95334283 Roadside Ditch Design
January 20, 2018 | Author: Bilal Ahmed Barbhuiya | Category: N/A
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Roadside Ditch Design...
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4/24/2012
CEE453 Urban Hydrology and Hydraulics Unit V: Design of Hydraulic Structures Lecture 26
Roadside Ditch Design Part I April 24, 2012 (Tue)
Roadside Ditch Design • Primary function to collect runoff from the highway right of way and tributary areas adjacent to the right of way, and to transport this accumulated water to an acceptable outlet point
• Secondary function to drain the base of the roadway to prevent saturation and loss of support for the pavement
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Roadside Ditch Design • Non‐traversable drainage ditches can be safety hazard.
http://safety.fhwa.dot.gov/local_rural/training/fhwasa09024/
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Roadside Ditch Design • Standard ditch ‐ Trapezoidal shape with a defined bottom width and sideslopes 1 ‐ d ‐ ‐
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Roadside Ditch Design • Hydrology ‐ Runoff is from primarily from overland flow → Rational method (peak runoff) ‐ Significant flow contribution from one or more defined watercourses → Other methods (regression equation from USGS, SCS method, HEC)
• Hydraulic ‐ Uniform flow assumption: Manning’s equation and Continuity equation 1.49
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Roadside Ditch Design • General Steps 1. Determine the standard or typical ditch cross sections. 2. Establish a ditch plan, which shows the proposed ditch flow patterns. 3. Determine the gradients to be used on all proposed ditches. 4. Investigate the capacity of the typical ditch with the proposed gradients and enlarge any ditch found to be inadequate. 5. Determine the limits and degree of protection necessary to prevent erosion in the ditch system. 6. Determine any special measures necessary to prevent adverse effects at and downstream from ditch outlets points. 6
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Roadside Ditch Design 1. Determine the standard or typical ditch cross sections. ‐
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Side slopes should not exceed the angle of the repose. Generally it should be 3:1 or flatter without consideration of the stability of the side slopes. 4:1 or flatter for the safety. Slopes steeper than 2:1 will usually require special erosion control measures. Shoulder widths: 8 to 10 ft outside (more details in the tables in Chapter 44 through 50, BDE)
FREEWAY AND EXPRESSWAY NOMENCLATURE Figure 34‐1.A (BDE)
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Roadside Ditch Design 2. Establish a ditch plan, which shows the proposed ditch flow patterns. ‐ ‐
To determine based on topography, drainage divides, flow directions, etc. Otherwise some arbitrary value from 100 to 200 ft can be used.
Sample Ditch Plan Figure 9‐201 (IDOT Drainage Manual) 8
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Roadside Ditch Design 3. Determine the gradients to be used on all proposed ditches. ‐ ‐ ‐
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To minimize ponding and silt accumulation, a grade of 0.3 % should be provided. Between 0.4 and 0.6 % is desirable. Channel gradients greater than 2% (up to 10%) may require the use of flexible linings. There’s no upper limit on ditch grade but steeper grade would cause greater expense for erosion control (riprap, wire‐enclosed riprap). Frequent breaks in the grade usually reduce earthwork and ditch lining cost. The desirable distance between breaks in rugged terrain is 100 ft and several thousands ft for flat land.
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Roadside Ditch Design 4. Investigate the capacity of the typical ditch with the proposed gradients and enlarge any ditch found to be inadequate. 1) Compute the design discharge at the downstream end of the section of ditch. (50 year design storm) 2) Select a trial size for the ditch at this location (the standard ditch is normally used initially), and assign a roughness coefficient for the finished surface of the ditch. Use Table 9‐403 (IDOT Drainage Manual) for n values. 3) Determine the maximum allowable depth of flow in the ditch. The discharge should be confined within the ditch and provide 1’ freeboard below the shoulder of the highway. 4) Check the capacity of this ditch. 5) If the initial ditch is too small, select a larger size or increase the capacity by other means and return to step 2). 10
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Roadside Ditch Design 4. Example problem – ditch capacity Given: An unlined trapezoidal ditch with a 2' bottom width, 2:1 sides and a gradient of 0.5 percent (0.005 ft/ft) is proposed to carry a discharge of 95 ft3/sec. The maximum allowable depth of flow has been set at 2.5 ft. Find: 1. The maximum capacity of this ditch. 2. The bottom width which should be used to meet the depth of flow limitation.
