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

Chapter 5: Manufactured (RECP) Lining Design

Manufacturers have developed a variety of rolled erosion control products (RECPs) for erosion protection of channels. These products consist of materials that are stitched or bound into a fabric. Table 5.1 summarizes the range of RECP linings that are available from the erosion control industry. Selection of a particular product depends on the overall performance requirements for the design. RECPs offer ease of construction in climate regions where vegetation establishes quickly.

Table 5.1. Manufactured (RECP) Linings
Type Description
Open-Weave Textile A temporary degradable RECP composed of processed natural or polymer yarns woven into a matrix, used to provide erosion control and facilitate vegetation establishment. Examples: jute net, woven paper net, straw with net.
Erosion Control Blankey A temporary degradable RECP composed of processed natural or polymer fibers mechanically, structurally or chemically bound together to form a continuous matrix to provide erosion control and facilitate vegetation establishment. Example: curled wood mat.
Turf-reinforcement Mat (TRM) A non-degradable RECP composed of UV stabilized synthetic fibers, filaments, netting and/or wire mesh processed into a three-dimensional matrix. TRMs provide sufficient thickness, strength and void space to permit soil filling and establishment of grass roots within the matrix. Example: synthetic mat.

5.1 RECP Properties

The density, stiffness and thickness of light-weight manufactured linings known as rolled erosion control products (RECPs) are the main properties that relate to flow resistance and erosion control performance. There are a series of standard tests referred to as index tests that measure these physical properties. The AASHTO National Transportation Product Evaluation Program (NTPEP) (AASHTO/NTPEP, 2002) has identified a set of test methods applicable to RECPs. Research on RECPs has not resulted in a relationship between these index tests and hydraulic properties. Hydraulic properties must be determined by full scale testing in laboratory flumes using defined testing protocols (ASTM D 6460). Table 5.2 summarizes index tests that relate to the physical properties of density, stiffness and thickness.

Qualitatively, denser linings prevent soil from entering into the higher-velocity flow above the liner (Gharabaghi, et al., 2002; Cotton, 1993). Linings with higher tensile strength and flexural rigidity have less deformation due to shear and uplift forces of the flow and remain in closer contact with the soil. Linings with more thickness have a larger moment of inertia, which further reduces the deformation of the lining.

NTPEP also includes two bench tests developed by the Erosion Control Technology Council (ECTC) that relate to channel erosion. Table 5.3 briefly describes the bench scale test methods applicable to RECP channel linings. The values generated from bench-scale tests are intended for qualitative comparison of products and product quality verification. These values should not be used to design a channel lining.Because of their small scale, these tests do not reflect larger scale currents that are generated in full scale testing in laboratory flumes using defined testing protocols (AASHTO/NTPEP, 2002; Robeson, et al., 2003).

Table 5.2. Index Tests for RECPs
Property Index Test Description
Density ASTM D 6475 Standard Test Method for Mass per Unit Area for Erosion Control Blankets
ASTM D 6566 Standard Test Method for Measuring Mass per Unit Area of Turf Reinforcement Mats
ASTM D 6567 Standard Test Method for Measuring the Light Penetration of Turf Reinforcement Mats
Stiffness ASTM D 4595 Test Method for Tensile Properties of Geotextile by the Wide-Width Strip Method
Thickness ASTM D 6525 Standard Test Method for Measuring Nominal Thickness of Erosion Control Products

Table 5.3. Bench-Scale Tests for RECPs
Bench Test Description
ECTC - Draft Method 3 Channel Erosion Standard test method for determination of RECP ability to protect soil from hydraulically induced shear stresses under bench-scale conditions.
ECTC - Draft Method 4 Germination and Plant Growth Standard test method for determination of RECP performance in encouraging seed germination and plant growth.

Proper installation of RECPs is critical to their performance. This includes the stapling of the lining to the channel perimeter, the lapping of adjacent fabric edges and the frequency of cutoff trenches. Each manufacturer provides guidelines on installation, which should be reviewed and incorporated into installation specifications. Construction inspection should verify that all installation specifications have been met prior to acceptance.

5.2 Manning's Roughness

There is no single n value formula for RECPs. The roughness of these linings must be determined by full-scale testing in laboratory flumes using defined testing protocols. As with vegetated linings, the n value varies significantly with the applied shear due to the displacement of the lining by shear and uplift forces.

The designer will need to obtain from the RECP manufacturer a table of n value versus applied shear. Three n values, with the corresponding applied shear values need to be provided by the manufacturer as shown in Table 5.4. The upper shear stress should equal or exceed the lining shear, τl. The upper and lower shear stress values must equal twice and one-half of the middle value, respectively.

