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Design For Fish Passage at Roadway - Stream Crossings: Synthesis Report8 Case Studies and Design ExamplesHow to use this chapter
8.1 Geomorphic simulation8.1.1 USFS Stream Simulation Design ExampleA survey of the existing channel, and a surface pebble count conducted on a representative reference reach, determined the following channel characteristics:
Channel width (W_{ch}) = 1.95 m (6.4 ft) The culvert is sized assuming that bank margins are desirable.
Culvert bed width (W_{culv}) = W_{ch} + 4*D_{100} The culvert should span a minimum of 2.70 m (8.9 ft), which would be rounded up to 2.75 m (9 ft). Bed mix gradation includes D_{100}-D_{50} determined from the surface pebble count, with D_{16} and D_{5} determined by the Fuller-Thompson equation (6.5). (Equation 6.5)P=(d/D_{100})^{n} The Fuller-Thompson 'n' value can be varied approximately between 0.45 and 0.7 to control gradation until an appropriate proportion of fines (5-10%) has been attained. To start, compare the effects of an n value of 0.7 vs. and n value of 0.45. The results of these calculations have been plotted in Figure 8.1. Using n = 0.7 D_{16} = 0.32^{1/n} * D_{50}
D_{5} = 0.10^{1/n} * D_{50} Using n = 0.45 D_{16} = 0.32^{1/n} * D_{50}
D_{5} = 0.10^{1/n} * D_{50}
It can be seen that an n value of 0.45 will lead to gradation of approximately 12-13% fines (2 mm or less). Refining further, using n = 0.55
D_{16} = 0.32^{1/n} * D_{50}
D_{5} = 0.10^{1/n} * D_{50} This distribution is plotted in Figure 8.2
An n value of 0.55 leads to a bed mix gradation with between 5-10% fines (smaller than 2 mm). The following gradation should be used for design.
D_{100} = 180 mm (0.59 ft) 8.1.2 USFS Stability Check Design ExampleThe following stability check example is taken (almost verbatim) from Bates et al. 2006. It is included here for clarification of the USFS Stream Simulation Design. Determining if D_{84} moves at bankfull flow (example from Bates et al. 2006) Channel parameters are as follows:
D_{84} = 120 mm (0.39 ft) Determine whether the D_{84} particle moves at bankfull flow in the stream using the modified critical shear stress equation for D_{84} (Equation 6.7).
τ_{ci} = τ^{*}_{D50} (γ_{s} - γ)D_{i}^{0.3}D_{50}^{0.7} (Equation 6.7) Find the average boundary shear stress in the reference reach at bankfull flow ( τ_{bf}) using Equation 6.6 with a hydraulic radius of 0.30 m (1 ft).
τ_{bf} = γRS (Equation 6.6) Therefore, the D_{84} particle size is stable bankfull flow How well does the modified critical shear stress equation apply here?
Conclusion: The modified critical shear stress equation is applicable to this stream Critical unit discharge equation Find the critical unit discharge for D_{50} (q_{cD50}) using Equation 6.9.
Calculate b (which quantifies the range in particle sizes) using Equation 6.11.
Find critical unit discharge for D_{84} (q_{cD84}) using Equation 6.10.
Both D_{50} and D_{84} are stable at bankfull flow in this example. These results agree with those of the modified critical shear stress equation. Is the Bathurst equation appropriate for this stream?
