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

Chapter 7: Gabion Lining Design

Gabions (rock filled wire containers) represent an approach for using smaller rock size than would be required by riprap. The smaller rock is enclosed in larger wire units in the form of mattresses or baskets. Gabion baskets are individual rectangular wire mesh containers filled with rock and frequently applied for grade control structures and retaining walls. Gabion mattresses are also rock filled wire mesh containers. The mattresses are composed of a series of integrated cells that hold the rock allowing for a greater spatial extent in each unit. Potential roadside applications for the gabion mattress include steep channels and rundowns.

The thickness of the gabion mattress may be less than the thickness of an equivalently stable riprap lining. Therefore, gabion mattresses represent a trade-off between less and smaller rock versus the costs of providing and installing the wire enclosures. Gabion mattresses are rarely cost effective on mildly sloped channels.

7.1 Manning's Roughness

Roughness characteristics of gabion mattresses are governed by the size of the rock in the baskets and the wire mesh enclosing the rock. For practical purposes, the effect of the mesh can be neglected. Therefore, Manning's roughness should be determined using the D50 of the basket rock as applied to the relationships provided for riprap and gravel linings. (See Section 6.1.)

7.2 Permissible Shear Stress

Values for permissible shear stress for gabion mattresses are based on research conducted at laboratory facilities and in the field. However, reports from these studies are difficult to reconcile. Simons, et al. (1984) reported permissible shear stresses in the range of 140 to 190 N/m2 (3 to 4 lb/ft2) while Clopper and Chen (1988) reported values approaching 1700 N/m2 (35 lb/ft2). Simons, et al. tested mattresses ranging in depth from 152 to 457 mm (6 to 18 in) and on slopes of up to 2 percent. Since the objective was to test embankment overtopping, Clopper and Chen tested 152 mm (6 in) mattresses on 25 and 33 percent slopes.

The difference in reported permissible shear stresses may be partly due to the definition of failure. In the Clopper and Chen report, failure was noted after rocks within the basket had shifted to the downstream end of the baskets and an undulating surface was formed leaving part of the embankment exposed. Although this may be an appropriate definition for a rare embankment-overtopping event, such failure is not appropriate for the more frequently occurring roadside design event. For this reason as well as to provide for conservative guidance, the Simons et al. results are emphasized in this guidance.

Permissible shear stress for gabions may be estimated based on the size of the rock fill or based on gabion mattress thickness. Both estimates are determined and the largest value is taken as the permissible shear stress.

Equation 7.1 provides a relationship for permissible shear stress based on rock fill size (Simons, et al., 1984). This shear stress exceeds that of loose riprap because of the added stability provided by the wire mesh. The equation is valid for a range of D50 from 0.076 to 0.457 m (0.25 to 1.5 ft)

tau sub p equals F sub * times the quantity (gamma sub s minus gamma) times D sub 50 (7.1)

where,

τp= permissible shear stress, N/m2 (lb/ft2)
F*= Shields' parameter, dimensionless
D50= median stone size, m (ft)

In the tests reported by Simons, et al. (1984), the Shields' parameter for use in Equation 7.1 was found to be equal to 0.10.

A second equation provides for permissible shear stress based on mattress thickness (Simons, et al., 1984). It is applicable for a range of mattress thickness from 0.152 to 0.457 m (0.5 to 1.5 ft).

tau sub p equals 0.0091 times the quantity (gamma sub s minus gamma) times the quantity (MT plus MT sub c) (7.2)

where,

MT= gabion mattress thickness, m (ft)
MTC= thickness constant, 1.24 m (4.07 ft)

The limits on Equations 7.1 and 7.2 are based on the range of laboratory data from which they are derived. Rock sizes within mattresses typically range from 0.076 to 0.152 m (0.25 to 0.5 ft) rock in the 0.152 m (0.5 ft) thick mattresses to 0.116 to 0.305 m (0.33 to 1 ft) rock in the 0.457 m (1.5 ft) thick mattresses.

When comparing, the permissible shear for gabions with the calculated shear on the channel, a safety factor, SF is required for Equation 3.2. The guidance found in Table 6.1 is applicable to gabions. Since, the Shields parameter in Equation 7.1 is 0.10, the appropriate corresponding safety factor is 1.25. Alternatively, the designer may compute the particle Reynolds number and, using Table 6.1, determine both a Shields' parameter and SF corresponding to the Reynolds number.

