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COASTAL AND HARBOR ENGINEERING
Rubble Mound Breakwater Design Given: Design Conditions Water depth (SWL)
: 5.0 m
Beach slope
: 1:20
Structure slope
: 1:2
Design high water (DHW) : 1.5 m Design wave;
Hs = 2 m H1/10 = 2.5 m Tm = 6 sec
Allowable overtopping Armor unit
:0.4 m3/sec/m : Dolos stone
Soil Data;
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COASTAL AND HARBOR ENGINEERING
Assume: Armor and under layer material is quarry stone: γ a = 2.5
t m3
Structure slope: 1:2 Structure will be symmetric (this may be changed to reduce structure size in necessary) Specify Design Condition: SWL = 5.0 m, DHW = 1.5 m; h = 5.0 + 1.5 = 6.5 m Assume listed conditions are at structure toe. Significance Wave Height, Hs = H1/3 = 2 m; Period, T = 6 sec; Deep Water Length, Lo = 100 m Lo = 1.56 T2 = 1.56 (6.0)2 = 56.16 m ~ 57 m, h/Lo = d/Lo = 6.5/57 = 0.114; From Table of functions; d/Lo = 0.114, so get
d/L = 0.153 (Deep Water); L = 6.5/0.153 = 42.5 m ~ 43 m
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COASTAL AND HARBOR ENGINEERING Wave Conditions Method 1 h/L ≤ 1/7 tanh kd = For wave criteria being Non-Breaking h/L ≥ 1/7 tanh kd = For wave criteria being Breaking So, Take Hs = 2 m, = 2.0/43 ≤ 1/7 x 0.7450 0.0465 ≤ 0.1060 , hence non-breaking wave condition.
Method 2 Calculate depth limited breaking wave height at structure site, compare with the unbroken storm wave height, and use the lesser of the two as the design wave; Hb/ hb = hb/gT2 = 6.5/ (9.81 X 36) = 0.018 ~ 0.02 Given slope, m = 1:20 so Hb/hb = 0.88; Hs = 2m at DHW at SWL
: Hb = 0.88 x 6.5 = 5.72 m ~ 5.7 m : Hb = 0.88 x 5.0 = 4.44 m ~ 4.4 m
Both wave heights are greater than Hs so waves are not breaking and design H = Hs = 2 m.
Set Break Water Dimensions (controlled by height & slope): Set-up: waves are not breaking per the previous calculations no set-up 3
COASTAL AND HARBOR ENGINEERING NOTE: there will be a set-down, but this will be neglected and considered an added factor of safety unless required to reduce the structure size,
=0
Allowable Overtopping Discharge Hs = Significant Wave Height Top = Wave period associated with the spectral peak in deep water. Rc = Free Board ϒr = Surface roughness reduction factor Based on Owen Model; Subjected to structures via following specifications such impermeable, smooth, rough, straight and bermed slope. Overtopping model, Q = a exp (-bR) Dimensionless discharge, Q = q/ (gHs tom) Dimensionless freeboard, Rc = (Rc/Hs)(S om/2∏)0.5 (1/ϒ) Straight & bermed impermeable slopes ( Figure VI – 5 – 14a & b), Irregular and Head on waves. Owen model in Table VI-5-8;
q Rc = a exp − b gH S TOM HS
S OM 1 2π γ r
From Table VI-5-8: 4
COASTAL AND HARBOR ENGINEERING Slope 1:2;
a = 0.013, b = 22 , q = 0.4 m 3/s per m
Rock Riprap with thickness greater than 2D 50, ϒr ~ 0.55 Solving equation Owen Model; Rc = R/Hs √(Sm/2∏) , Sm = Hs/ Lo Rc = R/Hs √(Hs/gTm2 Rc = (-ϒr/b) Ln (q/agHs tom) ; relative freeboard = (-0.55/22) Ln (0.4/0.013 x 9.81 x 2 x 6) = 0.0355 R = Hs Rc √(gTm2/Hs) = 2.0 x 0.0355 x √((9.81x62)/2) = 0.99 m ~ 0.9 m Allowable Rrun-up = 0.9
Wave Run-up
Surf similarity parameter also referred as breaker parameter of Iribarran number, indicates the type of breaker regarding the wave run-up and run-down on a structure.
