Skip to contentUnited States Department of Transportation - Federal Highway AdministrationFHWA HomeFeedback

Hydraulics Engineering

 
<< Previous Contents Next >>

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

Chapter 3: General Design Procedure

This chapter outlines the general design procedure for flexible channel linings based on design concepts presented in Chapter 2. The simplest case of the straight channel is described first. Subsequent sections consider variations to the straight channel including side slope stability, composite linings, and bends. The final two sections address additional considerations for channels with a steep longitudinal slope and determination of a maximum discharge for a given channel. This chapter is intended to apply to all flexible linings. Subsequent chapters provide more detailed guidance on specific flexible lining types.

3.1 Straight Channels

The basic design procedure for flexible channel linings is quite simple. The computations include a determination of the uniform flowdepth in the channel, known as the normal depth, and determination of the shear stresson the channel bottom at this depth. Both concepts were discussed in Chapter 2. Recalling Equation 2.8, the maximum shear stress is given by:


tau sub d equals gamma times d times S sub o (3.1)

where,

τd= shear stress in channel at maximum depth, N/m2 (lb/ft2)
γ= unit weight of water, N/m3 (lb/ft3)
d= depth of flow in channel, m (ft)
So= channel bottom slope, m/m (ft/ft)

The basic comparison required in the design procedure is that of permissible to computed shear stress for a lining. If the permissible shear stressis greater than or equal to the computed shear stress, including consideration of a safety factor, the lining is considered acceptable. If a lining is unacceptable, a lining with a higher permissible shear stress is selected, the discharge is reduced (by diversion or retention/detention), or the channel geometry is modified. This concept is expressed as:

tau sub p is greater than or equal to SF times tau sub d (3.2)

where,

τp= permissible shear stress for the channel lining, N/m2 (lb/ft2)
SF= safety factor (greater than or equal to one)
τd= shear stress in channel at maximum depth, N/m2 (lb/ft2)

The safety factor provides for a measure of uncertainty, as well as a means for the designer to reflect a lower tolerance for failure by choosing a higher safety factor. A safety factor of 1.0 is appropriate in many cases and may be considered the default. The expression for shear stress at maximum depth (Equation 3.1) is conservative and appropriate for design as discussed in Chapter 2. However, safety factors from 1.0 to 1.5 may be appropriate, subject to the designer's discretion, where one or more of the following conditions may exist:

  • critical or supercritical flows are expected
  • climatic regions where vegetation may be uneven or slow to establish
  • significant uncertainty regarding the design discharge
  • consequences of failure are high

The basic procedure for flexible liningdesign consists of the following steps and is summarized in Figure 3.1. (An alternative process for determining an allowable discharge given slope and shape is presented in Section 3.6.)

flow chart outlining the design steps for flexible channel lining design as discussed below
Figure 3.1. Flexible Channel Lining Design Flow Chart


Step 1. Determine a design discharge and select the channel slope and channel shape.

Step 2. Select a trial lining type. Initially, the designer may need to determine if a long-term lining is needed and whether or not a temporary or transitional lining is required. For determining the latter, the trial lining type could be chosen as the native material (unlined), typically bare soil. For example, it may be determined that the bare soil is insufficient for a long-term solution, but vegetation is a good solution. For the transitional period between construction and vegetative establishment, analysis of the bare soil will determine if a temporary lining is prudent.

Step 3. Estimate the depth of flow, di in the channel and compute the hydraulic radius, R. The estimated depth may be based on physical limits of the channel, but this first estimate is essentially a guess. Iterations on steps 3 through 5 may be required.

Step 4. Estimate Manning's n and the discharge implied by the estimated n and flow depth values. See Chapters 4 through 7 depending on lining type of interest, Table 2.1, or Table 2.2 for Manning's n values. Calculate the discharge, Qi.

Step 5. Compare Qi with Q. If Qi is within 5 percent of the design Q then proceed on to Step 6. If not, return to step 3 and select a new estimated flow depth, di+1. This 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

Step 6. Calculate the shear stress at maximum depth, τd (Equation 3.1), determine the permissible shear stress, τp, and select an appropriate safety factor. Permissible shear stress is determined based on guidance in Chapters 4 through 7, as applicable to the chosen lining, or Table 2.3. A safety factor of 1.0 is usually chosen, but may be increased as discussed earlier.

Step 7. Compare the permissible shear stress to the calculated shear stress from step 6 using Equation 3.2. If the permissible shear stress is adequate then the lining is acceptable. If the permissible shear is inadequate, then return to step 2 and select an alternative lining type with greater permissible shear stress. As an alternative, a different channel shape may be selected that results in a lower depth of flow.

The selected lining is stable and the design process is complete. Other linings may be tested, if desired, before specifying the preferred lining.

