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

2. 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

3. Del Rio Example

4. Pictures of Existing Culvert

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

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

7. Step 1 Cont.) Site Specific Data
Calculate Tail Water, TW
2.5’ btm, class 50 rip-rap lined FBD with 2:1 side 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’

8. 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

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

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

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

12. 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

13. 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 Outlet)
< 2’ , so I will not use the nomograph

14. 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)

15. 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
1.7’ = 1.7’ ,use 1.7’

16. 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

17. 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

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

19. Step 4) Try Another Size 24” pipe is acceptable.
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.
Step 6) Determine Invert Protection
Follow guidance in Chapter 5 of the hydraulics manual to determine if & what type of invert protection is required.

21. 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