11}-\lambda &\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}-\lambda &\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}-\lambda \\\end{vmatrix}}=0} Expanding the determinant leads to the characteristic equation
| σ i j − λ δ i j | = − λ 3 + I 1 λ 2 − I 2 λ + I 3 = 0 {\displaystyle \left|\sigma _{ij}-\lambda \delta _{ij}\right|=-\lambda ^{3}+I_{1}\lambda ^{2}-I_{2}\lambda +I_{3}=0} where
I 1 = σ 11 + σ 22 + σ 33 = σ k k = tr ( σ ) I 2 = | σ 22 σ 23 σ 32 σ 33 | + | σ 11 σ 13 σ 31 σ 33 | + | σ 11 σ 12 σ 21 σ 22 | = σ 11 σ 22 + σ 22 σ 33 + σ 11 σ 33 − σ 12 2 − σ 23 2 − σ 31 2 = 1 2 ( σ i i σ j j − σ i j σ j i ) = 1 2 [ ( tr ( σ ) ) 2 − tr ( σ 2 ) ] I 3 = det ( σ i j ) = det ( σ ) = σ 11 σ 22 σ 33 + 2 σ 12 σ 23 σ 31 − σ 12 2 σ 33 − σ 23 2 σ 11 − σ 31 2 σ 22 {\displaystyle {\begin{aligned}I_{1}&=\sigma _{11}+\sigma _{22}+\sigma _{33}\\&=\sigma _{kk}={\text{tr}}({\boldsymbol {\sigma }})\\[4pt]I_{2}&={\begin{vmatrix}\sigma _{22}&\sigma _{23}\\\sigma _{32}&\sigma _{33}\\\end{vmatrix}}+{\begin{vmatrix}\sigma _{11}&\sigma _{13}\\\sigma _{31}&\sigma _{33}\\\end{vmatrix}}+{\begin{vmatrix}\sigma _{11}&\sigma _{12}\\\sigma _{21}&\sigma _{22}\\\end{vmatrix}}\\&=\sigma _{11}\sigma _{22}+\sigma _{22}\sigma _{33}+\sigma _{11}\sigma _{33}-\sigma _{12}^{2}-\sigma _{23}^{2}-\sigma _{31}^{2}\\&={\frac {1}{2}}\left(\sigma _{ii}\sigma _{jj}-\sigma _{ij}\sigma _{ji}\right)={\frac {1}{2}}\left[\left({\text{tr}}({\boldsymbol {\sigma }})\right)^{2}-{\text{tr}}\left({\boldsymbol {\sigma }}^{2}\right)\right]\\[4pt]I_{3}&=\det(\sigma _{ij})=\det({\boldsymbol {\sigma }})\\&=\sigma _{11}\sigma _{22}\sigma _{33}+2\sigma _{12}\sigma _{23}\sigma _{31}-\sigma _{12}^{2}\sigma _{33}-\sigma _{23}^{2}\sigma _{11}-\sigma _{31}^{2}\sigma _{22}\\\end{aligned}}} The characteristic equation has three real roots λ i {\displaystyle \lambda _{i}} , i.e. not imaginary due to the symmetry of the stress tensor. The σ 1 = max ( λ 1 , λ 2 , λ 3 ) {\displaystyle \sigma _{1}=\max \left(\lambda _{1},\lambda _{2},\lambda _{3}\right)} , σ 3 = min ( λ 1 , λ 2 , λ 3 ) {\displaystyle \sigma _{3}=\min \left(\lambda _{1},\lambda _{2},\lambda _{3}\right)} and σ 2 = I 1 − σ 1 − σ 3 {\displaystyle \sigma _{2}=I_{1}-\sigma _{1}-\sigma _{3}} , are the principal stresses, functions of the eigenvalues λ i {\displaystyle \lambda _{i}} . The eigenvalues are the roots of the characteristic polynomial . The principal stresses are unique for a given stress tensor. Therefore, from the characteristic equation, the coefficients I 1 {\displaystyle I_{1}} , I 2 {\displaystyle I_{2}} and I 3 {\displaystyle I_{3}} , called the first, second, and third stress invariants , respectively, always have the same value regardless of the coordinate system's orientation.
For each eigenvalue, there is a non-trivial solution for n j {\displaystyle n_{j}} in the equation ( σ i j − λ δ i j ) n j = 0 {\displaystyle \left(\sigma _{ij}-\lambda \delta _{ij}\right)n_{j}=0} . These solutions are the principal directions or eigenvectors defining the plane where the principal stresses act. The principal stresses and principal directions characterize the stress at a point and are independent of the orientation.
A coordinate system with axes oriented to the principal directions implies that the normal stresses are the principal stresses and the stress tensor is represented by a diagonal matrix:
σ i j = [ σ 1 0 0 0 σ 2 0 0 0 σ 3 ] {\displaystyle \sigma _{ij}={\begin{bmatrix}\sigma _{1}&0&0\\0&\sigma _{2}&0\\0&0&\sigma _{3}\end{bmatrix}}} The principal stresses can be combined to form the stress invariants, I 1 {\displaystyle I_{1}} , I 2 {\displaystyle I_{2}} , and I 3 {\displaystyle I_{3}} . The first and third invariant are the trace and determinant respectively, of the stress tensor. Thus,
I 1 = σ 1 + σ 2 + σ 3 I 2 = σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 I 3 = σ 1 σ 2 σ 3 {\displaystyle {\begin{aligned}I_{1}&=\sigma _{1}+\sigma _{2}+\sigma _{3}\\I_{2}&=\sigma _{1}\sigma _{2}+\sigma _{2}\sigma _{3}+\sigma _{3}\sigma _{1}\\I_{3}&=\sigma _{1}\sigma _{2}\sigma _{3}\\\end{aligned}}} Because of its simplicity, the principal coordinate system is often useful when considering the state of the elastic medium at a particular point. Principal stresses are often expressed in the following equation for evaluating stresses in the x and y directions or axial and bending stresses on a part.[14] : p.58–59 The principal normal stresses can then be used to calculate the von Mises stress and ultimately the safety factor and margin of safety.
