📘 Theory of Elasticity – Complete Notes from Drawings

Built for beginners: every formula shows what each symbol means, when it applies, and how to check your units.

1) Introduction

Theory of Elasticity concerns the deformation of elastic solids under forces. In the elastic range, the body returns to its original shape when the load is removed (no permanent, plastic strain).

Context: Mechanics of Materials (ME110).

2) Fundamental Concepts & Diagrams

Cylinder under tension: an initial length \(L\) is stretched to \(L+\Delta L\). Cross-section and radius may change slightly: \(A\!\to\!A+\Delta A,\; r\!\to\!r+\Delta r\).

Small deformation assumption: \(\Delta L \ll L\) — geometry is nearly constant in basic 1D formulas.

3) Stress and Strain

Stress

\[ \sigma \;=\; \frac{F}{A}, \qquad F=\text{force }\mathrm{(N)},\;\; A=\text{area }\mathrm{(m^{2})}. \]

Engineering strain

\[ \varepsilon \;=\; \frac{\Delta L}{L}, \qquad \Delta L=\text{change in length }\mathrm{(m)},\;\; L=\text{original length }\mathrm{(m)}. \]

Why average stress \(F/A\) is OK here

In uniform axial loading with small deformations, stress is approximately uniform across the cross-section. For high-gradient cases (holes/notches), use elasticity solutions or FEA.

4) Stress–Strain Relationship

Hooke’s law (linear elasticity) :

\[ \sigma \;=\; E\,\varepsilon, \qquad E=\text{Young’s modulus }\mathrm{(Pa)}. \]

In the linear (initial straight) region of the stress–strain curve, \(E\) is constant. Beyond yield, plastic strain \(\varepsilon_p\) appears and unloading does not return to zero strain.

5) Focus of Class

This course focuses on the theory of linear elasticity (small strains, linear stress–strain relation, typically isotropic materials).

6) Example – Bar in Tension

6.1 Elongation

From \(\sigma=\tfrac{F}{A}\) and \(\sigma=E\varepsilon=E\,\tfrac{\Delta L}{L}\) :

Relationship

\[ \frac{F}{A}=E\frac{\Delta L}{L} \;\Rightarrow\; \Delta L=\frac{F\,L}{E\,A}, \qquad A=\pi R^{2}. \]

Area (R = 0.01 m)

\[ A=\pi(0.01)^{2}\;\approx\;3.1416\times 10^{-4}\ \mathrm{m^{2}} \]

Elongation

\[ \Delta L=\frac{10{,}000\cdot 1}{(70\times 10^{9})(3.1416\times 10^{-4})} \;\approx\;4.55\times 10^{-4}\ \mathrm{m}=0.455\ \mathrm{mm}. \]

Unit check

\[ [\mathrm{N\!\cdot\! m}]\;/\;([\mathrm{Pa}]\,\mathrm{m^{2}}) =[\mathrm{N\!\cdot\! m}]\;/\;([\mathrm{N/m^{2}}]\,\mathrm{m^{2}}) =\mathrm{m}\ \checkmark \]

6.2 Lateral Contraction (Poisson’s Effect)

Definition

\[ \nu=-\frac{\varepsilon_y}{\varepsilon_x},\qquad \varepsilon_x=\frac{\Delta L}{L},\; \varepsilon_y=\frac{\Delta R}{R} \;\Rightarrow\; \Delta R=-\,\nu\,\frac{\Delta L}{L}\,R. \]

Numerical result

\[ \Delta R\;\approx\;-0.3\times \frac{4.55\times 10^{-4}}{1}\times 0.01 =-1.37\times 10^{-6}\ \mathrm{m}=-1.37\ \mu\mathrm{m}. \]

Interpretation: tiny radius shrinkage relative to the axial elongation.

7) Classical Problems in Mechanics of Materials

(a) Beam under distributed load \(q\)

Simply supported beam, length \(L\), rectangular cross-section \(b\times h\) \(\bigl(I=\tfrac{b h^{3}}{12}\bigr)\). Governing equation (Euler–Bernoulli):

\[ E I\, w''''(x) = q(x). \]

Uniform load \(q\)

\[ M_{\max}=\frac{q L^{2}}{8},\qquad w_{\max}=\frac{5 q L^{4}}{384\,E I},\qquad \sigma_{\max}=\frac{M_{\max}\,c}{I}\ \ (c=h/2). \]

Tasks you’ll see: sketch supports & loads → write boundary conditions → solve for \(w(x)\) → compute moment \(M(x)\) and stresses.

8) General Elasticity Problems

Unknowns: displacement fields \(u_x(x,y)\), \(u_y(x,y)\). The (small) strain tensor is built from displacement gradients.

\[ \renewcommand{\arraystretch}{1.35} \begin{array}{c@{\qquad\qquad}rl} \displaystyle \boldsymbol{\varepsilon} \;=\; \begin{bmatrix} \varepsilon_{xx} & \varepsilon_{xy} \\ \varepsilon_{yx} & \varepsilon_{yy} \end{bmatrix} & \varepsilon_{xx} & = \dfrac{\partial u_x}{\partial x},\\[6pt] & \varepsilon_{yy} & = \dfrac{\partial u_y}{\partial y},\\[6pt] & \varepsilon_{xy}=\varepsilon_{yx} & = \tfrac12\!\left(\dfrac{\partial u_x}{\partial y}+\dfrac{\partial u_y}{\partial x}\right). \end{array} \]

\[ \renewcommand{\arraystretch}{1.45} \newcommand{\rowstrut}{\vphantom{\dfrac{\partial u_x}{\partial y}}} \boldsymbol{\varepsilon} = \left[ \begin{array}{cc} \rowstrut \displaystyle\dfrac{\partial u_x}{\partial x} & \rowstrut \displaystyle\tfrac12\!\left(\dfrac{\partial u_x}{\partial y}+\dfrac{\partial u_y}{\partial x}\right)\\[6pt] \rowstrut \displaystyle\tfrac12\!\left(\dfrac{\partial u_x}{\partial y}+\dfrac{\partial u_y}{\partial x}\right) & \rowstrut \displaystyle\dfrac{\partial u_y}{\partial y} \end{array} \right] \]

Geometry may be, e.g., a triangular domain (base \(a\), height \(b\)) under shear; material properties: \(E,\ \nu\).

Constitutive law in 2D (isotropic, linear)

3D form: \(\boldsymbol{\sigma}=2\mu\,\boldsymbol{\varepsilon} + \lambda\,\mathrm{tr}(\boldsymbol{\varepsilon})\,\mathbf I\), with \(\mu=\dfrac{E}{2(1+\nu)}\) and \(\lambda=\dfrac{E\,\nu}{(1+\nu)(1-2\nu)}\). In plane stress/strain, the equivalent 2D matrices are used.

9) Stress Concentration & Fracture

Plate with circular hole under tension

Applied far stress \(\sigma\), hole radius \(R\), outer size \(a\) with \(R\ll a\). The peak hoop stress at the hole edge is

\[ \sigma_{\max} = 3\,\sigma \quad\Rightarrow\quad K_t=\frac{\sigma_{\max}}{\sigma}=3. \]

Fracture Mechanics – near a crack tip

As distance to the tip \(r\to 0\), stresses grow like

\[ \sigma_{ij}(r,\theta) \sim \frac{K_I}{\sqrt{2\pi r}}\,f_{ij}(\theta)\quad(\text{Mode I}). \]

Key idea: stress intensity factor \(K_I\) captures geometry + load effects near the crack tip.

10) Broader Context of Mechanics

Engineering Mechanics: Statics & Dynamics (ME010 + ME103).
Solid Mechanics track: Mechanics of materials → Elasticity → Plasticity → Fracture mechanics.
Fluid Mechanics: Continuum mechanics of liquids/gases.

Quick symbols & units

\(\sigma\)

Stress \(\mathrm{(Pa = N/m^{2})}\)

\(\varepsilon\)

Strain (dimensionless)

\(E\)

Young’s modulus (Pa)

\(\nu\)

Poisson’s ratio (0–0.5 for isotropic stable solids)

\(I\)

Second moment of area \(\mathrm{(m^{4})}\)

\(q\)

Distributed load (N/m)

\(M\)

Bending moment (N·m)

Study flow: 1D bars → beams → 2D elasticity. Always sketch the body, supports, and loads; write boundary conditions before equations.

Vector Notes — Complete Breakdown

A quick, beginner-friendly path: definitions → components → index rules → dot/cross → δ/ε identities → triple products → a tiny worked example.

1) What is a Vector?

A vector has both magnitude (length) and direction. We draw it as an arrow. Common notations: \(\vec{a}\) (Gibbs) or bold \(\mathbf{a}\).

2) Vectors in Cartesian Coordinates

With orthonormal basis \(\{\vec e_1,\vec e_2,\vec e_3\}\):

Components

\[ \vec a = a_1 \vec e_1 + a_2 \vec e_2 + a_3 \vec e_3 \qquad\text{(components \(a_1,a_2,a_3\))}. \]

Einstein Summation

\[ \vec a = a_i\,\vec e_i \quad (i=1,2,3). \]

Dummy indices: repeated indices are summed and can be renamed freely: \(a_i \vec e_i \equiv a_m \vec e_m\).

3) Index Operations

\[ \text{Dot: } a_i b_i = a_1 b_1 + a_2 b_2 + a_3 b_3. \]

\[ \text{Outer (all pairs): } a_i b_j \ \text{→ 9 terms.} \]

\[ \text{Componentwise: } c_i = a_i \pm b_i. \]

Invalid mix

\(a_i + b_j\) (free indices don’t match) — no meaning.

\[ a_i b_i \, c_j d_j = (\vec a\!\cdot\!\vec b)\,(\vec c\!\cdot\!\vec d). \]

Index hygiene:

Each term: every index is either repeated exactly twice (summed) or appears once (free).
The same free indices must appear on both sides of an equation.

4) Vector Operations

(a) Addition / Subtraction

\[ \vec c = \vec a \pm \vec b, \qquad c_i = a_i \pm b_i. \]

Geometric view: triangle/parallelogram law.

(b) Dot Product

\[ \vec a\cdot\vec b = a b \cos\theta = a_i b_i,\qquad \lVert \vec a \rVert = \sqrt{a_i a_i}. \]

\[ \vec e_i\cdot\vec e_j = \delta_{ij}\Rightarrow a_i b_j \delta_{ij} = a_i b_i. \]

Angles & projections

\(\displaystyle \cos\theta = \frac{\vec a\cdot\vec b}{\lVert\vec a\rVert\,\lVert\vec b\rVert}\), \(\displaystyle \text{proj}_{\vec b}\vec a = \frac{\vec a\cdot\vec b}{\lVert\vec b\rVert^2}\,\vec b\), component along unit \(\hat n\): \(\vec a\cdot\hat n\).

5) Kronecker Delta \(\delta_{ij}\)

\[ \delta_{ij}=\begin{cases}1,& i=j\\ 0,& i\neq j\end{cases} \qquad\Rightarrow\qquad \vec e_i\cdot\vec e_j=\delta_{ij}. \]

\[ \delta_{ii}=3,\quad \delta_{ij}\delta_{ij}=3,\quad a_i b_i = \vec a\cdot\vec b. \]

6) Cross Product

\(\vec c=\vec a\times\vec b\) has magnitude \(c=ab\sin\theta\) and direction given by the right-hand rule; \(\vec c\perp\vec a,\vec b\).

\[ c_k = \epsilon_{kij}\,a_i b_j,\qquad \vec e_1\times\vec e_2=\vec e_3,\ \vec e_2\times\vec e_3=\vec e_1,\ \vec e_3\times\vec e_1=\vec e_2,\ \vec e_i\times\vec e_i=\vec 0. \]

\[ \vec a\times\vec b = \begin{vmatrix} \vec e_1 & \vec e_2 & \vec e_3\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}. \]

7) Levi-Civita Symbol \(\epsilon_{ijk}\)

\[ \epsilon_{ijk}= \begin{cases} +1,& (i,j,k)\ \text{even permutation of }(1,2,3)\\ -1,& (i,j,k)\ \text{odd permutation}\\ 0,& \text{otherwise} \end{cases} \]

\[ \epsilon_{123}=1,\quad \epsilon_{213}=-1,\quad \epsilon_{231}=1, \qquad \vec e_i\times\vec e_j=\epsilon_{ijk}\,\vec e_k. \]

8) Epsilon–Delta Relation

\[ \epsilon_{kij}\,\epsilon_{kmn} \;=\; \delta_{im}\,\delta_{jn} - \delta_{in}\,\delta_{jm}. \]

\[ \epsilon_{ijk}\,\epsilon_{pqk} = \delta_{ip}\,\delta_{jq} - \delta_{iq}\,\delta_{jp}. \]

Proof sketch (visible, no toggles): both sides are antisymmetric in \((i\leftrightarrow j)\) and \((m\leftrightarrow n)\) and agree on representative index choices (e.g., \(i=j\Rightarrow0\); \(i=m,\ j=n\Rightarrow+1\); \(i=n,\ j=m\Rightarrow-1\)). Linearity then fixes the identity.

9) Vector Triple Product

\[ \vec a\times(\vec b\times\vec c) = (\vec a\cdot\vec c)\,\vec b - (\vec a\cdot\vec b)\,\vec c. \]

\[ (\vec a\times\vec d)_k = \epsilon_{kij}\,a_i d_j. \]

\[ \text{Scalar triple: } \ \vec a\cdot(\vec b\times\vec c) = \epsilon_{ijk}\,a_i b_j c_k = \det\!\begin{bmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{bmatrix}. \]

\(\vec a\cdot(\vec b\times\vec c)\) is the oriented volume of the parallelepiped formed by \(\vec a,\vec b,\vec c\).

Mini Example (numbers you can check)

Let \(\vec a=(1,2,2)\), \(\vec b=(2,-1,1)\).

\[ \vec a\cdot\vec b = 1\cdot2 + 2\cdot(-1) + 2\cdot1 = 2,\quad \lVert\vec a\rVert = \sqrt{1^2+2^2+2^2}=3,\quad \lVert\vec b\rVert = \sqrt{2^2+(-1)^2+1^2}=\sqrt{6}. \]

\[ \cos\theta = \frac{2}{3\sqrt6}\approx 0.272 \Rightarrow \theta\approx 74.2^\circ. \]

\[ \vec a\times\vec b = \begin{vmatrix} \vec e_1 & \vec e_2 & \vec e_3\\ 1 & 2 & 2\\ 2 & -1 & 1 \end{vmatrix} = 4\,\vec e_1 + 3\,\vec e_2 - 5\,\vec e_3. \]

\[ \lVert \vec a\times\vec b \rVert=\sqrt{4^2+3^2+(-5)^2}=\sqrt{50}\approx 7.071,\quad ab\sin\theta \approx 3\cdot\sqrt6\cdot 0.962 \approx 7.07\ (\checkmark). \]

Projection of \(\vec a\) onto \(\vec b\)

\[ \frac{\vec a\cdot\vec b}{\lVert\vec b\rVert^2}\,\vec b = \frac{2}{6}\,\vec b = \Big(\tfrac{2}{3}, -\tfrac{1}{3}, \tfrac{1}{3}\Big). \]

Bonus: Component extraction & basis changes

Component along unit \(\hat n\)

\(a_{\parallel} = \vec a\cdot\hat n\), vector part \( (\vec a\cdot\hat n)\hat n\); perpendicular part \( \vec a - (\vec a\cdot\hat n)\hat n\).

Orthonormal change of basis

if \(\vec e'_i=\alpha_{im}\,\vec e_m\) with \(\alpha\alpha^\top=I\), then \(a'_i=\alpha_{im}\,a_m\).

Common pitfalls

Don’t mix free indices: \(a_i+b_j\) is invalid.
Use \(\vec e_i\cdot\vec e_j=\delta_{ij}\) and \(\vec e_i\times\vec e_j=\epsilon_{ijk}\vec e_k\) only in orthonormal bases.
Norm uses a square root: \(\lVert \vec a\rVert=\sqrt{a_i a_i}\).

Matrix Notes — Organized

Beginner roadmap: definitions → vectors-as-matrices → add/sub → transpose & symmetry → multiplication (with shapes!) → inverses (when they exist) → tiny numeric checks.

1) Definition of a Matrix

An \(m\times n\) matrix has \(m\) rows and \(n\) columns:

\[ A=\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \equiv [a_{ij}]_{m\times n}. \]

Shapes & special matrices:

Square: \(n\times n\)
Zero matrix: all entries 0
Identity \(I_n\): diagonal ones, else 0; \((I_n)_{ij}=\delta_{ij}\)
Diagonal: only diagonal entries possibly nonzero

2) Vector as a Matrix

A (column) vector is an \(n\times 1\) matrix:

\[ \vec a=\begin{bmatrix} a_1\\ a_2\\ a_3 \end{bmatrix} \quad\text{(also written as }(a_1,a_2,a_3)\text{ when listing values).} \]

3) Matrix Addition & Subtraction

Same shape required. Entrywise operation:

\[ A\pm B = [a_{ij}] \pm [b_{ij}] = [\,a_{ij}\pm b_{ij}\,]. \]

Properties: commutative \(A+B=B+A\), associative \((A+B)+C=A+(B+C)\). The zero matrix is the additive identity.

4) Transpose of a Matrix

Flip rows/columns: \((A^T)_{ij}=a_{ji}\).

\[ \begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix}^{\!T} = \begin{bmatrix} a_{11}&a_{21}&a_{31}\\ a_{12}&a_{22}&a_{32}\\ a_{13}&a_{23}&a_{33} \end{bmatrix}. \]

If \(A^T=A\) then \(A\) is symmetric.

Useful transpose identities:

\((A^T)^T=A\)
\((A+B)^T=A^T+B^T\)
\((AB)^T=B^T A^T\)
\((\alpha A)^T=\alpha A^T\) for scalar \(\alpha\)

5) Symmetric & Antisymmetric Decomposition

Any square \(A\) can be split as:

\[ A=\tfrac12(A+A^T)\;+\;\tfrac12(A-A^T) \]

First term is symmetric, second term is antisymmetric (skew-symmetric).

6) Matrix Multiplication

Defined when inner dimensions match: if \(A\) is \(m\times n\) and \(B\) is \(n\times p\), then \(C=AB\) is \(m\times p\) with

\[ C=[c_{ij}],\quad c_{ij}=\sum_{m=1}^{n} a_{i m}\,b_{m j} \quad\text{(Einstein: }c_{ij}=a_{im}b_{mj}\text{)}. \]

Algebra: associative \((AB)C=A(BC)\), not commutative in general (\(AB\neq BA\)). Distributive: \(A(B+C)=AB+AC\).

Mini numeric check:

\[ A=\begin{bmatrix}1&2\\3&4\end{bmatrix},\ B=\begin{bmatrix}2&0\\1&-1\end{bmatrix} \Rightarrow AB=\begin{bmatrix} 4 & -2\\ 10 & -4 \end{bmatrix}. \]

7) Inverse of a Matrix

Only for square matrices. If \(A^{-1}\) exists (i.e., \(A\) is invertible / nonsingular):

\[ A^{-1}A=AA^{-1}=I=\big[\delta_{ij}\big]. \]

When does the inverse exist?

\(\det(A)\neq 0\)
Columns (and rows) are linearly independent
\(\text{rank}(A)=n\) (full rank)

2×2 inverse (handy formula):

\[ A=\begin{bmatrix}a&b\\ c&d\end{bmatrix},\quad \det(A)=ad-bc\neq 0 \Rightarrow A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\ -c&a\end{bmatrix}. \]

Small identities:

\((A^T)^{-1}=(A^{-1})^T\)
\((AB)^{-1}=B^{-1}A^{-1}\)
\((\alpha A)^{-1}=\alpha^{-1}A^{-1}\) for nonzero scalar \(\alpha\)

Identity & δ-symbol

\[ I_n= \begin{bmatrix} 1&0&\cdots&0\\ 0&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&1 \end{bmatrix}, \qquad (I_n)_{ij}=\delta_{ij}. \]

Common beginner pitfalls

Trying to add or subtract matrices of different shapes.
Multiplying with incompatible dimensions (inner sizes must match).
Assuming \(AB=BA\) — generally false.
Symmetric/antisymmetric split requires a square matrix.
Not every square matrix is invertible — check \(\det(A)\neq 0\).

