What is the difference between stress and pressure? Are there any intuitive examples that explain the difference between the two? How about an example of when pressure and stress are not equal?
Stress vs Pressure – Understanding the Differences
definitionnewtonian-mechanicspressurestress-strain
Related Solutions
Generalities
The problem has spherical symmetry, so it makes sense to use spherical coordinates ($r$, $\theta$, $\phi$). We can divide the vessel into differential elements like the one shown in this post.
Deformation and strain
Only radial deformations are allowed by the spherical symmetry, so let's parametrize the deformation by
$$r \rightarrow r + u(r)$$
Assuming small deformations:
$$dr \rightarrow dr(1 + u'(r))$$
$$r\,d\theta \rightarrow (r + u(r))\,d\theta$$
$$r\cos\theta\,d\phi \rightarrow (r + u(r))\cos\theta\,d\phi$$
The associated strains:
$$\epsilon_{rr} = \frac{dr(1 + u'(r)) - dr}{dr} = u'(r)$$
$$\epsilon_{\theta\theta} = \frac{(r + u(r))\,d\theta - r\,d\theta}{r\,d\theta} = \frac{u(r)}{r}$$
$$\epsilon_{\phi\phi} = \frac{(r + u(r))\cos\theta\,d\phi - r\cos\theta\,d\phi}{r\cos\theta\,d\phi} = \frac{u(r)}{r}$$
Symmetry
By spherical symmetry we have:
$$\epsilon_{\theta\theta} = \epsilon_{\phi\phi} = \epsilon_{tt}$$
$$\sigma_{r\theta} = \sigma_{r\phi} = \sigma_{\theta\phi} = 0$$
$$\sigma_{\theta\theta} = \sigma_{\phi\phi} = \sigma_{tt}$$
$$\frac{\partial\,\sigma_{rr}}{\partial\,\theta} = \frac{\partial\,\sigma_{\theta\theta}}{\partial\,\theta} = \frac{\partial\,\sigma_{\phi\phi}}{\partial\,\theta} = \frac{\partial\,\sigma_{rr}}{\partial\,\phi} = \frac{\partial\,\sigma_{\theta\theta}}{\partial\,\phi} = \frac{\partial\,\sigma_{\phi\phi}}{\partial\,\phi} = 0$$
Equilibrium condition
The equilibrium condition can be expressed in spherical coordinates as
$$2\sigma_{rr} + r\frac{\partial\,\sigma_{rr}}{\partial\,r}-\sigma_{\theta\theta}-\sigma_{\phi\phi} = 0$$
or, using the symmetry conditions,
$$2\sigma_{rr}(r) + r\sigma_{rr}'(r) - 2\sigma_{tt} = 0$$
Hooke's law
Applying Hooke's law for isotropic materials we get
$$\epsilon_{rr} = \frac{1}{E}(\sigma_{rr} - 2 \nu \sigma_{tt})$$
$$\epsilon_{tt} = \frac{1}{E}(\sigma_{tt} - \nu (\sigma_{rr} + \sigma_{tt}))$$
where $E$ is the elastic modulus and $\nu$ is the Poisson's ratio.
Math
Combining the previous results we get the following equations:
$$u'(r) = \frac{1}{E}(\sigma_{rr} - 2 \nu \sigma_{tt})$$
$$\frac{u(r)}{r} = \frac{1}{E}(\sigma_{tt} - \nu (\sigma_{rr} + \sigma_{tt}))$$
$$2\sigma_{rr}(r) + r\sigma_{rr}'(r) - 2\sigma_{tt} = 0$$
From there we can get the following differential equation for the deformation (by a tedious but quite straightforward path):
$$2 u'(r) - 2 \frac{u(r)}{r} + r\,u''(r) = 0$$
From there, it's easy to get
$$u(r) = A\,r + \frac{B}{r^2}$$
and, as a consequence,
$$\sigma_{tt} = E \frac{A}{1 - 2\nu} + E\frac{B}{1 + \nu}\frac{1}{r^3}$$
$$\sigma_{rr} = E \frac{A}{1 - 2\nu} - 2 E\frac{B}{1 + \nu}\frac{1}{r^3}$$
where $A$ and $B$ are integration constants determined to be consistent with the boundary conditions.