ξ om =
tan α S OM
= 2.79
Where is ; α = Slope angle 5
COASTAL AND HARBOR ENGINEERING So = Deep water wave steepness (Ho/Lo) Ho = Deep water height Lo = Deep water wave length (gT2/ 2∏) T = Wave period G = Acceleration due to gravity
= 1.9 x 0.55 x 1 x 1 x 1 = 1.045
= 1.17 x 2.8
0.46
= 1.9 m ~ 2m
R = Hs = 2m
Since calculation of Rover-run, the Rover-top also to be recalculated in order to indicate actual (over-topping height and allowable over-topping discharge); q = 0.1 m3/sec/m; assume total settlement as 0.1 m; ῃ = 0 ~no wave setup since no breaking has been encountered. The smallest set-down will be neglected. The design elevation is determined to Rdesign = DHW + ῃ + R + ptotal = 6.5 + 0 + 2 +0.1 = 8.6 m
Design of Structure Cross-Section
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COASTAL AND HARBOR ENGINEERING A rubber mound structure is normally composed of a bedding layer and a core of quarry-run stone covered by one or more layers of larger stone and an exterior layer or layers of large quarrystone or concrete armor units. For non-breaking wave & 0~5% damage, random placement. Ns = H/ (SG-1)(W/ϒa)1/3 W = ϒa H3/ (kD)(SG-1)3 cot α Where is; W = Median weight of armor unit D = Diameter of armor unit ϒa = Unit Weight of armor H = Design wave height Kd = Stability coefficient α = Slope angle of horizontal
W = 2.5 X 23 / (14 X (2.5-1)3 X 2) = 20/94.5 = 0.2 ton
So,
SG = ϒa/ϒw = 2.5 t/m3
Armor thickness, t = nk∆ (W/ϒa)1/3 ; Where is; Number of concrete armor units in the thickness, n = 2 ϒa = Weight of individual armor unit W = Specific weight of armor unit k∆ = Layer coefficient thickness
So, t = 2 x 0.94 x (0.2/2.5)1/3 = 0.81m ~ 0.9 m 7
COASTAL AND HARBOR ENGINEERING Crest width, B = nk∆ (W/ ϒa)1/3 = 3 x 0.94 x (0.9/2.5)1/3 = 2 m Where is; B = Crest width N = Number of stones/ armor concrete (min = 3) k∆ = Layer coefficient W = Primary armor unit weight ϒa = Specific Height of armor unit materials
Number of armor units per unit surface area; P = average porosity for Dolos = 56 % Na/A = nk∆ (1-(P/100))(ϒa/W)2/3 = 2 X 0.94 X (1-0.56)(2.5/0.2)2/3 = 4.45 ~ 5 units/m2 Volume of armor per unit length; V/L = t(B +2(h + R) cot α = 0.9(2 + 2(8.6+2)2) = 39.96 ~ 40 m3/m
Design First under Layer (using Quarry stone) Minimum two stone thick (n =2) Under layer unit weight = W/10 = 0.2t/10 = 0.02 x 1000 = 20 kg 8
COASTAL AND HARBOR ENGINEERING Next larger available size is = 22.68 kg ~ 22.7 kg 1/ 3
W γ a
Thickness, t = nk∆
1/ 3
0.0277 = 2 ×1× 2.5
= 0.42m
Volume per unit of breakwater; H = 8.5 m – t armor = 8.5m - 0.9m = 7.6 m
t ul1 =
t armor = 0.45m 2
A =B crest = 2m, cot α = 2 a = A + 2 T (cot c=h
α
- csc
1+ cot 2 α = 7.6
α
) = 2 + 2(0.9)(2-2.2) = 1.7 m
1 + 4 = 17 m
V = t (a + 2c) = 0.5 (1.7 +2(17)) = 18 m 3 /m L
First under layer ul1; W 10 = 22.7 kg t ul1 = 0.5 m V = 18 m 3 /m Lul1
Second Under Layer Minimum two stone thick (n=2) Under layer unit weight =
W of the layer above; 20 9
