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typical steps of a rouble mound breakwater desing
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Rubble Mound Breakwater Design Example Given:
Design Conditions Water depth: 5.5 m Beach slope: 1:20 Design high water: 1.7 m Design wave Hs = 2 m
H1/10 = 2.5 m Tm = 8 sec Lo = 100 m
Allowable overtopping: 0.4 m3/sec/m Armor unit: rough quarry stone
Soil data:
0 m
5.5 m
SM (sand)
fine to medium loose
γ = 17 kN/m3 φ = 30° c = 0
21.5 m
CH (clay) soft
over-consolidated
γ = 14 kN/m3 φ = 0° c = 50 kPa eo = 2.2 k = 10-5 cm/s av = 3x10-3 m2/kN Cc = 0.3
limestone
B
crown/cap ocean side bay/harbor side crest
armor layer, W
hc R first underlayer DHW
SWL α hb αb
h second underlayer t toe
core/base
bedding and/or filter Bt
Assume: Armor and underlayer material is quarry stone: γa = 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 = 5.5 m, DHW = 1.7 m h = 5.5 + 1.7 = 7.2 m h = 7.2 m
Assume listed conditions are at structure toe. Hs = H1/3 = 2 m T = 8 sec Lo = 100 m
π
π
=
π h
L2tanh
L2g
T2 2
at h = 7.2
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 ~ 0.78 at DHW: Hb = 0.78×7.2 = 5.6 m at SWL: Hb = 0.78×5.5 = 4.3 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=η
Overtopping Discharge (CEM VI-5, pp. 19-33)
using the Owen model(Table VI-5-8):
γ
−=r
m
ms
bRexpaTgH
q *
where
π=
2s
HRR m
s
*m is the relative freeboard
2m
s
o
sm T
Hg
2LHs π
== 2m
s
s
*m gT
HHRR = •
from Table VI-5-8: slope 1:2 a = 0.013, b = 22 rock riprap > 2D thick γr ~ 0.55
•
• solving:
041.0828.9013.0
4.0ln2255.0
TagHqln
bR
ms
r*m =
×××−=
γ−=
m 45.12
88.9041.02H
gTRHR2
s
2m*
ms =×
×==
Rovertop = 1.45 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 = 0.96, B = 1.17, C = 0.46, D = 1.97 •
• (D/B)1/C = (1.97/1.17)(1/0.46) = 3.1, from above ξm = 2.7 for 1.5 < ξm ≤ (D/B)1/C C
mS%i,u BHR ξ=
( ) 85.17.217.1BHR 46.0CmS%2,u ==ξ=
Reduced Run-up assume 11155.0anglewave
watershallowberm
roughnesssurface ×××=γγγγ
( ) 155.085.1anglewave
watershallowberm
roughnesssurfaceHRHR S%2,uSuR =×=γγγγ=
•
R = Hs = 2m •
Rrun-up = 2 m
Choose the run-up requirement (purpose has not been specified, simpler)
actual overtopping 0565.088.9
222
gTH
HRR 22
m
s
s
*m =
×==
( )( )( ) msec//m 2.055.00565.022exp828.9013.0bRexpTagHq 3
r
*m
ms ≈
×
−=
γ
−=
•
R = 2 m q = 0.2 m3/sec/m
Settlement: must be determined later assume ρtotal = 0.1 m ρtotal = 0.1 m
Design elevation = DHW + η + R + ρtotal = 7.2 + 0 + 2 + 0.1 = 9.3 m h + R = 9.3 m
BW Dimension Summary: Assumed
structure is symmetric, α = αb • • • •
no set-down no crown, hc = R total settlement = 0.1 m (adjust later)
h = 7.2 m h + R = 9.3 m tan α = 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 = γa/γw = (2.5 t/m3)/(1 t/m3) = 2.5
