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

Chapter 10: Riprap Basins And Aprons

Riprap is a material that has long been used to protect against the forces of water. The material can be pit-run (as provided by the supplier) or specified (standard or special). State DOTs have standard specifications for a number of classes (sizes or gradations) of riprap. Suppliers maintain an inventory of frequently used classes. Special gradations of riprap are produced on-demand and are therefore more expensive than both pit-run and standard classes.

This chapter includes discussion of both riprap aprons and riprap basin energy dissipators. Both can be used at the outlet of a culvert or chute (channel) by themselves or at the exit of a stilling basin or other energy dissipator to protect against erosion downstream. Section 10.1 provides a design procedure for the riprap basin energy dissipator that is based on armoring a pre-formed scour hole. The riprap for this basin is a special gradation. Section 10.2 includes discussion of riprap aprons that provide a flat armored surface as the only dissipator or as additional protection at the exit of other dissipators. The riprap for these aprons is generally from State DOT standard classes. Section 10.3 provides additional discussion of riprap placement downstream of energy dissipators.

10.1 Riprap Basin

The design procedure for the riprap basin is based on research conducted at Colorado State University (Simons, et al., 1970; Stevens and Simons, 1971) that was sponsored by the Wyoming Highway Department. The recommended riprap basin that is shown on Figure 10.1 and Figure 10.2 has the following features:

  • The basin is pre-shaped and lined with riprap that is at least 2D50 thick.
  • The riprap floor is constructed at the approximate depth of scour, hs, that would occur in a thick pad of riprap. The hs/D50 of the material should be greater than 2.
  • The length of the energy dissipating pool, Ls, is 10hs, but no less than 3Wo; the length of the apron, LA, is 5hs, but no less than Wo. The overall length of the basin (pool plus apron), LB, is 15hs, but no less than 4Wo.
  • A riprap cutoff wall or sloping apron can be constructed if downstream channel degradation is anticipated as shown in Figure 10.1.

Figure 10.1. Profile of Riprap Basin

Summarizes dimensions described in the text.

Figure 10.2. Half Plan of Riprap Basin

Summarizes dimensions described in the text.

10.1.1 Design Development

Tests were conducted with pipes from 152 mm (6 in) to 914 mm (24 in) and 152 mm (6 in) high model box culverts from 305 mm (12 in) to 610 mm (24 in) in width. Discharges ranged from 0.003 to 2.8 m3/s (0.1 to 100 ft3/s). Both angular and rounded rock with an average size, D50, ranging from 6 mm (1.4 in) to 177 mm (7 in) and gradation coefficients ranging from 1.05 to 2.66 were tested. Two pipe slopes were considered, 0 and 3.75%. In all, 459 model basins were studied. The following conclusions were drawn from an analysis of the experimental data and observed operating characteristics:

  • The scour hole depth, hs; length, Ls; and width, Ws, are related to the size of riprap, D50; discharge, Q; brink depth, yo; and tailwater depth, TW.
  • Rounded material performs approximately the same as angular rock.
  • For low tailwater (TW/yo < 0.75), the scour hole functions well as an energy dissipator if hs/D50 > 2. The flow at the culvert brink plunges into the hole, a jump forms and flow is generally well dispersed.
  • For high tailwater (TW/yo > 0.75), the high velocity core of water passes through the basin and diffuses downstream. As a result, the scour hole is shallower and longer.
  • The mound of material that forms downstream contributes to the dissipation of energy and reduces the size of the scour hole. If the mound is removed, the scour hole enlarges somewhat.

Plots were constructed of hs/ye versus Vo/ (gye)1/2 with D50/ye as the third variable. Equivalent brink depth, ye, is defined to permit use of the same design relationships for rectangular and circular culverts. For rectangular culverts, ye = yo (culvert brink depth). For circular culverts, ye = (A/2)1/2, where A is the brink area.

Anticipating that standard or modified end sections would not likely be used when a riprap basin is located at a culvert outlet, the data with these configurations were not used to develop the design relationships. This assumption reduced the number of applicable runs to 346. A total of 128 runs had a D50/ye of less than 0.1. These data did not exhibit relationships that appeared useful for design and were eliminated. An additional 69 runs where hs/D50<2 were also eliminated by the authors of this edition of HEC 14. These runs were not considered reliable for design, especially those with hs = 0. Therefore, the final design development used 149 runs from the study. Of these, 106 were for pipe culverts and 43 were for box culverts. Based on these data, two design relationships are presented here: an envelope design and a best fit design.

To balance the need for avoiding an underdesigned basin against the costs of oversizing a basin, an envelope design relationship in the form of Equation 10.1 and Equation 10.2 was developed. These equations provide a design envelope for the experimental data equivalent to the design figure (Figure XI-2) provided in the previous edition of HEC 14 (Corry, et al., 1983). Equations 10.1 and 10.2, however, improve the fit to the experimental data reducing the root-mean-square (RMS) error from 1.24 to 0.83.

(10.1)

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o

where,

hs = dissipator pool depth, m (ft)

ye = equivalent brink (outlet) depth, m (ft)

D50 = median rock size by weight, m (ft

Co = tailwater parameter

The tailwater parameter, Co, is defined as:

(10.2)
Co = 1.4TW/ye < 0.75
Co = 4.0(TW/ye) -1.60.75 < TW/ye < 1.0
Co = 2.41.0 < TW/ye

A best fit design relationship that minimizes the RMS error when applied to the experimental data was also developed. Equation 10.1 still applies, but the description of the tailwater parameter, Co, is defined in Equation 10.3. The best fit relationship for Equations 10.1 and 10.3 exhibits a RMS error on the experimental data of 0.56.

