A simplified description

Traction vector

• The combination of the above stress components constitute a vector, called a traction vector \[\begin{align} \vec{T} =&\, \vec{T}_n + \vec{T}_s \\ \end{align}\]
• The tractions, once integrated over an area, becomes the forces on (or passing through) that plane \[\vec{F} = A \, \vec{T}\]

• The traction on each surface depends on the orientation of the surface.
• A complete description of the stress state requires that the traction be known for all \(\vec{n}\)

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:

• 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

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

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)

Normal stress (normal to a plane)

• 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

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

Example: tractions on 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 \({\sigma_{ns}}\)

• \({\vec{S}}= \vec{T}-{\vec{N}}\)
• \({\vec{S}}= \frac{1}{\sqrt{2}} {\vec{i}}+ 0 {\vec{j}}+ 0 {\vec{k}}- (\frac{1}{2\sqrt{2}} {\vec{i}}+ \frac{1}{2\sqrt{2}} {\vec{j}}+ 0 {\vec{k}})\)
• \({\vec{S}}= \frac{1}{2\sqrt{2}} {\vec{i}}- \frac{1}{2\sqrt{2}} {\vec{j}}+ 0 {\vec{k}}\)

Example: arbitrary traction computation

Given a stress: compute the normal and shear stress on a specified plane: \[[\sigma] = \left[ \begin{array}{ccc} 50 & 20 & -10 \\ 20 & -30 & 0 \\ -10 & 0 & 40 \\ \end{array} \right]\]

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

\[\begin{align} | {\vec{S}}| =& \, ? \\ =& \, (17.97)^2 + (-20.23)^2 + (-4.65)^2 \\ =& \, 27.46 \end{align}\]