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Design of Roadside Channels with Flexible Linings
Hydraulic Engineering Circular Number 15, Third Edition

Chapter 6: Riprap, Cobble, and Gravel Lining Design

Riprap, cobble, and gravel linings are considered permanent flexible linings. They may be described as a noncohesive layer of stone or rock with a characteristic size, which for the purposes of this manual is the D50. The applicable sizes for the guidance in this manual range from 15 mm (0.6 in) gravel up to 550 mm (22 in) riprap. For the purposes of this manual, the boundary between gravel, cobble, and riprap sizes will be defined by the following ranges:

  • Gravel: 15 - 64 mm (0.6 - 2.5 in)
  • Cobble: 64 - 130 mm (2.5 - 5.0 in)
  • Riprap: 130 - 550 mm (5.0 - 22.0 in)

Other differences between gravels, cobbles, and riprap may include gradation and angularity. These issues will be addressed later.

Gravel mulch, although considered permanent, is generally used as supplement to aid in the establishment of vegetation (See Chapter 4). It may be considered for areas where vegetation establishment is difficult, for example, in arid-region climates. For the transition period before the establishment of the vegetation, the stability of gravel mulch should be assessed using the procedures in this chapter.

The procedures in this chapter are applicable to uniform prismatic channels (as would be characteristic of roadside channels) with rock sizes within the range given above. For situations not satisfying these two conditions, the designer is referred to another FHWA circular (No. 11) "Design of Riprap Revetment" (FHWA, 1989).

6.1 Manning's Roughness

Manning's roughness is a key parameter needed for determining the relationships between depth, velocity, and slope in a channel. However, for gravel and riprap linings, roughness has been shown to be a function of a variety of factors including flow depth, D50, D84, and friction slope, Sf. A partial list of roughness relationships includes Blodgett (1986a), Limerinos (1970), Anderson, et al. (1970), USACE (1994), Bathurst (1985), and Jarrett (1984). For the conditions encountered in roadside and other small channels, the relationships of Blodgett and Bathurst are adopted for this manual.

Blodgett (1986a) proposed a relationship for Manning's roughness coefficient, n, that is a function of the flow depth and the relative flow depth (da/D50) as follows:

n equals alpha times d sub a to the one-sixth power divided by (2.25 plus 5.23 times logarithm (d sub a divided by D sub 50)) (6.1)

where,

n= Manning's roughness coefficient, dimensionless
da= average flow depth in the channel, m (ft)
D50 = median riprap/gravel size, m (ft)
α= unit conversion constant, 0.319 (SI) and 0.262 (CU)

Equation 6.1 is applicable for the range of conditions where 1.5 ≤ da/D50 ≤ 185. For small channel applications, relative flow depth should not exceed the upper end of this range.

Some channels may experience conditions below the lower end of this range where protrusion of individual riprap elements into the flow field significantly changes the roughness relationship. This condition may be experienced on steep channels, but also occurs on moderate slopes. The relationship described by Bathurst (1991) addresses these conditions and can be written as follows (See Appendix D for the original form of the equation):

n equals alpha times d sub a to the one-sixth power divided by the square root of g divided by function of Fr divided by function of REG divided by function of CG (6.2)

where,

da= average flow depth in the channel, m (ft)
g= acceleration due to gravity, 9.81 m/s2 (32.2 ft/s2)
Fr= Froude number
REG= roughness element geometry
CG= channel geometry
α= unit conversion constant, 1.0 (SI) and 1.49 (CU)

Equation 6.2 is a semi-empirical relationship applicable for the range of conditions where 0.3<da/D50<8.0. The three terms in the denominator represent functions of Froude number, roughness element geometry, and channel geometry given by the following equations:

 function of Fr equals (0.28 times Fr divided by b) raised to the power of logarithm (0.755 divided by b) (6.3)
function of REG equals 13.434 times (T divided by D sub 50) to the 0.492 power times b raised to (1.025 times (T divided by D sub 50) to the 0.118 power) (6.4)
function of CG equals (T divided by d sub a) to the minus b power (6.5)

where,

T= channel top width, m (ft)
b= parameter describing the effective roughness concentration

The parameter b describes the relationship between effective roughness concentration and relative submergence of the roughness bed. This relationship is given by:

b equals 1.14 times (D sub 50 divided by T) to the 0.453 power times (d sub a divided by D sub 50) raised to the 0.814 power (6.6)

Equations 6.1 and 6.2 both apply in the overlapping range of 1.5 ≤ da/D50 ≤ 8. For consistency and ease of application over the widest range of potential design situations, use of the Blodgett equation (6.1) is recommended when 1.5 ≤ da/D50. The Bathurst equation (6.2) is recommended for 0.3<da/D50<1.5.

