1 r0 v2

∆Z =

∆r

g ri r

Total superelevation is 0.040 meters.

Based on relationships, V = Q/A = 1.10 m/s, the following are obtained:

(1.10)

2

2

∆Z = V (r0 - ri) =

(156.4 - 106.7) = 0.047 m

9.81(131.4)

g rc

2

1.10 2

2W

2 x 49.7

1

1

∆Z = V

= 0.042 m

=

x

2

2

2g rc W

2 x 9.81

131.4

49.7

1 - 2r

1 -

2 x 131.4

c

2

2

= 1.10 x 2 x 49.7

2W

1

1

V

= 0.041 m

∆Z =

2g rc

2

4.97 2

W 2 x 9.81

131.4

1 -

1 -

2

12 x 131.4 2

12 rc

Vmax = 1.49 m/s

(

)

rc 1.49 2 106.7 131.4

2

2

2

2

2

ri

V max

- =

2-

= 0.0718 m

r 2 x 9.81 2 - 131.4 - 156.4

∆Z =

r

2g

c

o

The equations give comparable results (0.040 to 0.0718 m). Equation 2.158 which integrates

across the section using the velocity distribution is the most exact. But using the value 0.072 m

provides a safety factor.

A stream is rectangular in shape and 30.48 m wide. The design discharge is 141.6 cms and the

uniform depth for this discharge is 3.05 m. Neglecting energy losses what is the maximum

amount of constriction that a bridge can impose without causing backwater.

2.75

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