Design of Culverts

1
Design of Culverts

• CE154 Hydraulic Design
• Lectures 8-9

2
Culverts

• Definition - A structure used to convey surface
  runoff through embankments.
• It may be a round pipe, rectangular box, arch,
  ellipse, bottomless, or other shapes.
• And it may be made of concrete, steel, corrugated
  metal, polyethylene, fiberglass, or other
  materials.

3
Culverts

• End treatmentincludes projected, flared, head
  and wing walls

4
(No Transcript)
5
Concrete Box Culvert
6
Box culvert with fish passage
7
Corrugated metal horseshoe culvert
8
Bottomless culvert USFW
9
Some culvert, huh?
10
Culvert or Bridge?
11
Study materials

• Design of Small Dams (DSD) pp. 421429 (culvert
  spillway), 582-589 (hydraulic calculation charts)
• US Army Drainage Manual (ADM),TM 5-820-4/AFM
  88-5, Chapter 4, Appendix B - Hydraulic Design
  Data for Culverts

12
Study Objectives

• Recognize different culvert flow conditions
• Learn the steps to analyze culvert hydraulics
• Learn to design culverts

13
Definition Sketch
14
Definition Sketch
15
Relevant technical terms

• Critical depthThe depth at which the specific
  energy (yv2/2g) of a given flow rate is at a
  minimum
• Soffit or crownThe inside top of the culvert
• Invert thalwegChannel bottom lowest point of
  the channel bottom
• Headwater The water body at the inlet of a
  culvert

16
Relevant technical terms

• TailwaterThe water body at the outlet of a
  culvert
• Submerged outletAn outlet is submerged when the
  tailwater level is higher than the culvert
  soffit.

17
Relevant technical terms

• Inlet controlOccurs when the culvert barrel can
  convey more flow than the inlet will accept. The
  flow is only affected by headwater level, inlet
  area, inlet edge configuration, and inlet shape.
  Factors such as roughness of the culvert barrel,
  length of the culvert, slope and tailwater level
  have no effect on the flow when a culvert is
  under inlet control.

18
Relevant technical terms

• Outlet controlOccurs when the culvert barrel can
  not convey more flow than the inlet can accept.
  The flow is a function of the headwater
  elevation, inlet area, inlet edge configuration,
  inlet shape, barrel roughness, barrel shape and
  area, slope, and tailwater level.

19
Relevant technical terms

• Normal depthOccurs in a channel reach when the
  flow, velocity and depth stay constant. Under
  normal flow condition, the channel slope, water
  surface slope and energy slope are parallel.
• Steep slopeOccurs when the normal depth is less
  than the critical depth. The flow is called
  supercritical flow.

20
Relevant technical terms

• Mild slopeOccurs when the normal depth is higher
  than the critical depth. The flow is called
  subcritical flow.
• Submerged inletAn inlet is submerged when the
  headwater level is higher than approximately 1.2
  times the culvert height D. (Why is it not simply
  higher than 1.0 times D?)

21
Relevant Technical Terms

• FreeboardSafety margin over design water level
  before overflow occurs (in a unit of length)
• Free outletAn outlet condition at which the
  tailwater level is below the critical depth,
  whence further lowering of the tailwater will not
  affect the culvert flow

22
Design Setting

• a river
• a plan to build a road crossing
• need to design the road crossing - given river
  slope, geometry, design flood - given
  desirable roadway elevation - design culvert
  (unknown size) to pass Design Flood with
  suitable freeboard (design criteria)

23
Analysis Setting

• An existing culvert or bridge (known size)
• a river passing underneath
• determine water level under certain flood
  condition or vice versa

24
Inlet control (1)
25
Inlet control (2)
gt
26
Inlet control (3) sharp edge inlet
27
Outlet control (1)
28
Outlet control (2)
29
Outlet control (3)
30
Outlet control (4)
31
Intermittent control
32
Key Approaches

• Critical flow does not occur on mild slopes,
  except under certain special, temporary condition
  such as inlet control (3)
• Critical flow always occurs at the inlet of a
  steep slope, except when the inlet is deeply
  submerged H/D gt 1.2-1.5
• On mild slopes, most likely its outlet control

33
Approaches

• For unsubmerged inlet control, - for culvert on
  steep slope, use critical flow condition to
  determine the discharge- for culvert on mild
  slope, use weir equation to compute flow
• For submerged inlet control, use orifice flow
  equation to compute discharge
• For outlet control, perform energy balance
  between inlet and outlet