As a practical problem, both Equations 6.1 and 6.2 require depth to estimate n while n is needed to determine depth setting up an iterative process.

6.2 Permissible Shear Stress

Values for permissible shear stress for riprap and gravel linings are based on research conducted at laboratory facilities and in the field. The values presented here are judged to be conservative and appropriate for design use. Permissible shear stress is given by the following equation:

tau sub p equals F sub * times (gamma sub s minus gamma) times D sub 50 (6.7)

where,

τp= permissible shear stress, N/m2 (lb/ft2)
F*= Shield's parameter, dimensionless
γs= specific weight of the stone, N/m3 (lb/ft3)
γ= specific weight of the water, 9810 N/m3 (62.4 lb/ft3)
D50= mean riprap size, m (ft)

Typically, a specific weight of stone of 25,900 N/m3 (165 lb/ft3) is used, but if the available stone is different from this value, the site-specific value should be used.

Recalling Equation 3.2,

tau sub p is greater than or equal to SF times tau sub d

and Equation 3.1,

tau sub d equals gamma times d times S sub o

Equation 6.7 can be written in the form of a sizing equation for D50 as shown below:

D sub 50 is greater than or equal to SF times d times S sub o divided by F sub * divided by (SG minus 1) (6.8)

where,

d= maximum channel depth, m (ft)
SG= specific gravity of rock (γs/γ), dimensionless

Changing the inequality sign to an equality gives the minimum stable riprap size for the channel bottom. Additional evaluation for the channel side slope is given in Section 6.3.2.

Equation 6.8 is based on assumptions related to the relative importance of skin friction, form drag, and channel slope. However, skin friction and form drag have been documented to vary resulting in reports of variations in Shield's parameter by different investigators, for example Gessler (1965), Wang and Shen (1985), and Kilgore and Young (1993). This variation is usually linked to particle Reynolds number as defined below:

R sub e equals V sub * times D sub 50 divided by nu (6.9)

where,

Re= particle Reynolds number, dimensionless
V*= shear velocity, m/s (ft/s)
ν= kinematic viscosity, 1.131x10-6 m2/s at 15.5 deg C (1.217x10-5 ft2/s at 60 deg F)

Shear velocity is defined as:

V sub * equals the square root of the quantity (g times d times S) (6.10)

where,

g= gravitational acceleration, 9.81 m/s2 (32.2 ft/s2)
d= maximum channel depth, m (ft)
S= channel slope, m/m (ft/ft)

Higher Reynolds number correlates with a higher Shields parameter as is shown in Table 6.1. For many roadside channel applications, Reynolds number is less than 4x104 and a Shields parameter of 0.047 should be used in Equations 6.7 and 6.8. In cases for a Reynolds number greater than 2x105, for example, with channels on steeper slopes, a Shields parameter of 0.15 should be used. Intermediate values of Shields parameter should be interpolated based on the Reynolds number.

Table 6.1. Selection of Shields' Parameter and Safety Factor
Reynolds Number F* SF
≤4x104 0.047 1.0
4x104<Re<2x105 linear interpolation linear interpolation
≥2x105 0.15 1.5

Higher Reynolds numbers are associated with more turbulent flow and a greater likelihood of lining failure with variations of installation quality. Because of these conditions, it is recommended that the Safety Factor be also increased with Reynolds number as shown in Table 6.1. Depending on site-specific conditions, safety factor may be further increased by the designer, but should not be decreased to values less than those in Table 6.1.