(10.3)
Co = 2.0TW/ye < 0.75
Co = 4.0(TW/ye) -1.00.75 < TW/ye < 1.0
Co = 3.01.0 < TW/ye

Use of the envelope design relationship (Equations 10.1 and 10.2) is recommended when the consequences of failure at or near the design flow are severe. Use of the best fit design relationship (Equations 10.1 and 10.3) is recommended when basin failure may easily be addressed as part of routine maintenance. Intermediate risk levels can be adopted by the use of intermediate values of Co.

10.1.2 Basin Length

Frequency tables for both box culvert data and pipe culvert data of relative length of scour hole (Ls/hs < 6, 6 < Ls/h s< 7, 7 < Ls/hs <8 . . . 25 < Ls/hs < 30), with relative tailwater depth TW/ye in increments of 0.03 m (0.1 ft) as a third variable, were constructed using data from 346 experimental runs. For box culvert runs Ls/hs was less than 10 for 78% of the data and Ls/hs was less than 15 for 98% of the data. For pipe culverts, Ls/hs was less than 10 for 91% of the data and, Ls/hs was less than 15 for all data. A 3:1 flare angle is recommended for the basins walls. This angle will provide a sufficiently wide energy dissipating pool for good basin operation.

10.1.3 High Tailwater

Tailwater influenced formation of the scour hole and performance of the dissipator. For tailwater depths less than 0.75 times the brink depth, scour hole dimensions were unaffected by tailwater. Above this the scour hole became longer and narrower. The tailwater parameter defined in Equations 10.2 and 10.3 captures this observation. In addition, under high tailwater conditions, it is appropriate to estimate the attenuation of the flow velocity downstream of the culvert outlet using Figure 10.3. This attenuation can be used to determine the extent of riprap protection required. HEC 11 (Brown and Clyde, 1989) or the method provided in Section 10.3 can be used for sizing riprap.

Figure 10.3. Distribution of Centerline Velocity for Flow from Submerged Outlets

Provides design curves to determine V sub L (average) divided by V sub AVE as a function of the distance downstream from the outlet, L, divided by the equivalent cirular diameter of the culvert, D sub e. Curves are provided for rectangular and circular culverts.

10.1.4 Riprap Details

Based on experience with conventional riprap design, the recommended thickness of riprap for the floor and sides of the basin is 2D50 or 1.50Dmax, where Dmax is the maximum size of rock in the riprap mixture. Thickening of the riprap layer to 3D50 or 2Dmax on the foreslope of the roadway culvert outlet is warranted because of the severity of attack in the area and the necessity for preventing undermining and consequent collapse of the culvert. Figure 10.1 illustrates these riprap details. The mixture of stone used for riprap and need for a filter should meet the specifications described in HEC 11 (Brown and Clyde, 1989).

10.1.5 Design Procedure

The design procedure for a riprap basin is as follows:

Step 1. Compute the culvert outlet velocity, Vo, and depth, yo.

For subcritical flow (culvert on mild or horizontal slope), use Figure 3.3 or Figure 3.4 to obtain yo/D, then obtain Vo by dividing Q by the wetted area associated with yo. D is the height of a box culvert or diameter of a circular culvert.

For supercritical flow (culvert on a steep slope), Vo will be the normal velocity obtained by using the Manning's Equation for appropriate slope, section, and discharge.

Compute the Froude number, Fr, for brink conditions using brink depth for box culverts (ye=yo) and equivalent depth (ye = (A/2)1/2) for non-rectangular sections.

Step 2. Select D50 appropriate for locally available riprap. Determine Co from Equation 10.2 or 10.3 and obtain hs/ye from Equation 10.1. Check to see that hs/D50≥ 2 and D50/ye≥ 0.1. If hs/D50 or D50/ye is out of this range, try a different riprap size. (Basins sized where hs/D50 is greater than, but close to, 2 are often the most economical choice.)

Step 3. Determine the length of the dissipation pool (scour hole), Ls, total basin length, LB, and basin width at the basin exit, WB, as shown in Figures 10.1 and 10.2. The walls and apron of the basin should be warped (or transitioned) so that the cross section of the basin at the exit conforms to the cross section of the natural channel. Abrupt transition of surfaces should be avoided to minimize separation zones and resultant eddies.

Step 4. Determine the basin exit depth, yB = yc, and exit velocity, VB = Vc and compare with the allowable exit velocity, Vallow. The allowable exit velocity may be taken as the estimated normal velocity in the tailwater channel or a velocity specified based on stability criteria, whichever is larger. Critical depth at the basin exit may be determined iteratively using Equation 7.14:

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc) by trial and success to determine yB.

Vc = Q/Ac

z = basin side slope, z:1 (H:V)

If Vc≤ Vallow, the basin dimensions developed in step 3 are acceptable. However, it may be possible to reduce the size of the dissipator pool and/or the apron with a larger riprap size. It may also be possible to maintain the dissipator pool, but reduce the flare on the apron to reduce the exit width to better fit the downstream channel. Steps 2 through 4 are repeated to evaluate alternative dissipator designs.

Step 5. Assess need for additional riprap downstream of the dissipator exit. If TW/yo≤0.75, no additional riprap is needed. With high tailwater (TW/yo≥ 0.75), estimate centerline velocity at a series of downstream cross sections using Figure 10.3 to determine the size and extent of additional protection. The riprap design details should be in accordance with specifications in HEC 11 (Brown and Clyde, 1989) or similar highway department specifications.

Two design examples are provided. The first features a box culvert on a steep slope while the second shows a pipe culvert on a mild slope.

