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## Design For Fish Passage at Roadway - Stream Crossings: Synthesis Report

### 8 Case Studies and Design Examples

#### How to use this chapter

• Study worked out examples of culverts designs using selected design methods
• Review case studies for completed projects

#### 8.1 Geomorphic simulation

##### 8.1.1 USFS Stream Simulation Design Example

A survey of the existing channel, and a surface pebble count conducted on a representative reference reach, determined the following channel characteristics:

Channel width (Wch) = 1.95 m (6.4 ft)
Channel slope (Sch) = 2.0%
Culvert length (Lculv) = 30.5 m (100 ft)
D100 = 180 mm (0.591 ft)
D84 = 85 mm (0.279 ft)
D50 = 50 mm (0.164 ft)
The stream slope is constant in a reach extending more than 20 channel widths up- and downstream from the crossing site.

The culvert is sized assuming that bank margins are desirable.

Culvert bed width (Wculv) = Wch + 4*D100
Wculv = 2.67 m (8.76 ft)

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 D100-D50 determined from the surface pebble count, with D16 and D5 determined by the Fuller-Thompson equation (6.5).

(Equation 6.5)

P=(d/D100)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

D16 = 0.321/n * D50
D16 = 0.321/0.7 * 50 mm
D16 = 10 mm (6.5 x 10-3 ft)

D5 = 0.101/n * D50
D5 = 0.101/0.7 * 50 mm
D5 = 2 mm (6.1 x 10-3 ft)

Using n = 0.45

D16 = 0.321/n * D50
D16 = 0.321/0.45 * 50 mm
D16 = 4 mm (0.013 ft)

D5 = 0.101/n * D50
D5 = 0.101/0.45 * 50 mm
D5 = 0.3 mm (9.8 x 10-4 ft)

Figure 8.1 Example cumulative distribution curve for bed mix gradation using Fuller-Thompson method using n = 0.7 and n = 0.45

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

D16 = 0.321/n * D50
D16 = 0.321/(0.55) * 50 mm
D16 = 6.3 mm (0.021 ft)

D5 = 0.101/n * D50
D5 = 0.101/(0.55) * 50 mm
D5 = 0.75 mm (2.49x10-3 ft)

This distribution is plotted in Figure 8.2

Figure 8.2 Example cumulative distribution curve for the Fuller-Thompson method using n = 0.55 and n = 0.45

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.

D100 = 180 mm (0.59 ft)
D84 = 85 mm (0.28 ft)
D50 = 50 mm (0.164 ft)
D16 = 6.4 mm (0.021 ft)
D5 = 0.76 mm (2.5 x10-3 ft)

##### 8.1.2 USFS Stability Check Design Example

The 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 D84 moves at bankfull flow (example from Bates et al. 2006)

Channel parameters are as follows:

D84 = 120 mm (0.39 ft)
D50 = 52 mm (0.17 ft)
D16 = 27 mm (0.089 ft)
Bankfull flow (Qbf) = 3 m3/s (106 cfs)
Bankfull width (Wbf) = 5.7 m (18.7 ft)
Active channel width (W) = 5.0 m (15.3 ft)
Slope (S) = 0.0142 m/m (ft/ft)

Determine whether the D84 particle moves at bankfull flow in the stream using the modified critical shear stress equation for D84 (Equation 6.7).

τci = τ*D50s - γ)Di0.3D500.7 (Equation 6.7)
τ*D50 = 0.050 (From Table 6.4) with γs = 2.65γ and γ = 9810 N/m3
τcD84 = 16817(0.050)(0.12 mm)0.3(0.052)0.7
τcD84 = 54 Pa (1.12 lb/ft2)

Find the average boundary shear stress in the reference reach at bankfull flow ( τbf) using Equation 6.6 with a hydraulic radius of 0.30 m (1 ft).

τbf = γRS (Equation 6.6)
τbf = (9810 N/m3)(0.30 m)(0.0142)
τbf = 42 Pa < 54 Pa (0.886 lb/ft2 < 1.12 lb/ft2)

Therefore, the D84 particle size is stable bankfull flow

How well does the modified critical shear stress equation apply here?

• D84/D50 = 2.3, which is much less than 30
• Slope < 5%
• Channel unit is a riffle
• D84 particle size of 120 mm is between the range of 10 and 250 mm.