Table 5.4. Standard n value versus Applied Shear
Applied Shear, N/m2 (lb/ft2) n value
τlower = τmid/2 nlower
τmid nmid
τupper = 2 τmid nupper

This information is used to determine the following n value relationship:

n equals a times tau sub o to the b power (5.1)

where,

n= Manning's roughness value for the specific RECP
a= coefficient based on Equation 5.2
b= exponent based on Equation 5.3
τo= mean boundary shear stress, N/m2 (lb/ft2)

The coefficient "a" is based on the n value at the mid-range of applied shear.

a equals n sub mid divided by t sub mid to the b power (5.2)

The exponent "b" is computed by the following equation:

b equals negative of the square root of (natural logarithm of (n sub mid divided by n sub lower) times the natural logarithm of (n sub upper divided by n sub mid)) divided by 0.693 (5.3)

Note that exponent "b" should be a negative value.

5.3 Permissible Shear Stress

The permissible shear stress of an RECP lining is determined both by the underlying soil properties as well as those of the RECP. In the case of TRMs, the presence of vegetation also influences erosion resistance properties.

5.3.1 Effective Shear Stress

RECPs dissipate shear stress before it reaches the soil surface. When the shear stress at the soil surface is less than the permissible shear for the soil surface, then erosion of the soil surface will be controlled. RECPs provide shear reduction primarily by providing cover for the soil surface. As the hydraulic forces on the RECP lining increase, the lining is detached from the soil, which permits a current to establish between the lining and the soil surface. Turbulent fluctuations within this current eventually erode the soil surface. This process model for RECP shear on the soil surface is given by the following (See Appendix F for derivation):

tau sub e equals (tau sub d minus tau sub l divided by 4.3) times alpha divided by tau sub l (5.4)

where,

τe= effective shear stress on the soil, N/m2 (lb/ft2)
τd = design shear stress, N/m2 (lb/ft2)
τl= shear stress on the RECP that results in 12.5 mm (0.5 in) of erosion
α= unit conversion constant, 6.5 (SI), 0.14 (CU)

The value of τl is determined based on a standard soil specified in the testing protocol. Permissible shear stress for the underlying soil has been presented in Section 4.3.2. The reader is referred to that section for that discussion.

5.3.2 Permissible RECP/Soil Shear Stress

The combined effects of the soil permissible shear stress and the effective shear stress transferred through the RECP lining results in a permissible shear stress for the RECP lining. Taking Equation 5.4 and substituting the permissible shear stress for the soil for the effective shear stress on the soil, τe, gives the following equation for permissible shear stress for the RECP lining:

tau sub p equals tau sub l divided by alpha times the quantity (t sub p,soil plus alpha divided by 4.3) (5.5)

where,

τp= permissible shear stress on the RECP lining, N/m2 (lb/ft2)
τl= shear stress on the RECP that results in 12.5 mm (0.5 in) of erosion
τp,soil= permissible soil shear stress, N/m2 (lb/ft2)
α= unit conversion constant, 6.5 (SI), 0.14 (CU)

Design Example: Manufactured Lining Design (SI)

Evaluate a temporary channel lining for a roadside channel. Two alternative RECPs are available. Alternative A costs less.

Given:

Shape: Trapezoidal, B = 0.9 m, Z = 3

Soil: Clayey sand (SC classification), PI = 16, e = 0.5

Grade: 3.0 percent

Flow: 0.30 m3/s

RECP Product A:

Erosion Control Blanket, ECB, Manufacturers performance data

τl = 60 N/m2 (Shear on lining at 12.5 mm soil loss)

Roughness rating:

Applied Shear, N/m2 n value
35 0.038
70 0.034
140 0.031

RECP Product B:

Erosion Control Blanket, ECB, Manufacturers performance data

τl = 100 N/m2 (Shear on lining at 12.5 mm soil loss)

Roughness rating:

Applied Shear, N/m2 n value
50 0.040
100 0.036
200 0.033

Solution

First, try the less expensive "Product A." The solution is accomplished using procedure given in Section 3.1 for a straight channel.

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

Step 2. Select erosion ECB A.

