30 Min Culvert Design30 Min Culvert Design

Jered CarpenterJered Carpenter

Region 3 Roadway DesignRegion 3 Roadway Design

Design StepsDesign Steps

Step 1) Step 1) Gather hydrological & site Gather hydrological & site specific dataspecific data

Step 2)Step 2) Assume culvert sizeAssume culvert size Step 3)Step 3) Find headwater depth for trial sizeFind headwater depth for trial size Step 4)Step 4) Try another sizeTry another size Step 5)Step 5) Compute outlet velocityCompute outlet velocity Step 6)Step 6) Determine Invert ProtectionDetermine Invert Protection Step 7)Step 7) Document selectionDocument selection

Del Rio ExampleDel Rio Example

Pictures of Existing CulvertPictures of Existing Culvert

Step 1) Gather Hydrological DataStep 1) Gather Hydrological Data

Q = cia = (.35*52*.82) = 14.9 cfsQ = cia = (.35*52*.82) = 14.9 cfs

Culvert Location

Step 1 Cont.) Site Specific DataStep 1 Cont.) Site Specific Data

Invert In El. = 564.9’Invert In El. = 564.9’ Invert Out El. = 564.5’Invert Out El. = 564.5’ Pipe Length = 75’Pipe Length = 75’ Slope = -0.5%Slope = -0.5%

Step 1 Cont.) Site Specific DataStep 1 Cont.) Site Specific Data

Calculate Tail Water, TWCalculate Tail Water, TW• 2.5’ btm, class 50 rip-rap lined FBD with 2:1 side slopes @ outlet2.5’ btm, class 50 rip-rap lined FBD with 2:1 side slopes @ outlet• Manning’s Equation for dischargeManning’s Equation for discharge

Q = (1.486/n)*A*RQ = (1.486/n)*A*R2/32/3SS1/21/2

• Using a spreadsheet & setting Using a spreadsheet & setting Q as a function of depth, I get Q as a function of depth, I get the following results when the following results when d=1.7 ft:d=1.7 ft:

15.0 cfs = (1.486/0.07)*10.03ft15.0 cfs = (1.486/0.07)*10.03ft22*0.99ft*0.99ft2/32/30.050.051/21/2

TW = 1.7’TW = 1.7’

Key #15186 – Del Rio “DR” 92+15 Jered Carpenter 11 Feb 08 1 of 1

Rational

52 Ac

-0.5%

Trapezoid

50

100

14.9

16.4

564.9

75564.5

1.7

1.8

Step 2) Assume Culvert SizeStep 2) Assume Culvert Size

Start With 18” Circular CMP Start With 18” Circular CMP

Step 3) Find HW Depth For Trial SizeStep 3) Find HW Depth For Trial Size

Assume inlet controlAssume inlet control• From Nomograph, From Nomograph,

HW/D = 3HW/D = 3

HW depth w/ inlet controlHW depth w/ inlet control

approx. 569.4 ft., whichapprox. 569.4 ft., which

is too highis too high

I’ll try a 24” pipeI’ll try a 24” pipe

Step 3) Find HW Depth For Trial SizeStep 3) Find HW Depth For Trial Size

Assume inlet controlAssume inlet control• From Nomograph, From Nomograph,

HW/D = 1.2HW/D = 1.2

HW depth w/ inlet controlHW depth w/ inlet control

approx. 566.7 ft., whichapprox. 566.7 ft., which

is OK.is OK.

The 24” pipe looks The 24” pipe looks

good so fargood so far

Key #15186 – Del Rio “DR” 92+15 Jered Carpenter 11 Feb 08 1 of 1

Rational

52 Ac

-0.5%

Trapezoid

50

100

14.9

16.4

564.9

75564.5

CMP-Circ.-18”-Mitered 14.9 3.0 569.4

1.7

1.8

4.2

CMP-Circ.-24”-Mitered 14.9 1.2 567.32.4

Step 3 Step 3 ContinuedContinued

Need to calculate H, total outlet Need to calculate H, total outlet control head loss control head loss

Outlet control nomographs can Outlet control nomographs can be used if outlet crown is be used if outlet crown is submerged.submerged.