( ) ( )t74.0
215.2425.2
cot1sgKHW 3
3
3D
3a
50 =−×
=α−
γ=
Table VI-5-50 gives rock sizes: W ~ 0.77 t •
W50 = 0.77 t
Armor thickness
n = 2; k∆ = 1.0, P = 37% from Table VI-5-51 •
• m 4.15.2
77.012Wnkt3/13/1
a
=
××=
γ
= ∆
tarmor = 1.4 m
Crest width (B) (minimum n = 3): m 25.2
77.013Wk3B3/13/1
a
=
××=
γ
= ∆
B = 2 m
Number of armor units per unit surface area
( ) 8.277.05.237.0112
W100P1nk
AN 3/23/2
aa =
−××=
γ
−= ∆
Na/A = 2.8 units/m2
Volume of armor per unit length
( )[ ] [ 9.5423.9224.1cotRh2BtLV
=××+=α++= ]
V/L = 54.9 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
½ 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
+≈
( )α+=
α+=2
22
cot1h
cothhc
( )α−α+= sinTcotH2Ab
( )α−α+= csccotT2Aa
a A
α c
h 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 = 0.77 t/10 = 0.077 t × 1000 = 77 kg next larger available size is 90.7 kg
thickness m 66.05.2
091.012Wnkt3/13/1
a
=
××=
γ
= ∆
Volume per unit length of breakwater referring to diagram: h = 9.2 m – tarmor = 9.2 – 1.4 = 7.8 m; t = tul1 = 0.7 m, T = tarmor = 1.4 A = Bcrest = 2 m, cot α = 2
( ) ( ) m 4.1)2.22(4.122csccotT2Aa =−+=α−α+=
m 15418.6cot1hc 2 =+=α+=
( ) ( ) mm 221524.17.0c2atLV 3=×+=+≈
•
First Under-Layer W10 = 91 kg tul1 = 0.7 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 = 0.75 t/200 = 0.004 t × 1000 = 4 kg next larger available size is 4.5 kg
thickness m 24.05.2
0045.012Wnkt3/13/1
a
=
××=
γ
= ∆
Volume per unit length of breakwater referring to diagram: h = 9.2 m – tarmor – tul1 = 9.2 – 1.4 – 0.7 = 6.1 m t = tul2 = 0.24 m, T = tul1 = 0.7 A = aul1 = 1.4 m, cot α = 2
( ) ( ) m 1.1)2.22(7.024.1csccotT2Aa =−+=α−α+=
m 6.13411.6cot1hc 2 =+=α+=
( ) ( ) mm 8.66.1321.124.0c2atLV 3=×+=+≈
•
Second Under-Layer W200 = 4.5 kg tul2 = 0.24 m V/Lul1 = 6.8 m3/m
Core dynamic load requirement: 25 to15wW core ≤ W = 4.5 kg wcore ≥ 4.5/25 – 4.5/15 = 0.18 – 0.3 kg
•
W4000 = 0.75 t/4000 = 0.00019 t × 1000 = 0.2 kg • •
•
next larger available size is 0.23 kg
thickness m 24.05.2
0045.012Wnkt3/13/1
a
=
××=
γ
= ∆
Volume per unit length of breakwater referring to diagram: h = 9.2 m – tarmor – tul1 – tul2 = 9.2 – 1.4 – 0.7 – 0.24 = 5.9 m H = hul2 = 6.1 m, T = tul2 = 0.24 A = aul2 = 1.1 m; cot α = 2
( ) ( ) m 1)2.22(24.021.1csccotT2Aa =−+=α−α+=( )
( ) m 4.242.224.021.621.1cscTcotH2Ab =×−×+=α−α+=
trapezoid: ( ) ( ) mm 754.2419.5bahLV 3
21
21 =+=+≈
•
Core W4000 = 0.23 kg V/L = 75 m3/m
Toe Design: Toe Berm Width (Bt) should be the maximum of Bt = 2H or Bt = 0.4h, and at least 3
stones wide: 2H = 4 m, 0.4h = 0.4×5.5 = 2.2 m (use lower water level)
•
•
•
•
assume Bt = 4 m
assume height of toe = 1.4 m (guess) hb = 5.5 – 1.4 = 4.1 m (use lower water level)
