Forces on Breakwaters and Breakwater Design

Vertical breakwaters Rubble mound breakwaters Reference:

J. William Kamphuis (2000), Introduction to Coastal Engineering and management, World Scientific

Vertical Break Waters Type of break waters

Concrete caissonSeaward face

Fig. 1 Major component of a vertical breakwater: the Concrete caisson

Seaward face

Typical Failure Mechanisms

Fig. 2 Failure Modes

Forces for Non-Breaking Waves Design philosophy

Breakwater has to be stable under the maximum wave: Hdes = Hmax= KmaxHs (1)

pd

pb

pupd = pb

pv = pb + pu

Fig. 3 Force definitions

Forces on the seaward face:

hydrostatic force from the still water level;

hydrostatic force due to the wave generated water level rise H; Force due to the standing wave against the wall (approximated as static force)

kdkHrmsH coth2

2

= (2)

dessw gHkdp

)cosh(1= (3)

))cosh(

(kd

Hdgp desHv ++= (4)Forces on the landward face:

hydrostatic force from the still water level;

gdpd = (5)

Fig.4 Wave Force Definitions

Forces on the bottom of the caisson:

Buoyancy

uplift due to the seepage flow ;

Resultant Forces

deswHdesw HCHh =+= (6)des

Hw H

C +=1where

011 )( pCHgp Hdes =+=

(7)des

Hw H

CC +== 11desgHp =0 and

where

Define:

Pressure at still water level:

0212 )1( pCphfpw

v ==Pressure at the top of the structure:

wv hf (8)if < else 02 =p

and

=

w

v

hfCC 112

Pressure at the bottom of the structure:

033 )cosh(1 pC

Hkdgp

des

H =

+=

+=des

H

HkdC

)cosh(1

3where (9)

The uplift pressure:

vb gdp = (10)and 033 pCppu ==The sum of horizontal forces:

vvh fppdppF

222131 +++= (11)

The sum of vertical forces:

)2

( ubvmvubmvppBFdFFFF +== (12)

Where Fm is the weight of the structure

The overturning moment around the landward bottom corner:

3)

3(

2)

2(

3)(

2

221

2

2

31

2

30v

uv

vvv

vvvv BpfdfppfdfpdppdpM ++++++=

(13)

The restoring moment around the landward bottom corner:

22

2v

hv

mrBpBFM = (13)

Forces for Breaking Waves Impact force due to

breaking waves is very large but of very short duration (slamming)

Should avoid building structures in wave breaking zone

How to consider the impact of a breaking wave?

Fig. 5 Impact force history

Minikin Method:

a parabolic pressure distribution centered on the still water level is assumed

Fig. 6 Minikin Wave Forces

destss

td gHddLd

dp )(100 += (14)

=

2

21100des

vwdesdd H

fhHpF (15)for hw > fv

desdd HpF 100= (16)for hw fvvdd dFM =~ (17)

The structure based on this design will be very large and costly;

Method by Goda (2000) for breaking waves are used more often

Goda (2000) method is a pseudo-static design assuming wave breaking is a slow process;

Godas method uses Fig. 4 with different definitions for the pressures

Non-Breaking Goda Minikin

Cw 1+H/Hdes 0.75(1+cos ) 0.5C1 1+H/Hdes 0.5(1+cos )(1+4cos2) 0.5/cosh(kd)C2 C1(1-fv/hw)

C3 1/cosh(kd)+ H/Hdes 3 C1Cu C3 0.5(1+cos )13 C31 0.6+0.5((2d/k)/sinh(2d/k))2

2 Min(((d5H-ds)/3d5H)(Hdes/dt)2, 2dt/Hdes)3 1-((hv-fv)/ds(1-(1/cosh(ds/k))4 Max(2, (56))

Table 1: Design parameters for vertical breakwaters (Goda method)

The other quantities in Godas method are given below:

Eq. (18)

Stability Design Basics of risk analysis

Deterministic design SR = (19)

Where R is the resistance (or strength) of the structure, S is the design load related to a certain return period TR and is the factor of safety.

Probability of failure Limit state equation

SRG = (20)Where G is called the failure function. Eq. (20) is a design equation and design equation fails if G < 0.

