__Example 1__

Find the local scour depth at a bridge pier and abutment for the data shown below:

[Note that the online calculation will perform one pier calculation at a time, either P1 or P2]

**Hydraulic data (flow velocities,
channel geometry, Manning n, and sediment data):**

In the main channel:
V = 1.0 m/s, y = 8 m, n = 0.020

In the flood channel (labeled with a *): V* = 0.4 m/s,
L* = 110 m, y* = 2 m, n* = 0.040

The sediment is uniform with d_{50} = 1 mm, and σ_{g} = 1.3.

**Abutment A1 (left):
** Spans the flood channel and extends into the main channel. It is a wing-wall abutment, with L = 120 m, θ = 60^{o}.

**Abutment A2 (right):** It is situated in a rectangular channel having flood channel velocity (V*) and depth (y*).
It is a wing-wall abutment, with
L = 10 m, θ = 120^{o}.

**Pier P1 (in main channel): ** It has the main channel velocity (V) and depth (y).
It is nonuniform and square-nosed.

**Pier P2 (in flood channel):**
It has the flood channel velocity (V*) and depth (y*).
It is uniform and square-nosed.

**For both piers P1 and P2:** Upper pier, l = 8 m, b = 2 m, and θ = 15^{o};
lower pier, l = 9 m, b = 3 m, and θ = 15^{o}.

**Solution:**

**Step1:** Determine the velocity parameters in the main and flood channels.

Main channel:

Eq. (6a): u_{*c} = 0.024 m/s

Eq. (5a): V_{c} = 0.64 m/s

Therefore: V/V_{c} = 1.56 > 1, and live-bed scour occurs.

Flood channel:

Eq. (6a): u_{*c} = 0.024

Eq. (5a): V_{*c} = 0.56 m/s

Therefore: V*/V*_{c} = 0.72 < 1, and clear-water scour occurs.

**Step 2:** Evaluate the K factors

**Depth size factor K**_{yW} (K_{yL} in abutment and K_{yb} in pier):

For A1, L/y = 120/8 = 15, and from Eq. (2b): K_{yL} = 62.0 m

For A2, L/y = 10/2 = 5, and
from Eq. (2b): K_{yL} = 8.94 m

For P1 (nonuniform), D = 2 m, D* = 3 m, Z = 1, y = 8 m; then from Eq. (10): D_{e} = 2.36 m = b;
then: b/y = 0.3 < 0.7, and from Eq. (4a): K_{yb} = 5.7 m

For P2 (uniform), b/y = 1. Since 0.7 < b/y < 5; then from Eq. (4b): K_{yb} = 4 m

**Flow intensity factor K**_{I}:

Main channel A1 and P1: V/V_{c} = 1.56 > 1; therefore from Eq. (8b): K_{I}= 1

Flood channel A2 and P2: V*/V*_{c} = 0.7 < 1; therefore from Eq. (8a): K_{I} = 0.7

**Sediment size factor K**_{d}:

A1 and A2: L/d_{50}> 25; from Eq. (9b): K_{d} = 1

P1 and P2: b/d_{50} > 25; from Eq. (9b): K_{d} = 1

**Foundation shape factor K**_{s}:

A1, wing-wall, from Table 2, K_{s} = 0.75, L/y = 120/8 = 15, 10 < L/y < 25,
from Eq. (12b): K*_{s} = 0.833

A2, wing-wall, from Table 2, K_{s} = 0.75, L/y = 10/2 = 5, L/y < 10,
from Eq. (12a): K*_{s} = 0.75

P1 and P2: K_{s} = K*_{s} = 1 because θ > 5^{o}.

**Foundation alignment factor K**_{θ}:

A1, θ = 60^{o}, from Table 1, K_{θ} = 0.97, L/y = 120/8 = 15 > 3; then from Eq. (13a): K*_{θ} = 0.97

A2, θ = 120^{o}, from Table 1, K_{θ} = 1.06, L/y = 10/2 = 5 > 3; then from Eq. (13a): K*_{θ} = 1.06

P1 and P2, l/b = 8/2 = 4; and θ = 15^{o} from Table 1: K_{θ} = K*_{θ} = 1.5

**Channel geometry factor K**_{G}:

A1, spans in the flood channel, from Eq. (14): K_{G} = 0.36

A2, P1, P2: K_{G} = 1

**Step 3:** Scour depth,
d_{s} = K_{yw} K_{I} K_{d} K_{s} K_{&theta} K_{G}

A1: d_{s} = 18 m

A2: d_{s} = 5.1 m

P1: d_{s} = 8.5 m

P2: d_{s} = 4.3 m