As channel slope increases, the balance of resisting, sliding, and overturning forces is altered slightly. Simons and Senturk (1977) derived a relationship that may be expressed as follows:

D sub 50 is greater than or equal to SF times d times S times delta divided by F sub * divided by the quantity (SG minus 1) (6.11)

where,

Δ= function of channel geometry and riprap size

The parameter Δ can be defined as follows (see Appendix D for the derivation):

delta equals K sub 1 times the quantity (1 plus sine of the quantity (alpha plus beta)) times tangent of phi divided by 2 divided by the quantity (cosine theta times tangent phi minus quantity (SF times sine theta times cosine beta)) (6.12)

where,

α= angle of the channel bottom slope
β= angle between the weight vector and the weight/drag resultant vector in the plane of the side slope
θ= angle of the channel side slope
φ= angle of repose for the riprap

Finally, β is defined by:

beta equals inverse tangent of the quantity (cosine alpha divided by the quantity (2 times sine theta divided by eta divided by tangent phi plus sine alpha)) (6.13)

where,

η= stability number

The stability number is calculated using:

eta equals tau sub s divided by F sub * divided by the quantity (gamma sub s minus gamma) divided by D sub 50 (6.14)

Riprap stability on a steep slope depends on forces acting on an individual stone making up the riprap. The primary forces include the average weight of the stones and the lift and drag forces induced by the flow on the stones. On a steep slope, the weight of a stone has a significant component in the direction of flow. Because of this force, a stone within the riprap will tend to move in the flow direction more easily than the same size stone on a milder gradient. As a result, for a given discharge, steep slope channels require larger stones to compensate for larger forces in the flow direction and higher shear stress.

The size of riprap linings increases quickly as discharge and channel gradient increase. Equation 6.11 is recommended when channel slope is greater than 10 percent and provides the riprap size for the channel bottom and sides. Equation 6.8 is recommended for slopes less than 5 percent. For slopes between 5 percent and 10 percent, it is recommended that both methods be applied and the larger size used for design. Values for safety factor and Shields parameter are taken from Table 6.1 for both equations.

6.3 Design Procedure

In this section a design procedure for riprap and gravel linings is outlined. First, the basic design procedure for selecting the riprap/gravel size for the bottom of a straight channel is given. Subsequent sections provide guidance for sizing material on the channel side slopes and adjusting for channel bends.

6.3.1 Basic Design

The riprap and gravel lining design procedure for the bottom of a straight channel is described in the following steps. It is iterative by necessity because flow depth, roughness, and shear stress are interdependent. The procedure requires the designer to specify a channel shape and slope as a starting point and is outlined in the eight-step process identified below. In this approach, the designer begins with a design discharge and calculates an acceptable D50 to line the channel bottom. An alternative analytical framework is to use the maximum discharge approach described in Section 3.6. For the maximum discharge approach, the designer selects the D50, and determines the maximum depth and flow permitted in the channel while maintaining a stable lining. The following steps are recommended for the standard design.

Step 1. Determine channel slope, channel shape, and design discharge.

Step 2. Select a trial (initial) D50, perhaps based on available sizes for the project. (Also, determine specific weight of proposed stone.)

Step 3. Estimate the depth. For the first iteration, select a channel depth, di. For subsequent iterations, a new depth can be estimated from the following equation or any other appropriate method.

d sub (i plus 1) equals d sub i times (Q divided by Q sub i) to the 0.4 power

Determine the average flow depth, da in the channel. da = A/T

Step 4. Estimate Manning's n and the implied discharge. First, calculate the relative depth ratio, da/D50. If da/D50 is greater than or equal to 1.5, use Equation 6.1 to calculate Manning's n. If da/D50 is less than 1.5 use Equation 6.2 to calculate Manning's n. Calculate the discharge using Manning's equation.

Step 5. If the calculated discharge is within 5 percent of the design discharge, then proceed to step 6. If not, go back to step 3 and estimate a new flow depth.

Step 6. Calculate the particle Reynolds number using Equation 6.6 and determine the appropriate Shields parameter and Safety Factor values from Table 6.1. If channel slope is less than 5 percent, calculate required D50 from Equation 6.8. If channel slope is greater than 10 percent, use Equation 6.11. If channel slope is between 5 and 10 percent, use both Equations 6.8 and 6.11 and take the largest value.

Step 7. If the D50 calculated is greater than the trial size in step 2, then the trial size is too small and unacceptable for design. Repeat procedure beginning at step 2 with new trial value of D50. If the D50 calculated in step 6 is less than or equal to the previous trial size, then the previous trial size is acceptable. However, if the D50 calculated in step 6 is sufficiently smaller than the previous trial size, the designer may elect to repeat the design procedure at step 2 with a smaller, more cost-effective, D50.

Design Example: Riprap Channel (Mild Slope) (SI)

Design a riprap lining for a trapezoidal channel. Given:

Q = 1.13 m3/s

B = 0.6 m

Z = 3

So = 0.02 m/m

Solution

Step 1. Channel characteristics and design discharge are given above.

