Question 6.1: The seating in a football stadium is mounted on precast pres...

The generalized mass m^{*}, generalized stiffness k^{*}, and the generalized load p^{*} are given by

\begin{aligned}m^{*} &=\int_{0}^{L} \bar{m}\{\psi(x)\}^{2} d x=\frac{31}{630} \bar{m} L \\k^{*} &=\int_{0}^{L} E I\left\{\psi^{\prime \prime}(x)\right\}^{2} d x=\frac{24}{5} \frac{E I}{L^{3}} \\p^{*} &=\int_{0}^{L} \bar{p} \psi(x) d x=\frac{\bar{p} L}{5}\end{aligned}\qquad (a)

Mass per unit length, \bar{m}

Due to mass of the beam: 0.1587 \times 2400=380.9 \mathrm{~kg} / \mathrm{m}

Due to seats and spectators: \frac{2400 \times 0.762}{9.81}=186.4 \mathrm{~kg} / \mathrm{m}

\begin{aligned}m &=567.3 \mathrm{~kg} / \mathrm{m} \\m^{*} &=\frac{31 \times 567.3 \times 11.73}{630}=326.6 \mathrm{~kg} \\k^{*} &=\frac{24 \times 3.7 \times 10^{7} \times 6.327 \times 10^{-3}}{5 \times 11.7^{3}} \\&=701.6  \mathrm{kN} / \mathrm{m}\end{aligned}

The uniformly distributed harmonic load \bar{p} caused by stamping and clapping is obtained from

\bar{p}=0.40 \times 0.762 \sin \Omega t=0.3048 \sin \Omega t \mathrm{kN} / \mathrm{m} \qquad (b)

where \Omega=2 \pi \times 3  \mathrm{rad} / \mathrm{s}. The effective force p^{*} is thus given by

p^{*}=p_{0} \sin \Omega t=\frac{0.3048 \times 11.7}{5} \sin \Omega t=0.7137 \sin \Omega t \mathrm{kN}

The equivalent single-degree-of-freedom system is shown in Figure E6.1b. Its motion is governed by the following equations

\begin{aligned}&u=z(t) \psi(x) \qquad \qquad (c) \\&m^{*} \ddot{z}(t)+c^{*} \dot{z}(t)+k^{*} z(t)=p^{*}=p_{0} \sin  \Omega t \qquad (d)\end{aligned}

The solution of Equation \mathrm{d} is obtained from Equation 6.27.

u = \rho \sin (\Omega t-\phi)      (6.27)

Thus the displacement of the beam is given by

\begin{aligned}u(x, t) &=z(t) \psi(x) \\&=\rho \sin (\Omega t-\phi) \psi(x)\end{aligned}\qquad (e)

where

\rho=\frac{p_{0}}{k^{*}}\left[\left(1-\beta^{2}\right)^{2}+(2 \beta \xi)^{2}\right]^{-1 / 2}\qquad (f)

The frequency ratio \beta is obtained from the exciting frequency and the natural frequency.

\begin{aligned}\text { Natural frequency: } \quad \omega &=\sqrt{\frac{k^{*}}{m^{*}}}=\sqrt{\frac{701.6 \times 1000}{326.6}} \\&=46.35 \mathrm{rad} / \mathrm{s}=7.38 \mathrm{~Hz} \\\text { Frequency ratio: } \quad \beta &=\frac{\Omega}{\omega}=\frac{3}{7.38}=0.4065\end{aligned}

Substitution for \beta and \xi in Equation \mathrm{f} gives

\rho=\frac{p_{0}}{k^{*}} 1.197=\frac{0.7137}{701.6} \times 1.197=1.218 \times 10^{-3} \mathrm{~m}

The midspan deflection is obtained from Equation e with x=L / 2, so that

\begin{aligned}u\left(\frac{L}{2}, t\right) &=\rho \sin (\Omega t-\phi)\left\{\frac{x}{L}-2\left(\frac{x}{L}\right)^{3}+\left(\frac{x}{L}\right)^{4}\right\}_{x=L / 2} \\&=1.218 \times 10^{-3} \times 0.3125 \sin (\Omega t-\phi) \\&=3.805 \times 10^{-4} \sin (\Omega t-\phi) \mathrm{m}\end{aligned}

The maximum deflection at midspan is 3.805 \times 10^{-4} \mathrm{~m}. The maximum acceleration at midspan is equal to \Omega^{2} u_{\max }=(6 \pi)^{2} \times 3.805 \times 10^{-4}=0.135 \mathrm{~m} / \mathrm{s}^{2}=1.38 \% \mathrm{~g}, where g=9.81 \mathrm{~m} / \mathrm{s}^{2} is the acceleration due to gravity. Experience has shown that during a sports event, the spectators will not have a feeling of discomfort provided that the maximum acceleration experienced by them is below 5 \% \mathrm{~g}. The design is satisfactory from this point of view.

If w  \mathrm{kN} / \mathrm{m}^{2} is the equivalent static load that will produce a deflection of 3.805 \times 10^{-4},

\frac{5}{384} \frac{w \times 0.762 L^{4}}{E I}=3.805 \times 10^{-4}

from which we obtain

w=1.197 \times 0.4=0.4788  \mathrm{kN} / \mathrm{m}^{2}

The precast beams should therefore be designed to carry a total superimposed load of 2.4+0.479=2.88  \mathrm{kN} / \mathrm{m}^{2} in addition to their self-weight.