Question 16.4: A precast concrete T-beam used to support fixed seating in a...

The problem is similar to Example 6.1, where it was solved by assuming a vibration shape. The natural frequencies and mode shapes of the loaded beam are given by

\begin{aligned}\omega_{n} &=n^{2} \pi^{2} \sqrt{\frac{E I}{m L^{4}}} \\\phi_{n}(x) &=\sqrt{\frac{2}{m L}} \sin \frac{n \pi x}{L}\end{aligned} \qquad (a)

Since the mode shapes have been mass-orthonormalized, the generalized mass m_{n}^{*}=1 and the generalized stiffness k_{n}^{*}=\omega_{n}^{2}. The uniformly distributed harmonic load is given by

p=p_{0} \sin \Omega t \qquad (b)

where p_{0}=0.305  \mathrm{kN} / \mathrm{m} and \Omega=6 \pi \mathrm{rad} / \mathrm{s}. The effective modal force is obtained from

\begin{aligned}p_{n} &=\int_{0}^{L} p \phi_{n}(x) d x \\&=\sqrt{\frac{2}{m L}} \frac{2 L}{n \pi} p_{0} \sin \Omega t \quad n=\text { odd } \\&=0 \quad n=\text { even }\end{aligned} \qquad (c)

The uncoupled modal equations are given by

\ddot{y}_{n}(t)+2 \xi_{n} \omega_{n} \dot{y}_{n}(t)+\omega_{n}^{2} y_{n}(t)=\sqrt{\frac{2}{m L}} \frac{2 L}{n \pi} p_{0} \sin \Omega t \qquad(d)

Equation d has the solution

y_{n}=\sqrt{\frac{2}{m L}} \frac{2 L}{n \pi} p_{0} \frac{1}{\omega_{n}^{2}} D_{n} \sin \left(\Omega t-\theta_{n}\right) \qquad(e)

where

\begin{aligned}& D_{n}=\frac{1}{\sqrt{\left(1-\beta_{n}^{2}\right)^{2}+\left(2 \xi_{n} \beta_{n}\right)^{2}}} \\& \theta_{n}=\tan ^{-1} \frac{2 \xi_{n} \beta_{n}}{1-\beta_{n}^{2}} \\& \beta_{n}=\frac{\Omega}{\omega_{n}}\end{aligned} \qquad(f)

The displacement is given by

\begin{aligned}u(x, t) &=\sum_{n=1}^{\infty} y_{n} \phi_{n}(x) \\&=\sum_{n=1}^{\infty} \frac{4 p_{0} L^{4}}{\pi^{5} E I} \frac{D_{n}}{n^{5}} \sin \left(\Omega t-\theta_{n}\right) \sin \frac{n \pi x}{L}\end{aligned}

From the data supplied,

\begin{aligned}E &=3.7 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2} \\I &=6.32 \times 10^{-3} \mathrm{~m}^{4} \\L &=11.7 \mathrm{~m}\end{aligned}

Mass per unit length:

Due to mass of the beam 0.1587 \times 2.4=0.3809 \times 10^{3} \mathrm{~kg} / \mathrm{m}

Due to seats and spectators \frac{1.829}{9.81}=0.1864 \times 10^{3} \mathrm{~kg} / \mathrm{m}

Total =0.5673 \times 10^{3} \mathrm{~kg} / \mathrm{m} First mode response:

\begin{aligned}\omega_{1} &=\pi^{2} \sqrt{\frac{E I}{m L^{4}}} \\&=46.315  \mathrm{rad} / \mathrm{s} \\&=7.374 \mathrm{~Hz} \\\xi_{1} &=0.03 \\\beta_{1} &=\frac{3}{7.374}=0.4068 \\D_{1} &=\left\{\left(1-\beta_{1}^{2}\right)^{2}+\left(2 \xi \beta_{1}\right)^{2}\right\}^{-1 / 2} \\&=1.1978 \\\{u(L / 2)\}_{\max } &=\frac{4 p_{0} L^{4}}{\pi^{5} E I} D_{1} \\&=\frac{4 \times 0.305 \times(11.7)^{4} \times 1.1978}{\pi^{5} \times 3.7 \times 10^{7} \times 6.32 \times 10^{-3}} \\&=3.284 \times 10^{-4} \mathrm{~m}\end{aligned}