Coordinate Transformation Notes — Organized

Goal: represent the same geometric vector in two orthonormal bases, derive the component change law, build the rotation matrix, and verify orthogonality step-by-step.

1) Position Vector Representation

Let \(\{\vec e_1,\vec e_2,\vec e_3\}\) be the original orthonormal basis. Any vector (e.g., position) can be written as

\[ \vec x = x_1 \vec e_1 + x_2 \vec e_2 + x_3 \vec e_3 \;=\; x_i\,\vec e_i. \]

\[ \vec x = x'_1 \vec e'_1 + x'_2 \vec e'_2 + x'_3 \vec e'_3 \;=\; x'_i\,\vec e'_i. \]

What stays the same? What changes?

\(\vec x\) (the geometric arrow) is invariant.
The numbers \(x_i\) vs. \(x'_i\) change because the basis vectors change.

2) Transformation Using Dot Products

Start from equality of the two expansions and dot with a new basis vector \(\vec e'_j\):

\[ x_i \vec e_i = x'_m \vec e'_m \ \Longrightarrow\ \vec e'_j\!\cdot\!(x_i \vec e_i)=\vec e'_j\!\cdot\!(x'_m \vec e'_m) = x'_m\,\delta_{jm}=x'_j. \]

Direction cosines

\[ \alpha_{ji}=\vec e'_j\cdot\vec e_i. \]

Component transform

\[ x'_j=\alpha_{ji}\,x_i \qquad\text{(sum on \(i\))}. \]

Matrix form & index placement

\[ \mathbf{x}'=\mathbf{A}\mathbf{x},\quad \mathbf{A}=[\alpha_{ji}]_{j\text{ row},\,i\text{ col}}, \quad \begin{bmatrix}x'_1\\x'_2\\x'_3\end{bmatrix} = \begin{bmatrix} \alpha_{11}&\alpha_{12}&\alpha_{13}\\ \alpha_{21}&\alpha_{22}&\alpha_{23}\\ \alpha_{31}&\alpha_{32}&\alpha_{33} \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}. \]

Row \(j\) selects \(\vec e'_j\cdot(\cdot)\); column \(i\) pairs with \(x_i\).

Inverse relation

\[ \mathbf{x}=\mathbf{A}^T\mathbf{x}'\quad\Longleftrightarrow\quad x_i=\alpha_{ji}x'_j. \]

3) Example: Rotation in the \(x_1\)–\(x_2\) Plane

Rotate the basis by \(\theta\) about \(x_3\) (right-handed). Standard rotation matrix:

\[ \mathbf{A}(\theta)= \begin{bmatrix} \cos\theta & \sin\theta & 0\\ -\sin\theta& \cos\theta & 0\\ 0&0&1 \end{bmatrix}, \qquad \mathbf{x}'=\mathbf{A}\mathbf{x}. \]

\[ x'_1=\cos\theta\,x_1+\sin\theta\,x_2,\quad x'_2=-\sin\theta\,x_1+\cos\theta\,x_2,\quad x'_3=x_3. \]

Where do the entries come from?

\(\alpha_{11}=\vec e'_1\!\cdot\!\vec e_1=\cos\theta,\ \alpha_{12}=\vec e'_1\!\cdot\!\vec e_2=\sin\theta\), etc.; \(\alpha_{33}=1\) for rotation about \(x_3\).

4) Worked Example — rotate \((1,1,0)\) by \(\theta=\pi/4\)

Use \(\cos\theta=\sin\theta=\tfrac{\sqrt2}{2}\approx 0.7071\).

\[ \mathbf{A}\!\Big(\tfrac{\pi}{4}\Big)= \begin{bmatrix} \tfrac{\sqrt 2}{2} & \tfrac{\sqrt 2}{2} & 0\\ -\tfrac{\sqrt 2}{2} & \tfrac{\sqrt 2}{2} & 0\\ 0&0&1 \end{bmatrix},\quad \mathbf{x}=\begin{bmatrix}1\\[2pt]1\\[2pt]0\end{bmatrix},\quad \mathbf{x}'=\mathbf{A}\mathbf{x}. \]

\[ x'_1=\sqrt2,\qquad x'_2=0,\qquad x'_3=0 \ \Rightarrow\ \mathbf{x}'=\begin{bmatrix}\sqrt2\\[2pt]0\\[2pt]0\end{bmatrix}. \]

Numerical sanity checks

\(\|\vec x\|=\sqrt2=\|\vec x'\|\) (rigid rotation).
\(\mathbf{x}=\mathbf{A}^T\mathbf{x}'\) reproduces \((1,1,0)\).

5) Orthogonality Condition (rotations preserve dot products)

Rows/columns of \(\mathbf A\) are orthonormal:

\[ \alpha_{im}\alpha_{jm}=\delta_{ij} \;\Longleftrightarrow\; \mathbf{A}\mathbf{A}^T=\mathbf{I} \;\Longleftrightarrow\; \mathbf{A}^T\mathbf{A}=\mathbf{I}. \]

Why \(\mathbf{A}^{-1}=\mathbf{A}^T\)?

If \(\mathbf{A}^T\mathbf{A}=\mathbf{I}\), right-multiply by \(\mathbf{A}^{-1}\) to get \(\mathbf{A}^T=\mathbf{A}^{-1}\). Orthogonal matrices preserve inner products/lengths.

Determinant

Proper rotations have \(\det\mathbf A=+1\); orthogonal matrices satisfy \(|\det\mathbf A|=1\).

Bonus: How to build \(\mathbf{A}\) from basis vectors

Using \(\alpha_{ji}=\vec e'_j\!\cdot\!\vec e_i\): row \(j\) of \(\mathbf A\) holds components of \(\vec e'_j\) in the old basis; equivalently, columns of \(\mathbf A^T\). Hence \(\mathbf{x}'=\mathbf{A}\mathbf{x}\).

Common pitfalls (and fixes)

Row/column mixups: keep \(x'_j=\alpha_{ji}x_i\) (row = primed index).
Forgetting orthogonality: check \(\mathbf{A}^T\mathbf{A}=\mathbf{I}\).
Angle sign: standard \(x_1\!\!-\!x_2\) rotation above is \(+\theta\) about \(x_3\) (RH rule).
\(\vec x\) “changing”: only components change; the geometric vector is the same.

Tensor Notes — Organized

Conventions (3D unless stated): indices i,j,k,ℓ,m,n,p,q,r,s,t run over {1,2,3}. Repeated indices are summed (Einstein). Bold/arrows = vectors; plain indexed symbols = components.

Index & Symbol Legend (read me first)

\(x_i\) component \(i\) of vector \(\vec x\) in basis \(\{\vec e_i\}\).

\(a_i\) components of a 1st-order tensor (vector), shape \(3\times1\).

\(a_{ij}\) components of a 2nd-order tensor (matrix), shape \(3\times3\).

\(a_{ijk}\) components of a 3rd-order tensor, shape \(3\times3\times3\).

\(\alpha_{ij}\) direction cosines \(\alpha_{ij}=\vec e'_i\!\cdot\!\vec e_j\); rows/cols form orthogonal \(\mathbf A\).

\(\delta_{ij}\) Kronecker delta (identity components): \(1\) if \(i=j\), else \(0\).

\(\epsilon_{ijk}\) Levi–Civita symbol (permutation), used for cross products/curls.

Free vs. repeated indices

Repeated indices (e.g., \(a_i b_i\)) are summed and vanish from the result.
Free indices (e.g., \(c_j=\alpha_{ji} a_i\)) must appear on both sides with the same letters.
Use any index at most twice per term (we keep all as subscripts here).

1) Tensor Definition (by transformation law)

A quantity is a tensor if its components transform with the basis by the appropriate product of \( \alpha \)’s.

First-order (vector)

\[ a'_i=\alpha_{ij}a_j \quad\text{(free \(i\); \(j\) summed).} \]

Second-order

\[ a'_{ij}=\alpha_{im}\alpha_{jn}a_{mn} \quad\text{(one \(\alpha\) per index).} \]

Third-order

\[ a'_{ijk}=\alpha_{ir}\alpha_{js}\alpha_{kt}a_{rst}. \]

Shapes & counts. 1st: 3; 2nd: \(3^2=9\); 3rd: 27; 4th: 81 (symmetries reduce).
Matrix view: \(a_{ij}\!\leftrightarrow\!\mathbf A\) with \( \mathbf A'=\mathbf R\,\mathbf A\,\mathbf R^{\!T} \), \( \mathbf R=[\alpha_{ij}] \).

2) Example — Dyadic (outer) Product

Given vectors \(a_i\) and \(b_j\), define the dyad (a 2nd-order tensor)

\[ t_{ij} = a_i\,b_j. \]

Matrix form (rows \(i\), columns \(j\)):

\[ \mathbf T = \begin{bmatrix} a_1 b_1 & a_1 b_2 & a_1 b_3\\ a_2 b_1 & a_2 b_2 & a_2 b_3\\ a_3 b_1 & a_3 b_2 & a_3 b_3 \end{bmatrix}. \]

Proof that \(t_{ij}\) is a tensor

\[ t'_{ij} = a'_i b'_j = (\alpha_{i m} a_m)(\alpha_{j n} b_n) = \alpha_{i m}\,\alpha_{j n}\,(a_m b_n) = \alpha_{i m}\,\alpha_{j n}\,t_{mn}. \]

How it acts on a vector (contractions)

\(t_{ij} b_j = (b_k b_k)\, a_i\) (vector). Free index \(i\) remains.
\(a_i t_{ij} = (a_k a_k)\, b_j\) (vector). Free index \(j\) remains.
\(t_{ii} = a_i b_i = \vec a\cdot \vec b\) (scalar trace).

3) Isotropic Tensors (unchanged by any rotation)

\(t'_{ij}=t_{ij}\) for all orthogonal \(\alpha\) ⇒ components are the same in every orthonormal basis.

2nd order: \(\delta_{ij}\) (identity). Check: \(\delta'_{ij}=\alpha_{i m}\alpha_{j n}\delta_{mn}=\alpha_{i m}\alpha_{j m}=\delta_{ij}\).
3rd order: \(\epsilon_{ijk}\) (permutation). Using \(\det(\alpha)=\pm 1\): \(\epsilon'_{ijk}=\det(\alpha)\,\epsilon_{ijk}\), unchanged for proper rotations (\(+1\)).

Epsilon–delta relations

\[ \epsilon_{kij}\,\epsilon_{kmn}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm},\qquad \epsilon_{ijk}\,\epsilon_{pqk}=\delta_{ip}\delta_{jq}-\delta_{iq}\delta_{jp}. \]

4) Fourth-Order Isotropic Tensor

General isotropic form (common in linear elasticity):

\[ S_{ijkl} = \lambda\,\delta_{ij}\delta_{kl} + \mu\,(\delta_{ik}\delta_{j\ell}+\delta_{i\ell}\delta_{jk}) + \kappa\,(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}). \]

Constants \(\lambda,\mu,\kappa\) are material parameters. In classical elasticity for symmetric stresses/strains, the antisymmetric part vanishes (\(\kappa=0\)); we use the Lamé constants \(\lambda,\mu\) (\(\mu\) is the shear modulus \(G\)).

Why three terms?

\(\delta_{ij}\delta_{kl}\): couples traces (volumetric part).
\(\delta_{ik}\delta_{j\ell}+\delta_{i\ell}\delta_{jk}\): symmetric identity on 2nd-order tensors.
\(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}\): antisymmetric part (drops for symmetric fields).

5) How Tensors Act — Contractions & Shapes

Operation	Index form	Shape / comment
2nd-order on vector	\( c_i = a_{ij} b_j \)	\(\mathbf A\vec b\): \(3\times3\) with \(3\times1 \to 3\times1\)
Vector on 2nd-order	\( d_j = b_i a_{ij} \)	row \(\vec b^{\!\top}\mathbf A\): \(1\times3\) with \(3\times3 \to 1\times3\)
Double contraction	\( a_{ij} b_{ij} = \mathbf A\!:\!\mathbf B \)	scalar; also \(a_{ii}=\mathrm{tr}(\mathbf A)\)
4th-order on 2nd-order	\( c_{ij}=S_{ijkl}\,a_{kl} \)	stiffness/identity-type on \(\mathbf A\); free \(i,j\) preserved

Index hygiene: each dummy index appears exactly twice; free indices match on both sides. Examples: \(a_{ij} b_{ij}\) (dummy \(i,j\)); \(c_{ij}=S_{ijkl}a_{kl}\) (free \(i,j\)).

6) Physical Examples (names + indices)

Displacement gradient: \(u_{i,j}=\partial u_i/\partial x_j\) (2nd-order, not necessarily symmetric).
Small strain: \(\varepsilon_{ij}=\tfrac12(u_{i,j}+u_{j,i})\) (symmetric).
Stress: \(\sigma_{ij}\) (symmetric for no body couples).
Hooke (isotropic): \(\sigma_{ij}=\lambda\,\varepsilon_{kk}\,\delta_{ij}+2\mu\,\varepsilon_{ij}\).

Here \(\varepsilon_{kk}=\mathrm{tr}(\boldsymbol\varepsilon)\) and \(G=\mu\), \(E=\mu\,\frac{3\lambda+2\mu}{\lambda+\mu}\), \(\nu=\frac{\lambda}{2(\lambda+\mu)}\).

7) Tiny Numeric Sanity (2D slice)

Setup: let \(\vec a=(2,1,0)\), \(\vec b=(1,-1,0)\). Then \(t_{ij} = a_i b_j\):

\[ T = \begin{bmatrix} 2\cdot 1 & 2\cdot(-1) & 0 \\ 1\cdot 1 & 1\cdot(-1) & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

Now multiply by \(\vec b\):

\[ T \vec b = \begin{bmatrix} 2 & -2 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \\ 0 \end{bmatrix} = (\|\vec b\|^2)\,\vec a \]

Matches the contraction rule: \(t_{ij} b_j = (b_k b_k)\,a_i\).

Quick Crib (who’s who in a formula?)

\( a'_i = \alpha_{ij} a_j \) free: \(i\); summed: \(j\). (Vector change of basis.)
\( a'_{ij} = \alpha_{im}\alpha_{jn} a_{mn} \) free: \(i,j\); summed: \(m,n\). (Matrix change of basis.)
\( c_i = a_{ij} b_j \) free: \(i\); summed: \(j\). (Matrix–vector product.)
\( s = a_{ij} b_{ij} \) all summed. (Scalar double contraction.)
\( c_{ij} = S_{ij k\ell}\, a_{k\ell} \) free: \(i,j\); summed: \(k,\ell\). (4th on 2nd order.)

How \( \epsilon_{ijk} \) & \( \delta_{ij} \) Power Linear Elasticity

This section connects Levi–Civita \( \epsilon_{ijk} \) and Kronecker delta \( \delta_{ij} \) directly to kinematics, constitutive laws, vector identities, and the Navier–Cauchy equations.

Variable & Index Legend

Indices

\(i,j,k,\ell,m,n,p,q,r,s\in\{1,2,3\}\). Repeated indices are summed (Einstein).

\(u_i(x)\)

Displacement components of a solid point at position \(x\).

\(u_{i,j}\)

Partial derivative \( \partial u_i/\partial x_j \) (displacement gradient).

\(\epsilon_{ijk}\)

Levi–Civita symbol (permutation): \(+1\) for even, \(-1\) for odd permutations of \(123\), \(0\) otherwise.

\(\delta_{ij}\)

Kronecker delta: \(1\) if \(i=j\), else \(0\). Identity matrix entries.

\(\varepsilon_{ij}\)

Small (engineering) strain tensor: symmetric part of \(u_{i,j}\).

\(W_{ij}\)

Infinitesimal rotation tensor: antisymmetric part of \(u_{i,j}\).

\(\omega_k\)

Rotation (vorticity) vector associated with \(W_{ij}\).

\(\sigma_{ij}\)

Cauchy stress tensor (symmetric for no body couples).

\(\lambda,\mu\)

Lamé constants (material parameters). \( \mu=G \) is shear modulus.

\(f_i\)

Body force per unit volume (e.g., gravity components).

Kinematics: Splitting the Displacement Gradient

Start with the gradient of displacement, \(L_{ij}:=u_{i,j}\). Split it into symmetric (strain) and antisymmetric (rotation) parts:

\[ L_{ij} = u_{i,j} = \underbrace{\tfrac12\big(u_{i,j}+u_{j,i}\big)}_{\varepsilon_{ij}\ \text{(strain)}} + \underbrace{\tfrac12\big(u_{i,j}-u_{j,i}\big)}_{W_{ij}\ \text{(infinitesimal rotation)}}. \]

Why this split? Any square matrix equals its symmetric part \((A+A^T)/2\) plus its antisymmetric part \((A-A^T)/2\). Apply with \(A=L=[u_{i,j}]\).

The Antisymmetric Part Encoded with \( \epsilon_{ijk} \)

The curl and the rotation tensor are tied together by \( \epsilon_{ijk} \):

\[ (\nabla\times \mathbf u)_i \equiv (\mathrm{curl}\,u)_i = \epsilon_{ijk}\,u_{k,j}. \] \[ W_{ij}=\tfrac12\big(u_{i,j}-u_{j,i}\big) \quad\Longleftrightarrow\quad W_{ij}=\tfrac12\,\epsilon_{ijk}\,\omega_k, \ \ \omega_k=\epsilon_{k\ell m}\,W_{\ell m}. \]

Derivation sketch

Define \( \omega_k := \epsilon_{kij}\,u_{j,i} \) (equals \( (\mathrm{curl}\,u)_k \)).
\( \tfrac12 \epsilon_{ijk}\,\omega_k = \tfrac12 \epsilon_{ijk}\epsilon_{k\ell m} u_{m,\ell} \).
Use \( \epsilon_{ijk}\epsilon_{k\ell m}=\delta_{i\ell}\delta_{j m}-\delta_{i m}\delta_{j\ell} \).
Get \( \tfrac12(u_{j,i}-u_{i,j}) = -W_{ji}=W_{ij} \) since \(W\) is antisymmetric.

Memorize: \(\epsilon\epsilon=\delta\delta-\delta\delta\). It turns curls/rotations into deltas (and vice versa).

Isotropic Elasticity Tensor (Hooke’s Law)

Rotational invariance + minor symmetries \( C_{ijkl}=C_{jikl}=C_{ijlk} \) restrict the 4th-order stiffness \(C\) to the form:

\[ C_{ijkl}=\lambda\,\delta_{ij}\delta_{kl} +\mu\big(\delta_{ik}\delta_{j\ell}+\delta_{i\ell}\delta_{jk}\big). \]

Hooke’s law for small-strain isotropic elasticity is then

\[ \sigma_{ij}=C_{ijkl}\,\varepsilon_{kl} = \lambda\,\delta_{ij}\,\varepsilon_{kk} + 2\mu\,\varepsilon_{ij}. \]

Meaning of the two terms

Volumetric (hydrostatic): \( \lambda\,\varepsilon_{kk}\,\delta_{ij}\) depends only on the trace \( \varepsilon_{kk} \) (dilation).
Deviatoric (shear): \( 2\mu\,\varepsilon_{ij}\) controls shape change at fixed volume.

