Example 1

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

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: V* = 0.4 m/s, L* = 110 m, y* = 2 m, n* = 0.040
The sediment is uniform with d₅₀ = 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 nonuniform 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/d₅₀> 25; from Eq. (9b): K_d = 1
P1 and P2: b/d₅₀ > 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

Eq. (1)   d_s= K_yW K_I K_d K_s K_θ K_G

Eq. (2a)   K_yL= 2L, L/y <1

Eq. (2b)   K_yL = 2(yL)^0.5, 1< L/y < 25

Eq. (2c)   K_yL = 10y, L/y > 25

Eq. (4a)   K_yb = 2.4b, b/y< 0.7

Eq. (4b)    K_yb = 2(yb)^0.5, 0.7 < b/y < 5

Eq. (4c)   K_yb = 4.5y, b/y> 5

Eq. (5a)    V_c/u_*c = 5.75 log (5.53 * y/d₅₀ )

Eq. (5b)    V_ca/u_*ca= 5.75 log (5.53 * y/d_50a)

Eq. (6a)   u_*c = 0.0115 + 0.0125d^1.4, 0.1 mm < d < 1 mm, and d = d₅₀ or d_50a in mm

Eq. (6b)    u_*c = 0.0305d^0.5 - 0.0065d^-1, 1 mm < d < 100 mm, and d = d₅₀ or d_50a in mm, and u_*c = u_*ca in m/s

Eq. (7)   d_50a = d_max/1.8

Eq. (8a)    K_I = [V- (V_a- V_c)]/ V_c, when: [V- (V_a - V_c)]/ V_c< 1

Eq. (8b)    K_I = 1, when: [V- (V_a - V_c)]/ V_c ≥ 1

Eq. (9a)   K_d = 0.57 log (2.24 W/d₅₀), W/d₅₀ ≤ 25

Eq. (9b)   K_d= 1, W/d₅₀ > 25

Eq. (10))    D_e= D [(y - Z)/(y + D*)] + D*[(D* +Z)/(D* +y)]

For debris effects:

Eq. (11)    D_e= [0.52 T_dD_d +(y - 0.52T_d)D]/ y

Eq. (12a)   K*_s = K_s, L/y ≤ 10

Eq. (12b)   K*_s = K_s + [0.667 (1 - K_s)] [(0.1 L/y) - 1], 10 < L/y < 25

Eq. (12c)   K*_s=1, L/y ≥ 25

Eq. (13a)   K*_θ = K_θ, L/y ≥ 3

Eq. (13b)    K*_θ = K_θ + (1- K_θ)(1.5 - 0.5 L/y), 1 < L/y < 3

Eq. (13c)    K*_θ= 1, L/y ≤ 1

Eq. (14)   K_G = {1- (L*/L) [1-(y*/y)^5/3( n/n*)]}^0.5