Conclusion: The modified critical shear stress equation is applicable to this stream

Critical unit discharge equation

Find the critical unit discharge for D50 (qcD50) using Equation 6.9.

 qcD50 = 0.15g0.5D501.5 (Equation 6.9) S1.12
 qcD50 = 0.15g0.5m1.5 = 0.65 m2/s (7 ft2/s) 0.01421.12

Calculate b (which quantifies the range in particle sizes) using Equation 6.11.

 b = 1.5 D84 -1 Equation 6.11 D16
 b = 1.5 120 mm -1 = 0.338 27 mm

Find critical unit discharge for D84 (qcD84) using Equation 6.10.

 qci = qcD50 Di b Equation 6.10 D50
 qcD84 = 0.65 m2/s 120 mm 0.338 = 1.07 m2/s (11.6 ft2/s) 27 mm
 q = Qbf = 3 m2/s = 0.6 m2/s (6.46 ft2/s) W 5.0 m

Both D50 and D84 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%
D84 is small cobble
Rbf/D50 = 5.9, which is < 10 (low relative submergence)

Predicting the range of potential particle movement
Find the average boundary shear stress in the reference reach at bankfull flow (τbf) using Equation 6.6.

τbf = γRS (Equation 6.6)
τbf = (9810 N/m3)(0.30 m)(0.0142)
τbf = 42 Pa (0.88 lb/ft2)

Find the upper critical shear stress for the D84 particle size using Equation 6.12.

τci-u = 0.0814Di (Equation 6.12)

τcD84.u = 0.0814 (120 mm) = 468 Pa (9.77 lb/ft2)

Find the lower critical shear stress for the D84 particle size using Equation 6.13.

τci-i = 0.00355Di (Equation 6.13)

τcD84.l = 0.00355(120 mm) = 20 Pa (0.426 lb/ft2)

τbf = 42 Pa (0.90 lb/ft2) is less than τcD84-u = 468 Pa (9.77 lb/ft2) and greater than τcD84-I = 20 Pa (0.426 lb/ft2), indicating that the D84 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 D84 will be stable at bankfull conditions. The Williams equations (Equations 6.12 and 6.13) indicate potential movement of the D84.

Judgment: D84 is likely stable at bankfull conditions.

##### 8.1.3 WDFW Stream Simulation Design Example

Stream properties are determined from a channel survey and analysis of multiple representative cross sections.

Channel width (Wch) = 1.95 m (6.4 ft)
Channel slope (Sch) = 2.0%
Culvert Slope (Sculv) = 2.2%
Culvert length (Lculv) = 30.5 (100 ft)

Check Applicability

Sch = 2% < 6.0%

 Slope Ratio = Sculv = 1.1 Sch

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 (Wculv) = 1.2 Wch + 0.6 m
Wculv =3 m (9.64 ft)

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 ∙Wch and 0.24 m/Sculv, or

5 Wch = 9.75 mm (32 ft)

 0.24 m = 10.9 m (35.8 ft) Sculv

Therefore, spacing will be 9.75 m (32 ft).

Bands are separated from the entrance and exit by the lesser of:

2Wch = 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.

D100 = 180 mm (0.591 ft)
D84 = 85 mm (0.279 ft)
D50 = 50 mm (0.164 ft)

Rock bands are comprised of well-graded material within the following range.

D100 = 180 mm (0.59 ft) to 2D100 = 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.

D84 = 85 mm (0.279 ft) (from above)
V = 9.57(D84)0.487 (Equation 6.17, customary units)
V = 5.14 ft/s (1.57 m/s)

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)
Area =0.37 m2 (3.94 ft2)

Using the proposed cross-sectional area, this corresponds to a flow of

Q=A·V
Q= 0.58 m3/s (20.5 cfs)

##### 8.1.4 Unit Discharge Design Example

When 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 (Q100) = 3.54 m3/s (125 cfs)
Culvert slope (Sculv) =5.0 %
Channel width (Wch) =2.44 m (8.0 ft)

Solving for Critical Discharge (qc):

qc = Q100/Wch
qc = 1.45 m2/s (15.6 ft2/s)

Using the Critical Discharge equation (6.16) to solve for D84:

D84 = 3.45S0.747(1.25qc)2/3/g1/3 (Equation 6.16)
D84 = 3.45Sculv0.747(1.25qc)2/3/g1/3 = 256 mm (0.84 ft)

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 Analysis

D84 = 256 mm (0.84 ft)
V = 9.57(D84)0.487 (Equation 6.17, customary units)
V = 8.79 ft/s (2.68 m/s)

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)
Area = 1.44 m2 (15.5 ft2)
Q=V·A
Q=3.86 m3/s (136.0 cfs)

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 D84 to be stable at Q100 (125 cfs).