R = fr (R1, R2, R3, ., Rm) and S = fs (S1, S2, S3, ., Sm)

)0( = GPP rF (21)

Probability of failure

Note G < 0 does not imply failure of structure and it is only a failure of the design condition.

PF depends on all the factors that affect R and S

Levels of probabilistic design

Level III design considers actual probability density functions of all Ri and Si and perform a large number of calculations to determine PF;

Level II assumes normal distribution for all probability density functions and perform the same calculations as in Level III to determine PF;

Level I design is similar to deterministic design but uses partial safety factors taking into account the effects of the probability distributions and a target PF;

Level I design is often used

Level II Demonstration

0== chsr

ch SRG Where Rch and Sch are characteristic values of resistance and load; r and s are partial safety coefficients pertaining to resistance and load.

Assume Rch and Sch follow normal distributions with a mean value and a standard deviation as:

(22)

Design Equation:

rrrch ZR += (23)sssch ZS += (24)

Where Zr and Zs denote the number of standard deviations that Rch and Sch are removed from their means respectively.

describes how well or poorly we know R and S; In conservative designs, Zr is normally negative and Zs positive;

For example, Zr = -1.64, Rch represents the resistance that is exceeded 95% of the time by the structure to which Eq. (22) is applied.

Zs = 1.64, Sch represents the load that is exceeded 5% of the time.

The exceedence levels of the characteristic resistance Rch and load Schcorresponding to different Zr and Zs are given in the following table:

Exceedence Level for Rch (%) 90 95 98 99

Zr -1.28 -1.64 -2.05 -2.33

Exceedence Level for Sch (%) 10 5 2 1

Zs 1.28 1.64 2.05 2.33

r is often called the performance factor and s the load factor; Eq. (22) can also be written as

chchsrch SSR == )( (25)Where =rs is called the (global) factor of safety. Combination of Eq. (23) to (25) results in:

)( sssrrr ZZ ++= (26)

If the probability functions (normal distribution) for resistance r and load s are:

]/)(21exp[

21)( 22 rr

r

rrp = (27)]/)(

21exp[

21)( 22 ss

s

ssp = (28)

Since the load and resistance distributions are independent of each other, the probability of failure for the design condition, PF can be calculated as

== 000 )()()()( dsrPspdrdsrpspP sF (29)Where Ps(r) is the cumulative distribution function of the resistance r at any value of s.

Example 1: Level II design calculation

Determine the required strength of a structure for a characteristic load Sch. Assume Sch has a mean value s = 80 kN and a standard deviation s = 18 kN. The standard deviation of Rch is r = 20 kN.

One Single Distribution

To avoid the integration in Eq. (29), a single probability distribution has been developed by combining the two probability distributions for r and s:

{ }

=

=

= 11

g

g

g

gFP (30)

Where 22sr

sr

g

g

+

== (31)

srg = 222 srg += (32)Where denotes the cumulative standard normal distribution function and can be determined from published table or computed using expressions from Abramowitz and Stegun (1965). One of the published tables is given below:

3

Encounter Probability

1/1 ))1(1( = LNER PT (33)

Purpose: Determine probability of failure for a design condition during the lifetime of a project (PL)

Encounter probability PE:

Where TR is the return periods in years, PE is the encounter probability and NL is the project design life in years.

Lifetime probability of failure of the design condition may be computed from:

LN

RE TP )11(1 = (34)

OR

FEL PPP = (35)

Example 2:

Determine the lifetime probability of failure for a structure designed for NL = 50 years with a wave height derived from a 50-year return period storm. Assume the probability of failure PF is 0.111.

Level I Design Most practical designs at the present time is Level I design;

Level I design uses Eq. (22) with values being formulated to take into account the uncertainties and being derived from field experience and model studies, or from Level II and Level III analyses;

Level I design expression is a pseudo-deterministic design formula that has a level of safety based on available prototype and model information.

Example 3: Level I design example 1

Develop a level I design formula that uses the mean values of S and R as the characteristic values and represents a design condition with an inherent target PF = 0.1 and determine:

1. The lifetime probability of failure if the project design life is 50 years and the expression with mean load based on a return period of 50 years;

2. The return period for the mean load if the life probability of failure is set at 0.001.

Use Level I design approach For sliding:

Stability Design of Vertical Breakwaters

0)()( =+ dww

s

uwbmf FFFFFf

(36)

Where ff is the friction coefficient between the structure and the sub-base, s and w are the partial coefficients for resistance against sliding and wave loading, and Fd is the dynamic force that only applies to the Minikin design.

the partial coefficients s = 1.3 and w =1.25 for PF = 0.01 are suggested based on extensive Level II analysis of stability of existing vertical breakwaters and breakwater models (Burcharth and Sorensen 1998);

Typical friction coefficients may be found in CERC (1984); for concrete on rock or gravel ff = 0.6 and on sand ff = 0.4.