COASTAL AND HARBOR ENGINEERING W W X 20 10
=
W 0.2t of armor layer = = 1 kg 200 200
Next larger available size is 1.13 kg Thickness, t = nk ∆(
W
γa
)1 / 3 = 2 x 1 x (
0.001 1 / 3 ) = 0.15 m 2.5
Volume per unit length of breakwater; H = 8.5m - t ul1 −t armor = 8.5 m -0.9m – 0.5 m = 7.1 m a = A + 2T (cot α − csc α) = 0.9m + 2(0.5)(2 − 2.2) = 0.7 m c = h 1 +cot 2 α = 7.1 1 +4 =16m
V = t (a + 2c) = 0.15(0.7 + 32) = 4.9m 3 / m L Second under layer ul2; W 20 = 1 kg t ul 2 = 0.15 m V = 4.9 m 3 /m Lul 2
Core Design Dynamic load requirement;
W ≤ 15 ≈ 25 Wcore
W = 1 kg → W core =1.5kg ≈ 2.5kg W 4000 =
0.2t = 0.05kg 4000
Next larger available size is 0.68 kg 10
COASTAL AND HARBOR ENGINEERING W Thickness, t = nk∆ γ A
1/ 3
= 2 ×1×
0.001 = 0.15 2.5
Volume per unit length of breakwater; h = 8.5m −t armor −t ul1 −t ul 2 = 8.5m − 0.9m − 0.5m − 0.15m = 6.95m
A = a ul 2 = 0.7 m, cot α = 2; H = hul 2 = 7.1m, T = Tul 2 = 0.15m a = A + 2T (cot α − csc α) = 0.7 m + 2(0.15)(2 − 2.2) = 0.64m b = A + 2( H cot α −T csc α) = 0.7 m + 2(7.1× 2 − 0.15 × 2.2) = 28.44m
Trapezoidal; V 1 1 = h( a + b ) = × 6.95 × ( 0.64 + 28.44) = 102m 3 / m d 2 2 Core; W 4000 = 0.05 kg V = 102 m 3 /m L
Toe Design B t = toe berm width
≈
max (2H, 0.4h) ; h = SWL
2H = 2 X 2 =4 0.4 h = 0.4 x 5.0 = 2
Bt = 4
Assume B t = 4 m; Assume height of toe = t armor = 0.9m hb = SWL − Height of toe = 5m – 0.9 m
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COASTAL AND HARBOR ENGINEERING
h b 4 .1 3 = = 0.82 → N S ≈ 60 h 5
W=
γ SH3
N S ( SG − 1) 3
W D = γ S
3
=
2.5 × 2 3
60( 2.5 − 1)
3
= 0.1t → Nearest size are 136 ≈ 0.14t
1/ 3
= 0.38 → 2 Stone height = 2 × 0.38 = 0.76 < 1.4m
2 1 − K h ' ( 1− K) N S = Max 1.8, 1.3 1 / 3 + 1.8 exp − 1.5 HS K K 1/ 3
h' H
Where is;
K = K1 K 2
K1 =
2K ' h ' sin 2 K ' h '
{
K 2 = max 0.455 sin 2 θ cos 2 θ ( KB cos θ ) , cos 2 θ sin 2 θ ( KB cos θ )
}
h ‘ = Water depth on top of toe berm (excluding armor layer) B = Width of toe berm, K ‘ = Wave number; K =
2π ; θ = Wave incident angle ( LP
θ = 0 0 for head on)
K = K1 K 2 K' =
2π 2π = = 0.9 L 57
K1 =
2 × 0.9 × ( 6.5m −1.4m ) = 57.5 sin ( 2 × 0.9 × ( 6.5 −1.4 ) )
Assume height of toe = 1.4 m (guess)
h ' = 6.5m −1.4m = 5.1m
{
}
K 2 = max 0.455 sin 2 θ cos 2 θ ( KB cos θ ) , cos 2 0 sin 2 ( 0.9 × 4 × cos 0 ) = 3.9 × 10 −3 K = 0.23 12
COASTAL AND HARBOR ENGINEERING 2 1 − 0.23 5.1 ( 1 − 0.23) N S = Max 1.8, 1.3 + 1.8 exp − 1.5 1/ 3 2 0 . 23 0.231 / 3
5 .1 2
N S = Max{1.8, ( 4.17 + 0.04 )} = {1.8,4.21} ⇒ 4.21
W=
γ SHS
N S ( SG − 1) 3
3
=
2.5 3
( 4.21) ( 2.5 − 1) 3
3
=
20 = 0.08ton 252
Used W = 0.14t and recalculate with;
h = 5.0 − 0.8 = 4.2m →
h b 4 .2 = = 0.84 h 5
3
So → N S ≈ 60
Toe ; W toe = 136 kg Toe height = 0.8m Bt = 4m
Toe Volume Assume slope is 1:2
→ base length =
Bt + 2( SWL − hb ) cot α
= 4 + 2 (0.8) x 2 = 7.2 m Assume trapezoidal; V = ( SWL − hb )( Bt + base ) = 0.8( 4 + 7.2) = 9m 3 / m L
Toe-to toe width
W = 2 Bt + 2( SWL − hb ) cot α + B + 2(hb + DHW + R + ρ ) cot α 13