The maximum response in the first mode occurs when

\sin \left(\Omega t-\theta_{1}\right)=1

so that

\Omega t-\theta_{1}=\frac{\pi}{2}

Also,

\tan \theta_{1}=\frac{2 \xi_{1} \beta_{1}}{1-\beta_{1}^{2}}

or

\theta_{1}=0.02925  \mathrm{rad}

Hence the time at maximum is given by

\begin{aligned}t_{m} &=\frac{1}{\Omega}\left(\frac{\pi}{2}+\theta_{1}\right) \\&=0.0849 \mathrm{~s}\end{aligned}

Second mode response:

\begin{aligned}\omega_{2} &=4 \pi^{2} \sqrt{\frac{E I}{m L^{4}}} \\&=185.3  \mathrm{rad} / \mathrm{s} \\&=29.5 \mathrm{~Hz}\end{aligned}

Since the effective force in the second mode is zero, that mode does not contribute anything to the response.

Third mode response:

\begin{aligned}\omega_{3} &=9 \pi^{2} \sqrt{\frac{E I}{m L^{4}}} \\&=416.8  \mathrm{rad} / \mathrm{s} \\&=66.37 \mathrm{~Hz}\end{aligned}

The damping is mass proportional; hence from Equation 16.43,

u(L, t)=\frac{8 F L}{\pi^2 A E} \sum_{n=1}^{\infty} \frac{1}{(2 n-1)^2}\left(1-\cos \omega_n t\right)          (16.43)

\begin{aligned}&a_{0}=2 \xi_{1} \omega_{1} \\&a_{0}=2 \xi_{3} \omega_{3}\end{aligned}

or

\begin{aligned}\xi_{3} &=\xi_{1} \frac{\omega_{1}}{\omega_{3}} \\&=\frac{1}{9} \times 0.03=0.0033 \\\beta_{3} &=\frac{3}{66.37}=0.0452 \\D_{3} &=\left\{\left(1-\beta_{3}^{2}\right)^{2}+\left(2 \xi_{3} \beta_{3}\right)^{2}\right\}^{-1 / 2} \\&=1.002\end{aligned}

The absolute value of the maximum midspan deflection in the third mode is given by

\begin{aligned}\{u(L / 2)\}_{\max } &=\frac{4 p_{0} L^{4}}{3^{5} \pi^{5} E I} D_{3} \\&=1.317 \times 10^{-6} \mathrm{~m}\end{aligned}

which is quite small compared to the deflection in the first mode.

An upper bound estimate of the displacement in the first three modes can be obtained by adding the absolute values in the first and third modes. This gives

\begin{aligned}\Delta_{\max } &=3.824 \times 10^{-4}+1.317 \times 10^{-6} \\&=3.837 \times 10^{-4} \mathrm{~m}\end{aligned}

A more precise estimate of the deflection is likely to be obtained by calculating the deflection in the third mode at t_{m}=0.0849 \mathrm{~s}, at which time the deflection in the first mode is a maximum, and adding the calculated third-mode value to that obtained in the first mode. The midspan deflection in the third mode at t=t_{m}=0.0849 \mathrm{~s} is given by

u_{3}\left(L / 2, t_{m}\right)=\frac{4 p_{0} L^{4}}{3^{5} \pi^{5} E I} D_{3} \sin \left(\Omega t_{m}-\theta_{3}\right) \sin \frac{3 \pi}{2}

where

\begin{aligned}\tan \theta_{3} &=\frac{2 \xi_{3} \beta_{3}}{1-\beta_{3}^{2}} \\&=\frac{2 \times 0.0033 \times 0.0452}{1-0.0452^{2}} \\&=2.989 \times 10^{-4} \\\theta_{3} &=2.989 \times 10^{-4}  \mathrm{rad}\end{aligned}

On substituting this value of \theta_{3} and t_{m}=0.0849, we get

u_{3}\left(L / 2, t_{m}\right)=-1.316 \times 10^{-6} \mathrm{~m}

Hence the total deflection in the first three modes works out to

\begin{aligned}\Delta_{\max } &=3.824 \times 10^{-4}-1.316 \times 10^{-6} \\&=3.811 \times 10^{-4} \mathrm{~m}\end{aligned}