This document provides steps to calculate wave loads on a structure using random wave theory. It includes: 1) Defining mass and stiffness matrices for the structure. 2) Calculating natural frequencies and mode shapes. 3) Determining wave frequency distribution using Pierson-Moskowitz spectrum. 4) Dividing the frequency range into intervals to calculate wave loads at different points on the structure.

1. Vi DU TiNH TOAN BAI TAP LON
MON: BONG LUC HOC NGAU NHIEN
He duct rot rat hod nhtr hinh ve
a = 13 m 14 m
/
14 m
d = 29 m
/
14m
Trinh to tinh town duct tiEn hanh theo cat bok sau:
1. Roi rat hod ket cgu, xac dinh cdc ma tr4n kh6i luting M.
in 0 0 325.4 0 0
M
=
0 M2 0 = 0 650.8 0
[
0 0 3 0 0 650.8
Lenh MATLAB nhu sau:
M= [325.4 0 0;0 650.8 0;0 0 650.8];
2. Mc dinh ma tan de ming:
Cho cac (Ai tong P=1 dat lan loot tai cac nut duqc lan loot cac vec to chuyen vi
(c6 the dung chuang trinh Sap2000 de thirc hien). Ket qua to duct ma irk c10
mem D:

2. 811 812 813 6.571e - 6 3.432e-6 1.005e-6
D = 821 8 22 823
3.432e-6 2.010e-6 6.493e-7
83 1 832 833
1.005e-6 6.493e-7 2.937e-7
Lenh MATLAB nhu sau:
D= [6.571e-6 3.432e-6 1.005e-6;3.432e-6 2.010e-6 6.493e-7;1.005e-6 6.493e-7 2.937e-71;
Nghich dao D duce ma tan dO cling K
L4nh MATLAB nhu sau:
K = inv(D);
Ta co:
k11 ku k13 0.1715 -0.3613 0.2118
K= = k21 k2 2 k23 = -0.3613 0.9350 -0,8308 *1.0e+007
k3 1 k32 k3 3 0.2118 -0.8308 1.4526
3. Xac dinh cac tan so dao dOng rieng va sac clang dao do, ng
Ta da biet M va K. De dang tim duce tri rieng (TR) va vec to rieng (VR) cua K
va M bang chuong trinh MATLAB, veri lenh:
[VR,TR]= eig(K,M)
Ta co ket qua nhu sau:
VR =
0.5299 1.0000 -1.0000
-0.7932 0.5486 0.6655
1.0000 0.1702 0.7928
TR =
1.0e+004 *
3.4171 0 0
0 0.0288 0
0 0 0.7500
Vectu rieng cling el-A(1h la cac clang dao dOng, to co:
0 = VR =
0.5299 1.0000 -1.0000
-0.7932 0.5486 0.6655
1.0000 0.1702 0.7928
Dang I II III

3. Tan s6 dao Ong rieng bang can bac hai cua tri rieng, ta co:
ca = .1772 =
184.8545 0 0
0 16.9641 0
0 0 86.6019
4. Xac dinh cac ma tan d6 cimg K* va ma tan lchoi luong Mt
4.1. Ma tran do Ong IC'
10` = e. K. Co (K* trong MATLAB duce ky hi'e'u la Kl)
Sir dung MATLAB ta co:
K1 = (DT * K* (1) =
0.5299 -0.7932 1.000 0.1715 -0.3613 0.2118 0.5299 1.0000 -1.0000
1.0000 0.5486 0.1702 -0.3613 0.9350 -0.8308x -0.7932 0.5486 0.6655 x1.0e+007=
-1.0000 0.6655 0.7928 0.2118 -0.8308 1.4526 1.0000 0.1702 0.7928
3.9353 0.0000 0.0000
K* = K1 = 0.0000 0.0155 - 0.0000 x1.0e + 007
- 0.0000 - 0.0000 0.7670
4.2. Ma tran kit& lugng Mt
M* 0r . M. 0 (M* trong MATLAB dugc ky hi'eu la Ml)
Sir dung MATLAB ta c6:
M1= V.* M* (1) =
0.5299 -0.7932 1.000 11{ 325.4000 0 0.5299 1.0000 -1.0000
1.0000 0.5486 0.1702 0 650.8000 0 -0.7932 0.5486 0.6655
-1.0000 0.6655 0.7928 0 0 650.800 1.0000 0.1702 0.7928

4. 1.1516 0.0000 0.0000
M* = MI = 0.0000 0.5401 0.0000 x0.1e + 003
0.0000 0 1.0226
Chu K* fa Ise la the ma trAn chap.
5. Xac dinh ck gia tri w La (72
+Ta co, ph6 song theo Pierson – Moskowitz:
S —expi– co 4n
05
A B
Vgi:
3 hs 3 29
A= = 4z . = 0,6(m / s4 )
To4 (8,8) 4
16z 3 16;r3
B– = – 0,083(11 s 4 )
To4 (8,8) 4
Vay to co ph6 song co dang sau:
0.6 0
( .083)
o = —5- exp
So CO 00
Le", nh MATLAB dE ye dia thi ph6 song P-M nhu sau:
w=0.1:0.2:4;
y=0.6*exp(-0.083Jw.^4)./w. ^ 5;
plot(w,y) Dacia Rie,e, Voity
grid on
Khi do co dugc hinh ye d6 thi nhu sau:
Tir do co thE xac dinh dugc (mOt each
Wong d6i, Ichong can chinh xk) gia
tri :
= 0.25
C72 = 2.5