Design Example: Basic Channel (SI)

Evaluate a proposed gravel mulch lining on a trapezoidal channel for stability. Given:

Q = 0.42 m3/s

B = 0.4 m

Z = 3

So = 0.008 m/m

D50 = 25 mm

Solution

Step 1. Channel slope, shape and discharge have been given.

Step 2. Proposed lining type is a gravel mulch with D50 = 25 mm.

Step 3. Assume that the depth of flow, di in the channel is 0.5 m. Compute R. The equations in Appendix B may be used for this.

A = Bd+Zd2 = 0.4(0.5)+3(0.5)2 = 0.950 m2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.4 plus 2 times 0.5 times square root of (3 squared plus 1) equals 3.56 m

R = A/P = 0.950/3.56 = 0.267 m

Step 4. From Table 2.2, Manning's n equals 0.033. (Equations 6.1 or 6.2 should be used for this specific site, but for ease of illustration the value from Table 2.2 is used in this example.) The discharge is calculated using Manning's equation (Equation 2.1):

Q equals alpha divided by n times A times R to the two-thirds power times S sub f to the one-half power equals 1 divided by 0.033 times 0.950 times 0.267 to the two-thirds power times 0.008 to the one-half power equals 1.07 cubic meters per second

Step 5. Since this value is more than 5 percent different from the design flow, we need to go back to step 3 to estimate a new flow depth.

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 0.500 times (0.42 divided by 1.07) to the 0.4 power equals 0.344 meters

Compute new hydraulic radius.

A = Bd+Zd2 = 0.4(0.344)+3(0.344)2 = 0.493 m2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.4 plus 2 times 0.344 times square root of (3 squared plus 1) equals 2.58 meters

R = A/P = 0.493/2.58 = 0.191 m

Step 4 (2nd iteration). Table 2.2 does not have a 0.344 m depth so Equation 6.1 is used for estimating Manning's n. Manning's n equals 0.035. The discharge is calculated using Manning's equation:

Q equals alpha divided by n times A times R to the two-thirds power times S sub f to the one-half power equals 1 divided by 0.035 times 0.493 times 0.191 to the two-thirds power times 0.008 to the one-half power equals 0.42 cubic meters per second

Step 5 (2nd iteration). Since this value is within 5 percent of the design flow (we hit it right on), we can now proceed to step 6.

Step 6. The shear stress at maximum depth from Equation 3.1 is:

τd = γdSo = 9810(0.344)(0.008) = 27 N/m2

From Table 2.3, the permissible shear stress, τp = 19 N/m2.

For this channel, a SF = 1.0 is chosen.

Step 7. Compare calculated shear to permissible shear using Equation 3.2:

τp ≥ SF τd

19 ≥ 1.0 (27) No. The lining is not stable! Go back to step 2. Select an alternative lining type with greater permissible shear stress. Try the next larger size of gravel. If the lining had been stable, the design process would be complete.

Design Example: Basic Channel (CU)

Evaluate a proposed gravel mulch lining on a trapezoidal channel for stability. Given:

Q = 15 ft3/s

B = 1.3 ft

Z = 3

So = 0.008 ft/ft

D50 = 1 in

Solution

Step 1. Channel slope, shape and discharge have been given.

Step 2. Proposed lining type is a gravel mulch with D50 = 1 in.

Step 3. Assume that the depth of flow, di in the channel is 1.6 ft. Compute R. The equations in Appendix B may be used for this.

A = Bd+Zd2 = 1.3(1.6)+3(1.6)2 = 9.76 ft2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 1.3 plus 2 times 1.6 times square root of (3 squared plus 1) equals 11.4 feet

R = A/P = 9.76/11.4 = 0.856 ft

Step 4. From Table 2.2, Manning's n equals 0.033. (Equations 6.1 or 6.2 should be used for this specific site, but for ease of illustration the value from Table 2.2 is used in this example.) The discharge is calculated using Manning's equation (Equation 2.1):

Q equals alpha divided by n times A times R to the two-thirds power times S sub f to the one-half power equals 1.49 divided by 0.033 times 9.76 times 0.856 to the two-thirds power times 0.008 to the one-half power equals 35.5 cubic feet per second

Step 5. Since this value is more than 5 percent different from the design flow, we need to go back to step 3 to estimate a new flow depth.

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 1.6 times (15 divided by 35.5) to the 0.4 power equals 1.13 feet

Compute a new hydraulic radius.

A = Bd+Zd2 = 1.3(1.13)+3(1.13)2 = 5.30 ft2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 1.3 plus 2 times 1.13 times square root of (3 squared plus 1) equals 8.45 ft

R = A/P = 5.30/8.45 = 0.627 ft

Step 4 (2nd iteration). Table 2.2 does not have a 1.13 ft depth so Equation 6.1 is used for estimating Manning's n. Manning's n equals 0.035. The discharge is calculated using Manning's equation:

Q equals alpha divided by n times A times R to the two-thirds power times S sub f to the one-half power equals 1.49 divided by 0.035 times 5.3 times 0.627 to the two-thirds power times 0.008 to the one-half power equals 14.8 cubic feet per second

Step 5. (2nd iteration). Since this value is within 5 percent of the design flow, we can now proceed to step 6.

