Rubble Mound Breakwater Design [PDF]

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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;



1



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



2



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



14



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



15



COASTAL AND HARBOR ENGINEERING



Type of Armor Stones



16



COASTAL AND HARBOR ENGINEERING



Dimension for Dolos Stone



Wave Height vs Max. flexural tensile stress for several Dolos waist ratios



17



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



26



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



27