Rubble Mound Breakwater Design Equations - ESSIE

1 Rubble Mound Breakwater Design Example Given: Design Conditions Water depth: m Beach slope: 1:20 Design high water: m Design wave Hs = 2 m H1/10 = m Tm = 8 sec Lo = 100 m Allowable overtopping: m3/sec/m Armor unit: rough quarry stone Soil data: 0 m m SM (sand) fine to medium loose = 17 kN/m3 = 30 c = 0 m CH (clay) soft over-consolidated = 14 kN/m3 = 0 c = 50 kPa eo = k = 10-5 cm/s av = 3x10-3 m2/kN Cc = limestone B crown/cap ocean side bay/harbor side crest armor layer, W hc R first underlayer DHW SWL hb b h second underlayer t toe core/basebedding and/or filter Bt Assume: Armor and underlayer material is quarry stone: a = t/m3 Structure slope: 1:2 Structure will be symmetric (this may be changed to reduce structure size in necessary) Specify Design Condition: SWL = m, DHW = m h = + = m h = m Assume listed conditions are at structure toe. Hs = H1/3 = 2 m T = 8 sec Lo = 100 m = hL2tanhL2gT22 at h = Lm = 62 m 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 ~ at DHW: Hb = = m at SWL: Hb = = m Alternate methods in CEM II-4 Both wave heights in (1) are greater than Hs waves are not breaking and Design H = Hs = 2 m H = 2 m Set BW Dimensions (controlled by height & slope): Set-up: waves are not breaking per the previous calc no set-up 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= Overtopping Discharge (CEM VI-5, pp.)

2 19-33) using the Owen model(Table VI-5-8): =rmmsbRexpaTgHq* where =2sHRRms*m is the relative freeboard 2msosmTHg2 LHs == 2mss*mgTHHRR= from Table VI-5-8: slope 1:2 a = , b = 22 rock riprap > 2D thick r ~ solving: *m= = = m *ms= == Rovertop = m Run-up based on surf zone parameter at the structure, using CEM equation VI-5-13 Coefficients from VI-5-5: 2% run-up A = , B = , C = , D = (D/B)1/C = ( )(1 ) = , from above m = for < m (D/B)1/C CmS%i,uBHR = () ,u== = Reduced Run-up assume = () ,uSuR= = = R = Hs = 2m Rrun-up = 2 m Choose the run-up requirement (purpose has not been specified, simpler) actual overtopping *m= == ()()()msec//m *mms = = R = 2 m q = m3/sec/m Settlement: must be determined later assume total = m total = m Design elevation = DHW + + R + total = + 0 + 2 + = m h + R = m BW Dimension Summary: Assumed structure is symmetric, = b no set-down no crown, hc = R total settlement = m (adjust later) h = m h + R = m tan = 1/2 Armor Unit Design : Assume Armor unit is rough quarry stone, 2 layers, no overtopping Table VI-5-22 applies non-breaking waves, 0-5% damage, random placement: KD = 4 sg = a/ w = ( t/m3)/(1 t/m3) = ()() = = Table VI-5-50 gives rock sizes: W ~ t W50 = t Armor thickness n = 2.

3 K = , P = 37% from Table VI-5-51 m = = tarmor = m Crest width (B) (minimum n = 3): m = = B = 2 m Number of armor units per unit surface area () = = Na/A = units/m2 Volume of armor per unit length ()[][ += ++=] V/L = m3/m Under-layer Design : The goal to reduce the size of the stone to at point where W/wcore 15-25, where W is the stone in the layer covering the core. Roughly, this gives a size of ~W/4000 for the core lb stones, with 2 inch diameter. If some other size is readily available, that might be the goal. Must check to ensure the W/wcore 15-25 is met once the core over-layer is known. Diagram for Volume calculations (quarry stone is sold by unit weight & total volume) ()c2atLV+ () += +=222cot1hcothhc () +=sinTcotH2Ab () +=csccotT2Aa a A ch H t T b B First Under-Layer minimum two stone thick (n = 2) under-layer unit weight = W/10 since cover layer and first underlayer are both stone W10 = t/10 = t 1000 = 77 kg next larger available size is kg thickness m = = Volume per unit length of Breakwater referring to diagram: h = m tarmor = = m.