Design Example: Riprap Basin (Culvert on a Steep Slope) (SI)

Determine riprap basin dimensions using the envelope design (Equations 10.1 and 10.2) for a 2440 mm by 1830 mm reinforced concrete box (RCB) culvert that is in inlet control with supercritical flow in the culvert. Allowable exit velocity from the riprap basin, Vallow, is 2.1 m/s. Riprap is available with a D50 of 0.50, 0.55, and 0.75 m. Consider two tailwater conditions: 1) TW = 0.85 m and 2) TW = 1.28 m. Given:

  • Q = 22.7 m3/s
  • yo = 1.22 m (normal flow depth) = brink depth

Solution

Step 1. Compute the culvert outlet velocity, Vo, depth, yo, and Froude number for brink conditions. For supercritical flow (culvert on a steep slope), Vo will be Vn

yo = ye = 1.22 m

Vo = Q/A = 22.7/ [1.22 (2.44)] = 7.63 m/s

Fr = Vo / (9.81ye)1/2 = 7.63/ [9.81(1.22)]1/2 = 2.21

Step 2. Select a trial D50 and obtain hs/ye from Equation 10.1. Check to see that hs/D50≥ 2 and D50/ye≥ 0.1.

Try D50 = 0.55 m; D50/ye = 0.55/1.22 = 0.45 (≥ 0.1 OK)

Two tailwater elevations are given; use the lowest to determine the basin size that will serve the tailwater range, that is, TW = 0.85 m.

TW/ye = 0.85/1.22 = 0.7, which is less than 0.75. Therefore, from Equation 10.2, Co = 1.4

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.45) to the -0.55 power times (2.21) minus 1.4 equals 1.55

hS = (hS /ye)ye = 1.55 (1.22) = 1.89 m

hS/D50 = 1.89/0.55 = 3.4 and hS/D50≥ 2 is satisfied

Step 3. Size the basin as shown in Figures 10.1 and 10.2.

LS = 10hS = 10(1.89) = 18.9 m

LS min = 3Wo = 3(2.44) = 7.3 m, use LS = 18.9 m

LB = 15hS = 15(1.89) = 28.4 m

LB min = 4Wo = 4(2.44) = 9.8 m, use LB = 28.4 m

WB = Wo + 2(LB/3) = 2.44 + 2(28.4/3) = 21.4 m

Step 4. Determine the basin exit depth, yB = yc, and exit velocity, VB = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc)

22.72/9.81 = 52.5 = [yc(21.4 + 2yc)]3/ (21.4 + 4yc)

By trial and success, yc = 0.48 m, Tc = 23.3 m, Ac = 10.7 m2

VB = Vc = Q/Ac = 22.7/10.7 = 2.1 m/s (acceptable)

The initial trial of riprap (D50 = 0.55 m) results in a 28.4 m basin that satisfies all design requirements. Try the next larger riprap size to test if a smaller basin is feasible by repeating steps 2 through 4.

Step 2 (2nd iteration). Select riprap size and compute basin depth.

Try D50 = 0.75 m; D50/ye = 0.75/1.22 = 0.61 (≥ 0.1 OK)

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.61) to the -0.55 power times (2.21) minus 1.4 equals 1.09

hS = (hS /ye)ye = 1.09 (1.22) = 1.34 m

hS/D50 = 1.34/0.75 = 1.8 and hS/D50≥ 2 is not satisfied. Although not available, try a riprap size that will yield hS/D50 close to, but greater than, 2. (A basin sized for smaller riprap may be lined with larger riprap.) Repeat step 2.

Step 2 (3rd iteration). Select riprap size and compute basin depth.

Try D50 = 0.71 m; D50/ye = 0.71/1.22 = 0.58 (≥ 0.1 OK)

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.58) to the -0.55 power times (2.21) minus 1.4 equals 1.16

hS = (hS /ye)ye = 1.16 (1.22) = 1.42 m

hS/D50 = 1.42/0.71 = 2.0 and hS/D50≥ 2 is satisfied.

Step 3 (3rd iteration). Size the basin as shown in Figures 10.1 and 10.2.

LS = 10hS = 10(1.42) = 14.2 m

LS min = 3Wo = 3(2.44) = 7.3 m, use LS = 14.2 m

LB = 15hS = 15(1.42) = 21.3 m

LB min = 4Wo = 4(2.44) = 9.8 m, use LB = 21.3 m

WB = Wo + 2(LB/3) = 2.44 + 2(21.3/3) = 16.6 m

However, since the trial D50 is not available, the next larger riprap size (D50 = 0.75 m) would be used to line a basin with the given dimensions.

Step 4 (3rd iteration). Determine the basin exit depth, yB = yc, and exit velocity, VB = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc)

22.72/9.81 = 52.5 = [yc(16.6 + 2yc)]3/ (16.6 + 4yc)

By trial and success, yc = 0.56 m, Tc = 18.8 m, Ac = 9.9 m2

VB = Vc = Q/Ac = 22.7/9.9 = 2.3 m/s (greater than 2.1 m/s; not acceptable). If the apron were extended (with a continued flare) such that the total basin length was 28.4 m, the velocity would be reduced to the allowable level.

Two feasible options have been identified. First, a 1.89 m deep, 18.9 m long pool, with a 9.5 m apron using D50 = 0.55 m. Second, a 1.42 m deep, 14.2 m long pool, with a 14.2 m apron using D50 = 0.75 m. Because the overall length is the same, the first option is likely to be more economical.

Step 5. For the design discharge, determine if TW/yo ≤0.75.