Slope > 1% Predicting the range of potential particle movement
τ_{bf} = γRS (Equation 6.6) Find the upper critical shear stress for the D_{84} particle size using Equation 6.12. τ_{ci-u} = 0.0814D_{i} (Equation 6.12) τ_{cD84.u} = 0.0814 (120 mm) = 468 Pa (9.77 lb/ft^{2}) Find the lower critical shear stress for the D_{84} particle size using Equation 6.13. τ_{ci-i} = 0.00355D_{i} (Equation 6.13) τ_{cD84.l} = 0.00355(120 mm) = 20 Pa (0.426 lb/ft^{2}) τ_{bf} = 42 Pa (0.90 lb/ft^{2}) is less than τ_{cD84-u} = 468 Pa (9.77 lb/ft^{2}) and greater than τ_{cD84-I} = 20 Pa (0.426 lb/ft^{2}), indicating that the D_{84} particle has the potential to be mobile at bankfull flow. Summary: Both the modified critical shear stress and critical unit discharge equations predict that the D_{84} will be stable at bankfull conditions. The Williams equations (Equations 6.12 and 6.13) indicate potential movement of the D_{84}. Judgment: D_{84} is likely stable at bankfull conditions. 8.1.3 WDFW Stream Simulation Design ExampleStream properties are determined from a channel survey and analysis of multiple representative cross sections.
Channel width (W_{ch}) = 1.95 m (6.4 ft) Check Applicability S_{ch} = 2% < 6.0%
Channel has been assessed to have little susceptibility to vertical changes Conclusion: WDFW Stream Simulation is applicable in this situation Culvert span is determined according to Equation 7.1. Culvert bed width (W_{culv}) = 1.2 W_{ch} + 0.6 m Culvert should span a minimum of 3000 mm, which would likely be rounded up to 3048 mm (10 ft). Culvert bed configuration is based on slope scenarios. Since slope is less than 4%, design scenario I is employed, meaning that rock bands will be used to control the initial channel shape. This creates a situation that may be more adequately described as Hydraulic Simulation. Bands spacing should be the lesser of 5 ∙W_{ch} and 0.24 m/S_{culv}, or 5 W_{ch} = 9.75 mm (32 ft)
Therefore, spacing will be 9.75 m (32 ft). Bands are separated from the entrance and exit by the lesser of: 2W_{ch} = 3.9 m (12.7 ft ) or 7.62 m (25 ft) Therefore, spacing should be at least 3.9 m (13 ft) from culvert inlet and outlet. With a 30.5 m (100 ft) structure this leaves room for 3 rock weirs at a spacing of 9.75 m (32 ft) apart, and 5.5 m (18 ft) from the culvert entrance and exit. Sizing of rock band material is based on a surface pebble count of the reference reach.
D_{100} = 180 mm (0.591 ft) Rock bands are comprised of well-graded material within the following range. D_{100} = 180 mm (0.59 ft) to 2D_{100} = 360 mm (1.2 ft) Since channel slope is less than 4%, Paleohydraulic Analysis can be used to check the bed changing flow, ensuring that bed mix gradation is adequate.
D_{84} = 85 mm (0.279 ft) (from above) Using Table 6.5, slope (2.2%) and particle size 85 mm (0.28 ft) are used to find depth of flow Depth = 0.25 m (0.81 ft) With known depth, cross-sectional area can be computed from the proposed triangular cross section with 6:1 side slopes. (Area of a triangle is 0.5*base*height)
Area =0.5·Depth· (12·Depth) Using the proposed cross-sectional area, this corresponds to a flow of
Q=A·V 8.1.4 Unit Discharge Design ExampleWhen slopes are greater than 4%, the Unit-Discharge method is suggested for finding a stable bed material gradation. Necessary parameters include:
100 year exceedance flow (Q_{100}) = 3.54 m^{3}/s (125 cfs) Solving for Critical Discharge (q_{c}): q_{c} = Q_{100}/W_{ch} Using the Critical Discharge equation (6.16) to solve for D_{84}: D_{84} = 3.45S^{0.747}(1.25q_{c})^{2/3}/g^{1/3} (Equation 6.16) So a D84 of 256 mm (0.84 ft) will create the necessary stability, and a gradation can be created based on D84. This can also be checked using the Paleohydraulic analysis shown above. 8.1.4.1 Paleohydraulic AnalysisD_{84} = 256 mm (0.84 ft) Using Table 6.5, find flow depth Depth = 1.6 ft (0.49 m) Using the proposed channel dimensions (6:1 side slope, triangular channel) Area = 0.5·Depth·(12·Depth) This is consistent with the trend of Co'ta's equation to predict smaller particle sizes than Bathu'st's equation at higher slopes (Bates et. al 2003). Both equations show this D_{84} to be stable at Q_{100} (125 cfs). 8.1.5 WDFW No-Slope Design ExampleStream Properties Needed Channel width (W_{ch}) = 1.95 m (6.4 ft) Channel Type and Size
Culvert bed width (W_{culv}) = 1.25·W_{ch} Culvert should span a minimum of 2.44 m (8 ft). To check the applicability of No Slope Design, ensure that the product of channel slope times length is less than 0.2D.