Step 6. The shear stress at maximum depth from Equation 3.1 is:

τd = γdSo = 62.4(1.13)(0.008) = 0.56 lb/ft2

From Table 2.3, permissible shear stress, τp = 0.4 lb/ft2.

For this channel, a SF = 1.0 is chosen.

Step 7. Compare calculated shear to permissible shear using Equation 3.2:

τp ≥ SFτd

0.40 ≥ 1.0 (0.56) No. The lining is not stable! Go back to step 2. Select an alternative lining type with greater permissible shear stress. Try the next larger size of gravel. If the lining had been stable, the design process would be complete.

3.2 Side Slope Stability

As described in Chapter 2, shear stress is generally reduced on the channel sides compared with the channel bottom. The maximum shear on the side of a channel is given by the following equation:

tau sub s equals K sub 1 times tau sub d (3.3)

where,

τs= side shear stress on the channel, N/m2 (lb/ft2)
K1= ratio of channel side to bottom shear stress
τd= shear stress in channel at maximum depth, N/m2 (lb/ft2)

The value K1 depends on the size and shape of the channel. For parabolic or V-shape with rounded bottom channels there is no sharp discontinuity along the wetted perimeter and therefore it can be assumed that shear stress at any point on the side slope is related to the depth at that point using Equation 3.1.

For trapezoidal and triangular channels, K1 has been developed based on the work of Anderson, et al. (1970). The following equation may be applied.

K sub 1 equals 0.77 Z ≤ 1.5
K sub 1 equals 0.066 times Z plus 0.67 1.5 < Z < 5 (3.4)
K sub 1 equals 1.0 5 ≤ Z

The Z value represents the horizontal dimension 1:Z (V:H). Use of side slopes steeper than 1:3 (V:H) is not encouraged for flexible linings other than riprapor gabions because of the potential for erosion of the side slopes. Steep side slopes are allowable within a channel if cohesive soil conditions exist. Channels with steep slopes should not be allowed if the channel is constructed in non-cohesive soils.

For riprap and gabions, the basic design procedure is supplemented for channels with side slopes steeper than 1:3 in Chapters 6 and 7, respectively.

3.3 Composite Lining Design

Composite linings use two lining types in a single channel rather than one. A more shear resistant lining is used in the bottom of the channel while a less shear resistant lining protects the sides. This type of design may be desirable where the upper lining is more cost-effective and/or environmentally benign, but the lower lining is needed to resist bottom stresses.

Another important use of a composite lining is in vegetative channels that experience frequent low flows. These low flows may kill the submerged vegetation. In erodible soils, this leads to the formation of a small gully at the bottom of the channel. Gullies weaken a vegetative lining during higher flows, causing additional erosion, and can result in a safety hazard. A solution is to provide a non-vegetative low-flow channel lining such as concrete or riprap. The dimensions of the low-flow channel are sufficient to carry frequent low flows but only a small portion of the design flow. The remainder of the channel is covered with vegetation.

It is important that the bottom lining material cover the entire channel bottom so that adequate protection is provided. To insure that the channel bottom is completely protected, the bottom lining should be extended a small distance up the side slope.

Computation of flow conditions in a composite channel requires the use of an equivalent Manning's n value for the entire perimeter of the channel. For determination of equivalent roughness, the channel area is divided into two parts of which the wetted perimeters and Manning's n values of the low-flow section and channel sides are known. These two areas of the channel are then assumed to have the same mean velocity. The following equation is used to determine the equivalent roughness coefficient, ne.

n sub e equals ( ( n sub s divided by n sub L) raised to the 3/2 power times (1 minus P sub L divided by P) + P sub L divided by P) raised to the 2/3 power (3.5)

where,

ne= effective Manning's n value for the composite channel
PL= low flow lining perimeter, m (ft)
P= total flow perimeter, m (ft)
ns= Manning's n value for the side slope lining
nL= Manning's n value for the low flow lining

When two lining materials with significantly different roughness values are adjacent to each other, erosion may occur near the boundary of the two linings. Erosion of the weaker lining material may damage the lining as a whole. In the case of composite channel linings with vegetation on the banks, this problem can occur in the early stages of vegetative establishment. A transitional lining should be used adjacent to the low-flow channel to provide erosion protection until the vegetative lining is well established.

The procedure for composite liningdesign is based on the design procedure presented in Section 3.1 with additional sub-steps to account for the two lining types. Specifically, the modifications are:

Step 1. Determine design discharge and select channel slope and shape. (No change.)

Step 2. Need to select both a low flow and side slope lining.