For overturning (around the landward bottom corner of the structure):

0)~~~()~~( =++ udww

O

bmf MMMMMf (37)

Where O and w are the partial coefficients for resistance against overtuning and wave loading;

the partial coefficients O = 1.3 and w =1.25 for PF = 0.01 are suggested based on extensive Level II analysis of stability of existing vertical breakwaters and breakwater models (Burcharth and Sorensen 1998);

Safety factors are a functionn of the quality of the available data;

The suggested safety coefficients can be reduced when model studies have been performed;

It is possible to use different coefficients for design based on model study and no model study.

Geotechnical Stability The stress acting on the soil is transmitted through the granular rock berm at an angle of approximately 45o;

Different soil stability analysis methods can be used to design for geotechnical stability;

The simplest method is to calculate the stress on the column of soil below the structure and compare it with a critical value;

Insert Fig. 9.7

The extreme stress at the harbour side of the soil column may be computed as:

2

~62~

c

c

c

v

c

cc

c

vc B

MBF

I

BM

BF +=+= (38)

Bvc dBB 2+= (39)Where Fv is the total vertical force, Mc is the moment about the center of the top of the soil column and Ic is the moment of inertia of the soil column section.

The allowable soil pressure on a sandy bottom for a column of width Bc may be defined as a function of the blow count Nb:

cbb BN )006.000016.0(2 = (40)(MPa)

Practice in Japan limits b to 0.6 MPa.For structures placed on a rock berm, an additional allowable stress is present by simply considering the berm as a surcharge of height dB:

3)047.0026.0( 2 BbddN += (41)(MPa)

The maximum allowable stress in the sandy soil is the sum of b and d. Because the soil is underwater, the soil density is about half the density in air. Therefore

)(21

dbu += (42)

Other Design Considerations The discussed design considers forces per unit length of the structure and the

total force is obtained by multiplying the force by the length of the structure. This may be conservative as seldom the structure will be subjected to the force simultaneously along its length, considering short-crested waves and with waves arriving at an angle;

Wave transmission over the structure results in wave agitation behind the structure and possibly in damage to ships and facilities in the habour. The transmitted wave height can be estimated using (Goda 2000):

des

v

s

v

s

vT

Hf

dd

dd

HH 32.058.0)(4.0)(2.0 2 += (42)for 11

Design Example 1:Vertical breakwater in 12 m of water with a

short fetchDesign a vertical breakwater in 12 m of water with a short fetch and a foreshore slope m = 0.1.

Design conditions:

Hs = 1.2 m, Tp = 3.2 s, wave incidence = 0o and Kmax = 1.8;Rock berm height 3 m; freeboard = 1 m;

Foundation: sand with Nb = 20

Design Example 2:Vertical Breakwater on an open coast

Design a vertical breakwater in 12 m of water on an open coast with a foreshore slope m = 0.2.

Design conditions:

Deep water Hs = 6.9 m, Tp = 11.5 s, wave incidence = 30o and Kmax = 1.8;Use Level I design expression with safety factors that approximate PF = 0.01 and use PE = 0.1 or TR = 475 years.

Rubble Mound BreakwatersExamples:

Fig. 9.8

Filter Requirements

)(5)( 8515 layerlowerDlayerupperD < (43)The nominal armour unit diameter is defined as

3/150 )(

a

aa

MDD == (44)

Where Ma is the armour unit mass and a is the armour unit density.The toe filter is necessary to prevent the base material being removed if the breakwater is built on erodible material.

