EXPEDIENT DRAINAGE - tecom.marines.mil

EXPEDIENT DRAINAGE - tecom.marines.mil

EXPEDIENT DRAINAGE OVERVIEW Plan and design adequate drainage Types of drainage systems Purpose of adequate drainage Maintaining a drainage system

OBJECTIVES Terminal Learning Objectives Enabling Learning Objectives METHOD / MEDIA

Lecture method Power point Demonstration Practical application EVALUATION Written exam

SAFETY / CEASE TRAINING Classroom Instruction No safety concerns for this period of instruction Inclement weather plan

Fire exit plan QUESTIONS? Are there any questions concerning: What will be taught? How it will be taught? How the student will be evaluated? SOURCES OF WATER

Precipitation Interception Infiltration

Ground Water PRECIPITATION Rain Fall Snow Fall/Melt

Humidity INTERCEPTION Interception is the process of vegetation absorbing the water before it reaches the soil.

Once the holding capacity of the vegetation has been reached, the soil will then start receiving water. INFILTRATION

Infiltration is the waters ability to penetrate the soil surface. The following factors affect the process of infiltration: Vegetation presence or lack there of. Soil type. (some soil types retain water more than others.) Slope of terrain. GROUND WATER

Surface water: Surface water is retained in the top soil. (depended upon vegetation and soil type.) Sub-surface water: Water that is present below the ground. (water table). Capillary water: The water that

seeps to the surface. QUESTIONS? Any questions? Questions for you!! ESTIMATING WATER RUNOFF

Methods of estimating water runoff Hasty Field Estimate HASTY METHOD The hasty method is used when an existing stream crosses or interferes

with your construction site. Certain measures must be taken to avoid possible water damage to your construction site. Using the following formula, we can determine the Area of Waterway (AW) HASTY METHOD AW = WI + W2 x 2 H

AW = Area of the waterway W1 = Width of the channel bottom W2 = Width at the high water mark HASTY METHOD W 2 W1

HT DRAINAGE SAFETY FACTOR ADES = 2AW ADES = Design cross section 2 = Safety Factor AW = Area of the waterway that was previously computed EXAMPLE # 1 7 + 9 x 4 = 32 Sqft (AW) 2

32 Sqft x 2 = 64 Sqft (Ades) EXAMPLE # 2 5 + 7 x 3 = 18 Sqft (AW) 2 18 Sqft x 2 = 36 Sqft (Ades) COMPLETE HANDOUTS 1 & 2 PRACTICAL APPLICATION REVIEW

Review handouts #1 and #2 Take a Break FIELD ESTIMATE METHOD The field estimate method is used to estimate the peak volume of storm water runoff.

Results of this method are adequate for determining the size of drainage structures for temporary drainage in areas of 100 acres or less. FORMULA Q = 2xAxRxC Q = peak volume of storm water runoff, in cubic feet per second

2 = safety factor (constant) A = area of drainage basin, in acres R = design rainfall intensity based on the one hour, two year frequency rainstorm, in inches per hour C = coefficient representing a ration of runoff to rainfall DRAINAGE AREA The fastest and most preferred method for determining the size of the drainage area is the stripper method



Measure the length of each line in the drainage area. Add all the lengths together This is the map area in square inches 2 1 . 2 2 6 . 2

.50 2.12 + 2.62 + .50 = 5.25 square inches CONVERSION (INCHES TO ACRES) For a more accurate determination, you

can draw the lines or apart from the base line. If spacing is used, you must take total length of lines and divide by 4. If spacing is used, you must take total length of lines and divide by 2.

CONVERSION (INCHES TO ACRES) Determine how many feet are in one inch on the map. Example: MAP Scale: 1 : 5,000 5000 12 = 416.67 ft.