Relating \(E,\nu\) and \(\lambda,\mu\)

\[ \mu=G=\frac{E}{2(1+\nu)},\qquad \lambda=\frac{E\,\nu}{(1+\nu)(1-2\nu)}. \]

Useful \( \epsilon\)–\( \delta \) Contractions (Keep These Handy)

\[ \epsilon_{ijk}\epsilon_{ijk}=6, \qquad \epsilon_{ijk}\epsilon_{ijl}=2\,\delta_{kl}, \qquad \epsilon_{ijk}\epsilon_{mnk}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}. \]

Example: vector identity for \( \nabla\times(\nabla\times u) \)

\[ (\nabla\times(\nabla\times u))_i = \epsilon_{ijk}\partial_j(\epsilon_{k\ell m}\partial_\ell u_m) = (\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell})\partial_j\partial_\ell u_m = \partial_i(\partial_m u_m) - \partial_\ell\partial_\ell u_i. \] \[ \Rightarrow\ \nabla\times(\nabla\times u)=\nabla(\nabla\!\cdot u)-\nabla^2 u. \]

Navier–Cauchy Equations (Equilibrium with Body Force \(f_i\))

Start with equilibrium (no inertia): \( \sigma_{ij,j}+f_i=0 \). Substitute Hooke’s law \( \sigma_{ij}=\lambda\varepsilon_{kk}\delta_{ij}+2\mu\varepsilon_{ij} \), and use \( \varepsilon_{ij}=\tfrac12(u_{i,j}+u_{j,i})\):

\[ \sigma_{ij,j} = \lambda\,\partial_j(\varepsilon_{kk}\delta_{ij}) + 2\mu\,\partial_j\varepsilon_{ij} = \lambda\,\partial_i \varepsilon_{kk} + \mu\,\partial_j(u_{i,j}+u_{j,i}) = \lambda\,\partial_i u_{k,k} + \mu\,u_{i,jj} + \mu\,u_{j,ij}. \] \[ \Rightarrow\quad \mu\,u_{i,kk} + (\lambda+\mu)\,u_{k,ki} + f_i = 0. \]

Vector form

\[ \mu\,\nabla^2 \mathbf u + (\lambda+\mu)\,\nabla(\nabla\!\cdot \mathbf u) + \mathbf f = \mathbf 0. \]

Using the identity above, you can also express it with curls if convenient.

Rotation Vectors & Moments Use \( \epsilon_{ijk} \)

Moments of a force \( \mathbf F \) about position \( \mathbf x \) are encoded via Levi–Civita:

\[ M_i = \epsilon_{ijk}\,x_j\,F_k \quad\Longleftrightarrow\quad \mathbf M = \mathbf x \times \mathbf F. \]

Same symbol appears in angular momentum, torque, vorticity, and many skew-symmetric constructions.

Common Pitfalls (and quick fixes)

Index mismatch: Free indices must match on both sides; each dummy index appears exactly twice per term.
Sign errors with \( \epsilon_{ijk} \): Swap two indices → change sign. Check permutations against \(123\).
Forgetting symmetry: In small-strain isotropic elasticity, \(\sigma_{ij}=\sigma_{ji}\) and \(\varepsilon_{ij}=\varepsilon_{ji}\).
Using wrong identity: Keep \( \epsilon_{ijk}\epsilon_{mnk}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm} \) handy; it drives most reductions.

General → Isotropic Hooke, Strain–Displacement, & Navier–Cauchy

General linear elasticity → isotropic reduction → Hooke’s law in \((\lambda,\mu)\) → strain from displacement → equilibrium → Navier–Cauchy PDE.

1) Hooke’s Law (General Linear Elasticity)

The most general linear relation between stress and strain is

\[\sigma_{ij}=C_{ijkl}\,\varepsilon_{kl}.\]

\(\sigma_{ij}\)

Stress tensor (force/area). Index \(i\): force direction; \(j\): plane normal.

\(\varepsilon_{kl}\)

Strain tensor (from displacement gradients). \(k\): displacement component; \(l\): derivative direction.

\(C_{ijkl}\)

4th-order elasticity tensor (linear map: strain → stress).

Einstein summation: any index repeated exactly twice is summed over \(1,2,3\).

2) Specializing to an Isotropic Solid

Isotropy forces \(C\) to combine Kronecker deltas:

\[ C_{ijkl}=\lambda\,\delta_{ij}\delta_{kl} +\mu\big(\delta_{ik}\delta_{j\ell}+\delta_{i\ell}\delta_{jk}\big). \]

\(\delta_{ij}\)

Kronecker delta: \(1\) if \(i=j\), else \(0\).

\(\lambda,\mu\)

Lamé constants. \(\displaystyle \mu=\frac{E}{2(1+\nu)}\), \(\displaystyle \lambda=\frac{E\nu}{(1+\nu)(1-2\nu)}\).

3) Hooke’s Law in Isotropic Form

Using \(\varepsilon_{kk}=\nabla\!\cdot u\):

\[\sigma_{ij}=\lambda\,\varepsilon_{kk}\,\delta_{ij}+2\mu\,\varepsilon_{ij}.\]

4) Strain from Displacement

\[ \varepsilon_{ij}=\tfrac12\!\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right) =\tfrac12\,(u_{i,j}+u_{j,i}). \]

5) Equilibrium (Navier–Cauchy)

\[ \sigma_{ij,j}+f_i=0,\qquad \sigma_{ij,j}=\frac{\partial\sigma_{ij}}{\partial x_j}. \]

\[ \mu\,\nabla^2 u_i + (\lambda+\mu)\,\partial_i(\nabla\!\cdot u)+f_i=0, \qquad \mu\nabla^2\mathbf u+(\lambda+\mu)\nabla(\nabla\!\cdot\mathbf u)+\mathbf f=\mathbf0. \]

6) Variable Glossary

\(\sigma_{ij}\)

Stress tensor.

\(\varepsilon_{ij}\)

Strain tensor.

\(u_i\)

Displacement components.

\(\delta_{ij}\)

Kronecker delta.

\(\lambda,\mu\)

Lamé constants.

\(E,\nu\)

Young’s modulus, Poisson’s ratio.

\(\nabla\!\cdot u\)

Volumetric strain \((\varepsilon_{kk})\).

\(\nabla^2 u\)

Laplacian.

\(f_i\)

Body force per volume.

From 1D Bar Law → Full 3D Hooke’s Law

Start simple, add Poisson coupling, then arrive at the compact tensor form.

1) The “Simple” 1D Hooke’s Law

\[ \sigma=E\,\varepsilon,\qquad \sigma=\tfrac{F}{A},\;\varepsilon=\tfrac{\Delta L}{L}. \]

\(\sigma\)

Normal stress (force/area).

\(\varepsilon\)

Normal strain (elongation/original length).

\(E\)

Young’s modulus.

2) Including Lateral Contraction (Poisson)

\[\varepsilon_{\perp}=-\nu\,\varepsilon_{\parallel}.\]

3) 3D Generalization (Isotropic Tensor Form)

\[\sigma_{ij}=\lambda\,\varepsilon_{kk}\delta_{ij}+2\mu\,\varepsilon_{ij}.\]

4) Why It Feels Different

Intro courses show \(\sigma=E\varepsilon\). Elasticity needs indices + Lamé constants to handle combined normal/shear states and equilibrium in 3D.

1D Hooke’s Law → 3D with Poisson & Shear

Both “strain from stress” and “stress from strain,” plus shear with \(G\).

1) 1D Hooke (recap)

\[\sigma_x=E\,\varepsilon_x,\qquad \sigma_x=\frac{F}{A},\;\varepsilon_x=\frac{\Delta L}{L}.\]

2) 3D (strain from stress)

\[ \varepsilon_x = \frac{1}{E}\big(\sigma_x-\nu(\sigma_y+\sigma_z)\big),\; \varepsilon_y = \frac{1}{E}\big(\sigma_y-\nu(\sigma_x+\sigma_z)\big),\; \varepsilon_z = \frac{1}{E}\big(\sigma_z-\nu(\sigma_x+\sigma_y)\big). \]

\[ \gamma_{xy}=\frac{\tau_{xy}}{G},\quad \gamma_{yz}=\frac{\tau_{yz}}{G},\quad \gamma_{zx}=\frac{\tau_{zx}}{G},\qquad G=\frac{E}{2(1+\nu)}. \]

3) 3D (stress from strain)

\[ \sigma_x=\frac{E}{(1+\nu)(1-2\nu)}\Big[(1-\nu)\varepsilon_x+\nu(\varepsilon_y+\varepsilon_z)\Big], \] \[ \sigma_y=\frac{E}{(1+\nu)(1-2\nu)}\Big[(1-\nu)\varepsilon_y+\nu(\varepsilon_x+\varepsilon_z)\Big], \] \[ \sigma_z=\frac{E}{(1+\nu)(1-2\nu)}\Big[(1-\nu)\varepsilon_z+\nu(\varepsilon_x+\varepsilon_y)\Big]. \]

\[\tau_{xy}=G\,\gamma_{xy},\quad \tau_{yz}=G\,\gamma_{yz},\quad \tau_{zx}=G\,\gamma_{zx}.\]

4) Variable Glossary

Symbol	Meaning
\(E\)	Young’s modulus.
\(\nu\)	Poisson’s ratio.
\(G=\dfrac{E}{2(1+\nu)}\)	Shear modulus.
\(\sigma_i\)	Normal stresses.
\(\tau_{ij}\)	Shear stresses.
\(\varepsilon_i\)	Normal strains.
\(\gamma_{ij}\)	Engineering shear strains.

Hooke’s Law: 1D vs 3D (Isotropic Linear Elasticity)

Side-by-side comparison using the same symbols as above.

Concept	1D Bar	3D Isotropic Solid
Stress–Strain Law	\[\sigma=E\,\varepsilon\]	\[\sigma_{ij}=\lambda\,\varepsilon_{kk}\,\delta_{ij}+2\mu\,\varepsilon_{ij}\] Normal components (using \(E,\nu\)): \[ \sigma_x=\frac{E}{(1+\nu)(1-2\nu)}\big[(1-\nu)\varepsilon_x+\nu(\varepsilon_y+\varepsilon_z)\big], \] and cyclic permutations.
Axial strain	\[\varepsilon=\frac{\Delta L}{L}\]	\[\varepsilon_x=\frac{1}{E}\big(\sigma_x-\nu(\sigma_y+\sigma_z)\big)\]
Stress definition	\[\sigma=\frac{F}{A}\]	\(\sigma_{ij}\): Cauchy stress tensor.
Strain definition	\[\varepsilon=\frac{\Delta L}{L}\]	\[\varepsilon_{ij}=\tfrac12(u_{i,j}+u_{j,i})\]
Poisson effect	\[\varepsilon_{\perp}=-\nu\,\varepsilon\]	Built-in via coupling among directions.
Shear	Not in 1D	\[\gamma_{ij}=\tau_{ij}/G,\; G=E/2(1+\nu)\]
Material constants	\(E\)	\(E,\nu,G,\lambda,\mu\)

Key Takeaways

1D: \( \sigma=E\varepsilon \).
3D: \( \sigma_{ij}=\lambda\varepsilon_{kk}\delta_{ij}+2\mu\varepsilon_{ij} \) with Poisson + shear.

Calculus of Vector & Tensor Fields

Scalar fields → \(d\vec s\) → gradient & directional derivative → divergence theorem.

1) Scalar Field

\[ f(x,y,z)\equiv f(x_i). \]

2) Infinitesimal Vector Element

\[ d\vec s = dx_i\,\vec e_i,\qquad ds=\sqrt{dx_i\,dx_i},\qquad \hat n=\frac{d\vec s}{ds}=\frac{dx_i}{ds}\,\vec e_i. \]

3) Gradient & Directional Derivative

\[ df=\frac{\partial f}{\partial x_i}\,dx_i =\Big(\frac{\partial f}{\partial x_i}\vec e_i\Big)\!\cdot d\vec s,\qquad \nabla f=\frac{\partial f}{\partial x_i}\,\vec e_i,\qquad \frac{df}{ds}=\nabla f\cdot \hat n. \]

4) Divergence Theorem

\[ \int_V \frac{\partial f}{\partial x_i}\,dV=\int_S f\,n_i\,dS, \qquad \int_V \nabla\!\cdot \vec v \, dV=\int_S \vec v\cdot \hat n\, dS = \int_S v_i n_i\, dS. \]

📖 Variable Glossary

Symbol	Meaning
\(x_i\)	Cartesian coordinates (\(i=1,2,3\)).
\(\vec e_i\)	Orthonormal basis vectors.
\(f(x_i)\)	Scalar field.
\(d\vec s, ds\)	Infinitesimal displacement vector & magnitude.
\(\hat n\)	Unit tangent/normal along the path.
\(\nabla f\)	Gradient of \(f\).
\(\vec v=v_i\vec e_i\)	Vector field.
\(\nabla\!\cdot\vec v=\partial v_i/\partial x_i\)	Divergence.
\(n_i, V, S\)	Surface normal components, volume, boundary surface.

Deformation in Continuum Mechanics – Explanation

1. Viewpoint of Continuum Mechanics

Assumption 1: Materials are continuously divisible → no atomic/grain reference.
Assumption 2: Properties are defined at a mathematical point.

This enables use of calculus and field equations to describe materials.

2. Rigid vs. Deformable

If distance \(d\) between two points is constant → body is rigid.

In reality, all materials are deformable → distances change under loading.

3. Displacement Definition

Material point with reference coordinates:

\[ x_i \;\to\; x_i' \quad\Rightarrow\quad u_i \,=\, x_i' - x_i . \]

In general (fields defined on the reference coordinates \(x_m\)):

\[ u_i \;=\; u_i(x_m), \qquad x_i' \;=\; x_i'(x_m). \]

Displacement depends on original position \(x_m\).

4. Admissible Deformation

Integral transformation under coordinate change:

\[ \int f(x',y',z')\, dV' \;=\; \int f\!\big(x'(x,y,z),\,y'(x,y,z),\,z'(x,y,z)\big)\, J\, dV . \]

Jacobian determinant:

\[ J \;=\; \det\!\left(\frac{\partial x_i'}{\partial x_j}\right). \]

\(J > 0\): admissible (no overlap).
\(J = 1\): volume preserved.
\(J \neq 1\): compression or expansion.

5. Example: Simple Shear Deformation

Mapping:

\[ x_1' = x_1 + \alpha\, x_2, \qquad x_2' = x_2 + \alpha\, x_1, \qquad x_3' = x_3 . \]

Jacobian matrix and determinant:

\[ J \;=\; \begin{vmatrix} 1 & \alpha & 0 \\ \alpha & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \;=\; 1 - \alpha^{2}. \]

Admissibility: \(J>0 \Rightarrow -1<\alpha<1\).

6. Worked Example

Take \(\alpha=\tfrac{1}{2}\):

\[ x_1' = 1 + \tfrac{1}{2}\cdot 1 = 1.5, \qquad x_2' = 1 + \tfrac{1}{2}\cdot 1 = 1.5 . \]

So point \((1,1)\) maps to \((1.5,1.5)\).

Continuity is preserved because \(J>0\).

📖 Variable Glossary

Symbol	Meaning
\(x_i\)	Reference (undeformed) coordinates of a material point.
\(x_i'\)	Deformed (current) coordinates after loading.
\(u_i = x_i'-x_i\)	Displacement vector components.
\(d\)	Distance between two material points.
\(f(x,y,z)\)	Scalar field defined on the domain.
\(dV,\; dV'\)	Volume element in undeformed / deformed configuration.
\(J=\det\!\big(\partial x_i'/\partial x_j\big)\)	Jacobian determinant (volume scaling).
\(\alpha\)	Shear parameter (controls deformation intensity).

Key Takeaways

Jacobian \(J\) links undeformed to deformed volumes.
Displacement \(u_i\) tracks motion of material points.
Admissibility requires \(J>0\).
Shear mapping example shows geometric limits \((-1<\alpha<1)\).

From Deformation Mapping → Small Strain (and Finite Strain)

We go from the kinematic map \(x' = x + u(x)\) to the small (engineering) strain \( \varepsilon = \tfrac12(\nabla u + \nabla u^{T}) \), show the finite-strain alternative via Green–Lagrange \( \mathbf E \), and evaluate both for the symmetric shear example.

1) Kinematics

Mapping (current coordinates as functions of reference coordinates):

\[ x_i' = x_i + u_i(x), \qquad u_i = x_i' - x_i. \]

Deformation gradient and displacement gradient:

\[ \mathbf F = \frac{\partial x'}{\partial x} = \mathbf I + \nabla u, \qquad F_{ij} = \delta_{ij} + u_{i,j}, \quad u_{i,j} \equiv \frac{\partial u_i}{\partial x_j}. \]

2) Small (Engineering) Strain

For small deformations/rotations \( \|\nabla u\| \ll 1 \):

\[ \varepsilon_{ij} = \tfrac12\,(u_{i,j} + u_{j,i}) \quad\Longleftrightarrow\quad \boldsymbol\varepsilon = \tfrac12\left(\nabla u + (\nabla u)^T\right). \]

Volumetric strain (dilatation) and rotation (skew) split:

\[ \varepsilon_v = \mathrm{tr}\,\boldsymbol\varepsilon = \varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}, \qquad \omega_{ij} = \tfrac12\,(u_{i,j} - u_{j,i}), \qquad \nabla u = \boldsymbol\varepsilon + \boldsymbol\omega. \]

Jacobian (volume change) to first order:

\[ J = \det \mathbf F \;\approx\; 1 + \mathrm{tr}(\nabla u) \;=\; 1 + \varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}. \]

3) Finite-Strain Alternative (no smallness assumption)

Right Cauchy–Green tensor and Green–Lagrange strain:

\[ \mathbf C = \mathbf F^T \mathbf F, \qquad \mathbf E = \tfrac12(\mathbf C - \mathbf I) = \tfrac12\!\left(\nabla u + (\nabla u)^T + (\nabla u)^T\nabla u\right). \]

If quadratic terms are negligible, \( \mathbf E \to \boldsymbol\varepsilon \).

4) Plug-in Your Mapping (Symmetric Shear)

Mapping:

\[ x_1' = x_1 + \alpha x_2,\qquad x_2' = x_2 + \alpha x_1,\qquad x_3' = x_3. \]

Displacement and gradients:

\[ u_1=\alpha x_2,\quad u_2=\alpha x_1,\quad u_3=0; \qquad u_{1,1}=0,\; u_{1,2}=\alpha,\; u_{2,1}=\alpha,\; u_{2,2}=0,\; u_{3,j}=0. \]

Small-strain tensor

\[ \varepsilon_{11}=0,\quad \varepsilon_{22}=0,\quad \varepsilon_{33}=0,\quad \varepsilon_{12}=\varepsilon_{21}=\tfrac12(\alpha+\alpha)=\alpha. \] \[ \gamma_{12}=2\varepsilon_{12}=2\alpha,\qquad \omega_{12}=\tfrac12(\alpha-\alpha)=0,\qquad \varepsilon_v=\mathrm{tr}\,\boldsymbol\varepsilon=0. \]

Infinitesimally incompressible (\( \varepsilon_v=0 \)) and no rigid rotation to first order.

Finite (Green–Lagrange) strain

With \( \nabla u = \begin{bmatrix} 0 & \alpha & 0\\ \alpha & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\), we have \( (\nabla u)^T\nabla u = \begin{bmatrix} \alpha^2 & 0 & 0\\ 0 & \alpha^2 & 0\\ 0 & 0 & 0 \end{bmatrix}\).

\[ \mathbf E = \tfrac12 \begin{bmatrix} 0 & 2\alpha & 0\\ 2\alpha & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} + \tfrac12 \begin{bmatrix} \alpha^2 & 0 & 0\\ 0 & \alpha^2 & 0\\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \tfrac{\alpha^2}{2} & \alpha & 0\\ \alpha & \tfrac{\alpha^2}{2} & 0\\ 0 & 0 & 0 \end{bmatrix}. \]

Finite shear produces normal strains \(E_{11}=E_{22}=\alpha^2/2\). Dropping \(O(\alpha^2)\) terms recovers the small-strain result.

Jacobian for this mapping: \( \mathbf F = \begin{bmatrix} 1 & \alpha & 0\\ \alpha & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \Rightarrow J=\det\mathbf F=1-\alpha^2 \approx 1 \) for small \(|\alpha|\).

