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
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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
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243
Gary P. Merkley
II. A Culvert Design Solution
1. Determine the Horizontal Distance
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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
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Use a maximum average barrel velocity of 3.0 m/s
Then, for the design discharge of 2.4 m3/s:
D=
4Q
=
πV
4(2.4)
= 1.01 m
π(3.0)
(1)
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The largest available pipe size is 90 cm; therefore, two or more pipes are needed
in parallel for this culvert design
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For half the design discharge, 1.2 m3/s, the required diameter is:
D=
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4Q
=
πV
4(1.2)
= 0.71 m
π(3.0)
(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
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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
Sf =
Gary P. Merkley
Q2n2 Wp4 / 3
A10 / 3
2
2
4/3
1.2 ) ( 0.015 ) ( 2.51)
(
=
( 0.503 )10 / 3
244
= 0.011 m / m
(3)
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
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For critical flow, the Froude number is equal to unity:
Fr2
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=
Q 2T
gA 3
= 1.0
(4)
For circular pipes, the following definitions apply:
⎛ 2h ⎞
β = 2cos−1 ⎜ 1 −
⎟
D⎠
⎝
⎛β⎞
T = D sin ⎜ ⎟
⎝2⎠
and,
(5)
Wp =
βD
2
D2
A=
(β − sin β )
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(7)
Solve for depth, h, such that Fr2 = 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
1.2 ) ( 0.015 ) (1.83 )
(
( Sf )crit =
( 0.445 )10 / 3
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(6)
= 0.011 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
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245
Gary P. Merkley
5. Determine the Minimum Upstream Pipe Slope
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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
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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]
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246
BIE 5300/6300 Lectures
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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
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The approximate length of the upstream (steep) pipe is:
Lus =
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= 17.07 m
(9)
The approximate length of the downstream pipe is:
Lds =
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(16.60 )2 + (132.12 − 128.14 )2
( 31.23 − 16.60 )2 + (128.14 − 128.06 )2
= 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
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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)
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247
Gary P. Merkley
9. Specify Collar Placement and Size
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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