online_scour_melville:   Pier and abutment scour using the Melville equation (1997)

ds = KyW KI Kd Ks Kθ KG

For abutment: KyW = KyL

For pier: KyW = Kyb

Reference

EXAMPLES OF THE ESTIMATION OF LOCAL SCOUR DEPTH AT BRIDGE PIER AND ABUTMENT

Original data by Bruce W. Melville (1997)

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 d50 = 1 mm, and σg = 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 = 60o.

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 = 120o.

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 = 15o; lower pier, l = 9 m, b = 3 m, and θ = 15o.

Solution:

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

Main channel:

Eq. (6a): u*c = 0.024 m/s

Eq. (5a): Vc = 0.64 m/s

Therefore: V/Vc = 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 KyW (KyL in abutment and Kyb in pier):
For A1, L/y = 120/8 = 15, and from Eq. (2b): KyL = 62.0 m
For A2, L/y = 10/2 = 5, and from Eq. (2b): KyL = 8.94 m
For P1 (nonuniform), D = 2 m, D* = 3 m, Z = 1, y = 8 m; then from Eq. (10): De = 2.36 m = b; then: b/y = 0.3 < 0.7, and from Eq. (4a): Kyb = 5.7 m
For P2 (uniform), b/y = 1. Since 0.7 < b/y < 5; then from Eq. (4b): Kyb = 4 m

Flow intensity factor KI:
Main channel A1 and P1: V/Vc = 1.56 > 1; therefore from Eq. (8b): KI= 1
Flood channel A2 and P2: V*/V*c = 0.7 < 1; therefore from Eq. (8a): KI = 0.7

Sediment size factor Kd:
A1 and A2: L/d50> 25; from Eq. (9b): Kd = 1
P1 and P2: b/d50 > 25; from Eq. (9b): Kd = 1

Foundation shape factor Ks:
A1, wing-wall, from Table 2, Ks = 0.75, L/y = 120/8 = 15, 10 < L/y < 25, from Eq. (12b): K*s = 0.833
A2, wing-wall, from Table 2, Ks = 0.75, L/y = 10/2 = 5, L/y < 10, from Eq. (12a): K*s = 0.75
P1 and P2: Ks = K*s = 1 because θ > 5o.

Foundation alignment factor Kθ:
A1, θ = 60o, from Table 1, Kθ = 0.97, L/y = 120/8 = 15 > 3; then from Eq. (13a): K*θ = 0.97
A2, θ = 120o, from Table 1, Kθ = 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 = 15o from Table 1: Kθ = K*θ = 1.5

Channel geometry factor KG:
A1, spans in the flood channel, from Eq. (14): KG = 0.36
A2, P1, P2: KG = 1

Step 3: Scour depth, ds = Kyw KI Kd Ks Kθ KG

A1: ds = 18 m
A2: ds = 5.1 m
P1: ds = 8.5 m
P2: ds = 4.3 m

Example 2

Find the pier and abutment scour depths at the bridge crossing shown in Fig. 9(b). Mean velocities are 3.0 and 1.5 m/s in the main and flood channels, respectively, and the sediment is nonuniform. Abutment A1 is considered to be situated in a rectangular channel having flood channel flow velocity (V*) and depth (y*); it is assumed that the abutment is set back a relatively large distance from the main channel. The flow velocity and depth for pier P1 are the flood channel velocity and depth, while the main channel velocity and depth are considered appropriate to pier P2. Both piers are subject to debris contamination as shown. The given data are:

Flow:
V = 3.0 m/s, V* = 1.5 m/s, y = 6 m, n = 0.020

Sediment:
d50 = 12 mm, dmax = 100 mm, σ = 4

Fundation geometry:
A1: Spill-through abutment (1:1), L = 10 m, θ = 90o
P1, P2: Circular piers D = 1m, Dd = 5 m, Td = 1.5 m, θ = 0o

Solution:

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

Main channel:

Eq. (6b): u*c = 0.105 m/s

Eq. (5a): Vc = 2.08 m/s

Eq. (7): d50a = 55.6 mm

Eq. (6b): u*ca = 0.227 m/s

Eq. (5b): Vca = 3.63 m/s

Va = 0.8Vca = 2.90 m/s

Therefore: [V - (Va - Vc)] / Vc = 1.05 > 1, and live-bed scour occurs.