5) Quick Checklist

\(\mathbf F = \mathbf I + \nabla u\).
\(\boldsymbol\varepsilon = \tfrac12(\nabla u + \nabla u^{T})\) (small strain).
\(\boldsymbol\omega = \tfrac12(\nabla u - \nabla u^{T})\) (rotation).
\(J \approx 1 + \mathrm{tr}\,\boldsymbol\varepsilon\) (to first order).
Symmetric shear: \(\varepsilon_{12}=\alpha\), \(\varepsilon_v=0\), \(J\approx 1\).

📖 Variable Glossary

Symbol	Meaning
\(x_i\)	Reference (undeformed) coordinates.
\(x_i'\)	Current (deformed) coordinates.
\(u_i = x_i'-x_i\)	Displacement components.
\(u_{i,j}=\partial u_i/\partial x_j\)	Displacement gradient components.
\(\nabla u\)	Displacement gradient tensor \([\;u_{i,j}\;]\).
\(\mathbf F=\partial x'/\partial x=\mathbf I+\nabla u\)	Deformation gradient.
\(F_{ij}=\delta_{ij}+u_{i,j}\)	Components of \(\mathbf F\).
\(\delta_{ij}\)	Kronecker delta (identity components).
\(\boldsymbol\varepsilon\)	Small (engineering) strain tensor, \(\varepsilon_{ij}=\tfrac12(u_{i,j}+u_{j,i})\).
\(\varepsilon_v=\mathrm{tr}\,\boldsymbol\varepsilon\)	Volumetric strain (dilatation).
\(\boldsymbol\omega\)	Infinitesimal rotation tensor, \(\omega_{ij}=\tfrac12(u_{i,j}-u_{j,i})\).
\(J=\det\mathbf F\)	Jacobian determinant (volume change factor).
\(\mathbf C=\mathbf F^T\mathbf F\)	Right Cauchy–Green tensor.
\(\mathbf E=\tfrac12(\mathbf C-\mathbf I)\)	Green–Lagrange strain tensor.
\(\alpha\)	Shear parameter in the symmetric shear mapping.
\(\gamma_{ij}=2\varepsilon_{ij}\)	Engineering shear strain.
\(\mathrm{tr}(\cdot)\)	Trace operator (sum of diagonal entries).
\((\cdot)^T\)	Transpose of a tensor/matrix.

Displacement Gradient → Strain & Rotation → Finite Strains

We build from the displacement field \(u_i(x)\) to its gradient \(\nabla u\), split it into strain (symmetric) and rotation (antisymmetric), derive strain from line-element changes, and then present finite (geometric) strain measures.

1) Displacement Gradient

Think of the gradient operator acting on each component of the displacement vector:

\[ \nabla \mathbf u \;=\; \vec e_1\,\frac{\partial}{\partial x_1} +\vec e_2\,\frac{\partial}{\partial x_2} +\vec e_3\,\frac{\partial}{\partial x_3} \quad \text{applied to} \quad \mathbf u=u_j\,\vec e_j. \] \[ \Rightarrow\quad \nabla \mathbf u \;=\; \vec e_i \vec e_j \,\frac{\partial u_j}{\partial x_i} \;\;\equiv\;\; L_{ij}\,\vec e_i \vec e_j,\qquad L_{ij}=\frac{\partial u_j}{\partial x_i}. \]

Here \(L_{ij}\) are the components of the (second-order) displacement-gradient tensor \(\nabla \mathbf u\).

2) Decomposition into Strain and Rotation

Any second-order tensor splits uniquely into symmetric + antisymmetric parts:

\[ D_{ij}=\tfrac12\,(L_{ij}+L_{ji}) \quad\text{(symmetric, strain-like)}, \qquad W_{ij}=\tfrac12\,(L_{ij}-L_{ji}) \quad\text{(antisymmetric, rotation-like)}. \] \[ \Rightarrow\quad \nabla \mathbf u \;=\; \mathbf D + \mathbf W, \qquad D_{ij}=D_{ji},\; W_{ij}=-W_{ji}. \]

In components (matrices):

\[ \mathbf D = \begin{bmatrix} u_{1,1} & \tfrac12(u_{1,2}+u_{2,1}) & \tfrac12(u_{1,3}+u_{3,1})\\[4pt] \tfrac12(u_{2,1}+u_{1,2}) & u_{2,2} & \tfrac12(u_{2,3}+u_{3,2})\\[4pt] \tfrac12(u_{3,1}+u_{1,3}) & \tfrac12(u_{3,2}+u_{2,3}) & u_{3,3} \end{bmatrix}, \quad \mathbf W = \begin{bmatrix} 0 & \tfrac12(u_{1,2}-u_{2,1}) & \tfrac12(u_{1,3}-u_{3,1})\\[4pt] \tfrac12(u_{2,1}-u_{1,2}) & 0 & \tfrac12(u_{2,3}-u_{3,2})\\[4pt] \tfrac12(u_{3,1}-u_{1,3}) & \tfrac12(u_{3,2}-u_{2,3}) & 0 \end{bmatrix}. \]

Interpretation: \(\mathbf D\) measures shape/volume change (strain); \(\mathbf W\) encodes rigid-body rotation (no strain energy in linear theory).

Examples: \(u_{1,1}=\varepsilon_{11}\) (normal strain along \(x_1\)); \(\varepsilon_{12}=\tfrac12(u_{2,1}+u_{1,2})\) (engineering shear/2); \(\omega_3=\tfrac12(u_{2,1}-u_{1,2})\) (small rotation about \(x_3\)).

3) Strain from Line-Element Differences

Let \((ds)^2=dx_i\,dx_i\) be the squared length in the reference configuration and \((ds')^2=dx_i' dx_i'\) in the deformed configuration with \(x_i'=x_i+u_i\).

\[ (ds')^2-(ds)^2 = d(x_i+u_i)\,d(x_i+u_i)-dx_i\,dx_i = 2\,u_{i,j}\,dx_j\,dx_i \;+\; u_{i,m}u_{i,n}\,dx_m dx_n. \] \[ = \big(u_{j,i}+u_{i,j}+u_{k,i}u_{k,j}\big)\,dx_i\,dx_j. \]

This motivates the geometric strain tensor (exact, no linearization):

\[ \varepsilon_{ij}^{\text{geom}} = \tfrac12\Big(u_{j,i}+u_{i,j}+u_{k,i}\,u_{k,j}\Big). \]

Neglecting quadratic terms \(u_{k,i}u_{k,j}\) yields the familiar small (engineering) strain \(\varepsilon_{ij}=\tfrac12(u_{i,j}+u_{j,i})\).

4) Finite Strain Measures (Beyond Small-Strain)

Introduce the deformation gradient \(\mathbf F = \partial x'/\partial x = \mathbf I + \nabla u\).

Green–Saint-Venant (reference-based)

\[ \mathbf C = \mathbf F^T \mathbf F,\qquad \mathbf E = \tfrac12(\mathbf C - \mathbf I) = \tfrac12\!\left(\nabla u + (\nabla u)^T + (\nabla u)^T\nabla u\right). \]

\(\mathbf E\) measures strain relative to the reference configuration; reduces to \(\mathbf D\) when \(\|\nabla u\|\ll1\).

Euler–Almansi (current-based)

\[ \mathbf b = \mathbf F\,\mathbf F^T,\qquad \mathbf e = \tfrac12\big(\mathbf I - \mathbf b^{-1}\big). \]

\(\mathbf e\) measures strain relative to the current configuration. For small strains \(\mathbf e\approx \mathbf D\).

Equivalence to line-element result: \(\varepsilon^{\text{geom}}_{ij}=\tfrac12(u_{j,i}+u_{i,j}+u_{k,i}u_{k,j})\) is exactly Green–Saint-Venant \(\mathbf E\) in components with \(F_{ij}=\delta_{ij}+u_{i,j}\).

5) Quick Checklist

\(\nabla \mathbf u = \mathbf D+\mathbf W\) with \( \mathbf D=\tfrac12(\nabla u + \nabla u^T)\), \( \mathbf W=\tfrac12(\nabla u - \nabla u^T)\).
Small (engineering) strain: \(\boldsymbol\varepsilon=\mathbf D\).
Line element: \( (ds')^2-(ds)^2 = 2\,\varepsilon_{ij}^{\text{geom}}\,dx_i dx_j\).
Finite strains: \( \mathbf E=\tfrac12(\mathbf F^T\mathbf F-\mathbf I)\), \( \mathbf e=\tfrac12(\mathbf I-\mathbf b^{-1})\).
For \(\|\nabla u\|\ll1\): \( \mathbf E \approx \mathbf e \approx \mathbf D \).

📖 Variable Glossary

Symbol	Meaning
\(x_i,\,x_i'\)	Reference and current (deformed) Cartesian coordinates of a material point.
\(\vec e_i\)	Orthonormal basis unit vectors along \(x_i\).
\(u_i(x)\)	Displacement components; \(\mathbf u = u_i \vec e_i\).
\(u_{i,j}=\partial u_i/\partial x_j\)	Displacement-gradient components.
\(\nabla \mathbf u=[u_{i,j}]\)	Displacement-gradient tensor (second order).
\(L_{ij}\)	Alternate notation for \(u_{j,i}\) (used in some texts).
\(\mathbf D\)	Symmetric part of \(\nabla u\); small (engineering) strain tensor when linearized.
\(\mathbf W\)	Antisymmetric part of \(\nabla u\); infinitesimal rotation tensor.
\(\varepsilon_{ij}\)	Small (engineering) strain: \(\tfrac12(u_{i,j}+u_{j,i})\).
\(\omega_{ij}\)	Infinitesimal rotation: \(\tfrac12(u_{i,j}-u_{j,i})\).
\(\mathbf F=\partial x'/\partial x=\mathbf I+\nabla u\)	Deformation-gradient tensor.
\(\mathbf C=\mathbf F^T\mathbf F\)	Right Cauchy–Green tensor (reference measure).
\(\mathbf b=\mathbf F\,\mathbf F^T\)	Left Cauchy–Green tensor (current measure).
\(\mathbf E=\tfrac12(\mathbf C-\mathbf I)\)	Green–Saint-Venant finite strain tensor.
\(\mathbf e=\tfrac12(\mathbf I-\mathbf b^{-1})\)	Euler–Almansi finite strain tensor.
\(\delta_{ij}\)	Kronecker delta (identity tensor components).
\((\cdot)^T\)	Transpose operator; \(\mathrm{tr}(\cdot)\): trace (sum of diagonal entries).
\(\\|\nabla u\\|\)	Norm of displacement gradient; “\(\ll1\)” means “much less than one.”

Engineering Stress, Cauchy Stress Tensor & Traction (with FCC example)

1) Engineering Stress

Average (engineering) normal stress on a cross-section of area \(A\) carrying force \(F\):

\[ \sigma = \frac{F}{A}. \]

Local (true) stress components are limits as the patch shrinks to a point:

\[ \text{Normal:}\quad \sigma = \lim_{\Delta A\to 0}\frac{F_n}{\Delta A}, \qquad \text{Shear:}\quad \tau = \lim_{\Delta A\to 0}\frac{F_t}{\Delta A}. \]

Same units as pressure (Pa = N/m\(^2\)).

2) Cauchy Stress Tensor

At a point, stress is a second-order tensor \(\boldsymbol\sigma=[\sigma_{ij}]\):

\[ \boldsymbol\sigma = \begin{bmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{bmatrix}, \qquad \sigma_{ij}=\sigma_{ji}\ \text{(no couples → symmetry)}. \]

Index \(i\): component (direction) of force. Index \(j\): normal direction of the plane on which that force acts.

3) Stress Vectors on Coordinate Planes

Stress vector on the plane normal to \(\vec e_1,\vec e_2,\vec e_3\):

\[ \vec\sigma_{1}=\sigma_{11}\vec e_1+\sigma_{12}\vec e_2+\sigma_{13}\vec e_3,\quad \vec\sigma_{2}=\sigma_{21}\vec e_1+\sigma_{22}\vec e_2+\sigma_{23}\vec e_3,\quad \vec\sigma_{3}=\sigma_{31}\vec e_1+\sigma_{32}\vec e_2+\sigma_{33}\vec e_3. \]

Because \(\boldsymbol\sigma\) is symmetric, there are 6 independent components.

4) Traction Vector (Cauchy’s Formula)

The traction (stress) vector on a plane with unit normal \(\vec n\) is linear in \(\vec n\):

\[ \vec T = \boldsymbol\sigma\,\vec n \quad\Longleftrightarrow\quad T_i = \sigma_{ij}\,n_j. \]

From equilibrium on an infinitesimal tetrahedron: summing face forces and using \(n_j=A_j/A\) yields \(T_i=\sigma_{ij}n_j\).

5) Application: Resolved Shear Stress in an FCC Crystal

Loading: Uniaxial tension along \([001]\) (the \(x_3\) axis). The Cauchy stress is

\[ \boldsymbol\sigma = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & \sigma \end{bmatrix}. \]

Slip system: \(\{111\}\langle 110\rangle\). Take plane normal and slip direction as

\[ \vec n = \frac{1}{\sqrt{3}}(1,1,1),\qquad \vec m = \frac{1}{\sqrt{2}}(-1,0,1), \]

Both \(\vec n\) and \(\vec m\) are unit vectors (required for Schmid’s law). Note \(n_3=1/\sqrt{3}\), \(m_3=1/\sqrt{2}\).

Resolved shear on the slip direction

Traction on the slip plane: \(\vec T=\boldsymbol\sigma\,\vec n\). Resolved shear on \(\vec m\):

\[ \tau = \vec T\cdot \vec m = (\boldsymbol\sigma\,\vec n)\cdot \vec m = n_j\,\sigma_{ji}\,m_i = \sigma_{33}\,n_3\,m_3 = \sigma\;\frac{1}{\sqrt{3}}\;\frac{1}{\sqrt{2}} = \frac{\sigma}{\sqrt{6}}. \]

Normal stress on the slip plane

\[ p = \vec n\cdot \boldsymbol\sigma\,\vec n = n_j\,\sigma_{ji}\,n_i = \sigma_{33}\,n_3^2 = \sigma\,\frac{1}{3}. \]

If you (incorrectly) use non-unit \(\vec n=(1/3)(1,1,1)\), \(\vec m=(1/2)(-1,0,1)\), you’d get \(\tau=\sigma/6\). The unit-vector form above (\(\tau=\sigma/\sqrt{6}\)) is the correct Schmid resolved shear.

📖 Variable Glossary

Symbol	Meaning
\(F\)	Resultant force on a cross-section (N).
\(A\)	Cross-sectional area (m\(^2\)).
\(F_n, F_t\)	Normal and tangential components of force on an infinitesimal patch.
\(\sigma\)	Scalar normal (engineering) stress, \(F/A\) (Pa).
\(\tau\)	Scalar shear (engineering) stress (Pa).
\(\boldsymbol\sigma=[\sigma_{ij}]\)	Cauchy stress tensor at a point (Pa).
\(\sigma_{ij}\)	\(i\)-component of traction acting on the plane whose outward normal is along \(j\).
\(\vec e_i\)	Orthonormal basis unit vectors along \(x_i\).
\(\vec n=(n_1,n_2,n_3)\)	Unit normal to a plane (for traction); satisfies \(n_i n_i=1\).
\(\vec T\)	Traction vector on plane with normal \(\vec n\): \(\vec T=\boldsymbol\sigma\,\vec n\).
\(T_i=\sigma_{ij}n_j\)	Component form of Cauchy’s traction relation.
[001]	Crystallographic direction along the cube \(z\)-axis.
\(\{111\}\langle110\rangle\)	FCC slip systems: planes of type (111) with directions of type \(\langle110\rangle\).
\(\vec n=\tfrac{1}{\sqrt{3}}(1,1,1)\)	Unit normal to the (111) plane.
\(\vec m=\tfrac{1}{\sqrt{2}}(-1,0,1)\)	Unit slip direction in \(\langle110\rangle\) family.
\(\tau=\vec T\cdot \vec m\)	Resolved shear stress (Schmid stress) on the slip direction.
\(p=\vec n\cdot \boldsymbol\sigma\,\vec n\)	Normal stress acting on the slip plane.

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Mean & Deviatoric Stress, Plane-Stress Transformations, and Mohr’s Circle

1) Mean (Hydrostatic) and Deviatoric Stress

Cauchy stress (symmetric):

\[ \boldsymbol\sigma=\{\sigma_{ij}\}= \begin{bmatrix} \sigma_{11}&\sigma_{12}&\sigma_{13}\\ \sigma_{21}&\sigma_{22}&\sigma_{23}\\ \sigma_{31}&\sigma_{32}&\sigma_{33} \end{bmatrix}, \qquad \sigma_{ij}=\sigma_{ji}. \]

Mean (hydrostatic) stress and deviatoric stress:

\[ \sigma_m=\frac{\sigma_{11}+\sigma_{22}+\sigma_{33}}{3} =\frac{\sigma_1+\sigma_2+\sigma_3}{3}, \qquad s_{ij}=\sigma_{ij}-\sigma_m\,\delta_{ij}. \]

\[ s_{ii}=0, \qquad \sigma_{ij}=s_{ij}+\sigma_m\,\delta_{ij}. \]

Hydrostatic part changes volume (pressure); deviatoric part changes shape (distortion).

2) Plane Stress Setup

Plane stress (thin plate, in-plane loading):

\[ \sigma_{33}=\sigma_{13}=\sigma_{23}=0. \]

\[ \boldsymbol\sigma_{2D}= \begin{bmatrix} \sigma_{11}&\sigma_{12}\\ \sigma_{21}&\sigma_{22} \end{bmatrix} \equiv \begin{bmatrix} \sigma_x&\tau_{xy}\\ \tau_{xy}&\sigma_y \end{bmatrix}. \]

Rotate axes in the \(x\!-\!y\) plane by angle \(\theta\) about \(x_3\) to get \((x',y')\).

3) Stress Transformation (Derivation)

Rotation (direction cosines) and tensor transformation:

\[ \boldsymbol\alpha= \begin{bmatrix} \cos\theta&\sin\theta&0\\ -\sin\theta&\cos\theta&0\\ 0&0&1 \end{bmatrix}, \qquad \sigma'_{ij}=\alpha_{im}\alpha_{jn}\sigma_{mn} \quad(\boldsymbol\sigma'=\boldsymbol\alpha\,\boldsymbol\sigma\,\boldsymbol\alpha^{\!\top}). \]

Carry out multiplication for plane stress; using \(\cos2\theta=\cos^2\theta-\sin^2\theta\) and \(\sin2\theta=2\sin\theta\cos\theta\), we obtain the Mohr form:

\[ \sigma_{11}'= \tfrac12(\sigma_x+\sigma_y)+\tfrac12(\sigma_x-\sigma_y)\cos2\theta+\tau_{xy}\sin2\theta, \]

\[ \sigma_{22}'= \tfrac12(\sigma_x+\sigma_y)-\tfrac12(\sigma_x-\sigma_y)\cos2\theta-\tau_{xy}\sin2\theta, \]

\[ \tau_{x'y'}'= -\tfrac12(\sigma_x-\sigma_y)\sin2\theta+\tau_{xy}\cos2\theta. \]

Special case (start in principal axes: \((\sigma_x,\sigma_y,\tau_{xy})=(\sigma_1,\sigma_2,0)\)):

\[ \sigma_{11}'=\tfrac12(\sigma_1+\sigma_2)+\tfrac12(\sigma_1-\sigma_2)\cos2\theta, \qquad \sigma_{22}'=\tfrac12(\sigma_1+\sigma_2)-\tfrac12(\sigma_1-\sigma_2)\cos2\theta, \]

\[ \tau_{x'y'}'=-\tfrac12(\sigma_1-\sigma_2)\sin2\theta. \]

4) Mohr’s Circle (Interpretation)

Build the circle from in-plane components \((\sigma_x,\sigma_y,\tau_{xy})\):

\[ C=\left(\frac{\sigma_x+\sigma_y}{2},\,0\right), \qquad R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^{\,2}}. \]

\[ (\sigma_n,\tau_n)=(\sigma_{11}',\,\tau_{x'y'}') \ \text{lies on the circle at geometric angle }2\theta\ \text{from the }x\text{-plane point}. \]

\[ \sigma_{1,2}=C\pm R\quad(\tau=0\ \text{at these points}), \qquad \tau_{\max}=R\ \text{at }90^\circ\ \text{around the circle},\ \ \sigma_n=C\ \text{there}. \]

Link to mean/deviatoric: circle center \(C\) is the in-plane mean; radius \(R\) reflects the in-plane deviatoric part. In 3D, the full mean is \(\sigma_m=\tfrac13\mathrm{tr}(\boldsymbol\sigma)\), and deviatoric \( \mathbf s=\boldsymbol\sigma-\sigma_m\mathbf I\) governs distortion.

