Question 18.6: Figure 18.11 shows a shear frame (i.e. rigid beams) and its ......
Figure 18.11 shows a shear frame (i.e. rigid beams) and its floor masses and storey stiffness. This structure is subjected to horizontal harmonic force at the top floor. p_0=500 ~kN.
(a) Determine the equation for steady state displacement of the structure.
(b) Determine the direct solution of coupled equations.
(c) Determine the modal analysis (neglect damping).
(d) If k = 63 600 kN/m, m = 45 413 kg and ω = 0.75 ω_{1} split the load into modal components and obtain the solution using Chopra’s method.
(a) Equation for steady state displacement
\left[\begin{array}{cc}m & \\& m / 2\end{array}\right]\left\{\begin{array}{l}\ddot{U}_1 \\\ddot{U}_2\end{array}\right\}+\left[\begin{array}{cc}2k & -k \\-k & k\end{array}\right]\left\{\begin{array}{l}U_1\\U_2\end{array}\right\}=\left\{\begin{array}{c}0 \\p_0\end{array}\right\} \sin \omega t
(b) Direct solution of coupled equations, using the MATHEMATICA package
\omega_1=0.765~ \sqrt{\frac{k}{m}}
\omega_2=1.848~ \sqrt{\frac{k}{m}}
To obtain a particular solution assume
\begin{aligned}& U_1=U_{10}~ \sin w t \\ \\& U_2=U_{20}~ \sin ~w t\end{aligned}
We get
\left[\begin{array}{cc}\left(2 k-m \omega^2\right) & -k \\-k & \left(k-\frac{m}{2} \omega^2\right)\end{array}\right]\left\{\begin{array}{l}U_{10} \\U_{20}\end{array}\right\}=\left\{\begin{array}{c}0 \\p_0 \sin \omega t\end{array}\right\}
solving
U_{10}=\frac{p_o / k}{\left[1~-~\left(\frac{\omega}{\omega_1}\right)^2\right]\left[1~-~\left(\frac{\omega}{\omega_2}\right)^2\right]}
U_{20}=\frac{2~ p_o\left(2 k~-~m w^2\right)}{m^2 ~\omega_1^2 \omega_2^2\left[1~-~\left(\frac{\omega}{\omega_1}\right)^2\right]\left[1~-~\left(\frac{\omega}{\omega_2}\right)^2\right]}
(c) Modal analysis
\text { Eigenvector }=[\psi]=\left[\begin{array}{cc}0.5773 & -0.5735 \\0.81649 & 0.81649\end{array}\right]
To obtain the unit eigenvector
\psi^{ T } m \psi=m\left[\begin{array}{cc}0.666&0\\0&0.666\end{array}\right]=\left[\begin{array}{cc}M_1^2 & \\& M_2^2\end{array}\right]
M_1=M_2=0.816 \sqrt{m}
[\phi]=\frac{1}{\sqrt{m}}\left[\begin{array}{cc}0.707 & -0.707 \\1 & 1\end{array}\right]
\phi^{ T }~ m \phi ~\ddot{y}+\phi^{ T }~ k~ \phi ~y=\phi^{ T }~ F
\{\ddot{y}\}+\frac{k}{m}\left[\begin{array}{cc}0.5852 & \\& 3.415\end{array}\right]\{y\}=\frac{1}{\sqrt{m}}\left\{\begin{array}{l}p_0 \\p_0\end{array}\right\} \sin \omega t
Solving the above two coupled equations we get
y_1=\frac{p_0 ~\sqrt{m}}{0.58529~ k} \frac{1}{\left[1~-~\left(\frac{\omega}{\omega_1}\right)^2\right]} \sin~ \omega t
y_2=\frac{p_0~ \sqrt{m}}{3.415~ k} \frac{1}{\left[1~-~\left(\frac{\omega}{\omega_2}\right)^2\right]} \sin~ \omega t
The physical coordinate {U} can be written as
{U} = [φ]{y}
=\frac{1}{\sqrt{m}}\left[\begin{array}{cc}0.707 & -0.707 \\1 & 1\end{array}\right]\left\{\begin{array}{l}y_1 \\y_2\end{array}\right\}
U_1(t)=\frac{p_0}{k}\left\{\frac{1.207}{\left[1~-~\left(\frac{\omega}{\omega_1}\right)^2\right]}-\frac{0.207}{\left[1~-~\left(\frac{\omega}{\omega_2}\right)^2\right]}\right\} \sin~ \omega t
U_2(t)=\frac{p_0}{k}\left\{\frac{1.707}{\left[1-\left(\frac{\omega}{\omega_1}\right)^2\right]}+\frac{0.293}{\left[1-\left(\frac{\omega}{\omega_2}\right)^2\right]}\right\} \sin \omega t
By algebraic manipulation it can be shown that these results are the same as those obtained by solving coupled equation (k = 63600kN/m m = 45413kg).