Step 3. Initial depth is estimated at 0.30 m

From the geometric relationship of a trapezoid (see Appendix B):

A equals B times d plus Z times d squared equals 0.9 times 0.3 plus 3 times 0.3 squared equals 0.540 square meters

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. To estimate n, the applied shear stress on the lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 9810 times 0.193 times 0.03 equals 56.8 Newtons per square meter

Determine a Manning's n value from Equation 5.1 with support from Equations 5.2 and 5.3.

b equals minus square root of (natural logarithm of (n sub mid divided by n sub lower) times natural logarithm of (n sub upper divided by n sub mid)) divided by 0.693 equals minus square root of (natural logarithm of (0.034 divided by 0.038) times natural logarithm of (0.031 divided by 0.034)) divided by 0.693 equals -0.146

a equals n sub mid divided by tau sub mid to the b power equals 0.034 divided by 70 to the -0.146 power equals 0.0632

n equals a times tau sub o to the b power equals 0.0632 times 56.8 to the -0.146 power equals 0.035

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 0.035 times 0.540 times 0.193 to the two-thirds power times 0.03 to the one-half power equals 0.88 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 solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth:

d = 0.18 m

Revised hydraulic radius.

A equals B times d plus Z times d squared equals 0.9 times 0.18 plus 3 times 0.18 squared equals 0.259 square meters

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.9 plus 2 times 0.18 times square root of (3 squared plus 1) equals 2.04 meters

R = A/P = (0.259)/(2.04) = 0.127 m

Step 4 (2nd iteration). To estimate n, the applied shear stress on the lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 9810 times 0.127 times 0.03 equals 37.3 Newtons per square meter

Determine a Manning's n value from Equation 5.1. The exponent, b, and the coefficient, a, are unchanged from the earlier calculation.

n equals a times tau sub o to the b power equals 0.0632 times 37.3 to the -0.146 power equals 0.037

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 0.037 times 0.259 times 0.127 to the two-thirds power times 0.03 to the one-half power equals 0.31 cubic meters per second

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

Step 6. The maximum shear on the lining of the channel bottom is.

tau sub d equals gamma times d times S sub o equals 9810 times 0.18 times 0.03 equals 52.9 Newtons per square meter

Determine the permissible soil shear stress from Equation 4.6 and Table 4.6.

tau sub p,soil equals (c sub 1 times PI squared plus c sub 2 times PI plus c sub 3) times (c sub 4 + c sub 5 times e) squared times c sub 6 equals (1.07 times 16 squared plus 14.3 times 16 plus 47.7) times (1.42 minus 0.61 times 0.5) squared times 0.0048 equals 3.28 Newtons per square meter

Equation 5.5 gives the permissible shear on the RECP.

tau sub p equals tau sub l divided by alpha times (tau sub p,soil plus alpha divided by 4.3) equals 60 divided by 6.5 times (3.28 plus 6.5 divided by 4.3) equals 44.2 Newtons per square meter

Safety factor for this channel is selected to be equal to 1.0.

Step 7. Product A (ECB lining) is not acceptable since the maximum shear on the RECP surface is greater than the permissible shear of the RECP.

Now try the alternative "Product B." The flow and channel configuration as well as the permissible shear stress are the same. Also, it is reasonable to assume an initial depth equal to the last depth we calculated for Product A. Therefore, using the area and hydraulic radius from that calculation, we can start with Step 4.

Step 4. To estimate n, the applied shear stress on the lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 9810 times 0.127 times 0.03 equals 37.4 Newtons per square meter

Determine a Manning's n value from Equation 5.1 with support from Equations 5.2 and 5.3.

b equals minus square root of (natural logarithm of (n sub mid divided by n sub lower) times natural logarithm of (n sub upper divided by n sub mid)) divided by 0.693 equals minus square root of (natural logarithm of (0.036 divided by 0.040) times natural logarithm of (0.033 divided by 0.036)) divided by 0.693 equals -0.138

a equals n sub mid divided by tau sub mid to the b power equals 0.036 divided by 100 to the -0.138 power equals 0.0680

n equals a times tau sub o to the b power equals 0.0680 times 37.4 to the -0.138 power equals 0.041

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 0.041 times 0.259 times 0.127 to the two-thirds power times 0.03 to the one-half power equals 0.28 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 solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth:

d = 0.19 m

Revise hydraulic radius.

A equals B times d plus Z times d squared equals 0.9 times 0.19 plus 3 times 0.19 squared equals 0.279 square meters

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) = 0.132 m

Step 4 (2nd iteration). To estimate n, the applied shear stress on the lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 9810 times 0.132 times 0.03 equals 38.8 Newtons per square meter

Determine a Manning's n value from Equation 5.1. The exponent, b, and the coefficient, a, are unchanged from the earlier calculation.

n equals a times tau sub o to the b power equals 0.0680 times 38.8 to the -0.138 power equals 0.041

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 0.041 times 0.279 times 0.132 to the two-thirds power times 0.03 to the one-half power equals 0.305 cubic meters per second

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

Step 6. The maximum shear on the lining of the channel bottom is.

tau sub d equals gamma times d times S sub o equals 9810 times 0.19 times 0.03 equals 55.9 Newtons per square meter

Equation 5.5 gives the permissible shear on the RECP.

tau sub p equals tau sub l divided by alpha times (tau sub p,soil plus alpha divided by 4.3) equals 100 divided by 6.5 times (3.28 plus 6.5 divided by 4.3) equals 73.7 Newtons per square meter

Step 8. Product B (ECB lining) is an acceptable temporary lining since the maximum shear on the RECP surface is less than the permissible shear of the RECP.Choose Product B. (Remember the permanent vegetative lining must also be evaluated.)