Flood channel:

V*c = 1.90 m/s

V*a = 2.58 m/s

Therefore: [V* - (V*a - V*c)] / V*c = 0.43 < 1, and clear-water scour occurs.

Step 2: Evaluate the K factors

Depth size factor KyW (KyL in abutment and Kyb in pier):
For A1, L/y = 10/3 = 3.33, and from Eq. (2b): KyL = 11.0 m
For P1 (uniform), De = 2.04 m, De/y = 2.04/3 = 0.68; then from Eq. (4a): Kyb = 4.9 m
For P2 (uniform), De = 1.52 m, De/y = 1.52/6 = 0.25; then from Eq. (4a): Kyb = 3.65 m

Flow intensity factor KI:
Main channel P2: [V - (Va - Vc)] / Vc = 1.05 > 1; therefore from Eq. (8b): KI= 1
Flood channel A1 and P1: V*a = 2.58 m/s, [V* - (V*a - V*c)] / V*c = 0.43 < 1; therefore from Eq. (8a): KI = 0.43

Sediment size factor Kd:
A1: L/d50> 25; from Eq. (9b): Kd = 1
P1 and P2: De/d50 > 25; from Eq. (9b): Kd = 1

Foundation shape factor Ks:
A1, Spill-through abutment (1:1), from Table 2, Ks = 0.5, L/y = 3.33, from Eq. (12a): K*s = 0.5
P1 and P2: Ks = K*s = 1

Foundation alignment factor Kθ:
A1, θ = 90o, from Table 1, Kθ = 1.0, L/y = 3.33; then from Eq. (13a): K*θ = 1.0
P1 and P2, Kθ = K*θ = 1.0 because θ = 0o

Channel geometry factor KG:
A1, P1, P2: KG = 1

Step 3: Scour depth, ds = Kyw KI Kd Ks Kθ KG

A1: ds = 2.4 m
P1: ds = 2.1 m
P2: ds = 3.65 m

Scour depth equations

Eq. (1)   ds= KyW KI Kd Ks Kθ KG

Eq. (2a)   KyL= 2L, L/y <1

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

Eq. (2c)   KyL = 10y, L/y > 25

Eq. (4a)   Kyb = 2.4b, b/y< 0.7

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

Eq. (4c)   Kyb = 4.5y, b/y> 5

Eq. (5a)    Vc/u*c = 5.75 log (5.53 * y/d50 )

Eq. (5b)    Vca/u*ca= 5.75 log (5.53 * y/d50a)

Eq. (6a)   u*c = 0.0115 + 0.0125d1.4, 0.1 mm < d < 1 mm, and d = d50 or d50a in mm

Eq. (6b)    u*c = 0.0305d0.5 - 0.0065d-1, 1 mm < d < 100 mm, and d = d50 or d50a in mm, and u*c = u*ca in m/s

Eq. (7)   d50a = dmax/1.8

Eq. (8a)    KI = [V- (Va- Vc)]/ Vc, when: [V- (Va - Vc)]/ Vc< 1

Eq. (8b)    KI = 1, when: [V- (Va - Vc)]/ Vc ≥ 1

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

Eq. (9b)   Kd= 1, W/d50 > 25

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

For debris effects:

Eq. (11)    De= [0.52 TdDd +(y - 0.52Td)D]/ y

Eq. (12a)   K*s = Ks, L/y ≤ 10

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

Eq. (12c)   K*s=1, L/y ≥ 25

Eq. (13a)   K*θ = Kθ, L/y ≥ 3

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

Eq. (13c)    K*θ= 1, L/y ≤ 1

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

 Table 1.  Kθ values Foundation type(1) θo 0(2) 15(3) 30 (4) 45(5) 60(6) 90 (7) 120(8) 150(9) Pier l/b 4.0 1.0 1.5 2.0 2.3 2.4 2.5 2.5 2.5 8.0 1.0 2.0 2.75 3.3 3.6 3.9 3.9 3.9 12.0 1.0 2.5 3.5 4.3 4.65 5.0 5.0 5.0 Abutment 0.9 0.9 0.9 0.935 0.97 1.0 1.06 1.08

 Table 2.  Shape Factors for Piers and Abutments Foundation type(1) Shape(2) Ks(3) Pier Circular cylinderRound nosedSquare nosedSharp nosed 1.01.01.10.9 Abutment Vertical wallWing wallSpill through 0.5:1(H:V)Spill through 1:1 Spill through 1.5:1 1.00.750.60.5 0.45

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