σ 1 , σ 2 = σ x + σ y 2 ± ( σ x − σ y 2 ) 2 + τ x y 2 {\displaystyle \sigma _{1},\sigma _{2}={\frac {\sigma _{x}+\sigma _{y}}{2}}\pm {\sqrt {\left({\frac {\sigma _{x}-\sigma _{y}}{2}}\right)^{2}+\tau _{xy}^{2}}}} Using just the part of the equation under the square root is equal to the maximum and minimum shear stress for plus and minus. This is shown as:
τ max , τ min = ± ( σ x − σ y 2 ) 2 + τ x y 2 {\displaystyle \tau _{\max },\tau _{\min }=\pm {\sqrt {\left({\frac {\sigma _{x}-\sigma _{y}}{2}}\right)^{2}+\tau _{xy}^{2}}}} Maximum and minimum shear stresses edit The maximum shear stress or maximum principal shear stress is equal to one-half the difference between the largest and smallest principal stresses, and acts on the plane that bisects the angle between the directions of the largest and smallest principal stresses, i.e. the plane of the maximum shear stress is oriented 45 ∘ {\displaystyle 45^{\circ }} from the principal stress planes. The maximum shear stress is expressed as
τ max = 1 2 | σ max − σ min | {\displaystyle \tau _{\max }={\frac {1}{2}}\left|\sigma _{\max }-\sigma _{\min }\right|} Assuming σ 1 ≥ σ 2 ≥ σ 3 {\displaystyle \sigma _{1}\geq \sigma _{2}\geq \sigma _{3}} then
τ max = 1 2 | σ 1 − σ 3 | {\displaystyle \tau _{\max }={\frac {1}{2}}\left|\sigma _{1}-\sigma _{3}\right|} When the stress tensor is non zero the normal stress component acting on the plane for the maximum shear stress is non-zero and it is equal to
σ n = 1 2 ( σ 1 + σ 3 ) {\displaystyle \sigma _{\text{n}}={\frac {1}{2}}\left(\sigma _{1}+\sigma _{3}\right)} Derivation of the maximum and minimum shear stresses[8] : p.45–78 [11] : p.1–46 [13] [15] : p.111–157 [16] : p.9–41 [17] : p.33–66 [18] : p.43–61 The normal stress can be written in terms of principal stresses ( σ 1 ≥ σ 2 ≥ σ 3 ) {\displaystyle (\sigma _{1}\geq \sigma _{2}\geq \sigma _{3})} as σ n = σ i j n i n j = σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 {\displaystyle {\begin{aligned}\sigma _{\mathrm {n} }&=\sigma _{ij}n_{i}n_{j}\\&=\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\\\end{aligned}}} Knowing that ( T ( n ) ) 2 = σ i j σ i k n j n k {\displaystyle \left(T^{(n)}\right)^{2}=\sigma _{ij}\sigma _{ik}n_{j}n_{k}} , the shear stress in terms of principal stresses components is expressed as
τ n 2 = ( T ( n ) ) 2 − σ n 2 = σ 1 2 n 1 2 + σ 2 2 n 2 2 + σ 3 2 n 3 2 − ( σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 ) 2 = ( σ 1 2 − σ 2 2 ) n 1 2 + ( σ 2 2 − σ 3 2 ) n 2 2 + σ 3 2 − [ ( σ 1 − σ 3 ) n 1 2 + ( σ 2 − σ 3 ) n 2 2 + σ 3 ] 2 = ( σ 1 − σ 2 ) 2 n 1 2 n 2 2 + ( σ 2 − σ 3 ) 2 n 2 2 n 3 2 + ( σ 1 − σ 3 ) 2 n 1 2 n 3 2 {\displaystyle {\begin{aligned}\tau _{\text{n}}^{2}&=\left(T^{(n)}\right)^{2}-\sigma _{\text{n}}^{2}\\&=\sigma _{1}^{2}n_{1}^{2}+\sigma _{2}^{2}n_{2}^{2}+\sigma _{3}^{2}n_{3}^{2}-\left(\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\right)^{2}\\&=\left(\sigma _{1}^{2}-\sigma _{2}^{2}\right)n_{1}^{2}+\left(\sigma _{2}^{2}-\sigma _{3}^{2}\right)n_{2}^{2}+\sigma _{3}^{2}-\left[\left(\sigma _{1}-\sigma _{3}\right)n_{1}^{2}+\left(\sigma _{2}-\sigma _{3}\right)n_{2}^{2}+\sigma _{3}\right]^{2}\\&=(\sigma _{1}-\sigma _{2})^{2}n_{1}^{2}n_{2}^{2}+(\sigma _{2}-\sigma _{3})^{2}n_{2}^{2}n_{3}^{2}+(\sigma _{1}-\sigma _{3})^{2}n_{1}^{2}n_{3}^{2}\\\end{aligned}}} The maximum shear stress at a point in a continuum body is determined by maximizing τ n 2 {\displaystyle \tau _{\mathrm {n} }^{2}} subject to the condition that
n i n i = n 1 2 + n 2 2 + n 3 2 = 1 {\displaystyle n_{i}n_{i}=n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1} This is a constrained maximization problem, which can be solved using the Lagrangian multiplier technique to convert the problem into an unconstrained optimization problem. Thus, the stationary values (maximum and minimum values)of τ n 2 {\displaystyle \tau _{\text{n}}^{2}} occur where the gradient of τ n 2 {\displaystyle \tau _{\text{n}}^{2}} is parallel to the gradient of F {\displaystyle F} .
The Lagrangian function for this problem can be written as
F ( n 1 , n 2 , n 3 , λ ) = τ 2 + λ ( g ( n 1 , n 2 , n 3 ) − 1 ) = σ 1 2 n 1 2 + σ 2 2 n 2 2 + σ 3 2 n 3 2 − ( σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 ) 2 + λ ( n 1 2 + n 2 2 + n 3 2 − 1 ) {\displaystyle {\begin{aligned}F\left(n_{1},n_{2},n_{3},\lambda \right)&=\tau ^{2}+\lambda \left(g\left(n_{1},n_{2},n_{3}\right)-1\right)\\&=\sigma _{1}^{2}n_{1}^{2}+\sigma _{2}^{2}n_{2}^{2}+\sigma _{3}^{2}n_{3}^{2}-\left(\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\right)^{2}+\lambda \left(n_{1}^{2}+n_{2}^{2}+n_{3}^{2}-1\right)\\\end{aligned}}} where λ {\displaystyle \lambda } is the Lagrangian multiplier (which is different from the λ {\displaystyle \lambda } use to denote eigenvalues).
The extreme values of these functions are
∂ F ∂ n 1 = 0 ∂ F ∂ n 2 = 0 ∂ F ∂ n 3 = 0 {\displaystyle {\frac {\partial F}{\partial n_{1}}}=0\qquad {\frac {\partial F}{\partial n_{2}}}=0\qquad {\frac {\partial F}{\partial n_{3}}}=0} thence
∂ F ∂ n 1 = n 1 σ 1 2 − 2 n 1 σ 1 ( σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 ) + λ n 1 = 0 {\displaystyle {\frac {\partial F}{\partial n_{1}}}=n_{1}\sigma _{1}^{2}-2n_{1}\sigma _{1}\left(\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\right)+\lambda n_{1}=0} ∂ F ∂ n 2 = n 2 σ 2 2 − 2 n 2 σ 2 ( σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 ) + λ n 2 = 0 {\displaystyle {\frac {\partial F}{\partial n_{2}}}=n_{2}\sigma _{2}^{2}-2n_{2}\sigma _{2}\left(\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\right)+\lambda n_{2}=0} ∂ F ∂ n 3 = n 3 σ 3 2 − 2 n 3 σ 3 ( σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 ) + λ n 3 = 0 {\displaystyle {\frac {\partial F}{\partial n_{3}}}=n_{3}\sigma _{3}^{2}-2n_{3}\sigma _{3}\left(\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\right)+\lambda n_{3}=0} These three equations together with the condition n i n i = 1 {\displaystyle n_{i}n_{i}=1} may be solved for λ , n 1 , n 2 , {\displaystyle \lambda ,n_{1},n_{2},} and n 3 {\displaystyle n_{3}}
By multiplying the first three equations by n 1 , n 2 , {\displaystyle n_{1},\,n_{2},} and n 3 {\displaystyle n_{3}} , respectively, and knowing that σ n = σ i j n i n j = σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 {\displaystyle \sigma _{\text{n}}=\sigma _{ij}n_{i}n_{j}=\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}} we obtain
n 1 2 σ 1 2 − 2 σ 1 n 1 2 σ n + n 1 2 λ = 0 {\displaystyle n_{1}^{2}\sigma _{1}^{2}-2\sigma _{1}n_{1}^{2}\sigma _{\text{n}}+n_{1}^{2}\lambda =0} n 2 2 σ 2 2 − 2 σ 2 n 2 2 σ n + n 2 2 λ = 0 {\displaystyle n_{2}^{2}\sigma _{2}^{2}-2\sigma _{2}n_{2}^{2}\sigma _{\text{n}}+n_{2}^{2}\lambda =0} n 3 2 σ 3 2 − 2 σ 1 n 3 2 σ n + n 3 2 λ = 0 {\displaystyle n_{3}^{2}\sigma _{3}^{2}-2\sigma _{1}n_{3}^{2}\sigma _{\text{n}}+n_{3}^{2}\lambda =0} Adding these three equations we get
[ n 1 2 σ 1 2 + n 2 2 σ 2 2 + n 3 2 σ 3 2 ] − 2 ( σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 ) σ n + λ ( n 1 2 + n 2 2 + n 3 2 ) = 0 [ τ n 2 + ( σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 ) 2 ] − 2 σ n 2 + λ = 0 [ τ n 2 + σ n 2 ] − 2 σ n 2 + λ = 0 λ = σ n 2 − τ n 2 {\displaystyle {\begin{aligned}\left[n_{1}^{2}\sigma _{1}^{2}+n_{2}^{2}\sigma _{2}^{2}+n_{3}^{2}\sigma _{3}^{2}\right]-2\left(\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\right)\sigma _{\mathrm {n} }+\lambda \left(n_{1}^{2}+n_{2}^{2}+n_{3}^{2}\right)&=0\\\left[\tau _{\mathrm {n} }^{2}+\left(\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\right)^{2}\right]-2\sigma _{\mathrm {n} }^{2}+\lambda &=0\\\left[\tau _{\mathrm {n} }^{2}+\sigma _{\mathrm {n} }^{2}\right]-2\sigma _{\mathrm {n} }^{2}+\lambda &=0\\\lambda &=\sigma _{\mathrm {n} }^{2}-\tau _{\mathrm {n} }^{2}\end{aligned}}} this result can be substituted into each of the first three equations to obtain
