## Governing Equations for 3D Elastic Bodies ### Equations of Motion Relates acceleration and forces to stress $ \sum_{j=1}^{3} \frac{\partial \sigma_{ij}}{\partial x_j} + \rho g_i = \rho a_i \quad \text{for } i = 1,2,3 $ #### Expanded form $ \begin{align} \frac{\delta\sigma_{11}}{\delta x}+\frac{\delta\sigma_{12}}{\delta y}+\frac{\sigma_{13}}{\delta z}+\rho g_{1}&=\rho a_{1} \\ \frac{\delta\sigma_{21}}{\delta x}+\frac{\delta\sigma_{22}}{\delta y}+\frac{\sigma_{23}}{\delta z}+\rho g_{2}&=\rho a_{2} \\ \frac{\delta\sigma_{31}}{\delta x}+\frac{\delta\sigma_{32}}{\delta y}+\frac{\sigma_{33}}{\delta z}+\rho g_{3}&=\rho a_{3} \end{align} $ ### Strain-Displacement Relation Relates strain and displacement $ [\underline{\epsilon}]=\frac{1}{2}([\nabla \underline{u}]+[\nabla \underline{u}]^T) $ ### Elastic Constitutive Relation Relates stress and strain $ [\underline{\sigma}]=K(\text{tr}\underline{\epsilon})[\underline{1}] + 2G[\underline{\epsilon}'] $ #### Compliance relation Inverting the above equation $ \begin{align} [\underline{\epsilon}]&=\frac{1}{9K}(\text{tr}\underline{\sigma})[\underline{1}]+\frac{1}{2G}[\underline{\sigma}'] \\ \\ &=\frac{1}{E}(-\nu(\text{tr}\underline{\sigma})[\underline{1}] + (1+\nu)[\underline{\sigma}]) \end{align} $ #### Expanded form $ \begin{align} \epsilon_{11} &= \frac{1}{E} (\sigma_{11} - \nu (\sigma_{22} + \sigma_{33})) \\ \epsilon_{22} &= \frac{1}{E} (\sigma_{22} - \nu (\sigma_{33} + \sigma_{11}) )\\ \epsilon_{33} &= \frac{1}{E} (\sigma_{33} - \nu (\sigma_{11} + \sigma_{22}) ) \\ \epsilon_{12} &=\frac{1+\nu}{E}\sigma_{12}\\ \epsilon _{23}&=\frac{1+\nu}{E}\sigma_{23}\\ \epsilon _{31}&=\frac{1+\nu}{E}\sigma_{31}\\ \end{align} $ ## Stress ### Cauchy's Result $ [\underline{t}(\underline{x},\underline{n})]= [\underline{\sigma}(\underline{x})][\underline{n}] $ ### Stress Tensor $\underline{\sigma}$: tensor, physical load state, inputs vector $\underline{n}$ and outputs vector $\underline{t}$ $[\underline{\sigma}]$: matrix, representation of $\underline{\sigma}$ in a given basis $ [\underline{\sigma}] = \begin{bmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{11} & \sigma_{12} & \sigma_{13} \\\end{bmatrix} $ $ [\underline{\sigma}]=[\underline{\sigma}]^T $ ### Stress Decomposition $ [\underline{\sigma}]=\sigma_{M}[\underline{1}]+[\underline{\sigma}'] $ #### Spherical Part Tries to make the material change volume $ \sigma_{M}= \frac{1}{3}tr(\underline{\sigma}) $ #### Deviatoric Part Tries to make the material change shape $ [\underline{\sigma}']= [\underline{\sigma}]-\sigma_{M}[\underline{1}] $ $ tr([\underline{\sigma}'])=0 $ ### Principle Stress $ [\underline{\sigma}]^P= \begin{bmatrix} \sigma_{1}^P & 0 & 0 \\ 0 & \sigma_{2}^P & 0 \\ 0 & 0 & \sigma_{3}^P \end{bmatrix} $ No off-diags, no shear stress Each of $\sigma_{i}^P$ and $\underline{e}_{i}^P$ are eigenvalue eigenvector pairs of $[\underline{\sigma}]$ ### Common Stress States #### Simple Tension $ [\underline{\sigma}] = \begin{bmatrix} 0 & 0 & 0 \\ 0 & \sigma & 0 \\ 0 & 0 & 0 \end{bmatrix} $ #### Simple Shear $ [\underline{\sigma}] = \begin{bmatrix} 0 & 0 & \tau \\ 0 & 0 & 0 \\ \tau & 0 & 0 \end{bmatrix} $ #### Hydrostatic Pressure $ [\underline{\sigma}]=\begin{bmatrix} -p & 0 & 0 \\ 0 & -p & 0 \\ 0 & 0 & -p \end{bmatrix} $ #### Plane Stress One of $\sigma_{i}^P=0$ Can simplify to a 2D problem Does not lead to plane strain in most cases ## Strain ### Displacement Field $ \underline{u}(\underline{X})=\underline{x}-\underline{X} $ $ [\nabla u] = \begin{bmatrix} \frac{\delta u_{1}}{\delta x_{1}} & \frac{\delta u_{1}}{\delta x_{2}} & \frac{\delta u_{1}}{\delta x_{3}} \\ \frac{\delta u_{2}}{\delta x_{1}} & \frac{\delta u_{2}}{\delta x_{2}} & \frac{\delta u_{2}}{\delta x_{3}} \\ \frac{\delta u_{3}}{\delta x_{1}} & \frac{\delta u_{3}}{\delta x_{2}} & \frac{\delta u_{3}}{\delta x_{3}} \end{bmatrix} $ ### Strain Matrix $ \begin{align} [\underline{\epsilon}]&=\frac{1}{2}([\nabla \underline{u}]+[\nabla \underline{u}]^T) \\ \\ &= \begin{bmatrix} \frac{du_{1}}{dX_{2}} & \frac{1}{2}\left( \frac{du_{1}}{dX_{2}}+\frac{du_{2}}{dX_{1}} \right) & \dots\\ \frac{1}{2}\left( \frac{du_{1}}{dX_{2}} +\frac{du_{2}}{dX_{1}}\right) & \frac{du_{2}}{dX_{2}} & \dots \\ \dots & \dots & \dots \end{bmatrix} \end{align} $ - Strain matrix is symmetric - Off-diags reflect angle change, diags reflect length change - Diagonal reflects - Pure rotations and translations give $[\underline{\epsilon}]=0$ $ \epsilon_{ij} \begin{cases} i\neq j : & (\text{angle change between } \underline{e}_{i} \text{ and } \underline{e}_{j})/2 \\ i=j: & \text{relative length change} \end{cases} $ $[\underline{\epsilon}]$ relates the dot product from $\underline{a}$ and $\underline{b}$ to $\underline{\alpha}$ and $\underline{\beta}$ $ [\underline{a}]\cdot([\underline{\epsilon}][\underline{b}])=\frac{1}{2}(\underline{\alpha}\cdot\underline{\beta}-\underline{a}\cdot \underline{b}) $ ### Decomposing Strain $ \begin{align} [\underline{\epsilon}] &= \underbrace{ \frac{1}{3}\text{tr}(\underline{\epsilon})[\underline{1}] }_{ spherical \ part }+ \underbrace{ [\underline{\epsilon}'] }_{ deviatoric \ part } \\ \\ &=\frac{1}{9K}(\text{tr}\underline{\sigma})[\underline{1}]+\frac{1}{2G}[\underline{\sigma}'] \end{align} $ $ [\underline{\epsilon}']=\frac{1}{3}\text{tr}(\underline{\epsilon})[\underline{1}]-[\underline{\epsilon}] $ - Deviatoric part causes shape change - Spherical part causes volume change ### Common Deformations #### Simple Shear $ \underline{\epsilon}=\begin{bmatrix} 0 & \frac{\gamma}{2} & 0 \\ \frac{\gamma}{2} & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $ #### Uniform Compaction $ \underline{\epsilon}=\begin{bmatrix} -\frac{\Delta}{3} & 0 & 0 \\ 0 & -\frac{\Delta}{3} & 0 \\ 0 & 0 & -\frac{\Delta}{3} \end{bmatrix} $ $ \text{Volumetric Strain}=tr\underline{\epsilon}= \frac{\text{volume change}}{\text{initial vol}}=-\Delta $ ## Material Properties ### Young's Modulus Resistance to stretching $ E = \frac{\sigma_{11}}{\epsilon_{11}}=\frac{9KG}{3K+G} $ ### Shear Modulus Resistance to shear $ G=\frac{E}{2(1+\nu)} $ $ [\underline{\sigma}']=2G[\underline{\epsilon}'] $ ### Bulk Modulus Resistance to volume change $ K = \frac{E}{3(1-2\nu)} $ $ \sigma_{m}=3K\epsilon_{m}=K(tr\underline{\epsilon})[\underline{1}] $ ### Poisson's Ratio Ratio of lateral contraction to longitudinal strain $ \nu= -\frac{\epsilon_{22}}{\epsilon_{11}}=\frac{3K-2G}{6K+2G} $ As material becomes incompressible, $\frac{K}{G} \to \infty$ and $\nu\to \frac{1}{2}$ ### Lamé Coefficient $ \lambda=K-\frac{2}{3}G $ ## Stress Concentration :sob: ### Thick-Walled Cylinders ![[Pasted image 20240519042131.png|400]] ![[Pasted image 20240519042320.png|295]] Solution after $ \begin{align} \sigma_{rr} &= \frac{ p_{i}-p_{o}\left( \frac{b}{a} \right)^{2} - (p_{i}-p_{o})\left( \frac{b}{r} \right) ^{2}}{\left( \frac{b}{a} \right)^{2}-1} \\ \sigma_{\theta\theta} &= \frac{ p_{i}-p_{o}\left( \frac{b}{a} \right)^{2} +(p_{i}-p_{o})\left( \frac{b}{r} \right) ^{2}}{\left( \frac{b}{a} \right)^{2}-1} \\ \sigma_{zz} &= \frac{ 2\nu\left( p_{i}-p_{o}\left( \frac{b}{a} \right)^{2} +E\epsilon_{zz} \right)}{\left( \frac{b}{a} \right)^{2}-1} \end{align} $ $ u_{r} = \frac{1+\nu}{E} \frac{r}{\left( \frac{b}{a} \right)^{2}-1}((1-2\nu)\left( p_{i}-p_{o}\left( \frac{b}{a} \right)^{2} \right)+(p_{i}-p_{o})\left( \frac{b}{r}^{2} \right)) - \nu r\epsilon_{0} $ ### Stress Concentration Factor ### Failure Criterion ### Von Mises $ \sqrt{ \frac{3}{2} }\sqrt{ (\sigma_{1}')^{2}+(\sigma_{2}')^{2}+(\sigma_{3}')^{2} }=\bar{\sigma}\leq\sigma _{y} $ $ \bar{\sigma}=\sqrt{ \frac{1}{2}((\sigma_{11}-\sigma_{22})^{2}+(\sigma_{22}-\sigma_{33})^{2}+(\sigma_{11}-\sigma_{33})^{2})+3(\sigma_{12}^{2}+\sigma_{23}^{2}+\sigma_{13}^{2}) } $ ### Tresca $ f([\underline{\sigma}])=\frac{\sigma_{1}^P-\sigma_{2}^P}{2}=\tau_{max}\leq \frac{\sigma_{y}}{2}=\tau_{y,\text{tresca}} $ $\tau_{max}$ is maximal shear stress $\tau_{max}=\text{max}|\underline{m}\cdot (\underline{\sigma} \underline{n})|$ $\underline{m}\perp \underline{n}$ ### Fast Fracture Fast fracture occurs when $ \sigma_{1}^P\leq\sigma_{c}=\text{Critical Fracture Stress} $ We use $\sigma_{1}^P$ not von mises because micro cracks are so small that only local tension is relevant ## Large Deformations ### Stretch Alternative way of expressing strain that is more convenient for large deformations $ \lambda_{i}=\frac{dx_{i}}{dX_{i}} $ ![[Pasted image 20240328140355.png|350]] ### Volume Ratio $ J=\frac{\text{deformed volume}}{\text{original volume}} =\frac{dx_{1}dx_{2}dx_{3}}{dX_{1}dX_{2}dX_{3}}=\lambda_{1}\lambda_{2}\lambda_{3} $ If incompressible: $J=1$ ### Strain-Energy Density $ \psi(\lambda_{1},\lambda_{2},\lambda_{3},{\text{material properties}}) =\frac{\text{energy stored}}{\text{original volume}} =S_{1}\dot{\lambda}_{1}+S_{2}\dot{\lambda}_{2}+S_{3}\dot{\lambda}_{3} $ ### Engineering Stress $ S = \frac{F}{A_{0}}=\frac{ \partial \psi }{ \partial \lambda} $ ### Engineering Strain $ e = \frac{L-L_{0}}{L_{0}}=\lambda-1 $ ### True Stress $ \sigma = \frac{F}{A} = \frac{\lambda}{J} \frac{ \partial \psi }{ \partial \lambda} =s(1+e) $ ### True Strain $ \epsilon =\ln \frac{L}{L_{0}} = \ln(1+e) $ ### Incompressible Rubber Constitutive Relation Due to incompressibility, arbitrary hydrostatic pressure $p$ does no work ($\psi$ is unaffected) and $J=\lambda_{1}\lambda_{2}\lambda_{3}=1$ $ \sigma_{i}=\lambda_{i}\frac{ \partial \psi }{ \partial \lambda i } -p $ ### Neo-Hookean Simplest and most common strain-energy function for rubber $ \psi_{\text{NH}} = C\cdot (\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}-3) $ $C$ can be determined from a small deformation stretch or shear test: $C = \frac{E}{6}=\frac{G}{2}$ Experiments show that rubber is stiffer than neo-hookean beyond $\lambda=1.5$ ### Stretch Invariants Constants that are unaffected by rotation $ \begin{align} I_{1} &=\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2} \\ I_{2}&=\lambda_{1}^{2}\lambda_{2}^{2}+\lambda_{2}^{2}\lambda_{3}^{2}+\lambda_{1}^{2} \lambda_{3}^{2} \\ I_{3}&= \lambda_{1}^{2}\lambda_{2}^{2}\lambda_{3}^{2} \end{align} $ ### Other Material Models Neo-Hookean: $\psi=C\cdot(I_{1}-3)$ Mooney-Rivlin: $\psi = C_{1} \cdot (I_{1}-3) + C_{2}\cdot(I_{2}-3)$ ## Viscoelasticity ### Math functions #### Heaviside function $ h(t) =\begin{cases} 0 & t\leq 0 \\ 1 & t>0 \end{cases} $ #### Dirac function $ \delta(t) = \dot{h}(t) = \begin{cases} 0 & \text{for } t \neq 0 \\ \infty & \text{for } t=0 \end{cases} $ $ \int_{-\infty}^{\infty} \delta(t) \, dt=1 $ For any function $g(t)$, continuous at $t=0$: $ \int_{-\infty}^{\infty} g(t)\delta(t) \, dt=g(0) $ ### Material Properties #### Relaxation time $ \tau_{R} = \frac{\eta}{E_{2}} $ $\eta=\text{viscosity}$ #### Creep retardation time $ \tau_{C} = \frac{E_{1}+E_{2}}{E_{1}}\tau_{R} $ ### Standard Linear Solid Model for viscoelastic behavior consisting of springs and dampers $ (\eta+E_{1}\tau_{R})\dot{\epsilon}-\tau_{R}\dot{\sigma}=\sigma-E_{1}\epsilon $ ![[Pasted image 20240416173556.png|250]] ### Stress Relaxation Stress spikes and relaxes over time when a sudden strain is applied: $\epsilon=\epsilon_{0}h(t)$ e.g. clamping in a vice $ \begin{align} E_{r}(t) &=\frac{\sigma(t)}{\epsilon_{0}} = E_{2}e^{-t/\tau_{R}} + E_{1} \\ &=E_{\text{re}}+(E_{\text{rg}}+E_\text{re})e^{-t/\tau_{R}} \end{align} $ ### Creep How a material elongates over time after application of sudden stress: $\sigma = \sigma_{0}h(t)$ e.g. hanging a weight $ \epsilon(t)=C