COASTAL AND HARBOR ENGINEERING = 2 × 4 + 2( 5 − 4.1) × 2 + 2 + 2( 4.1 +1.5 + 2 + 0.1) × 2 = 44.4m
Cross-Section of Breakwater Design
2.0 m
2.0 m =
= 1.7 m =
4.1 m = 6.5 m =
4.0 m
DOLOS STONE DESIGN Breakage formula for Dolos
→ Burcharth, 1993b and Liu, 1995 B = C 0 M C1 f t
C2
HS
C3
Where is; B = Relative Breakage M = Armor unit mass in ton, 2.5 ≤ m ≤ 50
FT = Concrete static tensile strength in Mpa , 2 ≤ FT ≤ 4 H S = Significance wave height
C 0 , C1 , C 2 , C 3 =Fitted Parameters
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COASTAL AND HARBOR ENGINEERING
∴This formula considered the effect of static and impact stress.
Co
C1
C2
C3
Waist Ratio
Trunk of Dolosses
0.00973 0.00546 0.01306
-0.749 -0.1782 -0.507
-2.58 -1.22 -0.507
4.143 3.147 2.871
0.325 0.37 0.42
Round end Dolos
0.025
-0.65
-0.66
2.42
0.37
Design Method
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COASTAL AND HARBOR ENGINEERING
Type of Armor Stones
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COASTAL AND HARBOR ENGINEERING
Dimension for Dolos Stone
Wave Height vs Max. flexural tensile stress for several Dolos waist ratios
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COASTAL AND HARBOR ENGINEERING
Dolos mass vs Max. flexural tensile stress for several Dolos waist ratios
Dolos Dimension Assumption(inches) A = 36. 43 in, B = 58.28 in, C = 182.13 in, D = 10.38 in n(σ S ) P < f T
Where is; 18
COASTAL AND HARBOR ENGINEERING N = Model Scale Factor,
(σ S ) P = Static principal stress in model Dolos with probability of an exceedence, p f T=Prototype concrete static tensile strength (Mpa) So,
( σ S ) P = 10 ( log( σ
S
) est ) + ( 0.31[ φ − 1( P ) ] )
Where is ; Dv Log (σ S ) est = −2.28 + 0.91α + 0.3 − 0.45 + 0.34 ; n W n = 9.43 0.1549W a
1/ 3
;
α = tangent of seaward armor slope;
=Layer ( 0 for top, 1 for bottom) = in here since top = 0; DV = Vertical distance from crest to stressed Dolos location;
(φ ) −1
p
=Tabulated inverse normal variation
W = Prototype armor unit weight W a =Armor concrete specific weight −1 Assume the probability of an exceedence = 0.1 → (φ ) p = 1.28
Checking Criteria For Mcr and Tcr
γ SM K M σ 1 ≤ φ ( 0.7 M cr ) γ ST K T σ 1 < φ ( 0.7Tcr + TS )
Where is; 19
COASTAL AND HARBOR ENGINEERING
S M = 0.1053 ( rc ) 3 → Section modules for flexure 3 S T = 0.2105( rc ) → Section module for torsion
V = Dolos waist ratio; C= Dolos fluke length;
K M = K T = 0.6 → Moment and torsion distribution factor M cr = Tcr = 0.7 f ct → Critical strength of concrete in moment and torsion TS =Strength contribution from the torsion steel
σ1 = Principal stress reinforcement Check Steel Reinforcement For Torsion
AS > γ ( S T K T σ 1 ) − φ ( 0.7Tcr ) φf y Rh Where is; As = Total area of steel intersecting the crack Rh = Distance to the center of the section Fy = Yield Strength of the steel
∴Bending reinforcement design needs to be calculated by Whitney rectangular stress that is out of this calculation for this project.