34
Critical Flow Condition

• yc (q2/g)1/3
• Fr vc/(gyc)1/2 1
• vc (gyc)1/2
• Ec yc vc2/2g 3/2 yc
• q unit discharge Q/width (for non-circular
  conduit for circular pipe use table to find
  critical condition)Fr Froude numberE
  specific energyy depthc subscript denotes
  critical flow condition

35
Weir Flow

• Weir flow equation
• B culvert widthCw weir discharge
  coefficient, aninitial estimate may be 3.0note
  that this eq. is similar to equations for ogee
  crest weir, broadcrested weir, sharp crest weir

36
Orifice Flow

• Inlet control with submerged inlet,
• Cd orifice discharge coefficient, an initial
  estimate ? 0.60
• b culvert height
• HW-b/2 average head over the culvert

37
Outlet control hydraulics

• Energy balance between inlet and outlet

38
Outlet control hydraulics

• Entrance loss coefficient on p.B-12 of ADM and p.
  426 454 of Design of Small Dam
• Exit loss coefficient as a function of area
  change from the culvert (a1) to downstream
  channel (a2)Kex (1- a1/a2)2 1 for outlet
  into reservoir
• Friction loss coefficient may be computed using
  Darcy-Weisbach or Manning equation

39
Outlet control hydraulics

• Darcy-Weisbach equation for circular
  pipesfriction head loss hf f L/D V2/2gor
  for non-circular channels, using hydraulic radius
  RA/PD/4 to replace D hf f L/(4R)
  V2/2g kf f L/(4R)

40
Outlet control hydraulics

• Mannings equation to compute friction lossv
  (1.49 R2/3 S1/2) / nS v2 n2 / (2.22 R4/3)hf
  SL v2/2g (29.1 n2L/R4/3)kf 29.1 n2L/R4/3
  - see Eq. on p. B-1

41
Design Procedure

1. Establish design criteria - Q, HWmax, and other
   design data L, S, TW, etc.
2. Determine trial size (e.g., AQ/10)
3. Assume inlet control, compute HW-unsubmerged,
   weir flow eq.-submerged, orifice flow eq.
4. Assume outlet control, compute HW
5. Compare results of 3 4. The higher HW governs.
6. Try a different size until the design criteria
   are met

42
Example (1)

• A circular corrugated metal pipe culvert, 10 in
  diameter, 50 long, square edge with headwall, on
  slope of 0.02, Mannings n0.024, is to convey
  flood flow of 725 cfs. Tailwater is at the
  center of the culvert outlet. Determine the
  culvert flow condition.
• Assuming first if the slope is steep, inlet
  control. If mild, outlet control.
• Determine if the slope is steep or mild by
  comparing normal and critical flow depth, e.g.
  tables from Design of Small Dams (DSD)

43
(No Transcript)
44
(No Transcript)
45
Example (1)

• Q 725, n 0.024, D 10 ft, S 0.02
• Qn/(D8/3S1/2) 0.265
• Table B-3, it corresponds to d/D 0.541, or the
  normal depth dn 5.41 ft
• Q/D2.5 2.293
• From Table B-2, find d/D 0.648, or the critical
  depth dc 6.48 ft
• dc gt dn, so the 0.02 slope is steep ? inlet
  control
• Critical flow occurring at the culvert entrance
• Use Figure 9-68 (or Figure B-8 of DSD p.585) for
  circular culverts on steep slope to determine
  headwater depth

46
Example (1)
47
Example (1)

• For Q/D2.5 2.293, and square edge inlet, Curve
  A on figure 9-68 shows
• H/D 1.0
• The headwater is at the culvert soffit level, and
  it drops to 6.48 ft at the inlet and continues to
  drop to 5.41 ft to flow through the culvert,
  before dropping to 5 ft at the outlet.

48
(No Transcript)
49
Example (2)

• Concrete pipe (n0.015) culvert 10 ft in
  diameter, 0.02 slope, square edge, vertical
  headwall, Q 1550 cfs, tailwater at pipe center
  at outlet. Determine the culvert flow condition.
• Q/D2.5 1550/(10)2.5 4.90
• Qn/(D8/3S1/2) 0.35
• dn determined from Table B-3, d/D0.65
• dc determined from Table B-2, d/D 0.913

50
Example (2)

• The culvert will run open-channel, same as in
  Example (1) and the water level drops to the pipe
  center level at the outlet.
• To compute headwater level, Figure 9-68 shows
  that H/D 2.15
• The culvert entrance will be submerged, with
  water level dropping to dc 9.13 ft at the inlet
  and continues dropping to dn 6.5 ft for the
  bulk length of the pipe.