∂ F ∂ n 1 = n 1 σ 1 2 − 2 n 1 σ 1 ( σ 1 n 1 2 + σ 2 n 2 2 + σ 3 n 3 2 ) + ( σ n 2 − τ n 2 ) n 1 = 0 n 1 σ 1 2 − 2 n 1 σ 1 σ n + ( σ n 2 − τ n 2 ) n 1 = 0 ( σ 1 2 − 2 σ 1 σ n + σ n 2 − τ n 2 ) n 1 = 0 {\displaystyle {\begin{aligned}{\frac {\partial F}{\partial n_{1}}}=n_{1}\sigma _{1}^{2}-2n_{1}\sigma _{1}\left(\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}\right)+\left(\sigma _{\text{n}}^{2}-\tau _{\text{n}}^{2}\right)n_{1}&=0\\n_{1}\sigma _{1}^{2}-2n_{1}\sigma _{1}\sigma _{\text{n}}+\left(\sigma _{\text{n}}^{2}-\tau _{\text{n}}^{2}\right)n_{1}&=0\\\left(\sigma _{1}^{2}-2\sigma _{1}\sigma _{\text{n}}+\sigma _{\text{n}}^{2}-\tau _{\text{n}}^{2}\right)n_{1}&=0\\\end{aligned}}} Doing the same for the other two equations we have
∂ F ∂ n 2 = ( σ 2 2 − 2 σ 2 σ n + σ n 2 − τ n 2 ) n 2 = 0 {\displaystyle {\frac {\partial F}{\partial n_{2}}}=\left(\sigma _{2}^{2}-2\sigma _{2}\sigma _{\text{n}}+\sigma _{\text{n}}^{2}-\tau _{\text{n}}^{2}\right)n_{2}=0} ∂ F ∂ n 3 = ( σ 3 2 − 2 σ 3 σ n + σ n 2 − τ n 2 ) n 3 = 0 {\displaystyle {\frac {\partial F}{\partial n_{3}}}=\left(\sigma _{3}^{2}-2\sigma _{3}\sigma _{\text{n}}+\sigma _{\text{n}}^{2}-\tau _{\text{n}}^{2}\right)n_{3}=0} A first approach to solve these last three equations is to consider the trivial solution n i = 0 {\displaystyle n_{i}=0} . However, this option does not fulfill the constraint
cauchy, stress, tensor, continuum, mechanics, symbol, displaystyle, boldsymbol, sigma, named, after, augustin, louis, cauchy, also, called, true, stress, tensor, simply, stress, tensor, completely, defines, state, stress, point, inside, material, deformed, sta. In continuum mechanics the Cauchy stress tensor symbol s displaystyle boldsymbol sigma named after Augustin Louis Cauchy also called true stress tensor 1 or simply stress tensor completely defines the state of stress at a point inside a material in the deformed state placement or configuration The second order tensor consists of nine components s i j displaystyle sigma ij and relates a unit length direction vector e to the traction vector T e across an imaginary surface perpendicular to e Cauchy stress tensorComponents of stress in three dimensionsCommon symbolssSI unitpascal Pa Other unitsPound per square inch psi barIn SI base unitsPa kg m 1 s 2Behaviour undercoord transformationtensorDimensionL 1 M T 2 displaystyle mathsf L 1 mathsf M mathsf T 2 T e e s or T j e i s i j e i displaystyle mathbf T mathbf e mathbf e cdot boldsymbol sigma quad text or quad T j e sum i sigma ij e i a The SI base units of both stress tensor and traction vector are newton per square metre N m2 or pascal Pa corresponding to the stress scalar The unit vector is dimensionless The Cauchy stress tensor obeys the tensor transformation law under a change in the system of coordinates A graphical representation of this transformation law is the Mohr s circle for stress The Cauchy stress tensor is used for stress analysis of material bodies experiencing small deformations it is a central concept in the linear theory of elasticity For large deformations also called finite deformations other measures of stress are required such as the Piola Kirchhoff stress tensor the Biot stress tensor and the Kirchhoff stress tensor According to the principle of conservation of linear momentum if the continuum body is in static equilibrium it can be demonstrated that the components of the Cauchy stress tensor in every material point in the body satisfy the equilibrium equations Cauchy s equations of motion for zero acceleration At the same time according to the principle of conservation of angular momentum equilibrium requires that the summation of moments with respect to an arbitrary point is zero which leads to the conclusion that the stress tensor is symmetric thus having only six independent stress components instead of the original nine However in the presence of couple stresses i e moments per unit volume the stress tensor is non symmetric This also is the case when the Knudsen number is close to one K n 1 displaystyle K n rightarrow 1 or the continuum is a non Newtonian fluid which can lead to rotationally non invariant fluids such as polymers There are certain invariants associated with the stress tensor whose values do not depend upon the coordinate system chosen or the area element upon which the stress tensor operates These are the three eigenvalues of the stress tensor which are called the principal stresses Contents 1 Euler Cauchy stress principle stress vector 1 1 Cauchy s postulate 1 2 Cauchy s fundamental lemma 2 Cauchy s stress theorem stress tensor 2 1 Transformation rule of the stress tensor 2 2 Normal and shear stresses 3 Balance laws Cauchy s equations of motion 3 1 Cauchy s first law of motion 3 2 Cauchy s second law of motion 4 Principal stresses and stress invariants 5 Maximum and minimum shear stresses 6 Stress deviator tensor 6 1 Invariants of the stress deviator tensor 7 Octahedral stresses 8 See also 9 Notes 10 ReferencesEuler Cauchy stress principle stress vector editMain article Continuum mechanics nbsp Figure 2 1a Internal distribution of contact forces and couple stresses on a differential d S displaystyle dS nbsp of the internal surface S displaystyle S nbsp in a continuum as a result of the interaction between the two portions of the continuum separated by the surface nbsp Figure 2 1b Internal distribution of contact forces and couple stresses on a differential d S displaystyle dS nbsp of the internal surface S displaystyle S nbsp in a continuum as a result of the interaction between the two portions of the continuum separated by the surface nbsp Figure 2 1c Stress vector on an internal surface S with normal vector n Depending on the orientation of the plane under consideration the stress vector may not necessarily be perpendicular to that plane i e parallel to n displaystyle mathbf n nbsp and can be resolved into two components one component normal to the plane called normal stress s n displaystyle sigma mathrm n nbsp and another component parallel to this plane called the shearing stress t displaystyle tau nbsp The Euler Cauchy stress principle states that upon any surface real or imaginary that divides the body the action of one part of the body on the other is equivalent equipollent to the system of distributed forces and couples on the surface dividing the body 2 and it is represented by a field T n displaystyle mathbf T mathbf n nbsp called the traction vector defined on the surface S displaystyle S nbsp and assumed to depend continuously on the surface s unit vector n displaystyle mathbf n nbsp 3 4 p 66 96 To formulate the Euler Cauchy stress principle consider an imaginary surface S displaystyle S nbsp passing through an internal material point P displaystyle P nbsp dividing the continuous body into two segments as seen in Figure 2 1a or 2 1b one may use either the cutting plane diagram or the diagram with the arbitrary volume inside the continuum