Design of Open Channels and Culverts

1. Design of Open Channels and Culverts

2. Transverse Slopes • Removes water from pavement surfaces in shortest amount of time possible

3. Longitudinal Slopes • Gradient longitudinal direction of highway to facilitate movement of water along roadway

4. Drains • Along ROW • Collect surface water A typical intercepting drain placed in the impervious zone http://www.big-o.com/constr/hel-cor.htm

5. Drainage Channels (Ditches) • Design • Adequate capacity • Minimum hazard to traffic • Hydraulic efficiency • Ease of maintenance • Desirable design (for safety): flat slopes, broad bottom, and liberal rounding

6. Ditch Shape • Trapezoidal – generally preferred considering hydraulics, maintenance, and safety • V-shaped – less desirable from safety point of view and maintenance

8. Flow Velocity • Depends on lining type • Typically 1 to 5% slopes used • Should be high enough to prevent deposit of transported material (sedimentation) • For most linings, problem if S < 1% • Should be low enough to prevent erosion (scour) • For most types of linings, problem if S > 5%

9. Use spillway or chute if Δelev is large

10. Rip Rap for drainage over high slope

11. Side Ditch/Open Channel Design-Basics • Find expected Q at point of interest (see previous lecture) • Select a cross section for the slope, and any erosion control needed • Manning’s formula used for design • Assume steady flow in a uniform channel

12. Manning’s Formula V = R2/3*S1/2(metric) V = 1.486 R2/3*S1/2 n n where: V = mean velocity (m/sec or ft/sec) R = hydraulic radius (m, ft) = area of the cross section of flow (m2, ft2) divided by wetted perimeter (m,f) S = slope of channel n = Manning’s roughness coefficient

13. Side Ditch/Open Channel Design-Basics Q = VA Q = discharge (ft3/sec, m3/sec) A = area of flow cross section (ft2, m2) FHWA Hydraulic Design Charts FHWA has developed charts to solve Manning’s equation for different cross sections

15. Open Channel Example • Runoff = 340 ft3/sec (Q) • Slope = 1% • Manning’s # = 0.015 • Determine necessary cross-section to handle estimated runoff • Use rectangular channel 6-feet wide

16. Open Channel Example Q = 1.486 R2/3*S1/2 n • Hydraulic radius, R = a/P • a = area, P = wetted perimeter P

17. Open Channel Example • Flow depth = d • Area = 6 feet x d • Wetted perimeter = 6 + 2d Flow depth (d) 6 feet

18. Example (continued) Q = 1.486 a R2/3*S1/2 n 340 ft3/sec = 1.486 (6d) (6d)2/3 (0.01)1/2 (6 + 2d) 0.015 d  4 feet Channel area needs to be at least 4’ x 6’

19. Example (continued) Find flow velocities. V = 1.486 R2/3*S1/2 n with R = a/P = 6 ft x 4 ft = 1.714 2(4ft) + 6ft so, V = 1.486(1.714)2/3 (0.01)1/2 = 14.2 ft/sec 0.015 If you already know Q, simpler just to do V=Q/A = 340/24 = 14.2)

21. Example (continued) Find critical velocities. From chart along critical curve, vc 13 ft/sec Critical slope = 0.007 Find critical depth: yc = (q2/g)1/3 g = 32.2 ft/sec2 q = flow per foot of width = 340 ft3/sec /6 feet = 56.67ft2/sec yc = (56.672/32.2)1/3 = 4.64 feet > depth of 4’

22. Check lining for max depth of flow …

24. Rounded

25. A cut slope with ditch

26. A fill slope

27. Inlet or drain marker

28. Ditch treatment near a bridge US 30 – should pier be protected?

29. A fill slope

32. Median drain

33. Culvert Design - Basics • Top of culvert not used as pavement surface (unlike bridge), usually less than 20 foot span • > 20 feet use a bridge • Three locations • Bottom of Depression (no watercourse) • Natural stream intersection with roadway (Majority) • Locations where side ditch surface drainage must cross roadway

34. Hydrologic and Economic Considerations • Alignment and grade of culvert (with respect to roadway) are important • Similar to open channel • Design flow rate based on storm with acceptable return period (frequency)

35. Culvert Design Steps • Obtain site data and roadway cross section at culvert crossing location (with approximation of stream elevation) – best is natural stream location, alignment, and slope (may be expensive though) • Establish inlet/outlet elevations, length, and slope of culvert

36. Culvert Design Steps • Determine allowable headwater depth (and probable tailwater depth) during design flood – control on design size – f(topography and nearby land use) • Select type and size of culvert • Examine need for energy dissipaters

37. Headwater Depth • Constriction due to culvert creates increase in depth of water just upstream • Allowable level of headwater upstream usually controls culvert size and inlet geometry • Allowable headwater depth depends on topography and land use in immediate vicinity

38. Types of culvert flow • Type of flow depends on total energy available between inlet and outlet • Inlet control • Flow is controlled by headwater depth and inlet geometry • Usually occurs when slope of culvert is steep and outlet is not submerged • Supercritical, high v, low d • Most typical • Following methods ignore velocity head

40. Ans. Example: Design ElevHW = 230.5 Stream bed at inlet = 224.0 Drop = 6.5’ Flow = 250cfs 5x5 box HW/D = 1.41 HW = 1.41x5 = 7.1’ Need 7.1’, have 6.5’ Drop box 0.6’ below stream - OK

41. Types of culvert flow • Outlet control • When flow is governed by combination of headwater depth, entrance geometry, tailwater elevation, and slope, roughness, and length of culvert • Subcritical flow • Frequently occur on flat slopes • Concept is to find the required HW depth to sustain Q flow • Tail water depth often not known (need a model), so may not be able to estimate for outlet control conditions

43. Example: Design ElevHW = 230.5 Flow = 250cfs 5x5 box Stream elev at inlet = 240 200’ culvert Outlet invert = 240-0.02x200 = 220.0’ Given tail water depth = 6.5’ Check critical depth = 4.3’ from fig. 17.23 (next page) Depth to hydraulic grade line = (dc+D)/2 = 4.7 < 6.5, use 6.5’ Head drop = 3.3’ (from chart) 220.0+6.5+3.3 = 229.8’<230.5 OK