51
Example (3)

• Same condition as in Example (1), with corrugated
  pipe 10 ft diameter, S0.02, L300 ft, tailwater
  level at pipe center, Q2000 cfs. Determine flow
  condition.
• Q/D2.5 2000/(10)2.5 6.32
• Qn/(D8/3S1/2) 20000.024/(65.4) 0.73
• Critical depth at 9.65 ft, practically full flow
• Normal depth shows full flow since data is out
  of range of table ? outlet control

52
Example (3)

• Calculate entrance loss coefficient square edge
  flush with vertical headwall(p.426) Ken 0.5
• Calculate exit loss coefficient tailwater at
  pipe centerline, outlet channel is not supported,
  full exit velocity head is lost Kex 1.0

53
Example (3)

• Calculate friction loss coefficient R A/P
  D/4 2.5n 0.024Kf 29.1 n2 L / R4/3 1.48
• Eq. (32) on p. 425 shows thatH/D L/D So 0.5
  0.0252(kex ken kf)(Q/D5/2)2H/10 300.02
  0.5 0.0252 (10.51.48)(6.32)2
• H 29 ft
• Check using Figure B-10 of Design of Small Dams
  or Figure B-13 of Reader graphical solution
  shows H32

54
(No Transcript)
55
Example (4)

• Design a culvert for the following condition-
  Design Flow Q 800 cfs- culvert length L 100
  ft- Allowable headwater depth HW 15 ft-
  Concrete pipe culvert- Slope S 0.01 (1.0)-
  Tailwater level (TW) at 0.8D above invert at
  outlet

56
Example (4)

1. Select a trial culvert pipe sizeAssuming culvert
   flow velocity V 10 fpsA Q/V 800/10
   80 ft2D sqrt(80?4/?) 10.1 ftSay D 10 ft

57
Example (4)

• Assuming inlet control- using rounded inlet to
  reduce headloss- Q/D5/2 2.53- From Figure
  9-68 of DSD, H/D 1
• This is a conservative design. Reasonably H/D
  could be designed as high as 1.2 to maintain
  un-submerged inlet condition.
• Check by using Figure B-7 of DSD. The rounded
  inlet is similar to groove inlet (see Table B-1
  of ADM)

58
Example (4)

1. Assuming outlet control- First determine the
   outlet flow condition. From Table B-2 of DSD, at
   Q800 cfs, Q/D5/22.53, the critical depth
   dc0.682D. Hence, TW0.8D is above the critical
   level.The normal flow is determined from Table
   B-3 of DSD. Use n0.018 for aged concrete.
   Qn/D8/3S1/20.31dn0.6DThe normal flow depth is
   6.0 ft in the culvert

59
Example (4)

• The normal flow condition is- from Table B-3
  again, A/D2 0.492- An 49.2 ft2- Vn Q/An
  16.3 fps- R hydraulic radius 0.278D 2.78
  ft- Fr Froude number V/(gR)1/2 1.7- This
  shows that flow is supercritical in the culvert.
  It transitions to the tailwater depth at the
  outlet (S3 or jump). TW flow may be
  supercritical or subcritical, depending on the
  downstream slope.

60
Example (4)

• To compute the headloss of the outlet-control
  condition HW SoL HL TW HL (Ken Kex
  Kf)V2/2g Ken 0.2 for rounded edge with
  headwall Kex 1.0 being conservative since not
  all the velocity head is lost (draw
  profile) Kf 29n2L/R1.333 0.24

61
Example (4)

• HL (1 0.24 0.2) V2/2g 1.44
  (16.3)2/64.4 5.94 ft
• The energy balance equation becomes HW HL TW
  SoL 5.94 8 0.01?100 12.94
• HW/D 12.94/10 1.3
• Compare the headwater depth for inlet and outlet
  conditions, select the higher value for design.

62
Example (4)

• The governing headwater depth is 13 ft
• This is less than the maximum of 15 ft of the
  allowable headwater depth. Hence, it is
  acceptable. The culvert size may be reduced
  slightly to reduce cost and still meets design
  criteria.Hence, use 10 ft diameter concrete
  pipe rounded edge at inlet maximum headwater
  depth 13 ft

63
Culvert failure modes along forest roads in
northern CA
64
Design Considerations

• Flared ends improve efficiency
• Use culverts as wide as stream width
• Use same gradient as stream channel
• Use same alignment as stream channel
• Single large culvert is better for debris passage
  than several small ones