1. Crossing Structures
Crossing structures are those constructed at intersections of:
1. Waterway-road (culvert or bridge).
2. Waterway-Waterway (syphon or aqueduct).
3. Waterway ending at another waterway (tail escape).

2. Culvert or Bridge
Road
Waterway
Syphon or Aqueduct
Waterway
Tail escape
Waterway

3. Culvert

4. The culvert is a closed conduit (pipe or box section) constructed to carry the discharge of a waterway under a road.
The selection of the type of the culvert is based on the discharge as follows:
For Q ≤ 3.0 m3/sec use Pipe culvert.
For Q ≤ 12.0 m3/sec use Box section.

5. ≤ .15 m d
≥ .3 m D H S

6. Hydraulic Design
The purpose of the hydraulic design is to select the culvert’s dimensions.
The dimensions are selected based on:
Velocity through culvert between 1.0 to 2.0 m/sec.
Head losses is less then 15 cm.
Entrance submergence is at least 30 cm.

7. The only equation available is the continuity equation:
Q = A . V
Where Q is the discharge in m3/sec, A is the culvert’s cross section area in m2, and V is the water velocity through the culvert in m/sec. Since the discharge is known, the velocity must be assumed to determine the dimensions of the culvert. The velocity is usually assumed between 1.0 and 2.0 m/sec for practical reasons.

8. Once the dimensions are determined, then head losses must be checked
Once the dimensions are determined, then head losses must be checked. Head losses are calculated as follows,
Where Hl is the head losses in m
V is the velocity in m/sec
g is the acceleration of gravity

9. Ce is the entrance coefficient of
loss (= 0.5 )
Cf is the friction coefficient
Co is the outlet coefficient of losses
( = 1.0 )
Friction coefficient can be calculated as follows,

10. Where f is coefficient depends on the culvert material and dimension, L is the length of the culvert and m is the hydraulic radius.
The coefficient a and b are constant and depends on the material of the conduit.
For Steel: a = , b =
For Concrete: a = , b=

11. Example 1:
Design a culvert under a road. The discharge is 3 m3/sec and the water depth is 1.8 m. The culvert length is 20 m.
Sol.
Q = 3 m3/sec * use pipe culvert
assume V=1.3 m/sec
A = Q/V = 2.31 m2
A = ∏ D2/ * D = 1.7 m
However, D ≤ 1.8 – … 1.5 m

12. Then use 2 pipes …one pipe A = 1.155 m2
D = 1.3 m …. V = 1.13 m/sec … OK
Check of heading up
m = D/4 = m
f =
Cf = 0.32
Hu = (1.13)2/(2*9.81) *( )
= 0.12 m … OK

13. Example 2:
design a culvert under a road. The discharge is 5 m3/sec and the water depth is 2.2 m. The culvert length is 20 m.
Sol.
Q = 5 m3/sec … Use Box culvert
Assume V = 1.3 m/sec
A = Q/V = 3.85 m2
H = 2.2 – 0.3 = 1.9 m
S = 3.85/1.9 = 2.0 m

14. S < 1.5 H … OK
V = 5/(1.9*2) = 1.32 m/sec
Check for Heading Up
m = 2*1.9/(2+1.9)*2= 0.49 m
f =
Cf = 0.14
Hu = m … OK