4 T = tul1 = m, T = tarmor = A = Bcrest = 2 m, cot = 2 ()()m ) ( += += m += += ()()mm +=+ First Under-Layer W10 = 91 kg tul1 = m V/Lul1 = 22 m3/m Second Under-Layer minimum two stone thick (n = 2) under-layer unit weight = W/20 of the layer above W/200 of armor W200 = t/200 = t 1000 = 4 kg next larger available size is kg thickness m = = Volume per unit length of Breakwater referring to diagram: h = m tarmor tul1 = = m t = tul2 = m, T = tul1 = A = aul1 = m, cot = 2 ()()m ) ( += += m += += ()()mm +=+ Second Under-Layer W200 = kg tul2 = m V/Lul1 = m3/m Core dynamic load requirement: 25 to15wWcore W = kg wcore = kg W4000 = t/4000 = t 1000 = kg next larger available size is kg thickness m = = Volume per unit length of Breakwater referring to diagram: h = m tarmor tul1 tul2 = = m H = hul2 = m, T = tul2 = A = aul2 = m.

5 Cot = 2 ()()m 1) ( += +=() ()m += += trapezoid: ()()mm +=+ Core W4000 = kg V/L = 75 m3/m Toe Design : Toe Berm Width (Bt) should be the maximum of Bt = 2H or Bt = , and at least 3 stones wide: 2H = 4 m, = = m (use lower water level) assume Bt = 4 m assume height of toe = m (guess) hb = = m (use lower water level) Table VI-5-45 with hb/h = = Ns3 ~ 60 ()() = = nearest size is 136 kg = t m = = 2 stones height = 2 = m < m Table VI-5-48 k = 2 /Lm = 2 /62 = m-1 2khb = 2 = ()() == ()() + = + =() Ns3 ~ 103 () = =W use W = t and recalculate with ht = = m hb/h = = this is not on the chart Ns3 ~ 60 keep previous calculation Wtoe = 136 kg hb = m (below SWL) toe height = m Bt = 4 m Toe volume assume slope is 1:2 base length = Bt + 2(SWL-hb)cot = 4 + 2 2 = m assume trapezoidal V/L = (SWL-hb)(Bt + base) = (4 + ) = 9 m3/m V/Ltoe = 9 m3/m Toe-to-Toe Width: W = 2Bt + 2(SWL-hb)cot t + B + 2(hb + DHW + R + )cot = 2 4 + 2 2 + 2 + 2 ( + + 2 + ) 2 = m Filter/Bed Design : To prevent material from leaching out: 02 to15WW)core(50)bed(50< Wcore = kg Wbed > kg dbed ~ 12 cm cobble General guidelines for stability against wave attack, bedding Layer thickness should be: o 2-3 times the diameter for large stone o 10 cm for coarse sand o 20 cm for gravel For foundation stability Bedding Layer thickness should be at least 2 feet Bedding Layer should extend 5 feet horizontally beyond the toe cover stone.

6 Bedding layer should be m thick, d50 ~ 12 cm (cobble) Extent: toe-to-toe width + 2 m = + 3 = m Structure Summary: total height (h + R): m slope (tan ): 1:2 Crest Width (B): 2 m Freeboard (R): 2 m Estimated Overtopping (q) m3/sec/m Settlement ( ): m (assumed) Toe-to-Toe width: m Armor: W50 = t n = 2, t = m Na/A = units/m2 V/L = m3/m First Under-Layer: W50 = 91 kg n = 2, t = m V/L = 22 m3/m Second Under-Layer: W50 = kg n = 2, t = m V/L = m3/m Core: W50 = kg V/L = 75 m3/m Toe: W50 = 136 kg hb = m below SWL toe height = m Bt = 4 m toe base width = m V/L = 9 m3/m Bedding: W50 = kg thickness = m horizontal length = m V/L = m3/m Settlement & Bearing Capacity: BW Load Volume & Weight above SWL (dry, unsubmerged load): Height = = m B = 2 Width at WL = B + 2hcot = 2 + 2 2 = m V/L = (2 + ) = m3/m Weight of material = Wabove WL = (1-P/100) V/L = (1 ) = 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 = 55 + 22 + + 75 + 9 + = 198 m3/m V/Lsubmerged = 198 = 162 m3/m W = [ (1 P/100) + w(P/100)] V/Lsubmerged = [ ( ) + 1 ]162 Wbelow WL = 315 t/m Total Load = (Wabove WL + Wbelow WL)/(foundation width) Sand Layer.