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Roadside Ditch Design 5. Determine the limits and degree of protection necessary to prevent erosion in the ditch system. • 10 year as design storm frequency is used for ditch lining. (More details in 9‐5 of IDOT Drainage Manual)
http://www.interstatelandscaping.com/Page_4.html
6. Determine any special measures necessary to prevent adverse effects at and downstream from ditch outlets points.
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Reference • IDOT Drainage Manual, Chapter 9. Roadside Ditches. • Mays, L.W., Hydraulic Design Handbook, Chapter 13, Hydraulic Design of Drainage for Highways. • Illinois Bureau of Design and Environmental Manual (BDE), Chapter 34. Cross Section Elements.
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CEE453 Urban Hydrology and Hydraulics Unit V: Design of Hydraulic Structures Lecture 26
Roadside Ditch Design Part II April 24, 2012 (Tue)
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Roadside Ditch Design – Scour Design
http://safety.fhwa.dot.gov/local_rural/training/fhwasa09024/
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Maximum Velocity Method
© 2010 Arthur R. Schmidt All Rights Reserved
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Procedure 1. 2. 3. 4. 5. 6. 7. 8.
Estimate n Determine Vmax Calculate R from Mannings to give V = Vmax Guess b, calculate y, A Calculate Q = VA If Q ≠ Qdesign back to 4 Estimate required freeboard Summarize results with dimensioned sketch
© 2010 Arthur R. Schmidt All Rights Reserved
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Example—Maximum velocity Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.005
1 3 b
From table, n = 0.020, Vmax = 3.5 ft/sec
Solve for b and y that gives V=Vmax and Q = 30 b =11.8, y = 0.63, giving A = 8.57
© 2010 Arthur R. Schmidt All Rights Reserved
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Maximum Tractive Force
Force on bed = ALtan() or, for small , ALSo Force per unit bed area = o = ALSo / PL = RSo For wide channel (R ≈ y), o = ySo © 2010 Arthur R. Schmidt All Rights Reserved
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Tractive force—bottom ySo
© 2010 Arthur R. Schmidt All Rights Reserved
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Tractive force on sides
≈0.76 ySo
© 2010 Arthur R. Schmidt All Rights Reserved
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Resisting Force Bottom Tractive force = AeL where Ae is the exposed area and L is the unit tractive force on a level surface. Resisting force = Ws tan() where Ws is the submerged weight and is the angle of repose. AeL = Ws tan() or
L = (Ws / Ae )tan()
© 2010 Arthur R. Schmidt All Rights Reserved
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Resisting Force Sides
Ws cos tan Critical shear: s Tractive Force Ratio:
K
W sin A 2
s
2
e s
tan 2 Ws cos tan 1 Ae tan 2
tan 2 sin 2 s cos 1 1 2 2 L tan sin
© 2010 Arthur R. Schmidt All Rights Reserved
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© 2010 Arthur R. Schmidt All Rights Reserved
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© 2010 Arthur R. Schmidt All Rights Reserved
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© 2010 Arthur R. Schmidt All Rights Reserved
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© 2010 Arthur R. Schmidt All Rights Reserved
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Procedure 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
Estimate n, angle of repose () Estimate channel sinuosity and correction factor Assume bottom width, b Assume depth y Determine R, Determine =RS If > max go to 4 Calculate Q from Mannings If Q ≠ Qdesign back to 3 Calculate tractive force on sides S Calculate maximum permissible tractive force on sides smax If S > Smax go to 3 Estimate required freeboard Summarize results with dimensioned sketch
© 2010 Arthur R. Schmidt All Rights Reserved
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Example—Maximum Shear Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.005 (assume angle of repose = 25o) 1 3 b From table, n = 0.020, max = 0.15 lb/ft2
Solve for b and y that gives = max and Q = 30 b =16.01, y = 0.53, giving A = 9.31, V = 3.22 b/y = 30 so Level ≈ ySo; sides ≈ 0.76 ySo = 0.11 lb/ft2 Permissible side shear = 0.667; max = 0.10 lb/ft2 Side shear > permissible—try new solution © 2010 Arthur R. Schmidt All Rights Reserved
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Example—Maximum Shear Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.005 (assume angle of repose = 25o) 1 3 b From table, n = 0.020, max = 0.15 lb/ft2; max side shear = 0.10 lb/ft2
