5.2: Solution for a Beam on Roller Support
- Page ID
- 21498
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider first case 1. From the constitutive equation, zero axial force beams that there is no extension of the beam axis, \(\epsilon^{\circ} = 0\). Then, from Equation (5.1.1)
\[\frac{du}{dx} = \frac{1}{2}\left(\frac{dw}{dx}\right)^2 \label{5.2.1}\]
At the same time, the nonlinear term in the vertical equilibrium vanishes and the beam response is governed by the linear differential equation
\[EI\frac{d^4w}{dx^4} = q(x)\]
which is identical to the one derived for the infinitesimal deflections. As an example, consider the pin-pin supported beam under mid-span point load. From Equations (4.3.7) and (4.3.8), the deflection profile is
\[w(x) = w_o \left[ 3 \frac{x}{l} − 4\left(\frac{x}{l}\right)^3\right]\]
and the slope is
\[\frac{dw}{dx} = \frac{w_o}{l}\left[3 − 12\left(\frac{x}{l}\right)^2\right]\]
where \(w_o\) is the central deflection of the beam. Now, Equation \ref{5.2.1} can be used to calculate relative horizontal displacement \(\Delta u\). Integrating Equation \ref{5.2.1} in the limits \((0, l)\) gives
\[\int_{0}^{l} \frac{du}{dx} dx= u|_{0}^{l} = u(l) − u(0) = \Delta u = − \int_{0}^{l} \frac{1}{2} \left( \frac{dw}{dx}\right)^2 dx\]
The result of the integration is
\[\Delta u \approx 7 \frac{w_{o}^2}{l}\]
In order to get a physical sense of the above result, the vertical and horizontal displacements are normalized by the thickness \(h\) of the beam
\[\frac{\Delta u}{h} = \frac{7}{l/h} \left(\frac{w_{o}}{h}\right)^2\]
For a beam with \(\frac{l}{h} = 21\), the result
\[\frac{\Delta u}{h} = \frac{1}{3} \left(\frac{w_{o}}{h}\right)^2\]
is ploted in Figure (\(\PageIndex{1}\)).
It is seen that the amount of sliding in the horizontal direction can be very large compared to the thickness.
To summarize the results, the roller supported beam can be treated as a classical beam even though the displacements and rotations are large (moderate). The solution of the linear differential equation can then be used a posteriori to determine the magnitude of sliding. The analysis of fully restrained beam is much more interesting and difficult. This is the subject of the next section.