For the first tailwater condition, TW/yo = 0.85/1.22 = 0.70, which satisfies TW/yo ≤ 0.75. No additional riprap needed downstream.

For the second tailwater condition, TW/yo = 1.28/1.22 = 1.05, which does not satisfy TW/yo ≤ 0.75. To determine required riprap, estimate centerline velocity at a series of downstream cross sections using Figure 10.3.

Compute equivalent circular diameter, De, for brink area:

A = π De2 /4 = (yo)(Wo) = (1.22)(2.44) = 3.00 m2

De = [3.00(4)/ π ]1/2 = 1.95 m

Rock size can be determined using the procedures in Section 10.3 (Equation 10.6) or other suitable method. The computations are summarized below.

Computation Summary for Design Example (SI)
L/DeL (m)VL/Vo (Figure 10.3)VL (m/s)Rock Size D50 (m)
1019.50.594.500.43
1529.30.423.200.22
2039.00.302.290.11
2141.00.282.130.10

The calculations above continue until VL ≤ Vallow. Riprap should be at least the size shown. As a practical consideration, the channel can be lined with the same size rock used for the basin. Protection must extend at least 41.0 m downstream from the culvert brink, which is 12.6 m beyond the basin exit. Riprap should be installed in accordance with details shown in HEC 11.

Design Example: Riprap Basin (Culvert on a Steep Slope) (CU)

Determine riprap basin dimensions using the envelope design (Equations 10.1 and 10.2) for an 8 ft by 6 ft reinforced concrete box (RCB) culvert that is in inlet control with supercritical flow in the culvert. Allowable exit velocity from the riprap basin, Vallow, is 7 ft/s. Riprap is available with a D50 of 1.67, 1.83, and 2.5 ft. Consider two tailwater conditions: 1) TW = 2.8 ft and 2) TW = 4.2 ft. Given:

  • Q = 800 ft3/s
  • yo = 4 ft (normal flow depth) = brink depth

Solution

Step 1. Compute the culvert outlet velocity, Vo, depth, yo, and Froude number for brink conditions. For supercritical flow (culvert on a steep slope), Vo will be Vn.

yo = ye = 4 ft

Vo = Q/A = 800/ [4 (8)] = 25 ft/s

Fr = Vo / (32.2ye)1/2 = 25/ [32.2(4)]1/2 = 2.2

Step 2. Select a trial D50 and obtain hs/ye from Equation 10.1. Check to see that hs/D50 ≥ 2 and D50/ye ≥ 0.1.

Try D50 = 1.83 ft; D50/ye = 1.83/4 = 0.46 (≥ 0.1 OK)

Two tailwater elevations are given; use the lowest to determine the basin size that will serve the tailwater range, that is, TW = 2.8 ft.

TW/ye = 2.8/4 = 0.7, which is less than 0.75. From Equation 10.2, Co = 1.4

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.46) to the -0.55 power times (2.2) minus 1.4 equals 1.50

hS = (hS /ye)ye = 1.50 (4) = 6.0 ft

hS/D50 = 6.0/1.83 = 3.3 and hS/D50≥ 2 is satisfied

Step 3. Size the basin as shown in Figures 10.1 and 10.2.

LS = 10hS = 10(6.0) = 60 ft

LS min = 3Wo = 3(8) = 24 ft, use LS = 60 ft

LB = 15hS = 15(6.0) = 90 ft

LB min = 4Wo = 4(8) = 32 ft, use LB = 90 ft

WB = Wo + 2(LB/3) = 8 + 2(90/3) = 68 ft

Step 4. Determine the basin exit depth, yB = yc, and exit velocity, VB = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc)

8002/32.2 = 19,876 = [yc(68 + 2yc)]3/ (68 + 4yc)

By trial and success, yc = 1.60 ft, Tc = 74.4 ft, Ac = 113.9 ft2

VB = Vc = Q/Ac = 800/113.9 = 7.0 ft/s (acceptable)

The initial trial of riprap (D50 = 1.83 ft) results in a 90 ft basin that satisfies all design requirements. Try the next larger riprap size to test if a smaller basin is feasible by repeating steps 2 through 4.

Step 2 (2nd iteration). Select riprap size and compute basin depth.

Try D50 = 2.5 ft; D50/ye = 2.5/4 = 0.63 (≥ 0.1 OK)

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.63) to the -0.55 power times (2.2) minus 1.4 equals 1.04

hs = (hs /ye)ye = 1.04 (4) = 4.2 ft

hs/D50 = 4.2/2.5 = 1.7 and hS/D50≥ 2 is not satisfied. Although not available, try a riprap size that will yield hs/D50 close to, but greater than, 2. (A basin sized for smaller riprap may be lined with larger riprap.) Repeat step 2.

Step 2 (3rd iteration). Select riprap size and compute basin depth.

Try D50 = 2.3 ft; D50/ye = 2.3/4 = 0.58 (≥ 0.1 OK)

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.58) to the -0.55 power times (2.2) minus 1.4 equals 1.15

hs = (hs /ye)ye = 1.15 (4) = 4.6 ft

hs/D50 = 4.6/2.3 = 2.0 and hs/D50≥ 2 is satisfied.

Step 3 (3rd iteration). Size the basin as shown in Figures 10.1 and 10.2.

LS = 10hS = 10(4.6) = 46 ft

LS min = 3Wo = 3(8) = 24 ft, use LS = 46 ft

LB = 15hS = 15(4.6) = 69 ft

LB min = 4Wo = 4(8) = 32 ft, use LB = 69 ft

WB = Wo + 2(LB/3) = 8 + 2(69/3) = 54 ft

However, since the trial D50 is not available, the next larger riprap size (D50 = 2.5 ft) would be used to line a basin with the given dimensions.

