4.3 Stress tensor and its components
Having recognized that the internal planes have associated traction vectors as loads pass through them, these vectors must be related to a more fundamental set of quantities which we will now refer to as the stress tensor.
Consider now a parallelepiped conveniently aligned with a Cartesian coordinate system.
Examine one face:
Positive \(x\) normal components of stress
- There is a traction on the face, with three components aligned with the coordinate system
- Each component is named using two subscripts.
- The first describes the normal direction to the face
- The second describes the direction in which the stress acts
- The components have stress units (Force/Area), and will be subsequently called components of the stress tensor
- Positive stress is defined when the stress acts in the
positive direction on the positive face or when the stress acts in
the negative direction on the negative direction face.
- Normal stresses
- Shear stresses
- Since tractions exist on each face of the cube, and the faces are aligned with the Cartesian directions, the components on these faces indicate that stress is “multi-axial” (i.e., is not generally aligned only with one axis).
- The traction vectors on the three positive faces are:
\[\begin{align} \vec{T}_{x} =& {\sigma_{xx}}\, i + {\sigma_{xy}}\, j + {\sigma_{xz}}\, k \\ \vec{T}_{y} =& {\sigma_{yx}}\, i + {\sigma_{yy}}\, j + {\sigma_{yz}}\, k \\ \vec{T}_{z} =& {\sigma_{zx}}\, i + {\sigma_{zy}}\, j + {\sigma_{zz}}\, k \\ \end{align}\]
Observe that at this point in space, there are three different planes with three different traction vectors. These vectors have a total of 9 components.
- The 9 components, with their associated face normals and component directions, combine to be a tensor at the point. The tensor is called stress (with units of force/area) and be written as:
\[\left[\sigma\right] = \left[ \begin{array}{ccc} {\sigma_{xx}}& {\sigma_{xy}}& {\sigma_{xz}}\\ {\sigma_{yx}}& {\sigma_{yy}}& {\sigma_{yz}}\\ {\sigma_{zx}}& {\sigma_{zy}}& {\sigma_{zz}}\\ \end{array} \right]\]
- Each stress component is named using two subscripts.
The first describes the normal direction to the face
The second describes the direction in which the component acts on that face
\({\sigma_{ij}}\) – stress on plane \(i\) in direction \(j\)
- \({\sigma_{xx}}\) is normal (sometimes abbreviated as \(\sigma_{\mathrm{x}}\) or \(\sigma\) (if uni-axial)
- \({\sigma_{xz}}\equiv {\tau_{xz}}\), \({\sigma_{xy}}\equiv {\tau_{xy}}\) are shear stresses (sometimes abbreviated as \(\tau\) if 2D stress field)
- Positive stresses are shown
4.3.1 Tensor representation
- A tensor describes a linear relationship in geometric space, where the components are associated with directions
- The rank of a tensor describes the number of directions that must
be assigned when describing a quantity. By example:
- Scalar – Rank 0 tensor (magnitude and 0 directions – \(3^0=1\) component)
- Vector – Rank 1 tensor (magnitude and 1 direction – \(3^1=3\) components)
- Dyad – Rank 2 tensor (magnitude and 2 directions – \(3^2=9\) components)
- Triad – Rank 3 tensor (magnitude and 3 directions – \(3^3=27\) components)
- Rank \(n\) tensor (magnitude and \(n\) directions – \(3^n\) components)
- Thus, a tensor can be thought of as a generalization of a vector (in actuality, vectors are a subset of tensors)
- A tensor follows transformation (mathematical) rules
- i.e., vectors must conserve the magnitudes and directions regardless of the coordinate system assigned
- Higher rank tensors must conserve that and more
4.3.2 Traction vector on arbitrary plane (Cauchy’s formula)
- The traction vector (force vector per area) on an arbitrary plane can be determined from the tensor by passing a cutting plane through the equilibrium element and summing forces.
- Let the normal be described by \({\vec{n}}= l \, {\vec{i}}+ m \, {\vec{j}}+ n \, {\vec{k}}\)
- \(l,m,n\) are direction cosines. \(\cos \theta_{nx}, \cos \theta_{ny}, \cos \theta_{nz}\).