##### 8.1.5 WDFW No-Slope Design Example

Stream Properties Needed

Channel width (Wch) = 1.95 m (6.4 ft)
Channel slope (Sch) = 2%
Culvert Slope (Sculv) = 2.2%
Culvert length (Lculv) = 30.5 m (100 ft)

Channel Type and Size

Culvert bed width (Wculv) = 1.25·Wch
Wculv = 2.44 m (8 ft)

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.

Lculv·Sculv = 0.67 m (2.2 ft)
0.2·D= 0.49 m (1.6 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 History

The 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

• Mad River Basin, Northern California
• Mather Creek

Project Type

• Embedded Structural Plate Pipe
• Geomorphic Simulation

Pre-Project Barrier

• Undersized Corrugated Metal Pipe (Overtopped at 5-yr flow)
• 1800 mm (6 ft) diameter CMP
• 41.1 m (135 ft) long at 0.4 % slope
• Cascade over rock apron at outlet

Figure 8.3 Pre-project barrier culvert (United States Forest Service 2006b)

Channel Characteristics

• 100-year Flow: 16.1 cms (570 cfs)
• Drainage Area: 4.4 km2 (1.7 mi2)
• Bankfull Width: 3.4 m (11 ft)

Ecological Value

• Provide access to 4.2 km (2.6 mi) of rearing habitat for coho salmon, steelhead and cutthroat trout. Upstream habitat is low gradient, marshy, and maintains good year-round flows.

Project Characteristics

• Culvert Diameter: 4.9 m (16 ft)
• Length: 32.0 m (130 ft)
• Depth Embedded: 0.6-0.9 m (2-2.5 ft)
• Slope of Bed in Culvert: 0.75 %

Figure 8.4 Replacement culvert (United States Forest Service 2006b)
(Note: headwall recommended for State DOT projects)

Challenges

• Protecting buried water line
• Stabilizing side slopes during excavation to set culvert at desired depth for embedding

Project Funding

• Humboldt County
• California Dept. Fish and Game

Completion Date

• October, 2002

Total Project Cost

• \$234,544

Project Description
When installed in the 1970s, the downstream channel was realigned and channelized. Subsequently, a rock apron spanning the channel had been placed below the culvert outlet. A fish passage assessment conducted in 1999 found the sloping rock apron created a complete barrier to juvenile salmonids and a low-flow barrier to larger fish. The original culvert also had inadequate flood capacity and was in poor condition, with the bottom rusted-through.

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 Simulation

##### 8.2.1 Culvert with Floodplain Relief Case History

The following case history was provided by Andrzej ("Andy") Kosicki of the Maryland State Highway Administration.

Location

• MD Route 25 over Beaverdam Run, Baltimore County, Maryland, USA

Project Type

• Main channel Structure Plate Pipe Arch (SPPA)
• Floodplain culverts (one SPPa. and on SPP)
• Hydraulic Simulation

Pre-project Barrier

• Single span slab bridge with a 6.1 m (20 ft) long invert which was paved in the 1960s due to scour and poor structural condition. A single 3.05 m (10 ft) diameter structural plate pipe was added in 1972 after hurricane Agnes washed away a roadway approach on the north side. See Figures 8.5-8.7.
• Fish blockages included an upstream earth and debris dam, a 0.15 m (6 in) drop at the downstream outlet, and a 0.025-0.05 m (1-2 in) flow depth under low flow conditions. No aquatic life has been observed within 15.2 m (50 ft) upstream of the bridge.

Channel Characteristics

• 100-year Design Flow: 70.3 m3/s (2482 cfs)
• High Flow Velocity: 3.05 m/s (10 ft/s)
• Mannings n: 0.034
• Drainage Area: 16.4 km2 (5.9 mi2)
• Low Flow: 0.2 m3/s (7 cfs)
• Low Flow Velocity: 0.58 m/s (1.9 ft/s)
• Mannings n: 0.030

Ecological Value

• Department of Natural Resources stream classification is a Class III (Natural Trout Stream)

Project Characteristics

• 2-12'4"x7'9" Structural Plate Pipe Arches (SPPA)
• 1-10'0" Structural Plate Pipe (SPP) with end walls
• Culvert length: 10.76 m (35.5 ft)
• Culvert slope: 0.56%
• One of the two SPPAs was placed in the channel 0.6 m (2.0 ft) below the existing stream invert (low flow cell). The other SPPA and the round pipe were placed at bankfull elevations, approximately 0.9 m (3.0 ft) higher than the low flow cell.
• Buried riprap aprons, each 7.62 m (25 ft) long were placed at both upstream and downstream ends.