Step 4 (3rd iteration). Determine the basin exit depth, yB = yc, and exit velocity, VB = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc)

8002/32.2 = 19,876 = [yc(54 + 2yc)]3/ (54 + 4yc)

By trial and success, yc = 1.85 ft, Tc = 61.4 ft, Ac = 106.9 ft2

VB = Vc = Q/Ac = 800/106.9 = 7.5 ft/s (not acceptable). If the apron were extended (with a continued flare) such that the total basin length was 90 ft, the velocity would be reduced to the allowable level.

Two feasible options have been identified. First, a 6-ft-deep, 60-ft-long pool, with a 30-ft-apron using D50 = 1.83 ft. Second, a 4.6-ft-deep, 46-ft-long pool, with a 44-ft-apron using D50 = 2.5 ft. Because the overall length is the same, the first option is likely to be more economical.

Step 5. For the design discharge, determine if TW/yo ≤ 0.75.

For the first tailwater condition, TW/yo = 2.8/4.0 = 0.70, which satisfies TW/yo ≤ 0.75. No additional riprap needed downstream.

For the second tailwater condition, TW/yo = 4.2/4.0 = 1.05, which does not satisfy TW/yo ≤ 0.75. To determine required riprap, estimate centerline velocity at a series of downstream cross sections using Figure 10.3.

Compute equivalent circular diameter, De, for brink area:

A = π De2 /4 = (yo)(Wo) = (4)(8) = 32 ft2

De = [32(4)/ π ]1/2 = 6.4 ft

Rock size can be determined using the procedures in Section 10.3 (Equation 10.6) or other suitable method. The computations are summarized below.

Computation Summary for Design Example (CU)
L/DeL (ft)VL/Vo (Figure 10.3)VL (ft/s)Rock Size D50 (ft)
10640.5914.71.42
15960.4210.50.72
201280.307.50.37
211350.287.00.32

The calculations above continue until VL ≤ Vallow. Riprap should be at least the size shown. As a practical consideration, the channel can be lined with the same size rock used for the basin. Protection must extend at least 135 ft downstream from the culvert brink, which is 45 ft beyond the basin exit. Riprap should be installed in accordance with details shown in HEC 11.

Design Example: Riprap Basin (Culvert on a Mild Slope) (SI)

Determine riprap basin dimensions using the envelope design (Equations 10.1 and 10.2) for a pipe culvert that is in outlet control with subcritical flow in the culvert. Allowable exit velocity from the riprap basin, Vallow, is 2.1 m/s. Riprap is available with a D50 of 0.125, 0.150, and 0.250 m. Given:

  • D = 1.83 m CMP with Manning's n = 0.024
  • So = 0.004 m/m
  • Q = 3.82 m3/s
  • yn = 1.37 m (normal flow depth in the pipe)
  • Vn = 1.80 m/s (normal velocity in the pipe)
  • TW = 0.61 m (tailwater depth)

Solution

Step 1. Compute the culvert outlet velocity, Vo, and depth, yo.

For subcritical flow (culvert on mild slope), use Figure 3.4 to obtain yo/D, then calculate Vo by dividing Q by the wetted area for yo.

Ku Q/D2.5 = 1.81 (3.82)/1.832.5 = 1.53

TW/D = 0.61/1.83 = 0.33

From Figure 3.4, yo/D = 0.45

yo = (yo/D)D = 0.45(1.83) = 0.823 m (brink depth)

From Table B.2, for yo /D = 0.45, the brink area ratio A/D2 = 0.343

A = (A/D2)D2 = 0.343(1.83)2 = 1.15 m2

Vo = Q/A = 3.82/1.15 = 3.32 m/s

ye = (A/2)1/2 = (1.15/2)1/2 = 0.76 m

Fr = Vo / [9.81(ye)]1/2 = 3.32/ [9.81(0.76)]1/2 = 1.22

Step 2. Select a trial D50 and obtain hs/ye from Equation 10.1. Check to see that hs/D50≥ 2 and D50/ye≥ 0.1.

Try D50 = 0.15 m; D50/ye = 0.15/0.76 = 0.20 (≥ 0.1 OK)

TW/ye = 0.61/0.76 = 0.80. Therefore, from Equation 10.2,

Co = 4.0(TW/ye) -1.6 = 4.0(0.80) -1.6 = 1.61

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.20) to the -0.55 power times (1.22) minus 1.61 equals 0.933

hS = (hS /ye)ye = 0.933 (0.76) = 0.71 m

hS/D50 = 0.71/0.15 = 4.7 and hS/D50≥ 2 is satisfied

Step 3. Size the basin as shown in Figures 10.1 and 10.2.

LS = 10hS = 10(0.71) = 7.1 m

LS min = 3Wo = 3(1.83) = 5.5 m, use LS = 7.1 m

LB = 15hS = 15(0.71) = 10.7 m

LB min = 4Wo = 4(1.83) = 7.3 m, use LB = 10.7 m

WB = Wo + 2(LB/3) = 1.83 + 2(10.7/3) = 9.0 m

Step 4. Determine the basin exit depth, yB = yc and exit velocity, VB = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc)

3.822/9.81 = 1.49 = [yc(9.0 + 2yc)]3/ (9.0 + 4yc)

By trial and success, yc = 0.26 m, Tc =10.0 m, Ac = 2.48 m2

Vc = Q/Ac = 3.82/2.48 = 1.5 m/s (acceptable)

The initial trial of riprap (D50 = 0.15 m) results in a 10.7 m basin that satisfies all design requirements. Try the next larger riprap size to test if a smaller basin is feasible by repeating steps 2 through 4.

