waves solution

7/28/2019 waves Solution

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Waves and wave forces:

In this task some considerations on the wave and current conditions and their associated loads on a

circular bridge support will be addressed. A bridge is planned across the sound between the mainland

A and an island B. To support the bridge a 13 m diameter circular caisson is to be placed midway in

the crossing. A flat sea bottom at the location of the caisson is built up of gravel and sand of various

grain diameters. The water depth at the caisson is to be 26 meters. The design waves are coming from

a sector 315 -345 and there is a semidiurnal tidal current of 1.0 m/s amplitude as indicated by the

double-ended arrow on the sketch. In an earlier project wave statistics was provided at an open water

location C, North-West of the site where the water depth was 150 m. To determine wave loads on the

structure it is found appropriate to use those data, but using only data for the waves coming from the

sector 315 -345. Tables for joint occurrences of sea state parameters are provided on page 6 and 7.

A

B

waves

C


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a) what will be the probability that an arbitrary individual wave exceeds 3.0 m in height ?

Using the formula for the Rayleigh distribution and remember that in its form presented gives

the probability that the wave height is less than a given value. Ssince we want the probability

of exceeding a certain value this is given by 1 F(h) and for the given values it becomes

0.135.

b) what would be the expected highest wave in this sea-state assuming the duration of the sea

state is 2 hours?

Here we use the formula for Hmax. N is the number of waves which is easily found by using

the information given by the mean wave period. It is defined as the total time divided by

number of waves so N = 2 hrs/7 s = 1028 waves. Result is Hmax = 5.8 m.

c) what wave period would you recommend to use for wave kinematics calculation for the

expected highest wave you determined in 1.b)? Justify your answer.

Different answers were validated, however no score if you suggested a shorter period,

Information of figure 1 (cf home assignment) was the best source in this exam set, indicating

that the wave period for the higher waves in sea state is approx 1.2 times the mean wave

period, so I would say that T = 8.4 s was the best answer.

3) A long-term statistical analysis is required to determine the design wave conditions at location C.

a) Use the joint occurrence table Hm0-Mean Direction for the sector 315 -345 to establish atable for plotting the observed distribution of significant wave height on the attached Weibull-

scaled paper. UseHm0 intervals of 0.4 m. Use the table to estimate the remaining parameters of

the 3-parameter Weibull distribution that fits the data in a best way by using the Weibull

plotting paper. AssumeH0=0 in the 3-parameter Weibull distribution. Include the plotting

table in your answer.

00 00

Pr ( ) 1 exp ,mc

h HH h F h h HH H

See enclosed diagram.

b) Thedatainthejointoccurrencetablesarebasedondatasampledevery 3hoursover 1490


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10 yr: 0,1028.7

( ) 1 0.9996710 365 24

mF H

30 yr: 0,1028.7

( ) 1 0.9998930 365 24

mF H

c) Determine the 10 and 30 year significant wave height at location C. (If you did not find a

solution in 3.b) use =1.5, Hc=1.2 m andH0=0 m.)

1/0

0 00( ) 1 exp ln 1 ( )

cc

h H

F h h H H H F hH H

10 yr: 1/1.5

0,10 1.2 ln 1 0.99967 4.8mh H m m

30 yr: 1/1.5

0,10 1.2 ln 1 0.99989 5.2mh H m m

4) Wave and current forces must also be considered. Assume that the design wave condition is

H =3.0 m andT =7.0 s and the design current speed is 1.0 m/s. Assume that the caisson is smooth.

a) Determine what set of formulas you would apply to estimate the total horizontal force on the

caisson in the case of waves only. Give reasoning for your choice.

