---
title: "Euler-Bernoulli law"
date: 2021-06-03
categories:
- Physics
layout: "concept"
---

**Euler-Bernoulli beam theory** concerns itself with the bending of beams
(e.g. the metal beams used in large buildings),
subject to certain simplifying assumptions,
which are generally valid for beams that are narrow,
i.e. longitudinally much larger than transversely.

Consider a beam of length $L$, placed upright
on the $z = 0$ plane, above the origin.
If we pull the top of this beam in the postive $y$-direction,
we assume that it bends uniformly,
i.e. with constant radius of [curvature](/know/concept/curvature/) $R$.
We also assume that the bending is **shear-free**:
if we treat the beam as a bundle of elastic strings,
then there is no friction between them.

The central string has its length unchanged (i.e. still $L$),
while an arbitrary non-central string is extended or compressed to $L'$.
The [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) element $u_{zz}$ is then:

$$\begin{aligned}
    u_{zz}
    = \frac{L' - L}{L}
\end{aligned}$$

Because the bending is uniform, the central string
is an arc with radius $R$ and central angle $\theta$,
where $L = \theta R$.
The non-central string has $L' = \theta R'$,
where $R'$ is geometrically shown to be $R' = R - y$,
with $y$ being the $y$-coordinate of that string at the beam's base.
So:

$$\begin{aligned}
    u_{zz}
    = \frac{R' - R}{R}
    = - \frac{y}{R}
\end{aligned}$$

By assumption, there are no shear stresses
and no forces acting on the beam's sides,
so the only nonzero component of the
[Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
is $\sigma_{zz}$, given by [Hooke's law](/know/concept/hookes-law/):

$$\begin{aligned}
    \sigma_{zz} = E u_{zz}
\end{aligned}$$

Where $E$ is the elastic modulus of the material.
By Hooke's inverse law,
the other nonzero strain components are as follows,
where $\nu$ is Poisson's ratio:

$$\begin{aligned}
    u_{xx}
    = u_{yy}
    = - \frac{\nu}{E} \sigma_{zz}
    = - \nu u_{zz}
    = \nu \frac{y}{R}
\end{aligned}$$

For completeness, we turn the strain tensor $\hat{u}$
into a full displacement field $\va{u}$:

$$\begin{aligned}
    \boxed{
        u_x = \nu \frac{x y}{R}
        \qquad
        u_y = \frac{z^2}{2 R} + \nu \frac{y^2 - x^2}{2 R}
        \qquad
        u_z = - \frac{y z}{R}
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-field"/>
<label for="proof-field">Proof</label>
<div class="hidden">
<label for="proof-field">Proof.</label>
By integrating the above strains $u_{ii} = \ipdv{u_i}{i}$,
we get the components of $\va{u}$:

$$\begin{aligned}
    u_x
    = \nu \frac{x y}{R} + f_x(y, z)
    \qquad
    u_y
    = \nu \frac{y^2}{2 R} + f_y(x, z)
    \qquad
    u_z
    = - \frac{y z}{R} + f_z(x, y)
\end{aligned}$$

Where $f_x$, $f_y$ and $f_z$ are integration constants,
which we find by demanding that the off-diagonal strains $u_{ij}$ are zero.
Starting with $u_{xz} = 0$:

$$\begin{aligned}
    0
    = u_{xz}
    = \frac{1}{2} \Big( \pdv{u_x}{z} + \pdv{u_z}{x} \Big)
    = \frac{1}{2} \Big( \pdv{f_x}{z} + \pdv{f_z}{x} \Big)
\end{aligned}$$

Here, only $f_x$ may depend on $z$,
and only $f_z$ may depend on $x$.
This equation thus tell us:

$$\begin{aligned}
    f_x(y, z)
    = z \: g(y)
    \qquad \quad
    f_z(x, y)
    = - x \: g(y)
\end{aligned}$$

Where $g(y)$ is an unknown integration constant.
Moving on to $u_{xy} = 0$:

$$\begin{aligned}
    0
    = \frac{1}{2} \Big( \pdv{u_x}{y} + \pdv{u_y}{x} \Big)
    = \frac{1}{2} \Big( \nu \frac{x}{R} + \pdv{f_x}{y} + \pdv{f_y}{x} \Big)
\end{aligned}$$

Only $f_x$ may contain $y$,
so its $y$-derivative must be a constant,
so $g(y) = C y$. Therefore:

$$\begin{aligned}
    f_x(y, z)
    = C y z
    \qquad
    f_y(x, z)
    = - \nu \frac{x^2}{2 R} - C x z + h(z)
    \qquad
    f_z(x, y)
    = - C x y
\end{aligned}$$