Step 3. Estimate the depth of flowin the channel and compute the hydraulic radius. (No change.)

Step 4. After determining the Manning's n for the low flow and side slope linings, use Equation 3.5 to calculate the effective Manning's n.

Step 5. Compare implied discharge and design discharge. (No change.)

Step 6. Determine the shear stress at maximum depth, τd (Equation 3.1), and the shear stress on the channel side slope, τs (Equation 3.3).

Step 7. Compare the shear stresses, τd and τs, to the permissible shear stress, τp, for each of the channel linings. If τd or τs is greater than the τp for the respective lining, a different combination of linings should be evaluated.

Design Example: Composite Lining Design (SI)

Evaluate the channel design for the composite concrete and vegetation lining given in Figure 3.2. Given:

Q = 0.28 m3/s

B = 0.9 m Concrete low flow channel

Z = 3

So = 0.02 m/m

Vegetation: Class C, height = 0.2 m (mixed with good cover)

The underlying soil is a clayey sand (SC) soil with a plasticity index of 16 and a porosity of 0.5. (See Section 4.3.2.)

Solution

Step 1. Channel slope, shape and discharge have been given.

Step 2. Low flow lining is concrete. Side slope lining is Class C vegetation.

Step 3. Assume that the depth of flow, di in the channel is 0.30 m. Determine R. Assume that the concrete portion is essentially flat.

A = Bd+Zd2 = 0.9(0.3)+3(0.3)2 = 0.540 m2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.9 plus 2 times 0.3 times square root of (3 squared plus 1) equals 2.80 meters

R = A/P = 0.540/2.80 = 0.193 m

Step 4. From the methods in Chapter 4 (Table 4.4 and Equation 4.2) the n value for Class C vegetation is determined to be ns = 0.043. For concrete, the typical n value for concrete from Table 2.1 is nL = 0.013.

Effective Manning's n is calculated using Equation 3.5.

n sub e equals 0.035 (solution of equation 3.5)

Calculate flow using Manning's equation (Equation 2.1):

Q equals alpha divided by n times A times R to the two-thirds power times S sub f to the one-half power equals 1 divided by 0.035 times 0.540 times 0.193 to the two-thirds power times 0.02 to the one-half power equals 0.73 cubic meters per second

Step 5. This flow is larger than the design flow of 0.28 m3/s by more than 5 percent. Go back to step 3.

Step 3 (2nd iteration). Estimate a new depth from an appropriate method. Try d=0.19 m.

Determine R with the new depth.

A = Bd+Zd2 = 0.9(0.19)+3(0.19)2 = 0.279 m2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.9 plus 2 times 0.19 times square root of (3 squared plus 1) equals 2.10 meters

R = A/P = 0.279/2.10 = 0.133 m

Step 4 (2nd iteration). From the methods in Chapter 4 (Table 4.4 and Equation 4.2) the n value for Class C vegetation is determined to be ns = 0.052. For concrete, the n value is the same nL = 0.013.

Effective Manning's n is calculated using Equation 3.5.

n sub e equals 0.038 (solution of equation 3.5

Calculate flow using Manning's equation:

Q equals 1 divided by 0.038 times 0.279 times 0.133 to the two-thirds power times 0.02 to the one-half power equals 0.27 cubic meters per second

Step 5 (2nd iteration). This flow is within 5 % of the design discharge, therefore, proceed to step 6.

Step 6. Calculate maximum shear stress, determine permissible shear stress, and select SF.

Bottom shear stress is calculated from Equation 3.1:

tau sub d equals gamma times d times S sub o equals 9810 times 0.19 times 0.02 equals 37 newtons per square meter

Maximum side shear stress is calculated from Equation 3.3 after calculating K1:

K sub 1 equals 0.066 times Z plus 0.67 equals 0.066 times 3 plus 0.67 equals 0.87

Concrete is a non-erodible, rigid lining so it has a very high rigid permissible shear stress. By using the techniques in Section 4.3 (Equation 4.7 and Table 4.5) the permissible shear stress of the vegetative portion of the lining is 123 N/m2.

tau sub s equals K sub 1 times tau sub d equals 0.87 times 37 equals 32 newtons per square meter

A safety factor of 1.0 is chosen for this situation.

Step 7. The maximum shear stress on the channel side slopes (32 N/m2) is less than permissible shear stress on the vegetation (123 N/m2) so the lining is acceptable. (The concrete bottom lining is non-erodible.)

Sketch showing the design example composite lining with a concrete low flow channel and class C vegetation side slopes. Variables are defined in the description of the design example
Figure 3.2. Composite Lining Design Example

Design Example: Composite Lining Design (CU)

Evaluate the channel design for the composite concrete and vegetation lining given in Figure 3.2. Given:

Q = 9.9 ft3/s

B = 3 ft Concrete low flow channel

Z = 3

So = 0.02 ft/ft

Vegetation: Class C, height = 0.66 ft (mixed with good cover)

The underlying soil is a clayey sand (SC) soil with a plasticity index of 16 and a porosity of 0.5 (See Section 4.3.2)

Solution

Step 1. Channel slope, shape and discharge have been given.