Design Example: Manufactured Lining Design (CU)

Evaluate a temporary channel lining for a roadside channel. Two alternative RECPs are available. Alternative A costs less.

Given:

Shape: Trapezoidal, B = 3.0 ft, Z = 3

Soil: Clayey sand (SC classification), PI = 16, e = 0.5

Grade: 3.0 percent

Flow: 10 ft3/s

RECP Product A:

Erosion Control Blanket, ECB, Manufacturers performance data

τl = 1.25 lb/ft2 (Shear on lining at 0.5 in soil loss)

Roughness rating:

Applied Shear, lb/ft2 n value
0.75 0.038
1.5 0.034
3.0 0.031

RECP Product B:

Erosion Control Blanket, ECB, Manufacturers performance data

τl = 2.0 lb/ft2 (Shear on lining at 0.5 in soil loss)

Roughness rating:

Applied Shear, lb/ft2 n value
1.0 0.038
2.0 0.034
4.0 0.031

Solution

First, try the less expensive "Product A." The solution is accomplished using procedure given in Section 3.1 for a straight channel.

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

Step 2. Try ECB Product A.

Step 3. Initial depth is estimated at 1.0 ft

From the geometric relationship of a trapezoid (see Appendix B):

A equals B times d plus Z times d squared equals 3.0 times 1.0 plus 3 times 1.0 squared equals 6.00 square feet

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) = 0.644 ft

Step 4. To estimate n, the applied shear stress on the lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 62.4 times 0.644 times 0.03 equals 1.21 pounds per square foot

Determine a Manning's n value from Equation 5.1 with support from Equations 5.2 and 5.3.

b equals minus square root of (natural logarithm of (n sub mid divided by n sub lower) times natural logarithm of (n sub upper divided by n sub mid)) divided by 0.693 equals minus square root of (natural logarithm of (0.034 divided by 0.038) times natural logarithm of (0.031 divided by 0.034)) divided by 0.693 equals -0.146

a equals n sub mid divided by tau sub mid to the b power equals 0.034 divided by 1.5 to the -0.146 power equals 0.0361

n equals a times tau sub o to the b power equals 0.0361 times 1.21 to the -0.146 power equals 0.035

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 0.035 times 6.0 times 0.644 to the two-thirds power times 0.03 to the one-half power equals 33.0 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 solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth:

d = 0.57 ft

Revise hydraulic radius.

A equals B times d plus Z times d squared equals 3.0 times 0.57 plus 3 times 0.57 squared equals 2.68 square feet

P equals B plus 2 times d times square root of (Z squared plus 1) equals 3.0 plus 2 times 0.57 times square root of (3 squared plus 1) equals 6.60 feet

R = A/P = (2.68)/(6.60) = 0.406 ft

Step 4 (2nd iteration). To estimate n, the applied shear stress on the lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 62.4 times 0.406 times 0.03 equals 0.76 pounds per square foot

Determine a Manning's n value from Equation 5.1. The exponent, b, and the coefficient, a, are unchanged from the earlier calculation.

n equals a times tau sub o to the b power equals 0.0361 times 0.76 to the -0.146 power equals 0.037

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 0.037 times 2.68 times 0.406 to the two-thirds power times 0.03 to the one-half power equals 10.2 cubic feet per second

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

Step 6. The maximum shear on the channel bottom is:

tau sub d equals gamma times d times S sub o equals 62.4 times 0.57 times 0.03 equals 1.07 pounds per square foot

Determine the permissible soil shear stress from Equation 4.6 and Table 4.6.

tau sub p,soil equals (c sub 1 times PI squared plus c sub 2 times PI plus c sub 3) times (c sub 4 + c sub 5 times e) squared times c sub 6 equals (1.07 times 16 squared plus 14.3 times 16 plus 47.7) times (1.42 minus 0.61 times 0.5) squared times 0.00001 equals 0.068 pounds per square foot

Equation 5.5 gives the permissible shear on the RECP.

tau sub p equals tau sub l divided by alpha times (tau sub p,soil plus alpha divided by 4.3) equals 1.25 divided by 0.14 times (0.068 plus 0.14 divided by 4.3) equals 0.90 pounds per square foot

Safety factor for this channel is selected to be equal to 1.0.

