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Hydraulic Design of Energy Dissipators for Culverts and Channels
Hydraulic Engineering Circular Number 14, Third Edition

Chapter 5: Estimating Scour At Culvert Outlets

This chapter presents a method for predicting local scour at the outlet of culverts based on discharge, culvert shape, soil type, duration of flow, culvert slope, culvert height above the bed, and tailwater depth. In addition to this local scour, channel degradation (discussed in Chapter 2) should be evaluated. The procedures in this chapter provide a good method for estimating the extent of the local scour hole. The designer should also review maintenance history, site reconnaissance and data on soils, flows, and flow duration to determine the best estimate of the potential scour hazard at a culvert outlet.

The prediction equations presented in this chapter are intended to serve along with field reconnaissance as guidance for determining the need for energy dissipators at culvert outlets. The designer should remember that the equations do not include long-term channel degradation of the downstream channel. The equations are based on tests that were conducted to determine maximum scour for the given condition and therefore represent what might be termed worst-case scour geometries. The equations were derived from tests conducted by the Corps of Engineers (Bohan, 1970) and Colorado State University (Abt, et al., 1985; Abt, et al., 1987; Abt, 1996; Doehring, 1994; Donnell and Abt, 1983; Ruff, et al., 1982).

5.1 Cohesionless Soils

The general expression for determining scour geometry in a cohesionless soil at the culvert outlet is:

(5.1)

h sub s divided by R sub c or W sub s divided by R sub c or L sub s divided by R sub c or V sub s divided by R sub c cubed equals (depending on the selection of coefficients in Table 5.1) C sub s times C sub h times (alpha divided by sigma to the one-third power) times (Q divided by the square-root of g divided by R sub c to the 2.5 power) to the beta power times (t divided by 316) to the theta power

where,

hs = depth of scour, m (ft)

Ws = width of scour, m (ft)

Ls = length of scour, m (ft)

Vs = volume of scour, m3 (ft3)

Rc = hydraulic radius at the end of the culvert (assuming full flow)

Q = discharge, m3/s (ft3/s)

g = acceleration of gravity, 9.81 m/s2 (32.2 ft /s2)

t = time in minutes

σ = (D84/D16)0.5, material standard deviation

α, β, θ are coefficients, see Table 5.1

Cs = slope correction coefficient, see Table 5.2

Ch = drop height adjustment coefficient, see Table 5.3

The bed-material grain-size distribution is determined by performing a sieve analysis (ASTM DA22-63). The values of D84 and D16 are extracted from the grain size distribution. If σ < 1.5, the material is considered to be uniform. If σ > 1.5, the material is classified as graded. Typical values for σ are 2.10 for gravel and 1.87 for sand.

5.1.1 Scour Hole Geometry

Investigators Bohan (1970) and Fletcher and Grace (1972) indicate that the scour hole geometry varies with tailwater conditions with the maximum scour geometry occurring at tailwater depths less than half the culvert height (Bohan, 1970); and that the maximum depth of scour, hs, occurs at a location approximately 0.4 Ls downstream of the culvert outlet (Fletcher and Grace, 1972) where Ls is the length of scour. The α, β, θ coefficients to determine scour geometry are shown in Table 5.1.

Table 5.1. Coefficients for Culvert Outlet Scour in Cohesionless Soils
 αβθ
Depth, hs2.270.390.06
Width, Ws6.940.530.08
Length, Ls17.100.470.10
Volume, Vs127.081.240.18
5.1.2 Time of Scour

The time of scour is estimated based upon knowledge of peak flow duration. Lacking this knowledge, it is recommended that a time of 30 minutes be used in Equation 5.1. The tests indicate that approximately 2/3 to 3/4 of the maximum scour depth occurs in the first 30 minutes of the flow duration. The exponents for the time parameter in Table 5.1 reflect the relatively flat part of the scour-time relationship (t > 30 minutes) and are not applicable for the first 30 minutes of the scour process.

5.1.3 Headwalls

Installation of a perpendicular headwall at the culvert outlet moves the scour hole downstream (Ruff, et al., 1982). However, the magnitude of the scour geometry remains essentially the same as for the case without the headwall. If the culvert is installed with a headwall, the headwall should extend to a depth equal to the maximum depth of scour.

5.1.4 Drop Height

The scour hole dimensions will vary with the height of the culvert invert above the bed. The scour hole shape becomes deeper, wider, and shorter, as the culvert invert height is increased (Doehring, 1994). The coefficients, Ch, are derived from tests where the pipe invert is adjacent to the bed. In order to compensate for an elevated culvert invert, Equation 5.1 can be modified to where Ch, expressed in pipe diameters, is a coefficient for adjusting thecompound scour hole geometry. The values of Ch are presented in Table 5.2.

