Modeling with higher order linear differential equations, boundary-value problems
Deflection of a Beam
In civil engineering, analyzing structures for their internal forces and deflections is one of the most important topic.
The above table depicts $\theta$ and $y$ for several types of structures and/or their loads. The problem of deflections was observed and described by means of differential equations. In structure analysis we usually work either with precomputed results (see the table above) or we work routinelly with simple DE equations of higher order.
There are some rules or a guideline worth to mention. The deflection $\boldsymbol y$ is a linear displacement measured from the beam's axis. Angular displacement $\theta$ can be derived from deflection $y$. In structural analysis we typically start from load (either $w(x)$ or a single external force $P$), then we are able to compute internal shear forces $V(x)$ on the beam and internal moments $M(x)$ on the beam. From $M(x)$ we are able to get $\theta(x)$ by integration or we are able to get linear displacement $y(x)$ by double integration of $M(x)$.
- The load $w(x)$ is typically given.
- The internal shear forces $V(x)$ can be found from equilibrium of forces (and also $V(x) = -\intop w(x)\ dx$).
- The internal moments $M(x)$ can be found from equilibrium of forces (and also $M(x) = \intop V(x)\ dx$).
- The angular displacement $\theta$ can be found by integrating
$$\theta(x) = \frac {1}{EI} \intop M(x)\ dx, \hskip1em \text{because} \hskip1em \frac{d\theta}{dx} = \frac{M}{EI}$$. - The deflection $y$ can be found by double integrating
$$y(x) = \frac{1}{EI} \iint M(x)\ dx, \hskip1em \text{because} \hskip1em \frac{d^2y}{dx^2} = \frac{M}{EI}$$.
In short we integrate $w(x) \to V(x) \to M(x) \to \theta(x) \to y(x)$. The task is then to work on a boundary-value problem. The boundary value problem is a differential equation with a set of additional restraints.
Example
(The first case from the above table: simple beam with a concentrated load $P$ in the middle)
Assuming we have already computed moment function as $M(x) = Px/2$ for $0\lt x \lt L/2$, find deflection $y(x)$
and $y_{\max}$ at $x=L/2$.
From the differential equation, describing deflection of the beam, we know, that we need to integrate $M(x)$ two times to get desired deflection.
$$ \frac{d^2y}{dx^2} = \frac{M}{EI} \implies y(x) = \frac{1}{EI} \iint M(x)\ dx \implies \color{red}{y(x) \cdot EI= \frac{Px^3}{12} + c_1x + c_2} $$The task remains to find constants $c_1,\ c_2$. These will be obtained by means of boundary value conditions.
It can be observed that
- at $\boldsymbol{x=0}$ the deflection is $\boldsymbol{y=0}$. The second condition is
- at the middle of the span ($\boldsymbol{x=L/2}$) the slope $\boldsymbol{\theta =0}$.
Note that these are not initial values, since they are in different points. From the first condition we get
$$ \begin{align} y(x) \cdot EI &= \frac{Px^3}{12} + c_1x + c_2 \nonumber \\ 0 &= P\cdot 0 + c_1 \cdot 0 + c_2 \implies c_2 = 0 \nonumber \end{align} $$
For the second condition: we have to differentiate $y(x)$ and then we can get $c_1$.
$$ \begin{align} y'(x) \cdot EI &= \frac{Px^3}{12} + c_1\nonumber \\ 0 &= \frac{P (\frac{L}{2})^3}{12} + c_1 \frac{L}{2} \implies c_1 = -\frac{PL^2}{16}\nonumber \\ \end{align} $$
So the deflection curve on the left half of the beam is
$$ \underline{\underline{y(x) \cdot EI = \frac{Px^3}{12} - \frac{PL^2}{16} x}} \nonumber \\ $$The maximum deflection at the middle of the span is
$$ \underline{\underline{y_{\max} \cdot EI = \frac{P(\frac L 2)^3}{12} - \frac{PL^2}{16} \frac L 2 = -\frac 1 {48} \frac P {L^3}}} \nonumber \\ $$
So we obtained the same results which are written within the above table for civil engineers.
Example
(The third case from the above table: simple beam with an uniformly distributed load $w$)
Assuming we have already computed moment function as $M(x) = wLx/2 - x^2w/2$ for $0\lt x \lt L$, find deflection $y(x)$
and $y_{\max}$ at $x=L/2$.
From the differential equation, describing deflection of the beam, we know, that we need to integrate $M(x)$ two times to get desired deflection.
$$ \frac{d^2y}{dx^2} = \frac{M}{EI} \implies y(x) = \frac{1}{EI} \iint M(x)\ dx \implies \color{red}{y(x) \cdot EI= \frac{wLx^3}{12} - \frac{x^4w}{24}+ c_1x + c_2} $$The task remains to find constants $c_1,\ c_2$. These will be obtained by means of boundary value conditions.
It can be observed that
- at $\boldsymbol{x=0}$ the deflection is $\boldsymbol{y=0}$. The second condition is
- at the end of the span ($\boldsymbol{x=L}$) the deflection has to be also $\boldsymbol{y =0}$.
Note that these are not initial values, since they are in different points. From the first condition we get
$$ \begin{align} y(x) \cdot EI&= \frac{wLx^3}{12} - \frac{x^4w}{24}+ c_1x + c_2 \nonumber \\ 0 &= 0 - 0+ c_1\cdot 0 + c_2 \implies c_2 = 0 \nonumber \end{align} $$
From the second condition we can get $c_1$.
$$ \begin{align} y(x) \cdot EI&= \frac{wLx^3}{12} - \frac{x^4w}{24}+ c_1x \nonumber \\ 0 &= \frac{wL L^3 }{12} - \frac{L^4w}{24}+ c_1 L \nonumber \implies c_1 = -\frac{wl^3}{24}\\ \end{align} $$
So the deflection curve of the beam is
$$ \underline{\underline{y(x) \cdot EI= \frac{wLx^3}{12} - \frac{x^4w}{24} -\frac{wL^3}{24}x}} \nonumber \\ $$The maximum deflection at the middle of the span $(x=L/2)$ is
$$ \underline{\underline{y_{\max} \cdot EI= \frac{wL^4}{12\cdot 8} - \frac{L^4w}{24\cdot 16} -\frac{wL^4}{24\cdot2} = -\frac{5}{384}wL^4}} \nonumber \\ $$
So we obtained the same results which are written within the above table for civil engineers.