Step 7. Product A (ECB lining) is not acceptable since the maximum shear on the RECP surface is greater than the permissible shear of the RECP.

Now try the alternative "Product B." The flow and channel configuration as well as the permissible shear stress are the same. Also, it is reasonable to assume an initial depth equal to the last depth we calculated for Product A. Therefore, using the area and hydraulic radius from that calculation, we can start with Step 4.

Step 4. To estimate n, the applied shear stress on the lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 62.4 times 0.406 times 0.03 equals 0.76 pounds per square foot

Determine a Manning's n value from Equation 5.1 with support from Equations 5.2 and 5.3.

b equals minus square root of (natural logarithm of (n sub mid divided by n sub lower) times natural logarithm of (n sub upper divided by n sub mid)) divided by 0.693 equals minus square root of (natural logarithm of (0.036 divided by 0.040) times natural logarithm of (0.033 divided by 0.036)) divided by 0.693 equals -0.138

a equals n sub mid divided by tau sub mid to the b power equals 0.036 divided by 2.0 to the -0.138 power equals 0.0396

n equals a times tau sub o to the b power equals 0.0396 times 0.76 to the -0.138 power equals 0.041

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 0.041 times 2.68 times 0.406 to the two-thirds power times 0.03 to the one-half power equals 9.25 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 solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth:

d = 0.59 ft

Revise hydraulic radius.

A equals B times d plus Z times d squared equals 3.0 times 0.59 plus 3 times 0.59 squared equals 2.81 square feet

P equals B plus 2 times d times square root of (Z squared plus 1) equals 3.0 plus 2 times 0.59 times square root of (3 squared plus 1) equals 6.73 feet

R = A/P = (2.81)/(6.73) = 0.418 ft

Step 4 (2nd iteration). To estimate n, the applied shear stress on the lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 62.4 times 0.418 times 0.03 equals 0.78 pounds per square foot

Determine a Manning's n value from Equation 5.1. The exponent, b, and the coefficient, a, are unchanged from the earlier calculation.

n equals a times tau sub o to the b power equals 0.0396 times 0.78 to the -0.138 power equals 0.041

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 0.041 times 2.81 times 0.418 to the two-thirds power times 0.03 to the one-half power equals 9.89 cubic feet per second

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

Step 6. The maximum shear on the lining of the channel bottom is.

tau sub d equals gamma times d times S sub o equals 62.4 times 0.59 times 0.03 equals 1.10 pounds per square foot

Equation 5.5 gives the permissible shear on the RECP.

tau sub p equals tau sub l divided by alpha times (tau sub p,soil plus alpha divided by 4.3) equals 2.0 divided by 0.14 times (0.068 plus 0.14 divided by 4.3) equals 1.44 pounds per square foot

Step 7. Product B (ECB lining) is an acceptable temporary lining since the maximum shear on the RECP is less than the permissible shear of the RECP. Choose Product B. (Remember the permanent vegetative lining must also be evaluated.)

5.4 Turf Reinforcement with RECPS

Turf reinforcement integrates soil, lining material and grass/stems roots within a single matrix (Santha and Santha, 1995). Since turf reinforcement is a long-term solution, the lining consists of non-degradable materials. Turf Reinforcement Mats (TRMs), a subset of RECPs, are integrated with soil, and subsequently vegetation, by either covering the mat with soil or through surface application (no soil filling) allowing the vegetation to grow up through the TRM.

In the initial unvegetated state, the linings respond according to Equation 5.4. However, stability of the TRMs is achieved through proper installation per the manufacturer's recommended methods and use of proper length and quantity of fasteners.

As grass roots/stems develop within or through the TRM matrix, the lining becomes more integrated with the vegetation and soil. In the case of TRM linings, the plant roots/stems bind the mat, which prevents the detachment of the mat from the soil surface - significantly reducing the formation of under currents between the mat and soil. Grass growth further deflects turbulence away from the soil surface, establishing a positive relationship between lining and grass growth.

Where lining material is placed on top of the seed bed, the plant stem will grow through the lining (Lancaster, 1996). In this type of placement, the lining material offers more initial protection for the seed bed and provides stem reinforcement of the vegetation. However, the grass stem may be less effective at securing the lining to the soil surface than the plant roots, permitting the lining to be displaced by hydraulic forces. Specific system performance is determined by the interaction of the vegetation and the soil-filled or surface applied TRM.

When lining material remains in place for the long-term, roadside maintenance activities need to be considered, particularly mowing. Use of proper installation methods, including a sufficient quantity and size of fasteners is necessary to prevent potential problems with mowing and other vegetation maintenance equipment. Additionally, proper maintenance techniques are required to avoid damage to the installation and ensure the integrity of the system over time.