(TW>Top of Pipe @ Outlet)(TW>Top of Pipe @ Outlet)

TW @1.7’ < D @ 2’ , so I will not TW @1.7’ < D @ 2’ , so I will not use the nomographuse the nomograph

Check for outlet controlCheck for outlet control

Step 3 ContinuedStep 3 Continued Alternatively, H can be calculated by following Alternatively, H can be calculated by following

the steps on the culvert design sheetthe steps on the culvert design sheet• Calculate critical depth, dCalculate critical depth, dcc

• Critical depth = 1.4’ (from graph located in Chapter Critical depth = 1.4’ (from graph located in Chapter 8, appendix B)8, appendix B)

Step 3 ContinuedStep 3 Continued

Following Steps 4 thru 5 on the culvert Following Steps 4 thru 5 on the culvert design sheet:design sheet:

• (d(dcc+D)/2 = (1.4’+2’)/2 = 1.7’+D)/2 = (1.4’+2’)/2 = 1.7’

• hhoo = TW or (d = TW or (dcc+D)/2 , whichever is greater+D)/2 , whichever is greater

TW @ 1.7’ = (dTW @ 1.7’ = (dcc+D)/2 @ 1.7’ ,use 1.7’+D)/2 @ 1.7’ ,use 1.7’

Key #15186 – Del Rio “DR” 92+15 Jered Carpenter 11 Feb 08 1 of 1

Rational

52 Ac

-0.5%

Trapezoid

50

100

14.9

16.4

564.9

75564.5

CMP-Circ.-24”-Mitered 14.9 1.2 567.3

1.7

1.8

2.4 1.7 1.4 1.7 1.7

CMP-Circ.-18”-Mitered 14.9 3.0 569.44.2

Step 3 ContinuedStep 3 Continued

Calculate Energy losses, H for the Calculate Energy losses, H for the culvert From the culvert design sheet:culvert From the culvert design sheet:• H = (1+kH = (1+kee+(29*n+(29*n22*L/R*L/R1.331.33))*(V))*(V22/2g)/2g) KKe e = .7 (from table 9-3, mitered to slope)= .7 (from table 9-3, mitered to slope)

n = .024 (from table 8-A-2)n = .024 (from table 8-A-2)L = 75’L = 75’Use values of V and R under full flow (d=2’)Use values of V and R under full flow (d=2’)R = 0.5’R = 0.5’V = 4.74 ft/s (Average Velocity)V = 4.74 ft/s (Average Velocity)

• H = 1.52H = 1.52

Step 3) ContinuedStep 3) Continued

Inlet Control HW Elevation = 567.3’ Inlet Control HW Elevation = 567.3’

Outlet Control HW Elevation = 567.9’Outlet Control HW Elevation = 567.9’

The pipe flowing under outlet control.The pipe flowing under outlet control.

Step 4) Try Another SizeStep 4) Try Another Size

24” pipe is acceptable.24” pipe is acceptable.

Remainder of the Culvert design sheet is Remainder of the Culvert design sheet is completed.completed.

Step 5) Compute Outlet VelocityStep 5) Compute Outlet Velocity Previously calculated value. Previously calculated value.

Determine if channel protection is Determine if channel protection is required following the guidance in required following the guidance in Chapter 11 of the hydraulics manual.Chapter 11 of the hydraulics manual.

Step 6) Determine Invert ProtectionStep 6) Determine Invert Protection Follow guidance in Chapter 5 of the Follow guidance in Chapter 5 of the

hydraulics manual to determine if & what hydraulics manual to determine if & what type of invert protection is required.type of invert protection is required.

Key #15186 – Del Rio “DR” 92+15 Jered Carpenter 11 Feb 08 1 of 1

Rational

52 Ac

-0.5%

Trapezoid

50

100

14.9

16.4

564.9

75564.5

CMP-Circ.-24”-Mitered 14.9 1.2 567.3

1.7

1.8

2.4 1.7 1.4 1.7 1.7

CMP-Circ.-18”-Mitered 14.9 3.0 569.44.2

1.71.7 567.9 567.9 4.74 Outlet Control

Trap. Channel:W=2.5’2:1 side slopesN= .07 (class 50 rip rap)S=0.5%Dc=1.7’

Metal

Circ.24”

Mitered0.024

Step 7) Document SelectionStep 7) Document Selection