Question 18.13: A uniform two span continuous bridge shown in Fig. 18.35 wit......

Formulation stiffness matrix 10 × 10. Assuming translational displacement as master and other degrees of freedom as slaves we get reduced stiffness matrix of size 5 × 5.

                 [K]=\frac{E I}{L^3}\left[\begin{array}{ccccc}236.5 & 92.4 & -87.6 & -226.2 & -15.6 \\92.4 & 236.7 & -15.3 & -226.2 & -83.3 \\-87.6 & -15.3 & 38.7 & 61.5 & 2.58 \\-226.2 & -226.2 & 61.5 & 329.1 & 61.5 \\-15.6 & -83.3 & 2.58 & 61.5 & 38.7\end{array}\right]

                         =\left[\begin{array}{cc}k & k_g \\k_g^{ T } & k_{g g}\end{array}\right]

where

                 \left[k_g\right]=\frac{E I}{L^3}\left[\begin{array}{lll}-87.6 & -226.2 & -15.6 \\-15.3 & -226.2 & -83.3\end{array}\right]

                \left[k_{g g}\right]=\frac{E I}{L^3}\left[\begin{array}{ccc}38.7 & 61.5 & 2.58 \\61.5 & 329.1 & 61.5 \\2.58 & 61.5 & 38.7\end{array}\right]

                k\left\lgroup\begin{array}{l}u_1 \\u_2\end{array}\right\rgroup+k g\left\lgroup\begin{array}{l}u_3 \\u_4 \\u_5\end{array}\right\rgroup=0

                                              \left\lgroup\begin{array}{l}u_1\\u_2\end{array}\right\rgroup=-[k]^{-1} k g\left\lgroup\begin{array}{l}u_3 \\u_4 \\u_5\end{array}\right\rgroup

                                              \left\{\begin{array}{l}U_1\\U_2\end{array}\right\}=\left[\begin{array}{ccc}0.406 & 0.687 & -0.09375 \\-0.09375&0.687&0.406\end{array}\right]\left\{\begin{array}{l}U_3 \\U_4 \\U_5\end{array}\right\}

The influence vectors associated with each support are

                 i_1=\left[\begin{array}{c}0.406  25 \\-0.093  75\end{array}\right]

                 i_2=\left[\begin{array}{l}0.687  50 \\0.687  50\end{array}\right]

                 i_3=\left[\begin{array}{c}-0.093  75 \\0.406  25\end{array}\right]

We could obtain the above influence vector using a strength of materials approach, as shown below.

Apply unit load at 3 (to get influence vector i_1)

Moment at 4 = L

Rotation at 4 = \frac{L^2}{3EI}

Hence u_{2}=-\frac{M L^{2}}{{16}E I}=-\frac{L^{3}}{{16}E I}

Displacement at the free end u_{3}=\frac{L^{3}}{3E I}+\frac{L^{3}}{3E I}=\frac{2L^{3}}{3E I}

Calculate u_2 when u_3 = 1

That is given by

-\frac{L^{3}\ \ 3E I}{16E I\ \ 2L^{3}}=\frac{3}{32}= \mathrm{-}0.093~ 75

The – sign shown the deflection at 2 is downward.

Now calculate deflection at 1 due to unit load at 3= deflection at 3 due to unit load at 1.

Apply unit load at 1

Rotation at 4=\frac{L^{2}}{2\times3E I}=\frac{L^{2}}{6E I}

Deflection at 1={\frac{(0.5L)^{3}}{3E I}}+{\frac{L^{2}}{6E I}}\,{\frac{L}{{2}}}={\frac{L^{3}}{8E I}}

Slope at 1=\frac{L^{2}}{6EI}+\frac{L^{2}}{8E I}=\frac{7L^{2}}{24E I}

Deflection at 3=\frac{L^{3}}{8E I}+\frac{7L^{2}}{24E I}\frac{L}{2}=\frac{13L^{3}}{48E I}

Both deflections at 3 and 1 are upwards and hence positive.

If u_3 = 1 let us calculate what is u_1. It is given by

\frac{13L^{3}}{48E I}\frac{3E I}{2L^{3}}=\frac{13}{32}=0.406\,25

Hence

{i}_{1}\,=\,\left\{\begin{array}{l}{{0.406\,25}}\\ {{-\displaystyle0.093\,75}}\end{array}\right\}

To find the influence vector i_2, apply unit load at 4 and find the deflection at 1 and 4. The deflection at 4 due to unit load at 4 is given by L^3 /6EI. If the load is at a distance of ×2 and the deflection is to be calculated at ×1, then deflection at ×1 is given by

u_{1}=u_{2}\,=\,\frac{x_{1}\,\,\,x_{2}}{6E I\,\,2L}(4L^{2}\,-\,x_{1}^{\,2}\,-\,x_{2}^{\,2}\,)\\

= \frac{(L/2)(L)}{6E I\ 2L}(4L^{2}-L^{2} – L^{2}/4)=\frac{11L^{3}}{96E I}

When the deflection at 4 is equal to 1 what is the deflection at 1 and 2 which may be calculated as

\frac{11L^{3}}{96E I}\frac{6\,E I}{L^{3}}=0.6875

Hence the influence vector i_2 is given as

{i}_{2}\,=\,\left\{\begin{array}{l}{{0.6875}}\\ {{\displaystyle 0.6875}}\end{array}\right\}

Once i_1 vector is known, i_3 vector is written as

{i}_{3}\,=\,\left\{\begin{array}{l}{{-0.093\,75}}\\ {{\displaystyle0.406\,25}}\end{array}\right\}

The equations of motion are

(m)\ddot{u}+ku = P_g \\ m\ddot{u} + ku = -m \left[\begin{matrix} i_1 & i_2 & i_3  \end{matrix} \right] \begin{pmatrix} \ddot{u}_{g1}   \\ \ddot{u}_{g_2}\\\ddot{u}_{g_3}  \end{pmatrix}