5.4.1 Testing Data and Protocols

Unlike other RECPs there is no broadly accepted protocol for the testing of TRMs. This places an additional burden on the designer of a TRM lining to review and understand how each manufacturer has tested its products. The following checklist (Table 5.5) is based on ASTM D 6460 with the addition of requirements for TRM testing (Lipscomb, et al., 2003). It can be used as minimum standard to evaluate manufacturer's testing protocols. Products that are based on a testing protocol for TRMs that do not meet these minimum elements should only be used cautiously.

Test data consisting of the following is recommended:

  1. Permissible lining shear stress for a vegetated lining alone,
  2. Permissible lining shear stress for a vegetated lining with turf reinforcement.
  3. A plant density factor (fractional increase or decrease in plant density) that is attributable to the lining material as a fraction of the cover.

Items (1) and (2) will vary depending on the soil and vegetation used in the testing. However, this is mitigated by their use in a relative, not absolute, manner as will be described in the next section. In addition, the definition of instability provided in Table 5.5 is qualitative and may be expected to vary from researcher to researcher. As long as a researcher maintains consistency within a set of tests for a particular TRM, the results should be acceptable.

5.4.2 Turf-Reinforcement Mat Cover Factor

A TRM modifies the cover factor for vegetated linings (Equation 4.3). The adjusted cover factor is determined by the following equation.

C sub f,TRM equals 1 minus (tau sub p,VEG-test divided by tau sub p,TRM-test) times (1 minus C sub f,VEG) (5.6)

where,

τp,VEG-test= permissible shear stress on the vegetative lining, N/m2 (lb/ft2) as reported by Manufacturer's testing
τp,TRM-test= permissible shear stress on the turf-reinforced vegetative lining, N/m2 (lb/ft2) as reported by manufacturer's testing
Cf,VEG = grass cover factor (see Table 4.5)
Cf,TRM = TRM cover factor

If the manufacturer notes that the TRM affects plant cover density, this information should be used in the selection of Cf,veg from Table 4.5.

Table 5.5. TRM Protocol Checklist
Element Protocol Description Check
Test Channel Preparation In accordance with ASTM D 6460, except that soil type may vary.
  1. Soil should be a suitable medium for plant material.
  2. Soil should be of the same type and properties (plasticity index and D75) for the test on vegetation alone and for test on vegetation with turf reinforcement.
 
Calibration In accordance with ASTM D 6460.  
Pre-Test Documentation In accordance with ASTM D 6460. In addition, the vegetation type and density should be determined.
  1. The type of plant material and habit (sod, bunched, mixed) should be identified.
  2. A vegetation stem density count should be performed using a minimum 58 sq. cm. (9 sq. in.) frame at the beginning of each test.
 
Test Setup In accordance with ASTM D 6460. In addition:
  1. Vegetation should be grown from seed for a minimum period of one year.
  2. Prior to testing, the vegetation should be mowed to a standard height not to exceed 0.20 meters (8 inches).
 
Test Operation and Data Collection In accordance with ASTM D 6460, except that the test should not be conducted to catastrophic failure.
  1. The channel surface should be inspected after hydraulic conditions have been maintained for no less than one-hour.
  2. A vegetation stem density count should be performed and bed elevations measured.
  3. The channel surface should then be inspected to determine the stability of the system. Instability is defined as: " Loss of vegetation sufficient to expose the roots and subject the underlying soil to significant erosion."
  4. Upon inspection, if instability of the channel surface is observed then testing should be terminated. If instability is not noted, testing should be continued with the next target discharge.
 

Design Example: Turf Reinforcement with a Turf Reinforcement Mat (SI)

Evaluate the following proposed lining design for a vegetated channel reinforced with turf reinforcement mat (TRM). The TRM will be placed into the soil and secured to channel boundary following manufacturer's recommendations. The permissible shear stress values for the TRM were developed from testing that meets the minimum requirements of Table 5.5.

Given:

Shape: Trapezoidal, B = 0.6 m, Z = 3

Soil: Silty sand (SM classification), PI = 17, e= 0.6

Grass: Sod, good condition, h = 0.150 m

Grade: 10.0 percent

Flow: 0.25 m3/s

TRM Product Information from manufacturer:

τp,TRM-test= 550 N/m3

τp,VEG-test= 425 N/m3

Effect on plant density is negligible.

Solution

First, we will check to see if the channel is stable with a grass lining alone.

The solution is accomplished using procedure given in Section 3.1 for a straight channel.