Table 5.2. Coefficient Ch for Outlets above the Bed
Hd1DepthWidthLengthVolume
01.001.001.001.00
11.221.510.731.28
21.261.540.731.47
41.341.660.731.55
1Hd is the height above bed in pipe diameters.
5.1.5 Slope

The scour hole dimensions will vary with culvert slope. The scour hole becomes deeper, wider, and longer as the slope is increased (Abt, 1985). The coefficients presented are derived from tests where the pipe invert is adjacent to the bed. In order to compensate for a sloped culvert, Equation 5.1 can be adjusted with a coefficient, Cs, adjusting for scour hole geometry. The values of Cs are shown in Table 5.3.

Table 5.3. Coefficient Cs for Culvert Slope
Slope %DepthWidthLengthVolume
01.001.001.001.00
21.031.281.171.30
51.081.281.171.30
>71.121.281.171.30
5.1.6 Design Procedure

Step 1. Determine the magnitude and duration of the peak discharge. Express the discharge in m3/s (ft3/s) and the duration in minutes.

Step 2. Compute the full flow hydraulic radius, Rc

Step 3. Compute the culvert invert height above the bed ratio, Hd, for slopes > 0%.

H sub d equals Drop Height divided by Diameter

Step 4. Determine scour coefficients from Table 5.1 and coefficients for culvert drop height, Ch, from Table 5.2 and slope, Cs, from Table 5.3.

Step 5. Determine the material standard deviation, σ = (D84/D16)0.5 from a sieve analysis of a soil sample at the proposed culvert location.

Step 6. Compute the scour hole dimensions using Equation 5.1.

Step 7. Compute the location of maximum scour, Lm = 0.4 Ls.

Design Example: Estimating Scour Hole Geometry in a Cohesionless Soil (SI)

Determine the scour geometry-maximum depth, width, length and volume of scour. Given:

  • D = 457 mm CMP Culvert
  • S = 2%
  • Drop height = 0.914 m from channel degradation
  • Q = 0.764 m3/s
  • σ = 1.87 for downstream channel which is graded sand

Solution

Step 1. Determine the magnitude and duration of the peak discharge: Q = 0.764 m3/s and the peak flow duration is estimated to be 30 minutes.

Step 2. Compute the full flow hydraulic radius, Rc:

R sub c equals D divided by 4 equals 0.457 divided by 4 equals 0.114 meters

Step 3. Compute the height above bed ratio, Hd, for slopes > 0%:

H sub d equals Drop Height divided by Diameter equals 0.914 divided by 0.457 equals 2

Step 4. The Coefficients of scour obtained from Table 5.1, Table 5.2, and Table 5.3 are:

αβθCsCh
Depth of scour2.270.390.061.031.26
Width of scour6.940.530.081.281.54
Length of scour17.100.470.101.170.73
Volume of scour127.081.240.181.301.47

Step 5. Determine the material standard deviation. σ = 1.87

Step 6. Compute the scour hole dimensions using Equation 5.1:

h sub s divided by R sub c or W sub s divided by R sub c or L sub s divided by R sub c or V sub s divided by R sub c cubed equals (depending on the selection of coefficients in Table 5.1) C sub s times C sub h times (alpha divided by sigma to the one-third power) times (Q divided by the square-root of g divided by R sub c to the 2.5 power) to the beta power times (t divided by 316) to the theta power

h sub s equals C sub s times C sub h times (alpha divided by 1.87 to the one-third power) times (Q divided by the square-root of g divided by R sub c to the 2.5 power) to the beta power times (30 divided by 316) to the theta power times R sub c

h sub s equals 1.03 times 1.26 times (2.27 divided by 1.23) times (0.764 divided by the square-root of 9.81 divided by 0.114 to the 2.5 power) to the 0.39 power times (0.095) to the 0.06 power times 0.114 equals 1.14 meters

Similarly,

W sub s equals 1.28 times 1.54 times (6.94 divided by 1.23) times (0.764 divided by the square-root of 9.81 divided by 0.114 to the 2.5 power) to the 0.53 power times (0.095) to the 0.08 power times 0.114 equals 8.82 meters

L sub s equals 1.17 times 0.73 times (17.10 divided by 1.23) times (0.764 divided by the square-root of 9.81 divided by 0.114 to the 2.5 power) to the 0.47 power times (0.095) to the 0.10 power times 0.114 equals 7.06 meters

V sub s equals 1.30 times 1.47 times (127.08 divided by 1.23) times (0.764 divided by the square-root of 9.81 divided by 0.114 to the 2.5 power) to the 1.24 power times (0.095) to the 0.18 power times 0.114 cubed equals 27.8 cubic meters

Step 7. Compute the location of maximum scour. Lm = 0.4 Ls = 0.4 (7.06) = 2.82 m downstream of the culvert outlet.