Variable Glossary

Symbol	Meaning
\(\boldsymbol\sigma=\{\sigma_{ij}\}\)	Cauchy stress tensor (Pa); symmetric (\(\sigma_{ij}=\sigma_{ji}\)).
\(\sigma_{ii}\)	Normal stresses along axes; \(\sigma_{12}=\sigma_{21}=\tau_{xy}\) etc. are shear stresses.
\(\sigma_1,\sigma_2,\sigma_3\)	Principal stresses (eigenvalues of \(\boldsymbol\sigma\)).
\(\sigma_m\)	Mean (hydrostatic) stress \(=\tfrac13(\sigma_{11}+\sigma_{22}+\sigma_{33})\).
\(s_{ij}\)	Deviatoric stress \(=\sigma_{ij}-\sigma_m\delta_{ij}\); zero trace.
\(\delta_{ij}\)	Kronecker delta (identity tensor components).
\(\sigma_x,\sigma_y,\tau_{xy}\)	In-plane stress components in the \(x\!-\!y\) frame.
\(\theta\)	Rotation angle of axes in the plane (positive CCW).
\(\boldsymbol\alpha\)	Rotation (direction-cosine) matrix.
\(\boldsymbol\sigma'\)	Stress tensor in the rotated frame; \(\sigma'_{ij}=\alpha_{im}\alpha_{jn}\sigma_{mn}\).
\(\sigma_{11}',\sigma_{22}',\tau_{x'y'}'\)	In-plane stresses after rotation by \(\theta\).
\(C\)	Mohr circle center \(=(\tfrac{\sigma_x+\sigma_y}{2},\,0)\) (in-plane mean stress).
\(R\)	Mohr circle radius \(=\sqrt{((\sigma_x-\sigma_y)/2)^2+\tau_{xy}^2}\).
\((\sigma_n,\tau_n)\)	Normal/shear on a plane at angle \(\theta\); mapped by \(2\theta\) on the circle.
\(\sigma_{1,2}=C\pm R\)	In-plane principal stresses; occur where \(\tau_n=0\).
\(\tau_{\max}=R\)	Maximum in-plane shear; \(90^\circ\) from principal points on Mohr’s circle.
\(\mathrm{tr}(\cdot),\ \mathbf I\)	Trace; identity tensor.

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Principal Stresses as an Eigenvalue Problem (with invariants & worked example)

Page 1 – Principal Stress as Eigenvalue Problem

Start from Cauchy’s traction relation on a plane with unit normal \( \vec n \):

\[ T_i=\sigma_{ji}\,n_j, \]

Seek planes where traction is parallel to the normal (no shear on that plane):

\[ \sigma_{ji}\,n_j=\lambda\,n_i. \]

Rearrange to the homogeneous linear system

\[ (\sigma_{ji}-\lambda\,\delta_{ij})\,n_j=0, \]

which component-wise is

\[ \begin{cases} (\sigma_{11}-\lambda)n_1+\sigma_{21}n_2+\sigma_{31}n_3=0,\\[4pt] \sigma_{12}n_1+(\sigma_{22}-\lambda)n_2+\sigma_{32}n_3=0,\\[4pt] \sigma_{13}n_1+\sigma_{23}n_2+(\sigma_{33}-\lambda)n_3=0. \end{cases} \]

Non-trivial solutions exist iff the characteristic determinant vanishes:

\[ \begin{vmatrix} \sigma_{11}-\lambda & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22}-\lambda & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33}-\lambda \end{vmatrix}=0. \]

This yields a cubic (the characteristic polynomial):

\[ \lambda^3-I\,\lambda^2+II\,\lambda-III=0. \]

Stress invariants (frame-invariant combinations):

\[ I=\sigma_{ii}=\sigma_{11}+\sigma_{22}+\sigma_{33}, \qquad II=\tfrac12\big(\sigma_{ii}\sigma_{jj}-\sigma_{ij}\sigma_{ij}\big), \qquad III=\det(\sigma_{ij}). \]

Page 2 – Principal Stresses, Directions & Orthogonality

The three roots of the characteristic cubic are the principal stresses:

\[ \lambda_1,\ \lambda_2,\ \lambda_3. \]

For each \(\lambda_k\), the associated unit normal \(\vec n^{(k)}\) (principal direction) satisfies

\[ \sigma_{ji}\,n^{(k)}_j=\lambda_k\,n^{(k)}_i. \]

With a symmetric stress tensor (\(\sigma_{ij}=\sigma_{ji}\)), the principal directions are mutually orthogonal:

\[ \vec n^{(1)}\!\perp\!\vec n^{(2)},\quad \vec n^{(1)}\!\perp\!\vec n^{(3)},\quad \vec n^{(2)}\!\perp\!\vec n^{(3)}. \]

Proof sketch (inline)

Take \(\sigma_{ji}n^{(1)}_j=\lambda_1 n^{(1)}_i\) and \(\sigma_{ji}n^{(2)}_j=\lambda_2 n^{(2)}_i\). Multiply the first by \(n^{(2)}_i\) and the second by \(n^{(1)}_i\), subtract, and use symmetry \(\sigma_{ij}=\sigma_{ji}\) to obtain \((\lambda_1-\lambda_2)\,n^{(1)}_i n^{(2)}_i=0\). If \(\lambda_1\neq\lambda_2\) then \(n^{(1)}_i n^{(2)}_i=0\).

Page 3 – Worked Example (find principal stresses & directions)

Given the (symmetric) stress tensor

\[ \boldsymbol\sigma=\begin{bmatrix} 3&1&1\\ 1&0&2\\ 1&2&0 \end{bmatrix}, \]

Characteristic equation (expand \(\det(\boldsymbol\sigma-\lambda\mathbf I)=0\)):

\[ \lambda^3-3\lambda^2-6\lambda+8=0 \quad\Rightarrow\quad (\lambda+2)(\lambda-1)(\lambda-4)=0. \]

Principal stresses (eigenvalues):

\[ \lambda_1=-2,\qquad \lambda_2=1,\qquad \lambda_3=4. \]

Find corresponding unit eigenvectors \(\vec n^{(k)}\) (principal directions):

For \(\lambda_1=-2\): solve \((\boldsymbol\sigma+2\mathbf I)\vec n^{(1)}=\vec 0\). One solution is proportional to \((0,1,1)\), normalize to get
\[ \vec n^{(1)}=\big(0,\ \tfrac{\sqrt2}{2},\ \tfrac{\sqrt2}{2}\big). \]
For \(\lambda_2=1\): solve \((\boldsymbol\sigma-\mathbf I)\vec n^{(2)}=\vec 0\). One solution is proportional to \((1,-1,-1)\), normalize to
\[ \vec n^{(2)}=\big(\tfrac{\sqrt3}{3},\ -\tfrac{\sqrt3}{3},\ -\tfrac{\sqrt3}{3}\big). \]
For \(\lambda_3=4\): solve \((\boldsymbol\sigma-4\mathbf I)\vec n^{(3)}=\vec 0\). One solution is proportional to \((-2,1,1)\), normalize to
\[ \vec n^{(3)}=\big(-\tfrac{\sqrt6}{3},\ \tfrac{\sqrt6}{6},\ \tfrac{\sqrt6}{6}\big). \]

Verify orthogonality: \( \vec n^{(i)}\cdot \vec n^{(j)}=0\) for \(i\neq j\), and unit length \( \|\vec n^{(k)}\|=1\).

Page 4 – Transformation & Diagonalization (principal basis)

Form the orthogonal matrix whose columns are the principal directions:

\[ \boldsymbol\alpha=\begin{bmatrix} 0 & \tfrac{\sqrt3}{3} & -\tfrac{\sqrt6}{3}\\[4pt] \tfrac{\sqrt2}{2} & -\tfrac{\sqrt3}{3} & \tfrac{\sqrt6}{6}\\[4pt] \tfrac{\sqrt2}{2} & -\tfrac{\sqrt3}{3} & \tfrac{\sqrt6}{6} \end{bmatrix}, \qquad \boldsymbol\alpha^{\!\top}\boldsymbol\alpha=\mathbf I. \]

Transform the stress tensor to the principal frame:

\[ \boldsymbol\sigma'=\boldsymbol\alpha^{\!\top}\,\boldsymbol\sigma\,\boldsymbol\alpha = \begin{bmatrix} -2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 4 \end{bmatrix}. \]

Diagonal entries are exactly the principal stresses; shear terms vanish in the principal frame.

Summary

Principal stresses solve \(\det(\boldsymbol\sigma-\lambda\mathbf I)=0\): \(\lambda^3-I\lambda^2+II\lambda-III=0\).
Eigenvectors (principal directions) satisfy \(\sigma_{ji}n^{(k)}_j=\lambda_k n^{(k)}_i\); they are orthonormal.
Rotate into the principal basis to diagonalize \(\boldsymbol\sigma\): \(\boldsymbol\sigma'=\text{diag}(\lambda_1,\lambda_2,\lambda_3)\).

Variable Glossary

Symbol	Meaning
\(\boldsymbol\sigma=[\sigma_{ij}]\)	Cauchy stress tensor (symmetric; units Pa).
\(\sigma_{ij}\)	\(i\)-component of traction acting on the plane whose outward normal is along \(j\).
\(T_i\)	Traction on a plane with unit normal \(\vec n\): \(T_i=\sigma_{ji}n_j\).
\(\vec n=(n_1,n_2,n_3)\)	Unit normal to a plane; satisfies \(n_i n_i=1\).
\(\delta_{ij}\)	Kronecker delta (\(\delta_{ij}=1\) if \(i=j\), else \(0\)).
\(\lambda\)	Eigenvalue of \(\boldsymbol\sigma\); equals a principal stress.
\(\lambda_1,\lambda_2,\lambda_3\)	The three principal stresses (roots of the characteristic cubic).
\(\vec n^{(k)}\)	Principal direction (unit eigenvector) associated with \(\lambda_k\).
\(I,\,II,\,III\)	Stress invariants: \(I=\sigma_{ii}\), \(II=\tfrac12(\sigma_{ii}\sigma_{jj}-\sigma_{ij}\sigma_{ij})\), \(III=\det(\sigma_{ij})\).
\(\mathbf I\)	Identity tensor (matrix).
\(\boldsymbol\alpha\)	Orthogonal matrix of principal directions (columns); rotates to the principal frame.
\(\boldsymbol\sigma'=\boldsymbol\alpha^{\!\top}\boldsymbol\sigma\,\boldsymbol\alpha\)	Stress tensor expressed in the principal basis (diagonal).

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Mean/Deviatoric Stress, Plane-Stress Transformations & Mohr’s Circle (with variable definitions)

1) Mean (hydrostatic) and deviatoric stress

Start from the symmetric Cauchy stress tensor \( \boldsymbol\sigma=\{\sigma_{ij}\} \) with \( \sigma_{ij}=\sigma_{ji} \):

\[ \boldsymbol\sigma= \begin{bmatrix} \sigma_{11}&\sigma_{12}&\sigma_{13}\\ \sigma_{21}&\sigma_{22}&\sigma_{23}\\ \sigma_{31}&\sigma_{32}&\sigma_{33} \end{bmatrix},\qquad \sigma_{ij}=\sigma_{ji}. \]

Mean (hydrostatic) stress (one-third of the trace):

\[ \sigma_m=\frac{\sigma_{11}+\sigma_{22}+\sigma_{33}}{3} =\frac{\sigma_1+\sigma_2+\sigma_3}{3}. \]

Deviatoric stress tensor (pure shape-change part):

\[ s_{ij}=\sigma_{ij}-\sigma_m\,\delta_{ij},\qquad s_{ii}=0. \]

Additive split of stress into mean + deviatoric:

\[ \sigma_{ij}=s_{ij}+\sigma_m\,\delta_{ij}. \]

2) Plane stress setup

Plane stress (thin plate) assumptions:

\[ \sigma_{33}=\sigma_{13}=\sigma_{23}=0. \]

In-plane stress tensor (identify \(x\equiv 1\), \(y\equiv 2\)):

\[ \boldsymbol\sigma_{2D}= \begin{bmatrix} \sigma_{11}&\sigma_{12}\\ \sigma_{21}&\sigma_{22} \end{bmatrix} \equiv \begin{bmatrix} \sigma_x & \tau_{xy}\\ \tau_{xy} & \sigma_y \end{bmatrix}. \]

Rotate axes in the \(x\!-\!y\) plane by angle \( \theta \) about \(x_3\) to get \((x',y')\).

3) Stress transformation (derivation)

Rotation (direction-cosine) matrix for an in-plane rotation by \( \theta \):

\[ \boldsymbol\alpha= \begin{bmatrix} \cos\theta & \sin\theta & 0\\ -\sin\theta& \cos\theta & 0\\ 0&0&1 \end{bmatrix}. \]

Second-order tensor transformation:

\[ \sigma'_{ij}=\alpha_{im}\,\alpha_{jn}\,\sigma_{mn} \quad\text{(equivalently } \boldsymbol\sigma'=\boldsymbol\alpha\,\boldsymbol\sigma\,\boldsymbol\alpha^\top\text{)}. \]

Carrying out the multiplication for plane stress yields

\[ \begin{aligned} \sigma'_{11}&=\sigma_x\cos^2\theta+\sigma_y\sin^2\theta+2\,\tau_{xy}\sin\theta\cos\theta,\\ \sigma'_{22}&=\sigma_x\sin^2\theta+\sigma_y\cos^2\theta-2\,\tau_{xy}\sin\theta\cos\theta,\\ \sigma'_{12}&=(\sigma_y-\sigma_x)\sin\theta\cos\theta+\tau_{xy}(\cos^2\theta-\sin^2\theta). \end{aligned} \]

Using \( \cos^2\theta-\sin^2\theta=\cos(2\theta) \) and \( 2\sin\theta\cos\theta=\sin(2\theta) \), the Mohr form is

\[ \begin{aligned} \sigma'_{11}&=\tfrac12(\sigma_x+\sigma_y)+\tfrac12(\sigma_x-\sigma_y)\cos(2\theta)+\tau_{xy}\sin(2\theta),\\ \sigma'_{22}&=\tfrac12(\sigma_x+\sigma_y)-\tfrac12(\sigma_x-\sigma_y)\cos(2\theta)-\tau_{xy}\sin(2\theta),\\ \tau'_{x'y'}&=-\tfrac12(\sigma_x-\sigma_y)\sin(2\theta)+\tau_{xy}\cos(2\theta). \end{aligned} \]

Principal-frame special case \((\sigma_x,\sigma_y,\tau_{xy})=(\sigma_1,\sigma_2,0)\):

\[ \sigma'_{11}=\tfrac12(\sigma_1+\sigma_2)+\tfrac12(\sigma_1-\sigma_2)\cos(2\theta),\quad \sigma'_{22}=\tfrac12(\sigma_1+\sigma_2)-\tfrac12(\sigma_1-\sigma_2)\cos(2\theta),\quad \tau'_{x'y'}=-\tfrac12(\sigma_1-\sigma_2)\sin(2\theta). \]

4) Mohr’s circle (interpretation)

Define the circle in the \((\sigma_n,\tau_n)\) plane:

\[ C=\Big(\,\tfrac{\sigma_x+\sigma_y}{2}\,,\,0\Big),\qquad R=\sqrt{\Big(\tfrac{\sigma_x-\sigma_y}{2}\Big)^2+\tau_{xy}^2}. \]

A plane at physical angle \( \theta \) corresponds to the point

\[ (\sigma_n,\tau_n)=\big(\sigma'_{11},\,\tau'_{x'y'}\big) \quad\text{located at geometric angle }2\theta\text{ from the }x\text{-plane point.} \]

Principal stresses are where shear is zero (horizontal intercepts):

\[ \sigma_{1,2}=C\pm R,\qquad \tau=0. \]

Maximum in-plane shear occurs \(90^\circ\) around the circle from principal points:

\[ \tau_{\max}=R,\qquad \sigma_n=C. \]

Link to mean/deviatoric (2D): the circle’s center \(C\) is the in-plane mean; the radius \(R\) is set by the in-plane deviatoric content. In 3D, the mean is \(\sigma_m=\tfrac13\,\mathrm{tr}(\boldsymbol\sigma)\) and deviatoric governs distortion.

Quick checklist

\(\sigma_m=\tfrac13\,\mathrm{tr}(\boldsymbol\sigma)\), \( \mathbf s=\boldsymbol\sigma-\sigma_m\mathbf I\), \(s_{ii}=0\).
Plane stress → use \(2\times2\) in-plane tensor \([\sigma_x\ \ \tau_{xy};\ \tau_{xy}\ \ \sigma_y]\).
Rotate by \( \theta \): \( \boldsymbol\sigma'=\boldsymbol\alpha\,\boldsymbol\sigma\,\boldsymbol\alpha^\top\).
Mohr form (with \(2\theta\)) gives the circle: center \(C\), radius \(R\), principals \(C\pm R\), max shear \(R\).

Variable Glossary (every symbol defined)

Symbol	Meaning
\(\boldsymbol\sigma=\{\sigma_{ij}\}\)	Cauchy stress tensor at a point (symmetric; units Pa).
\(\sigma_{ij}\)	\(i\)-component of traction acting on the plane with outward normal along \(j\) (\(i,j\in\{1,2,3\}\)).
\(\sigma_x,\sigma_y,\tau_{xy}\)	In-plane stresses in the \(x\!-\!y\) frame (\(\sigma_{11}=\sigma_x,\ \sigma_{22}=\sigma_y,\ \sigma_{12}=\sigma_{21}=\tau_{xy}\)).
\(\sigma_{33},\sigma_{13},\sigma_{23}\)	Out-of-plane components (zero under plane-stress assumptions).
\(\sigma_m\)	Mean (hydrostatic) stress \(=\tfrac13(\sigma_{11}+\sigma_{22}+\sigma_{33})\) (equals \(-p\) in fluid convention).
\(s_{ij}\)	Deviatoric stress \(s_{ij}=\sigma_{ij}-\sigma_m\,\delta_{ij}\) (zero trace).
\(\delta_{ij}\)	Kronecker delta (\(\delta_{ij}=1\) if \(i=j\), else \(0\)).
\(\theta\)	Physical rotation angle of axes in the \(x\!-\!y\) plane (CCW positive).
\(\boldsymbol\alpha\)	Direction-cosine (rotation) matrix; orthogonal with \(\boldsymbol\alpha^\top\boldsymbol\alpha=\mathbf I\).
\(\boldsymbol\sigma'\)	Rotated stress tensor: \(\boldsymbol\sigma'=\boldsymbol\alpha\,\boldsymbol\sigma\,\boldsymbol\alpha^\top\).
\(\sigma'_{11},\sigma'_{22},\tau'_{x'y'}\)	Transformed in-plane normal and shear stresses on a plane at angle \(\theta\).
\((\sigma_n,\tau_n)\)	Normal/shear on a specific cut plane; equals \((\sigma'_{11},\tau'_{x'y'})\) for normal \(x'\).
\(C\)	Mohr circle center \(C=\big(\tfrac{\sigma_x+\sigma_y}{2},\,0\big)\) (in-plane mean stress).
\(R\)	Mohr circle radius \(R=\sqrt{\big(\tfrac{\sigma_x-\sigma_y}{2}\big)^2+\tau_{xy}^2}\).
\(\sigma_{1,2}\)	In-plane principal stresses \(=\ C\pm R\).
\(\tau_{\max}\)	Maximum in-plane shear \(=R\); occurs \(90^\circ\) around Mohr’s circle from principal points.