Solving the original problem
The original problem has zero external pressure, internal pressure $P$, interior radius $R_i$ and exterior radius $R_o$. For continuity, we must have
$$\sigma_{rr}(R_i) = -P$$
$$\sigma_{rr}(R_o) = 0$$
Getting the constants from these boundary conditions:
$$A = \frac{P(1 - 2\nu)}{E}\frac{R_i^3}{R_o^3 - R_i^3}$$
$$B = \frac{P(1 + \nu)}{2 E}\frac{R_o^3 R_i^3}{R_o^3 - R_i^3}$$
Getting the stresses:
$$\sigma_{rr} = P\frac{R_i^3}{R_o^3 - R_i^3}\left(1 - \frac{R_o^3}{r^3}\right)$$
$$\sigma_{tt} = P\frac{R_i^3}{R_o^3 - R_i^3}\left(1 + \frac{1}{2}\frac{R_o^3}{r^3}\right)$$
Consider a continuous body $C$ and a closed subset $V$ with boundary $\partial V$ completely included in $C$ (notice that $\partial V \subset V$). If $p \in \partial V$, the external part of $C$, namely $C \setminus V$ acts on $p$ (actually on a neighbourhood of $p$ in $\partial V$) by means of a superficial force $$d\vec{f} = \vec{s}(p, \vec{n}) dS(p)\:,$$ where $\vec{n}$ is the outgoing unit normal vector at $p$ and $dS(p)$ the infinitesimal surface around $p$ included in the larger surface made of the boundary $\partial V$. So $dS(p)$ is normal to $\vec{n}$.
The total force acting on $V$ due to the part of $C$ outside $V$ is: $$\vec{F}_V = \int_{+\partial V} \vec{s}(p, \vec{n}) dS(p)\quad (1)$$ The map $ C \times \mathbb{S}^2 \ni (p, \vec{n}) \mapsto \vec{s}(p, \vec{n})$ is called stress function.
Notice that a point $p \in C$ has several stress vectors applied on it, in view of the various choices of $\vec{n}$.
Moreover $ \vec{s}(p, \vec{n})$ is not necessarily parallel to $\vec{n}$, it could have components normal to $\vec{n}$ (so parallel to $dS(p)$) describing friction forces.
A fundamental theorem due to Cauchy establishes that, under some quite general hypotheses on the continuous body, the map $\mathbb{S}^2 \ni \vec {n} \mapsto \vec{s}(p, \vec{n})$ is the restriction of a linear map. As a consequence it is the action of a tensor field (I use Einstein's convention on indices below)
$$s^a(p, \vec{n}) = \sigma(p)^{ab} n_b$$
where $n_a$ are the covariant components of $\vec{n}$ and $s^a$ the contravariant components of $\vec{s}$ (actually using orthonormal frames there is no distinction between these two types of components).
The Cauchy stress-tensor $\sigma$ is always symmetric: $\sigma(p)^{ab}= \sigma(p)^{ba}$ (it can be proved by imposing the standard relation between momenta of forces and angular momentum).
The identity (1) can be re-written taking advantage of divergence theorem:
$$F^a_V = \int_{V} \sigma^{ab}, _b d^3x$$
where $,_b$ denotes the derivative with respect to $x^b$. This is the starting point to write down in local form "F=ma" for continuous bodies.
A very special case is that of an isotropic $\sigma$: $$\sigma(p)^{ab} = -P(p)\delta^{ab}\qquad (3)$$ where $P(p)\geq 0$ defines a scalar field. Here all the stress vectors are always normal to the boundary of $V$ no matter how you fix that portion of continuous body, and the total force is purely compressive. The scalar $P$ is the pressure. Non-viscous fluids have the stress tensor of the form (3). Even viscous fluids have that form if the fluid is in an equilibrium configuration.
The presence of viscous forces adds some non isotropic (i.e. non-diagonal) terms to the RHS of (3).
Best Answer
Pressure is defined as force per unit area applied to an object in a direction perpendicular to the surface. And naturally pressure can cause stress inside an object. Whereas stress is the property of the body under load and is related to the internal forces. It is defined as a reaction produced by the molecules of the body under some action which may produce some deformation. The intensity of these additional forces produced per unit area is known as stress (pretty picture from wikipedia):
EDIT PER COMMENTS
Overburden Pressure or lithostatic pressure is a case where the gravity force of the object's own mass creates pressure and results in stress on the soil or rock column. This stress increases as the mass (or depth) increases. This type of stress is uniform because the gravity force is uniform.
http://commons.wvc.edu/rdawes/G101OCL/Basics/earthquakes.html
Since this is a uniform force applied throughout the substance due to mostly to the substance itself, the terms pressure and stress are somewhat interchangeable because pressure can be viewed as both an external and internal force.
For a case where they are not equal, just look that the image of the ruler. If pressure is applied at the far end (top of image) it creates unequal stress inside the ruler, especially where the internal stress is high at the corners.