Rubble mound breakwaters consist of layers of stone with stone size increasing from core to outside layers. To prevent the small size stones from being removed through the openings of large stones in the outside layer, rubble mound breakwaters are often built up like filters. A typical example of filter relationship to prevent removal of the lower material through the upper layer is:

Rock Armour

cotcot13

33

aD

desa

aD

desaa K

H

K

HM =

= (45)

Where is the angle of the front slope of the structure with respect to horizontal. KD is an empirically determined damage coefficient depending on the type of armour, its shape, its location along the breakwater and the amount of damage considered to be acceptable. Typical KD values for a double layer rough angular stones, placed randomly in a double layer are shown below:

Stable rock armour mass can be calculated using the Hudson formula (CERC, 1984):

Non-Breaking Waves Breaking Waves

Structure Trunk 4.0 2.0Structure Head 3.2 1.9

Table 4 Damage Coefficients for Rock with Zero Damage

Combining Eq. (44) and Eq. (45) and rearranging Eq. (45), one gets the stability number:

3/1)cot( Daa

dess KD

HN == (46)

Van der Meer (1987) suggested the following expressions for the stability number:

For plunging breakers

5.0

2.0

18.02.6

== mw

ab

aa

dess N

SPD

HN (47)

For surging breakers

bPm

w

ab

aa

dess N

SPD

HN cot2.0

13.0

==

Where Pb is the overall porosity of the breakwater. Van de Meer suggests

Pb = 0.1 for an armour layer over an impermeable layer

(48)

Pb = 0.4 for for armour over a filter or a coarse core.

Armour damage Sa is defined as:

2a

ea D

AS = (49)Where Ae is the erosion area in the breakwater profile between the still water +/-one wave height. Van de Meer suggests Sa = 2 is equivalent to zero damage and Sa = 15 for failure of the breakwater (the secondary layer becomes exposed).

The surf similarity parameter is defined as:

mm s

tan= (50)Where sm is the mean wave steepness:

2,0

2

m

s

m

sm gT

HLHs == (51)

The transition from plunging to surging waves on the breakwater takes place at a critical value of surf similarity parameter (Van de Meer):

)5.0/(131.0 )tan2.6( += bPbmc P (52)Nw represents the number of times that the design condition is reached over the design life of the structure. Van de Meer (1993) recommended that the maximum value of Nw = 7500 be used in design.

Concrete Armour Concrete amour can be manufactured into various shapes to improve stability through interlocking;

Typical KD values for zero damage of concrete armours are given in Table 5

5

Armour unit density Increase of armour unit density will reduce the size of the armour unit;

Use of heavy aggregate such as blast furnace slag

Primary armour layer

On the seaward side, it is customary to extend the primary armour layer from the breakwater crest down to about 1.5Hs below the lowest water level;

Smaller armour units can be used below -1.5Hs;

Primary armour is placed on the back of the breakwater down to the lowest water level due to the possibility of wave overtopping;

The required armour layer thickness can be determined using:

aaaa Dknr = (52)Where na is the number of layers (usually taken as 2) and ka is a shape factor as specified in Table 6.

The number of armour units required per unit length of the structure is:

2)1(

a

aaaa D

eknAN = (54)Where Aa is the surface area to be covered by the armour units and e is the porosity of the armour layer as specified in Table 6.

6

Breakwater crest The crest of a rock armour breakwater is usually made up of the same rock as the rest of the armour layer (normally three stones wide);

The crest of a breakwater with concrete units is often a monolithic cap unit that can carry traffic and infrastructure;

There is often concentrated damage at the interface between the cap unit and the armour units because the cap is impermeable;

The first estimate of the design crest elevation is the limit of the runup of the largest wave, superposed on the highest water level;

Such an estimate is often too high for visual and economical reasons. Practical design will reduce this elevation by allowing overtopping of the larger waves. The actual breakwater crest elevation is determined by considering many factors such as safety in the harbour, navigation requirement, costs, etc..

Possible breakwater settlements after construction have to be taken into account in determining the crest elevation;

A simple estimate of vertical runup height above still water level is recommended (Van de Meer 1993):

25.1%2

Forces on offshore structures Slender objects (D/L < 0.2)

Piles, pipeline, risers and cables Flow separation and vortex shedding dominate the

forces on slender objects Forces on a cylinder in steady current Forces on a cylinder in regular waves

Little effect on waves Large objects (D/L > 0.2)

Flow separation insignificant Significant effect on waves; Wave diffraction becomes important

Reference

Sumer, B. M. and Fredsoe, J. (1997) Hydrodynamics around cylindrical structures, World Scientific

Sarpkaya, T. and Isaacson, M. (1981), Mechanics of wave forces on offshore structures, Van Nostrand Reinhold Company

Forces on a slender cylinder in steady current

Flow around a cylinder exerts a force on the cylinder.