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Cauchy Stress & Static Equilibrium (with variable definitions)

Page 1 — Cauchy Stress

Stress tensor representation

The (symmetric) Cauchy stress tensor at a point:

\[ \sigma_{ij}= \begin{bmatrix} \sigma_{11}&\sigma_{12}&\sigma_{13}\\ \sigma_{21}&\sigma_{22}&\sigma_{23}\\ \sigma_{31}&\sigma_{32}&\sigma_{33} \end{bmatrix},\qquad \sigma_{ij}=\sigma_{ji}. \]

Each component \( \sigma_{ij} \) is the i-component of traction (force per area) acting on the plane whose outward unit normal is along the j-direction.

Integral balance of forces on a body

\[ \int_V f_i\,\mathrm dV + \int_S T_i\,\mathrm dS = 0 \;\;\Longleftrightarrow\;\; \sum F_i=0. \]

Body force from stress divergence

By equilibrium inside the body, body forces balance stress divergence:

\[ f_i = -\,\frac{\partial \sigma_{ki}}{\partial x_k}. \]

Substitute into the volume integral:

\[ \int_V f_i\,\mathrm dV =-\int_V \frac{\partial \sigma_{ki}}{\partial x_k}\,\mathrm dV. \]

Surface traction (Cauchy’s formula)

Traction vector on a surface with unit normal \( n_k \):

\[ T_i=\sigma_{ki}\,n_k. \]

Differential equilibrium (static)

Local (pointwise) balance of linear momentum in the absence of inertia:

\[ \frac{\partial \sigma_{ki}}{\partial x_k}+f_i=0. \]

Page 2 — Equilibrium Expansion

Force balance on a differential element

Expanding in Cartesian coordinates \(x_1,x_2,x_3\):

\[ \begin{cases} \dfrac{\partial \sigma_{11}}{\partial x_1} +\dfrac{\partial \sigma_{21}}{\partial x_2} +\dfrac{\partial \sigma_{31}}{\partial x_3} +f_1=0,\\[6pt] \dfrac{\partial \sigma_{12}}{\partial x_1} +\dfrac{\partial \sigma_{22}}{\partial x_2} +\dfrac{\partial \sigma_{32}}{\partial x_3} +f_2=0,\\[6pt] \dfrac{\partial \sigma_{13}}{\partial x_1} +\dfrac{\partial \sigma_{23}}{\partial x_2} +\dfrac{\partial \sigma_{33}}{\partial x_3} +f_3=0. \end{cases} \]

Compact index form (Einstein summation)

\[ \frac{\partial \sigma_{ji}}{\partial x_j}+f_i=0 \quad\Longleftrightarrow\quad \sigma_{ji,j}+f_i=0. \]

Variable Glossary (every symbol defined)

Symbol	Meaning
\(\sigma_{ij}\)	Components of the Cauchy stress tensor (Pa). \(i\)=force-component direction; \(j\)=normal direction of the plane. Symmetry: \(\sigma_{ij}=\sigma_{ji}\).
\(\boldsymbol\sigma=\{\sigma_{ij}\}\)	Stress tensor ( \(3\times3\) matrix ).
\(T_i\)	Traction (stress) vector components on a surface; Cauchy’s formula: \(T_i=\sigma_{ki}n_k\).
\(n_k\)	Components of outward unit normal (\(\\|\mathbf n\\|=1\)).
\(f_i\)	Body-force density (N/m\(^3\)); e.g., gravity \(f_i=\rho g_i\).
\(x_k\)	Spatial coordinates; \(k,i,j\in\{1,2,3\}\) correspond to \(x_1=x,\ x_2=y,\ x_3=z\).
\(V\)	Control volume (region for volume integrals).
\(S\)	Boundary surface of \(V\); \(\mathrm dS\) is area element with outward normal \(\mathbf n\).
\(\sum F_i=0\)	Net force balance (static equilibrium) in direction \(i\).
\(\sigma_{ji,j}\)	Divergence of stress in index form: \(\sigma_{ji,j}\equiv \partial \sigma_{ji}/\partial x_j\).
Units	\(\sigma_{ij},\,T_i\): Pa (N/m\(^2\)); \(f_i\): N/m\(^3\); \(x_k\): m.

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Orthotropic, Cubic, and Isotropic Solids (with variable definitions)

Page 1: Orthotropic, Cubic, and Isotropic Solids (Voigt notation)

Voigt vector mapping (convention used here):

Stresses: \([\,\sigma_1,\sigma_2,\sigma_3,\sigma_4,\sigma_5,\sigma_6\,]^T \equiv [\,\sigma_{11},\sigma_{22},\sigma_{33},\sigma_{23},\sigma_{13},\sigma_{12}\,]^T\).
Strains (engineering shear strains): \([\,\varepsilon_1,\varepsilon_2,\varepsilon_3,\varepsilon_4,\varepsilon_5,\varepsilon_6\,]^T \equiv [\,\varepsilon_{11},\varepsilon_{22},\varepsilon_{33},\gamma_{23},\gamma_{13},\gamma_{12}\,]^T\), with \(\gamma_{ij}=2\,\varepsilon_{ij}\) for \(i\ne j\).

Orthotropic solids (three orthogonal material planes of symmetry)

\[ \begin{bmatrix} \sigma_1\\ \sigma_2\\ \sigma_3\\ \sigma_4\\ \sigma_5\\ \sigma_6 \end{bmatrix} = \begin{bmatrix} C_{11}&C_{12}&C_{13}&0&0&0\\ C_{21}&C_{22}&C_{23}&0&0&0\\ C_{31}&C_{32}&C_{33}&0&0&0\\ 0&0&0&C_{44}&0&0\\ 0&0&0&0&C_{55}&0\\ 0&0&0&0&0&C_{66} \end{bmatrix} \begin{bmatrix} \varepsilon_1\\ \varepsilon_2\\ \varepsilon_3\\ \varepsilon_4\\ \varepsilon_5\\ \varepsilon_6 \end{bmatrix}. \]

Cubic solids (three independent constants)

\[ \mathbf C_\text{cubic}= \begin{bmatrix} C_{11}&C_{12}&C_{12}&0&0&0\\ C_{12}&C_{11}&C_{12}&0&0&0\\ C_{12}&C_{12}&C_{11}&0&0&0\\ 0&0&0&C_{44}&0&0\\ 0&0&0&0&C_{44}&0\\ 0&0&0&0&0&C_{44} \end{bmatrix}. \]

Zener anisotropy factor (cubic crystals): \[ A=\frac{2\,C_{44}}{C_{11}-C_{12}}. \] Isotropy corresponds to \(A=1\).

Isotropic solids (two independent constants)

\[ \mathbf C_\text{iso}= \begin{bmatrix} C_{11}&C_{12}&C_{12}&0&0&0\\ C_{12}&C_{11}&C_{12}&0&0&0\\ C_{12}&C_{12}&C_{11}&0&0&0\\ 0&0&0&\dfrac{C_{11}-C_{12}}{2}&0&0\\ 0&0&0&0&\dfrac{C_{11}-C_{12}}{2}&0\\ 0&0&0&0&0&\dfrac{C_{11}-C_{12}}{2} \end{bmatrix}. \]

For isotropy, \(C_{11}=\lambda+2\mu\), \(C_{12}=\lambda\), and \(C_{44}=\mu\), where \(\lambda,\mu\) are the Lamé constants (see Page 2).

Tensor transformation (general)

Under a rotation with direction cosines \(\alpha_{im}\), a 4th-order tensor transforms as

\[ C_{ijkl}=\alpha_{im}\,\alpha_{jn}\,\alpha_{ks}\,\alpha_{lt}\,C_{mnst}. \]

For isotropic solids, \(C_{ijkl}\) reduces to an isotropic combination of Kronecker deltas (next page).

Page 2: Isotropic 4th-Order Tensor & Lamé Constants

General isotropic 4th-order tensor

Most general isotropic 4th-order tensor (often written with two Lamé constants for elasticity):

\[ C_{ijkl}=\lambda\,\delta_{ij}\,\delta_{kl} +\mu\left(\delta_{ik}\,\delta_{jl}+\delta_{il}\,\delta_{jk}\right), \]

(\(\delta_{ij}\) is the Kronecker delta; \(\lambda,\mu\) are Lamé constants; \(\mu\) is also the shear modulus \(G\)).

Hooke’s law (isotropic linear elasticity)

\[ \sigma_{ij}=C_{ijkl}\,\varepsilon_{kl} =\lambda\,\delta_{ij}\,\varepsilon_{kk}+2\mu\,\varepsilon_{ij}. \]

Expanded relations

Normal stress trace:

\[ \sigma_{ii}=\left(3\lambda+2\mu\right)\,\varepsilon_{ii}. \]

Volumetric strain:

\[ \varepsilon_{ii}=\frac{\sigma_{ii}}{3\lambda+2\mu} =\frac{\sigma_{kk}}{3\lambda+2\mu}. \]

General stress in terms of strain and mean stress:

\[ \sigma_{ij}=\lambda\,\delta_{ij}\,\frac{\sigma_{kk}}{3\lambda+2\mu}+2\mu\,\varepsilon_{ij}. \]

Inverse (strain in terms of stress)

\[ 2\mu\,\varepsilon_{ij} =-\frac{\lambda}{3\lambda+2\mu}\,\delta_{ij}\,\sigma_{kk}+\sigma_{ij}, \qquad \varepsilon_{ij} =-\frac{\lambda}{2\mu(3\lambda+2\mu)}\,\delta_{ij}\,\sigma_{kk} +\frac{1}{2\mu}\,\sigma_{ij}. \]

Useful modulus relations (isotropic)

\[ E=2\mu(1+\nu),\qquad \lambda=\frac{\nu E}{(1+\nu)(1-2\nu)},\qquad \mu=G=\frac{E}{2(1+\nu)},\qquad K=\lambda+\tfrac{2}{3}\mu, \]

where \(E\) = Young’s modulus, \(\nu\) = Poisson’s ratio, \(G\) = shear modulus, \(K\) = bulk modulus.

Variable Glossary

Symbol	Meaning
\(\sigma_{ij}\)	Cauchy stress components (Pa). Voigt map: \(\sigma_1=\sigma_{11}\), \(\sigma_2=\sigma_{22}\), \(\sigma_3=\sigma_{33}\), \(\sigma_4=\sigma_{23}\), \(\sigma_5=\sigma_{13}\), \(\sigma_6=\sigma_{12}\).
\(\varepsilon_{ij}\)	Small-strain components (–); engineering shears \(\gamma_{ij}=2\varepsilon_{ij}\) for \(i\ne j\). Voigt: \(\varepsilon_1=\varepsilon_{11}\), \(\varepsilon_2=\varepsilon_{22}\), \(\varepsilon_3=\varepsilon_{33}\), \(\varepsilon_4=\gamma_{23}\), \(\varepsilon_5=\gamma_{13}\), \(\varepsilon_6=\gamma_{12}\).
\(\mathbf C\)	Stiffness (Voigt \(6\times6\)) with entries \(C_{IJ}\) mapping \(\boldsymbol\varepsilon \mapsto \boldsymbol\sigma\).
Orthotropic	Three orthogonal symmetry planes; 9 independent elastic constants (matrix shown above).
Cubic	Crystal class with 3 constants \(C_{11},C_{12},C_{44}\); isotropy occurs when Zener \(A=1\).
Isotropic	Direction-independent properties; 2 constants (e.g., \(\lambda,\mu\) or \(E,\nu\)).
\(A\)	Zener anisotropy factor (cubic): \(A=2C_{44}/(C_{11}-C_{12})\) (–).
\(\lambda,\mu\)	Lamé constants (Pa); \(\mu=G\) is the shear modulus.
\(E,\ \nu,\ G,\ K\)	Young’s modulus (Pa), Poisson’s ratio (–), shear modulus (Pa), bulk modulus (Pa); \(K=\lambda+\tfrac{2}{3}\mu\).
\(\delta_{ij}\)	Kronecker delta (\(\delta_{ij}=1\) if \(i=j\), else 0).
\(\,C_{ijkl}\,\)	Elasticity tensor; isotropic form \(C_{ijkl}=\lambda\,\delta_{ij}\delta_{kl}+\mu(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})\).
\(\alpha_{im}\)	Direction cosines (rotation matrix entries) for tensor transformations.
Units	\(\sigma\) in Pa; moduli \(E,\lambda,\mu,G,K\) in Pa; \(\varepsilon,\gamma\) dimensionless.

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Isotropic Linear Elastic Solids — Tensile Testing, Poisson Relations, and Depth Stress

Page 1 — Isotropic Linear Elastic Solids (Tensile Testing)

General Hooke’s law (isotropic)

\[ \sigma_{ij}=\lambda\,\delta_{ij}\,\varepsilon_{kk}+2\mu\,\varepsilon_{ij} \]

Trace relation

\[ \sigma_{ii}=\lambda\,\delta_{ii}\,\varepsilon_{kk}+2\mu\,\varepsilon_{ii} =(3\lambda+2\mu)\,\varepsilon_{ii}. \]

Strain in terms of stress (inverse relation)

\[ \varepsilon_{ij} =-\frac{\lambda}{2\mu(3\lambda+2\mu)}\,\delta_{ij}\,\sigma_{kk} +\frac{1}{2\mu}\,\sigma_{ij}. \]

Uniaxial tensile test

Apply only \(\sigma_{11}\neq 0\) and set \(\sigma_{22}=\sigma_{33}=\sigma_{12}=\sigma_{23}=\sigma_{31}=0\).

Then the axial stress–strain reduces to

\[ \sigma_{11}=E\,\varepsilon_{11},\qquad \varepsilon_{11} =-\frac{\lambda}{2\mu(3\lambda+2\mu)}(\sigma_{11}+\sigma_{22}+\sigma_{33}) +\frac{1}{2\mu}\sigma_{11}. \]

With only \(\sigma_{11}\) nonzero, this simplifies to

\[ \varepsilon_{11} =\frac{\lambda+\mu}{\mu(3\lambda+2\mu)}\,\sigma_{11} \;\;\Rightarrow\;\; E=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu}. \]

Lamé parameter from \((E,\mu)\)

\[ \lambda=\frac{\mu\,(E-2\mu)}{3\mu-E}. \]

Variable Definitions (Page 1)

Symbol	Meaning
\(\sigma_{ij}\)	Cauchy stress components (Pa); \(\sigma_{11}\) is axial stress in the 1-direction.
\(\varepsilon_{ij}\)	Small (engineering) strains (–); \(\varepsilon_{11}\) is axial strain.
\(\delta_{ij}\)	Kronecker delta (1 if \(i=j\), else 0).
\(\lambda,\mu\)	Lamé constants (Pa); \(\mu\) is the shear modulus \(G\).
\(E\)	Young’s modulus (Pa).
Note	Einstein summation: \(\varepsilon_{kk}=\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\).

Page 2 — Relations with Poisson’s Ratio

Poisson’s ratio

\[ \nu=-\frac{\varepsilon_{22}}{\varepsilon_{11}} \quad(\text{for uniaxial tension along 1 with }\sigma_{22}=\sigma_{33}=0). \]

Stress–strain using \(\lambda,\mu\)

\[ \sigma_{11}=\lambda(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33})+2\mu\,\varepsilon_{11}. \]

Eliminating lateral strains using \(\nu\) gives the familiar forms:

Core identities (isotropic)

\[ E=2\mu(1+\nu),\qquad \lambda=\frac{\nu E}{(1+\nu)(1-2\nu)},\qquad \mu=\frac{E}{2(1+\nu)}. \]

Stiffness matrix (Hooke’s law) in Voigt notation

\[ \{\sigma\} =\frac{E}{(1+\nu)(1-2\nu)} \begin{bmatrix} 1-\nu & \nu & \nu & 0 & 0 & 0\\ \nu & 1-\nu & \nu & 0 & 0 & 0\\ \nu & \nu & 1-\nu & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1-2\nu}{2} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1-2\nu}{2} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{1-2\nu}{2} \end{bmatrix} \{\varepsilon\}. \]

Compliance matrix (strain from stress)

\[ \{\varepsilon\}=\frac{1}{E} \begin{bmatrix} 1 & -\nu & -\nu & 0 & 0 & 0\\ -\nu & 1 & -\nu & 0 & 0 & 0\\ -\nu & -\nu & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 2(1+\nu) & 0 & 0\\ 0 & 0 & 0 & 0 & 2(1+\nu) & 0\\ 0 & 0 & 0 & 0 & 0 & 2(1+\nu) \end{bmatrix} \{\sigma\}. \]

Variable Definitions (Page 2)

Symbol	Meaning
\(\nu\)	Poisson’s ratio (–): lateral contraction per unit axial extension (uniaxial stress).
\(\{\sigma\}\)	Voigt stress vector \([\,\sigma_{11},\sigma_{22},\sigma_{33},\sigma_{23},\sigma_{13},\sigma_{12}\,]^T\).
\(\{\varepsilon\}\)	Voigt strain vector \([\,\varepsilon_{11},\varepsilon_{22},\varepsilon_{33},\gamma_{23},\gamma_{13},\gamma_{12}\,]^T\), with \(\gamma_{ij}=2\varepsilon_{ij}\) for \(i\neq j\).

Page 3 — Stress State at 3000 m Depth (Oedometric / Uniaxial Strain Condition)

Setup

Depth: \(h=3000\ \text{m}\)
Density: \(\rho=2500\ \text{kg/m}^3\)
Gravity: \(g=9.81\ \text{m/s}^2\)
Poisson’s ratio: \(\nu=0.25\)
Lateral strains constrained: \(\varepsilon_{11}=\varepsilon_{22}=0\), only \(\varepsilon_{33}\neq 0\).

Vertical stress (overburden)

\[ \sigma_{33}=\rho\,g\,h \quad\Rightarrow\quad \sigma_{33}=2500\times 9.81\times 3000\ \text{Pa}\approx 73.6\ \text{MPa}. \]

Uniaxial strain relations (isotropic)

With \(\varepsilon_{11}=\varepsilon_{22}=0\), Hooke’s law gives

\[ \sigma_{11}=\sigma_{22}=\lambda\,\varepsilon_{33}, \qquad \sigma_{33}=(\lambda+2\mu)\,\varepsilon_{33}. \]

Hence

\[ \varepsilon_{33}=\frac{\sigma_{33}}{\lambda+2\mu} =\frac{(1+\nu)(1-2\nu)}{E(1-\nu)}\,\sigma_{33}, \qquad \sigma_{11}=\sigma_{22}=\frac{\nu}{1-\nu}\,\sigma_{33}. \]

Horizontal stress ratio

\[ \boxed{\ \sigma_h=\frac{\nu}{1-\nu}\,\sigma_v\ } \quad\text{with}\quad \sigma_h\equiv\sigma_{11}=\sigma_{22},\;\sigma_v\equiv\sigma_{33}. \]

Numerical example (\(\nu=0.25\))

\[ \sigma_{33}\approx 73.6\ \text{MPa},\qquad \sigma_{22}=\frac{0.25}{1-0.25}\,\sigma_{33}\approx 0.333\,\sigma_{33}\approx 24.5\ \text{MPa}. \]

Variable Definitions (Page 3)

Symbol	Meaning
\(h\)	Depth below ground surface (m).
\(\rho\)	Mass density of overburden (kg/m\(^3\)).
\(g\)	Gravitational acceleration (m/s\(^2\)).
\(\sigma_{33}\equiv\sigma_v\)	Vertical (overburden) stress (Pa or MPa).
\(\sigma_{11}=\sigma_{22}\equiv\sigma_h\)	Horizontal stresses under lateral strain constraint (Pa or MPa).
\(\varepsilon_{ij}\)	Small strains; here \(\varepsilon_{11}=\varepsilon_{22}=0\) (oedometer), \(\varepsilon_{33}\) vertical strain.
\(\lambda,\mu; E,\nu\)	Lamé constants and moduli with \(E=2\mu(1+\nu)\), \(\lambda=\nu E/((1+\nu)(1-2\nu))\).