Structure Summary: Total height (h + R)
: 8.6 m
Slope (tan α)
: 1:2
Crest Width (B)
:2m
Freeboard (R)
:2m
Estimated overtopping (q)
: 0.2 m3/sec/m
Settlement (ρ)
: 0.1 m (assumed)
Toe-to-Toe width
: 44.4 m 20
COASTAL AND HARBOR ENGINEERING
Armor:
W50 = 0.99 t n = 2, t = 2 m Na/A = 5 units/m2 V/L = 40 m3/m
First Under-Layer:
W50 = 22.7 kg n = 2, t = 0.5 m V/L = 18 m3/m
Second Under-Layer:
W50 = 1 kg n = 2, t = 0.15 m V/L = 4.9 m3/m
Core:
W50 = 0.05 kg V/L = 102 m3/m
Toe:
W50 = 136 kg hb = 4.1 m below SWL toe height = 0.8 m Bt = 4 m toe base width = 7.2 m V/L = 9 m3/m
Bedding:
W50 = 4.5 kg thickness = 0.6 m horizontal length = 47.4 m V/L = 30.1 m3/m
Check Settlement & Bearing Capacity: Breakwater Load
→ Volume & Weight above SWL (dry, unsubmerged load): Height = 8.6 – 5.0 = 3.6 m, B = 2 Width at WL = B + 2hcot α = 2 + 2×3.6×2 = 16.4 m V/L = ½ 3.6(2 + 17.2) = 34.6 m3/m Weight of material = Wabove WL = γ (1-P/100) V/L = 2.5 (1 – 0.37)34.6 = 54.5 t/m
→ Submerged Volume & Weight; Submerged, V/Ltotal = (V/L)armor + (V/L)ul1 + (V/L)ul2 + (V/L)core +(V/L)toe + (V/L)bed = 40 + 18 + 4.9 + 102 + 9 + 30.1 = 204 m3/m 21
COASTAL AND HARBOR ENGINEERING
∴V/Lsubmerged = 204 – 34.6 = 169.4 m3/m W = [γ(1 – P/100) + γw(P/100)] V/Lsubmerged = [2.5(1-0.37) + 1×0.37]169.4 Wbelow WL = 329 t/m
→ Total Load, Δσ = (Wabove WL + Wbelow WL)/(foundation width) Sand Layer: Δσ = (54.5 + 329)/44.4 = 8.6 t/m2 Clay Layer → correct for distribution of load through sand layer (see diagram) Clay Layer, Δσ = (54.5 + 329)/[44.4 + 2×(5.00.6)×2] = 6.19 t/m2
γ' = 4kN/m3 c = 50kPa
Bearing Capacity Evaluate the ultimate bearing capacity, qu, for each level (very conservative, but simple) For saturated, submerged soils;
∴Strip foundations:
q ult = q c + q q + qγ = cN c + qN q + 0.5γ ' BN γ
NOTE: This formula is not for multiple layer soils. This calculation will only give a rough approximation. Sand Layer: γ = 17 kN/m3, φ = 30°, c = 0; Terzaghi Table: Nc = 37.16, Nq = 22.46, Nγ = 19.13 22
COASTAL AND HARBOR ENGINEERING
Df = Foundation depth (bedding layer thickness) = 0.6 m Assume γw = 10 kN/m3 Breakwater foundation width (neglect bed) = 44.4 m qc = cNc = 0 q= γ'DfNq = (17-10)×0.6×22.46 = 94 kN/m2 q qγ = ½ γ'BNγ = ½ ×(17-10) ×44.4×19.13 = 2973 kN/m2 qu = 0 + 94 + 2973 = 3067 kN/m2 = 325 t/m2 Δσ = 8.6 t/m2 FS = qu/ Δσ = 307/8.6 = 35.7 FSsand = 36
Clay Layer: γ = 14 kN/m3, φ = 0, c = 50 kN/m2 Terzaghi Table: Nc = 5.7, Nq = 1, Nγ = 0 Df = 0 qc = cNc = 50×5.7 = 285 kN/m2 q= γ'DfNq = 0
q