7 = ( + 315) = t/m2 Clay Layer correct for distribution of load through sand layer (see diagram) = ( + 315)/[ + 2 ( ) 2] = t/m2 DHW SWL Sand H1 ' = 7 kN/m3 = 35 Clay BB ' = 4 kN/m3 c = 20 kPa BB + 2H1cot Bearing Capacity Evaluate the ultimate bearing capacity, qu, for each level (very conservative, but simple) For saturated, submerged soils strip foundations: ++=++= NOTE: This formula is not for multiple layer soils. This calculation will only give a rough approximation. Sand Layer: = 17 kN/m3, = 30 , c = 0 Terzaghi Table: Nc = , Nq = , N = Df = Foundation depth (bedding layer thickness) = m Assume w = 10 kN/m3 BW foundation width (neglect bed) = m qc = cNc = 0 qq = 'DfNq = (17-10) = 94 kN/m2 q = 'BN = (17-10) = 3160 kN/m2 qu = 0 + 94 + 3160 = 3254 kN/m2 = 325 t/m2 = t/m2 FS = qu/ = 325 = FSsand = 21 Clay Layer: = 14 kN/m3, = 0, c = 50 kN/m2 Terzaghi Table: Nc = , Nq = 1, N = 0 Df = 0 qc = cNc = 50 = 285 kN/m2 qq = 'DfNq = 0 q = 'BN = 0 qu = 285 + 0 + 0 = 285 kN/m2 = t/m2 clay layer also supports the sand layer: sand = t/m2 = t/m2 = t/m2 + t/m2 = t/m2 FS = qu/ = = FSclay = Preliminary Safety Factor FS = Settlement Sand Layer: = t/m2 Clay Layer: = t/m2 Settlement in Sand: Assume L/B > 10 Iz = Iz10 = depth of Izp.

8 Z = z10 = Z = 1 'zp = zp u = 'ZB = ( 1) B = = 33 t/m2 'z = q - '0 = - ( 1) = t/m2 '' += += depth of influence: z = 4B = 4 = 190 m assume one layer z = m z = = m += += assume qc/N60 ~ 5 bar = 50 t/m2 (see table in notes) L/B = 10 E = = 50 = 175 t/m2 (note: E table in notes gives E 10 higher for loose sand) () '' = = log log += +=, assume 25 yr life m zE CC iin1iz21= = = = sand = m Settlement in Clay: Primary Consolidation Settlement ( c) = 14 kN/m3, = 0 , c = 50 kPa, eo = , k = 10-5 cm/s, av = 3x10-3 m2/kN, Cc = = t/m2 '0 = ( 1) + ( ) = t/m2 assume CR = = Over-consolidated: m ++ = Consider time to consolidate: k = 10-5cm/s 10-2m/cm 3600s/hr 24hrs/day 365days/yr = m/yr ()()yr/m += += N = 1, Tv (95%) = ()2vvNHtcT= yrs Secondary Consolidation Settlement ( s) Assume C /Cc ~ C ~ assume tp = 2 yrs and the Breakwater lifetime is 25 yrs m log log e1HC PF0s= + = += m ++= + + /= clay = m Total Settlement = sand + clay = + = m total = m should recalculate Design with ~ m vice m