Applied Mechanics

Table of contents

The Rod Problem - a second order boundary value problem

Figure 1 shows a model of an elastic rod with a constant cross sectional area, $A\text{ [mm}^{2}\text{]}$, exposed to a boundary force $F\text{ [N]}$ at the far end of the rod at the length $L\text{ [mm]}$, as well as a body load $f(x)\text{[N/mm]}$ acting along the entirety of the rod. The body load can be seen as a gravitational force.

Problem definition

The problem is to derive an expression for the displacement, $u$, at every point $x$ with respect to the load case, which in this case consists of the two boundary values: $u(0)=0$ as well as the load $F$ at $x=L$.

Solution

We shall begin by examining an infinitesimal element, a cut of size $\Delta x$, of the rod at an arbitrary location $x$, see the figure below

Force equilibrium at $x$ along the rod yields:

where $N$ denotes the normal force and the third term is the body load summed up over the length of the infinitesimal element. Note that at the boundary $x=L$ we have $N(L)=F$. If we now let $\Delta x \rightarrow 0$ then naturally $\int_{x}^{x+\Delta x}f(\xi)d\xi\rightarrow f(x)$ and using the definition of the derivative we get

which leads to the point-wise equilibrium equation

The relation between forces and stresses assumes a constant cross-sectional area:

where $\sigma$ denotes the stress with a unit of MPa or $\text{N/mm}^{2}$.

The relation between the stress and the strain is assumed to be linear, hence Hooke's law is used to describe it

The strain is assumed to be small such that it can be defined as

(this is also know as the small strain assumption) and with the displacements in our infinitesimal element this becomes

Now we insert (7) into (5) and the result into (4) and (3) and we get the point-wise equilibrium ordinary differential equation

We can set the body load $f(x)$ to be the point-wise gravity, i.e., $f(x) = \rho g$. This BVP can be easily solved in Matlab as follows:

syms rho g u(x) E A L F
f = rho*g;
du = diff(u);
DE = -diff(E*A*du) == f
BC = [u(0) == 0; E*A*du(L) == F]

u = dsolve(DE,BC)
u = expand(u)

and with $x=L$ we get

x = L;
subs(u)