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<br />4. Since the engineer is confident about the flow velocity and angle of attack, and the <br />channel is not expected to experience any vertical instability, a pad radius of R = 1.5b. <br />was chosen. <br /> <br />Pad Radius, R = 1.5(2.5) = 3.75 m. <br /> <br />The Toskanes will be installed around the pier, a horizontal distance of 3.75 m from the <br />wall of the pier. <br /> <br />5. From Figure 6.2 or 6.3, the number of <br />Toskanes per unit area for the 100 kg <br />Toskane size with a pad thickness of 2Du is <br />7.08 Toskanes/m2 Total area of the pad <br />(see Figure 6.5) is: <br /> <br />Area = 2(5(3.75)) + (1t(4.252 - 0.52)) = 93.5 <br />m2. <br /> <br />#Toskanes = 7.08(93.5) = 662 Toskanes. <br /> <br />The pad thickness is 2Du = 900 mm. <br /> <br />6. Since the bed material is cobbles and <br />gravel, a granular filter is added beneath <br />the pad of Toskanes. The dss of the filter <br />directly beneath the pad of Toskanes is 95 <br />mm. The cobbles and gravel are <br />sufficiently large so no additional filter <br />layers are required. <br /> <br />Area of Pad <br /> <br />~.- <br /> <br />0.5. <br />r- 3. 75m., <br />1 <br /> <br />5m <br /> <br />J <br />~ <br /> <br /> <br /> <br />Figure 6.5 Area of Pier Pad <br /> <br />Design Example for a Bridge Abutment (Fotherby & Ruff 1995) <br /> <br />The bridge at Blue Creek in Figure 6.4 has vertical wall abutments with wing walls. During <br />normal flows the west abutment extends 0.6 m into the flow, but during high flows it <br />obstructs 2.4 m of the flow (normal to the flow field). The embankment slope is at1 H:1V. <br />The east abutment does not obstruct the flow even during high flows. <br /> <br />1. Determine the velocity value, Vv (m/s). <br />The abutment is located near the bank, outside of the thalweg, C1 = 0.9. <br />Since the abutment has wing walls, Gs = 0.85. <br />The T oskane pad is installed so that the top of the pad is level with the bed, Gh = 1.0. <br />A randomly installed pad of Toskanes is selected, C; = 1.0. <br /> <br />Vv= 2.5(0.9)(0.85)(1.0)(1.0)(1.5) = 2.87 mls. <br /> <br />2. Calculate the adjusted structure width, b. (m). <br /> <br />6.10 <br />