L_{culv}·S_{culv} = 0.67 m (2.2 ft) Since slope times length is > 0.2D, 0.67 m > 0.49 m (2.2 ft > 1.6 ft), No-Slope method is not applicable in this situation due to the inability to meet embedment requirements. 8.1.6 Embedded Pipe Case HistoryThe following example of stream simulation is taken from the USFS FishXing website (United States Forest Service 2006b), maintaining the format and content developed by the authors. It is reproduced here with permission from Mike Furniss of the USFS. Location
Project Type
Pre-Project Barrier
Channel Characteristics
Ecological Value
Project Characteristics
Challenges
Project Funding
Completion Date
Total Project Cost
Project Description An embedded 4900 mm (16 ft) diameter culvert was selected as the replacement crossing. The new culvert is designed to pass a 100-year flood at Headwater-to-Diameter ratio (HW/D) of 0.6 and is 145% wider than the upstream bankfull channel. The appropriate slope and elevation for constructing the streambed within the culvert was determined from a 137 m (450 ft) long channel profile. Since the road was closed and no traffic bypass was needed during construction, the project took only four weeks to complete. This project experienced many construction challenges. Although originally designed to be embedded 1.8 m (6 ft), problems with buried utilities, groundwater and slope stability during excavation resulted in only embedding the culvert approximately 0.9 m (2.5 ft). 8.2 Hydraulic Simulation8.2.1 Culvert with Floodplain Relief Case HistoryThe following case history was provided by Andrzej ("Andy") Kosicki of the Maryland State Highway Administration. Location
Project Type
Pre-project Barrier
Channel Characteristics
Ecological Value
Project Characteristics
Post Project Observations and Lessons Learned No formal monitoring program was set up since monitoring was not required by the permitting agency. Periodic field trips showed beneficial changes in the channel and within the structure:
8.3 Hydraulic Design8.3.1 John Hatt Creek Case HistorySource
Location
Project Type
Pre-Project Conditions
Pre-Project Barrier
Hydrologic Characteristics
Ecological Value
Project Characteristics
Challenges and Lessons Learned
Project Description Baffles were designed to satisfy, as best as possible, State and Federal velocity and depth criteria for fish passage while avoiding excessive turbulence. Hydraulics of corner baffles at fish passage flows were modeled using empirical equations developed by Rajaratnam and Katopodis (1990) and provided by WDFW (2003). The energy dissipation factor (EDF) was calculated as a measure of turbulence.