Step 2. Available riprap sizes include Class 1: D50 = 125 mm, Class 2: D50 = 150 mm, Class 3: D50 = 250 mm. γs=25,900 N/m3 for all classes. Try Class 1 riprap for initial trial. D50=125 mm

Step 3. Assume an initial trial depth of 0.5 m

Using the geometric properties of a trapezoid, the maximum and average flow depths are found:

A = Bd+Zd2 = 0.6(0.5)+3(0.5)2 = 1.05 m2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.6 plus 2 times 0.5 times square root of (3 squared plus 1) equals 3.76 meters

R = A/P =1.05/3.76 = 0.279 m

T = B+2dZ = 0.6+2(0.5)(3) = 3.60 m

da = A/T = 1.05/3.60 = 0.292 m

Step 4. The relative depth ratio, da/D50 = 0.292/0.125 = 2.3. Therefore, use Equation 6.1 to calculate Manning's n.

n equals alpha times d sub a to the one-sixth power divided by (2.25 plus 5.23 times the logarithm of (d sub a divided by D sub 50)) equals 0.319 times 0.292 to the one-sixth power divided by (2.25 plus 5.23 times the logarithm of (0.292 divided by 0.125)) equals 0.062

Calculate Q using Manning's equation:

Q equals alpha divided by n times A times R to the two-thirds power times S to the one-half power equals 1.0 divided by 0.062 times 1.05 times 0.279 to the two-thirds power times 0.02 to the one-half power equals 1.02 cubic meters per second

Step 5. Since this estimate is more than 5 percent from the design discharge, estimate a new depth in step 3.

Step 3 (2nd iteration). Estimate a new depth estimate:

d sub (i plus 1) equals d sub i times (Q divided by Q sub i) to the 0.4 power equals 0.5 times (1.13 divided by 1.02) to the 0.4 power equals 0.521 meters

Using the geometric properties of a trapezoid, the maximum and average flow depths are found:

A = Bd+Zd2 = 0.6(0.521)+3(0.521)2 = 1.13 m2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.6 plus 2 times 0.521 times square root of (3 squared plus 1) equals 3.90 meters

R = A/P =1.13/3.90 = 0.289 m

T = B+2dZ = 0.6+2(0.521)(3) = 3.73 m

da = A/T = 1.13/3.73 = 0.303 m

Step 4. (2nd iteration). The relative depth ratio, da/D50 = 0.302/0.125 = 2.4. Therefore, use Equation 6.1 to calculate Manning's n.

n equals alpha times d sub a to the one-sixth power divided by (2.25 plus 5.23 times the logarithm of (d sub a divided by D sub 50)) equals 0.319 times 0.302 to the one-sixth power divided by (2.25 plus 5.23 times the logarithm of (0.302 divided by 0.125)) equals 0.061

Calculate Q using Manning's equation:

Q equals alpha divided by n times A times R to the two-thirds power times S to the one-half power equals 1.0 divided by 0.061 times 1.13 times 0.289 to the two-thirds power times 0.02 to the one-half power equals 1.14 cubic meters per second

Step 5 (2nd iteration). Since this estimate is within 5 percent of the design discharge, proceed to step 6 with the most recently calculated depth.

Step 6. Need to calculate the shear velocity and Reynolds number to determine Shields' parameter and SF.

Shear velocity, V sub * equals square root of (g times d times S) equals square root of (9.81 times 0.521 times 0.02) equals 0.320 meters per second

Reynolds number, R sub e equals V sub * times D sub 50 divided by nu equals 0.320 times 0.125 divided by 0.000001131 equals 3.5 times 10 to the fourth power

Since Re≤ 4x104, F*=0.047 and SF = 1.0

Since channel slope is less than 5 percent, use Equation 6.8 to calculate minimum stable D50.

SG = γsw = 25,900/9810 = 2.64

R sub e equals V sub * times D sub 50 divided by nu equals 0.320 times 0.125 divided by 0.000001131 equals 3.5 times 10 to the fourth power

Step 7. The stable D50 is slightly larger than the Class 1 riprap, therefore Class 1 riprap is insufficient. Class 2 should be specified. The suitability of Class 2 should be verified by repeating the design procedure starting at step 2.

Design Example: Riprap Channel (Mild Slope) (CU)

Design a riprap lining for a trapezoidal channel. Given:

Q = 40 ft3/s

B = 2.0 ft

Z = 3

So = 0.02 ft/ft

Solution

Step 1. Channel characteristics and design discharge are given above.