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:

• Aquatic life that was not seen before
• Various water bugs and good sediment movement resulting in clear water, whereas the pre-1994 structure passed water that was dark and murky
• Side cells have displayed wildlife tracks (probably small mammals)

Figure 8.5 Pre-project channel condition (1992)

Figure 8.6 Upstream of pre-existing structure looking downstream

Figure 8.7 Downstream of pre-existing structure looking upstream

Figure 8.8 Downstream of culvert, shortly after project completion in 1994

Figure 8.9 Upstream of current crossing in 2005

Figure 8.10 Upstream of current crossing during high-flow event

#### 8.3 Hydraulic Design

##### 8.3.1 John Hatt Creek Case History

Source

• FishXing Case Studies (United States Forest Service 2006b)
• Study from Sebastian Cohen P.E., California Dept. of Transportation

Location

• Navarro River Watershed, Northern California, USA

Project Type

• Culvert Rehabilitation with Metal Insert
• Corner Baffle Retrofit
• Hydraulic Design
• Placement of Concrete Weirs Below Outlet

Pre-Project Conditions

• 1700 mm (5.5 ft) diameter CSP, 52.4 m (172 ft) long, at 2.4% slope
• Culvert distorted (out of round) and deteriorating
• Culvert bottom lined with concrete
• Concrete drop structure at culvert inlet

Pre-Project Barrier

• Insufficient depth, high velocities, excessive leap (Figure 8.11)
• Partial barrier to adult steelhead trout
• Total barrier to juvenile salmonids

Hydrologic Characteristics

• Drainage Area: 1.6 km2 (0.6 mi2)
• 2-year Peak Flow: 1.7 cms (60 cfs)
• Design Capacity (100-year Flow): 7.5 cms (266 cfs)
• Headwater-to-diameter ratio at 7.5 cms (266 cfs) = 2.5
• Upper = 0.85 cms (30 cfs), 50% of 2-yr peak flow
• Lower = 0.08 cms (3 cfs)
• Juvenile Salmonid Passage Design Flows:
• Upper = 0.17 cms (6 cfs), 10% of 2-yr peak flow
• Lower = 0.03 cms (1 cfs)

Ecological Value

• Provide access to 0.9 km (0.6 miles) of upstream spawning and rearing habitat for steelhead trout

Project Characteristics

• Insert a 9.6 mm (3/8 in) thick welded steel pipe, 1500 mm (5 ft) diameter and 52.4 m (172 ft) long into existing culvert
• Weld 43 steel corner baffles into pipe insert
• Baffles 0.21 m (8.3 in) tall at center and spaced 1.2 m (4 ft) apart
• 3 precast concrete weirs with wooden low-flow notches below culvert outlet
• 0.23 m (9 in) drops between concrete weirs

Challenges and Lessons Learned

• Bedrock surrounding culvert made "jacking" a larger pipe through the fill impractical
• Existing culvert was out-of-round so smaller culvert had to be inserted
• Only 7.5 m (25 ft) right-of-way available below culvert outlet for grade control weirs
• Lack of rock armoring, and weirs not sufficiently keyed into banks resulted in flanking
• Wooden low flow notch in center of concrete weir causes plunging water to strike concrete lip at low flow.
• Need for inspection by personnel familiar with fish passage design concepts and objectives

Project Description
The existing 1700 mm (5.5 ft) diameter corrugated steel pipe (CSP) was deteriorated and identified as a depth barrier at low flow and a velocity barrier at high flow for adult and juvenile steelhead. The culvert required rehabilitation due to its deteriorated conditions. Retrofitting involved inserting a 1500 mm (5 ft) diameter, 52.4 m (172 ft) long, welded steel pipe (WSP) into the existing culvert at a 2.4% slope. This design was selected after removing the fill to replace the culvert was deemed too costly.

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.

Figure 8.11 Example detail of culvert rehabilitation and corner baffle retrofit, John Hatt Creek (customary units)
(Tops of the 9 in tall baffles were placed at 15 degrees to horizontal. The left and right edges are 4.3 in and 15.2 in above the invert, respectively.)

Table 8.1 Modeled Hydraulic Conditions at Fish Passage Design Flows for John Hatt Creek (Customary Units)
Species/Lifestage: Juvenile Salmonids Passage Flows Adult Steelhead Passage Flows
Fish Passage Flow: Lower Upper Lower Upper
Flow: 1 cfs 6 cfs 3 cfs 30 cfs
Water Depth: 0.6 ft 1.1 ft 0.8 ft 2.0 ft
Ave. Water Velocity: 0.9 ft/s 1.9 ft/s 1.4 ft/s 4.1 ft/s
Turbulence (EDF): 1.5 lb-ft/s/ft3 3.0 lb-ft/s/ft3 2.2 lb-ft/s/ft3 6.0 lb-ft/s/ft3

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
The baffles appear to be effective at reducing water velocities and increasing water depth within the pipe. The weir crest elevations below the outlet were placed within design tolerances.