Note: The uniaxial strain (oedometric) case used here differs from uniaxial stress. Under uniaxial strain, \(\varepsilon_{11}=\varepsilon_{22}=0\) and lateral stresses develop; under uniaxial stress, \(\sigma_{11}\neq 0\) with \(\sigma_{22}=\sigma_{33}=0\).

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Static Elastic Problem → Navier’s Equations (with full variable definitions)

Page 1 — Static Elastic Problem (small deformation, linear elasticity)

Unknown fields

Displacements: \(u_i\) → 3 unknown scalar fields (\(i=1,2,3\)).
Strains: \(\varepsilon_{ij}\) → 6 unknowns (symmetric in \(i,j\)).
Stresses: \(\sigma_{ij}\) → 6 unknowns (symmetric Cauchy stress).

Equilibrium (no acceleration)

\[ \sigma_{ji,j}+f_i=0 \quad\text{(equivalently } \sigma_{ij,j}+f_i=0\text{)} \]

Represents local force balance in the body.

Kinematics (small-strain definition)

\[ \varepsilon_{ij}=\tfrac12\,(u_{j,i}+u_{i,j}) \]

Constitutive law (linear elastic, general)

\[ \sigma_{ij}=C_{ijkl}\,\varepsilon_{kl} \]

\(C_{ijkl}\) is the (4th-order) stiffness tensor.

Boundary conditions

Displacement BC on \(S_u\): \(\;u_i=\bar{u}_i\) on \(S_u\) (prescribed displacements).
Traction BC on \(S_\sigma\): \(\;T_i=\sigma_{ji}n_j=\bar{T}_i\) on \(S_\sigma\) (prescribed tractions).

Displacement formulation (governing starting point)

\[ \sigma_{ji,j}+f_i=0 \quad \text{(1)} \]

Variable Definitions (Page 1)

Symbol	Meaning
\(x_i\)	Spatial coordinates; \(i=1,2,3\) (often \(x,y,z\)).
\(u_i(x)\)	Displacement components (length units).
\(\varepsilon_{ij}\)	Small (engineering) strain; symmetric tensor (dimensionless).
\(\sigma_{ij}\)	Cauchy stress; symmetric (Pa = N/m\(^2\)).
\(C_{ijkl}\)	Stiffness tensor (Pa); maps strain to stress.
\(f_i\)	Body force per unit volume (N/m\(^3\)); e.g., gravity.
\(T_i\)	Traction (surface force/area) on a plane with unit normal \(n_j\); \(T_i=\sigma_{ji}n_j\).
\(n_j\)	\(j\)-component of the outward unit normal to a surface.
\(S_u,\ S_\sigma\)	Boundary parts with prescribed displacement / traction.
\(\bar u_i,\ \bar T_i\)	Prescribed boundary values for displacement / traction.
Comma	Comma derivative: \(u_{i,j}\!\equiv\!\partial u_i/\partial x_j\); \(\sigma_{ij,k}\!\equiv\!\partial\sigma_{ij}/\partial x_k\).
Einstein	Repeated indices imply summation over \(\{1,2,3\}\).

Page 2 — Governing PDE (Navier’s Equations)

Start from equilibrium

\[ \sigma_{ji,j}+f_i=0 \quad \text{(1)} \]

Isotropic Hooke’s law

\[ \sigma_{ij}=\lambda\,\delta_{ij}\,\varepsilon_{kk}+2\mu\,\varepsilon_{ij} \quad \text{(2)} \]

Strain–displacement (small strain)

\[ \varepsilon_{ij}=\tfrac12\,(u_{j,i}+u_{i,j}) \quad \text{(3)} \]

Substitute (3) into (2)

\[ \sigma_{ij}=\lambda\,\delta_{ij}\,u_{k,k}+\mu\,(u_{j,i}+u_{i,j}) \quad \text{(4)} \]

Insert (4) into equilibrium (1) and simplify

\[ \mu\,u_{i,kk}+(\lambda+\mu)\,u_{k,ki}+f_i=0 \]

Vector form: \(\mu\nabla^2 \mathbf{u}+(\lambda+\mu)\nabla(\nabla\!\cdot\!\mathbf{u})+\mathbf{f}=0\).

Fundamental solution (point force)

\[ f_i=\delta(\mathbf{x}-\mathbf{x}_0)\,F_i, \qquad \int_V \delta(\mathbf{x}-\mathbf{x}_0)\,dV=1. \]

\(\delta(\cdot)\) is the Dirac delta; \(F_i\) is the concentrated force.

Compatibility (strain integrability)

\[ \varepsilon_{ij,kl}+\varepsilon_{kl,ij}-\varepsilon_{ik,jl}-\varepsilon_{jl,ik}=0 \]

Ensures that a continuous displacement field \(u_i\) exists producing \(\varepsilon_{ij}\).

Variable Definitions (Page 1)

Symbol	Meaning
\(x_i\)	Spatial coordinates; \(i=1,2,3\) (often \(x,y,z\)).
\(u_i(x)\)	Displacement components (length units).
\(\varepsilon_{ij}\)	Small (engineering) strain; symmetric tensor (dimensionless).
\(\sigma_{ij}\)	Cauchy stress; symmetric (Pa = N/m\(^2\)).
\(C_{ijkl}\)	Stiffness tensor (Pa); maps strain to stress.
\(f_i\)	Body force per unit volume (N/m\(^3\)); e.g., gravity.
\(T_i\)	Traction (surface force/area) on a plane with unit normal \(n_j\); \(T_i=\sigma_{ji}n_j\).
\(n_j\)	\(j\)-component of the outward unit normal to a surface.
\(S_u,\ S_\sigma\)	Boundary parts with prescribed displacement / traction.
\(\bar u_i,\ \bar T_i\)	Prescribed boundary values for displacement / traction.
Comma	Comma derivative: \(u_{i,j}\!\equiv\!\partial u_i/\partial x_j\); \(\sigma_{ij,k}\!\equiv\!\partial\sigma_{ij}/\partial x_k\).
Einstein	Repeated indices imply summation over \(\{1,2,3\}\).

Variable Definitions (Page 2)

Symbol	Meaning
\(\lambda,\ \mu\)	Lamé constants (Pa); \(\mu\) is shear modulus, \(\lambda\) controls volumetric response.
\(\delta_{ij}\)	Kronecker delta (1 if \(i=j\), else 0).
\(\nabla^2\mathbf u\)	Vector Laplacian; componentwise \(u_{i,kk}\).
\(\nabla(\nabla\!\cdot\!\mathbf u)\)	Gradient of divergence; component \(u_{k,ki}\).
\(\mathbf f\)	Body force vector field with components \(f_i\) (N/m\(^3\)).
\(\delta(\mathbf x-\mathbf x_0)\)	Dirac delta distribution centered at \(\mathbf x_0\).
\(F_i\)	Components of a concentrated point force at \(\mathbf x_0\) (N).
\(V\)	Control volume (region) for integration.
Indices	\(i,j,k,l\in\{1,2,3\}\) (Cartesian components).

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Method of Nveu — Airy Stress Function (with full variable definitions)

Page 1 — Airy Stress Function & Biharmonic Condition

Airy stress function (plane problems in x–y):

\[ \phi=\phi(x,y),\qquad \sigma_x=\frac{\partial^2\phi}{\partial y^2},\quad \sigma_y=\frac{\partial^2\phi}{\partial x^2},\quad \tau_{xy}=-\,\frac{\partial^2\phi}{\partial x\,\partial y}. \]

Series ansatz (most general polynomial form):

\[ \phi(x,y)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}A_{mn}\,x^m y^n. \]

Biharmonic equation (stress compatibility in 2D):

\[ \nabla^4\phi = 0. \]

Expanded coefficient relation (from \(\nabla^4\phi=0\)):

\[ \sum_{m=2}^{\infty}\sum_{n=2}^{\infty} \Big[(m+2)(m+1)m(m-1)A_{m+2,n-2} + 2m(m-1)n(n-1)A_{m,n} + (n+2)(n+1)n(n-1)A_{m-2,n+2}\Big]\,x^{m-2}y^{n-2}=0. \]

Variable Definitions (Page 1)

Symbol	Meaning
\(x,\ y\)	In-plane Cartesian coordinates (length).
\(\phi(x,y)\)	Airy stress function (units: Pa·m\(^2\)).
\(\sigma_x,\ \sigma_y\)	Normal Cauchy stresses in \(x\), \(y\) (Pa).
\(\tau_{xy}\)	Shear Cauchy stress on \(x\)–\(y\) planes (Pa).
\(A_{mn}\)	Series coefficients set by BCs (units consistent with \(\phi\)).
\(\nabla^4\)	Biharmonic operator: \(\nabla^2(\nabla^2\cdot)\), \(\nabla^2=\partial_{xx}+\partial_{yy}\).
\(m,n\)	Nonnegative integer indices for polynomial degrees.

Page 2 — Boundary Tractions & Trial Airy Function

Traction on a boundary with unit normal \(\mathbf{n}=(n_x,n_y)\)

\[ T_x=\sigma_{xx}n_x+\sigma_{xy}n_y,\qquad T_y=\sigma_{yx}n_x+\sigma_{yy}n_y\quad(\sigma_{xy}=\sigma_{yx}). \]

Boundary segments

Side \(OA\): \(x=0\), \(n_x=-1\), \(n_y=0\)
\[ T_x=-\sigma_{xx}=-p\,y,\qquad T_y=-\sigma_{xy}=0. \]
Side \(AB\): line with slope \(-\tan\theta\Rightarrow y=-x\tan\theta\), \(\;n_x=\sin\theta,\;n_y=\cos\theta\)
\[ T_x=\sigma_{xx}\sin\theta+\sigma_{xy}\cos\theta=0,\quad T_y=\sigma_{xy}\sin\theta+\sigma_{yy}\cos\theta=0. \]

Trial Airy stress function (cubic polynomial)

\[ \phi=\frac{A}{2}x^2+Bxy+\frac{C}{2}y^2+\frac{D}{6}x^3+\frac{E}{2}x^2y+\frac{F}{2}xy^2+\frac{G}{6}y^3. \]

Resulting stresses (from \(\phi\))

\[ \sigma_{xx}=\frac{\partial^2\phi}{\partial y^2}=C+Fx+Gy,\qquad \sigma_{yy}=\frac{\partial^2\phi}{\partial x^2}=A+Dx+Ey,\qquad \sigma_{xy}=-\frac{\partial^2\phi}{\partial x\,\partial y}=-(B+Ex+Fy). \]

Additionally, \(\phi\) must satisfy the biharmonic condition \(\nabla^4\phi=0\) (used to filter admissible coefficients).

Variable Definitions (Page 2)

Symbol	Meaning
\(T_x,\ T_y\)	Boundary traction components (Pa).
\(n_x,\ n_y\)	Outward unit normal components to the boundary (–).
\(p\)	Load parameter for linearly varying traction \(-p\,y\) on \(x=0\) (Pa/m).
\(\theta\)	Angle of slanted side \(AB\) w.r.t. \(x\)-axis (rad).
\(A,\dots,G\)	Unknown constants in trial \(\phi\); fixed by BCs and \(\nabla^4\phi=0\).
\(\sigma_{xx},\ \sigma_{yy},\ \sigma_{xy}\)	In-plane Cauchy stresses (Pa).

Page 3 — Enforcing Boundary Conditions & Final Stresses

Apply BCs on \(x=0\) (side \(OA\))

\[ -\sigma_{xx}\big|_{x=0}=-p\,y \Rightarrow C=0,\;G=p;\qquad -(-B-Ex-Fy)\big|_{x=0}=0 \Rightarrow B=0,\;F=0. \]

Apply BCs on \(AB\): \(y=-x\tan\theta,\; n_x=\sin\theta,\; n_y=\cos\theta\)

\[ \sigma_{xx}\sin\theta+\sigma_{xy}\cos\theta=0,\quad \sigma_{xy}\sin\theta+\sigma_{yy}\cos\theta=0 \Rightarrow E=-p\tan^2\theta,\; A=0,\; D=-2p\tan^3\theta. \]

Final stress field

\[ \sigma_{xx}=p\,y,\qquad \sigma_{yy}=-2p\,x\,\tan^3\theta - p\,y\,\tan^2\theta,\qquad \sigma_{xy}=p\,x\,\tan^2\theta. \]

Corresponding Airy stress function

\[ \phi(x,y)= -\frac{1}{3}\,p\,x^3\tan^3\theta - \frac{1}{2}\,p\,y\,x^2\tan^2\theta + \frac{1}{6}\,p\,y^3. \]

Notes & definitions (Page 3)

These constants satisfy both the traction BCs and the biharmonic equation (for the chosen cubic ansatz).
Sign convention: tensions positive; \(\sigma_{xy}=\sigma_{yx}\).
Geometry: the linear boundary \(AB\) carries zero traction; \(OA\) carries a linearly varying normal traction \(-p\,y\).

Page 4 — Strains & Displacements (Plane Stress, Isotropic)

Hooke’s law (plane stress form)

\[ \varepsilon_{xx}=\frac{1}{E}\big(\sigma_{xx}-\nu\,\sigma_{yy}\big),\quad \varepsilon_{yy}=\frac{1}{E}\big(\sigma_{yy}-\nu\,\sigma_{xx}\big),\quad \gamma_{xy}=\frac{\tau_{xy}}{G}=\frac{2(1+\nu)}{E}\,\sigma_{xy}, \] \[ G=\frac{E}{2(1+\nu)}. \]

Substitute the stresses:

\[ \varepsilon_{xx} =\frac{1}{E}\Big(py-\nu\big[-2p\,x\,\tan^3\theta - p\,y\,\tan^2\theta\big]\Big) =a\,x+b\,y, \] with \[ a=\frac{2\nu(1+\nu)}{E}\,p\,\tan^3\theta,\qquad b=\frac{1-\nu^2+(1+\nu)\nu\tan^2\theta}{E}\,p. \] \[ \varepsilon_{yy} =\frac{1}{E}\Big(-2p\,x\,\tan^3\theta - p\,y\,\tan^2\theta - \nu\,py\Big) =c\,x+d\,y, \] with \[ c=-\frac{2(1-\nu^2)}{E}\,p\,\tan^3\theta,\qquad d=-\frac{\tan^2\theta+\nu}{E}\,p. \] \[ \varepsilon_{xy}=\tfrac12\,\gamma_{xy} =\frac{1}{2}\frac{2(1+\nu)}{E}\,\sigma_{xy} =\frac{(1+\nu)}{E}\,p\,x\,\tan^2\theta =e\,x,\quad e=\frac{(1+\nu)}{E}\,p\,\tan^2\theta. \]

Recover displacements (choose a simply connected domain; constants absorb rigid motions):

\[ u_x(x,y)=\int \varepsilon_{xx}\,dx=\tfrac12 a\,x^2+b\,x\,y+F_1(y), \qquad u_y(x,y)=\int \varepsilon_{yy}\,dy=c\,x\,y+\tfrac12 d\,y^2+F_2(x). \]

Compatibility with shear strain: \(\;\varepsilon_{xy}=\tfrac12(u_{x,y}+u_{y,x})=e\,x\).

\[ \tfrac12\big(b\,x+F_1'(y)+c\,y+F_2'(x)\big)=e\,x \quad\Rightarrow\quad \begin{cases} b\,x+F_2'(x)=2e\,x\\ c\,y+F_1'(y)=0 \end{cases} \Rightarrow \begin{cases} F_2(x) = \tfrac12(2e-b)\,x^2 + K_1\\ F_1(y) = -\tfrac12 c\,y^2 + K_2 \end{cases} \]

Final displacement field (up to rigid motions \(K_1,K_2\))

\[ u_x=\tfrac12 a\,x^2+b\,x\,y-\tfrac12 c\,y^2+K_2,\qquad u_y=c\,x\,y+\tfrac12 d\,y^2-\tfrac12(2e-b)\,x^2+K_1. \]

Variable Definitions (Page 4)

Symbol	Meaning
\(E\)	Young’s modulus (Pa), tensile stiffness.
\(\nu\)	Poisson’s ratio (–), lateral contraction ratio.
\(G\)	Shear modulus (Pa), \(G=E/[2(1+\nu)]\).
\(\varepsilon_{xx},\ \varepsilon_{yy},\ \varepsilon_{xy}\)	Normal/shear engineering strains (–). Note \(\gamma_{xy}=2\varepsilon_{xy}\).
\(a,b,c,d,e\)	Coefficients (functions of \(p,\theta,E,\nu\)) making strain fields linear in \(x,y\).
\(u_x,\ u_y\)	In-plane displacement components (length).
\(F_1(y),\ F_2(x)\)	Integration functions; fixed by shear-compatibility & boundary \(u\).
\(K_1,\ K_2\)	Integration constants representing rigid translations.

Extra notes / interpretation

The Airy function automatically enforces in-plane equilibrium (\(\sigma_{xx,x}+\sigma_{xy,y}=0,\;\sigma_{xy,x}+\sigma_{yy,y}=0\)).
The biharmonic constraint \(\nabla^4\phi=0\) is the compatibility of stresses for linearly elastic, body-force-free, plane problems.
Boundary tractions translate to linear conditions on the stress components, hence linear constraints on the coefficients of \(\phi\).
Once stresses are known, strains follow from Hooke’s law (choose plane stress/strain), then integrate to obtain displacements and enforce shear compatibility.

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Elasticity in Polar Coordinates (with full variable definitions)

Page 1 — Equilibrium, Displacements, and Strains (Polar)

Equilibrium equations (plane problem in r–θ)

\[ \frac{\partial \sigma_r}{\partial r} +\frac{1}{r}\frac{\partial \tau_{r\theta}}{\partial \theta} +\frac{\sigma_r-\sigma_\theta}{r} +f_r=0, \qquad \frac{\partial \tau_{r\theta}}{\partial r} +\frac{1}{r}\frac{\partial \sigma_\theta}{\partial \theta} +\frac{2\,\tau_{r\theta}}{r} +f_\theta=0. \]

Cartesian equilibrium (reference)

\[ \frac{\partial \sigma_x}{\partial x} +\frac{\partial \tau_{xy}}{\partial y}+f_x=0,\qquad \frac{\partial \tau_{xy}}{\partial x} +\frac{\partial \sigma_y}{\partial y}+f_y=0. \]

Displacements

\[ u_r=u_r(r,\theta),\qquad u_\theta=u_\theta(r,\theta). \]

Small strains in polar coordinates

\[ \varepsilon_r=\frac{\partial u_r}{\partial r},\qquad \varepsilon_\theta=\frac{1}{r}\frac{\partial u_\theta}{\partial \theta}+\frac{u_r}{r},\qquad \gamma_{r\theta}=\frac{1}{r}\frac{\partial u_r}{\partial \theta}+\frac{\partial u_\theta}{\partial r}-\frac{u_\theta}{r}. \]

Variable Definitions (Page 1)

Symbol	Meaning
\(r\), \(\theta\)	Radial coordinate (length) and polar angle (radians).
\(\sigma_r\)	Radial normal Cauchy stress (Pa).
\(\sigma_\theta\)	Hoop (circumferential) normal stress (Pa).
\(\tau_{r\theta}\)	Shear stress on \(r\)-normal plane, tangent along \(\theta\) (Pa).
\(f_r,\ f_\theta\)	Body-force density components in \(r\), \(\theta\) (N/m\(^3\)).
\(u_r,\ u_\theta\)	Displacement components in \(r\), \(\theta\) (length).
\(\varepsilon_r,\ \varepsilon_\theta\)	Radial and hoop normal strains (–).
\(\gamma_{r\theta}\)	Engineering shear strain (–); \(\gamma_{r\theta}=2\,\varepsilon_{r\theta}\).
\(\sigma_x,\sigma_y,\tau_{xy}\)	Cartesian stresses (Pa); \(f_x,f_y\) Cartesian body forces (N/m\(^3\)).