A total of 43 corner baffles were welded into the pipe prior to insertion. Baffles constructed of 9.6 mm (3/8 in) thick steel and spaced 1.2 m (4 ft) apart. The 0.23 m (9 in) tall baffles were rotated 15 degrees from horizontal, resulting in the low and high sides of the baffle located 0.11 m and 0.39 m (4.3 and 15.2 in) above the invert, respectively. The gap between the existing and new pipes was filled with concrete slurry to prevent seepage. The existing culvert outlet was perched nearly 0.5 m (1.5 ft) above the downstream water surface and the channel below the culvert was steep. To improve fish passage conditions at the outlet, three precast concrete weirs were installed within the 7.5 m (25 ft) right-of-way below the outlet. The concrete weirs were spaced 2.5 m (8 ft) apart with 0.23 m (9 in) drops. The weirs were keyed into the bank approximately 0.6 m (2 ft). Although facing class rock was to be placed on both banks between the weirs for scour protection, the contractor only placed rock on the left bank. Post Project Observations and Lessons Learned Rock was only placed on the left bank below the outlet which allowed for rapid bank erosion, resulting in flanking of the weirs. The bank was rocked later to prevent further erosion. Placing rock along both banks, as designed, and keying the weirs further into the banks may have prevented flanking. A design problem with the wooden low-flow notch was also discovered. The wood is not set flush with the downstream edge of the weir. Instead of plunging directly into the downstream pool at low flows, the water strikes the lip of the concrete weir. Installing a steel low-flow notch flush with the downstream edge of the concrete weir would create the desired plunging conditions at low-flow. A steep slab of existing concrete at the culvert inlet was to be removed as part of the project. However, it was left in place. Using inspectors familiar with the project's fish passage objectives may have avoided some of these problems. Completion Date
Total Project Cost
8.3.2 WDFW Roughened Channel Design ExampleStream properties needed: Channel width (W_{ch}) = 2.13 m (7 ft) Slope ratio
This is a situation where slope ratio exceeds 1.25 (typical upper range for Stream Simulation Design in Washington). Culvert span is an iterative parameter beginning with channel bed width. Width of Culvert Bed (W_{culv}) = W_{ch}=2.13 m (7 ft) Culvert bed configuration by U.S. Army Corps of Engineers Riprap Design, requiring computation of unit discharge as follows:
q = 1.66 m^{2}/s (17.9 ft^{2}/s) This allows the D_{30} particle size to be calculated by Equation 6.14 for riprap sizing (other methods, such as those included in WDFW Stream Simulation design can also be used for bed sizing, and may be preferable over Equation 6.14, however, 6.14 is used here for illustrative purposes):
D_{30} = 183 mm (0.60 ft) Note - it may be pertinent to increase the factor of safety (1.25) since rock sizing is greater than 152 mm (0.5 ft). Use D_{30} to find D_{84}, using the approximate scaling factor provided for riprap, Equation 6.15: D_{84}=1.5·D_{30} (Equation 6.15) This particle size is checked to ensure that it does not exceed 1/4 of the culvert span 4·D_{84}=3.584 < 7ft A gradation can now be created based on D_{84} = 0.9 ft. 8.3.2.1 Fish Passage VelocityFish passage velocity is now calculated to ensure that fish are able to traverse the structure. In this case, design is for juvenile Coho salmon, and velocity cannot exceed 4 ft/s according to WDFW Hydraulic Design criteria (based on 90 ft structure). Additional parameters required include fish passage velocity and hydraulic radius: Allowable Velocity (V_{fish}) = 4 ft/s For use with Limerinos and Jarrett's equations, velocity will be based on a Manning's n value, and will be calculated according to Equation 7.7.
Limerinos equation is solved as follows (Equation 6.2)
which can be used into Equation 7.7 to solve for velocity
V_{l}=1.20 ft/s < 4.0 ft/s So, according to the Limerinos equation, this would be an acceptable velocity Jarrett's equation is solved as follows n =0.32·S_{culv}^{0.38}·R^{-0.16} Using n to solve for velocity
V_{j}=1.42 ft/s <4.0ft/s Mussetter's equation utilizes the Darcy-Weisbach friction factor, and is solved according to Equation 6.4.
For this equation D_{50} is needed, and can be solved for according to the relations provided in Washington's Stream Simulation Design.
Channel depth is also needed, taken from analysis based on a 6:1 triangular channel at the fish passage design flow. depth = 1.1 ft
V_{m}=1.48 ft/s < 4.0 ft/s, which is acceptable for fish passage 8.3.2.2 TurbulenceTurbulence is then checked through the calculation of channel EDF. EDF = γQ_{fp}S_{culv}/A (Equation 3.1) EDF = 2.13 ft·lb/(ft^{3}·s) < and is acceptatble for fish passage design
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Updated: 04/07/2011 |
Contact:Bert Bergendahl |