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

Step 2. A vegetative lining on silty sand soil will be evaluated

Step 3. Initial depth is estimated at 0.30 m

From the geometric relationship of a trapezoid (see Appendix B):

A equals B times d plus Z times d squared equals 0.6 times 0.3 plus 3 times 0.3 squared equals 0.450 square meters

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.6 plus 2 times 0.3 times square root of (3 squared plus 1) equals 2.50 meters

R = A/P = (0.540)/(2.80) = 0.180 m

Step 4. Estimate the Manning's n value appropriate for the lining type from Equation 4.2, first calculating the mean boundary shear.


tau sub o equals gamma times R times S sub o equals 9810 times 0.180 times 0.10 equals 177 Newtons per square meter
n equals alpha times C sub n times tau sub o to the -0.4 power equals 1.0 times 0.205 times 177 to the minus 0.4 power equals 0.026

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 0.026 times 0.450 times 0.180 to the two-thirds power times 0.10 to the one-half power equals 1.80 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 solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth:

d = 0.13 m

Revise the hydraulic radius

A equals B times d plus Z times d squared equals 0.6 times 0.13 plus 3 times 0.13 squared equals 0.129 square meters

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.6 plus 2 times 0.13 times square root of (3 squared plus 1) equals 1.42 meters

R = A/P = (0.129)/(1.42) = 0.091 m

Step 4 (2nd iteration). Estimate the Manning's n value appropriate for the lining type from Equation 4.2, first calculating the mean boundary shear.

tau sub o equals gamma times R times S sub o equals 9810 times 0.091 times 0.10 equals 89 Newtons per square meter

Determine a Manning's n value from Equation 4.2. From Table 4.3, Cn = 0.205

n equals alpha times C sub n times tau to the -0.4 power equals 1.0 times 0.205 times 89 to the minus 0.4 power equals 0.034

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 0.034 times 0.129 times 0.091 to the two-thirds power times 0.10 to the one-half power equals 0.25 cubic meters per second

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

Step 6. The maximum shear on the channel bottom is.

tau sub d equals gamma times d times S sub o equals 9810 times 0.13 times 0.10 equals 128 Newtons per square meter

Determine the permissible soil shear stress from Equation 4.6.

tau sub p,soil equals (c sub 1 times PI squared plus c sub 2 times PI plus c sub 3) times (c sub 4 + c sub 5 times e) squared times c sub 6 equals (1.07 times 17 squared plus 7.15 times 17 plus 11.9) times (1.42 minus 0.61 times 0.6) squared times 0.0048 equals 2.36 Newtons per square meter

Equation 4.7 gives the permissible shear stress on the vegetation. The value of Cf is found in Table 4.5.

tau sub p,VEG equals tau sub p,soil divided by (1 minus C sub f) times (n divided by n sub s) squared equals 2.36 divided by (1 minus 0.9) times (0.034 divided by 0.016) squared equals 107 Newtons per square meter

The safety factor for this channel is taken as 1.0.

Step 7. The grass lining is not acceptable since the maximum shear on the vegetation, 128 N/m2 is more than the permissible shear of grass lining, 107 N/m2.

Now try the same grass lining with turf reinforcement. The flow and channel configuration are the same. Therefore, we begin at step 6.

Step 6. The maximum shear on the channel bottom and the permissible soil shear are the same as in the previous iteration. A new cover factor is computed based on the TRM properties. Equation 4.7 gives the permissible shear stress on the vegetation. The value of Cf is computed using Equation 5.6

C sub f,TRM equals 1 minus tau sub p,VEG divided by tau sub p,TRM times (1 minus C sub f,VEG) minus 1 equals 1 minus 425 divided by 550 times (1 minus 0.9) equals 0.923

tau sub p,VEG equals tau sub p,soil divided by (1 minus C sub f) times (n divided by n sub s) squared equals 2.36 divided by (1 minus 0.923) times (0.034 divided by 0.016) squared equals 138 Newtons per square meter

The safety factor for this channel is taken as 1.0.

Step 7. The turf reinforced grass lining is acceptable since the maximum shear on the vegetation, 128 N/m2 is less than the permissible shear of the reinforced grass lining, 138 N/m2.

Design Example: Turf Reinforcement with a Turf Reinforcement Mat (CU)

Evaluate the following proposed lining design for a vegetated channel reinforced with turf reinforcement mat (TRM). The TRM will be placed into the soil and secured to channel boundary following manufacturer's recommendations. The permissible shear stress values for the TRM were developed from testing that meets the minimum requirements of Table 5.5.