enclosed by the surface S displaystyle S nbsp Following the classical dynamics of Newton and Euler the motion of a material body is produced by the action of externally applied forces which are assumed to be of two kinds surface forces F displaystyle mathbf F nbsp and body forces b displaystyle mathbf b nbsp 5 Thus the total force F displaystyle mathcal F nbsp applied to a body or to a portion of the body can be expressed as F b F displaystyle mathcal F mathbf b mathbf F nbsp Only surface forces will be discussed in this article as they are relevant to the Cauchy stress tensor When the body is subjected to external surface forces or contact forces F displaystyle mathbf F nbsp following Euler s equations of motion internal contact forces and moments are transmitted from point to point in the body and from one segment to the other through the dividing surface S displaystyle S nbsp due to the mechanical contact of one portion of the continuum onto the other Figure 2 1a and 2 1b On an element of area D S displaystyle Delta S nbsp containing P displaystyle P nbsp with normal vector n displaystyle mathbf n nbsp the force distribution is equipollent to a contact force D F displaystyle Delta mathbf F nbsp exerted at point P and surface moment D M displaystyle Delta mathbf M nbsp In particular the contact force is given by D F T n D S displaystyle Delta mathbf F mathbf T mathbf n Delta S nbsp where T n displaystyle mathbf T mathbf n nbsp is the mean surface traction Cauchy s stress principle asserts 6 p 47 102 that as D S displaystyle Delta S nbsp becomes very small and tends to zero the ratio D F D S displaystyle Delta mathbf F Delta S nbsp becomes d F d S displaystyle d mathbf F dS nbsp and the couple stress vector D M displaystyle Delta mathbf M nbsp vanishes In specific fields of continuum mechanics the couple stress is assumed not to vanish however classical branches of continuum mechanics address non polar materials which do not consider couple stresses and body moments The resultant vector d F d S displaystyle d mathbf F dS nbsp is defined as the surface traction 7 also called stress vector 8 traction 4 or traction vector 6 given by T n T i n e i displaystyle mathbf T mathbf n T i mathbf n mathbf e i nbsp at the point P displaystyle P nbsp associated with a plane with a normal vector n displaystyle mathbf n nbsp T i n lim D S 0 D F i D S d F i d S displaystyle T i mathbf n lim Delta S to 0 frac Delta F i Delta S dF i over dS nbsp This equation means that the stress vector depends on its location in the body and the orientation of the plane on which it is acting This implies that the balancing action of internal contact forces generates a contact force density or Cauchy traction field 5 T n x t displaystyle mathbf T mathbf n mathbf x t nbsp that represents a distribution of internal contact forces throughout the volume of the body in a particular configuration of the body at a given time t displaystyle t nbsp It is not a vector field because it depends not only on the position x displaystyle mathbf x nbsp of a particular material point but also on the local orientation of the surface element as defined by its normal vector n displaystyle mathbf n nbsp 9 Depending on the orientation of the plane under consideration the stress vector may not necessarily be perpendicular to that plane i e parallel to n displaystyle mathbf n nbsp and can be resolved into two components Figure 2 1c one normal to the plane called normal stresss n lim D S 0 D F n D S d F n d S displaystyle mathbf sigma mathrm n lim Delta S to 0 frac Delta F mathrm n Delta S frac dF mathrm n dS nbsp where d F n displaystyle dF mathrm n nbsp is the normal component of the force d F displaystyle d mathbf F nbsp to the differential area d S displaystyle dS nbsp and the other parallel to this plane called the shear stresst lim D S 0 D F s D S d F s d S displaystyle mathbf tau lim Delta S to 0 frac Delta F mathrm s Delta S frac dF mathrm s dS nbsp where d F s displaystyle dF mathrm s nbsp is the tangential component of the force d F displaystyle d mathbf F nbsp to the differential surface area d S displaystyle dS nbsp The shear stress can be further decomposed into two mutually perpendicular vectors Cauchy s postulate edit According to the Cauchy Postulate the stress vector T n displaystyle mathbf T mathbf n nbsp remains unchanged for all surfaces passing through the point P displaystyle P nbsp and having the same normal vector n displaystyle mathbf n nbsp at P displaystyle P nbsp 7 10 i e having a common tangent at P displaystyle P nbsp This means that the stress vector is a function of the normal vector n displaystyle mathbf n nbsp only and is not influenced by the curvature of the internal surfaces Cauchy s fundamental lemma edit A consequence of Cauchy s postulate is Cauchy s Fundamental Lemma 1 7 11 also called the Cauchy reciprocal theorem 12 p 103 130 which states that the stress vectors acting on opposite sides of the same surface are equal in magnitude and opposite in direction Cauchy s fundamental lemma is equivalent to Newton s third law of motion of action and reaction and is expressed as T n T n displaystyle mathbf T mathbf n mathbf T mathbf n nbsp Cauchy s stress theorem stress tensor editThe state of stress at a point in the body is then defined by all the stress vectors T n associated with all planes infinite in number that pass through that point 13 However according to Cauchy s fundamental theorem 11 also called Cauchy s stress theorem 1 merely by knowing the stress vectors on three mutually perpendicular planes the stress vector on any other plane passing through that point can be found through coordinate transformation equations Cauchy s stress theorem states that there exists a second order tensor field s x t called the Cauchy stress tensor independent of n such that T is a linear function of n T n n s or T j n s i j n i displaystyle mathbf T mathbf n mathbf n cdot boldsymbol sigma quad text or quad T j n sigma ij n i nbsp This equation implies that the stress vector T n at any point P in a continuum associated with a plane with normal unit vector n can be expressed as a function of the stress vectors on the planes perpendicular to the coordinate axes i e in terms of the components sij of the stress tensor s To prove this expression consider a tetrahedron with three faces oriented in the coordinate planes and with an infinitesimal area dA oriented in an arbitrary direction specified by a normal unit vector n Figure 2 2 The tetrahedron is formed by slicing the infinitesimal element along an arbitrary plane with unit normal n The stress vector on this plane is denoted by T n The stress vectors acting on the faces of the tetrahedron are denoted as T e1 T e2 and T e3 and are by definition the components sij of the stress tensor s This tetrahedron is sometimes called the Cauchy tetrahedron The equilibrium of forces i e Euler s first law of motion Newton s second law of motion gives T n d A T e 1 d A 1 T e 2 d A 2 T e 3 d A 3 r h 3 d A a displaystyle mathbf T mathbf n dA mathbf T mathbf e 1 dA 1 mathbf T mathbf e 2 dA 2 mathbf T mathbf e 3 dA 3 rho left frac h 3 dA right mathbf a nbsp nbsp Figure 2 2 Stress vector acting on a plane with normal unit vector n A note on the sign convention The tetrahedron is formed by slicing a parallelepiped along an arbitrary plane n So the force acting