Table VI-5-45 with hb/h = 4.1/5.5 = 0.75 Ns3 ~ 60
( ) ( ) t1.0
15.26025.2
1sgNHW 3
3
33s
3S =
−×
=−
γ= nearest size is 136 kg = 0.14 t
m 38.05.2
14.0WD3/13/1
s
=
=
γ
= 2 stones height = 2×0.38 = 0.76 m < 1.4 m
Table VI-5-48
k = 2π/Lm = 2π/62 = 0.101 m-1 2khb = 2×0.101×5.8 = 1.17
( ) ( ) 124.04101.0sin17.1sinh
17.1kBsinkh2sinh
kh2K 2t
2
b
b =×==
( )
( ) 7.421.4
124.0124.015.1exp8.1
21.4
124.0124.013.1
Hh
KK15.1exp8.1
Hh
KK13.1N
3/1
2
3/1
s
b3/1
2
s
b3/1s
=
−
−+
−
=
−−+
−
=
( )
Ns3 ~ 103
( ) t06.0
15.210325.2
1sgNH
3
3
33s
3S =
−×
=−
γ=W
•
use W = 0.14 t and recalculate with ht = 5.5 – 0.8 = 4.7 m hb/h = 4.7/5.5 = 0.85
this is not on the chart Ns3 ~ 60 keep previous calculation
•
Wtoe = 136 kg hb = 4.7 m (below SWL) toe height = 0.8 m Bt = 4 m
• Toe volume
assume slope is 1:2 base length = Bt + 2(SWL-hb)cot α = 4 + 2×0.8×2 = 7.2 m
assume trapezoidal V/L = (SWL-hb)(Bt + base) = 0.8(4 + 7.2) = 9 m3/m
V/Ltoe = 9 m3/m
Toe-to-Toe Width:
W = 2Bt + 2(SWL-hb)cot αt + B + 2(hb + DHW + R + ρ)cot α
= 2×4 + 2×0.8×2 + 2 + 2× (4.7 + 1.7 + 2 + 0.1) ×2 = 47.2 m
Filter/Bed Design:
To prevent material from leaching out: 02 to15WW
)core(50
)bed(50 < •
•
•
• •
Wcore = 0.23 kg Wbed > 3.5 – 4.6 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.
Bedding layer should be 0.6 m thick, d50 ~ 12 cm (cobble) Extent: toe-to-toe width + 2×1.5 m = 47.2 + 3 = 50.2 m
Structure Summary:
total height (h + R): 9.3 m
slope (tan α): 1:2
Crest Width (B): 2 m
Freeboard (R): 2 m
Estimated Overtopping (q) 0.2 m3/sec/m
Settlement (ρ): 0.1 m (assumed)
Toe-to-Toe width: 47.2 m
Armor: W50 = 0.77 t n = 2, t = 1.4 m Na/A = 2.8 units/m2 V/L = 54.9 m3/m
First Under-Layer: W50 = 91 kg n = 2, t = 0.7 m V/L = 22 m3/m
Second Under-Layer: W50 = 4.5 kg n = 2, t = 0.24 m V/L = 6.8 m3/m
Core: W50 = 0.23 kg V/L = 75 m3/m
Toe: W50 = 136 kg hb = 4.7 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 = 50.2 m V/L = 30.1 m3/m
Settlement & Bearing Capacity:
BW Load Volume & Weight above SWL (dry, unsubmerged load):
Height = 9.3 – 5.5 = 3.8 m B = 2 Width at WL = B + 2hcot α = 2 + 2×3.8×2 = 17.2 m V/L = ½ 3.8(2 + 17.2) = 36.5 m3/m Weight of material = Wabove WL = γ (1-P/100) V/L = 2.5 (1 – 0.37)36.5 = 57.5 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 = 55 + 22 + 6.8 + 75 + 9 + 30.1 = 198 m3/m V/Lsubmerged = 198 – 36.5 = 162 m3/m W = [γ(1 – P/100) + γw(P/100)] V/Lsubmerged = [2.5(1-0.37) + 1×0.37]162 Wbelow WL = 315 t/m
Total Load ∆σ = (Wabove WL + Wbelow WL)/(foundation width) Sand Layer: ∆σ = (57.5 + 315)/47.2 = 7.9 t/m2 Clay Layer correct for distribution of load through sand layer (see diagram) ∆σ = (57.5 + 315)/[47.2 + 2×(5.5-0.6)×2] = 5.5 t/m2
