Suggest edit — Cantilever Beams

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title: "Cantilever Beams"
type: concept
related: [Second Moment Of Area, Elastic Modulus, Numerical Cantileaver Simulations, Finite Difference Method, Cantilever Beams]
source: https://www.jemoka.com/posts/kbhcantilever_beams/
confidence: high
status: active
---

A Cantilever beam is a rigid structure which is extended horizontally and supported on one end.
Working with Cantilever Beams curvature Let&rsquo;s first define a function:
\begin{equation} w(x) \end{equation}
this represents the deflection of the beam at point \(x\). We will begin by taking its derivative by location:
\begin{equation} \Delta w = \pdv{w}{x} \end{equation}
is the change in deflection over location. &ldquo;How much deviation of the beam from the resting axi is there as you run along it?&rdquo;
We now take another derivative:
\begin{equation} k = \pdv[2]{w}{x} \end{equation}
\(k\) is defined as the &ldquo;Curvature&rdquo; of the beam: the &ldquo;change in the change of bentness.&rdquo; The intuition is essentially this:
a straight, flat beam fixed an one end has \(\Delta w=0\), \(k=0\). It does not change from its resting axis, and its rate of change from resting does not change a straight, slanted beam fixed at one end has \(\Delta w=C, k=0\). It changes from its resting axis with a linear rate, and its rate of change from resting does not change. a curved, slanted beam fixed at one end has \(\Delta \omega = f(x), k=C\). It changes from its resting axis non-linearly (hence curving at a function of \(x\)), and its rate of change from resting is changing at a constant \(c\). flexural rigidity Flexural Rigidity is the &ldquo;force couple&rdquo; (&ldquo;rate&rdquo;) which relates the Curvature of an non-rigid body and how much torque it actually generates given the object&rsquo;s properties.
Recall first our Elastic Modulus \(E\): it is a fraction of \(\frac{stress}{strain}\) measured in Pascals (force per unit area, i.e. \(\frac{N}{m^{2}} = \frac{kg}{ms^{2}}\)).
Find also second moment of area \(I\): a value in units \(m^{4}\) which is the sum (by area) of the squared displacement of each infinitesimal area to the axis of origin.
And we bam! we multiply the two things together, creating a value \(EI\) in units \(Nm^{2}\).
bending moment bending moment is the torque from bending. It is expressed usually in \(M\). As mentioned in the section about Flexural Rigidity, we can use that value to relate \(M\) with the actual Curvature of your object.
Specifically, that:
\begin{equation} M = -(EI)k = -EI\pdv[2]{w}{x} \end{equation}
&ldquo;bending moment is flexural rigidity times curvature&rdquo; =&gt; &ldquo;[how much force per distance you exert] is the result of [how bendy your thing is] times [how much you bent it].&rdquo;
There is a negative in front because if you pull out your lovely little right hand, point your thumb forward (+y), start curling your nice fingers around your nice hand (-z), you will notice that you are wrapping them downwards (the - part of the z) which is rather not positive. If we want \(\pdv[2]{w}{x}\) to be positive (bend up), we will need to chuck a negative in front of it to make both things positive.
This relation, while intuitive, is not from first-principles. In order to get such a derivation, you read Wikipedia.
magic We can take two derivatives by location&mdash;
\begin{equation} \pdv[2] x \qty(EI \pdv[2]{w}{x}) = -\mu \pdv{w}{t}+q(x) \end{equation}
where \(\mu\) is the mass density, \(q(x)\) is the force applied (in Newtons) by area. this is magic. Will come back to it.
Solving this? See Finite Difference Method
Actually attempting to solve it Numerical Cantileaver Simulations
Working on Deformation