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diff --git a/content/know/concept/euler-bernoulli-law/index.pdc b/content/know/concept/euler-bernoulli-law/index.pdc new file mode 100644 index 0000000..e9d3fdd --- /dev/null +++ b/content/know/concept/euler-bernoulli-law/index.pdc @@ -0,0 +1,314 @@ +--- +title: "Euler-Bernoulli law" +firstLetter: "E" +publishDate: 2021-06-03 +categories: +- Physics + +date: 2021-05-14T09:15:30+02:00 +draft: false +markup: pandoc +--- + +# Euler-Bernoulli law + +**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 $u_{ii} = \pdv*{u_i}{i}$, +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: + +$$\begin{aligned} + T_x + \approx - E I \dv[2]{y}{z} +\end{aligned}$$ + + + +## 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 $\dv*{y}{z}$ 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: + +$$\begin{aligned} + T_x + = - E I \dv[2]{y}{z} + \qquad + \dv{T_x}{z} + = F_y + \qquad + \dv{F_y}{z} + = - K_y +\end{aligned}$$ + +Which we can simply substitute into each other, +eventually leading to: + +$$\begin{aligned} + \boxed{ + K_y + = \dv[2]{z} \Big( E I \dv[2]{y}{z} \Big) + } +\end{aligned}$$ + +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: + +$$\begin{aligned} + K_y + = E I \dv[4]{y}{z} +\end{aligned}$$ + +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. |