Step 2. Low flow lining is concrete. Side slope lining is Class C vegetation.

Step 3. Assume that the depth of flow, di in the channel is 1.0 ft. Determine R. Assume that the concrete portion is essentially flat.

A = Bd+Zd2 = 3.0(1.0)+3(1.0)2 = 6.00 ft2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 3.0 plus 2 times 1.0 times square root of (3 squared plus 1) equals 9.32 feet

R = A/P = 6.00/9.32 = 0.643 ft

Step 4. From the methods in Chapter 4 (Table 4.4 and Equation 4.2) the n value for Class C vegetation is determined to be ns = 0.043. For concrete, the typical n value for concrete from Table 2.1 is nL = 0.013.

Effective Manning's n is calculated using Equation 3.5.

n sub e equals 0.035 (solution of equation 3.5)

Calculate flow using Manning's equation (Equation 2.1):

Q equals alpha divided by n times A times R to the two-thirds power times S sub f to the one-half power equals 1.49 divided by 0.035 times 6.00 times 0.643 to the two-thirds power times 0.02 to the one-half power equals 26.9 cubic feet per second

Step 5. This flow is larger than the design flow of 9.9 ft3/s by more than 5 percent. Go back to step 3.

Step 3 (2nd iteration). Estimate a new depth from an appropriate approach. Try d = 0.62 ft.

Determine R with the new depth.

A = Bd+Zd2 = 3.0(0.62)+3(0.62)2 = 3.01 ft2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 3.0 plus 2 times 0.62 times square root of (3 squared plus 1) equals 6.92 feet

R = A/P = 3.01/6.92 = 0.435 ft

Step 4 (2nd iteration). From the methods in Chapter 4 (Table 4.4 and Equation 4.2) the n value for Class C vegetation is determined to be ns = 0.052. For concrete, the n value is the same nL = 0.013.

Effective Manning's n is calculated using Equation 3.5.

n sub e equals 0.038 (solution of equation 3.5)

Calculate flow using Manning's equation:

Q equals 1.49 divided by 0.038 times 3.01 times 0.435 to the two-thirds power times 0.02 to the one-half power equals 9.6 cubic feet per second

Step 5 (2nd iteration). This flow is within 5% of the design discharge, therefore, proceed to step 6.

Step 6. Calculate maximum shear stress, determine permissible shear stress, and select SF.

Bottom shear stress is calculated from Equation 3.1:

tau sub d equals gamma times d times S sub o equals 62.4 times 0.62 times 0.02 equals 0.77 pounds per square foot

Maximum side shear stress is calculated from Equation 3.3 after calculating K1:

K sub 1 equals 0.066 times Z plus 0.67 equals 0.066 times 3 plus 0.67 equals 0.87

tau sub s equals K sub 1 times tau sub d equals 0.87 times 0.77 equals 0.67 pounds per square foot

Concrete is a non-erodible, rigid lining so it has a very high rigid permissible shear stress. By using the techniques in Section 4.3 (Equation 4.7 and Table 4.5) the permissible shear stress of the vegetative portion of the lining is 2.5 lb/ft2.

A safety factor of 1.0 is chosen for this situation.

Step 7. The maximum shear stress on the channel side slopes (0.67 lb/ft2) is less than permissible shear stress on the vegetation (2.5 lb/ft2) so the lining is acceptable. (The concrete bottom lining is non-erodible.)

3.4 Stability in Bends

Flow around a bend creates secondary currents, which impose higher shear stresses on the channel sides and bottom compared to a straight reach (Nouh and Townsend, 1979) as shown in Figure 3.3. At the beginning of the bend, the maximum shear stress is near the inside and moves toward the outside as the flow leaves the bend. The increased shear stress caused by a bend persists downstream of the bend.

Equation 3.6 gives the maximum shear stress in a bend.

tau sub b equals K sub b times tau sub d (3.6)

where,

τb= side shear stress on the channel, N/m2 (lb/ft2)
Kb= ratio of channel bend to bottom shear stress
τd= shear stress in channel at maximum depth, N/m2 (lb/ft2)

The maximum shear stress in a bend is a function of the ratio of channel curvature to the top (water surface) width, RC/T. As RC/T decreases, that is as the bend becomes sharper, the maximum shear stress in the bend tends to increase. Kb can be determined from the following equation from Young, et al., (1996) adapted from Lane (1955):

High shear stress zones highlighted on the inside of the bend and on the outside leaving the bend
Figure 3.3. Shear Stress Distribution in a Channel Bend (Nouh and Townsend, 1979)


