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Hi All, Module 6 of the TUFLOW tutorial is now available. This module introduces a large bridge structure that spans the creek and part of the floodplain. The model includes both a a 1D bridge channel and 2D flow constriction layer to represent the bridge. The TUFLOW tutorial can be found here: http://wiki.tuflow.com/index.php?title=Tutorial_Module06 The required module data and completed models are available on our TUFLOW downloads section: http://www.tuflow.co...ial Models.aspx If you have any questions, clarifications or feedback, please email email@example.com. Regards Phillip
Hi there, Im modelling an Arched bridge in a 1D network channel and trying to calculate the Loss Coefficients to form part of an LC table. Im using Figure 6 in the Hydraulics of Bridge Waterways from March 1978 Backwater coefficient to help me. I am using a 30 degree wingwall value. My Bridge is 4.42m wide at an elevation of 55.799mAOD and the cross section is 6.025m on the upstream face. Using this value gives me a M value of 0.73 and conversely a Kb value of 0.5. This loss seems a little high as this is the springing point of my arch. As I work out the LC table, nearing the top of the arch, I recalculate the coefficients. At an elevation of 57.449, the channel section is 12.796 and the arch is 3.552 wide, giving me a constriction of 9.244. This gives me an M value of 0.277 and a Kb value of 2.2 Finally, just below the soffit, the bridge width is 1.776 at the top. The channel width is 12.796 and hence the constriction is 11.02m. Which gives me an M value of 0.13 and a Kb value of 3 How do the Kb values listed above, fit in with an accepted loss value of 1.56 at the soffit. Is the 'M' opening ratio being calculated and applied correctly.
Q: Why is the loss coefficient set to 1.56 when flow is surcharging against a bridge deck in a 1D bridge channel? A: The loss coefficient is derived from AustRoads, Waterway Design. Refer specifically to Figure 5.18 (pg 47, 1994 edition) (attached) The discharge coefficient, Cd, for a surcharging deck is 0.8, as shown highlighted in the attached figure Assuming that V=Q/(bn*Z) and rearranging the equation given at the bottom of the figure, one arrives at V=Cd* (2g* dh)^(0.5). Where bn is the net waterway width. Also given the loss formula used by the ESTRY engine dh=k*(V^2/2g), the final formula is k=1/(Cd^2). Thus, using a Cd of 0.8 as stated in AustRoads, k=1.5625.