e^{-t/\tau_{C}} + \frac{\sigma_{0}}{E_{1}} $ $ J_{c}(t) = \frac{\epsilon(t)}{\sigma_{0}} = \left( \frac{1}{E_{1}+E_{2}}-\frac{1}{E_{1}} \right)e^{-t/\tau_{C}}+\frac{1}{E_{1}} $ ### Boltzmann Superposition Principle Relates stress history and strain history #### Discrete form Input: $ \sigma(t)=\sum_{i=1}^n h(t-t_{i})\Delta\sigma_{i} $ Output: $ \epsilon(t) = \sum_{i=1}^N J_{c}(t-t_{i})\Delta\sigma_{i} $ #### Stress relaxation integral form Input: $ \epsilon(t) = \int _{0^-}^t h(t-\tau)\frac{d\sigma(\tau)}{d\tau} \, d\tau $ Output: $ \sigma(t)=\int _{0^-}^t E_{r}(t-\tau) \frac{d\epsilon(t)}{d\tau} \, d\tau $ #### Creep integral form Input: $ \sigma(t) = \int _{0^-}^t h(t-\tau)\frac{d\epsilon(\tau)}{d\tau} \, d\tau $ Output: $ \epsilon(t)=\int _{0^-}^t J_{r}(t-\tau) \frac{d\sigma(\tau)}{d\tau} \, d\tau $ ### Correspondence Principle Pretend body is elastic and replace $J$ with $J_{c}$ or replace $E$ with $E_{r}$ $ \underline{u}(\underline{x},t) = \underline{u}_{0}(\underline{x})\cdot \frac{J_{c}(t)}{J} $ $ \underline{\sigma}(\underline{x},t) = \underline{\sigma}_{0}(\underline{x})\cdot \frac{E_{r}(t)}{E} $ ### Dynamic Mechanical Analysis (DMA) Viscoelastic response to oscillatory inputs #### Stress Input Stress Input: $ \sigma(t) = \sigma_{0}\cos \omega t $ Strain Output: $ \begin{align} \epsilon(t) &= \epsilon_{0}\cos(\omega t-\delta) \\ &=\sigma_{0} (J' \cos(\omega t)+J''\sin(\omega t)) \end{align} $ ![[Pasted image 20240408112504.png|425]] #### Strain Input Strain Input: $ \epsilon(t) = \epsilon_{0}\cos\omega t $ Stress Output: $ \begin{align} \sigma(t) &= \sigma_{0}\cos(\omega t-\delta) \\ &=\epsilon_{0}(E'\cos(\omega t) -E''\sin(\omega t)) \end{align} $ #### Material Properties ##### Loss Angle Phase lag of DMA, depends on frequency and temp: $\delta=\delta(\omega,T)$ Fully out of phase if $\delta=\frac{\pi}{2}$ ##### Loss Tangent $ \tan\delta = \frac{J''}{J'}=\frac{E''}{E'} $ ##### Storage compliance Measure of how in-phase the strain is with the stress, function of $\omega$ $ J' = \frac{\epsilon_{0}}{\sigma_{0}}\cos\delta $ ##### Loss compliance Measure of how out-of-phase the strain is with the stress, function of $\omega$ $ J'' = \frac{\epsilon_{0}}{\sigma_{0}}\sin\delta $ ##### Storage modulus Represents rate of energy absorbed by material $ E' = \frac{\sigma_{0}}{\epsilon_{0}}\cos\delta $ ##### Loss modulus Represents rate of energy dissipated by material $ E'' = \frac{\sigma_{0}}{\epsilon_{0}}\sin\delta $ #### Loss per cycle Power expended per volume: $ \begin{align} P &= \sigma \dot{\epsilon} \\ &=-\omega\sigma_{0}\epsilon_{0}\sin(\omega t)\cos(\omega t+\delta) \end{align} $ Work done in one cycle with period $T=\frac{2\pi}{\omega}$ (dissipation loss per cycle): $ \begin{align} W&=\int _{0}^T P \, dt \\ &=\pi\sigma_{0}\epsilon_{0}\sin\delta \\ &=\pi\sigma_{0}^{2}J'' \\ \end{align} $ Dissipation depends on $\delta, E'', \text{and } J''$ ![[Pasted image 20240416180522.png|375]] ## Plasticity ### True Strain and Stress In plasticity we use true strain and stress: $\epsilon = \ln \frac{L}{L_{0}}= \ln(1+e)$ $\sigma = \frac{P}{A} = s(1+e)$ $s = \frac{P}{A_{0}}$ $e = \frac{L}{L_{0}}-1$ ### Metal Plasticity Due to motion of dislocations (as opposed to stretching bonds) - Incompressible - Unaffected by hydrostatic pressure #### Hall-Petch Relation Grain boundaries inhibit dislocation motion. Smaller grains are stronger. $ \sigma_{y} = \sigma_{0} +\frac{k}{\sqrt{ D }} $ $\sigma_{0} =$ "large grain" yield strength $k = \text{fit parameter}$ $d = \text{grain size}$ ### Assumptions Elastic-Perfectly-Plastic: No strain hardening Rigid-Plastic: No elastic deformation Rigid-Perfectly-Plastic: No elastic deformation or strain hardening ### Elastic-Plastic Stress-Strain Response ![[Pasted image 20240518165338.png|300]] ### Elastoplasticity Stress is independent of strain rate in metals when $T<0.35T_{m}$ (melting point in K) | | 1D | 3D | | --------------------------------- | --------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------- | | Kinematic decomposition of strain | $\epsilon = \epsilon^{e} + \epsilon^{p}$ | $\underline{\epsilon}=\underline{\epsilon}^{e} + \underline{\epsilon}^{p}$ | | Elastic constitutive relation | $\sigma = E\epsilon^{e}$ | $\underline{\sigma}=2G(\underline{\epsilon}^{e})'+K(\text{tr}\underline{\epsilon})\underline{I}$ | | Equivalent tensile plastic strain | $\bar{\epsilon}^{p} =\int \lvert \dot{\epsilon}^{p} \rvert\, dt$ | $\bar{\epsilon}^{p}= \int \sqrt{ \frac{2}{3} } \cdot\lvert \dot{{\underline{\epsilon}}}^{p} \rvert\\ \, dt$ | | Yield Condition | $\lvert \sigma \rvert = Y(\bar{\epsilon}^{p})$ | $\bar{\sigma} = Y(\bar{\epsilon}^{p})$ | | Codirectionality | $\text{sign}(\sigma) = \text{sign}(d\epsilon^{p})$ | $\text{sign}(\underline{\sigma}') = \text{sign}(d \underline{e}^{p})$ | | Flow Rule | $\dot{\epsilon}^{p} = \dot{\bar{\epsilon}}^{p} \frac{\sigma}{\bar{\sigma}}$ | $\underline{\dot{\epsilon}}^{p}=\frac{3}{2}\dot{\bar{\epsilon}}^{p}\frac{\underline{\sigma}'}{\bar{\sigma}}$ | $\underline{\dot{\epsilon}}^{p} = \text{strain rate tensor}$ $\dot{\bar{\epsilon}}^{p} = \text{equivalent strain rate}$ $\bar{\sigma}= \text{von mises stress}$ $\underline{\sigma}' = \text{deviatoric stress state}$ $\lvert \underline{A} \rvert = \sqrt{ \sum_{ij} A_{ij}^{2} }$ $\text{sign}(\underline{A}) = \frac{\underline{A}}{\lvert \underline{A} \rvert }$ #### Incremental Form When you know the stress state, you can calculate the strain state as follows: $ d\epsilon_{ij} = d\epsilon_{ij}^{e} + d\epsilon_{ij}^{p} = \underbrace{ \frac{1}{E}((1+\nu)d\sigma_{ij} -\nu d\sigma_{kk}\delta_{ij}) }_{ d\epsilon_{ij}^{e} } +\underbrace{ \frac{3}{2}d\bar{\epsilon}^{P}\frac{\sigma_{ij}'}{\bar{\sigma}} }_{ d\epsilon_{ij}^{p} } $ or $ \dot{\epsilon}= \frac{1}{E}((1+\nu)\dot{\sigma} -\nu ~\text{tr}(\dot{\sigma})\underline{1}) + \frac{3}{2}\dot{\bar{\epsilon}}^{p}\frac{\sigma'}{\bar{\sigma}} $ #### Analog Model ![[Pasted image 20240518224943.png|300]] Slider has static friction $\sigma _y$ and Mr. Plasticity adds weights to it every time it moves (strain hardening). ### Viscoplasticity Metals exhibit viscoplasticity when $T>0.35T_{m}$ due to atomic diffusion affecting plastic flow - No yield strength: $\dot{\bar{\epsilon}}^{p} \neq 0$ whenever $\bar{\sigma} \neq 0$ - hot metals experience creep and stress relaxation - material response becomes rate dependent #### New Flow Rule: $ \dot{\bar{\epsilon}}^{p}=\dot{\epsilon}_{0}\left( \frac{\bar{\sigma}}{S} \right)^{1/m} $ If $T>0.5 T_{m}$, then we have $ \dot{\epsilon}_{0}= A e^{-Q/RT} $ $m =$ strain rate sensitivity parameter, $0<m<1$ $\dot{\epsilon}_{0} =$ reference strain rate $S =$ flow strength (like Y) #### Isothermal tensile loading: $ \dot{\epsilon} = \dot{\epsilon}^{e} + \dot{\epsilon}^{p} = \frac{\dot{\sigma}}{E} + B\sigma^{n} = \frac{\dot{\epsilon}_{0}}{S^{n}} $ #### Analog Model ![[Pasted image 20240506104915.png|400]] $ \dot{\bar{\epsilon}}^{p}=\dot{\epsilon}_{0}\left( \frac{<\bar{\sigma}-Y_{th}(\bar{\epsilon}^{P})>}{S} \right)^{1/m} $ $ <u> = \begin{cases} u & \text{if } u\geq0 \\ 0 & \text{if } u<0 \end{cases} $ $Y_{th}(\bar{\epsilon}^{p}) =$ threshold resistance ## Fracture ### Stress Intensity Factor $K_{I}$ is determined by geometry and loading conditions $ K_{I} = Q \sigma_{\infty}\sqrt{ \pi a } $ $Q=$ configuration factor, $Q=1$ for infinite plate $a=$ crack radius, $2a =$ crack width $\sigma_{\infty} =$ stress applied perpendicular to crack ### Critical Stress Intensity factor $K_{IC}$ is a function of material properties $ K_{IC} = \sigma_{c}\sqrt{ \frac{\pi a_{0}}{2} } $ $a_{0} =$ lattice spacing of atoms $\sigma_{c} =$ critical stress Crack propagates when $K_{I} = K_{IC}$ (if factor of safety, then $K_{I} = \frac{K_{IC}}{FoS}$) $K_{I}=K_{Ic}$ only if $r_{Ic} \ll \{a, (w-a), B, h\}$ ### Critical Stress $ \sigma_{c} = 2\sqrt{ \frac{a}{a_{0}} } \sigma^{\infty } $ $a_{0} =$ lattice spacing of atoms ### Local Stress $ t(r_{k},\theta) = \frac{K_{I}}{\sqrt{ 2\pi r }}[f(\theta)] \cdot \hat{\underline{n}} $ $ \sigma_{22}(r,\theta=0) = \frac{K_{I}}{\sqrt{ 2\pi r }} \leq \sigma_{y} $ $\sigma_{22} =$ stress component that tries to open the crack ### Small Scale Yielding Condition The plastic deformation zone radius is given by $ r_{Ip} = \frac{1}{2\pi}\left( \frac{K_{I}}{\sigma_{y}} \right)^{2} $ For linear elastic fracture mechanics to apply, specimen dimensions much be much greater than $r_{Ip}$: $ \{ a, w-a, h, B \} > 15r_{Ip} $ ![[Pasted image 20240519014614.png|300]] #### Superposition By linearity, we can superpose loading conditions and stress intensity factors: $ \sigma_{\infty} =\sigma_{\infty,1} + \sigma_{\infty,2} + \sigma_{\infty,3} + \dots + \sigma_{\infty, n} $ $ K_{I} = K_{I,1} + K_{I,2} + K_{I, 3} + \dots + K_{I,n} $ $ K_{Ic} = \sum_{i=1}^{n}K_{I,i} $ ## Fatigue Failure of a structure due to repeated cyclic loading that causes local plastic deformation ![[Pasted image 20240519023924.png|300]] $\sigma_{a} =$ stress amplitude $\Delta\sigma = \sigma _\text{max} - \sigma _\text{min} =$ stress range $\sigma_{m} =$ mean stress ### Defect-Free Approach Estimate number of cycles until crack is initiated, then assume failure is immediate #### S-N Curves Empirically determined plot of stress amplitude vs number of cycles to failure ![[Pasted image 20240519024026.png|300]] Endurance limit: some metals have a value $S=\sigma_{a}$ such that $N_{f}\to \infty$ Pseudo-endurance limit: $\sigma_{a}$ corresponding to $N_{f}=10^{7}$ #### Strain-Life Uses strain amplitude instead of stress amplitude to better model plastic deformations where failure occurs after only a few cycles $ \epsilon_{a} = \underbrace{ \frac{\sigma_{f}'}{E}(2N_{f})^{b} }_{ \text{Basquin} } + \underbrace{ \epsilon_{f}' \cdot(2N_{f})^{c} }_{ \text{Coffin-Manson }} $ High cycle fatigue — Basquins' Relation ($N_{f} > 10^{4} \text{ cycles}$) Low-cycle fatigue — Coffin-Manson relation ($N _{f}< 10^{4} \text{ cycles}$) $\sigma_{f}', \epsilon_{f}', b, c =$ material parameters ### Defect-Tolerant Approach Assume a crack of initial size $a_{i}$ exists at the most highly stressed location $a_{i}$ is set to either the largest measured crack or the minimum crack size that can be reliably measured #### Critical crack size Crack length that allows fracture to occur at maximum stress $ a_{c} =\frac{1}{\pi}\left( \frac{K_{Ic}}{Q\sigma _\text{max}} \right)^{2} $ $a_{c} =$ critical crack size ### Fatigue Crack Growth $ \Delta K_{I} = Q(\sigma _\text{max} - \sigma _\text{min})\sqrt{\pi a } $ $\Delta K_{I} =$ stress intensity range $\frac{da}{dN} =$ crack growth rate ![[Pasted image 20240519032836.png|375]] ![[Pasted image 20240519032311.png|300]] - Low $\Delta K_{I}$: Cracks don't grow beneath a threshold $\Delta K_{Ith}$ - Moderate $\Delta K_{I}$: Power law fit between $\frac{da}{dN}$ and $\Delta K_{I}$ - High $\Delta K_{I}$: Fracture occurs in a few cycles #### Paris's Law $ \frac{da}{dN} = \begin{cases} 0 & \text{if } \Delta K_{I}< \Delta K_{Ith} \\ C(\Delta K_{I})^{m} & \text{if } \Delta K_{I} > \Delta K_{Ith} \end{cases} $ $C, m =$ experimentally determined material constants So in the moderate $\Delta K_{I}$ region we have $ \frac{da}{dN} =C[Q(\sigma _\text{max} - \sigma _\text{min})\sqrt{ \pi a }]^{m} $ Integrating gives us If $m \neq 2$: $ N_{f} = \frac{2}{(m-2)C(Q\Delta\sigma\pi)^{m}}[a_{i}^{(2-m)/2}-a_{f}^{(2-m)/2}] $ If $m=2$: $ N_{f}=\frac{1}{C} \frac{1}{(Q\Delta\sigma \sqrt{ \pi })^{2}}\left[ \ln\left( \frac{a_{f}}{a_{i}} \right) \right] $