Design Example: Estimating Scour Hole Geometry in a Cohesionless Soil (CU)

Determine the scour geometry-maximum depth, width, length and volume of scour. Given:

  • D = 18 in CMP Culvert
  • S = 2%
  • Drop height = 3 ft from channel degradation
  • Q = 27 ft3/s
  • σ = 1.87 for downstream channel which is graded sand

Solution

Step 1. Determine the magnitude and duration of the peak discharge: Q = 27 ft3/s and the peak flow duration is estimated to be 30 minutes.

Step 2. Compute the full flow hydraulic radius, Rc:

R sub c equals D divided by 4 equals 1.5 divided by 4 equals 0.375 feet

Step 3. Compute the height above bed ratio, Hd, for slopes > 0%

H sub d equals Drop Height divided by Diameter equals 3 divided by 1.5 equals 2

Step 4. The coefficients of scour obtained from Table 5.1, Table 5.2, and Table 5.3 are:

 αβθCsCh
Depth of scour2.270.390.061.031.26
Width of scour6.940.530.081.281.54
Length of scour17.100.470.101.170.73
Volume of scour127.081.240.181.301.47

Step 5. Determine the material standard deviation. σ = 1.87

Step 6. Compute the scour hole dimensions using Equation 5.1:

h sub s divided by R sub c or W sub s divided by R sub c or L sub s divided by R sub c or V sub s divided by R sub c cubed equals (depending on the selection of coefficients in Table 5.1) C sub s times C sub h times (alpha divided by sigma to the one-third power) times (Q divided by the square-root of g divided by R sub c to the 2.5 power) to the beta power times (t divided by 316) to the theta power

h sub s equals C sub s times C sub h times (alpha divided by 1.87 to the one-third power) times (Q divided by the square-root of g divided by R sub c to the 2.5 power) to the beta power times (30 divided by 316) to the theta power times R sub c

h sub s equals 1.03 times 1.26 times (2.27 divided by 1.23) times (27 divided by the square-root of 32.2 divided by 0.375 to the 2.5 power) to the 0.39 power times (0.095) to the 0.06 power times 0.375 equals 3.7 feet

Similarly,

W sub s equals 1.28 times 1.54 times (6.94 divided by 1.23) times (27 divided by the square-root of 32.2 divided by 0.375 to the 2.5 power) to the 0.53 power times (0.095) to the 0.08 power times 0.375 equals 29.0 feet

L sub s equals 1.17 times 0.73 times (17.10 divided by 1.23) times (27 divided by the square-root of 32.2 divided by 0.375 to the 2.5 power) to the 0.47 power times (0.095) to the 0.10 power times 0.375 equals 23.2 feet

V sub s equals 1.30 times 1.47 times (127.08 divided by 1.23) times (27 divided by the square-root of 32.2 divided by 0.375 to the 2.5 power) to the 1.24 power times (0.095) to the 0.18 power times 0.375 cubed equals 987 cubic feet

Step 7. Compute the location of maximum scour. Lm = 0.4 Ls = 0.4 (23.2) = 9.2 ft downstream of the culvert outlet.

5.2 Cohesive Soils

If the soil is cohesive in nature, Equation 5.2 should be used to determine the scour hole dimensions. Shear number expressions, which relate scour to the critical shear stress of the soil, were derived to have a wider range of applicability for cohesive soils besides the one specific sandy clay that was tested. The sandy clay tested had 58 percent sand, 27 percent clay, 15 percent silt, and 1 percent organic matter; had a mean grain size of 0.15 mm (0.0059 in); and had a plasticity index, PI, of 15. The shear number expressions for circular culverts are:

(5.2)

h sub s divided by D or W sub s divided by D or L sub s divided by D or V sub s divided by D cubed (depending on the selection of coefficients in Table 5.4) equals C sub s times C sub h times alpha times (rho times V squared divided by tau sub c) to the beta power times (t divided by 316) to the theta power

and for other shaped culverts:

(5.3)

h sub s divided by y sub e or W sub s divided by y sub e or L sub s divided by y sub e or V sub s divided by y sub e cubed (depending on the selection of coefficients in Table 5.4) equals C sub s times C sub h times alpha sub e times (rho times V squared divided by tau sub c) to the beta power times (t divided by 316) to the theta power

where,

D = culvert diameter, m (ft)

ye = equivalent depth (A/2)1/2, m (ft)

A = cross-sectional area of flow, m2 (ft2)

V = mean outlet velocity, m/s (ft/s)

τc = critical tractive shear stress, N/m2 (lb/ft2)

ρ = fluid density of water, 1000 kg/m3 (1.94 slugs/ft3)

(ρV2)/τc is the modified shear number

αe = αe = α/0.63 for hs, Ws, and Ls and αe = α/(0.63)3 for Vs

α, β, θ, and αe are coefficients found in Table 5.4

Use 30 minutes for t in Equation 5.2 and Equation 5.3 if it is not known.

The critical tractive shear stress is defined in Equation 5.4 (Dunn, 1959; Abt et al., 1996). Equations 5.2 and 5.3 should be limited to sandy clay soils with a plasticity index of 5 to 16.

(5.4)

τc = 0.001 (Sν+ αu) tan (30 + 1.73 PI)

where,

τc = critical tractive shear stress, N/m2 (lb/ft2)

Sν = the saturated shear strength, N/m2 (lb/ft2)

αu = unit conversion constant, 8630 N/m2 (SI), 180 lb/ft2 (CU)

PI = Plasticity Index from the Atterberg limits

Table 5.4. Coefficients for Culvert Outlet Scour in Cohesive Soils
 αβθαe
Depth, hs0.860.180.101.37
Width, Ws3.550.170.075.63
Length, Ls2.820.330.094.48
Volume, Vs0.620.930.232.48

The design procedure for estimating scour in cohesive materials with PI from 5 to 16 may be summarized as follows.

Step 1. Determine the magnitude and duration of the peak discharge, Q. Express the discharge in m3/s (ft3/s) and the duration in minutes.

Step 2. Compute the culvert average outlet velocity, V.

Step 3. Obtain a soil sample at the proposed culvert location.

  1. Perform Atterberg limits tests and determine the plasticity index (PI) using ASTM D423-36.
  2. Saturate a sample and perform an unconfined compressive test (ASTM D211-66-76) to determine the saturated shear stress, S ν.

Step 4. Compute the critical tractive shear strength, τc, from Equation 5.4.

Step 5. Compute the modified shear number, Snm, at the peak discharge and height above bed ratio, Hd, for slopes > 0%.

S sub nm equals rho times V squared divided by tau sub candH sub d equals Drop Height divided by Diameter

Step 6. Determine scour coefficients from Table 5.4 and, if appropriate, coefficients for culvert drop height, Ch, from Table 5.2 and slope, Cs, from Table 5.3.

Step 7. Compute the scour hole dimensions using Equation 5.2 for circular culverts and Equation 5.3 for other shapes.

Step 8. Compute the location of maximum scour. Lm = 0.4 Ls.

Design Example: Estimating Scour Hole Geometry in a Cohesive Soil (SI)

Determine the scour geometry: maximum depth, width, length and volume of scour. Given:

  • D = 610 mm CMP Culvert
  • S = 0%
  • Drop height = 0 m
  • Q = 1.133 m3/s
  • PI = 12 and Sv = 23,970 N/m2 for downstream channel

Solution

Step 1. Determine the magnitude and duration of the peak discharge: Q = 1.133 m3/s and the peak flow duration is estimated to be 30 minutes.

Step 2. Compute the culvert average outlet velocity, V:

V equals Q divided by A equals 1.133 (divided by 3.14 times 0.61 squared divided by 4) equals 3.88 meters per second

Step 3. Obtain a soil sample at the proposed culvert location: The sandy-clay soil was tested and found to have:

  1. Plasticity index (PI) = 12
  2. Saturated shear stress, Sν = 23,970 N/m2.

Step 4. Compute the critical tractive shear strength, τc, from Equation 5.4.

τc = 0.001 (Sν+ αu) tan (30 + 1.73 PI)

τc = 0.001 (23970 + 8630) tan [30 + 1.73(12)]

τc = 0.001 (32600) tan (50.76) = 39.9 N/m2

Step 5. Compute the modified shear number, Snm, at the peak discharge and height above bed ratio, Hd, for slopes > 0%.