on the plane n is the reaction exerted by the other half of the parallelepiped and has an opposite sign where the right hand side represents the product of the mass enclosed by the tetrahedron and its acceleration r is the density a is the acceleration and h is the height of the tetrahedron considering the plane n as the base The area of the faces of the tetrahedron perpendicular to the axes can be found by projecting dA into each face using the dot product d A 1 n e 1 d A n 1 d A displaystyle dA 1 left mathbf n cdot mathbf e 1 right dA n 1 dA nbsp d A 2 n e 2 d A n 2 d A displaystyle dA 2 left mathbf n cdot mathbf e 2 right dA n 2 dA nbsp d A 3 n e 3 d A n 3 d A displaystyle dA 3 left mathbf n cdot mathbf e 3 right dA n 3 dA nbsp and then substituting into the equation to cancel out dA T n T e 1 n 1 T e 2 n 2 T e 3 n 3 r h 3 a displaystyle mathbf T mathbf n mathbf T mathbf e 1 n 1 mathbf T mathbf e 2 n 2 mathbf T mathbf e 3 n 3 rho left frac h 3 right mathbf a nbsp To consider the limiting case as the tetrahedron shrinks to a point h must go to 0 intuitively the plane n is translated along n toward O As a result the right hand side of the equation approaches 0 so T n T e 1 n 1 T e 2 n 2 T e 3 n 3 displaystyle mathbf T mathbf n mathbf T mathbf e 1 n 1 mathbf T mathbf e 2 n 2 mathbf T mathbf e 3 n 3 nbsp Assuming a material element Figure 2 3 with planes perpendicular to the coordinate axes of a Cartesian coordinate system the stress vectors associated with each of the element planes i e T e1 T e2 and T e3 can be decomposed into a normal component and two shear components i e components in the direction of the three coordinate axes For the particular case of a surface with normal unit vector oriented in the direction of the x1 axis denote the normal stress by s11 and the two shear stresses as s12 and s13 T e 1 T 1 e 1 e 1 T 2 e 1 e 2 T 3 e 1 e 3 s 11 e 1 s 12 e 2 s 13 e 3 displaystyle mathbf T mathbf e 1 T 1 mathbf e 1 mathbf e 1 T 2 mathbf e 1 mathbf e 2 T 3 mathbf e 1 mathbf e 3 sigma 11 mathbf e 1 sigma 12 mathbf e 2 sigma 13 mathbf e 3 nbsp T e 2 T 1 e 2 e 1 T 2 e 2 e 2 T 3 e 2 e 3 s 21 e 1 s 22 e 2 s 23 e 3 displaystyle mathbf T mathbf e 2 T 1 mathbf e 2 mathbf e 1 T 2 mathbf e 2 mathbf e 2 T 3 mathbf e 2 mathbf e 3 sigma 21 mathbf e 1 sigma 22 mathbf e 2 sigma 23 mathbf e 3 nbsp T e 3 T 1 e 3 e 1 T 2 e 3 e 2 T 3 e 3 e 3 s 31 e 1 s 32 e 2 s 33 e 3 displaystyle mathbf T mathbf e 3 T 1 mathbf e 3 mathbf e 1 T 2 mathbf e 3 mathbf e 2 T 3 mathbf e 3 mathbf e 3 sigma 31 mathbf e 1 sigma 32 mathbf e 2 sigma 33 mathbf e 3 nbsp In index notation this is T e i T j e i e j s i j e j displaystyle mathbf T mathbf e i T j mathbf e i mathbf e j sigma ij mathbf e j nbsp The nine components sij of the stress vectors are the components of a second order Cartesian tensor called the Cauchy stress tensor which can be used to completely define the state of stress at a point and is given by s s i j T e 1 T e 2 T e 3 s 11 s 12 s 13 s 21 s 22 s 23 s 31 s 32 s 33 s x x s x y s x z s y x s y y s y z s z x s z y s z z s x t x y t x z t y x s y t y z t z x t z y s z displaystyle boldsymbol sigma sigma ij left begin matrix mathbf T mathbf e 1 mathbf T mathbf e 2 mathbf T mathbf e 3 end matrix right left begin matrix sigma 11 amp sigma 12 amp sigma 13 sigma 21 amp sigma 22 amp sigma 23 sigma 31 amp sigma 32 amp sigma 33 end matrix right equiv left begin matrix sigma xx amp sigma xy amp sigma xz sigma yx amp sigma yy amp sigma yz sigma zx amp sigma zy amp sigma zz end matrix right equiv left begin matrix sigma x amp tau xy amp tau xz tau yx amp sigma y amp tau yz tau zx amp tau zy amp sigma z end matrix right nbsp where s11 s22 and s33 are normal stresses and s12 s13 s21 s23 s31 and s32 are shear stresses The first index i indicates that the stress acts on a plane normal to the Xi axis and the second index j denotes the direction in which the stress acts For example s12 implies that the stress is acting on the plane that is normal to the 1st axis i e X1 and acts along the 2nd axis i e X2 A stress component is positive if it acts in the positive direction of the coordinate axes and if the plane where it acts has an outward normal vector pointing in the positive coordinate direction Thus using the components of the stress tensor T n T e 1 n 1 T e 2 n 2 T e 3 n 3 i 1 3 T e i n i s i j e j n i s i j n i e j displaystyle begin aligned mathbf T mathbf n amp mathbf T mathbf e 1 n 1 mathbf T mathbf e 2 n 2 mathbf T mathbf e 3 n 3 amp sum i 1 3 mathbf T mathbf e i n i amp left sigma ij mathbf e j right n i amp sigma ij n i mathbf e j end aligned nbsp or equivalently T j n s i j n i displaystyle T j mathbf n sigma ij n i nbsp Alternatively in matrix form we have T 1 n T 2 n T 3 n n 1 n 2 n 3 s 11 s 12 s 13 s 21 s 22 s 23 s 31 s 32 s 33 displaystyle left begin matrix T 1 mathbf n amp T 2 mathbf n amp T 3 mathbf n end matrix right left begin matrix n 1 amp n 2 amp n 3 end matrix right cdot left begin matrix sigma 11 amp sigma 12 amp sigma 13 sigma 21 amp sigma 22 amp sigma 23 sigma 31 amp sigma 32 amp sigma 33 end matrix right nbsp The Voigt notation representation of the Cauchy stress tensor takes advantage of the symmetry of the stress tensor to express the stress as a six dimensional vector of the form s s 1 s 2 s 3 s 4 s 5 s 6 T s 11 s 22 s 33 s 23 s 13 s 12 T displaystyle boldsymbol sigma begin bmatrix sigma 1 amp sigma 2 amp sigma 3 amp sigma 4 amp sigma 5 amp sigma 6 end bmatrix textsf T equiv begin bmatrix sigma 11 amp sigma 22 amp sigma 33 amp sigma 23 amp sigma 13 amp sigma 12 end bmatrix textsf T nbsp The Voigt notation is used extensively in representing stress strain relations in solid mechanics and for computational efficiency in numerical structural mechanics software Transformation rule of the stress tensor edit It can be shown that the stress tensor is a contravariant second order tensor which is a statement of how it transforms under a change of the coordinate system From an xi system to an xi system the components sij in the initial system are transformed into the components sij in the new system according to the tensor transformation rule Figure 2 4 s i j a i m a j n s m n or s A s A T displaystyle sigma ij a im a jn sigma mn quad text or quad boldsymbol sigma mathbf A boldsymbol sigma mathbf A textsf T nbsp where A is a rotation matrix with components aij In matrix form this is s 11 s 12 s 13 s 21 s 22 s 23 s 31 s 32 s 33 a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 s 11 s 12 s 13 s 21 s 22 s 23 s 31 s 32 s 33 a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 displaystyle left begin matrix sigma 11 amp sigma 12 amp sigma 13 sigma 21 amp sigma 22 amp sigma 23 sigma 31 amp sigma 32 amp sigma 33 end matrix right left begin matrix a 11 amp a 12 amp a 13 a 21 amp a 22 amp a 23 a 31 amp a 32 amp a 33 end matrix right left begin matrix sigma 11 amp sigma 12 amp sigma 13 sigma 21 amp sigma 22 amp sigma 23 sigma 31 amp sigma 32 amp sigma 33 end matrix right left begin matrix a 11 amp a 21 amp a 31 a 12 amp a 22 amp a 32 a 13 amp a 23 amp a 33 end matrix right nbsp nbsp Figure 2 4 Transformation of the stress tensor Expanding the matrix operation and simplifying terms using the symmetry of the stress tensor gives s 11 a 11 2 s 11 a 12 2 s 22 a 13 2 s 33 2 a 11 a 12 s 12 2 a 11 a 13 s 13 2 a 12 a 13 s 23 s 22 a 21 2 s 11 a 22 2 s 22 a 23 2 s 33 2 a 21 a 22 s 12 