Step 2 (2nd iteration). Select a trial D50 and obtain hs/ye from Equation 10.1.

Try D50 = 0.25 m; D50/ye = 0.25/0.76 = 0.33 (≥ 0.1 OK)

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.33) to the -0.55 power times (1.22) minus 1.61 equals 0.320

hS = (hS /ye)ye = 0.320 (0.76) = 0.24 m

hS/D50 = 0.24/0.25 = 0.96 and hS/D50≥ 2 is not satisfied. Although not available, try a riprap size that will yield hS/D50 close to, but greater than 2. (A basin sized for smaller riprap may be lined with larger riprap.) Repeat step 2.

Step 2 (3rd iteration). Select a trial D50 and obtain hs/ye from Equation 10.1.

Try D50 = 0.205 m; D50/ye = 0.205/0.76 = 0.27 (≥ 0.1 OK)

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.27) to the -0.55 power times (1.22) minus 1.61 equals 0.545

hS = (hS /ye)ye = 0.545 (0.76) = 0.41 m

hS/D50 = 0.41/0.205 = 2.0 and hS/D50≥ 2 is satisfied. Continue to step 3.

Step 3 (3rd iteration). Size the basin as shown in Figures 10.1 and 10.2.

LS = 10hS = 10(0.41) = 4.1 m

LS min = 3Wo = 3(1.83) = 5.5 m, use LS = 5.5 m

LB = 15hS = 15(0.41) = 6.2 m

LB min = 4Wo = 4(1.83) = 7.3 m, use LB = 7.3 m

WB = Wo + 2(LB/3) = 1.83 + 2(7.3/3) = 6.7 m

However, since the trial D50 is not available, the next larger riprap size (D50 = 0.25 m) would be used to line a basin with the given dimensions.

Step 4 (3rd iteration). Determine the basin exit depth, yB = yc and exit velocity, VB = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc)

3.822/9.81 = 1.49 = [yc(6.7 + 2yc)]3/ (6.7 + 4yc)

By trial and success, yc = 0.31 m, Tc =7.94 m, Ac = 2.28 m2

Vc = Q/Ac = 3.82/2.28 = 1.7 m/s (acceptable)

Two feasible options have been identified. First, a 0.71 m deep, 7.1 m long pool, with an 3.6 m apron using D50 = 0.15 m. Second, a 0.41 m deep, 5.5 m long pool, with a 1.8 m apron using D50 = 0.25 m. The choice between these two options will likely depend on the available space and the cost of riprap.

Step 5. For the design discharge, determine if TW/yo ≤ 0.75

TW/yo = 0.61/0.823 = 0.74, which satisfies TW/yo ≤ 0.75. No additional riprap needed.

Design Example: Riprap Basin (Culvert on a Mild Slope) (CU)

Determine riprap basin dimensions using the envelope design (Equations 10.1 and 10.2) for a pipe culvert that is in outlet control with subcritical flow in the culvert. Allowable exit velocity from the riprap basin, Vallow, is 7.0 ft/s. Riprap is available with a D50 of 0.42, 0.50, and 0.83 ft. Given:

  • D = 6 ft CMP with Manning's n = 0.024
  • So = 0.004 ft/ft
  • Q = 135 ft3/s
  • yn = 4.5 ft (normal flow depth in the pipe)
  • Vn = 5.9 ft/s (normal velocity in the pipe)
  • TW = 2.0 ft (tailwater depth)

Solution

Step 1. Compute the culvert outlet velocity, Vo, depth, yo and Froude number.

For subcritical flow (culvert on mild slope), use Figure 3.4 to obtain yo/D, then calculate Vo by dividing Q by the wetted area for yo.

KuQ/D2.5 = 1.0(135)/62.5 = 1.53

TW/D = 2.0/6 = 0.33

From Figure 3.4, yo/D = 0.45

yo = (yo/D)D = 0.45(6) = 2.7 ft (brink depth)

From Table B.2 for yo/D = 0.45, the brink area ratio A/D2 = 0.343

A = (A/D2)D2 = 0.343(6)2 = 12.35 ft2

Vo = Q/A = 135/12.35 = 10.9 ft/s

ye = (A/2)1/2 = (12.35/2)1/2 = 2.48 ft

Fr = Vo / [32.2(ye)]1/2 = 10.9/ [32.2(2.48)]1/2 = 1.22

Step 2. Select a trial D50 and obtain hs/ye from Equation 10.1. Check to see that hs/D50≥ 2 and D50/ye≥ 0.1.

Try D50 = 0.5 ft; D50/ye = 0.5/2.48 = 0.20 (≥ 0.1 OK)

TW/ye = 2.0/2.48 = 0.806. Therefore, from Equation 10.2,

Co = 4.0(TW/ye) -1.6 = 4.0(0.806) -1.6 = 1.62

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.20) to the -0.55 power times (1.22) minus 1.62 equals 0.923

hS = (hS /ye)ye = 0.923 (2.48) = 2.3 ft

hS/D50 = 2.3/0.5 = 4.6 and hS/D50≥ 2 is satisfied

Step 3. Size the basin as shown in Figures 10.1 and 10.2.