Step 2. Available riprap sizes include Class 1: D50 = 5 in, Class 2: D50 = 6 in, Class 3: D50 = 10 in. γs=165 lb/ft3 for all classes. Try Class 1 riprap for initial trial. D50=(5/12)=0.42 ft

Step 3. Assume an initial trial depth of 1.5 ft.

Using the geometric properties of a trapezoid, the maximum and average flow depths are found:

A = Bd+Zd2 = 2.0(1.5)+3(1.5)2 = 9.75 ft2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 2.0 plus 2 times 1.5 times square root of (3 squared plus 1) equals 11.5 feet

R = A/P = 9.75/11.5 = 0.848 ft

T = B+2dZ = 2.0+2(1.5)(3) = 11.0 ft

da = A/T = 9.75/11.0 = 0.886 ft

Step 4. The relative depth ratio, da/D50 = 0.886/0.42 = 2.1. Therefore, use Equation 6.1 to calculate Manning's n.

n equals alpha times d sub a to the one-sixth power divided by (2.25 plus 5.23 times the logarithm of (d sub a divided by D sub 50)) equals 0.262 times 0.886 to the one-sixth power divided by (2.25 plus 5.23 times the logarithm of (0.886 divided by 0.42)) equals 0.065

Calculate Q using Manning's equation:

Q equals 1.49 divided by n times A times R to the two-thirds power times S to the one-half power equals 1.49 divided by 0.065 times 9.75 times 0.849 to the two-thirds power times 0.02 to the one-half power equals 28.3 cubic feet per second

Step 5. Since this estimate is more than 5 percent from the design discharge, estimate a new depth in step 3.

Step 3 (2nd iteration). Estimate a new depth estimate:

d sub (i plus 1) equals d sub i times (Q divided by Q sub i) to the 0.4 power equals 1.5 times (40 divided by 28.3) to the 0.4 power equals 1.72 feet

Using the geometric properties of a trapezoid, the maximum and average flow depths are found:

A = Bd+Zd2 = 2.0(1.72)+3(1.72)2 = 12.3 ft2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 2.0 plus 2 times 1.72 times square root of (3 squared plus 1) equals 12.9 feet

R = A/P = 12.3/12.9 = 0.953 ft

T = B+2dZ = 2.0+2(1.72)(3) = 12.3 ft

da = A/T = 12.3/12.3 = 1.0 ft

Step 4. (2nd iteration). The relative depth ratio, da/D50 = 1.0/0.42 = 2.4. Therefore, use Equation 6.1 to calculate Manning's n.

n equals alpha times d sub a to the one-sixth power divided by (2.25 plus 5.23 times the logarithm of (d sub a divided by D sub 50)) equals 0.262 times 1.0 to the one-sixth power divided by (2.25 plus 5.23 times the logarithm of (1.0 divided by 0.42)) equals 0.062

Calculate Q using Manning's equation:

Q equals 1.49 divided by n times A times R to the two-thirds power times S to the one-half power equals 1.49 divided by 0.062 times 12.3 times 0.956 to the two-thirds power times 0.02 to the one-half power equals 40.6 cubic feet per second

Step 5 (2nd (iteration). Since this estimate is within 5 percent of the design discharge, proceed to step 6 with the most recently calculated depth.

Step 6. Need to calculate the shear velocity and Reynolds number to determine Shields' parameter and SF.

Shear velocity, V sub * equals square root of (g times d times S) equals square root of (32.2 times 1.72 times 0.02) equals 1.05 feet per second

Reynolds number, R sub e equals V sub * times D sub 50 divided by nu equals 1.05 times 0.42 divided by 0.00001217 equals 3.6 times 10 to the fourth power

Since Re≤ 4x104, F*=0.047 and SF = 1.0

Since channel slope is less than 5 percent, use Equation 6.8 to calculate minimum stable D50.

SG = γsw = 165/62.4 = 2.64

D sub 50 is greater than or equal to SF times d times S sub o divided by F sub * divided by (SG minus 1) equals 1.0 times 1.72 times 0.02 divided by 0.047 divided by (2.64 minus 1) equals 0.45 feet

Step 7. The stable D50 is slightly larger than the Class 1 riprap, therefore Class 1 riprap is insufficient. Class 2 should be specified. The suitability of Class 2 should be verified by repeating the design procedure starting at step 2.