2 a 21 a 23 s 13 2 a 22 a 23 s 23 s 33 a 31 2 s 11 a 32 2 s 22 a 33 2 s 33 2 a 31 a 32 s 12 2 a 31 a 33 s 13 2 a 32 a 33 s 23 s 12 a 11 a 21 s 11 a 12 a 22 s 22 a 13 a 23 s 33 a 11 a 22 a 12 a 21 s 12 a 12 a 23 a 13 a 22 s 23 a 11 a 23 a 13 a 21 s 13 s 23 a 21 a 31 s 11 a 22 a 32 s 22 a 23 a 33 s 33 a 21 a 32 a 22 a 31 s 12 a 22 a 33 a 23 a 32 s 23 a 21 a 33 a 23 a 31 s 13 s 13 a 11 a 31 s 11 a 12 a 32 s 22 a 13 a 33 s 33 a 11 a 32 a 12 a 31 s 12 a 12 a 33 a 13 a 32 s 23 a 11 a 33 a 13 a 31 s 13 displaystyle begin aligned sigma 11 amp a 11 2 sigma 11 a 12 2 sigma 22 a 13 2 sigma 33 2a 11 a 12 sigma 12 2a 11 a 13 sigma 13 2a 12 a 13 sigma 23 sigma 22 amp a 21 2 sigma 11 a 22 2 sigma 22 a 23 2 sigma 33 2a 21 a 22 sigma 12 2a 21 a 23 sigma 13 2a 22 a 23 sigma 23 sigma 33 amp a 31 2 sigma 11 a 32 2 sigma 22 a 33 2 sigma 33 2a 31 a 32 sigma 12 2a 31 a 33 sigma 13 2a 32 a 33 sigma 23 sigma 12 amp a 11 a 21 sigma 11 a 12 a 22 sigma 22 a 13 a 23 sigma 33 amp a 11 a 22 a 12 a 21 sigma 12 a 12 a 23 a 13 a 22 sigma 23 a 11 a 23 a 13 a 21 sigma 13 sigma 23 amp a 21 a 31 sigma 11 a 22 a 32 sigma 22 a 23 a 33 sigma 33 amp a 21 a 32 a 22 a 31 sigma 12 a 22 a 33 a 23 a 32 sigma 23 a 21 a 33 a 23 a 31 sigma 13 sigma 13 amp a 11 a 31 sigma 11 a 12 a 32 sigma 22 a 13 a 33 sigma 33 amp a 11 a 32 a 12 a 31 sigma 12 a 12 a 33 a 13 a 32 sigma 23 a 11 a 33 a 13 a 31 sigma 13 end aligned nbsp The Mohr circle for stress is a graphical representation of this transformation of stresses Normal and shear stresses edit The magnitude of the normal stress component sn of any stress vector T n acting on an arbitrary plane with normal unit vector n at a given point in terms of the components sij of the stress tensor s is the dot product of the stress vector and the normal unit vector s n T n n T i n n i s i j n i n j displaystyle begin aligned sigma mathrm n amp mathbf T mathbf n cdot mathbf n amp T i mathbf n n i amp sigma ij n i n j end aligned nbsp The magnitude of the shear stress component tn acting orthogonal to the vector n can then be found using the Pythagorean theorem t n T n 2 s n 2 T i n T i n s n 2 displaystyle begin aligned tau mathrm n amp sqrt left T mathbf n right 2 sigma mathrm n 2 amp sqrt T i mathbf n T i mathbf n sigma mathrm n 2 end aligned nbsp where T n 2 T i n T i n s i j n j s i k n k s i j s i k n j n k displaystyle left T mathbf n right 2 T i mathbf n T i mathbf n left sigma ij n j right left sigma ik n k right sigma ij sigma ik n j n k nbsp Balance laws Cauchy s equations of motion edit nbsp Figure 4 Continuum body in equilibriumCauchy s first law of motion edit According to the principle of conservation of linear momentum if the continuum body is in static equilibrium it can be demonstrated that the components of the Cauchy stress tensor in every material point in the body satisfy the equilibrium equations s j i j F i 0 displaystyle sigma ji j F i 0 nbsp where s j i j j j s j i displaystyle sigma ji j sum j partial j sigma ji nbsp For example for a hydrostatic fluid in equilibrium conditions the stress tensor takes on the form s i j p d i j displaystyle sigma ij p delta ij nbsp where p displaystyle p nbsp is the hydrostatic pressure and d i j displaystyle delta ij nbsp is the kronecker delta Derivation of equilibrium equationsConsider a continuum body see Figure 4 occupying a volume V displaystyle V nbsp having a surface area S displaystyle S nbsp with defined traction or surface forces T i n displaystyle T i n nbsp per unit area acting on every point of the body surface and body forces F i displaystyle F i nbsp per unit of volume on every point within the volume V displaystyle V nbsp Thus if the body is in equilibrium the resultant force acting on the volume is zero thus S T i n d S V F i d V 0 displaystyle int S T i n dS int V F i dV 0 nbsp By definition the stress vector is T i n s j i n j displaystyle T i n sigma ji n j nbsp then S s j i n j d S V F i d V 0 displaystyle int S sigma ji n j dS int V F i dV 0 nbsp Using the Gauss s divergence theorem to convert a surface integral to a volume integral gives V s j i j d V V F i d V 0 displaystyle int V sigma ji j dV int V F i dV 0 nbsp V s j i j F i d V 0 displaystyle int V sigma ji j F i dV 0 nbsp For an arbitrary volume the integral vanishes and we have the equilibrium equations s j i j F i 0 displaystyle sigma ji j F i 0 nbsp Cauchy s second law of motion edit According to the principle of conservation of angular momentum equilibrium requires that the summation of moments with respect to an arbitrary point is zero which leads to the conclusion that the stress tensor is symmetric thus having only six independent stress components instead of the original nine s i j s j i displaystyle sigma ij sigma ji nbsp Derivation of symmetry of the stress tensorSumming moments about point O Figure 4 the resultant moment is zero as the body is in equilibrium Thus M O S r T d S V r F d V 0 0 S e i j k x j T k n d S V e i j k x j F k d V displaystyle begin aligned M O amp int S mathbf r times mathbf T dS int V mathbf r times mathbf F dV 0 0 amp int S varepsilon ijk x j T k n dS int V varepsilon ijk x j F k dV end aligned nbsp where r displaystyle mathbf r nbsp is the position vector and is expressed as r x j e j displaystyle mathbf r x j mathbf e j nbsp Knowing that T k n s m k n m displaystyle T k n sigma mk n m nbsp and using Gauss s divergence theorem to change from a surface integral to a volume integral we have 0 S e i j k x j s m k n m d S V e i j k x j F k d V V e i j k x j s m k m d V V e i j k x j F k d V V e i j k x j m s m k e i j k x j s m k m d V V e i j k x j F k d V V e i j k x j m s m k d V V e i j k x j s m k m F k d V displaystyle begin aligned 0 amp int S varepsilon ijk x j sigma mk n m dS int V varepsilon ijk x j F k dV amp int V varepsilon ijk x j sigma mk m dV int V varepsilon ijk x j F k dV amp int V varepsilon ijk x j m sigma mk varepsilon ijk x j sigma mk m dV int V varepsilon ijk x j F k dV amp int V varepsilon ijk x j m sigma mk dV int V varepsilon ijk x j sigma mk m F k dV end aligned nbsp The second integral is zero as it contains the equilibrium equations This leaves the first integral where x j m d j m displaystyle x j m delta jm nbsp therefore V e i j k s j k d V 0 displaystyle int V varepsilon ijk sigma jk dV 0 nbsp For an arbitrary volume V we then have e i j k s j k 0 displaystyle varepsilon ijk sigma jk 0 nbsp which is satisfied at every point within the body Expanding this equation we have s 12 s 21 displaystyle sigma 12 sigma 21 nbsp s 23 s 32 displaystyle sigma 23 sigma 32 nbsp and s 13 s 31 displaystyle sigma 13 sigma 31 nbsp or in general s i j s j i displaystyle sigma ij sigma ji nbsp This proves that the stress tensor is symmetricHowever in the presence of couple stresses i e moments per unit volume the stress tensor is non symmetric This also is the case when the Knudsen number is close to one K n 1 displaystyle K n rightarrow 1 nbsp or the continuum is a non Newtonian fluid which can lead to rotationally non invariant fluids such as polymers Principal stresses and stress invariants edit nbsp Stress components on a 2D rotating element Example of how stress components vary on the faces edges of a rectangular element as the angle of its orientation is varied Principal stresses occur when the shear stresses simultaneously disappear from all faces The orientation at which this