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+\documentclass[11pt,a4paper,dvipsnames]{article}
+\usepackage[utf8]{inputenc}
+
+\input{../preamble.tex}
+
+\title{Homework 4\\Continuum Mechanics}
+\author{Adrià Vilanova Martínez}
+\date{April 11, 2021}
+
+\begin{document}
+
+\maketitle
+
+\begin{Problem}
+ Consider a solid sphere of constant density $\rho$ and undeformed radius $R$, subjected to its own gravitational force. It is made of an elastically homogeneous and isotropic material of Young's modulus $Y$ and Poisson's ratio $\nu$.
+
+ \begin{enumerate}
+ \item Determine the displacement field of the material particles in the sphere.
+ \item Compute the corresponding strain and stress fields.
+ \end{enumerate}
+
+ \underline{Note:} The gravitational force per unit volume experienced by a material particle is: $\vec{f} = \rho \vec{g} = - \rho g \frac{r}{R} \hat{e}_r$, where $g = \frac{4}{3} \pi G \rho R$, with $G$ the gravitational constant.
+\end{Problem}
+
+\Solution
+The displacement field in spherical coordinates will be $\vec{u} = (u_r(r), 0, 0)$ due to the symmetry of the problem.
+
+Let's use Navier-Cauchy's equation:
+\[ \vec{f} + \mu \vec{\nabla}^2 \vec{u} + (\lambda + \mu) \vec{\nabla} (\vec{\nabla} \cdot \vec{u}) = 0 \]
+which in spherical coordinates becomes:
+\[ \begin{pmatrix}
+ - \rho g \frac{r}{R} \\
+ 0 \\
+ 0
+\end{pmatrix} + \mu \begin{pmatrix}
+ \partial^2_{rr} u_r + \frac{2}{r} \partial_r u_r - \frac{2}{r^2} u_r \\
+ 0 \\
+ 0
+\end{pmatrix} + (\lambda + \mu) \begin{pmatrix}
+ \partial^2_{rr} u_r + \frac{2}{r} \partial_r u_r - \frac{2}{r^2} u_r \\
+ 0 \\
+ 0
+\end{pmatrix} = 0 \iff \]
+\[ \iff - \rho g \frac{r}{R} + (\lambda + 2 \mu) \left( \partial^2_{rr} u_r + \frac{2}{r} \partial_r u_r - \frac{2}{r^2} u_r \right) = 0 \iff \]
+\[ \iff r^2 \partial^2_{rr} u_r + 2 r \partial_r u_r - u_r = \frac{\rho g}{(\lambda + 2 \mu) R} r^3 =: k r^3. \]
+
+If we take as an \textit{ansatz} the function
+\[ u_r(r) := a \cdot r^b, \]
+we can see that a particular solution to the ODE is
+\[ u_r(r) = \frac{k}{10} r^3. \]
+
+Due to the fact that the homogeneous part of the ODE has the solution space
+\[ \{ A r + B \frac{1}{r^2} : A, B \in \mathbb{R} \}, \]
+and the fact that since $u_r(r = 0) \neq \infty \implies B = 0$, we conclude that \[ \vec{u}(r) = \left( \frac{k}{10} r^3 + A r, 0, 0 \right). \]
+
+Since we have
+\[ \vec{\nabla} \vec{u} = \begin{pmatrix}
+ \frac{3}{10} k r^2 + A & 0 & 0 \\
+ 0 & \frac{1}{10} k r^2 + A & 0 \\
+ 0 & 0 & \frac{1}{10} k r^2 + A
+\end{pmatrix}, \]
+Cauchy's strain tensor is:
+\[ \TT{u} = \frac{1}{2} \left( \vec{\nabla} \vec{u} + (\vec{\nabla} \vec{u})^T \right) = \vec{\nabla} \vec{u}. \]
+
+Via Hooke's law, the only non-zero components of the stress vector are the ones in the diagonal:
+\[ \sigma_{ii} = 2 \mu u_{ii} + \lambda \Tr \TT{u} \implies \]
+\[ \TT{\sigma} = \begin{pmatrix}
+ \frac{6 \mu + 5 \lambda}{10} k r^2 + (3 \lambda + 2 \mu) A & 0 & 0 \\
+ 0 & \frac{2 \mu + 5 \lambda}{10} k r^2 + (3 \lambda + 2 \mu) A & 0 \\
+ 0 & 0 & \frac{2 \mu + 5 \lambda}{10} k r^2 + (3 \lambda + 2 \mu) A \\
+\end{pmatrix}. \]
+
+If we impose the boundary condition $\sigma_{rr} (r = R) = 0$ (free surface), we get:
+\[ (3 \lambda + 2 \mu) A = \frac{6 \mu + 5 \lambda}{10} k R^2. \]
+
+We can transform $Y$ and $\nu$ into the Lamé coefficients by applying the following transformation:
+\[ \begin{cases}
+ \lambda = \frac{Y \nu}{(1 - 2 \nu) (1 + \nu)}, \\
+ \mu = \frac{Y}{2 (1 + \nu)}.
