Lecture 22 Example Culvert Design Much of the following is based on the USBR technical publication “Design of Small Canal Structures” (1978) I. An Example Culvert Design

• Design a concrete culvert using the procedures given by the USBR (Design of Small Canal Structures, USBR 1978)

• The culvert will go underneath a concrete-line canal as shown in the figure below, perpendicular to the canal alignment (shortest path across)

• The outside slope of the canal banks is 1.5:1.0 on both sides (H:V) • The inside side slope of the concrete lining is 1:1 (both sides) • The concrete lining thickness is 0.05 m • Elevation of the top of the canal banks is given as 134.00 m, with elevations of

the original ground surface at intersections with canal banks given in the figure • The berm on both banks is 3.0 m wide • The depth of the canal from the bottom to the top of the berms is 4.0 m, with the

upper 30 cm unlined

• The barrel will be pre-cast circular concrete pipe, and the inlet and outlet structures can be specified as Type 1, 2, 3, or 4, as given by the USBR

• The available concrete pipe has an inside diameter of: 60, 70, 80, and 90 cm • You must select from one of these diameters for the culvert barrel • Length of the pipe is 1.5 m per section

• The hydrological assessment of the area came up with a maximum surface

runoff of 2.4 m3/s for a 20-year flood in the area uphill from the canal • This is the runoff rate that the culvert must be designed to carry

• The upstream and downstream natural channels are wide and poorly defined in

cross-section, and no effort will be made to develop prismatic channels upstream of the culvert inlet, nor downstream of the culvert outlet

BIE 5300/6300 Lectures 243 Gary P. Merkley

II. A Culvert Design Solution 1. Determine the Horizontal Distance

• The horizontal distance from the culvert inlet to the pipe outlet is (from left to

right):

1.5*(134.00-132.12)+3.0+4.0+5.5+4.0+3.0+1.5*(134.00-128.06) = 31.23 m

2. Determine the Required Pipe Size

• Use a maximum average barrel velocity of 3.0 m/s • Then, for the design discharge of 2.4 m3/s:

4Q 4(2.4)D 1

V (3.0)= = =

π π.01m (1)

• The largest available pipe size is 90 cm; therefore, two or more pipes are needed

in parallel for this culvert design • For half the design discharge, 1.2 m3/s, the required diameter is:

4Q 4(1.2)D 0

V (3.0)= = =

π π.71m (2)

• Then, we can use two 80-cm ID pipes at a full pipe flow velocity of 2.39 m/s • It would also be possible to use three 60-cm ID pipes at a full pipe flow velocity of

2.83, which is closer to the maximum velocity of 3.0 m/s • But, choose two 80-cm ID pipes because it will simplify installation, require less

excavation work, and may reduce the overall pipe cost

3. Determine the Energy Loss Gradient

• With the full pipe flow impending, the energy loss gradient can be estimated by the Manning equation for open-channel flow

• Use a Manning n value of 0.015 for new concrete pipe, with a slight safety factor for aging (typical useful life is estimated as 40 to 50 years)

• Use half the design discharge because two 80-cm ID pipes will be installed in parallel

( ) ( ) ( )

( )

2 2 4 / 32 2 4 / 3p

f 10 / 3 10 / 3Q n W 1.2 0.015 2.51

S 0.011 m/mA 0.503

= = = (3)

Gary P. Merkley 244 BIE 5300/6300 Lectures

where the wetted perimeter, Wp, for full pipe flow is πD; and the area, A, is πD2/4, for an inside diameter of 0.80 m and half the design discharge, 1.2 m3/s

4. Determine the Critical Slope

• For critical flow, the Froude number is equal to unity:

2

2r 3

Q TF 1gA

= = .0 (4)

• For circular pipes, the following definitions apply:

1 2h2cos 1D

− ⎛ ⎞β = −⎜ ⎟⎝ ⎠

(5)

T Dsin2β⎛ ⎞= ⎜ ⎟

⎝ ⎠ and, p

DW2β

= (6)

( )2DA sin

8= β − β (7)

• Solve for depth, h, such that Fr

2 = 1.0 for Q = 1.2 m3/s and D = 0.80 m

• Using the Newton method, hc = 0.663 m

• Calculate the energy loss gradient (critical slope) corresponding to this depth

• For a depth of 0.663 m, the flow area is 0.445 m2, and the wetted perimeter is 1.83 m

• Applying the Manning equation:

( ) ( ) ( ) ( )( )

2 2 4 / 3

f crit 10 / 31.2 0.015 1.83

S 00.445

= = .011 m/m (8)