- If the cut plane has an area of \(dA\), then the faces have areas
\[\begin{align} dA_x =&\, dA \, \vec{n} \cdot {\vec{i}}= dA \,l\\ dA_y =&\, dA \, \vec{n} \cdot {\vec{j}}= dA \,m\\ dA_z =&\, dA \, \vec{n} \cdot {\vec{k}}= dA \,n\\ \end{align}\] where \({\vec{i}},{\vec{j}},{\vec{k}}\) are the unit vectors in the Cartesian frame.
- Summing forces: \[{\vec{T}_n}A - {\vec{T}_x}A_x - {\vec{T}_y}A_y - {\vec{T}_z}A_z = 0\]
- From this: \[{\vec{T}_n}= l \, {\vec{T}_x}+ m \, {\vec{T}_y}+ n \, {\vec{T}_z}\]
- Equivalently it can be written as components: \[{\vec{T}_n}= T_{nx} {\vec{i}}+ T_{ny} {\vec{j}}+ T_{nz} {\vec{k}}\]
- The components of \({\vec{T}_n}\) are: \[\begin{align} T_{\mathrm{nx}} =& {\sigma_{xx}}\, l + {\sigma_{xy}}\, m + {\sigma_{xz}}\, n \\ T_{\mathrm{ny}} =& {\sigma_{yx}}\, l + {\sigma_{yy}}\, m + {\sigma_{yz}}\, n \\ T_{\mathrm{nz}} =& {\sigma_{zx}}\, l + {\sigma_{zy}}\, m + {\sigma_{zz}}\, n \\ \end{align}\]
- Or, more easily, \[\left\{T\right\} = [\sigma] \cdot \left\{n\right\}\]
This is called Cauchy’s formula (1827)
4.3.2.1 Normal stress (normal to a plane)
Recall tensor components of stress
- It is now established that the stress tensor and the traction vector are related: \[\begin{align} {\vec{T}_x}=& {\sigma_{xx}}{\vec{i}}+ {\sigma_{xy}}{\vec{j}}+ {\sigma_{xz}}{\vec{k}}\\ {\vec{T}_y}=& {\sigma_{yx}}{\vec{i}}+ {\sigma_{yy}}{\vec{j}}+ {\sigma_{yz}}{\vec{k}}\\ {\vec{T}_z}=& {\sigma_{zx}}{\vec{i}}+ {\sigma_{zy}}{\vec{j}}+ {\sigma_{zz}}{\vec{k}}\\ \end{align}\]
- What is the value of the stress component normal to a surface?
- The stress normal to the cutting plane \({\sigma_{nn}}\) is: \[\begin{align} {\sigma_{nn}}=& \, {\vec{T}_n}\cdot {\vec{n}}= l \, T_{nx} + m \, T_{ny} + n \, T_{nz} \\ {\sigma_{nn}}=& \, l^2 {\sigma_{xx}}+ m^2 {\sigma_{yy}}+ n^2 {\sigma_{zz}}\\ & \, + m \, n \, {\sigma_{yz}}+ l \, n \, {\sigma_{xz}}+ l \, m \, {\sigma_{xy}}\\ & \, + m \, n \, {\sigma_{zy}}+ l \, n \, {\sigma_{zx}}+ l \, m \, {\sigma_{yx}} \end{align}\]
- The maximum value and direction of \({\sigma_{nn}}\) will often control
the design of a member.
- Examples:
- Brittle material fracture
- Crack propagation
- Examples:
4.3.2.2 Shear stress (perpendicular to a plane)
- The shear stress \({\sigma_{ns}}\) can be obtained from:
- \({\sigma_{ns}}= \sqrt{{\vec{T}_n}^2 - {\sigma_{nn}}^2}\)
- \({\sigma_{ns}}= \sqrt{T_{\mathrm{nx}}^2 + T_{\mathrm{ny}}^2 + T_{\mathrm{nz}}^2 - {\sigma_{nn}}^2}\)
- The maximum value of the shear stress is also important in design.