6.3.2 Side Slopes

As was explained in Chapter 3, the shear stress on the channel side is less than the maximum shear stress occurring on the channel bottom as was described in Equation 3.3 repeated below.

tau sub s equals K sub 1 times tau sub d

However, since gravel and riprap linings are noncohesive, as the angle of the side slopeapproaches the angle of repose of the channel lining, the lining material becomes less stable. The stability of a side slope lining is a function of the channel side slope and the angle of repose of the rock/gravel lining material. This essentially results in a lower permissible shear stress on the side slope than on the channel bottom.

These two counterbalancing effects lead to the following design equation for specifying a stone size for the side slope given the stone size required for a stable channel bottom. The following equation is used if Equation 6.8 is used to size the channel bottom stone. If Equation 6.11 was used, the D50 from Equation 6.11 is used for the channel bottom and sides.

D sub 50,s equals K sub 1 divided by K sub 2 times D sub 50,b (6.15)

where,

D50,s= D50 required for a stable side slope, m (ft)
D50,b= D50 required for a stable channel bottom, m (ft)
K1= ratio of channel side to bottom shear stress (see Section 3.2)
K2= tractive force ratio (Anderson, et al., 1970)

K2 is a function of the side slope angle and the stone angle of repose and is determined from the following equation.

 K sub 2 equals the square root of ( 1 minus (sine theta divided by sine phi) squared) (6.16)

where,

θ= angle of side slope
φ= angle of repose

When the side slope is represented as 1:Z (vertical to horizontal), the angle of side slope is:

theta equals the inverse tangent of 1 divided by Z (6.17)

Angle of repose is a function of both the stone size and angularity and may be determined from Figure 6.1.

Channels lined with gravel or riprap on side slopessteeper than 1:3 may become unstable and should be avoided where feasible. If steeper side slopes are required they should be assessed using both Equation 6.11 and Equation 6.8 (in conjunction with Equation 6.15) taking the largest value for design.

Design Example: Riprap Channel Side Slope Assessment (SI)

Consider the stability of the side slopes for the design example in Section 6.3.1. Recall that a class 1 riprap (D50=0.125 m) was evaluated and found to be unstable. Stable D50 was determined to be 0.135 m and a class 2 riprap (D50=0.150 m) was recommended. Assess stability on the side slope of the trapezoidal channel.

Given:

Riprap angle of repose = 38 degrees

Z = 3

Solution

Step 1. Calculate K1. From Equation 3.4, K1 is a function of Z (for Z between 1.5 and 5)

K1 = 0.066Z + 0.67 = 0.066(3)+0.67 = 0.87

Step 2. Calculate K2 from Equation 6.16 with θ calculated from Equation 6.17.

theta equals inverse tangent of (one divided by Z) equals inverse tangent of (one divided by 3) equals 18.4 degrees

K sub 2 equals square root of (1 minus (sine theta divided by sine phi) squared) equals square root of (1 minus (sine 18.4 divided by sin 38) squared) equals 0.86

Step 3. Calculate the stable D50 for the side slope using Equation 6.15.

D sub 50,s equals K sub 1 divided by K sub 2 times D sub 50,b equals 0.87 divided by 0.86 times 0.135 equals 0.137 meters

Since 0.137 m is greater than the Class 1 D50 selected the side slope is also unstable. However, like for the channel bottom, Class 2 (D50=0.150 m) would provide a stable side slope.

Design Example: Riprap Channel Side Slope Assessment (CU)

Consider the stability of the side slopes for the design example in Section 6.3.1. Recall that a class 1 riprap (D50=5 in) was evaluated and found to be unstable. Stable D50 was determined to be 5.4 in and a class 2 riprap (D50=6 in) was recommended. Assess stability on the side slope of the trapezoidal channel.

Given:

Riprap angle of repose = 38 degrees

Z = 3

Solution

Step 1. Calculate K1. From Equation 3.4, K1 is a function of Z (for Z between 1.5 and 5)

K1 = 0.066Z + 0.67 = 0.066(3)+0.67 = 0.87

Step 2. Calculate K2 from Equation 6.16 with θ calculated from Equation 6.17.

theta equals inverse tangent of (one divided by Z) equals inverse tangent of (one divided by 3) equals 18.4 degrees

K sub 2 equals square root of (1 minus (sine theta divided by sine phi) squared) equals square root of (1 minus (sine 18.4 divided by sin 38) squared) equals 0.86

Step 3. Calculate the stable D50 for the side slope using Equation 6.15.

