Let us consider the simplest solid mechanics problem that we can think of - the axial loading of a bar (see Figure 1).

Given:

The bar is of length ${\displaystyle L\,}$ and has an area of cross-section ${\displaystyle A\,}$. The Young's modulus of the material is ${\displaystyle E\,}$.

Find:

We want to find the stresses and deformations in the bar due to the concentrated axial load ${\displaystyle \mathbf {R} }$ at the end.

Assumptions:

2. The material is linear elastic, isotropic, and homogeneous.
4. End-effects are not of interest to us.

### Strength of Materials Approach

From the free-body diagram of the bar, a balance of forces gives

${\displaystyle \mathbf {f} _{\text{wall}}=\mathbf {R} =\mathbf {f} (\mathbf {x} )~.}$

The stress in the bar is given by

${\displaystyle {\boldsymbol {\sigma }}(\mathbf {x} ):={\cfrac {\mathbf {f} (\mathbf {x} )}{A}}\qquad \implies \qquad {{\boldsymbol {\sigma }}(\mathbf {x} )={\cfrac {\mathbf {R} }{A}}}~.}$

The constitutive model for the bar is given by Hooke's law (${\displaystyle {\boldsymbol {\sigma }}=E{\boldsymbol {\varepsilon }}}$). Therefore, the strain in the bar is

${\displaystyle {\boldsymbol {\varepsilon }}(\mathbf {x} )={\cfrac {{\boldsymbol {\sigma }}(\mathbf {x} )}{E}}\qquad \implies \qquad {{\boldsymbol {\varepsilon }}(\mathbf {x} )={\cfrac {\mathbf {R} }{AE}}}~.}$

To get the deformation of the bar, we use the strain-displacement relations

${\displaystyle {\boldsymbol {\varepsilon }}:={\cfrac {\Delta l}{l}}={\cfrac {\delta _{(\mathbf {x} =L)}}{L}}\qquad \implies \qquad {\delta _{(\mathbf {x} =L)}={\cfrac {\mathbf {R} L}{AE}}}~.}$

We can get the deformation of any point in the bar in a similar manner:

${\displaystyle {\boldsymbol {\varepsilon }}(\mathbf {x} )={\cfrac {\delta (\mathbf {x} )}{\mathbf {x} }}\qquad \implies \qquad {\delta (\mathbf {x} )={\cfrac {\mathbf {R} \mathbf {x} }{AE}}}~.}$

## A more complicated axial load

Now, suppose that we have a linear distributed axial load ${\displaystyle \mathbf {q} (\mathbf {x} )=a\mathbf {x} }$ in addition to the load ${\displaystyle \mathbf {R} }$ (see Figure 2(a)). This load has units of force per unit length and is also called a body load. Examples are gravity and inertial forces due to rotation around the ${\displaystyle y}$ axis.

${\displaystyle \mathbf {f} _{\text{wall}}=\mathbf {R} +\int _{0}^{L}\mathbf {q} (\mathbf {x} )~dx\qquad \implies \qquad {\mathbf {f} _{\text{wall}}=\mathbf {R} +{\cfrac {aL^{2}}{2}}}~.}$
To do this, we can make a cut at any point along the length and use the free body diagram to compute a reaction force ${\displaystyle \mathbf {f} (\mathbf {x} )}$ (see Figure 2(b)). But this is tedious since we have to repeat the process for every point in the bar.