4.4: Continuity Conditions, an Example
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- 21493
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In Section 4.4 the continuity requirements were formulated, but the system of eight algebraic equations was not solved. Here a complete solution will be presented for a beam loaded by a point force acting at an arbitrary location \(x = a\).
The reaction forces are calculated from moment equilibrium:
\[R_A = P \frac{l − a}{l}\]
\[R_B = P \frac{a}{l}\]
The sum of the reaction forces is equal to \(P\). The corresponding bending moments and shear forces are
\[M(x) = \begin{cases} R_Ax = \frac{P(l − a)x}{l}, \\ R_B(l-x) = \frac{Pa(l − x)}{l}, \end{cases}, \quad V (x) = \begin{cases} \frac{P(l − a)}{l}, & 0 < x < a \\ −\frac{P a}{l}, & a < x < l \end{cases} \label{4.4.3}\]
The jump in the shear force across the discontinuity point \(x = a\) is
\[[V] = V^+ − V^− = \frac{P(l − a)}{l} − (−\frac{P a}{l}) = P\]
The bending moments are continuous on both sides, \([M] = 0\). Therefore the static continuity conditions are automatically satisfied at \(x = a\). The kinematic continuity conditions, formulated in Equations (4.2.6-4.2.7) require displacements and slopes to be continuous. Integrating the governing equations (4.3.1) with \ref{4.4.3} in two regions gives
\[−EIw^{\mathrm{I}} = \frac{P(l − a)x^3}{6l} + C_1x + C_2 \quad 0 < x < a\]
\[−EIw^{\mathrm{II}} = \frac{P a}{l} (\frac{lx^2}{2} − \frac{x^3}{6}) + C_3x + C_4 \quad a < x < l\]
The four integration constants are found from two boundary condition and two continuity condition
\[w(0) = w(l) = 0, \; w^{\mathrm{I}} (a) = w^{\mathrm{II}}(a), \; \left. \frac{dw^{\mathrm{I}}}{dx}\right|_{x=a} = \left. \frac{dw^{\mathrm{II}}}{dx}\right|_{x=a}\]
This gives rise to the system of four linear inhomogeneous algebraic equations for \(C_1\), \(C_2\), \(C_3\), and \(C_4\)
\[\begin{cases} C_2 = 0 \\ \frac{Pal^2}{3} + C_3l + C_4 = 0 \\ \frac{Pba^3}{6l} + C_1a = \frac{Pa}{l} \left(\frac{la^2}{2} - \frac{a^3}{6} \right) + C_3a + C_4 \\ \frac{Pba^2}{2l} + C_1 = \frac{Pa}{l} \left(la - \frac{1}{2}a^2 \right) + C_3 \end{cases} \label{4.4.8}\]
A simple problem has led to a quite complex algebra. Now, you understand why the previous example with eight unknown coefficients was only formulated but not solved. The solution to the system \ref{4.4.8} is
\[C_1 = − \frac{P a(a^2 − 3al + 2l^2)}{6l}\]
\[C_2 = 0 \]
\[C_3 = -\frac{P a(a^2 + 2l^2)}{6l} \]
\[C_4 = \frac{P a^3}{6} \]
and the final solution of unsymmetrically loaded beam is
\[w^{\mathrm{I}}(x) = \frac{P x [a^3 - 3a^2l - lx^2 + a(2l^2 + x^2)]}{6EIl} \quad 0 < x < a\]
\[w^{\mathrm{II}}(x) = - \frac{P a (l - x) [a^2 + x(-2l + x)]}{6EIl} \quad a < x < l \]
One can easily check that the continuity conditions are met at \(x = a\). The above example teaches us that symmetry in nature and engineering not only means beauty, but also brings simplicity.