Introduction:
A culvert is a structure that allows water to flow under a road, railroad, trail, or
similar obstruction from one side to the other side. A culvert may be made from a
pipe, reinforced concrete or other material, but herein we use culvert made of
reinforced concrete. Culverts are commonly used both as cross-drains for ditch
relief, and to pass water under a road at natural drainage and stream crossings.
Culverts come in many sizes and shapes including round, elliptical, flat-bottomed,
open-bottomed, pear-shaped, and box-like constructions. Culvert type and shape
selection is based on number of factors including requirements for hydraulic
performance, limitations on upstream water surface elevation, and roadway
embankment height.
Failures:
Culvert failures can occur for a wide variety of reasons including maintenance,
environmental and installation related failures, functional or process failures
related to capacity and volume causing the erosion of the soil around or under
them, and structural or material failures that cause culvert to fail due to collapse
or corrosion of the materials from which they are made.
Resources:
We use USGS Quad sheet for Mendham, N.J. for drainage area and hydraulic path.
Quad sheet shows the wooded areas, streams and cultural features, thus the pink
areas represent residential districts with ½-acre average lot size.
Site parameters:
Hydraulic soil group: C
Stream channel: straight, regular, some grass and weeds
Left overbank: Light brush and trees
Design Criteria:
Compute peak runoff by NRCS Method
Design storm: 100-years storm
Allowable upstream water elevation: top of road (water cannot overtop road).
First of all we delineate the watershed tributary to the culvert for the designated
projected area, which includes, wooded areas, residential area, streams and
roads.
**Drainage area is measured by triangles geometrical shapes.
1: triangle 1.
A: 1/ 2bh= 1/2 (4000)(3200)= 147 acres
2: Triangle 2.
A: ½ bh= ½ (2700)(3700)= 114 acres
3: triangle 3.
A: ½ bh= ½ (3500)(4200)= 169 aces
4: Triangle 4.
A : ½ bh = ½ (2000)(2500)= 57 acres
5: Triangle 5.
A: ½ bh = ½ (2100)(1900) = 45 acres
6: Triangle 6.
A: ½ bh = ½ (2600)(3500)= 104 acres
TOTAL AREA= A1+A2+A3+A4+A5+A6= 636 Acres
**Now we also need to calculate the Residential and impervious areas.
Residential area = 191 acres
Impervious area = 4.2 acres
**As we have total area 636 acres, and sub total residential 191 acres and roads
4.2 acres, we make following calculation to subtract residential and roads from
the whole delineated are which is 636 acres, thus we obtain area for each phase
of land.
Woods
Residential area
roads
440.8
191
4.2
** now we have to calculate the CN number base on the soils delineated as
impervious, residential and wooded. We refer to Appendix-1 on page-238 and
look for soil group C. in order to have clear understanding, we make a table.
cover
Impervious
Residential
wooded
Total
Area
4.2
191
440.8
636
CN
98
80
76
Product
411.6
15280
33500.8
49192.4
** 636 acres = 0.993 square miles
** now can calculate the composite c:
CN= 49192.4/636= 77.34
Now that CN = 77.34, and we don’t have Ia for this, we have two options either to
consider CN = 77 OR CN = 78, as I see in my above calculation CN value is between
77 and 78, so I prefer to take CN= 78.
As we refer to table 11-2, Ia = 0.564 for CN = 78.
Ia/p = 0.564/7.6= 0.0742
S = (1000/CN)-10= 1000/78-10= 2.82 in
Base on table 11-9, to find Q as we have p = 7.6 and CN=78 so we get Q = 4.95
inches.
Q= 4.95 inches
** now to compute time of Concentration:
A: overland flow:
L = 100 ft
n= 0.40
slope = (1300-1000)/100 = 3%
T1 = 0.007((0.40)(100))0.8/(7.6)0.5(3)0.4= 0.03h
B: Shallow concentrated flow:
d= 2300
V = 4.4
T2 = d/V = 2300/4.4= 522.7 min=522.7/3600= 0.14 h
C: stream flow:
By using manning’s equation.
d = 8100
A = 42
P = 63
n = 0.032
R = a/p= 42/63= 0.6667
V = 1.49/n R2/3 S1/2 = (1.49/0.032) x (0.67)2/3(0.03)1/2= 6.2
T3 = d/v= 8100/6.2= 1306.5=0.36 h
Tc= 0.03+0.14+0.36= 0.53h
** using appendix D-5 RAIN TYPE-III
Qu= 325
So, peak runoff = qp=quAmQ= 325x4.95x0.993= 1597.5 cfs= 329 inches
** computing normal depth:
LOB
n= 0.043
S= 2%=0.02
CHANNEL
n= 0.033
S= 3%=0.03
ROB
n= 0.043
S= 0.02
** allowable head-water elevation:
As we have existing culvert 4’x 8’and we can check this for adequacy.
Inlet control is assumed to be as following, base on appendix-B-1.
Q/B= 1597.5/8 = 199.7.
And we check Q/B= 199.7 for the height 4’ of the existing culvert it is off
headwater measurement (gauge).
** finding normal depth:
Dn a
LOB
p
R
R
1
0
1
0
2
0
1
0
3
14.5 30
0.483 0.616 44.56 48.5 15.1 3.21
4
33.7 48
5
0.703 0.791 133.2 58.5 15
2/3
p
Channel
R
R2/3
Q
a
0
0
14.5 12
1.2
0
0
29
2.07
14
3.9
Q
a
RO
Total
2/3
B
R R
Q Q
p
1.13 179. 0
4
1.62 514. 0
5
2.17 1152 0
1
0 0
0 179.4
1
0 0
0 514.5
1
0 0
0 1196
2.47 1583 2.
5
10
. 0.3 4 1718
2 97 .
5
9
**Approximately Normal depth = 3.45 ft
Culvert design:
Now we are going to consider suitable culvert for the project designated with
keep in mind the allowable head-water elevation which is 349 ft.
So we have,
Road elevation = 349 ft
Normal depth is assumed to be = 3.45 ft
Trial 1: culvert-box = 6’x8’
HW/D = 3 x 8= 24ft
As per elevation of the bottom-stream 340 ft
340 ft + 24 ft= 360 ft ----------- no good because it is above the road elevation.
Trial 2: culvert-box = 8’ X 10’
HW/D = 2.8 ft
HW = 2.8’ X 8’= 22.4’
As per elevation of the bottom-stream 340 ft
340 ft + 22.4 ft = 362.4 ft -----------no good, it is also above the road elevation.
Note: for the third trial, we have to consider a double concrete culvert 8’ x 12’
and to see if this double or twin culvert will work.
Trial3: HW/D = 1.1
Q/B = 1597.5/12= 133.12
HW = 1.1 x 8 = 8.8
As per elevation of the bottom-streem 340 ft
340 ft + 8.8= 348.8 --------- this one is good, because 348.8 < 349’ (road elevation).