30 Min Culvert Design

Jered CarpenterRegion 3 Roadway Design

Design StepsStep 1) Gather hydrological & site specific dataStep 2)Assume culvert sizeStep 3)Find headwater depth for trial sizeStep 4)Try another sizeStep 5)Compute outlet velocityStep 6)Determine Invert ProtectionStep 7)Document selection

Del Rio Example

Pictures of Existing Culvert

Step 1) Gather Hydrological DataQ = cia = (.35*52*.82) = 14.9 cfsCulvert Location

Step 1 Cont.) Site Specific DataInvert In El. = 564.9Invert Out El. = 564.5Pipe Length = 75Slope = -0.5%

Step 1 Cont.) Site Specific DataCalculate Tail Water, TW2.5 btm, class 50 rip-rap lined FBD with 2:1 side slopes @ outletMannings Equation for dischargeQ = (1.486/n)*A*R2/3S1/2

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

15.0 cfs = (1.486/0.07)*10.03ft2*0.99ft2/30.051/2

TW = 1.7

Key #15186 Del RioDR 92+15Jered Carpenter11 Feb 081 of 1Rational52 Ac-0.5%Trapezoid5010014.916.4564.975564.51.71.8

Step 2) Assume Culvert Size

Start With 18 Circular CMP

Step 3) Find HW Depth For Trial SizeAssume inlet controlFrom Nomograph, HW/D = 3

HW depth w/ inlet control approx. 569.4 ft., whichis too high

Ill try a 24 pipe

Step 3) Find HW Depth For Trial SizeAssume inlet controlFrom Nomograph, HW/D = 1.2

HW depth w/ inlet control approx. 566.7 ft., whichis OK.

The 24 pipe looks good so far

Key #15186 Del RioDR 92+15Jered Carpenter11 Feb 081 of 1Rational52 Ac-0.5%Trapezoid5010014.916.4564.975564.5CMP-Circ.-18-Mitered14.93.0569.41.71.84.2CMP-Circ.-24-Mitered14.91.2567.32.4

Step 3 ContinuedNeed to calculate H, total outlet control head loss

Outlet control nomographs can be used if outlet crown is submerged.(TW>Top of Pipe @ Outlet)

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

Check for outlet control

Step 3 ContinuedAlternatively, H can be calculated by following the steps on the culvert design sheetCalculate critical depth, dc

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

Step 3 ContinuedFollowing Steps 4 thru 5 on the culvert design sheet:

(dc+D)/2 = (1.4+2)/2 = 1.7

ho = TW or (dc+D)/2 , whichever is greaterTW @ 1.7 = (dc+D)/2 @ 1.7 ,use 1.7

Key #15186 Del RioDR 92+15Jered Carpenter11 Feb 081 of 1Rational52 Ac-0.5%Trapezoid5010014.916.4564.975564.5CMP-Circ.-24-Mitered14.91.2567.31.71.82.41.71.41.71.7CMP-Circ.-18-Mitered14.93.0569.44.2

Step 3 ContinuedCalculate Energy losses, H for the culvert From the culvert design sheet:H = (1+ke+(29*n2*L/R1.33))*(V2/2g) Ke = .7 (from table 9-3, mitered to slope)n = .024 (from table 8-A-2)L = 75Use values of V and R under full flow (d=2)R = 0.5V = 4.74 ft/s (Average Velocity)H = 1.52

Step 3) ContinuedInlet Control HW Elevation = 567.3

Outlet Control HW Elevation = 567.9

The pipe flowing under outlet control.

Step 4) Try Another Size24 pipe is acceptable.

Remainder of the Culvert design sheet is completed.

Step 5) Compute Outlet VelocityPreviously calculated value. Determine if channel protection is required following the guidance in Chapter 11 of the hydraulics manual.Step 6) Determine Invert ProtectionFollow guidance in Chapter 5 of the hydraulics manual to determine if & what type of invert protection is required.

Key #15186 Del RioDR 92+15Jered Carpenter11 Feb 081 of 1Rational52 Ac-0.5%Trapezoid5010014.916.4564.975564.5CMP-Circ.-24-Mitered14.91.2567.31.71.82.41.71.41.71.7CMP-Circ.-18-Mitered14.93.0569.44.21.71.7567.9567.94.74Outlet ControlTrap. Channel:W=2.52:1 side slopesN= .07 (class 50 rip rap)S=0.5%Dc=1.7MetalCirc.24Mitered0.024Step 7) Document Selection