DHW
SWL
Sand H1 γ' = 7 kN/m3
φ = 35°
Clay BB γ' = 4 kN/m3 c = 20 kPa
BB + 2H1cot φ
Bearing Capacity Evaluate the ultimate bearing capacity, qu, for each level (very conservative, but simple)
For saturated, submerged soils strip foundations: γγ γ′++=++= BN5.0qNcNqqqq qcqcu
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 Df = Foundation depth (bedding layer thickness) = 0.6 m Assume γw = 10 kN/m3 BW foundation width (neglect bed) = 47.2 m qc = cNc = 0 qq = γ'DfNq = (17-10)×0.6×22.46 = 94 kN/m2 qγ = ½ γ'BNγ = ½ ×(17-10) ×47.2×19.13 = 3160 kN/m2 qu = 0 + 94 + 3160 = 3254 kN/m2 = 325 t/m2 ∆σ = 7.9 t/m2 FS = qu/ ∆σ = 325/15.1 = 21.5
FSsand = 21 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
qq = γ'DfNq = 0 qγ = ½ γ'BNγ = 0 qu = 285 + 0 + 0 = 285 kN/m2 = 28.5 t/m2 clay layer also supports the sand layer: ∆σsand = 0.7×4.9 t/m2 = 3.4 t/m2
∆σ = 5.5 t/m2 + 3.4 t/m2 = 8.9 t/m2 FS = qu/ ∆σ = 28.5/8.9 = 3.2
FSclay = 3.2 Preliminary Safety Factor
FS = 3.2
Settlement
Sand Layer: ∆σ = 7.9 t/m2
Clay Layer: ∆σ = 5.5 t/m2
Settlement in Sand:
Assume L/B > 10
Iz = Iz10 = 0.2
depth of Izp: z = z10 = 1.0B Z = 1
σ'zp = σzp – u = γ'ZB = (1.7 – 1) B = 0.7×47.2 = 33 t/m2
∆σ'z = q - σ'0 = 7.9 - (1.7 – 1)×0.6 = 7.5 t/m2
55.033
5.71.05.0''1.05.0I
zp
zzp =+=
σσ∆
+=
depth of influence: z = 4B = 4×47.2 = 190 m
assume one layer ∆z = 4.9 m
z = 4.9/2 = 2.45 m 22.045.22.47
2.055.02.0zz
2.0I2.0I
p
zpz =
−+=
−+=
assume qc/N60 ~ 5 bar = 50 t/m2 (see table in notes)
L/B = 10 E = 3.5qc = 3.5×50 = 175 t/m2 (note: E table in notes gives E 10× higher for loose sand)
( ) 97.05.7
6.017.15.01'
'5.01CZ
01 =
−
−=
σ∆
σ−=
5.11.0
25 log 0.211.0
t log 0.21C 10
yrs102 =
+=
+= , assume 25 yr life
m 01.09.41750.221.50.97 ∆z
EΙCCρ i
i
n
1i
z21 =
××=
σ∆= ∑
=
ρsand = 0.01 m
Settlement in Clay:
Primary Consolidation Settlement (ρc)
γ = 14 kN/m3, φ = 0°, c = 50 kPa, eo = 2.2, k = 10-5 cm/s, av = 3x10-3 m2/kN, Cc = 0.3
∆σ = 5.5 t/m2
σ'0 = (1.7 – 1)×4.9 + ½ (1.4-1)×21.5 = 7.7 t/m2
assume CR = 0.2Cc = 0.06
Over-consolidated: m 09.07.7
5.57.7log2.21
5.2106.0c =
+
+×
−=ρ
Consider time to consolidate:
k = 10-5cm/s × 10-2m/cm × 3600s/hr × 24hrs/day × 365days/yr = 3.15 m/yr
( ) ( ) yr/m 33610310
2.2115.3ae1kc 2
3vw
0v =
××+
=γ
+= −
N = 1, Tv (95%) = 1.129
( )2v
v NHtcT = yrs 55.1
3365.21129.1
cHTt
2
v
2
v ===
Secondary Consolidation Settlement (ρs) Assume Cα/Cc ~ 0.03 Cα ~ 0.01 assume tp = 2 yrs and the breakwater lifetime is 25 yrs
m 07.0225 log
2.215.1201.0
tt log
e1HCρ
P
F
0s =
+×
=
+
= α
m 16.007.009.00sci =++=ρ+ρ+ρ/=ρ
ρclay = 0.16 m
Total Settlement
ρ = ρsand + ρclay = 0.01 + 0.16 = 0.17 m
ρtotal = 0.17 m should recalculate design with ρ ~ 0.2 m vice 0.1 m