qγ = ½ γ'BNγ = 0 qu = 285 + 0 + 0 = 285 kN/m2 = 28.5 t/m2 clay layer also supports the sand layer: Δσsand = 0.7×8.6 t/m2 = 6.02 t/m2 Δσ = 6.19 t/m2 + 6.02 t/m2 = 12.21 t/m2 FS = qu/ Δσ = 28.5/12.21 = 3.2 FSclay = 2.3
∴Preliminary Safety Factor , FS = 2.3 23
COASTAL AND HARBOR ENGINEERING
Settlement Sand Layer:
∆σ = 8.6t / m 2
Clay Layer:
∆σ = 6.2t / m 2
Check settlement in Sand; Assume
L > 10 , I Z = I Z 10 = 0.2 ; B
Depth of I zp : z = z10 = 1.0 B → Z = 1
σ ' zp = σ zp − u = γ ' ZB = (1.7 − 1) B = 0.7 × 44.4 = 31t / m 2
∆ σ ' Z = q − σ ' 0 = 8.6 − (1.7 − 1) × 0.6 = 8.2t / m 2 I zp = 0.5 + 0.1
∆σ z
σ zp
'
= 0.5 + 0.1
8.2 = 0.55 31
Depth of influence: z = 4 B = 4 × 44.4 = 178m Assume one layer, ∆z = 4.9m 24
COASTAL AND HARBOR ENGINEERING
z=
I − 0.2 4.9 0.55 − 0.2 = 2.45m →I Z = 0.2 + ZP z = 0.22 + 2.45 = 0.24 2 zp 44.4
Assume
qc ~ 25bar = 50t / m 2 (see table in notes) N 60
L = 10 → E = 3.5q c = 3.5 × 50 = 175t / m 2 B
(Note: E table in notes gives E 10x higher for loose sand)
σ 0' = 1 − 0.5 (1.7 − 1) 0.6 = 0.97 ; C1 = 1 − 0.5 ∆σ ' 8.2 Z t yrs 25 = 1 + 0.2 log 10 C 2 = 1 + 0.2 log 10 = 1.5, assume 25-year life; 0.1 0.1 n I 0.22 ∴ρ = C1C 2 ∆σ ∑i =1 Z ∆z i = 0.97 ×1.5 × 4.9 = 0.01m 175 E i
Check settlement in clay;
⇒ Primary consolidation settlement ( ρ c )
a = 3 × 10 − 3 m 2 / kN ,
γ =14 kN / m 3 , φ = 0 0 , c = 50 kPa, e 0 = 2.2 , k = 10 −5 cm / s , v C C = 0.3
∆σ = 6.2t / m 2
σ 0 ' = (1.7 −1) × 4.9 +
1 (1.4 −1) × 21.5 = 7.7t / m 2 2
Assume C R = 0.2C C = 0.06 25
COASTAL AND HARBOR ENGINEERING
Over-consolidated; ρc =
− 0.06 × 21.5 7 .7 + 5 .0 log = 0 .9 m 1 + 2 .2 7 .7
Consider time to consolidate; k = 10 −5 cm / s ×10 −2 m / cm ×3600 s / hr × 24hrs / day ×365days / yr = 3.15m / yr
cV =
k (1 + e 0 ) 3.15(1 + 2.2 ) = = 336m 2 / yr −3 γ W aV 10 × 3 × 10
N = 1 , TV ( 95% ) = 1.129 TV =
cv t H N
2
→ t = TV
H2 21.5 2 = 1.129 = 1.55 yrs cv 336
⇒ Secondary consolidation settlement ( ρ S ) Assume
Ca ~ 0.03 → C a ~ 0.01 Cc
Assume t p = 2 yrs and the breakwater lifetime is 25 yrs; C H ρ S = α 1 + e0
t log F t p
0.01× 21.5 25 = log = 0.07 m 1 + 2.2 2
∴ ρ = ρ1 + ρ c + ρ s = 0 + 0.09 + 0.07 = 0.16m
So that, Total Settlement, ρ = ρsand + ρclay = 0.01 + 0.16 = 0.17 m ~ 0.2 m It should recalculate design with ρ ~ 0.2m vice 0.1 m
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COASTAL AND HARBOR ENGINEERING
REFERENCES - Coastal Engineering Manual - Part IV, USACE, 1 August 2008 - Iranian Coastal Engineering Code manual – NO.300-5 / IRAN Ministry of Transportation,2005
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