Question 16.1: A bridge with a single span of 80ft has a deck of uniform cr...
A bridge with a single span of 80 \mathrm{ft} has a deck of uniform cross section with mass m=200 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft}^{2} and flexural rigidity E I=1.328 \times 10^{10} \mathrm{lb} \cdot \mathrm{ft}^{2}. A single wheel load P travels across the bridge at a uniform velocity of 50 \mathrm{mph}, as shown in Figure E16.1a.
(a) Obtain the influence line for deflection at midspan of the bridge considering only the first mode of vibration.
(b) Obtain the impact factor defined as the ratio of maximum midspan deflection from part (a) above and the maximum static deflection.
(c) How much would the next-higher mode contribute to the maximum midspan deflection?
(a) From the data supplied, we get
\begin{aligned}\omega_{n} &=n^{2} \pi^{2} \sqrt{\frac{E I}{m L^{4}}} \\&=12.566 n^{2} \mathrm{rad} / \mathrm{s} \\v &=73.33 \mathrm{ft} / \mathrm{s} \\\frac{n \pi v}{L} &=2.88 n\end{aligned} \qquad (a)
Table 16.1 Mid-span deflection in the first mode due to traveling load.
The dynamic displacement caused by the traveling wheel load is obtained from Equation 16.31.
Substitution for \omega_{n} and n \pi \nu / L from Equation a in Equation 16.31 gives
For n=1 and x=L / 2, Equation \mathrm{b} reduces to
u\left(\frac{L}{2}, t\right)=\frac{P}{1,196,930}\{-0.2292 \sin 12.566 t+\sin 2.88 t\} \qquad (c)
The wheel load will take L / v=1.091 s to travel across the deck. Equation c is therefore valid for 0 \leq t \leq 1.091 \mathrm{~s}. The values of midspan deflection calculated from Equation \mathrm{c} at intervals of 0.1 \mathrm{~s} are shown in Table E16.1. (b) To obtain the magnitude of displacement, the figures in column 4 of Table E16.1 must be multiplied by P / 1,196,930. The resultant displacement at midspan has been plotted in Figure E16.1b. The maximum midspan displacement occurs at t=0.4 \mathrm{~s} when the wheel load is at a distance of x=v t=29.33 \mathrm{ft} from the left-hand support. The magnitude of maximum displacement is u_{\max }(L / 2)=1.315(P / 1,196,930). The maximum static midspan deflection is given by
\Delta_{\max }=\frac{P L^{3}}{48 E I}=\frac{P}{1,245,000} \qquad (d)
The ratio of maximum dynamic deflection to the maximum static deflection is
\frac{u_{\max }}{\Delta_{\max }}=1.177 \qquad (e)
The static load should thus be increased by 17.7 \% to account for dynamic effect.
(c) The midspan deflection in the second mode is zero because the shape function \sin (n \pi x / L) is zero for n=2 and x=L / 2. The contribution due to the third mode is obtained from Equation \mathrm{b} by substituting n=3 and x=L / 2.
u\left(\frac{L}{2}, t\right)=\frac{P}{101.7 \times 10^{6}}(0.0764 \sin 113.1 t-\sin 8.64 t) \qquad (f)
For t=0.4 \mathrm{~s} Equation \mathrm{f} gives
u\left(\frac{L}{2}, 0.4\right)=\frac{P}{101.7 \times 10^{6}} \times 0.3817 \qquad (g)
The additional impact factor at t=0.4 \mathrm{~s} becomes
\frac{u(L / 2,0.4)}{\Delta_{\max }}=0.0047 \qquad (h)
This may be added to the value obtained in Equation e to give a total impact factor of 1.182. The contribution from the third mode is comparatively very small.