7.3 Design Procedure

The design procedure for gabions is as follows. It uses the same roughness relationships developed for riprap.

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

Step 2. Select a trial (initial) mattress thickness and fill rock 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. 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.

Step 6. Calculate the permissible shear stress from Equations 7.1 and 7.2 and take the largest as the permissible shear stress.

Use Equation 3.1 to determine the actual shear stress on the bottom of the channel.

Select a safety factor.

Apply Equation 3.2 to compare the actual to permissible shear stress.

Step 7. If permissible shear is greater than computed shear, the lining is stable. If not, repeat the design process beginning at step 2.

Design Example: Gabion Design (SI)

Determine the flow depth and required thickness of a gabion mattress lining for a trapezoidal channel.

Given:

Q = 0.28 m3/s

S = 0.09 m/m

B = 0.60 m

Z = 3

Solution

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

Step 2. Try a 0.23 m thick gabion basket with a D50 = 0.15 m; γs= 25.9 kN/m3

Step 3. Assume an initial trial depth of 0.3 m

Using the geometric properties of a trapezoid:

A = Bd+Zd2 = 0.6(0.3)+3(0.3)2 = 0.450 m2

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 =0.45/2.50 = 0.180 m

T = B+2dZ = 0.6+2(0.3)(3) = 2.40 m

da = A/T = 0.45/2.40 = 0.188 m

Step 4. The relative depth ratio, da/D50 = 0.188/0.150 = 1.3. Therefore, use Equation 6.2 to calculate Manning's n.

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) to the 0.814 power equals 1.14 times (0.15 divided by 2.40) to the 0.453 power times (0.188 divided by 0.150) to the 0.814 power equals 0.390

Fr equals Q divided by A divided by square root of (g times d sub a) equals 0.28 divided by 0.450 divided by square root of (9.81 times 0.188) equals 0.458

function of Fr equals (0.28 times Fr divided by b) to the power logarithm (0.755 divided by b) equals (0.28 times 0.458 divided by 0.390) to the power logarithm (0.755 divided by 0.389) equals 0.726

function of REG equals 13.434 times (T divided by D sub 50) to the 0.492 power times b to the power (1.025 times (T divided by D sub 50) to the 0.118 power) equals 13.434 times (2.40 divided by 0.150) to the 0.492 power times 0.390 to the power (1.025 times (2.40 divided by 0.150) to the 0.118 power) equals 13.7

function of CG equals (T divided by d sub a) to the minus b power equals (2.4 divided by 0.188) to the -0.389 power equals 0.371

n equals alpha times d sub a to the one-sixth power divided by the square root of g divided by funtion of Fr divided by function of REG divided by function of CG equals 1.0 times 0.188 to the one-sixth power divided by the square root of 9.81 divided by 0.726 divided by 13.7 divided by 0.371 equals 0.065

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 0.065 times 0.45 times 0.180 to the two-thirds power times 0.09 to the one-half power equals 0.66 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 0.3 times (0.28 divided by 0.66) to the 0.4 power equals 0.21 meters

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

A = Bd+Zd2 = 0.6(0.21)+3(0.21)2 = 0.258 m2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 0.6 plus 2 times 0.21 times square root of (3 squared plus 1) equals 1.93 meters

R = A/P =0.258/1.93 = 0.134 m

T = B+2dZ = 0.6+2(0.21)(3) = 1.86 m

da = A/T = 0.258/1.86 = 0.139 m

Step 4. (2nd iteration). The relative depth ratio, da/D50 = 0.139/0.150 = 0.9. Therefore, use Equation 6.2 to calculate Manning's n.

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) to the 0.814 power equals 1.14 times (0.15 divided by 1.86) to the 0.453 power times (0.139 divided by 0.150) to the 0.814 power equals 0.343

Fr equals Q divided by A divided by square root of (g times d sub a) equals 0.28 divided by 0.258 divided by square root of (9.81 times 0.139) equals 0.929

function of Fr equals (0.28 times Fr divided by b) to the power logarithm (0.755 divided by b) equals (0.28 times 0.929 divided by 0.343) to the power logarithm (0.755 divided by 0.343) equals 0.910

function of REG equals 13.434 times (T divided by D sub 50) to the 0.492 power times b to the power (1.025 times (T divided by D sub 50) to the 0.118 power) equals 13.434 times (1.86 divided by 0.150) to the 0.492 power times 0.343 to the power (1.025 times (1.86 divided by 0.150) to the 0.118 power) equals 10.6

function of CG equals (T divided by d sub a) to the minus b power equals (1.86 divided by 0.139) to the -0.343 power equals 0.411

n equals alpha times d sub a to the one-sixth power divided by the square root of g divided by funtion of Fr divided by function of REG divided by function of CG equals 1.0 times 0.139 to the one-sixth power divided by the square root of 9.81 divided by 0.910 divided by 10.6 divided by 0.411 equals 0.058

