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+---
+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.
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