Jered Carpenter Region 3 Roadway Design 30 Min Culvert Design Jered Carpenter Region 3 Roadway Design

Design Steps Step 1) Gather hydrological & site specific data Step 2) Assume culvert size Step 3) Find headwater depth for trial size Step 4) Try another size Step 5) Compute outlet velocity Step 6) Determine Invert Protection Step 7) Document selection

Del Rio Example

Pictures of Existing Culvert

Step 1) Gather Hydrological Data Zone 6 IDF Curve Culvert Location Q = cia = (.35*52*.82) = 14.9 cfs

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

Step 1 Cont.) Site Specific Data Calculate Tail Water, TW 2.5’ btm, class 50 rip-rap lined FBD with 2:1 side slopes @ outlet Manning’s Equation for discharge Q = (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 Rio “DR” 92+15 Jered Carpenter 11 Feb 08 1 of 1 Rational 52 Ac -0.5% Trapezoid 564.9 75 564.5 50 14.9 1.7 100 16.4 1.8

Step 2) Assume Culvert Size Start With 18” Circular CMP

Step 3) Find HW Depth For Trial Size Assume inlet control From Nomograph, HW/D = 3 HW depth w/ inlet control approx. 569.4 ft., which is too high I’ll try a 24” pipe

Step 3) Find HW Depth For Trial Size Assume inlet control From Nomograph, HW/D = 1.2 HW depth w/ inlet control approx. 566.7 ft., which is OK. The 24” pipe looks good so far

Key #15186 – Del Rio “DR” 92+15 Jered Carpenter 11 Feb 08 1 of 1 Rational 52 Ac -0.5% Trapezoid 564.9 75 564.5 50 14.9 1.7 100 16.4 1.8 CMP-Circ.-18”-Mitered 14.9 3.0 4.2 569.4 CMP-Circ.-24”-Mitered 14.9 1.2 2.4 567.3

Step 3 Continued Check for outlet control Need 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

Step 3 Continued Alternatively, H can be calculated by following the steps on the culvert design sheet Calculate critical depth, dc Critical depth = 1.4’ (from graph located in Chapter 8, appendix B)

Step 3 Continued Following 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 greater TW @ 1.7’ = (dc+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 564.9 75 564.5 50 14.9 1.7 100 16.4 1.8 CMP-Circ.-18”-Mitered 14.9 3.0 4.2 569.4 CMP-Circ.-24”-Mitered 14.9 1.2 2.4 567.3 1.7 1.4 1.7 1.7

Step 3 Continued Calculate 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 = 75’ Use values of V and R under full flow (d=2’) R = 0.5’ V = 4.74 ft/s (Average Velocity) H = 1.52

Step 3) Continued Inlet Control HW Elevation = 567.3’ Outlet Control HW Elevation = 567.9’ The pipe flowing under outlet control.

Step 4) Try Another Size 24” pipe is acceptable. Remainder of the Culvert design sheet is completed.

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

Step 7) Document Selection Key #15186 – Del Rio “DR” 92+15 Jered Carpenter 11 Feb 08 1 of 1 Rational 52 Ac -0.5% Trapezoid 564.9 75 564.5 50 14.9 1.7 100 16.4 1.8 CMP-Circ.-18”-Mitered 14.9 3.0 4.2 569.4 CMP-Circ.-24”-Mitered 14.9 1.2 2.4 567.3 1.7 1.4 1.7 1.7 1.7 1.7 567.9 567.9 4.74 Outlet Control Trap. Channel: W=2.5’ 2:1 side slopes N= .07 (class 50 rip rap) S=0.5% Dc=1.7’ 24” Circ. Metal 0.024 Mitered