1 inch on the map is 416.67 ft CONVERSION (INCHES TO ACRES) Determine how many square feet are in one square inch on the map. 416.67 = 173,613.88 One square inch on a map

contains 173,613.88 square feet on the ground CONVERSION (INCHES TO ACRES) Total square feet in the drainage area? 5.25 x 173,613.88 = 911,472.87 SqFt Now convert square feet to acres. 911,472.87 43,560 = 20.92 or

FORMULA Q = 2xAxR xC A = 21 Q = 2 x 21 x R x C DEMONSTRATION Example on page 6 of the student handout

Follow along with the demonstration Practical Application Perform the Practical Exercise Worksheet #1 RAINFALL INTENSITY RAINFALL INTENSITY

The Project is in Eastern North Carolina It falls between 1.5 and 2.0, always use the larger number. Formula Q = 2 x 21 x 2 x C

RUNOFF COEFFICIENT The ratio of runoff to rainfall. The amount of water expected to drain from an area as the result of a specific amount of rainfall.

It is expressed as a decimal. There are three primary factors that affect the percentage; Soil type Surface cover slope SOIL TYPE Porous soil - A large portion of the soil will infiltrate leading to a smaller runoff coefficient

Man made surfaces Like asphalt, concrete, and compacted gravel or macadam will result in a higher runoff coefficient SURFACE COVER

To use table 6-1, you need to understand the following terms Without Turf Is ground that is completely bare With Turf Is ground that is covered with vegetation. If the area has some vegetation but is not completely covered, use the higher without turf value

SLOPE As terrain becomes steeper, water flows sooner and more rapidly. This allows less time for infiltration to occur and results in the C value becoming larger for the natural cover or soil categories. USCS

Use the Unified Soil Classification System (USCS) to select the PREDOMINANT soil type. This will be needed for the left column of table 6-1 (the next slide).

If the area is wooded or covered with asphalt, concrete, gravel or macadam simply lookup the C value in the left hand column. FINDING THE RUNOFF COEFFICIENT SLOPE PERCENTAGE

Indentify the slope on the map. Find the difference from the top to the bottom of the slope SLOPE PERCENTAGE 181 B 180 Elevation B =181m

Elevation A =100m Difference in 81m elevation (Vd) Horizontal Distance =4150m 81_ X 100 4150

160 140 120 100 100 A Vd Hd X100=

=1.9% % of Slope TURF/SAFETY TURF: If the soil is not covered, determine whether the area is with or without turf

SAFETY: In all cases where you have more than one possible runoff coefficient, use the highest value C VALUES Slope < 2 % Soil or Cover Classification GW, GP, SW, SP w/turf

.10 w/o turf .20 Slope > 2 & < 7% w/turf w/o turf

.15 .25 Slope > 7% w/turf .20 w/o turf .30

GMd, SMd, ML, MH, Pt .30 .40 .35 .45 .40

.50 GMu, GC, SMu, SC CL, OL, CH, OH .55 .65 .60 .70

.65 .75 Wooded area .20 .20 .20 .20

.20 .20 Asphalt Pavement .95 .95 .95

Concrete Pavement .90 .90 .90 Gravel/macadam .70

.70 .70 RUNOFF COEFFICIENT (EXAMPLE) Your drainage area is made up of ML soil,

with 49% turf and a slope of 2%. Looking at Table 6-1 you should come up with 0.40. Now in final formula from Q = 2 x 21 x 2 x .40 The Answer : Q = 33.6 CFS WATERWAY AREA

Expedient culvert and ditch design is based on the waterway area. The hasty method deals with waterway area. The field estimate method deals with


For expedient purposes, you will always use a velocity of 4 fps for design of expedient drainage structures. Example Q = V x Aw (divide both sides by V) The Results are: Q V = Aw (Using the previous calculation from

your handout of 33.6 cfs) Final answer 33.6 (cfs) 4 (constant) = 8.4 sqft (Aw, Area of waterway) SAFETY FACTOR

As with the hasty method, you rarely design a drainage system to flow completely full. You must apply a safety factor (Ades) Ades = 2 x Aw


