Question 2.6: A SDOF frame has a mass = 5000 kg, lateral stiffness k = 4 ×...

Undamped natural frequency \omega =\sqrt{\frac{k}{m} } =\sqrt{\frac{4\times 10^{6}}{5000} }  = 28.28 rad/sec

f=\frac{\omega }{2\pi } = 4.5 Hz

\omega _{D}=\omega \sqrt{1-\xi ^{2}}

\omega _{D} is damped frequency of vibration.

∴  \omega _{D}=28.28\sqrt{1-0.04^{2}} = 28.257 rad/sec or,  f_{D} = 4.496 Hz

Damped free response of a SDOF system is given by

x(t)=e^{-\xi \omega t}\left\{\frac{x(0)\xi \omega +\dot{x}(0) }{\omega _{D}} \sin \omega _{D} t +x(0)\cos \omega _{D}t\right\}      (i)

If the initial displacement x(0) =x_{0}, and initial velocity \dot{x}(0) = 0, damped free response can be written as:

x(t)=x_{0}e^{-\xi \omega t}(\cos \omega t+\xi \sin \omega t) if \omega _{D} ≈ ω      (ii)

Differentiating Equation (ii)

\frac{dx}{dt} =\dot{x} (t)= – x_{0}\omega e^{-\xi \omega t}((1+\xi ^{2})\sin \omega t)

For maximum displacement, velocity = 0,
or,                               sin ω t = 0
or,                                 ω t = n π
or,                                  t = n π/ω

For first five peak displacements, time of occurrence is given by

t = 0, 2π/ω, 4π/ω, 6π/ω, 8π/ω

and  x_{\max }=x_{0}e^{-2\pi \xi n}=25  e^{-2\pi \xi n}=25  e^{-0.251  n} where n = 0, 1, 2, 3, and 4

∴ Peak displacements are 25, 19.44, 15.12, 9.15 and 7.12 mm as shown in Figure 2.11.