K sub b = 2.00 RC/T ≤ 2
K sub b = 2.38 minus 0.206 times (R sub c divided by T) plus 0.0073 times (R sub c divided by T) squared 2 < RC/T < 10 (3.7)
K sub b = 1.05 10 ≤ RC/T

where,

Rc= radius of curvature of the bend to the channel centerline, m (ft)
T= channel top (water surface) width, m (ft)

The added stress induced by bends does not fully attenuate until some distance downstream of the bend. If added lining protection is needed to resist the bend stresses, this protection should continue downstream a length given by:

L sub p equals alpha times R to the seven-eighths power divided by n (3.8)

where,

Lp= length of protection, m (ft)
R= hydraulic radius of the channel, m (ft)
n= Manning's roughness for lining material in the bend
α= unit conversion constant, 0.74 (SI) and 0.60 (CU)

A final consideration for channel design at bends is the increase in water surface elevation at the outside of the bend caused by the superelevation of the water surface. Additional freeboard is necessary in bends and can be calculated use the following equation:

delta d equals V squared times T divided by g divided by R sub c (3.9)

where,

Δd= additional freeboard required because of superelevation, m (ft)
V= average channel velocity, m/s (ft/s)
T= water surface top width, m (ft)
g= acceleration due to gravity, m/s2 (ft/s2)
Rc= radius of curvature of the bend to the channel centerline, m (ft)

The design procedure for channel bends is summarized in the following steps:

Step 1. Determine the shear stress in the bend and check whether or not an alternative lining is needed in the bend.

Step 2. If an alternative lining is needed, select a trial lining type and compute the new hydraulic properties and bend shear stress.

Step 3. Estimate the required length of protection.

Step 4. Calculate superelevation and check freeboard in the channel.

Design Example: Channel Bends (SI)

Determine an acceptable channel lining for a trapezoidal roadside channel with a bend. Also compute the necessary length of protection and the superelevation. The location is shown in Figure 3.4. A riprap lining (D50 = 0.15 m) has been used on the approaching straight channel (τp = 113 N/m2 from Table 2.3).

Given:

Q = 0.55 m3/s

d = 0.371 m

T = 3.42 m

B = 1.2 m

Z = 3

So = 0.015 m/m

RC = 10 m

Shear stress in the approach straight reach, τd = 54.5 N/m2

Solution

Step 1. Determine the shear stress in the bend using Equation 3.6. First, calculate Kb from Equation 3.7:

K sub b equals 2.38 minus 0.206 times (R sub c divided by T) plus 0.0073 times (R sub c divided by T) squared equals 2.38 minus 0.206 times (10.0 divided by 3.42) plus 0.0073 times (10.0 divided by 3.42) squared equals 1.84

Bend shear stress is then calculated from Equation 3.6:

tau sub b equals K sub b times tau sub d equals 1.84 times 54.5 equals 100 newtons per square meter

Step 2. Compare the bend shear stress with the permissible shear stress. The permissible shear stress has not been exceeded in the bend. Therefore, the lining in the approach channel can be maintained through the bend. If the permissible shear stress had been exceeded, a more resistant lining would need to be evaluated. A new normal depth would need to be found and then Step 1 repeated.

Step 3. Calculate the required length of protection. Since the same lining is being used in the approach channel and the bend, the length of protection is not relevant in this situation. However, we will calculate it to illustrate the process. Using Equation 3.8 and the channel geometrics, the hydraulic radius, R is 0.24 m and n = 0.074 (using Equation 6.1).

L sub p equals alpha times (R to the seven-sixths power divided by n) = 0.74 times (0.24 to the seven-sixths power divided by 0.074 equals 1.9 meters

Step 4. Calculate the superelevation of the water surface. First, top width and cross-sectional area must be computed using the geometric properties,

T = B + 2Zd = 1.2 + 2(3)(0.371) = 3.42 m

A = Bd + Zd2 = 1.2(0.371) + 3(0.371)2 = 0.86 m2

The velocity in the channel found using the continuity equation,

V = Q/A = 0.55/0.86 = 0.64 m/s

Solving Equation 3.9,

delta d equals V squared times T divided by g divided by R sub c equals 0.64 squared times 3.42 divided by 9.81 divided by 10.0 equals 0.014 meters

The freeboard in the channel bend should be at least 0.014 meters to accommodate the super elevation of the water surface.

Sketch showing definitions of bend for example
Figure 3.4. Location Sketch of Flexible Linings for Bend Example

Design Example: Channel Bends (CU)

Determine an acceptable channel lining for a trapezoidal roadside channel with a bend. Also compute the necessary length of protection and the superelevation. The location is shown in Figure 3.4. A riprap lining (D50 = 0.5 ft) has been used on the approaching straight channel (τp = 2.4 lb/ft2 from Table 2.3).