TRIANGULAR V-DITCHES Triangular (V) ditches are used to move small amounts of water. Q 60 cfs or Aw 15 sqft TRIANGULAR V-DITCHES

SYMMETRICAL Both sides of the ditch are inclined equally NON_SYMMETRICAL Each side of the ditch are inclined differently

Ensure the appropriate side-slope ratio is selected to serve its designed purpose. If the side walls are too step it invites excessive corrosion and ditch clogging. TRIANGULAR V-DITCHES Ditches have two sloped sides, with each

having a respective slope ratio. This is expressed as horizontal feet to vertical feet. Example: 3 : 1 is a side slope of 3 feet horizontal to a 1 foot vertical. (1:1) (3:1) TRIANGULAR V-DITCHES

BACK SLOPE FRONT SLOPE ROAD The sidewall of a ditch located adjacent to the shoulder is called the front slope of the ditch. The far slope, called the back slope, is simple an extension of the cut face of the excavation.

TRIANGULAR V-DITCHES FORMULA (DEPTH) D= Ca x 2 X+Y + 0.5 D = Ditch depth in feet. Rounded to two decimal places. Ca = Channel area computed previously.

X =Horizontal run of the front slope ratio. Y = Horizontal run of the back slope ratio. 0.5 = Safety factor constant. (1/2 foot freeboard) TRIANGULAR V-DITCHES FORMULA (WIDTH) Ditch Width: W = D x (X + Y) W = Ditch width in feet. Rounded to two decimal places. D = Ditch depth in feet. X = Front slope ratio. Y = Back slope ratio.

EXAMPLE Using your previous Ades of 16.8 sqft and a front slope of 3 : 1 and a back slope 1 : 1, calculate the depth and width of the ditch. D= 16.8 + 0.5 3+1 W = D x (X +Y) W = 2.55' x (3 + 1)

D= D= 16.8 + 0.5 4 4.2 + 0.5 D = 2.05 + 0.5 D = 2.55 W = 2.55 x 4 W = 10.20


Installed for larger runoff requirements, usually 60 cfps / 15 aw or greater. The designer of the ditch determines the bottom width based upon the cutting edge of the equipment used. CUTTING DEPTH

WIDTH OF DITCH FORMULA Ditch Depth: D = Aw + 0.5 W D = Ditch Depth in feet. Rounded to two decimals Ca = Channel area in square feet. W = Width of ditch (bottom) in feet. 0.5 = Safety factor constant. (1/2 foot of freeboard)

EXAMPLE With an AW of 18.75, using a D7G to excavate the ditch, determine the ditch depth. 18.75 aw 7.25 (D7 width) + .5 (freeboard) = 3.1 deep PRACTICAL APPLICATION

Trapezodial Ditch worksheet EROSION CONTROL EROSION CONTROL METHODS There are several methods of erosion control.

The desirable gradient for a ditch is between 05 and 2%. Ditches larger than 2% will require erosion control. Examples: Ditch Linings Check Dams DITCH LINING

May be lined to prevent erosion. Examples: Concrete Asphalt Rock Mortor Does not decrease the flow but protects the soil. Expensive and not always readily available

Grass Protects the soil, slow the flow and is cheap EXAMPLES CHECK DAMS CHECK DAMS Constructed with 6-8 diameter timbers.

Set 2 into the sides of the ditch. Weir notch is 6 deep and a minimum of 12 long. 4 of rock apron for every 1 of dam height.

The top of the check dam should be at the high water mark, when high water mark is not visible, place check dam 1 below the top of the ditch. DAM SPACING Will have a minimum spacing of 50 feet.

