Transcript Crossing Structures

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).
Road
Water
way
Culvert or
Bridge
Water
way
Water
way
Water
way
Syphon or
Aqueduct
Water
way
Tail
escape
Culvert
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.
≤ .15 m
≤ .3 m
d
D
S
H
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.
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.
Once the dimensions are determined, then
head losses must be checked. Head
losses are calculated as follows,
V2
Hl 
(Ce  C f  Co )
2g
Where Hl is the head losses in m
V is the velocity in m/sec
g is the acceleration of gravity
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,
L
Cf  f
m
Where f is coefficient depends on the culvert
material and dimension, L is the length of
the culvert and m is the hydraulic radius.
b
f  a (1  )
m
The coefficient a and b are constant and
depends on the material of the conduit.
For Steel:
a = 0.0048, b = 0.0256
For Concrete: a = 0.00316, b= 0.0305
• 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 = ∏ D2/4
* D = 1.7 m
However, D ≤ 1.8 – 0.3 … 1.5 m
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 = 0.325 m
f = 0.0052
Cf = 0.32
Hu = (1.13)2/(2*9.81) *(0.5+0.32+1)
= 0.12 m … OK
• 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
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 = 0.00336
Cf = 0.14
Hu = 0.145 m … OK