Solve for b and y b =21.22, y = 0.45, giving A = 10.15 ft2, V = 2.95 ft/sec b/y = 47 so Level ≈ ySo = 0.13 lb/ft2; sides ≈ 0.76 ySo = 0.10 lb/ft2 Permissible side shear = 0.667; max = 0.10 lb/ft2 This is good solution © 2010 Arthur R. Schmidt All Rights Reserved
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Vegetated Channels Resistance n is a function of vegetation type, depth, and velocity Use maximum velocity approach
© 2010 Arthur R. Schmidt All Rights Reserved
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© 2010 Arthur R. Schmidt All Rights Reserved
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© 2010 Arthur R. Schmidt All Rights Reserved
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Procedure 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
Assume n, determine corresponding VR Select permissible velocity corresponding to slope, soil, vegetation Calculate R from 1 & 2 Using Manning eqn. and n compute Repeat steps 1‐4 until values for VR from steps 1 & 4 agree Determine A from design flow and permissible velocity, Determine channel proportions for calculated values of A & R Assume depth and compute A & R Calculate V = Q/A Compute VR Use results of 10 to compute n from figure Use n from 11 and R from 8 and Manning’s eqation to compute V Compare velocities from 9 and 12 and repeat steps 8‐12 until approx. equal Estimate required freeboard Summarize results with dimensioned sketch © 2010 Arthur R. Schmidt All Rights Reserved
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Example—Vegetated Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.04
1 3 b From Maximum velocity
From Maximum shear
Solve for b and y that gives V=Vmax and Q = 30 b =74, y = 0.12, giving A = 8.57, V = 3.5 ft/sec
max = 0.15 lb/ft2; max side shear = 0.10 lb/ft2 b =272, y = 0.05, giving A = 14.4, V = 2.09 ft/sec = 0.13, side = 0.10
© 2010 Arthur R. Schmidt All Rights Reserved
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Example—Vegetated Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.04
1 3 b Use Bermuda grass sometimes mowed to 2.5”, sometimes unmowed
© 2010 Arthur R. Schmidt All Rights Reserved
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Example—Vegetated Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.04 1 3 b Use Bermuda grass sometimes mowed to 2.5”, sometimes unmowed Maximum velocity = 6.0 ft/sec Gives A = Q/Vmax = 5 ft2 Start with mowed—Class D retardance; Guess n=0.05 R = 1.01 VR = 6.06, n = 0.033
© 2010 Arthur R. Schmidt All Rights Reserved
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Example—Vegetated Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.04 1 3 b Use Bermuda grass sometimes mowed to 2.5”, sometimes unmowed Maximum velocity = 6.0 ft/sec Gives A = Q/Vmax = 5 ft2 Start with mowed—Class D retardance; Guess n=0.05 R = 1.01 VR = 6.06, n = 0.033 R = 0.54, VR = 3.25, n = 0.038 R = 0.67, VR = 4.02, n = 0.036 R = 0.62, VR = 3.70, n = 0.037 R = 0.64, VR = 3.86, n = 0.037 © 2010 Arthur R. Schmidt All Rights Reserved
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Example—Vegetated Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.04 1 z=3 b Use Bermuda grass mowed to 2.5, Vmax = 6.0 ft/sec A = Q/Vmax = 5 ft2 ; R = 0.64
Solving gives b = 6.46 ft, y = 0.22 ft Check: this gives Q = 30 ft3/sec, V = 6 ft/sec © 2010 Arthur R. Schmidt All Rights Reserved
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Example—Vegetated Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.04 1 3 b Now solve for Bermuda grass unmowed, retardance class B Because unmowed, depth will be greater (more resistance) Use bottom width from mowed (b=6.46 ft) Guess depth, solve for A, R, V = Q / A, VR, & n = f(VR) y = 0.20, A =4.26, V = 7.04, R = 0.55, VR = 3.88, n = 0.068, Q = 12.6 CFS y = 0.30, A =8.61, V = 3.48, R = 1.03, VR = 3.59, n = 0.070, Q = 37.2 CFS y = 0.28, A =7.50, V = 4.00, R = 0.91, VR = 3.66, n = 0.070, Q = 30.0 CFS
© 2010 Arthur R. Schmidt All Rights Reserved
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Example—Vegetated Q = 30 CFS, flow containing solids, ordinary firm loam soil, slope So = 0.04 1 3 b
When grass is mowed, will flow at y = 0.22 ft, V = 6.0 ft/sec When unmowed, will flow at y = 0.28 ft, V = 4.0 ft/sec Add 1.0 ft freeboard, so need grass‐lined trapezoidal channel 1.3 ft deep, bottom width = 6.5 ft, top width = 14.3 ft
© 2010 Arthur R. Schmidt All Rights Reserved
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