D sub 50,s equals K sub 1 divided by K sub 2 times D sub 50,b equals 0.87 divided by 0.86 times 5.4 equals 5.46 inches

Since 5.46 inches is greater than the Class 1 D50 selected the side slope is also unstable. However, like for the channel bottom, Class 2 (D50= 6 in) would provide a stable side slope.

figure relating angle of repose to mean stone size, D sub 50 and rock angularity (crushed, very angular, very rounded)
Figure 6.1. Angle of Repose of Riprap in Terms of Mean Size and Shape of Stone

6.3.3 Bends

Added stresses in bends for riprap and gravel linings are treated as described in section 3.4. No additional considerations are required.

6.4 Additional Considerations

As with all lining types, the ability to deliver the expected channel protection depends on the proper installation of the lining. Additional design considerations for riprap linings include freeboard; proper specification of gradation and thickness; and use of a filter material under the riprap.

6.4.1 Freeboard and Slope Changes

Freeboard, as defined in Section 2.3.4, is determined based on the predicted water surface elevation in a channel. For channels on mild slopes the water surface elevation for freeboard considerations may be safely taken as the normal depth elevation. For steep slopes and for slope changes, additional consideration of freeboard is required.

For steep channels, freeboard should equal the mean depth of flow, since wave height will reach approximately twice the mean depth. This freeboard height should be used for both transitional and permanent channel installations. Extent of riprap on a steep gradient channel must be sufficient to protect transitions from mild to steep and from steep to mild sections.

The transition from a steep gradient to a culvert should allow for slight movement of riprap. The top of the riprap layer should be placed flush with the invert of a culvert while the riprap layer thickness should equal three to five times the mean rock diameter at the break between the steep slope and culvert entrance. The transition from a steep gradient channel to a mild gradient channel may require an energy dissipation structure such as a plunge pool. The transition from a mild gradient to a steep gradient should be protected against local scour upstream of the transition for a distance of approximately five times the uniform depth of flow in the downstream channel (Chow, 1959).

6.4.2 Riprap Gradation, Angularity, and Thickness

Riprap gradation should follow a smooth size distribution curve. Most riprap gradations will fall in the range of D100 /D50 and D50 /D20 between 3.0 to 1.5, which is acceptable. The most important criterion is a proper distribution of sizes in the gradation so that interstices formed by larger stones are filled with smaller sizes in an interlocking fashion, preventing the formation of open pockets. These gradation requirements apply regardless of the type of filter design used. More uniformly graded stone may distribute a higher failure threshold because it contains fewer smaller stones, but at the same time will likely exhibit a more sudden failure. Increasing the safety factor is appropriate when there are questions regarding gradation.

In general, riprap constructed with angular stones has the best performance. Round stones are acceptable as riprap provided they are not placed on side slopes steeper than 1:3 (V:H). Flat slab-like stones should be avoided since they are easily dislodged by the flow. An approximate guide to stone shape is that neither the breadth nor thickness of a single stone is less than one-third its length. Again, the safety factor should be increased if round stones are used. Permissible shear stress estimates are largely based on testing with angular rock.

The thickness of a riprap lining should equal the diameter of the largest rock size in the gradation. For most gradations, this will mean a thickness from 1.5 to 3.0 times the mean riprap diameter. It is important to note that riprap thickness is measured normal to ground surface slope.

6.4.3 Riprap Filter Design

When rock riprap is used, the need for an underlying filter material must be evaluated. The filter material may be either a granular filter blanket or a geotextile fabric.

To determine the need for a filter and, if one is required, to select a gradation for the filter blanket, the following criteria must be met (USACE, 1980). The subscripts "upper" and "lower" refer to the riprap and soil, respectively, when evaluating filter need; the subscripts represent the riprap/filter and filter/soil comparisons when selecting a filter blanket gradation.

the ratio of D sub 15 upper to D sub 85 lower is less than 5 (6.18a)
5 is less than the ratio of D sub 15 upper to D sub 15 lower is less than 40 (6.18b)
the ratio of D sub 50 upper to D sub 50 lower is less than 40 (6.18c)

In the above relationships, "upper" refers to the overlying material and "lower" refers to the underlying material. The relationships must hold between the filter blanket and base material and between the riprap and filter blanket.