The Keulegan Carpenter number is essential. K = u0T/D. Wave length as in problem 3.1:

73.5 m. This gives for the horizontal particle velocity at the surface:

00 cosh( ( ) 1 2 11.5 1.38 /2sinh( ) tanh( ) 7 tanh( 26)

73.5

zk z hu a a m m skh kh s

and 1.38 7/13 0.74K Since K is less than 2 potential theory must be used.With avertical circular cylinder ranging from the sea bed and through the sea surface MacCamy and

Fuchs te\ory is well suited. Alternatively the inertia term of Morison formula for force per

unit cylinder axis length can be applied:

2

( 0, , )4

x M M

Df f C u x z t with a CM dependent on the diameter to wavelength ratio

and2 cosh ( )cos( )

sinh

k z hu a t

kh

from linear wave theory. CM=2.0 would be a good

choiceheresinceD/L=018


4/6

0 22

022

202

2 22 2

2

cosh ( )cos( )

4 sinh

1cos( ) cosh ( )

4 sinh

1 1cos( ) sinh ( )

4 sinh

1 4 73.5cos( ) 1025 2 6.5 1.5 3846.7

4 7 2

X M

h

M

h

M h

M

D k z hF C a t dz

kh

DC a t k z h dz

kh

DC a t k z h

kh k

DC a t kN

k

0 2

2 cosh ( )cos( )4 sinh

M

d

D k z hM C a t d z dz

kh

Consider:

0 0

0 0

0 0

2 2

2 2

cosh( ) cosh cosh sinh sinh

cosh cosh sinh sinh

1 1cosh cosh sinh sinh sinh cosh

1 1cosh cosh sinh

d d

h h

h h

I kz kh zdz kz kh kz kh zdz

kh kz zdz kh kz zdz

z zkh kz kz kh kz kz

k k k k

hkh kh khk k k

2

2 2

2 2 2

2

1sinh sinh cosh

1 1 1cosh cosh sinh cosh sinh sinh cosh

11 cosh

hkh kh kzk k

h hkh kh kh kh kh kh kz

k k k k k

khk

Where Rottmann gave solution for the integrals needed:

2

1cosh( ) cosh sinh

xx ax dx ax ax C

a a

2

1sinh( ) sinh cosh

xx ax dx ax ax C

a a


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0 22

022

0 022

22

2 2

cosh ( )cos( )

4 sinh

1cos( ) cosh ( )

4 sinh

1cos( ) cosh ( ) cosh ( )

4 sinh

1 1 1 1

cos( ) sinh cosh4 sinh

M

h

M

h

M

h h

M

D k z hM C a t h z dz

kh

DC a t k z h h z dz

kh

DC a t k z h hdz k z h zdz

kh

D

C a t h khkh k k k

22 1 sinh 1 coshcos( )

4 sinh

sinh cosh 13846.7kN 16.6m 63855kNm; 16.6m

sinh

M

x

kh

D kh kh khC a t h

k kh kh

kh kh khF R where R h

kh kh

R is the moment arm( cf. with the same expression for MacCamy and Fuchs theory).

Using MacCamy and Fuchs theory:

22

2 2 1025 9.81 3 2 2tanh( ) ( ) tanh 26 6.5 cos( )

73.5 73.52

73.5

x

gHF kh A ka A t

k

From Table: A(0.5556)=?

By interpolation:

0.482 0.452 0.482

0.56 0.54 0.56 0.5556

0.482 1.5 0.0044 0.4754

A

A

This gives: Fx =3 834 kN

M=Fx

R =3 834

16.6=63643 kNm.sinh cosh 1

16.6msinh

kh kh khwhere R h

kh kh

The discrepancy between the values found using the two methods is negligible and a small

ti fC ld dj t f thi


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Hc =0.9 m

H F

0.4 0.08741

0.8 0.373697

1.2 0.653569

1.6 0.835605

2.0 0.935044

2.4 0.973536

2.8 0.987971

3.2 0.994387

3.6 0.996792

4.0 0.999198

= 1.35

Comment: In my choice of fitted line I gave more attentionto the highest wave heights than the lowest, since we areinterested in the extremes. Even so, it is not easy to selectthe best fit. Other lines may also be viewed as best fit. Thisdemonstrates that there are some uncertainties in estimatingthe best fit. Maybe if another distribution function was used,the data might fit better. In fact tryingH0 different from zerocould also help.

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waves solution

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