occurs gives the principal directions In this example when the rectangle is horizontal the stresses are given by s 11 s 12 s 21 s 22 10 10 10 15 displaystyle left begin matrix sigma 11 amp sigma 12 sigma 21 amp sigma 22 end matrix right left begin matrix 10 amp 10 10 amp 15 end matrix right nbsp At every point in a stressed body there are at least three planes called principal planes with normal vectors n displaystyle mathbf n nbsp called principal directions where the corresponding stress vector is perpendicular to the plane i e parallel or in the same direction as the normal vector n displaystyle mathbf n nbsp and where there are no normal shear stresses t n displaystyle tau mathrm n nbsp The three stresses normal to these principal planes are called principal stresses The components s i j displaystyle sigma ij nbsp of the stress tensor depend on the orientation of the coordinate system at the point under consideration However the stress tensor itself is a physical quantity and as such it is independent of the coordinate system chosen to represent it There are certain invariants associated with every tensor which are also independent of the coordinate system For example a vector is a simple tensor of rank one In three dimensions it has three components The value of these components will depend on the coordinate system chosen to represent the vector but the magnitude of the vector is a physical quantity a scalar and is independent of the Cartesian coordinate system chosen to represent the vector so long as it is normal Similarly every second rank tensor such as the stress and the strain tensors has three independent invariant quantities associated with it One set of such invariants are the principal stresses of the stress tensor which are just the eigenvalues of the stress tensor Their direction vectors are the principal directions or eigenvectors A stress vector parallel to the normal unit vector n displaystyle mathbf n nbsp is given by T n l n s n n displaystyle mathbf T mathbf n lambda mathbf n mathbf sigma mathrm n mathbf n nbsp where l displaystyle lambda nbsp is a constant of proportionality and in this particular case corresponds to the magnitudes s n displaystyle sigma mathrm n nbsp of the normal stress vectors or principal stresses Knowing that T i n s i j n j displaystyle T i n sigma ij n j nbsp and n i d i j n j displaystyle n i delta ij n j nbsp we have T i n l n i s i j n j l n i s i j n j l n i 0 s i j l d i j n j 0 displaystyle begin aligned T i n amp lambda n i sigma ij n j amp lambda n i sigma ij n j lambda n i amp 0 left sigma ij lambda delta ij right n j amp 0 end aligned nbsp This is a homogeneous system i e equal to zero of three linear equations where n j displaystyle n j nbsp are the unknowns To obtain a nontrivial non zero solution for n j displaystyle n j nbsp the determinant matrix of the coefficients must be equal to zero i e the system is singular Thus s i j l d i j s 11 l s 12 s 13 s 21 s 22 l s 23 s 31 s 32 s 33 l 0 displaystyle left sigma ij lambda delta ij right begin vmatrix sigma 11 lambda amp sigma 12 amp sigma 13 sigma 21 amp sigma 22 lambda amp sigma 23 sigma 31 amp sigma 32 amp sigma 33 lambda end vmatrix 0 nbsp Expanding the determinant leads to the characteristic equation s i j l d i j l 3 I 1 l 2 I 2 l I 3 0 displaystyle left sigma ij lambda delta ij right lambda 3 I 1 lambda 2 I 2 lambda I 3 0 nbsp where I 1 s 11 s 22 s 33 s k k tr s I 2 s 22 s 23 s 32 s 33 s 11 s 13 s 31 s 33 s 11 s 12 s 21 s 22 s 11 s 22 s 22 s 33 s 11 s 33 s 12 2 s 23 2 s 31 2 1 2 s i i s j j s i j s j i 1 2 tr s 2 tr s 2 I 3 det s i j det s s 11 s 22 s 33 2 s 12 s 23 s 31 s 12 2 s 33 s 23 2 s 11 s 31 2 s 22 displaystyle begin aligned I 1 amp sigma 11 sigma 22 sigma 33 amp sigma kk text tr boldsymbol sigma 4pt I 2 amp begin vmatrix sigma 22 amp sigma 23 sigma 32 amp sigma 33 end vmatrix begin vmatrix sigma 11 amp sigma 13 sigma 31 amp sigma 33 end vmatrix begin vmatrix sigma 11 amp sigma 12 sigma 21 amp sigma 22 end vmatrix amp sigma 11 sigma 22 sigma 22 sigma 33 sigma 11 sigma 33 sigma 12 2 sigma 23 2 sigma 31 2 amp frac 1 2 left sigma ii sigma jj sigma ij sigma ji right frac 1 2 left left text tr boldsymbol sigma right 2 text tr left boldsymbol sigma 2 right right 4pt I 3 amp det sigma ij det boldsymbol sigma amp sigma 11 sigma 22 sigma 33 2 sigma 12 sigma 23 sigma 31 sigma 12 2 sigma 33 sigma 23 2 sigma 11 sigma 31 2 sigma 22 end aligned nbsp The characteristic equation has three real roots l i displaystyle lambda i nbsp i e not imaginary due to the symmetry of the stress tensor The s 1 max l 1 l 2 l 3 displaystyle sigma 1 max left lambda 1 lambda 2 lambda 3 right nbsp s 3 min l 1 l 2 l 3 displaystyle sigma 3 min left lambda 1 lambda 2 lambda 3 right nbsp and s 2 I 1 s 1 s 3 displaystyle sigma 2 I 1 sigma 1 sigma 3 nbsp are the principal stresses functions of the eigenvalues l i displaystyle lambda i nbsp The eigenvalues are the roots of the characteristic polynomial The principal stresses are unique for a given stress tensor Therefore from the characteristic equation the coefficients I 1 displaystyle I 1 nbsp I 2 displaystyle I 2 nbsp and I 3 displaystyle I 3 nbsp called the first second and third stress invariants respectively always have the same value regardless of the coordinate system s orientation For each eigenvalue there is a non trivial solution for n j displaystyle n j nbsp in the equation s i j l d i j n j 0 displaystyle left sigma ij lambda delta ij right n j 0 nbsp These solutions are the principal directions or eigenvectors defining the plane where the principal stresses act The principal stresses and principal directions characterize the stress at a point and are independent of the orientation A coordinate system with axes oriented to the principal directions implies that the normal stresses are the principal stresses and the stress tensor is represented by a diagonal matrix s i j s 1 0 0 0 s 2 0 0 0 s 3 displaystyle sigma ij begin bmatrix sigma 1 amp 0 amp 0 0 amp sigma 2 amp 0 0 amp 0 amp sigma 3 end bmatrix nbsp The principal stresses can be combined to form the stress invariants I 1 displaystyle I 1 nbsp I 2 displaystyle I 2 nbsp and I 3 displaystyle I 3 nbsp The first and third invariant are the trace and determinant respectively of the stress tensor Thus I 1 s 1 s 2 s 3 I 2 s 1 s 2 s 2 s 3 s 3 s 1 I 3 s 1 s 2 s 3 displaystyle begin aligned I 1 amp sigma 1 sigma 2 sigma 3 I 2 amp sigma 1 sigma 2 sigma 2 sigma 3 sigma 3 sigma 1 I 3 amp sigma 1 sigma 2 sigma 3 end aligned nbsp Because of its simplicity the principal coordinate system is often useful when considering the state of the elastic medium at a particular point Principal stresses are often expressed in the following equation for evaluating stresses in the x and y directions or axial and bending stresses on a part 14 p 58 59 The principal normal stresses can then be used to calculate the von Mises stress and ultimately the safety factor and margin of safety s 1 s 2 s x s y 2 s x s y 2 2 t x y 2 displaystyle sigma 1 sigma 2 frac sigma x sigma y 2 pm sqrt left frac sigma x sigma y 2 right 2 tau xy 2 nbsp Using just the part of the equation under the square root is equal to the maximum and minimum shear stress for plus and minus This is shown as t max t min s x s y 2 2 t x y 2 displaystyle tau max tau min pm sqrt left frac sigma x sigma y 2 right 2 tau xy 2 nbsp