Should be placed as far apart as possible, while achieving the desired gradient. Spacing Calculations: S = 100 (H) AB S = Dam Spacing 100 = Constant H = Height of Dam A = Present Slope B = Desired Slope

DAM SPACING EXAMPLE What spacing will be needed for a 4 high check dam with a 10% slope. S = 4 x 100 10 2 S = 50 QUESTIONS CULVERTS

Two classifications Permanent (refer back to the Military Roads class) Expedient Different types of material used Corrugated metal Concrete Vitrified Clay (VC) Polyvinyl Chloride (PVC)

Timber Ect. CULVERTS Timber Box Good workmanship Large timber Strong enough to support heaviest vehicle traffic Minimum of 12 cover

Corrugated Metal Pipe Culvert (CMP) 8-72 diameter Shipped in 26 long half sections Bolted in every hole CULVERTS

Concrete pipe Comes in any size Comes in different shapes (circle, square, etc) Overall strength Smooth interior surface Higher amount of water flow Transportation considerations MAXIMUM ALLOWABLE CULVERT DIAMETER

Permanent culverts are selected based on their diameter. There are two maximum diameter (Dmax) equations. Fills greater than 36 inches Dmax = 2/3 x F

Fills less then 36 inches Dmax = F - 12 FILLS GREATER THAN 36 Dmax = 2/3 x Fill Dmax = Maximum culvert diameter in inches rounded to two decimal places. 2/3 = A constant that represents the minimum fill depth required for the maximum diameter of culvert to be calculated. Fill = Fill depth in inches rounded to two

decimal places. MAXIMUM ALLOWABLE CULVERT DIAMETER MAXIMUM ALLOWABLE CULVERT DIAMETER EXAMPLE Dmax = 2/3 x F F = 6 x 12 = 72 Dmax = 2/3 x 72 Dmax = 48 inches

PRACTICAL APPLICATION Complete the DMAX worksheet CULVERT MATERIALS Several Factors Economical Number Culvert

Order Diameter of pipe required Length Length ECONOMICAL DIAMETER

You want to save material. Put in the least amount of culverts. They need to equal or exceed the design area. Manpower requirements PIPES REQUIRED

To find the most economical size, you must divide the design area by the end area of several different pipe sizes. Use the largest pipe that satisfies the fill and cover requirements as a starting

point. Work your way down in size until the amount of pipes needed changes. Once changed, we have reached and passed our optimal design. Go back to the prior number and pipe demision. ECONOMICAL DIAMETER FORMULA N = Ades

PEA N = Number of Pipes Ades = Design Cross Section PEA = Pipe End Area, cross sectional end area of culvert in ft squared COMMON CULVERT SIZES Maximum Diameter () Cross Sectional Area (sqft)

12 -------------------------------------- 00.79 sqft 18 -------------------------------------- 01.77 sqft 24 -------------------------------------- 03.14 sqft 30 -------------------------------------- 04.91 sqft 36 -------------------------------------- 07.07 sqft 42 -------------------------------------- 09.62 sqft 48 -------------------------------------- 12.57 sqft 60 -------------------------------------- 19.64 sqft 72 -------------------------------------- 28.27 sqft EXAMPLE N48 N48

N48 N42 N42 N42 N36 N36 N36 = = = = =

= = = = Ades A48 17.5 12.57 = 1.4 or 2 (2) 48 Pipes Ades A42 17.5 9.62 = 1.8 or 2 (2) 42 Pipes Ades A36 17.5 7.07 = 2.5 or 3

(3) 36 Pipes CULVERT LENGTH Now that weve determined that we will need (2) 42 diameter culverts, we must now calculate the culvert length. Use the following formula to do so: (DL x SL) + ROADWAY WIDTH + (DR x SR) = CL Culvert Length Note: After calculating culvert length, ensure you round up to an even number.

EXAMPLE CL = ( 7 x 2 ) + 22 + ( 6 x 3 ) CL = 14 + 22 + 18 CL = 54 + 2 ( no headwalls on the exhaust end) CL = 56 ORDER FORMULA OL (order length) = CL x # of pipes x 1.15 (waste) OL = ( 56 X 2 ) 1.15 OL = ( 112 ) 1.15 OL = 128.8 or 130 of pipe needed


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