+\end{cases} \]
+
+\newpage
+
+\begin{Problem}
+ A cylindrical pipe of inner radius $R_0$, outer radius $R_1$, and Lame coefficients $\mu$ and $\lambda$, is subjected to an internal pressure $p_0$, an external pressure $p_1$, and a uniform tensile force per unit area $F_z/A$ along the symmetry axis of the pipe $z$. Consider the end of the pipe at $z = 0$ is clamped to a rigid wall.
+
+ \begin{enumerate}
+ \item Determine the displacement field corresponding to the elastic deformation of the pipe. Neglect gravity and corrections due to the finite length of the pipe.
+
+ \item Compute the stress and strain fields corresponding to such deformation.
+ \end{enumerate}
+\end{Problem}
+
+We'll consider the displacement field resulting from the hydrostatic pressure, and the one resulting from the tensile force.
+
+\textbf{Displacement field resulting from the hydrostatic pressure:}
+
+Due to the symmetry of the problem in this case, we have that $\vec{u} = (u_r(r), 0, 0)$ in cylindrical coordinates.
+
+By applying the Navier-Cauchy equation, we get:
+\[ (\lambda + 2 \mu) \frac{d}{dr} \left( \frac{1}{r} \frac{d}{dr} (r u_r) \right) \implies u_r = \alpha r + \beta \frac{1}{r}. \]
+
+Therefore, we have that the only non-zero components of the strain tensor are:
+\[ u_{rr} = \frac{d u_r}{dr} = \alpha - \beta \frac{1}{r^2}, \]
+\[ u_{\phi\phi} = \frac{u_r}{r} = \alpha + \beta \frac{1}{r^2}. \]
+
+By using Hooke's law, we get:
+\[ \sigma_{rr} = 2 \alpha (\lambda + \mu) - 2 \beta \mu \frac{1}{r^2}, \]
+\[ \sigma_{\theta \theta} = 2 \alpha (\lambda + \mu) + 2 \beta \mu \frac{1}{r^2}, \]
+\[ \sigma_{zz} = 2 \alpha \lambda. \]
+
+We can now impose the boundary conditions $\sigma_{rr}(z = R_0) = - p_0$, $\sigma_{rr}(z = R_1) = - p_1$, and we get:
+\[ \alpha = \frac{p_0 R_0^2 - p_1 R_1^2}{2 (\lambda + \mu) (R_1^2 - R_0^2)}, \]
+\[ \beta = \frac{(p_0 - p_1) R_0^2 R_1^2}{2 \mu (R_1^2 - R_0^2)}. \]
+
+\textbf{Displacement field resulting from the tensible force:}
+
+Due to the uniform force, we have that:
+\[ \sigma_{zz} = \frac{F_z}{A}, \]
+which is the only non-zero component of the stress tensor. Therefore, from Hooke's law we have that the only non-zero components of the strain tensor are:
+\[ u_{zz} = \frac{1}{Y} \sigma_{zz} = \frac{F_z}{A Y}, \]
+\[ u_{rr} = - \nu u_{zz} = - \frac{F_z \nu}{A Y}, \]
+\[ u_{\theta \theta} = - \nu u_{zz} = - \frac{F_z \nu}{A Y}. \]
+
+Now we can integrate (in cylindrical coordinates) the previous expressions to find the displacement field:
+\[ u_{zz} = \frac{du_z}{dz} \implies u_z(z) = \frac{F_z}{AY} z + c, \]
+and since $u_z(0) = 0$, we have:
+\[ u_z(z) = \frac{F_z}{AY} z. \]
+
+Also:
+\[ u_{rr} = \frac{du_r}{dr} \implies u_r(r) = - \frac{F_z \nu}{A Y} r + c', \]
+and since $u_{\theta \theta} = \frac{u_r(r)}{r} \implies c' = 0$, then:
+\[ u_r(r) = - \frac{F_z \nu}{A Y} r \]
+
+\textbf{Conclusion:}
+
+Thanks to the superposition principle, we can sum the previous displacement fields to calculate the total displacement field:
+\[ \vec{u}(r) = \begin{pmatrix}
+ \left( \alpha - \frac{F_z \nu}{A Y} \right) r + \beta \frac{1}{r} \\
+ 0 \\
+ \frac{F_z}{A Y} z
+\end{pmatrix}. \]
+
+As for the strain and stress tensors, we can do the same in order to get their total expressions:
+\[ \TT{\sigma} = \begin{pmatrix}
+ 2 \alpha (\lambda + \mu) - 2 \beta \mu \frac{1}{r^2} & 0 & 0 \\
+ 0 & 2 \alpha (\lambda + \mu) + 2 \beta \mu \frac{1}{r^2} & 0 \\
+ 0 & 0 & 2 \alpha \lambda + \frac{F_z}{A}
+\end{pmatrix}, \]
+\[ \TT{u} = \begin{pmatrix}
+ \alpha - \beta \frac{1}{r^2} - \frac{F_z \nu}{A Y} & 0 & 0 \\
+ 0 & \alpha + \beta \frac{1}{r^2} - \frac{F_z \nu}{A Y} & 0 \\
+ 0 & 0 & \frac{F_z}{A Y}
+\end{pmatrix}. \]
+
+\end{document}