1. 30 Min Culvert Design Jered Carpenter Region 3 Roadway Design

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

3. Del Rio Example

4. Pictures of Existing Culvert Pictures of Existing Culvert

5. Step 1) Gather Hydrological Data Q = cia = (.35*52*.82) = 14.9 cfs Culvert Location

6. Step 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%

7. Step 1 Cont.) Site Specific Data Calculate Tail Water, TW 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*R 2/3 S 1/2 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)*10.03ft 2 *0.99ft 2/3 0.05 1/2 TW = 1.7’ TW = 1.7’

8. Key #15186 – Del Rio“DR” 92+15Jered Carpenter11 Feb 081 of 1 Rational 52 Ac -0.5% Trapezoid 50 100 14.9 16.4 564.9 75 564.5 1.7 1.8

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

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

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

12. Key #15186 – Del Rio“DR” 92+15Jered Carpenter11 Feb 081 of 1 Rational 52 Ac -0.5% Trapezoid 50 100 14.9 16.4 564.9 75 564.5 CMP-Circ.-18”-Mitered 14.93.0 569.4 1.7 1.8 4.2 CMP-Circ.-24”-Mitered 14.91.2567.32.4

13. Step 3 Continued  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 TW @1.7’ < D @ 2’, so I will not use the nomograph Check for outlet control

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

15. Step 3 Continued Following Steps 4 thru 5 on the culvert design sheet: Following Steps 4 thru 5 on the culvert design sheet: (d c +D)/2 = (1.4’+2’)/2 = 1.7’(d c +D)/2 = (1.4’+2’)/2 = 1.7’ h o = TW or (d c +D)/2, whichever is greaterh o = TW or (d c +D)/2, whichever is greater TW @ 1.7’ = (d c +D)/2 @ 1.7’,use 1.7’

16. Key #15186 – Del Rio“DR” 92+15Jered Carpenter11 Feb 081 of 1 Rational 52 Ac -0.5% Trapezoid 50 100 14.9 16.4 564.9 75 564.5 CMP-Circ.-24”-Mitered 14.91.2 567.3 1.7 1.8 2.41.71.41.7 CMP-Circ.-18”-Mitered 14.93.0 569.44.2

17. Step 3 Continued Calculate Energy losses, H for the culvert From the culvert design sheet: Calculate Energy losses, H for the culvert From the culvert design sheet: H = (1+k e +(29*n 2 *L/R 1.33 ))*(V 2 /2g)H = (1+k e +(29*n 2 *L/R 1.33 ))*(V 2 /2g) K e =.7 (from table 9-3, mitered to slope) K e =.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 = 1.52H = 1.52

18. Step 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.

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

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

21. Key #15186 – Del Rio“DR” 92+15Jered Carpenter11 Feb 081 of 1 Rational 52 Ac -0.5% Trapezoid 50 100 14.9 16.4 564.9 75 564.5 CMP-Circ.-24”-Mitered 14.91.2567.3 1.7 1.8 2.41.71.41.7 CMP-Circ.-18”-Mitered 14.93.0 569.44.2 1.7 567.9 4.74Outlet Control Trap. Channel: W=2.5’ 2:1 side slopes N=.07 (class 50 rip rap) S=0.5% Dc=1.7’ Metal Circ.24” Mitered0.024 Step 7) Document Selection