The in-line (same as flow direction) component of the force is termed Drag

The cross-flow (or transverse) component is called Lift

Lift

DragDU

Vortex Shedding -Phenomenon

Vortex Shedding -Phenomenon

Vortex Shedding -Forces

Time history of the lift coefficient

-3

-2

-1

0

1

2

3

0 20 40 60 80Time

C L

H-type g rid

HOH-type g rid

Time history of the drag coefficient

0

0.4

0.8

1.2

1.6

2

0 20 40 60 80Time

CD

H-type grid

HOH-type grid

The mean drag and lift are often evaluated using the following formulae in engineering:

2

21 DUCF DD =

2

21 DUCF LL =

The drag and lift coefficients (CD and CL) depend on

Reynolds number (Re=UD/) Shape of the cylinder

Cylinder surface roughness

Turbulence intensity of the incoming flow

proximity of other objects

Angle of attack

etc.

Flow Regimes

Vortex Shedding -Mechanism Boundary layer

Flow separation

Interaction of vortices

Vortex Shedding -Forces Vortex shedding produces oscillating forces

(drag and lift) on cylinder Drag

Same direction as flow Non-zero mean Small oscillation amplitude but twice as fast as lift

Lift Transverse direction to flow Zero mean Larger oscillation amplitude Constant frequency

Vortex Shedding Frequency Normalized vortex shedding frequency is often used in practice and referred to as Strouhal number:

UDfS vt =

Where fv is the shedding frequency, D is the diameter of the cylinder and U is the flow velocity.

Stroughal number is found to depend on Reynolds number, roughness, cross-section shape, incoming flow turbulence, shear of incoming flow, wall proximity etc..

Effect of Reynolds Number

Example 9.1

Calculate vortex shedding frequency around a pipeline of 0.5 m diameter, subject to a uniform current of 0.8 m/s. Assume a smooth pipe.

Effect of Surface Roughness

Effect of Cross-Section Shape

Effect of Wall Proximity Vortex shedding is suppressed when e/D

is less than 0.2 Vortex shedding frequency increases

slightly as the gap ratio decreases.

Mean drag on a circular cylinder

Effect of surface roughness

Effect of cross-sectional shape

Effect of cross-sectional shape

Effect of cross-sectional shape

Effect of angle of attack

Oscillating drag and lift The drag and lift on a cylinder in steady current are oscillating at regular frequencies due to vortex shedding

Magnitudes of the force oscillations are represented by the corresponding RMS values of force oscillations

The frequency of the force oscillations is often described by the so-called Strouhalnumber

The oscillating forces are responsible for vortex-induced vibrations (VIV)

2''

21 DUCF DD =

2''

21 DUCF LL =

UDfS vt =

RMS forces

Strouhal number

Effect of Wall Proximity

Vortex shedding is suppressed for e/D

Wall Effect on Drag and Lift2

21

aDD DUCF =Where Ua is velocity at the same elevation as the pipe center above the seabed.

2

21

aLL DUCF =221 DUCF LL =

2

21 DUCF LL =

Fredsoe et al. (1985)

Example 9.2

Calculate the total drag on a pipeline of 1m in diameter laid on the seabed if the gap between the pipe and the seabed is 3 m, 1 m, 0.3 m and 0. The pipeline subject to a steady current of 0.77 m/s at 1 m above the seabed. The angle between the direction of the current and the pipeline is 72o. The water depth is 15 m.

Forces on a cylinder in regular waves

dtdUACUDUCF MD += 2

1

The Morison equation for in-line force on a stationary cylinder:

Where A is the cross-sectional area of the cylinder; CM (= Ca+1) is the inertia coefficient and Ca is the hydrodynamic mass coefficient

The Morison equation for in-line force on a moving cylinder:

dtdUAC

dtdUACUUUUDCF baMbbD += )(2

1

Where Ub is the velocity of the cylinder in the in-line direction.