Maximum and minimum shear stresses editThe maximum shear stress or maximum principal shear stress is equal to one half the difference between the largest and smallest principal stresses and acts on the plane that bisects the angle between the directions of the largest and smallest principal stresses i e the plane of the maximum shear stress is oriented 45 displaystyle 45 circ nbsp from the principal stress planes The maximum shear stress is expressed as t max 1 2 s max s min displaystyle tau max frac 1 2 left sigma max sigma min right nbsp Assuming s 1 s 2 s 3 displaystyle sigma 1 geq sigma 2 geq sigma 3 nbsp then t max 1 2 s 1 s 3 displaystyle tau max frac 1 2 left sigma 1 sigma 3 right nbsp When the stress tensor is non zero the normal stress component acting on the plane for the maximum shear stress is non zero and it is equal to s n 1 2 s 1 s 3 displaystyle sigma text n frac 1 2 left sigma 1 sigma 3 right nbsp Derivation of the maximum and minimum shear stresses 8 p 45 78 11 p 1 46 13 15 p 111 157 16 p 9 41 17 p 33 66 18 p 43 61 The normal stress can be written in terms of principal stresses s 1 s 2 s 3 displaystyle sigma 1 geq sigma 2 geq sigma 3 nbsp as s n s i j n i n j s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 displaystyle begin aligned sigma mathrm n amp sigma ij n i n j amp sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 end aligned nbsp Knowing that T n 2 s i j s i k n j n k displaystyle left T n right 2 sigma ij sigma ik n j n k nbsp the shear stress in terms of principal stresses components is expressed as t n 2 T n 2 s n 2 s 1 2 n 1 2 s 2 2 n 2 2 s 3 2 n 3 2 s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 2 s 1 2 s 2 2 n 1 2 s 2 2 s 3 2 n 2 2 s 3 2 s 1 s 3 n 1 2 s 2 s 3 n 2 2 s 3 2 s 1 s 2 2 n 1 2 n 2 2 s 2 s 3 2 n 2 2 n 3 2 s 1 s 3 2 n 1 2 n 3 2 displaystyle begin aligned tau text n 2 amp left T n right 2 sigma text n 2 amp sigma 1 2 n 1 2 sigma 2 2 n 2 2 sigma 3 2 n 3 2 left sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 right 2 amp left sigma 1 2 sigma 2 2 right n 1 2 left sigma 2 2 sigma 3 2 right n 2 2 sigma 3 2 left left sigma 1 sigma 3 right n 1 2 left sigma 2 sigma 3 right n 2 2 sigma 3 right 2 amp sigma 1 sigma 2 2 n 1 2 n 2 2 sigma 2 sigma 3 2 n 2 2 n 3 2 sigma 1 sigma 3 2 n 1 2 n 3 2 end aligned nbsp The maximum shear stress at a point in a continuum body is determined by maximizing t n 2 displaystyle tau mathrm n 2 nbsp subject to the condition that n i n i n 1 2 n 2 2 n 3 2 1 displaystyle n i n i n 1 2 n 2 2 n 3 2 1 nbsp This is a constrained maximization problem which can be solved using the Lagrangian multiplier technique to convert the problem into an unconstrained optimization problem Thus the stationary values maximum and minimum values of t n 2 displaystyle tau text n 2 nbsp occur where the gradient of t n 2 displaystyle tau text n 2 nbsp is parallel to the gradient of F displaystyle F nbsp The Lagrangian function for this problem can be written as F n 1 n 2 n 3 l t 2 l g n 1 n 2 n 3 1 s 1 2 n 1 2 s 2 2 n 2 2 s 3 2 n 3 2 s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 2 l n 1 2 n 2 2 n 3 2 1 displaystyle begin aligned F left n 1 n 2 n 3 lambda right amp tau 2 lambda left g left n 1 n 2 n 3 right 1 right amp sigma 1 2 n 1 2 sigma 2 2 n 2 2 sigma 3 2 n 3 2 left sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 right 2 lambda left n 1 2 n 2 2 n 3 2 1 right end aligned nbsp where l displaystyle lambda nbsp is the Lagrangian multiplier which is different from the l displaystyle lambda nbsp use to denote eigenvalues The extreme values of these functions are F n 1 0 F n 2 0 F n 3 0 displaystyle frac partial F partial n 1 0 qquad frac partial F partial n 2 0 qquad frac partial F partial n 3 0 nbsp thence F n 1 n 1 s 1 2 2 n 1 s 1 s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 l n 1 0 displaystyle frac partial F partial n 1 n 1 sigma 1 2 2n 1 sigma 1 left sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 right lambda n 1 0 nbsp F n 2 n 2 s 2 2 2 n 2 s 2 s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 l n 2 0 displaystyle frac partial F partial n 2 n 2 sigma 2 2 2n 2 sigma 2 left sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 right lambda n 2 0 nbsp F n 3 n 3 s 3 2 2 n 3 s 3 s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 l n 3 0 displaystyle frac partial F partial n 3 n 3 sigma 3 2 2n 3 sigma 3 left sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 right lambda n 3 0 nbsp These three equations together with the condition n i n i 1 displaystyle n i n i 1 nbsp may be solved for l n 1 n 2 displaystyle lambda n 1 n 2 nbsp and n 3 displaystyle n 3 nbsp By multiplying the first three equations by n 1 n 2 displaystyle n 1 n 2 nbsp and n 3 displaystyle n 3 nbsp respectively and knowing that s n s i j n i n j s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 displaystyle sigma text n sigma ij n i n j sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 nbsp we obtain n 1 2 s 1 2 2 s 1 n 1 2 s n n 1 2 l 0 displaystyle n 1 2 sigma 1 2 2 sigma 1 n 1 2 sigma text n n 1 2 lambda 0 nbsp n 2 2 s 2 2 2 s 2 n 2 2 s n n 2 2 l 0 displaystyle n 2 2 sigma 2 2 2 sigma 2 n 2 2 sigma text n n 2 2 lambda 0 nbsp n 3 2 s 3 2 2 s 1 n 3 2 s n n 3 2 l 0 displaystyle n 3 2 sigma 3 2 2 sigma 1 n 3 2 sigma text n n 3 2 lambda 0 nbsp Adding these three equations we get n 1 2 s 1 2 n 2 2 s 2 2 n 3 2 s 3 2 2 s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 s n l n 1 2 n 2 2 n 3 2 0 t n 2 s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 2 2 s n 2 l 0 t n 2 s n 2 2 s n 2 l 0 l s n 2 t n 2 displaystyle begin aligned left n 1 2 sigma 1 2 n 2 2 sigma 2 2 n 3 2 sigma 3 2 right 2 left sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 right sigma mathrm n lambda left n 1 2 n 2 2 n 3 2 right amp 0 left tau mathrm n 2 left sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 right 2 right 2 sigma mathrm n 2 lambda amp 0 left tau mathrm n 2 sigma mathrm n 2 right 2 sigma mathrm n 2 lambda amp 0 lambda amp sigma mathrm n 2 tau mathrm n 2 end aligned nbsp this result can be substituted into each of the first three equations to obtain F n 1 n 1 s 1 2 2 n 1 s 1 s 1 n 1 2 s 2 n 2 2 s 3 n 3 2 s n 2 t n 2 n 1 0 n 1 s 1 2 2 n 1 s 1 s n s n 2 t n 2 n 1 0 s 1 2 2 s 1 s n s n 2 t n 2 n 1 0 displaystyle begin aligned frac partial F partial n 1 n 1 sigma 1 2 2n 1 sigma 1 left sigma 1 n 1 2 sigma 2 n 2 2 sigma 3 n 3 2 right left sigma text n 2 tau text n 2 right n 1 amp 0 n 1 sigma 1 2 2n 1 sigma 1 sigma text n left sigma text n 2 tau text n 2 right n 1 amp 0 left sigma 1 2 2 sigma 1 sigma text n sigma text n 2 tau text n 2 right n 1 amp 0 end aligned nbsp Doing the same for the other two equations we have F n 2 s 2 2 2 s 2 s n s n 2 t n 2 n 2 0 displaystyle frac partial F partial n 2 left sigma 2 2 2 sigma 2 sigma text n sigma text n 2 tau text n 2 right n 2 0 nbsp F n 3 s 3 2 2 s 3 s n s n 2 t n 2 n 3 0 displaystyle frac partial F partial n 3 left sigma 3 2 2 sigma 3 sigma text n sigma text n 2 tau text n 2 right n 3 0 nbsp A first approach to solve these last three equations is to consider the trivial solution n i 0 displaystyle n i 0 nbsp However this option does not fulfill the constraint mstyle, wikipedia, wiki , book, books, library,
article , read, download, free, free download, mp3, video, mp4, 3gp, jpg, jpeg, gif, png, picture, music, song, movie, book, game, games.