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 0.058 times 0.258 times 0.134 to the two-thirds power times 0.09 to the one-half power equals 0.35 cubic meters per second

Since this estimate is also not within 5 percent of the design discharge, further iterations are required. Subsequent iterations will produce the following values:

d = 0.185 m

n = 0.055

Q = 0.29 m3/s

Proceed to step 6 with these values.

Step 6. Calculate the permissible shear stress from Equations 7.1 and 7.2 and take the largest as the permissible shear stress.

tau sub p equals F sub * times (gamma sub s minus gamma) times D sub 50 equals 0.10 times (25900 minus 9810) times 0.15 equals 241 Newtons per square meter

tau sub p equals 0.0091 times (gamma sub s minus gamma) times (MT plus MT sub c) equals 0.0091 times (25900 minus 9810) times (0.23 plus 1.24) equals 215 Newtons per square meter

Permissible shear stress for this gabion configuration is, therefore 241 N/m2.

Use Equation 3.1 to determine the actual shear stress on the bottom of the channel and apply Equation 3.2 to compare the actual to permissible shear stress.

tau sub d equals gamma times d times S sub o equals 9810 times 0.185 times 0.09 equals 163 Newtons per square meter

SF=1.25:

Step 7. From Equation 3.2: 241>1.25(163), therefore, the selected gabion mattress is acceptable.

Design Example: Gabion Design (CU)

Determine the flow depth and required thickness of a gabion mattress lining for a trapezoidal channel.

Given:

Q = 10 ft3/s

S = 0.09 ft/ft

B = 2.0 ft

Z = 3

Solution

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

Step 2. Try a 0.75 ft thick gabion basket with a D50 = 0.5 ft; γs= 165 lb/ft3

Step 3. Assume an initial trial depth of 1 ft.

Using the geometric properties of a trapezoid:

A = Bd+Zd2 = 2.0(1.0)+3(1.0)2 = 5.0 ft2

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 8.3 feet

R = A/P =5.0/8.3 = 0.601 ft

T = B+2dZ = 2.0+2(1.0)(3) = 8.0 ft

da = A/T = 5.0/8.0 = 0.625 ft

Step 4. The relative depth ratio, da/D50 = 0.625/0.50 = 1.3. Therefore, use Equation 6.2 to calculate Manning's n.

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) to the 0.814 power equals 1.14 times (0.50 divided by 8.0) to the 0.453 power times (0.625 divided by 0.50) to the 0.814 power equals 0.389

Fr equals Q divided by A divided by square root of (g times d sub a) equals 10 divided by 5.0 divided by square root of (32.2 times 0.625) equals 0.446

function of Fr equals (0.28 times Fr divided by b) to the power logarithm (0.755 divided by b) equals (0.28 times 0.446 divided by 0.389) to the power logarithm (0.755 divided by 0.389) equals 0.721

function of REG equals 13.434 times (T divided by D sub 50) to the 0.492 power times b to the power (1.025 times (T divided by D sub 50) to the 0.118 power)equals 13.434 times (8.0 divided by 0.50) to the 0.492 power times 0.389 to the power (1.025 times (8.0 divided by 0.50) to the 0.118 power) equals 13.7

function of CG equals (T divided by d sub a) to the minus b power equals (8.0 divided by 0.50) to the -0.389 power equals 0.371

n equals alpha times d sub a to the one-sixth power divided by the square root of g divided by funtion of Fr divided by function of REG divided by function of CG equals 1.49 times 0.625 to the one-sixth power divided by the square root of 32.2 divided by 0.721 divided by 13.7 divided by 0.371 equals 0.066

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 1.49 divided by 0.066 times 5.00 times 0.601 to the two-thirds power times 0.09 to the one-half power equals 24.1 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 1.0 times (10 divided by 19.8) to the 0.4 power equals 0.70 feet

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

A = Bd+Zd2 = 2.0(0.70)+3(0.70)2 = 2.87 ft2

P equals B plus 2 times d times square root of (Z squared plus 1) equals 2.0 plus 2 times 0.70 times square root of (3 squared plus 1) equals 6.43 feet

R = A/P =2.87/6.43 = 0.446 ft

T = B+2dZ = 2.0+2(0.70)(3) = 6.20 ft

da = A/T = 2.87/6.20 = 0.463 ft

Step 4. (2nd iteration). The relative depth ratio, da/D50 = 0.496/0.50 = 1.0. Therefore, use Equation 6.2 to calculate Manning's n.

