Offshore

CHAPTER 9
MAXIMUM WAVE FORCE ON THE ENTIRE STRUCTURE 9.1 Example

Obtain the variation with time of total horizontal wave force and moment for the entire offshore structure as shown below: Also calculate maximum force and moments. (Horizontal) Use linear theory (In-line) Given

H = 6m Pile dia. = 1.2m

L = 90m bracing dia. = 0.6m

d = 25m

CD = 1
Solution

CM = 2

γ = 10.06

kNm3

The members on which wave forces should be considered are: Group A: Vertical Piles : 1-3/ 7-9 /4-6/ 10-12 Group B: Horizontal Bracings : 2-8/ 5-11

9-1

Group C:

Diagonal Front Face Bracings:

2-9/ 5-12

Group D: Diagonal Side Face Bracings : 2-6/ 8-12 Calculate FT for each member at same time instant wt = 0, 1, 2, ……7 Example for wt = 6 [A] FT for pile 1-3:
k= 2π c = 0.0698 L m1 w = {gk tan h kd }2
1

= {9.81(0.0698) tan h 0.0698(25)}2 r = 0.8026 s
FTD = C D SD (WH ) {2k (d + z ) + sin h 2k (d + z )}z =η / cos θ / cos θ 32k sin h 2 kd H θ = kx – wt = cos θ 2 = -6 =3 cos (-6) x=0 = 2.881m
2

 0.8026   6  (10.06) 1.2   =1 {2(0.0698)(27.881) + sin h2(0.0698)(27.881)}/ cos(− 6) / cos(− 6) 2 9.81 32(0.0698) sin h 0.0698(25)

2

= 43.4 kN FTI = CM S

πD 24

(d + z (HW ){sin2hkksin h kd)}
2 2

z =η

sin θ

=2

10.16 π (1.2 ) sin h[0.0698(27.881)]}sin θ 2 { 6(0.8026 ) 9.81 4 2(0.0698) sin h[(0.0698)25]

(

)

= 22.17 kN
∴ FT

= 43.4 + 22.17 = 65.57 kN

9-2

IDENTICALLY: FT for pile FT for pile 7-9 = 65.57 kN 4-6: Here
θ = kx – wt

= 0.0698(15) – 6 = -4.953 rad and z = η =
H cos (− 4.953) = 0.7149m 2
2

FTD

C SD(WH ) = D {2k (d + z ) + sin h 2k (d + z )}z =0.7149 / cos θ / cos θ 32k sin h 2 kd

= 1.6583{2(0.0698)(25.715) + sin h2(0.0698)(25.715)}/ cos θ / cos θ

= 2.04 kN FTI = CM S

πD 2
4

)} d (HW ){sin h2k (sin+hzkd k
2

z = 0.715

sin θ
θ = - 4.953

= 8.9652 {sin h 0.0698(25.715)}sin θ = 65.75 kN

∴ Total FT on pile 4-6 = 2.04 + 65.75 = 67.8 kN

Identically FT on pile 10-12 = 67.8kN [B] FT on horizontal member 2-8: special case of inclined members
Vnx  anx   Fx         Fy  = C Vn Vny  + K any         Fz  Vnz  anz 

∴ Fx = C Vn Vnx + Kanx

9-3

2 Vnx  1 − C x − C x C y − C x C z  Vx       2 Vny  = − C x C y 1 − C y − C y C z  V y      2  Vnz  − C x C z − C y C z 1 − C z  Vz   

(Here C x = 0; C y = 1;C z = 0)
1 0 0 Vx  Vx      = 0 0 0  V y  = 0    0 0 1  Vz  Vz       anx  a x      any  = 0    a  anz   z 

Similarly,

∴ Vn = Vx2 + Vz2

&

Fx = C Vx2 + Vz2

Vx + Ka x

& & [Note: For calculating u , w, u , w

ORIGIN is AT SWL] ;
2π w 2π = = 7.82911 0.8026 T=

Vx = u =

πH cos h K (d + z )
T sin h kd

cos(kx − wt )

=

π(6 ) cos h[(0.0698)(15)]
7.829 sin h[0.0698(25)] m s

cos(− 6 )

= 1.333

Vz = w =

πH sin h k (d + z )
T sin h kd m s

sin (kx − wt )

= 0.303

& ax = u =

2π 2 H cos h k (d + z ) sin (kx − wt ) T2 sin h kd
m s2

= 0.311

9-4

C = CD S

D 10.06 0.6 =1 = 0.3076 2 9.81 2

K = CM S

πD
4

=2

10.06 (0.6 ) π = 0.58 9.81 4

∴ FX = 0.3076 1.3332 + 0.3032 1.333 +0.58(0.311)

= 0.741
∴ FT FT

kN m

for entire member length 2-8 = 0.741(15) = 11.16 kN for horizontal member 5-11:

Here x = 15m;

η

H cos(kx − wt ) 2

= 3 cos (0.0698(15)-6) = 0.715m
Vx =

πH cos h K (d + z )
T sin h kd

cos(kx − wt ) m s

=

π 6 cos h [0.0698(15)]
7.829 sin h[0.0698(25)]

cos(0.0698(15) − 6 ) = 0.331 m s

Vz =

πH sin h K (d + z )
T sin h kdsin (kx − wt ) = 1.052

ax =

2π 2 H cos h K (d + z ) m sin (kx − wt ) = 1.082 2 T sin h kd s

∴ FX = 0.3076 0.3312 + 1.0522 (0.331) + 0.58(1.082) = 0.74kN

Therefore, Total FT on entire member 5-11 = 15(0.74) = 11.1kN [C] Diagonal Front Face Member:

FT on member 2-9:

9-5

FX = C Vn Vnx + Kanx
2 Vnx  1 − C x − C x C y − C x C z  Vx       2 where; Vny  = − C x C y...