- It is equal to half the difference between the max and min normal stress.
- Ductile materials tend to fail by yield.
- Yield is driven by shear stress and slipping typically along long grain boundaries
4.3.2.3 Example: tractions on planes of an axial rod
Traction vectors on different planes of an axial rod
Given a stress, compute the normal and shear stress on a specified plane (Assume the cross sectional area is 1.): \[[\sigma] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right]\]
Required: \({\vec{N}}\), \({\vec{S}}\)
Solution:
\[\left\{T\right\} = [\sigma] \cdot \left\{n\right\}\]
- The normal is given by:
- \(\vec{n} = \frac{1}{\sqrt{2}} {\vec{i}}+ \frac{1}{\sqrt{2}} {\vec{j}}+ 0 {\vec{k}}\)
- The area of the angled plane is \(\sqrt{2}\)
Traction vector \[\left\{ \begin{array}{c} T_x \\ T_y \\ T_z \\ \end{array} \right\} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] \left\{ \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \\ \end{array} \right\}\]
\(T_x = \left(1\right) \left(\frac{1}{\sqrt{2}}\right) + 0 + 0 = \frac{1}{\sqrt{2}} \quad [\text{Pa}]\)
\(T_y = 0 \quad [\text{Pa}]\)
\(T_z = 0 \quad [\text{Pa}]\)
Therefore:
- \({\vec{T}_n}= \frac{1}{\sqrt{2}} {\vec{i}}+ 0 {\vec{j}}+ 0 {\vec{k}}\quad [\text{Pa}]\)
- The traction vector is a force per unit area
- The total forces is \(\left(\sqrt{2}\right) \left( \frac{1}{\sqrt{2}} \right) = 1\)
- That is consistent with expectation/intuition!
Normal vector (\({\vec{N}}\)) and \({\sigma_{nn}}\)
The component of \({\vec{T}_n}\) in the \({\vec{n}}\) direction is:
- \({\sigma_{nn}}= {\vec{T}_n}\cdot {\vec{n}}\)
- \({\sigma_{nn}}= [\frac{1}{\sqrt{2}} \; 0 \; 0] \cdot \left\{ \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \\ \end{array} \right\} = 0.5 \quad [\text{Pa}]\)
Therefore:
- \({\vec{N}}= {\sigma_{nn}}\, {\vec{n}}= \left(0.5\right) \left({\vec{n}}\right) = \frac{1}{2\sqrt{2}} \left\{ \begin{array}{c} 1 \\ 1 \\ 0 \\ \end{array} \right\} \quad [\text{Pa}]\)
Shear vector (\({\vec{S}}\)) and \({\sigma_{ns}}\)
- \({\vec{S}}= \vec{T}-{\vec{N}}\)
- \({\vec{S}}= \frac{1}{\sqrt{2}} {\vec{i}}+ 0 {\vec{j}}+ 0 {\vec{k}}- (\frac{1}{2\sqrt{2}} {\vec{i}}+ \frac{1}{2\sqrt{2}} {\vec{j}}+ 0 {\vec{k}})\)
- \({\vec{S}}= \frac{1}{2\sqrt{2}} {\vec{i}}- \frac{1}{2\sqrt{2}} {\vec{j}}+ 0 {\vec{k}}\)
4.3.2.4 Example: arbitrary traction computation
Given a stress: compute the normal and shear stress on a specified plane: \[[\sigma] = \left[ \begin{array}{ccc} 50 & 20 & -10 \\ 20 & -30 & 0 \\ -10 & 0 & 40 \\ \end{array} \right]\]
Credit: Dr. Judah Ari-Gur
Required: \({\vec{N}}\), \({\vec{S}}\)
Solution
\[\left\{T\right\} = [\sigma] \cdot \left\{n\right\}\]
The normal is given by the cross product of the vectors that define the plane.
