Continuum mechanics problem: Stresses

[MarkusNaslund19]

Homework Statement

Mechanics of Deformable Media (Bhatia and Singh), 5.2:

Consider a long rod of elastically isotropic material of L standing vertically in a vacuum in equilibrium under the gravitational field of the earth, then:

(i) What are the boundary conditions for $$\sigma_{ij}$$ on the various surfaces of the rod?

(ii) Solve for $$\sigma_{ij}$$.

Homework Equations

$$\sigma_{ij}$$ is defined as the stress on the ith surface in the jth direction.

Equilibrium suggests ($$\lambda$$ + 2$$\mu$$)grad(div$$\vec{s}$$) - $$\mu$$ curl(curl$$\vec{s}$$) + $$\rho$$ F = 0

where lambda and mu are constants, s is the displacement field, and F is the body force and rho the density of the material.

The Attempt at a Solution

The force IN the rod is the reactant force from it's weight exerted by the surface it rests on. It is purely in the z-axis, in cylindrical polar coordinates.

$$\sigma_{zz} = -\frac{mg}{A}$$ at z=0, the end of the rod touching the surface

$$\sigma_{zz} = 0$$ at z=L, the top of the rod since we're in a vacuum

All other stresses at the boundaries are 0.

I am not sure if my boundaries above are correct and it is not clear to me how to find the stress tensor for part (ii).

 

[mark726]

Can someone help me with this problem? it is important to approach problems with a clear understanding of the concepts and equations involved. In this case, we are dealing with the mechanics of deformable media, specifically a long rod made of elastically isotropic material.

First, let's define the boundary conditions for the stress tensor, \sigma_{ij}, on the various surfaces of the rod. Since the rod is standing vertically in a vacuum, we can assume that the only external force acting on it is the gravitational force from the Earth. Therefore, the boundary conditions for the stress tensor can be written as:

\sigma_{zz} = -\frac{mg}{A} at z=0, the end of the rod touching the surface

\sigma_{zz} = 0 at z=L, the top of the rod since we're in a vacuum

All other stresses at the boundaries are 0.

Next, let's look at the equation for equilibrium, which is given as (\lambda + 2\mu)grad(div\vec{s}) - \mu curl(curl\vec{s}) + \rho F = 0. This equation relates the stress tensor to the displacement field, s, and the body force, F. In this case, the body force is the gravitational force, which is given by F = \rho g, where \rho is the density of the material and g is the gravitational acceleration.

To solve for the stress tensor, we need to use the appropriate boundary conditions and the equation for equilibrium. Since we are dealing with a cylindrical polar coordinate system, we can use the appropriate expressions for the stress tensor \sigma_{ij} in terms of the displacement field, s, which are given as:

\sigma_{rr} = \lambda \frac{\partial s_r}{\partial r} + 2\mu \frac{\partial s_r}{\partial r}

\sigma_{\theta \theta} = \lambda \frac{s_r}{r} + \mu \left(\frac{\partial s_\theta}{\partial r} + \frac{s_\theta}{r}\right)

\sigma_{zz} = \lambda \frac{s_r}{r} + \mu \frac{s_\theta}{r}

\sigma_{r \theta} = \mu \left(\frac{\partial s_r}{\partial \theta} + \frac{1}{r}\frac{\partial s_\theta}{\