Given:

Q = 20 ft3/s

d = 1.24 ft

T = 11.35 ft

B = 3.9 ft

Z = 3

So = 0.015 ft/ft

RC = 33 ft

Shear stress in the approach straight reach, τd = 1.16 lb/ft2

Solution

Step 1. Determine the shear stress in the bend using Equation 3.6. First, calculate Kb from Equation 3.7:

K sub b equals 2.38 minus 0.206 times (R sub c divided by T) plus 0.0073 times (R sub c divided by T) squared equals 2.38 minus 0.206 times (33 divided by 11.35) plus 0.0073 times (33 divided by 11.35) squared equals 1.84

Bend shear stress is then calculated from Equation 3.6:

tau sub b equals K sub b times tau sub d equals 1.84 times 1.16 equals 2.13 pounds per square foot

Step 2. Compare the bend shear stress with the permissible shear stress. The permissible shear stress has not been exceeded in the bend. Therefore, the lining in the approach channel can be maintained through the bend. If the permissible shear stress had been exceeded, a more resistant lining would need to be evaluated. A new normal depth would need to be found and then Step 1 repeated.

Step 3. Calculate the required length of protection. Since the same lining is being used in the approach channel and the bend, the length of protection is not relevant in this situation. However, we will calculate it to illustrate the process. Using Equation 3.8 and the channel geometrics, the hydraulic radius, R is 0.805 ft and n = 0.075 (using Equation 6.1).

L sub p equals alpha times (R to the seven-sixths power divided by n) equals 0.60 times (0.805 to the seven-sixths power divided by 0.075 equals 6.2 feet

Step 4. Calculate the superelevation of the water surface. First, top width and cross-sectional area must be computed using the geometric properties,

T = B + 2Zd = 3.9 + 2(3)(1.24) = 11.3 ft

A = Bd + Zd2 = 3.9(1.24) + 3(1.24)2 = 9.45 ft2

The velocity in the channel found using the continuity equation,

V = Q/A = 20/9.45 = 2.12 ft/s

Solving Equation 3.9,

delta d equals V squared times T divided by g divided by R sub c equals 2.12 squared times 11.3 divided by 32.2 divided by 33 equals 0.048 feet

The freeboard in the channel bend should be at least 0.048 ft to accommodate the super elevation of the water surface.

3.5 Steep Slope Design

Intuitively, steep channel slopes may be considered a harsher environment than mild slopes for channel lining design. Furthermore, inspection of Equation 3.1 reveals that applied shear stress is directly proportional to channel slope. Therefore, it is appropriate to address the question of what, if any, additional consideration should be given to flexible channel lining design on steep slopes.

First, "steep" must be defined. From a hydraulic standpoint a steep slope is one that produces a supercritical normal depth (as opposed to a mild slope). Steep may also be defined as a fixed value such as 10 percent. Neither definition is appropriate for all circumstances and a single definition is not required. Two general questions arise when considering steep slopes for channel design.

First, are the same relationships for channel roughness (Manning's n) applicable over the entire range of slopes? For vegetative (Chapter 4) and manufactured (Chapter 5) linings the methodologies for determining roughness do apply to the full range of conditions suitable for these linings; that is, where Equation 3.2 is satisfied. However, for riprap (Chapter 6) and gabion mattress (Chapter 7) linings, slope may influence the approach for estimating roughness. As is discussed in Section 6.1, two roughness relationships are provided: one for relatively shallow flow (da/D50 < 1.5) and one for relatively moderate or deep flow (da/D50 > 1.5). Although slope plays an important role in determining depth; discharge, channel shape, and lining D50 also have an influence on the appropriate selection of roughness. (See Section 6.1 for more explanation.)

The second question is whether or not a steeper slope affects the development of the permissible shear stress used in Equation 3.2. Again, for vegetative and manufactured linings, the answer is no and the methods discussed in Chapters 4 and 5, respectively, apply to the full range of conditions where Equation 3.2 is satisfied. The same is also true for gabion linings. In fact, the permissible shear stress relationships presented in Section 7.2 were developed based on testing with slopes up to 33 percent. For riprap linings, Section 6.2 provides two alternative frameworks for evaluating permissible shear stress. One method is for slopes up to 10 percent and the other is for slopes equal to or greater than 10 percent.

Rigid channel linings may be a cost-effective alternative to flexible linings for steep slope conditions. Rigid linings could include asphalt, concrete, or durable bedrock. The decision to select a rigid or flexible lining may be based on other site conditions, such as foundation and maintenance requirements.

For flexible or rigid linings on steep slopes, bends should be avoided. A design requiring a bend in a steep channel should be reevaluated to eliminate the bend, modeled, or designed using an enclosed section.