Where $h(z)$ is an unknown integration constant.
Finally, we put everything in $u_{yz} = 0$:

$$\begin{aligned}
    0
    = \frac{1}{2} \Big( \pdv{u_y}{z} + \pdv{u_z}{y} \Big)
    = \frac{1}{2} \Big( \pdv{f_y}{z} - \frac{z}{R} + \pdv{f_z}{y} \Big)
    = \frac{1}{2} \Big( \!-\! 2 C x + \dv{h}{z} - \frac{z}{R} \Big)
\end{aligned}$$

Only the first term contains $x$, so to satisfy this equation, we must set $C = 0$.
The remaining terms then tell us that $h(z) = z^2 / (2 R)$.
Therefore:

$$\begin{aligned}
    f_x = 0
    \qquad
    f_y = - \nu \frac{x^2}{2 R} + \frac{z^2}{2 R}
    \qquad
    f_z = 0
\end{aligned}$$

Inserting this into the components $u_x$, $u_y$ and $u_z$
then yields the full displacement field.
</div>
</div>

In any case, the beam experiences a bending torque with an $x$-component $T_x$ given by:

$$\begin{aligned}
    T_x
    = - \int_A y \sigma_{zz} \dd{A}
    = - \frac{E}{R} \int_A y^2 \dd{A}
\end{aligned}$$

Where $A$ is the cross-section.
Th above integral is known as the **area moment**,
and is typically abbreviated by $I$.
This brings us to the **Euler-Bernoulli law**:

$$\begin{aligned}
    \boxed{
        T_x
        = - \frac{E I}{R}
    }
    \qquad \quad
    I
    \equiv \int_A y^2 \dd{A}
\end{aligned}$$

The product $E I$ is called the **flexural rigidity**,
i.e. the beam's "stiffness".
For a small deformation, i.e. a large radius of curvature $R$,
the law can be approximated by:

$$\begin{aligned}
    T_x
    \approx - E I \dvn{2}{y}{z}
\end{aligned}$$



## Slender rods

A beam that is very thin in the transverse directions ($x$ and $y$ in this case),
can be approximated as a single string or rod $y(z)$.
Each infinitesimal piece $(\dd{y}, \dd{z})$ of the rod
exerts forces $F_y$ and $F_z$ on the next piece,
and is feels external forces-per-length $K_y$ and $K_z$, e.g. gravity.
In order to have equilibrium, the total force must be zero:

$$\begin{aligned}
    0
    &= F_y(z + \dd{z}) - F_y(z) + K_y(z) \dd{z}
    \\
    0
    &= F_z(z + \dd{z}) - F_z(z) + K_z(z) \dd{z}
\end{aligned}$$

Rearranging these relations yields these equations for the internal forces $F_y$ and $F_z$:

$$\begin{aligned}
    \boxed{
        \dv{F_y}{z}
        = - K_y
    }
    \qquad \quad
    \boxed{
        \dv{F_z}{z}
        = - K_z
    }
\end{aligned}$$

Meanwhile, the rod also feels a torque with $x$-component $T_x$,
where equilibrium entails:

$$\begin{aligned}
    0
    = T_x(z + \dd{z}) - T_x(z) + F_z(z) \dd{y} - F_y(z) \dd{z}
\end{aligned}$$

This can be rearranged to get a differential equation for $T_x$, namely:

$$\begin{aligned}
    \boxed{
        \dv{T_x}{z}
        = F_y - F_z \dv{y}{z}
    }
\end{aligned}$$

If $F_z$ and $\idv{y}{z}$ are small, the last term can be dropped.
These equations are widely applicable,
but there is one especially important application,
so much so that it is usually what is meant by "Euler-Bernoulli law":
the shape of a laterally loaded rod.

Consider a beam along the $z$-axis, carrying a lateral load $K_y$,
e.g. its own weight $A g \rho$ or more.
Assuming there is no other load $K_z = 0$
and $F_y \ll F_z$, the above equations become:

$$\begin{aligned}
    T_x
    = - E I \dvn{2}{y}{z}
    \qquad
    \dv{T_x}{z}
    = F_y
    \qquad
    \dv{F_y}{z}
    = - K_y
\end{aligned}$$

Which we can simply substitute into each other,
eventually leading to:

$$\begin{aligned}
    \boxed{
        K_y
        = \dvn{2}{}{z}\Big( E I \dvn{2}{y}{z} \Big)
    }
\end{aligned}$$

This is often referred to as the **Euler-Bernoulli law** as well.
Typically the flexural rigidity $EI$ is a constant in $z$,
in which case we can reduce the equation to:

$$\begin{aligned}
    K_y
    = E I \dvn{4}{y}{z}
\end{aligned}$$

Which is clearly solved by a fourth-order polynomial,
given some boundary conditions.



## References
1.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.