The thickness of the granular filter blanket should approximate the maximum size in the filter gradation. The minimum thickness for a filter blanket should not be less than 150 mm (6 in).

In selecting an engineering filter fabric (geotextile), four properties should be considered (FHWA, 1998):

  • Soil retention (piping resistance)
  • Permeability
  • Clogging
  • Survivability

FHWA (1998) provides detailed design guidance for selecting geotextiles as a riprap filter material. These guidelines should be applied in situations where problematic soil environments exist, severe environmental conditions are expected, and/or for critical installations. Problematic soils include unstable or highly erodible soils such as non-cohesive silts; gap graded soils; alternating sand/silt laminated soils; dispersive clays; and/or rock flour. Severe environmental conditions include wave action or high velocity conditions. An installation would be considered critical where loss of life or significant structural damage could be associated with failure.

With the exception of problematic soils or high velocity conditions associated with steep channels and rundowns, geotextile filters for roadside applications may usually be selected based on the apparent opening size (AOS) of the geotextile and the soil type as shown in Table 6.2.

Table 6.2. Maximum AOS for Geotextile Filters (FHWA, 1998)
Soil Type Maximum AOS (mm)
Non cohesive, less than 15 percent passing the 0.075 mm (US #200) sieve 0.43
Non cohesive, 15 to 50 percent passing the 0.075 mm (US #200) sieve 0.25
Non cohesive, more than 50 percent passing the 0.075 mm (US #200) sieve 0.22
Cohesive, plasticity index greater than 7 0.30

Design Example: Filter Blanket Design

Determine if a granular filter blanket is required, and if so, find an appropriate gradation. Given:

Riprap Gradation

D85 = 400 mm (16 in)

D50 = 200 mm (8 in)

D15 = 100 mm (4in)

Base Soil Gradation

D85 =1.5 mm (0.1 in)

D50 = 0.5 mm (0.034 in)

D15 = 0.167 mm (0.0066 in)

Only an SI solution is provided.

Solution

Check to see if the requirements of Equations 6.18 a, b, and c are met when comparing the riprap (upper) to the underlying soil (lower):

D15 riprap /D85 soil < 5 substituting 100/1.5 = 67 which is not less than 5

D15 riprap /D15 soil > 5 substituting 100/0.167 = 600 which is greater than 5, OK

D15 riprap /D15 soil < 40 substituting 100/0.167 = 600 which is not less than 40

D50 riprap /D50 soil < 40 substituting 200/0.5 = 400 which is not less than 40

Since three out of the four relationships between riprap and the soil do not meet the recommended dimensional criteria, a filter blanket is required. First, determine the required dimensions of the filter with respect to the base material.

D15 filter /D85 soil < 5 therefore, D15 filter < 5 x 1.5 mm = 7.5 mm

D15 filter /D15 soil > 5 therefore, D15 filter > 5 x 0.167 mm = 0.84 mm

D15 filter /D15 soil < 40 therefore, D15 filter < 40 x 0.167 mm = 6.7 mm

D50 filter /D50 soil < 40 therefore, D50 filter < 40 x 0.5 mm = 20 mm

Therefore, with respect to the soil, the filter must satisfy:

0.84 mm <D15 filter < 6.7 mm

D50 filter < 20 mm

Determine the required filter dimensions with respect to the riprap,

D15 riprap / D85 filter < 5 therefore D85 filter > 100 mm/5 = 20 mm

D15 riprap / D15 filter > 5 therefore D15 filter < 100 mm/5 = 20 mm

D15 riprap / D15 filter < 40 therefore, D15 filter > 100 mm/40 = 2.5 mm

D50 riprap / D50 filter < 40 therefore D50 filter > 200 mm/40 = 5 mm

With respect to the riprap, the filter must satisfy:

2.5 mm < D15 filter < 20 mm

D50 filter > 5 mm

D85 filter > 20 mm

Combining:

2.5 mm < D15 filter < 6.7 mm

5 mm < D50 filter < 20 mm

D85 filter > 20 mm

A gradation satisfying these requirements is appropriate for this design and is illustrated in Figure 6.2.

Figure showing an example relationship between the base material and riprap gradation characteristics
Figure 6.2. Gradations of Granular Filter Blanket for Design Example


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Contact:

Dan Ghere
Resource Center (Matteson)
708-283-3557
dan.ghere@dot.gov

Updated: 04/07/2011
 

FHWA
United States Department of Transportation - Federal Highway Administration