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) to the 0.814 power equals 1.14 times (0.50 divided by 6.20) to the 0.453 power times (0.463 divided by 0.50) to the 0.814 power equals 0.342

Fr equals Q divided by A divided by square root of (g times d sub a) equals 10 divided by 2.87 divided by square root of (32.2 times 0.463) equals 0.902

function of Fr equals (0.28 times Fr divided by b) to the power logarithm (0.755 divided by b) equals (0.28 times 0.902 divided by 0.342) to the power logarithm (0.755 divided by 0.342) equals 0.901

function of REG equals 13.434 times (T divided by D sub 50) to the 0.492 power times b to the power (1.025 times (T divided by D sub 50) to the 0.118 power) equals 13.434 times (6.20 divided by 0.50) to the 0.492 power times 0.342 to the power (1.025 times (6.20 divided by 0.50) to the 0.118 power) equals 10.6

function of CG equals (T divided by d sub a) to the minus b power equals (6.20 divided by 0.463) to the -0.389 power equals 0.412

n equals alpha times d sub a to the one-sixth power divided by the square root of g divided by funtion of Fr divided by function of REG divided by function of CG equals 1.49 times 0.463 to the one-sixth power divided by the square root of 32.2 divided by 0.901 divided by 10.6 divided by 0.412 equals 0.059

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 1.49 divided by 0.059 times 2.87 times 0.446 to the two-thirds power times 0.09 to the one-half power equals 12.7 cubic feet per second

Step 5 (2nd iteration). Since this estimate is also not within 5 percent of the design discharge, further iterations are required. Subsequent iterations will produce the following values:

d = 0.609 ft

n = 0.055

Q = 10.2 ft3/s

Proceed to step 6 with these values.

Step 6. Calculate the permissible shear stress from Equations 7.1 and 7.2 and take the largest as the permissible shear stress.

tau sub p equals F sub * times (gamma sub s minus gamma) times D sub 50 equals 0.10 times (165 minus 62.4) times 0.5 equals 5.1 pounds per square foot

tau sub p equals 0.0091 times (gamma sub s minus gamma) times (MT plus MT sub c) equals 0.0091 times (165 minus 62.4) times (0.75 plus 4.07) equals 4.5 pounds per square foot

Permissible shear stress for this gabion configuration is, therefore 5.1 lb/ft2.

Use Equation 3.1 to determine the actual shear stress on the bottom of the channel and apply Equation 3.2 to compare the actual to permissible shear stress.

tau sub d equals gamma times d times S sub o equals 62.4 times 0.609 times 0.09 equals 3.4 pounds per square foot

SF=1.25:

Step 7. From Equation 3.2: 5.1>1.25(3.4), therefore, the selected gabion mattress is acceptable.

7.4 Additional Considerations

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

The stability of gabions depends on the integrity of the wire mesh. In streams with high sediment concentrations or with rocks moving along the bed of the channel, the wire mesh may be abraded and eventually fail. Under these conditions the gabion will no longer behave as a single unit but rather as individual stones. Applications of gabion mattresses and baskets under these conditions should be avoided. Such conditions are unlikely for roadside channel design.

Extent of gabions on a steep gradient (the most common roadside application for gabions) must be sufficient to protect transition regions of the channel both above and below the steep gradient section. The transition from a steep gradient to a culvert should allow for slumping of a gabion mattress.

Gabions should be placed flush with the invert of a culvert. The break between the steep slope and culvert entrance should equal three to five times the mattress thickness. 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).

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.

The rock gradation used in gabions mattress must be such that larger stones do not protrude outside the mattress and the wire mesh retains smaller stones.

When gabions are used, the need for an underlying filter material must be evaluated. The filter material may be either a granular filter blanket or geotextile fabric. See section 6.4.3 for description of the filter requirements.

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Contact:

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

Updated: 04/07/2011
 

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