Page 2 — Constitutive Relations (Plane Stress) & Airy Function in Polar

Isotropic plane-stress strain–stress relations (polar components)

\[ \varepsilon_r=\frac{1}{E}\big(\sigma_r-\nu\,\sigma_\theta\big),\quad \varepsilon_\theta=\frac{1}{E}\big(\sigma_\theta-\nu\,\sigma_r\big),\quad \gamma_{r\theta}=\frac{2(1+\nu)}{E}\,\tau_{r\theta}. \]

Stress–strain inversion

\[ \sigma_r=\frac{E}{1-\nu^2}\big(\varepsilon_r+\nu\,\varepsilon_\theta\big),\quad \sigma_\theta=\frac{E}{1-\nu^2}\big(\varepsilon_\theta+\nu\,\varepsilon_r\big),\quad \tau_{r\theta}=\frac{E}{2(1+\nu)}\,\gamma_{r\theta}. \]

Plane stress ↔ plane strain (effective moduli)

\[ E'=\frac{E}{1-\nu^2},\qquad 1+\nu'=\frac{1}{1-\nu}. \]

Airy stress function in polar

Introduce \(\Phi(r,\theta)\) so equilibrium is satisfied identically:

\[ \sigma_r=\frac{1}{r}\frac{\partial \Phi}{\partial r}+\frac{1}{r^2}\frac{\partial^2 \Phi}{\partial \theta^2},\qquad \sigma_\theta=\frac{\partial^2 \Phi}{\partial r^2},\qquad \tau_{r\theta}=-\,\frac{\partial}{\partial r}\!\left(\frac{1}{r}\frac{\partial \Phi}{\partial \theta}\right). \]

Biharmonic condition (compatibility in 2D, no body forces)

\[ \nabla^4\Phi= \Big(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}\Big) \Big(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}\Big)\Phi=0. \]

Separable/series solutions (Michell, 1899)

\[ \Phi(r,\theta)=\sum_{n=0}^{\infty} f_n(r)\cos(n\theta)+\sum_{n=0}^{\infty} g_n(r)\sin(n\theta), \] \[ \begin{aligned} \Phi &= A_{01}r^2+A_{02}r^2\ln r+A_{03}\ln r+A_{04}\theta\\ &\quad+\big(A_{11}r^3+A_{12}r\ln r+A_{13}r^{-1}\big)\cos\theta +\big(B_{11}r^3+B_{12}r\ln r+B_{14}r^{-1}\big)\sin\theta\\ &\quad+ A_{13}\,r\theta\sin\theta + B_{13}\,r\theta\cos\theta\\ &\quad+\sum_{n=2}^{\infty}\big(A_{n1}r^{n+2}+A_{n2}r^{-n+2}+A_{n3}r^{n}+A_{n4}r^{-n}\big)\cos(n\theta)\\ &\quad+\sum_{n=2}^{\infty}\big(B_{n1}r^{n+2}+B_{n2}r^{-n+2}+B_{n3}r^{n}+B_{n4}r^{-n}\big)\sin(n\theta). \end{aligned} \]

Variable Definitions (Page 2)

Symbol	Meaning
\(E\)	Young’s modulus (Pa).
\(\nu\)	Poisson’s ratio (–); \(G=E/[2(1+\nu)]\) is shear modulus.
\(\Phi(r,\theta)\)	Polar Airy stress function (Pa·m\(^2\)); generates \(\sigma_r,\sigma_\theta,\tau_{r\theta}\).
\(\nabla^4\)	Biharmonic operator in polar coordinates.
\(A_{\cdot}, B_{\cdot}\)	Series constants fixed by BCs/regularity.
\(f_n(r), g_n(r)\)	Radial functions in separable solutions (Michell series).
\(E',\ \nu'\)	Effective moduli for plane-stress/plane-strain recast.

Page 3 — Coordinate Transformations & Stress Relations

Geometry of the polar–Cartesian map

\[ r^2=x^2+y^2,\qquad x=r\cos\theta,\qquad y=r\sin\theta,\qquad \tan\theta=\frac{y}{x}. \]

Transforming derivatives (chain rule)

\[ \frac{\partial \Phi}{\partial y} =\frac{\partial \Phi}{\partial r}\frac{\partial r}{\partial y} +\frac{\partial \Phi}{\partial \theta}\frac{\partial \theta}{\partial y}. \] \[ r^2=x^2+y^2\Rightarrow 2r\,\frac{\partial r}{\partial y}=2y\Rightarrow \frac{\partial r}{\partial y}=\frac{y}{r}=\sin\theta,\qquad \tan\theta=\frac{y}{x}\Rightarrow \frac{\partial \theta}{\partial y}=\frac{1}{x}=\frac{\cos\theta}{r}. \] \[ \Rightarrow\quad \frac{\partial \Phi}{\partial y} =\frac{\partial \Phi}{\partial r}\sin\theta+\frac{1}{r}\frac{\partial \Phi}{\partial \theta}\cos\theta. \]

Second derivative (illustrative form)

\[ \frac{\partial^2 \Phi}{\partial y^2} =\frac{\partial^2 \Phi}{\partial r^2}\sin^2\theta +\frac{2}{r}\frac{\partial^2 \Phi}{\partial r\,\partial \theta}\sin\theta\cos\theta -\frac{2}{r^2}\frac{\partial \Phi}{\partial \theta}\sin\theta\cos\theta +\frac{1}{r}\frac{\partial}{\partial r}\!\Big(\frac{\partial \Phi}{\partial r}\Big)\cos^2\theta +\frac{1}{r^2}\frac{\partial^2 \Phi}{\partial \theta^2}\cos^2\theta. \]

Stress relations recovered from \(\Phi\)

Taking limits along axes gives the standard polar stress definitions:

\[ \theta\to 0:\quad \sigma_x\to \sigma_r,\ \cos\theta\to1,\ \sin\theta\to0 \Rightarrow \sigma_r=\frac{1}{r}\frac{\partial \Phi}{\partial r} +\frac{1}{r^2}\frac{\partial^2 \Phi}{\partial \theta^2}. \] \[ \theta=\frac{\pi}{2}:\quad \sigma_x\to \sigma_\theta \Rightarrow \sigma_\theta=\frac{\partial^2 \Phi}{\partial r^2}. \] \[ \tau_{r\theta}=-\frac{\partial^2 \Phi}{\partial x\,\partial y} =-\,\frac{1}{r}\frac{\partial}{\partial r}\!\Big(\frac{\partial \Phi}{\partial \theta}\Big). \]

Variable Definitions (Page 3)

Symbol	Meaning
\(x,\ y\)	Cartesian coordinates (length).
\(\partial(\cdot)/\partial r,\ \partial(\cdot)/\partial \theta\)	Partial derivatives in polar variables.
\(\sigma_x\)	Normal stress along \(x\) (Pa); used to relate limits to \(\sigma_r,\sigma_\theta\).

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Lame’s Solution — Thick-Walled Cylinder Under Internal/External Pressure

1) General setup (axisymmetric)

Geometry: inner radius a, outer radius b. Internal pressure p_i at r=a, external pressure p_o at r=b. Axisymmetric, no θ or z dependence.

Kinematics

\[ u_r=u(r),\quad u_\theta=0,\qquad \varepsilon_r=\frac{d u_r}{dr},\quad \varepsilon_\theta=\frac{u_r}{r},\quad \gamma_{r\theta}=0. \]

Constitutive (isotropic, linear)

For the stress field in an axisymmetric, traction–only loading, the closed-form stresses are independent of \(E,\nu\) (they enter the displacement). If needed (plane stress) one may use:

\[ \varepsilon_r=\frac{1}{E}(\sigma_r-\nu\sigma_\theta),\qquad \varepsilon_\theta=\frac{1}{E}(\sigma_\theta-\nu\sigma_r). \]

Radial equilibrium (no body force)

\[ \frac{d\sigma_r}{dr}+\frac{\sigma_r-\sigma_\theta}{r}=0. \]

Governing ODE for the displacement

Eliminating stresses using the strain relations yields the classical radial ODE:

\[ \frac{d}{dr}\!\Big(\frac{1}{r}\frac{d}{dr}(r u_r)\Big)=0 \quad\Longrightarrow\quad u_r(r)=A\,r+\frac{B}{r}, \]

with constants A,B set by boundary conditions (and material via \(E,\nu\)).

Variable Definitions

Symbol	Meaning
\(r\)	Radial coordinate (m).
\(a,\ b\)	Inner / outer radii (m).
\(p_i,\ p_o\)	Internal / external pressures (Pa), positive in compression on the wall.
\(u_r\)	Radial displacement (m).
\(\varepsilon_r,\ \varepsilon_\theta\)	Radial and hoop normal strains (–).
\(\sigma_r,\ \sigma_\theta\)	Radial and hoop (circumferential) Cauchy stresses (Pa).
\(E,\ \nu\)	Young’s modulus (Pa) and Poisson’s ratio (–).
\(A,\ B\)	Displacement integration constants in \(u_r=A r+B/r\) (depend on \(E,\nu,p_i,p_o,a,b\)).

2) Stress/displacement forms & boundary conditions

Lame stress form (axisymmetric, no body force)

From equilibrium one obtains the general stress field

\[ \sigma_r(r)=C-\frac{D}{r^{2}},\qquad \sigma_\theta(r)=C+\frac{D}{r^{2}}, \]

with constants C,D determined by the pressure boundary conditions.

Boundary conditions (pressures acting on the walls)

\[ \sigma_r(a) = -\,p_i,\qquad \sigma_r(b) = -\,p_o. \]

Constants (from boundary conditions)

Symbol	Definition
\(C\)	\(C=\dfrac{p_i a^2 - p_o b^2}{\,b^2-a^2\,}\)
\(D\)	\(D=\dfrac{a^2 b^2\,(p_o - p_i)}{\,b^2-a^2\,}\)

Radial displacement (if needed)

With \(u_r=A r+B/r\) and the plane-stress (or plane-strain) constitutive laws, \(A,B\) can be written in terms of \(E,\nu\) and the pressure data. (Stresses themselves do not depend on \(E,\nu\); only \(u_r\) does.)

Notes

C,D: stress integration constants fixed by \(\sigma_r(a)=-p_i,\ \sigma_r(b)=-p_o\).
Plane stress / plane strain: 2D idealizations for long cylinders (plane strain in z) or thin cylinders (plane stress); choice affects displacement, not the Lame stress field.

3) Final Lame stresses & useful evaluations

Closed-form stresses

\[ \boxed{\ \sigma_r(r)=\frac{p_i a^2 - p_o b^2}{b^2-a^2} -\frac{a^2 b^2\,(p_i - p_o)}{(b^2-a^2)\,r^2}\ } \] \[ \boxed{\ \sigma_\theta(r)=\frac{p_i a^2 - p_o b^2}{b^2-a^2} +\frac{a^2 b^2\,(p_i - p_o)}{(b^2-a^2)\,r^2}\ } \]

At the walls

\[ \sigma_r(a)=-p_i,\qquad \sigma_r(b)=-p_o, \] \[ \sigma_\theta(a)=\frac{p_i a^2 - p_o b^2}{b^2-a^2} +\frac{a^2 b^2\,(p_i - p_o)}{(b^2-a^2)\,a^2} =\frac{p_i(a^2+b^2)-2p_o b^2}{b^2-a^2}, \] \[ \sigma_\theta(b)=\frac{p_i a^2 - p_o b^2}{b^2-a^2} +\frac{a^2 b^2\,(p_i - p_o)}{(b^2-a^2)\,b^2} =\frac{2p_i a^2 - p_o(a^2+b^2)}{b^2-a^2}. \]

Inner-wall hoop stress is largest in magnitude

For internal pressure loading (\(p_i>0,\ p_o\approx 0\)), the maximum hoop (circumferential) stress occurs at \(r=a\):

\[ \sigma_\theta(a)=\frac{p_i\,(a^2+b^2)}{\,b^2-a^2\,},\qquad \sigma_\theta(b)=\frac{2p_i\,a^2}{\,b^2-a^2\,}. \]

Quick interpretations

\(\sigma_r\) balances the applied pressures; it varies from \(-p_i\) at \(r=a\) to \(-p_o\) at \(r=b\).
\(\sigma_\theta\) is larger in magnitude than \(\sigma_r\), peaking at the inner wall — critical for design.
Displacement grows with compliance (smaller \(E\), larger \(\nu\) effect); stresses do not depend on \(E,\nu\).

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Levi–Civita Contractions & Kronecker Selector — Assignments 1–3

Assignment 1 — Prove three identities

BAC–CAB: \( \mathbf a\times(\mathbf b\times \mathbf c)=(\mathbf a\!\cdot\!\mathbf c)\,\mathbf b-(\mathbf a\!\cdot\!\mathbf b)\,\mathbf c \).
Lagrange’s identity: \((\mathbf a\times\mathbf b)\!\cdot\!(\mathbf c\times\mathbf d)=(\mathbf a\!\cdot\!\mathbf c)(\mathbf b\!\cdot\!\mathbf d)-(\mathbf a\!\cdot\!\mathbf d)(\mathbf b\!\cdot\!\mathbf c)\).
Cross of two cross products: \((\mathbf a\times\mathbf b)\times(\mathbf c\times\mathbf d)=\mathbf c[(\mathbf a\times\mathbf b)\!\cdot\!\mathbf d]-\mathbf d[(\mathbf a\times\mathbf b)\!\cdot\!\mathbf c]\).

Conventions: \((\mathbf a\times\mathbf b)_i=\varepsilon_{ijk}a_j b_k,\quad \mathbf a\!\cdot\!\mathbf b=a_i b_i\).

Proof 1) BAC–CAB

\[ \begin{aligned} [\mathbf a\times(\mathbf b\times\mathbf c)]_i &=\varepsilon_{ijk}a_j\varepsilon_{kmn}b_m c_n =(\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm})a_j b_m c_n\\ &=b_i(a_j c_j)-c_i(a_j b_j) =(\mathbf a\!\cdot\!\mathbf c)\,b_i-(\mathbf a\!\cdot\!\mathbf b)\,c_i . \end{aligned} \]

Proof 2) Lagrange’s identity

\[ (\mathbf a\times\mathbf b)\!\cdot\!(\mathbf c\times\mathbf d) =\varepsilon_{ijk}a_j b_k\,\varepsilon_{ipq}c_p d_q =(\delta_{jp}\delta_{kq}-\delta_{jq}\delta_{kp})a_j b_k c_p d_q . \]

Proof 3) Cross of two cross products

Apply BAC–CAB with \(X=\mathbf a\times\mathbf b\), \(Y=\mathbf c\), \(Z=\mathbf d\):

\[ (\mathbf a\times\mathbf b)\times(\mathbf c\times\mathbf d) =\mathbf c\,[\,(\mathbf a\times\mathbf b)\!\cdot\!\mathbf d\,] -\mathbf d\,[\,(\mathbf a\times\mathbf b)\!\cdot\!\mathbf c\,]. \]

ε–δ Toolbox

Master contraction

\[ \boxed{\;\varepsilon_{ijk}\,\varepsilon_{imn} =\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km}\;} \]

Immediate consequences:

\(\varepsilon_{ijk}\varepsilon_{ijk}=6\).
\(\varepsilon_{ijk}\varepsilon_{imk}=2\,\delta_{jm}\).
\(\varepsilon_{ijk}\varepsilon_{abk}=\delta_{ia}\delta_{jb}-\delta_{ib}\delta_{ja}\).

Assignment 2 — Double ε–Contraction

Evaluate \(\displaystyle \sum_{s=1}^{3}\sum_{k=1}^{3}\varepsilon_{i s k}\,\varepsilon_{j s k}\).

Key identity

\[ \boxed{\;\varepsilon_{p q r}\,\varepsilon_{p s t} =\delta_{q s}\delta_{r t}-\delta_{q t}\delta_{r s}\;} \tag{★} \]

Evaluation

\[ \varepsilon_{i s k}\,\varepsilon_{j s k} =(\delta_{ij}\delta_{kk}-\delta_{ik}\delta_{kj}) =3\,\delta_{ij}-\delta_{ij} =2\,\delta_{ij}. \]

Result: \(\boxed{\sum_{s,k}\varepsilon_{i s k}\varepsilon_{j s k}=2\,\delta_{ij}}\).

Assignment 3 — Kronecker δ “selector”

Prove \(\displaystyle a_j=\sum_{i=1}^{3} a_i\,\delta_{ij}\) (Einstein: \(a_j=a_i\delta_{ij}\)).

Why it works

\(\delta_{ij}=\{1 \text{ if } i=j; 0 \text{ otherwise}\}\), so for fixed \(j\):

\[ \sum_{i=1}^{3} a_i\,\delta_{ij} = a_1\delta_{1j}+a_2\delta_{2j}+a_3\delta_{3j} = a_j. \]

Geometric viewpoint

Using \(\delta_{ij}=\mathbf e_i\!\cdot\!\mathbf e_j\):

\[ a_i\delta_{ij}=(a_i\mathbf e_i)\!\cdot\!\mathbf e_j =\mathbf a\!\cdot\!\mathbf e_j =a_j. \]

Assignment 4 — Orthogonality of direction cosines

Prove \(\alpha_{im}\alpha_{mj}=\delta_{ij}\).

Setup

\[ \mathbf e_i'=\alpha_{im}\mathbf e_m,\qquad \mathbf e_m=\alpha_{mi}\mathbf e_i',\qquad \mathbf e_i'\!\cdot\!\mathbf e_j'=\delta_{ij},\ \ \mathbf e_m\!\cdot\!\mathbf e_n=\delta_{mn}. \]

Derivation

\[ \delta_{ij} =(\alpha_{im}\mathbf e_m)\!\cdot\!(\alpha_{jn}\mathbf e_n) =\alpha_{im}\alpha_{jn}\delta_{mn} =\alpha_{im}\alpha_{jm}. \]

Therefore \(\alpha^{\!\top}\alpha=I\) and \(\alpha\alpha^{\!\top}=I\).

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ME 261 — Survival & Translation Guide

Course: Theory of Elasticity (Prof. Xu) • Focus: show progress, think in the professor’s notation, and adapt core elasticity tools to new contexts.

Grading Policy

Component	Weight
Homework Graded on effort & completeness	50%
Engagement Professor’s discretion	20%
Prelim-style exam Objective	30%

Letter scale (approx.): A = 80–100%, B = 60–80%, C = 40–60%.

Advice from a Former Student

Do all assignments — they show steady progress even when not graded for correctness.
Match the professor’s notation; write your work in his style.
Deflection problems are the trickiest; practice them early and often.
A bad exam day won’t sink you if homework and engagement are solid.

Final Exam Style — What to Expect

Friend’s note: “Problem involved semiconductors none of us had taken. Vaguely like thermal deflection but the example was a semiconductor plate.”

Core concept

Thermal stress / deflection (covered in class)

Context

Semiconductor plate (unfamiliar wrapper, not device physics)

Intent

Test adaptability: translate new stories into known equations

Safety net

Prof clarifies context on board if the class gets stuck

What This Means for You

Don’t panic about materials Focus on the method, not the label (Si, Mg, W…)
Map back to fundamentals Stress/strain, compatibility, thermal strain \( \alpha\Delta T \), BCs
Expect application framing Translate unfamiliar wrappers into standard elasticity steps

Strategy to Be Ready

1) Focus on Fundamentals

Strain–displacement relations
Hooke’s law (isotropic)
Equilibrium equations
Thermal strain: \( \varepsilon_\text{th}=\alpha\Delta T \)

2) Practice Generalization

Cylinder → composite rod
Beam deflection → plate with thermal loading
Same math, new story

3) Mindset

Don’t fixate on material system
Ask: “Which elasticity equations apply?”

How to Score Well (Given the Grading)

Homework (50%) — Complete, legible, in professor’s notation. Treat as exam rehearsal.
Engagement (20%) — Show up, visible notes, ask clarifying questions, show weekly progress.
Exam (30%) — If wrapper’s unfamiliar, slow down and translate to stress/strain + BCs.

Tip: build a one-page notation crib sheet (Voigt, index/tensor, thermal stress forms) and practice writing solutions in the professor’s exact style.

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