Keulegan Karpenter number (KC number): DTUKC wm=

Where Um is the maximum orbital velocity of water particles and Tw is the wave period

Hydrodynamic Mass Hydrodynamic mass (added mass) is defined as the mass of the fluid around an object which is accelerated with the acceleration of the object

Hydrodynamic mass is caused due to the relative acceleration between the object and the fluid

Hydrodynamic mass can be determined by the integration of pressure around the object and is often expressed by:

VCm a=Where Ca is the hydrodynamic mass coefficient and V is the volume of the object.

Example 9.3

Estimate the total force required to accelerate a long circular cylinder of diameter D through still water of density of ?

Drag versus Inertia Force for a Circular Cylinder

Drag and inertia force are 90o out of phase

The ratio between inertia force and drag force on a circular cylinder is

D

M

drag

inertia

CC

KCFF 2=

Inertia dominates the drag for small KC number but drag dominates for large KC number.

Measurements of Drag and Inertia Experiments often can only measure total in-line forces rather than drag and inertia force separately due to technical reasons

Drag and inertia force can be separated by utilizing the fact that they are 90o out of phase

Let Fm(t) be a time series of measured in-line force induced by a sinusoidal flow. One can get the following force coefficients by using the least square method:

)()cos()cos(138 2

02tdttF

DUC m

mD

=)()sin(12

2

023tdtF

DUKCC m

mM

=Where Um is the amplitude of oscillating velocity and is the angular frequency of the sinusoidal flow

Lift on a cylinder in oscillatory flow

Lift often oscillates at a different frequency from that of the oscillatory flow

The magnitude and frequency of the lift oscillations are important for design

2maxmax 2

1mLL DUCF =

2

21

mLrmsLrms DUCF =

LrmsL CC 2max =

Force coefficients for a smooth circular cylinder

Effect of roughness

Effect of roughness

Angle of attack

The maximum lift coefficient

Effect of Wall Proximity Waves

Breaking wave impact

Special attention needed if the structure is within wave breaking zone;

Impact forces due to wave breaking could be very high (2-3 times as high as for a vertical pile)

Refer to the studies done on different types of structures

Comments

Most of the data given previously were derived from lab tests of a plane oscillatory flows

The use of plane oscillatory flows to approximate motions by real waves tends to overestimate the forces

The over-prediction is not significant and acceptable in the design

Example 9.4

Calculate the total inline force on a pipeline of 1m in diameter laid on the seabed if the gap between the pipe and the seabed is 0.0. Plot the time dependent total inline force, drag component and inertia component of the total inline force respectively. The pipeline subject to a monochromatic wave of H = 2.44 m and T = 8 s. The angle between the direction of the wave propagation and the pipeline is 90o. The water depth is 15 m.

Wave Forces on Large Bodies

Vortex shedding is the major mechanism of hydrodynamic forces on small cylinders

Wave diffraction is the major mechanism for forces on large objects

Diffraction effect is considered important when D/L > 0.2 (Isaacson 1979), where D is the characteristic dimension of the structure and L is the wave length

Forces on Breakwaters and Breakwater DesignVertical Break WatersTypical Failure MechanismsForces for Non-Breaking WavesForces for Breaking WavesStability DesignStability Design of Vertical BreakwatersOther Design ConsiderationsDesign Example 1:Vertical breakwater in 12 m of water with a short fetchDesign Example 2:Vertical Breakwater on an open coastRubble Mound BreakwatersConcrete ArmourForces on offshore structuresReferenceForces on a slender cylinder in steady currentVortex Shedding -PhenomenonVortex Shedding -PhenomenonVortex Shedding -ForcesFlow RegimesVortex Shedding -MechanismVortex Shedding -ForcesVortex Shedding FrequencyEffect of Reynolds NumberEffect of Surface RoughnessEffect of Cross-Section ShapeEffect of Wall ProximityMean drag on a circular cylinderEffect of surface roughnessEffect of cross-sectional shapeEffect of cross-sectional shapeEffect of cross-sectional shapeEffect of angle of attackOscillating drag and liftEffect of Wall ProximityWall Effect on Drag and LiftForces on a cylinder in regular wavesHydrodynamic MassDrag versus Inertia Force for a Circular CylinderMeasurements of Drag and InertiaLift on a cylinder in oscillatory flowForce coefficients for a smooth circular cylinderEffect of roughnessEffect of roughnessAngle of attackThe maximum lift coefficientEffect of Wall Proximity WavesBreaking wave impactCommentsWave Forces on Large Bodies