1 r0 v2

∆Z =

∆r

g ri r

Total superelevation is 0.133 feet.

Based on relationships, V = Q/A = 3.61 ft/s, the following are obtained:

(3.61)

2

2

∆Z = V (r0 - ri) =

(513 - 350) = 0.153 ft

32.2 (431)

g rc

3.612

2

2W

2 x 163

1

1

∆Z = V

= 0.158 ft

=

x

2

2

2g rc W

2 x 32.2

431

163

1 - 2r

1 -

2 x 431

c

2

2

= 3.61 x 2 x 163

2W

1

1

V

= 0.151 ft

∆Z =

2g rc

2

2

W 2 x 32.2

431

163

1 -

1 -

2

2

12 rc

12 x 431

Vmax = 4.90 ft/s

(V )

rc

2

2

2

2

2

2

2 - ri

4.90 2 - 350 - 431

- =

= 0.236 ft

r 2 x 32.2 431 513

max

∆Z =

r

2g

c

o

The equations give comparable results (0.133 to 0.236 ft). Equation 2.158, which integrates

across the section using the velocity distribution is the most exact. But using the value 0.24 ft

provides a safety factor.

A stream is rectangular in shape and 100 ft wide. The design discharge is 5,000 cfs and the

uniform depth for this discharge is 10 ft. Neglecting energy losses what is the maximum amount

of constriction that a bridge can impose without causing backwater.

2.85

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