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

• October 2003

Total Project Cost

• Construction: \$140,000

Figure 8.12 Downstream view of culvert retrofit, John Hatt Creek
(A 1524 mm (5 ft) diameter welded steel pipe was inserted into the pre-existing culvert; concrete slurry was used to fill the gaps)

Figure 8.13 Pre-existing outlet of John Hatt Creek culvert perched at 0.46 m (1.5 ft) blocks migrating steelhead

Figure 8.14 Steel corner baffles welded to the pipe and spaced 1.22 m (4 ft) apart, John Hatt Creek
(Baffle height provides 0.15 m (6 in) of water depth at the juvenile low flow passage design flow of 0.028 cms (1 cfs))

Figure 8.15 Baffles slowing water velocities at high flows while producing minimal turbulence, John Hatt Creek
(Along the low side of the baffles, velocities are swift, improving passage of debris and sediment, while the high side of the baffle experiences slower velocities suitable for both adult and juvenile fish)

Figure 8.16 Culvert outlet after installation, John Hatt Creek
(Weirs below outlet were precast and lowered into place; weirs were keyed into the bank roughly 0.6 m (2 ft) and the contractor neglected to rock the left bank; the inspector failed to enforce this oversight)

##### 8.3.2 WDFW Roughened Channel Design Example

Stream properties needed:

Channel width (Wch) = 2.13 m (7 ft)
Channel slope (Sch) = 1.7%
Culvert slope (Sculv) = 2.3%
Culvert length (Lculv) =27.4 m (90 ft)
100 year exceedance flow (Q100) = 3.54 m3/s (125 cfs)
Fish passage flow (Qfp) = 0.27 m3/s (9.7 ft3/s)

Slope ratio

 Slope ratio = Sculv = 1.35 Sch

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 (Wculv) = Wch=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 = Q100 Wch

q = 1.66 m2/s (17.9 ft2/s)

This allows the D30 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):

 D30 = (1.95 · Sculv0.555) · (1.25 · q)2/3 g1/3

D30 = 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 D30 to find D84, using the approximate scaling factor provided for riprap, Equation 6.15:

D84=1.5·D30 (Equation 6.15)
D84=0.90 ft

This particle size is checked to ensure that it does not exceed 1/4 of the culvert span

4·D84=3.584 < 7ft

A gradation can now be created based on D84 = 0.9 ft.

###### 8.3.2.1 Fish Passage Velocity

Fish 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 (Vfish) = 4 ft/s
Hydraulic Radius (R) = 0.35 ft

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.

 V = 1.486 · R1/6 · (g · R · Sculv)0.5 (n · g)0.5

Limerinos equation is solved as follows (Equation 6.2)

 n= 0.0926R1/6 = 0.12 1.16 + 2log(R/D84)

which can be used into Equation 7.7 to solve for velocity

 Vl = 1.486R1/6 (gRSculv)0.5 ng0.5

Vl=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·Sculv0.38·R-0.16
n=0.10

Using n to solve for velocity

 Vj = 1.486R1/6 (gRSculv)0.5 ng0.5

Vj=1.42 ft/s <4.0ft/s

Mussetter's equation utilizes the Darcy-Weisbach friction factor, and is solved according to Equation 6.4.

 8 0.5 = 1.11 depth 0.46 D84 -0.85 Sculv-0.39 (Equation 6.4) f D84 D50

For this equation D50 is needed, and can be solved for according to the relations provided in Washington's Stream Simulation Design.

 D84 = 2.5 D50
 D50 = D84 = 0.36 ft 2.5

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
R = 0.52 ft

 8 0.5 = 1.11 depth 0.46 D84 -0.85 Sculv-0.39 f D84 D50
 Vm = 1.11 depth 0.46 D84 -0.85 Sculv-0.39 (gRSculv)0.5 D84 D50

Vm=1.48 ft/s < 4.0 ft/s, which is acceptable for fish passage

###### 8.3.2.2 Turbulence

Turbulence is then checked through the calculation of channel EDF.

EDF = γQfpSculv/A (Equation 3.1)
γ = 62.4 lb/ft3
Sculv = 0.023
Qfp = 9.7cfs
A = 6.53 ft2 (based on a triangular low flow channel with 6:1 side-slopes)

EDF = 2.13 ft·lb/(ft3·s) < and is acceptatble for fish passage design

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Updated: 04/07/2011

### Contact:

Bert Bergendahl
720-963-3754
Bart.Bergendahl@dot.gov

United States Department of Transportation - Federal Highway Administration