LS = 10hS = 10(2.3) = 23 ft

LS min = 3Wo = 3(6) = 18 ft, use LS = 23 ft

LB = 15hS = 15(2.3) = 34.5 ft

LB min = 4Wo = 4(6) = 24 ft, use LB = 34.5 ft

WB = Wo + 2(LB/3) = 6 + 2(34.5/3) = 29 ft

Step 4. Determine the basin exit depth, yB = yc and exit velocity, VB = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc)

1352/32.2 = 566 = [yc(29 + 2yc)]3/ (29 + 4yc)

By trial and success, yc = 0.86 ft, Tc =32.4 ft, Ac = 26.4 ft2

Vc = Q/Ac = 135/26.4 = 5.1 ft/s (acceptable)

The initial trial of riprap (D50 = 0.5 ft) results in a 34.5 ft basin that satisfies all design requirements. Try the next larger riprap size to test if a smaller basin is feasible by repeating steps 2 through 4.

Step 2 (2nd iteration). Select a trial D50 and obtain hs/ye from Equation 10.1.

Try D50 = 0.83 ft; D50/ye = 0.83/2.48 = 0.33 (≥ 0.1 OK)

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.33) to the -0.55 power times (1.22) minus 1.62 equals 0.311

hS = (hS /ye)ye = 0.311 (2.48) = 0.8 ft

hS/D50 = 0.8/0.83 = 0.96 and hS/D50≥ 2 is not satisfied. Although not available, try a riprap size that will yield hS/D50 close to, but greater than 2. (A basin sized for smaller riprap may be lined with larger riprap.) Repeat step 2.

Step 2 (3rd iteration). Select a trial D50 and obtain hs/ye from Equation 10.1.

Try D50 = 0.65 ft; D50/ye = 0.65/2.48 = 0.26 (≥ 0.1 OK)

From Equation 10.1,

h sub s divided by y sub e equals 0.86 times (D sub 50 divided by y sub e) to the -0.55 power times (V sub o divided by the square root of (g times y sub e)) minus C sub o equals 0.86 times (0.26) to the -0.55 power times (1.22) minus 1.62 equals 0.581

hS = (hS /ye)ye = 0.581 (2.48) = 1.4 ft

hS/D50 = 1.4/0.65 = 2.15 and hS/D50≥ 2 is satisfied. Continue to step 3.

Step 3 (3rd iteration). Size the basin as shown in Figures 10.1 and 10.2.

LS = 10hS = 10(1.4) = 14 ft

LS min = 3Wo = 3(6) = 18 ft, use LS = 18 ft

LB = 15hS = 15(1.4) = 21 ft

LB min = 4Wo = 4(6) = 24 ft, use LB = 24 ft

WB = Wo + 2(LB/3) = 6 + 2(24/3) = 22 ft

However, since the trial D50 is not available, the next larger riprap size (D50 = 0.83 ft) would be used to line a basin with the given dimensions.

Step 4 (3rd iteration). Determine the basin exit depth, yB = yc and exit velocity, VB = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + zyc)]3/ (WB + 2zyc)

1352/32.2 = 566 = [yc(22 + 2yc)]3/ (22 + 4yc)

By trial and success, yc = 1.02 ft, Tc =26.1 ft, Ac = 24.5 ft2

Vc = Q/Ac = 135/24.5 = 5.5 ft/s (acceptable)

Two feasible options have been identified. First, a 2.3-ft-deep, 23-ft-long pool, with an 11.5-ft-apron using D50 = 0.5 ft. Second, a 1.4-ft-deep, 18-ft-long pool, with a 6-ft-apron using D50 = 0.83 ft. The choice between these two options will likely depend on the available space and the cost of riprap.

Step 5. For the design discharge, determine if TW/yo ≤0.75

TW/yo = 2.0/2.7 = 0.74, which satisfies TW/yo ≤ 0.75. No additional riprap needed.

10.2 Riprap Apron

The most commonly used device for outlet protection, primarily for culverts 1500 mm (60 in) or smaller, is a riprap apron. An example schematic of an apron taken from the Federal Lands Division of the Federal Highway Administration is shown in Figure 10.4.

Figure 10.4. Placed Riprap at Culverts (Central Federal Lands Highway Division)

Example drawing showing installation with and without a standard end section (plan and profile for each). Parameters are defined in text.

They are constructed of riprap or grouted riprap at a zero grade for a distance that is often related to the outlet pipe diameter. These aprons do not dissipate significant energy except through increased roughness for a short distance. However, they do serve to spread the flow helping to transition to the natural drainage way or to sheet flow where no natural drainage way exists. However, if they are too short, or otherwise ineffective, they simply move the location of potential erosion downstream. The key design elements of the riprap apron are the riprap size as well as the length, width, and depth of the apron.

Several relationships have been proposed for riprap sizing for culvert aprons and several of these are discussed in greater detail in Appendix D. The independent variables in these relationships include one or more of the following variables: outlet velocity, rock specific gravity, pipe dimension (e.g. diameter), outlet Froude number, and tailwater. The following equation (Fletcher and Grace, 1972) is recommended for circular culverts:

(10.4)

D sub 50 equals 0.2 times D times (Q divided by (square root of g times D to the 2.5 power)) to the four-thirds power times (D divided by TW)

where,

D50 = riprap size, m (ft)

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

D = culvert diameter (circular), m (ft)

TW = tailwater depth, m (ft)

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

Tailwater depth for Equation 10.4 should be limited to between 0.4D and 1.0D. If tailwater is unknown, use 0.4D.

Whenever the flow is supercritical in the culvert, the culvert diameter is adjusted as follows:

(10.5)

D prime equals (D plus y sub n) divided by 2

where,

D' = adjusted culvert rise, m (ft)

yn = normal (supercritical) depth in the culvert, m (ft)

Equation 10.4 assumes that the rock specific gravity is 2.65. If the actual specific gravity differs significantly from this value, the D50 should be adjusted inversely to specific gravity.