Given:

Shape: Trapezoidal, B = 2.0 ft, Z = 3

Soil: Silty sand (SM classification), PI = 16, e= 0.6

Grass: Sod, good condition, h = 0.5 ft

Grade: 10.0 percent

Flow: 8.8 ft3/s

TRM Product Information from Manufacturer:

τp,TRM-test= 11.5 lb/ft3

τp,VEG-test= 8.9 lb/ft3

Effect on plant density is negligible.

Solution

First, we will check to see if the channel is stable with a grass lining alone.

The solution is accomplished using procedure given in Section 3.1 for a straight channel.

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

Step 2. A vegetative lining on silty sand soil will be evaluated

Step 3. Initial depth is estimated at 1.0 ft

From the geometric relationship of a trapezoid (see Appendix B):

A equals B times d plus Z times d squared equals 2.0 times 1.0 plus 3 times 1.0 squared equals 5.00 square feet

P equals B plus 2 times d times square root of (Z squared plus 1) equals 2.0 plus 2 times 1.0 times square root of (3 squared plus 1) equals 8.32 feet

R = A/P = (5.00)/(8.32) = 0.601 ft

Step 4. Estimate the Manning's n value appropriate for the lining type from Equation 4.2, first calculating the mean boundary shear.

tau sub o equals gamma times R times S sub o equals 62.4 times 0.601 times 0.10 equals 3.75 pounds per square foot

n equals alpha times C sub n times tau sub o to the -0.4 power equals 0.213 times 0.205 times 3.75 to the minus 0.4 power equals 0.026

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 0.026 times 5.0 times 0.601 to the two-thirds power times 0.10 to the one-half power equals 65 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 solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth:

d = 0.43 ft

Revise the hydraulic radius

A equals B times d plus Z times d squared equals 2.0 times 0.43 plus 3 times 0.43 squared equals 1.41 square feet

P equals B plus 2 times d times square root of (Z squared plus 1) equals 2.0 plus 2 times 0.43 times square root of (3 squared plus 1) equals 4.72 feet

R = A/P = (1.41)/(4.72) = 0.299 ft

Step 4 (2nd iteration). To estimate n, the applied shear stress on the grass lining is given by Equation 2.3

tau sub o equals gamma times R times S sub o equals 62.4 times 0.299 times 0.10 equals 1.87 pounds per square meter

n equals alpha times C sub n times tau sub o to the -0.4 power equals 0.213 times 0.205 times 1.87 to the minus 0.4 power equals 0.034

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 0.034 times 1.41 times 0.299 to the two-thirds power times 0.10 to the one-half power equals 8.7 cubic feet per second

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

Step 6. The maximum shear on the channel bottom is.

tau sub d equals gamma times d times S sub o equals 62.4 times 0.43 times 0.10 equals 2.7 pounds per square foot

Determine the permissible soil shear stress from Equation 4.6.

tau sub p,soil equals (c sub 1 times PI squared plus c sub 2 times PI plus c sub 3) times (c sub 4 + c sub 5 times e) squared times c sub 6 equals (1.07 times 17 squared plus 7.15 times 17 plus 11.9) times (1.42 minus 0.61 times 0.6) squared times 0.00001 equals 0.049 pounds per square foot

Equation 4.7 gives the permissible shear stress on the vegetation. The value of Cf is found in Table 4.5.

tau sub p,VEG equals tau sub p,soil divided by (1 minus C sub f) times (n divided by n sub s) squared equals 0.049 divided by (1 minus 0.9) times (0.034 divided by 0.016) squared equals 2.2 pounds per square foot

The safety factor for this channel is taken as 1.0.

Step 7. The grass lining is not acceptable since the maximum shear on the vegetation, 2.7 lb/ft2 is more than the permissible shear of grass lining, 2.2 lb/ft2.

Now try the same grass lining with turf reinforcement. The flow and channel configuration are the same. Therefore, we begin at step 6.

Step 6. The maximum shear on the channel bottom and the permissible soil shear are the same as in the previous iteration. A new cover factor is computed based on the TRM properties. Equation 4.7 gives the permissible shear stress on the vegetation. The value of Cf is found in Table 4.5.

C sub f,TRM equals 1 minus (tau sub p,VEG divided by tau sub p,TRM) times (1 minus C sub f,VEG) equals 1 minus (8.9 divided by 11.5) times (1 minus 0.9) equals 0.923

tau sub p,VEG equals tau sub p,soil divided by (1 minus C sub f) times (n divided by n sub s) squared equals 0.049 divided by (1 minus 0.923) times (0.034 divided by 0.016) squared equals 2.9 pounds per square foot

The safety factor for this channel is taken as 1.0.

Step 7. The turf reinforced grass lining is acceptable since the maximum shear on the vegetation, 2.7 lb/ft2 is less than the permissible shear of the reinforced grass lining, 2.9 lb/ft2.

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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