(d) Using Chopra’s method
\omega_1=0.765~ \sqrt{\frac{k}{m}}=0.765 ~\sqrt{\frac{63600 ~\times ~10^3}{45413}}=28.5 ~rad / s
\omega_2=1.848~ \sqrt{\frac{k}{m}}=1.848 ~\sqrt{\frac{63600~\times~ 10^3}{45413}}=69~ rad / s
Step 1: Determine the modal participation factors and spatial distribution of forces.
\left\{\begin{array}{l}\Gamma_1\\\Gamma_2\end{array}\right\}=\phi^{ T } F=\frac{1}{\sqrt{m}}\left[\begin{array}{cc}0.707 & 1 \\-0.707 & 1\end{array}\right]\left\{\begin{array}{c}0 \\p_0\end{array}\right\}=\left\{\begin{array}{l}p_0 / \sqrt{m} \\p_0 / \sqrt{m}\end{array}\right\}
\{F\}_1=\Gamma_1[m]\{\phi\}_1=\frac{p_0}{\sqrt{m}} m\left[\begin{array}{ll}1 & \\& 0.5\end{array}\right] \frac{1}{\sqrt{m}}\left\{\begin{array}{c}0.707\\1\end{array}\right\}=p_0\left\{\begin{array}{c}0.707 \\0.5\end{array}\right\}
\{F\}_2=\Gamma_2[m]\{\phi\}_2=\frac{p_0}{\sqrt{m}} m\left[\begin{array}{ll}1 & \\&0.5\end{array}\right] \frac{1}{\sqrt{m}}\left\{\begin{array}{c}-0.707 \\1\end{array}\right\}
=p_0\left\{\begin{array}{c}-0.707 \\0.5\end{array}\right\}
\{F\}_1 and \{F\}_2 are shown in Fig. 18.12.
Step 2: Determine the static displacements due to two modes.
Mode 1
\left\{\begin{array}{l}U_{11}^{s t} \\U_{21}^{s t}\end{array}\right\}=[k]^{-1}\left\{\begin{array}{c}0.707 \\0.5\end{array}\right\}p_0
=\frac{1}{63600}\left[\begin{array}{cc}2 & -1 \\-1 & 1\end{array}\right]^{-1}\left\{\begin{array}{c}0.707\\0.5\end{array}\right\} 500=\left\{\begin{array}{l}0.0095 \\0.0134\end{array}\right\}
\left\{\begin{array}{l}U_{12}^{s t}\\U_{22}^{s t}\end{array}\right\}=[k]^{-1}\left\{\begin{array}{c}-0.707\\0.5\end{array}\right\}p_0
=\frac{1}{63600}\left[\begin{array}{cc}2 & -1 \\-1 & 1\end{array}\right]^{-1}\left\{\begin{array}{c}-0.707 \\0.5\end{array}\right\} 500=\left\{\begin{array}{c}-0.0016 \\0.0023\end{array}\right\}
\left\{\begin{array}{l}U_1^{s t} \\U_2^{st}\end{array}\right\}=\left\{\begin{array}{l}0.0079 \\0.0157\end{array}\right\} ;\left\{\begin{array}{l}U_1 \\U_2\end{array}\right\}=\left\{\begin{array}{l}U_{11}^{s t} \\U_{21}^{s t}\end{array}\right\} \text { DLF1 }+\left\{\begin{array}{l}U_{12}^{s t} \\U_{22}^{s t}\end{array}\right\} \text { DLF2 }
To find the base shear V_{b 0}=k U_1. The calculations are shown in Table 18.3.
\frac{V_{b 0}}{p_0}=2.526
| Table 18.3 Dynamic base shear calculation | ||||
| Mode | \omega_n | \bar{R}=\frac{U_{1i}}{U_1^{\text {st }}} | R_d=\frac{1}{1~-~\left(\frac{\omega}{\omega_i}\right)^2} | \bar{\gamma}=\bar{R} ~R_d |
| 1 | 28.5 | \frac{0.0095}{0.0079}=1.21 | 2.28 | 2.758 |
| 2 | 69 | \frac{-0.0016}{0.0079}=-0.211 | 1.106 | –0.232 |
