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Concrete In Australia : June 2013
Concrete in Australia Vol 39 No 2 33 Span EI ΔAV = 0.500 (l1 -3.7) (L/π)4 (excl. UDL) EI/rib ΔAV xDLA Span/600 9.6 1670 32000000x0.0137/2 .0076x1.3 = .0099 .0160 11.6 3048 32000000x0.0234/2 .0082x1.3 = .0107 .0193 13.6 5672 32000000x0.0367/2 .0096x1.3 = .0125 .0227 15.6 9515 32000000x0.0536/2 .0110x1.3 = .0143 .0260 17.6 15416 32000000x0.0739/2 .0130x1.3 = .0169 .0293 19.6 23710 32000000x0.0981/2 .0152x1.3 = .0198 .0327 Table 2. Average deflection of the cross section. distribution (Figure 5) resemble the AS5100 M1600 loading for a traffic lane, the value of w1 has been chosen to make M(max) in (Figure 7), w1(L/π)2 , equal to M(max) as determined for M1600 loading. For the nominal deck beam lengths available in the system, the resulting values for w1 are shown in Table 1. Having determined the proportion of a lane load carried per beam, as described in this article, the distribution of shear force along a beam rib will be as per a shear force envelope determined from the M1600 live load rolling along the deck. is of course will vary from the shear force shown in Figure 6. However, variations to Figures 7, 8 and 9 will not be significant. at is, although the equivalent lane loading w1, applied as two lines of wheels, l1, is primarily used to determine the load distribution to individual beams or beam ribs, it may also be used directly to determine the bending moments, rotations and deflections. 3.0 SINGLE LANE BRIDGE In this example, the lane load is located to provide the maximum transverse LL eccentricity, as shown in Figure 10. e deflection of the beams is modified by equal and opposite shear transferred through the elastomeric shear key. e magnitude of this shear is such as to maintain compatibility at the deck-beam interface. In Figure 11, statically equivalent balancing loads located at the deck beam rib locations are shown on the basis that the St Venant torsional restraint is negligible compared with the torsional restraint provided by the equal and opposite restraints at the beam ribs. For a beam section like a hollow box, where the St Venant torsion is dominant, it is again possible to consider first term Fourier series diagrams, similar to Figures 5 to 9, but with: Applied torque/m = q1sin (xπ/L) Torque at section = q1(L/π)cos (xπ/L) GJ (rotation / m) = q1(L/π)cos (xπ/L) GJ (rotation at section) = q1(L/π)2 sin (xπ/L) GJ (beam edge deflection) = q1B(L/π)2 sin (xπ/L) where B = half the beam width. So as not to interrupt the current narrative concerning the twin rib modular bridge-beams, this process is taken to its conclusion in Appendix A. In Figure 12, statically equivalent loads balancing the elastomeric compatibility shear are shown. For convenience, k = k1sin (xπ/L) may be abbreviated as k1, and similarly w1 and l1 are used as abbreviations for the equivalent traffic lane loading and the equivalent line of wheels loading, where equivalent line of wheels l1 = (w1)/2. en, compatibility demands that: Δ(Figure 11) = Δ(Figure 12) at is, k1 = 0.330 l1 Figures 13, 14 and 15 show the loads, deflection and transverse moments respectively on the cross section. From Figure 14, the average LL deflection of the cross section, without the UDL component of the LL included (that is the LL deflection as defined in AS5100) is shown in Table 2, for comparison with its allowed value. 4.0 DOUBLE LANE BRIDGE Figure 16 shows the traffic lanes arranged on the two lane bridge deck, with 100% lane loading on the most eccentric lane and 80% lane loading on the accompanying lane, to provide the most eccentric LL effect on the bridge. Figure 17 shows the statically equivalent balancing loads at the rib locations, due to LL on the basis that the St Venant torsional restraint is not significant compared with that provided by the equal and opposite restraint in the deck- beam ribs. For the consideration of the opposite situation, see Appendix A. Figure 18 shows the statically equivalent balancing loads at rib locations, due to beam interface shears (k1, k2 and k3) transmitted through the elastomeric shear keys. Compatibility demands that Δ 1, Δ 2 & Δ3 from Figures 17 and 18 are identical. at is: 0.914 l1 = +3.250 k1 - 0.625 k2 0.938 l1 = -0.625 k1 + 3.250 k2 + 0.625 k3 0.871 l1 = +0.625 k2 + 3.250 k3 e solution of these equations provides:- k 1 = 0.3417 l1 k 2 = 0.3144 l1 k 3 = 0.2075 l1 en Figures 19, 20 and 21 represent the transverse loads, deflections and moments. e average serviceability LL deflections including the Dynamic Load Allowance (DLA), but excluding the UDL