3.6 Maximum Discharge Approach

As the discharge increases along a channel, the shear stress may at some point reach the permissible shear for the channel lining selected indicating the need to proceed with the design of another lining for the next section of channel or provide a relief inlet or culvert to divert the flow out of the channel. The methodology for determining the length or section of channel that a selected lining will remain stable is often referred to as the maximum discharge approach. By knowing the maximum discharge that a lining can sustain, the designer can determine the maximum length of lining for a selected lining type based on the hydrology of the site. This information can assist the designer in an economic evaluation of lining types and can determine relief inlet or culvert spacing.

The procedure presented is for both vegetative linings and non-vegetative linings. Applying the procedure for vegetative linings is particularly useful, since it does not involve a trial and error solution.

Combining Equations 3.1 and 3.2 in the following form can derive the maximum depth a channel lining can withstand:

d is less than or equal to tau sub p divided by SF divided by gamma divided by S sub o (3.10)

The analysis approach is applied as follows:

Step 1. Select a candidate lining and determine its permissible shear value and an appropriate safety factor.

Step 2. Use Equation 3.10 to calculate the maximum depth. Check that this depth does not exceed the depth (including freeboard) provided in the typical roadway section.

Step 3. Determine the area and hydraulic radius corresponding to the allowable depth based on the channel geometry

Step 4. Estimate the Manning's n value appropriate for the lining type and depth.

Step 5. Solve Manning's equation to determine the maximum discharge for the channel. The length of roadway and/or contributing drainage must be limited to an area generating less than or equal to this amount.

Design Example: Maximum Discharge Approach (SI)

Determine the maximum discharge for a median ditch lined with riprap (D50 = 0.150 m). The ditch has a depth of 0.9 m from the roadway shoulder.

Given:

So = 0.015 m/m

B = 1.0 m

Z = 4

SF = 1.0

Solution

Step 1. Riprap (D50 = 0.150 m) has a permissible shear stress of τp = 113 N/m2 (Table 2.3).

Step 2. Determine the allowable depth from Equation 3.10.

d is less than or equal to tau sub p divided by S sub f divided by gamma divided by S sub o equals 113 divided by 1.0 divided by 9810 divided by 0.015 equals 0.768 meters

The allowable depth is less than the depth of the ditch (0.9 m).

Step 3. Determine the flow area and hydraulic radiusfrom the geometric properties of a trapezoidal channel for the allowable depth:

A equals B times d plus Z times d squared equals 1 times 0.768 plus 4 times 0.768 squared equals 3.13 square meters

P equals B plus 2 times d times square root of (Z squared plus 1) equals 1 plus 2 times 0.768 times square root of (4 squared plus 1) equals 7.33 meters

R = A/P = 3.13/7.33 = 0.427 m

Step 4. From Equation 6.1, n = 0.059

Step 5. Solving Manning's equation (Equation 2.1):

Q equals alpha divided by n times A times R to the two-thirds power times S sub f to the one-half power equals 1 divided by 0.059 times 3.13 times 0.427 to the two-thirds power times 0.015 to the one-half power equals 3.7 cubic meters per second

Flow in this channel must be limited to the calculated flow by providing a relief inlet or culvert or by providing a lining with greater shear resistance.

Design Example: Maximum Discharge Approach (CU)

Determine the maximum discharge for a median ditch lined with riprap (D50 = 0.5 ft). The ditch has a depth of 3 ft from the roadway shoulder.

Given:

So = 0.015 ft/ft

B = 3.3 ft

Z = 4

SF = 1.0

Solution

Step 1. Riprap (D50 = 0.5 ft) has a permissible shear stress of τp = 2.4 lb/ft2 (Table 2.3).

Step 2. Determine the allowable depth from Equation 3.10.

d is less than or equal to tau sub p divided by S sub f divided by gamma divided by S sub o equals 2.4 divided by 1.0 divided by 62.4 divided by 0.015 equals 2.56 feet

The allowable depth is less than the depth of the ditch (3 ft).

Step 3. Determine the flow area and hydraulic radiusfrom the geometric properties of a trapezoidal channel for the allowable depth:

A equals B times d plus Z times d squared equals 3.3 times 2.56 plus 4 times 2.56 squared equals 34.7 square feet

P equals B plus 2 times d times square root of (Z squared plus 1) equals 3.3 plus 2 times 2.56 times square root of (4 squared plus 1) equals 24.4 feet

R = A/P = 34.7/24.4 = 1.42 ft

Step 4. From Equation 6.1, n = 0.060

Step 5. Solving Manning's equation (Equation 2.1):

Q equals alpha divided by n times A times R to the two-thirds power times S sub f to the one-half power equals 1.49 divided by 0.060 times 34.7 times 1.42 to the two-thirds power times 0.015 to the one-half power equals 133 cubic feet per second

Flow in this channel must be limited to the calculated flow by providing a relief inlet or culvert or by providing a lining with greater shear resistance.

<< Previous | Contents | Next >>

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