• This is essentially the same loss gradient as for impending full pipe flow, but

note that the critical flow depth is 83% of the pipe ID • If the slope of the pipe is 0.011 m/m or greater, critical flow can occur

BIE 5300/6300 Lectures 245 Gary P. Merkley

5. Determine the Minimum Upstream Pipe Slope

• The upstream pipe will be situated so as to begin at Elev 132.12, and just clear the canal base at the left side

• The elevation of the canal base is 134.00 - 4.0 = 130.00 m

• The horizontal distance from the culvert inlet to the left side of the canal base is 1.5*(134.00-132.12)+3.0+4.0 = 9.82 m

• The pipe must drop at least 132.12-130.00+0.05+0.2 = 2.37 m over this horizontal distance

• This corresponds to a pipe slope of 2.37/9.82 = 0.24 m/m (24%)

• The critical slope is 1.1% (< 24%), so the culvert will have inlet flow for the design discharge (and for lower discharge values)

• At the design discharge, we will expect a hydraulic jump in the pipe upstream of the bend, because the pipe slope will be lower in the remaining (downstream) portion of the culvert

• It is necessary to check that the slope of the downstream pipe does not

exceed the critical slope

6. Determine the Downstream Pipe Slope

• The downstream part of the culvert barrel will travel a horizontal distance of 31.23 - 9.82 = 21.41 m

• The change in elevation over this distance will be 129.75 - 128.06 = 1.69 m • Then, the slope of the downstream part of the pipe will be 1.69/21.41 = 0.079

m/m (7.9%)

• This slope is greater than the critical slope, and is not acceptable because it would cause supercritical flow throughout, from inlet to outlet, causing erosion downstream (unless erosion protection is used)

• Use the USBR recommended downstream slope of 0.005 (0.5%), which is less than the critical slope of 1.1%

• To accomplish this, the upstream (steep) portion of the culvert pipe can be

extended further in the downstream direction (to the right) • Equations can be written for the tops of the upstream and downstream pipes:

Upstream: .................................... y 0.24x 132.12= − +

Downstream:.............................. y 0.005x 128.22= − +

[Note: 128.06+(31.23)(0.005) = 128.22]

Gary P. Merkley 246 BIE 5300/6300 Lectures

• Solving the two equations for distance, a value of x = 16.60 m is obtained • This is the distance from the inlet at which the top of the upstream pipe

intersects the top of the downstream pipe • The elevation of the intersection point is y = -0.24(16.60) + 132.12 = 128.14

m • The minimum clearance of 0.2 m under the canal bed is still provided for, but

the excavation for the culvert will require more work • Note that in many locations the natural ground slope is insufficient to justify a

critical slope on the upstream side, and a subcritical slope on the downstream side of the culvert

7. Determine the Pipe Lengths

• The approximate length of the upstream (steep) pipe is:

( ) ( )= + − =2 2usL 16.60 132.12 128.14 17.07 m (9)

• The approximate length of the downstream pipe is:

( ) ( )= − + − =2 2dsL 31.23 16.60 128.14 128.06 14.63 m (10)

• At 1.5 m per pipe, this corresponds to (17.07 + 14.63)/1.5 ≈ 21 pipes • For a double-barreled culvert, there must be about 42 pipe lengths

8. Specify Inlet and Outlet Types

• The inlet and outlet can be USBR Type 2, 3, or 4 • Type 1 would not be appropriate because the upstream and downstream

channels are not well defined • An energy dissipation structure at the outlet is not needed (outlet velocity will be

< 15 fps)

BIE 5300/6300 Lectures 247 Gary P. Merkley

9. Specify Collar Placement and Size

• Use the standard USBR culvert design, calling for two collars under the downhill canal bank, and one collar under the uphill bank

• The distance between the two collars under the downhill bank will be approximately 3.0 m plus 2.0 ft, or 3.61 m

• Then, the Y value is X/1.2, or Y = 3.61/1.2 = 3.0 m

• This gives very large collars • There are other methods for determining collar size, but in this case the Y value

can be taken as 1.0 m, which would be only about one meter below the uphill canal berm

• More information about the site and soil would be required to verify the adequacy of the collar design

• Many culverts don’t have collars anyway, and in some cases they are problematic because they impede effective soil compaction – “piping” may be worse with the collars

References & Bibliography

USBR. 1978. Design of small canal structures. U.S. Government Printing Office, Washington, D.C.

435 pp.

Gary P. Merkley 248 BIE 5300/6300 Lectures