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4.4: Continuity Conditions, an Example

  • Page ID
    21493
  • 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.