- \(\vec{a} = 0 {\vec{i}}-4 {\vec{j}}+ 5 {\vec{k}}\)
- \(\vec{b} = 3 {\vec{i}}-4 {\vec{j}}+ 0 {\vec{k}}\)
- \({\vec{n}}= \frac{\vec{a} \times \vec{b}}{\left| \vec{a} \times \vec{b}\right|}\)
- \(\vec{a} \times \vec{b} = (\left(-4\right)\left(3\right))({\vec{j}}\times {\vec{i}}) + (\left(-4\right) \left(5\right)) ({\vec{k}}\times {\vec{j}}) + (\left(5\right) \left(3\right)) ({\vec{k}}\times {\vec{i}})\)
- \(= + 20 {\vec{i}}+ 12 {\vec{k}}+ 15 {\vec{j}}\)
- \(\left| \vec{a} \times \vec{b}\right| = \sqrt{\left(12^2 + 20^2 + 15^2 \right)} = \sqrt{769} = 27.73\)
\({\vec{n}}= \frac{+ 20 {\vec{i}}+ 15 {\vec{j}}+ 12 {\vec{k}}}{\sqrt{769}}\)
\[\left\{ \begin{array}{c} T_x \\ T_y \\ T_z \\ \end{array} \right\} = \left[ \begin{array}{ccc} 50 & 20 & -10 \\ 20 & -30 & 0 \\ -10 & 0 & 40 \\ \end{array} \right] \left\{ \begin{array}{c} 20 \\ 15 \\ 12 \\ \end{array} \right\} \frac{1}{\sqrt{769}}\]
\[T_x = \left(\left(50 \right) \left(20\right) + \left(20\right) \left(15\right) - \left(10\right) \left(12\right)\right) / \sqrt{769} = 42.55 \quad \text{MPa} \] \[T_y = \left(\left(20 \right) \left(20\right) - \left(30\right) \left(15\right) + \left(0 \right) \left(12\right)\right) / \sqrt{769} = -1.80 \quad \text{MPa} \] \[T_z = \left(\left(-10\right) \left(20\right) + \left(0 \right) \left(15\right) + \left(40\right) \left(12\right)\right) / \sqrt{769} = 10.10 \quad \text{MPa} \]
Therefore:
- \[{\vec{T}_n}= 42.55 {\vec{i}}- 1.80 {\vec{j}}+ 10.10 {\vec{k}}\quad \text{MPa} \]
The component of \({\sigma_{nn}}\) in the \({\vec{n}}\) direction is:
- \[{\sigma_{nn}}= {\vec{T}_n}\cdot {\vec{n}}\]
- \[{\sigma_{nn}}= [42.55 \; -1.80 \; 10.10] \cdot \left\{ \begin{array}{c} 20 \\ 15 \\ 12 \\ \end{array} \right\} \frac{1}{\sqrt{769}} = 34.08 \quad \text{MPa} \]
Therefore:
- \[{\vec{N}}= {\sigma_{nn}}\, {\vec{n}}= 34.08 \, {\vec{n}}\quad \text{MPa} \]
Shear
\[\begin{align} {\vec{S}}&= {\vec{T}_n}- {\vec{N}}\\ &=42.55 \, {\vec{i}}- 1.80 \, {\vec{j}}+ 10.10 \, {\vec{k}}- 34.08 \, \frac{\left( 20 \, {\vec{i}}+15 \, {\vec{j}}+ 12 \, {\vec{k}}\right)}{\sqrt{769}} \\ &= 17.97 \, {\vec{i}}- 20.23 \, {\vec{j}}-4.65 \, {\vec{k}}\\ |{\vec{S}}| =& \sqrt{17.97^2 + 20.23^2 + 4.65^2} = 27.46 \mathrm{[MPa]} \\ \end{align}\]
Thus, we know the magnitude and direction of the shear and normal components of the traction vector.
What’s left?
Check for orthogonality
\[\begin{align} {\vec{N}}\cdot {\vec{S}}=& \, ? \\ =& \, (17.97)(25.10)-(20.23)(18.82)-(4.65)(15.06) \\ =& \, 0 \\ \end{align}\]
Verify that the magnitude is correct
\[\begin{align} | {\vec{S}}| =& \, ? \\ =& \, (17.97)^2 + (-20.23)^2 + (-4.65)^2 \\ =& \, 27.46 \end{align}\]