OCE 3016 Class Handout

Coastal Structures In the design of jetties, breakwaters, etc. the size and density of armor stone must be determined. One method of estimating the “stability” of the armor stone is to use the so called Modified Hudson Stability Number, Ns

*, *Know this formula for Quiz*

( )( )

1/32s p*

ss

50w

H LN =

D 1 cotρ − α ρ

Where: Hs= significant wave height Lp= wave length corresponding to Tp Tp= wave period at peak of wave energy density spectrum D50= median stone diameter

=

1/ 350s

m ρ

Has dimensions of

1/3

3

massmass L

L

=

m50= median mass of stone ρs= mass density of rock (stone) ρw= mass density of water cot (α)= slop of structure And the Dimensionless Damage, DD,

( )d

250

ADDD

=

Where: Ad= cross sectional area of damage

An Example:

Before storm shape

After storm shape Ad= Area of damage

Recall:

Wave frequency ω

ωp

Wave Energy Density

pp

2T π=ω

And Lp = wave length of wave with period Tp

The relationship between DD and Modified Hudson Stability Number (sometimes called the Spectral Stability Number) is determined experimentally.

Thus for Ns

* < 6 the

Ns*

Dimensionless Damage

0 6 8

structure will be stable

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