4.5 Principal stress

• Suppose we are interested in finding which plane extremizes (maximizes or minimizes) the normal stress component?
  • Why would we want to do that?
  • For example, crack propagation tends to propagate tangentially to the maximum normal stress
• To find an orientation that extremizes the normal component, we find an orientation where the shear stress is zero.

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\[\begin{align} {\vec{S}}=& \; 0 \\ {\vec{T}_n}=& \; {\vec{N}}+ {\vec{S}}\\ {\vec{T}_n}=& \; {\vec{N}}\\ {\vec{T}_n}=& \; |{\vec{T}_n}| \, {\vec{n}}\\ {\vec{T}_n}=& \; {\sigma_{\mathrm{P}}}\, {\vec{n}}\\ \end{align}\]

We call \(\sigma_P\) the principal stress and the orientation (\({\vec{n}}\)) the principal orientation or principal axis

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4.5.1 Principal stress values

Recall the relationship between the stress tensor and the traction vector: \[\begin{align} [\sigma] \cdot {\vec{n}}= {\vec{T}_n}\\ [\sigma] \cdot {\vec{n}}= {\sigma_{\mathrm{P}}}\, {\vec{n}}\\ [\sigma] \cdot {\vec{n}}- {\sigma_{\mathrm{P}}}\, {\vec{n}}= \vec{0}\\ \left([\sigma] - {\sigma_{\mathrm{P}}}[I]\right) \cdot {\vec{n}}= \vec{0}\\ \end{align}\]

Do you recognize this class of problem?

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4.5.2 Solution to the eigenvalue problem

By more than coincidence, this is the very definition of an eigenvalue problem with  representing the eigenvalues of \([\sigma]\):

\[\left[ \begin{array}{ccc} {\sigma_{xx}}- {\sigma_{\mathrm{P}}}& {\sigma_{xy}}& {\sigma_{xz}}\\ {\sigma_{xy}}& {\sigma_{yy}}- {\sigma_{\mathrm{P}}}& {\sigma_{yz}}\\ {\sigma_{xz}}& {\sigma_{yz}}& {\sigma_{zz}}- {\sigma_{\mathrm{P}}}\\ \end{array} \right] \left\{ \begin{array}{c} l \\ m \\ n \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right\}\]

Non-trivial solutions can be obtained by setting the determinant to zero. \[\det \left([\sigma] - {\sigma_{\mathrm{P}}}[I]\right) = 0\]

• The result is: \[{\sigma_{\mathrm{P}}}^3 - I_1 {\sigma_{\mathrm{P}}}^2 + I_2 {\sigma_{\mathrm{P}}}- I_3 = 0\]
• The roots of the equation (\(\sigma_P\)) are called principal stresses and they are conventionally ordered \(\sigma_1 \ge \sigma_2 \ge \sigma_3\).

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4.5.3 Characteristic equation and invariants

• The I values are called “invariants” of the stress tensor.
  • Meaning they do not change as the stress tensor is transformed.
• The invariants are:
  • The trace of the tensor, also called the hydrostatic stress \[I_1 = {\sigma_{xx}}+ {\sigma_{yy}}+{\sigma_{zz}}\]
  • \(I_2\) (No common name), relates to the strength of the deviatoric stress (and thus to yield). See J2 plasticity theory. \[I_2 = {\sigma_{xx}}{\sigma_{yy}}+ {\sigma_{xx}}{\sigma_{zz}}+ {\sigma_{yy}}{\sigma_{zz}} - \left({\sigma_{xy}}^2+{\sigma_{xz}}^2+{\sigma_{yz}}^2\right)\]
  • The determinant of the tensor \[I_3 = \det [\sigma]\]

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4.5.4 Principal directions

• Having solved the eigenvalue problem for the principal stresses, we can now compute the principal orientations:
• Define the principal orientations as \({\vec{n}}_i\) \[{\vec{n}}_i = l_i {\vec{i}}+ m_i {\vec{j}}+ n_i {\vec{k}}\]
• These are determined by: \[\left[ \begin{array}{ccc} {\sigma_{xx}}- \sigma_i & {\sigma_{xy}}& {\sigma_{xz}}\\ {\sigma_{xy}}& {\sigma_{yy}}- \sigma_i & {\sigma_{yz}}\\ {\sigma_{xz}}& {\sigma_{yz}}& {\sigma_{zz}}- \sigma_i \\ \end{array} \right] \left\{ \begin{array}{c} l_i \\ m_i \\ n_i \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right\}\]

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4.5.5 Notes about principal stresses and orientations

• The principal orientation vectors are normalized to have unit length and are the eigenvectors of the stress tensor.

