Question 18.7: Figure 18.13 shows a mass-less simply suported beam with thr......

   Step 1

              U_M^{ T }=<U_1 ~U_2 ~U_3>; U_S^{ T }=<\theta_1~ \theta_2~ \theta_3>

   Step 2 Mass matrix

             m=\left[\begin{array}{lll}33665 & & \\& 33665 & \\& & 33665\end{array}\right]

   Step 3 Determine stiffness matrix (sym)

K=\frac{E I}{L^{\prime 3}}\left[\begin{array}{ccc|ccc}7 L^{\prime 2} & 2 L^{\prime 2} & 0 & 3 L^{\prime} & -6L^{\prime} & 0 \\& 8 L^{\prime 2} & 2 L^{\prime 2} & 6 L^{\prime} & 0 & -6 L^{\prime} \\& & 7 L^{\prime 2} & 0 & 6 L^{\prime} & -3 L^{\prime} \\& K s s & & K S M & & \\\hline & & & 15 & -12 & 0 \\& & & & 24 & -12 \\& & & & & 15\\&  K ms && K mm &  \end{array}\right]

Using the static condensation procedure the modified stiffness matrix for the master degree of freedom can be written as

             \frac{E I}{L^{\prime 3}}\left[\begin{array}{ccc}9.8501 & -9.4283 & 3.864 \\& 13.713 & -9.4283 \\\text { sym } & & 9.8501\end{array}\right]

or

     Step 4 Determine lateral stiffness

             =\frac{E I}{L^3}\left[\begin{array}{ccc}630.86 & -603.43 & 246.86 \\& 877.71 & -603.43 \\& &630.86\end{array}\right]

            \frac{E I}{L^3}=\frac{207.15~ \times ~10^9 ~\times~ 4.1623~\times ~10^7}{10^{12}~ \times~ 3.81^3}

                 = 155 899.02

            K=10^5\left[\begin{array}{ccc}983.5 & -940.74 & 384.45 \\& 1368.34 & -940.74 \\\text { Syn } & & 983.5\end{array}\right]

      Step 5 Determine natural frequencies solving using the MATHEMATICA package

            \omega_1=10.589 ; \omega_2=42.1824, \omega_3=89.55~ rad / s

            E V=\left[\begin{array}{ccc}0.50 & 0.707 & 0.50 \\0.707 & 0 & -0.707 \\0.50 & -0.707 & 0.50\end{array}\right]

Normalized eigenvector =\phi^{ T } m \varphi=I

where

            \varphi=\left[\begin{array}{ccc}0.0027 &0.00385 & 0.0027 \\0.00385 & 0 & -0.00385 \\0.0027 & -0.00385 & 0.0027\end{array}\right]

The mode shapes are shown in Fig. 18.14.

      Step 6 Determine modal expansion F

            F=\sum\limits_{n=1}^3 F_n

                =\Sigma~ \Gamma_n m \phi_n~ \text { where }~\Gamma_n=\phi_n^{ T }~ F

            \Gamma=\left\{\begin{array}{c}0.0027 \\0.00385 \\0.0027\end{array}\right\}

          F_1=\Gamma_1 m \phi_1

The modal expansion of F is shown graphically in Fig. 18.15 and tabulated in Table 18.4.

      Step 7 Determine modal static response as shown in Fig. 18.16. The value of moments due to forces F is determined by the linear combination to the above three load cases.

The resultants are as given in Table 18.5.

     Next we can determine

                  M_1^{s t}=\frac{3 L}{16}

                           = 0.1875L

We have demonstrated

             \sum\limits_{n=1}^3 ~M_{1 n}^{s t}=M_1^{s t}

      Step 8 Determine modal contribution factors, their cumulative values and error.

     The modal contribution factors and error are given in Table 18.6

     In the above the modal contribution factor is largest for first mode and progressively decreases for the second and third modes.

      Step 9 Determine response to the half cycle sine pulse. The peak modal response equation is specialized for R = M_{1} to obtain (see Table 18.7)

           \left(M_{1 n}\right)_0=p_0 M_1^{s t}~ \bar{M}_{1 n} ~R_{d n}

      Step 10 Comments

• For the given force, the modal response decreases for higher modes. The decrease is more rapid because of R_{dn}. It also decreases with mode n.

• The peak value of the total response cannot be determined from the peak modal response because the modal peaks occur at different time instants. Square root of sum of squares (SRSS) and complete quadratic combination (CQC) do not apply to pulse excitations.