The designer should calculate D50 using Equation 10.4 and compare with available riprap classes. A project or design standard can be developed such as the example from the Federal Highway Administration Federal Lands Highway Division (FHWA, 2003) shown in Table 10.1 (first two columns). The class of riprap to be specified is that which has a D50 greater than or equal to the required size. For projects with several riprap aprons, it is often cost effective to use fewer riprap classes to simplify acquiring and installing the riprap at multiple locations. In such a case, the designer must evaluate the tradeoffs between over sizing riprap at some locations in order to reduce the number of classes required on a project.

Table 10.1. Example Riprap Classes and Apron Dimensions
ClassD50 (mm)D50 (in)Apron Length1Apron Depth
112554D3.5D50
215064D3.3D50
3250105D2.4D50
4350146D2.2D50
5500207D2.0D50
6550228D2.0D50
1D is the culvert rise.

The apron dimensions must also be specified. Table 10.1 provides guidance on the apron length and depth. Apron length is given as a function of the culvert rise and the riprap size. Apron depth ranges from 3.5D50 for the smallest riprap to a limit of 2.0D50 for the larger riprap sizes. The final dimension, width, may be determined using the 1:3 flare shown in Figure 10.4 and should conform to the dimensions of the downstream channel. A filter blanket should also be provided as described in HEC 11 (Brown and Clyde, 1989).

For tailwater conditions above the acceptable range for Equation 10.4 (TW > 1.0D), Figure 10.3 should be used to determine the velocity downstream of the culvert. The guidance in Section 10.3 may be used for sizing the riprap. The apron length is determined based on the allowable velocity and the location at which it occurs based on Figure 10.3.

Over their service life, riprap aprons experience a wide variety of flow and tailwater conditions. In addition, the relations summarized in Table 10.1 do not fully account for the many variables in culvert design. To ensure continued satisfactory operation, maintenance personnel should inspect them after major flood events. If repeated severe damage occurs, the location may be a candidate for extending the apron or another type of energy dissipator.

Design Example: Riprap Apron (SI)

Design a riprap apron for the following CMP installation. Available riprap classes are provided in Table 10.1. Given:

  • Q = 2.33 m3/s
  • D = 1.5 m
  • TW = 0.5 m

Solution

Step 1. Calculate D50 from Equation 10.4. First verify that tailwater is within range.

TW/D = 0.5/1.5 = 0.33. This is less than 0.4D, therefore,

use TW = 0.4D = 0.4(1.5) = 0.6 m

D sub 50 equals 0.2 times D times (Q divided by (square root of g times D to the 2.5 power)) to the four-thirds power times (D divided by TW) equals 0.2 times 1.5 times (2.33 divided by (square root of 9.81 times 1.5 to the 2.5 power)) to the four-thirds power times (1.5 divided by 0.6) equals 0.13 meters

Step 2. Determine riprap class. From Table 10.1, riprap class 2 (D50 = 0.15 m) is required.

Step 3. Estimate apron dimensions.

From Table 10.1 for riprap class 2,

Length, L = 4D = 4(1.5) = 6 m

Depth = 3.3D50 = 3.3 (0.15) = 0.50 m

Width (at apron end) = 3D + (2/3)L = 3(1.5) + (2/3)(6) = 8.5 m

Design Example: Riprap Apron (CU)

Design a riprap apron for the following CMP installation. Available riprap classes are provided in Table 10.1. Given:

  • Q = 85 ft3/s
  • D = 5.0 ft
  • TW = 1.6 ft

Solution

Step 1. Calculate D50 from Equation 10.4. First verify that tailwater is within range.

TW/D = 1.6/5.0 = 0.32. This is less than 0.4D, therefore,

use TW = 0.4D = 0.4(5) = 2.0 ft

D sub 50 equals 0.2 times D times (Q divided by (square root of g times D to the 2.5 power)) to the four-thirds power times (D divided by TW) equals 0.2 times 5.0 times (85 divided by (square root of 32.2 times 5.0 to the 2.5 power)) to the four-thirds power times (5.0 divided by 2.0) equals 0.43 feet equals 5.2 inches

Step 2. Determine riprap class. From Table 10.1, riprap class 2 (D50 = 6 in) is required.

Step 3. Estimate apron dimensions.

From Table 10.1 for riprap class 2,

Length, L = 4D = 4(5) = 20 ft

Depth = 3.3D50 = 3.3 (6) = 1.65 ft

Width (at apron end) = 3D + (2/3)L = 3(5) + (2/3)(20) = 28.3 ft

10.3 Riprap Aprons After Energy Dissipators

Some energy dissipators provide exit conditions, velocity and depth, near critical. This flow condition rapidly adjusts to the downstream or natural channel regime; however, critical velocity may be sufficient to cause erosion problems requiring protection adjacent to the energy dissipator. Equation 10.6 provides the riprap size recommended for use downstream of energy dissipators. This relationship is from Searcy (1967) and is the same equation used in HEC 11 (Brown and Clyde, 1989) for riprap protection around bridge piers.

(10.6)

D sub 50 equals 0.692 divided by (S minus 1) times (V squared divided by (2 times g))

where,

D50 = median rock size, m (ft)

V = velocity at the exit of the dissipator, m/s (ft/s)

S = riprap specific gravity

The length of protection can be judged based on the magnitude of the exit velocity compared with the natural channel velocity. The greater this difference, the longer will be the length required for the exit flow to adjust to the natural channel condition. A filter blanket should also be provided as described in HEC 11 (Brown and Clyde, 1989).

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Contact:

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

Updated: 04/07/2011
 

FHWA
United States Department of Transportation - Federal Highway Administration