• Note: The maximum shear stress is on an octahedral plane (The planes that bisect the principal planes) and be computed from the difference of the maximum principal stresses. (Mohr’s Circle)

\[\tau_{\mathrm{Max}} = \frac{1}{2}(\sigma_1 - \sigma_3 )\]

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4.5.6 Example

• Given a stress state: \[[\sigma]= \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right]\]

• Find the principal stresses, principal directions, and maximum shear stress:

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Solution

Invariants and principal stresses

First, find the invariants of the tensor: \[\begin{align} I_1 =& 3 \\ I_2 =& (1) (1) + (1) (1) + (1) (1) - (1^2 + 0^2 + 0^2) = 2 \\ I_3 =& \det [\sigma] = (1) (1-0) - (1) (1-0)+ (0) (0-0) =0 \\ \end{align}\]

Then put them into the equation and solve: \[{\sigma_{\mathrm{P}}}^3-3 {\sigma_{\mathrm{P}}}^2+2 {\sigma_{\mathrm{P}}}+ 0 =0\]

\[{\sigma_{\mathrm{P}}}= 2, 1, 0\]

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Solution

First principal direction

The directions are important; find them. \[\left[ \begin{array}{ccc} 1 -2 & 1 & 0 \\ 1 & 1 -2 & 0 \\ 0 & 0 & 1 -2\\ \end{array} \right] \left\{ \begin{array}{c} l_1 \\ m_1 \\ n_1 \\ \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right\}\] \[\left[ \begin{array}{ccc} -1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & -1 \\ \end{array} \right] \left\{ \begin{array}{c} l_1 \\ m_1 \\ n_1 \\ \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right\}\]

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Inspection of equation 3: \[\begin{align} -n_1 = 0 \\ \end{align}\]

Now equation 1: \[\begin{align} -l_1 + m_1 = 0 \\ l_1 = m_1 \\ \end{align}\]

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Magnitudes need be unitized, therefore:

\[\begin{align} l_1^2 + l_1^2 + 0 = 1\\ 2 l_1^2 = 1\\ l_1 = \sqrt{\frac{1}{2}}\\ \vec{n}_1 = \sqrt{\frac{1}{2}} \left[ \begin{array}{c} 1 \\ 1 \\ 0 \\ \end{array} \right] \end{align}\]

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Second principal direction

\[\left[ \begin{array}{ccc} 1 -1 & 1 & 0 \\ 1 & 1 -1 & 0 \\ 0 & 0 & 1 -1\\ \end{array} \right] \left\{ \begin{array}{c} l_2 \\ m_2 \\ n_2 \\ \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right\}\] \[\left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] \left\{ \begin{array}{c} l_2 \\ m_2 \\ n_2 \\ \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right\}\]

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Inspection of equation 1 and 2: \[\begin{align} l_2 = 0 \\ m_2 = 0 \\ \end{align}\]

Magnitudes need be unitized, therefore:

\[\begin{align} n_2^2 = 1\\ \vec{n}_2 = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ \end{array} \right] \end{align}\]

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Third principal direction

\[\left[ \begin{array}{ccc} 1 -0 & 1 & 0 \\ 1 & 1 -0 & 0 \\ 0 & 0 & 1 -0 \\ \end{array} \right] \left\{ \begin{array}{c} l_3 \\ m_3 \\ n_3 \\ \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right\}\]

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Inspection of equation 3: \[\begin{align} n_3 = 0 \\ \end{align}\]

Now equation 1: \[\begin{align} l_3 + m_3 = 0 \\ l_3 = -m_3 \\ \end{align}\]

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Magnitudes need be unitized, therefore: \[\begin{align} l_3^2 + (-l_3)^2 + 0 = 1\\ 2 l_3^2 = 1\\ l_3 = \sqrt{\frac{1}{2}}\\ \vec{n}_3 = \sqrt{\frac{1}{2}} \left[ \begin{array}{c} 1 \\ -1 \\ 0 \\ \end{array} \right] \end{align}\]

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In summary, the unit vectors are:

\[\sqrt{\frac{1}{2}} \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & \sqrt{2} & 0 \\ \end{array} \right]\]

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Sanity check: We can check orthogonality:

\[\vec{n}_1 \times \vec{n}_2 = \vec{n}_3\]

\[\mathbf{a}\times\mathbf{b}= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{vmatrix} = \left| \begin{array}{ccc} {\vec{i}}& {\vec{j}}& {\vec{k}}\\ \sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} & 0 \\ 0 & 0 & 1 \\ \end{array} \right|\]

which is:

\[\vec{n}_3 = \frac{i}{\sqrt{2}}-\frac{j}{\sqrt{2}}\]

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4.5.7 Additional thoughts on principal stresses

• The principal stresses are defined as the normal stresses which can occur when all shear stresses vanish.
  • These coordinate directions are called the principal axes.
• If two principal stresses are equal, then two axes are arbitrary in a plane.
• If all three principal stresses are equal, then all three axes are arbitrary.
  • (See example in the text for the procedure to follow).

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