S sub nm equals rho times V squared divided by tau sub c equals 1000 times 3.88 squared divided by 39.9 equals 377.3

Step 6. Determine scour coefficients from Table 5.4 and coefficients for culvert drop height, Ch, from Table 5.2 and slope from Table 5.3: Ch = 1 and Cs = 1

 αβθ
Depth, hs0.860.180.10
Width, Ws3.550.170.07
Length, Ls2.820.330.09
Volume, Vs0.620.930.23

Step 7. Compute the scour hole dimensions using Equation 5.2 for circular culverts:

h sub s divided by D or W sub s divided by D or L sub s divided by D or V sub s divided by D cubed (depending on the selection of coefficients in Table 5.4) equals C sub s times C sub h times alpha times (rho times V squared divided by tau sub c) to the beta power times (t divided by 316) to the theta power

h sub s equals 1.0 times 1.0 times alpha times 377.3 to the beta power times (30 divided by 316) to the theta power times D

hs = (1.0)(1.0)(0.86)(377.3)0.18(0.09)0.10(0.61) = 1.2 m

Similarly,

Ws = (1.0)(1.0)(3.55)(377.3)0.17(0.09)0.07(0.61) = 5.02 m

Ls = (1.0)(1.0)(2.82)(377.3)0.33(0.09)0.09(0.61) = 9.81 m

Vs = (1.0)(1.0)(0.62)(377.3)0.93(0.09)0.23(0.61)3 = 20.15 m3

Step 8. Compute the location of maximum scour. Lm = 0.4 Ls = 0.4(9.81) = 3.92 m downstream of culvert outlet.

Design Example: Estimating Scour Hole Geometry in a Cohesive Soil (CU)

Determine the scour geometry: maximum depth, width, length and volume of scour. Given:

  • D = 24 in CMP Culvert
  • S = 0%
  • Drop height = 0 ft
  • Q = 40 ft3/s
  • PI = 12 and Sv = 500 lb/ft2 for downstream channel

Solution

Step 1. Determine the magnitude and duration of the peak discharge: Q = 40 ft3/s and the peak flow duration is estimated to be 30 minutes.

Step 2. Compute the culvert average outlet velocity, V:

V equals Q divided by A equals 40 (divided by 3.14 times 2 squared divided by 4) equals 12.74 ft per second

Step 3. Obtain a soil sample at the proposed culvert location: The sandy-clay soil was tested and found to have:

  1. Plasticity index, PI = 12
  2. Saturated shear stress, Sν = 500 lb/ft2.

Step 4. Compute the critical tractive shear strength, τc, from Equation 5.4.

τc = 0.001 (Sν+ αu) tan (30 + 1.73 PI)

τc = 0.001 (500 + 180) tan [30 + 1.73(12)]

τc = 0.001 (680) tan (50.76) = 0.83 lb/ft2

Step 5. Compute the modified shear number, Snm, at the peak discharge and height above bed ratio, Hd, for slopes > 0%.

S sub nm equals rho times V squared divided by tau sub c equals 1.94 times 12.74 squared divided by 0.83 equals 379.4

Step 6. Determine scour coefficients from Table 5.4 and coefficients for culvert drop height, Ch, from Table 5.2 and slope from Table 5.3: Ch = 1 and Cs = 1

 αβθ
Depth, hs0.860.180.10
Width, Ws3.550.170.07
Length, Ls2.820.330.09
Volume, Vs0.620.930.23

Step 7. Compute the scour hole dimensions using Equation 5.2 for circular culverts:

h sub s divided by D or W sub s divided by D or L sub s divided by D or V sub s divided by D cubed (depending on the selection of coefficients in Table 5.4) equals C sub s times C sub h times alpha times (rho times V squared divided by tau sub c) to the beta power times (t divided by 316) to the theta power

h sub s equals 1.0 times 1.0 times alpha times 379.4 to the beta power times (30 divided by 316) to the theta power times D

hs = (1.0)(1.0)(0.86)(379.4)0.18(0.09)0.10(2) = 3.9 ft

Similarly,

Ws = (1.0)(1.0)(3.55)(379.4)0.17(0.09)0.07(2) = 16.5 ft

Ls = (1.0)(1.0)(2.82)(379.4)0.33(0.09)0.09(2) = 32.2 ft

Vs = (1.0)(1.0)(0.62)(379.4)0.93(0.09)0.23(2)3 = 713.7 ft3

Step 8. Compute the location of maximum scour. Lm = 0.4 Ls = 0.4(32.2) = 12.9 ft downstream of culvert outlet.

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Cynthia Nurmi
Resource Center (Atlanta)
404-562-3908
cynthia.nurmi@dot.gov

Updated: 04/07/2011
 

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