5. pLJA
6. Trong khoang to c7), den FO 2 ta chia ra lam 10 doan, taco:
w2 - (71 2.5 - 0.25
- 0.225= A
10 10
Sir clang MATLAB de chia khoang diem chia tin so va gia tri cua phis P - M twang
w = 0.25:0.225:2.5;
S=0.6*exp(-0.083./w. ^4)Jw. ^ 5;
Khi chay MATLAB, ta co:
BAm w tren MATLAB, xuAt hien gia tri cac tan so:
w=
0.2500 0.4750 0.7000 0.9250 1.1500 1.3750 1.6000 1.8250 2.0500 2.2750 2.5000
DAM S tren MATLAB, xuat hien gia trj cac phis twang img voi tin so w:
S=
0.0000 4.8595 2.5266 0.7911 0.2845 0.1193 0.0565 0.0294 0.0165 0.0098 0.0061:
Tinh so song k khi cid biet w, ta co:
co 2 = gktanh(kd)
Ta tinh cho timg gia tri w, vi du vai w = 0.25, le", nh MATLAB nhu sau:
k =solve(10.2.5A2=9.81*k*tanh(k*29)')
MATLAB cho ket qua:
k= .15294175491692857744421253097448e-1
LAy tree: k = 0.0153
Twang to nhu vay, ta co tat ca cac gia tri so song k twang img voi w:
k=
0.0153 0.0317 0.0544 0.0883 0.1349 0.1927 0.2610 0.3395 0.4284 0.5276 0.6371
ting vai rnOi gia tri cua w, S, k (co tat ca 11 gia tri) ta tinh ducc tai hong q theo (4) tai
cac diem A, B, C va D, img voi cac dO sau twang img 0, 14, 28 va 29 met, khi tat thira
so chung la mat cat song ri(t).
taco:
z=0

6. w = 0.25 ; H s = 9 m vei diem A co z = 0 met Tir bleu thirc (1) to tim duqc cac
tri so theo breu thirc:
ch(kz)
v=
sh(kd)
nhu sau
ch(kz)
h.
=co 2
sh(kd)
z=0;
d=29;
2.2750 2.5000];
w10.2500 0.4750 0.7000 0.9250 1.1500 1.3750 1.6000 1.8250 2.0500
0.5276 0.6371];
k=(0.0153 0.0317 0.0544 0.0883 0.1349 0.1927 0.2610 0.3395 0.4284
v=w.*cosh(k.*z)Jsinh(k. *d)
v1=L*w. ^2. *cosh(k.*z)Jsinh(k. *d)
(Chti , vl biiu gia tri
ket qua:
0.0000
v = 0.5454 0.4505 0.3019 0.1438 0.0460 0.0103 0.0017 0.0002 0.0000 0.0000
0.00001 0.00001
vl = 0.13631 0.2140i 0.2114i 0.1330i 0.0529i 0.0141i 0.0026i 0.0004i 0.0000i
Tinh Cluqc g(w) irng voi cac gia tri w theo bik thirc:
ch(kz) 2 c
g(w) (a) )
sh(kd) "
Ta co:
22750 2.5000];
w=10.2500 0.4750 0.7000 0.9250 1.1500 1.3750 1.6000 1.8250 2.0500
0.0098 0.0061];
S=10.0000 4.8595 2.5266 0.7911 0.2845 0.1193 0.0565 0.0294 0.0165
gr-v. ^2. *5'
K et qua nhu sau:
.
g = 0 0.9863 0.2303 0.0164 0.0006 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
Nhu vay, (mg voi z = 0, to có gia tri
6-,,, 2 = g(co)d (co) = ig(co)d (w) = - 52 ( g ° + g 1 + + g + )
N 2 2
2.5 -0.25 ( 0/2 + 0.9863 + 0.2303+ 0.0164 + 0.0006 + 0.0000 + 0.0000+ 0.0000+ 0.0000
10
+0.0000+0.0000/2)
= 0.2776
o- 0.5268
Tinh c , theo MATLAB nhu sau:
X- g(1)/2
s

7. for j=2:10
X=X+g(j)
end
X=(X+g(11)/2) *0.225
XIMA=sqrt(X)
Tir bieu thirc (4), gia thief C.& chuyen vi nhe so veri sang, khi do: rx =
Ta co:
qA op * rx +C1 * ix
tong do:
1r.D2 3.14159x5.52
- p.C,. 1.025x2x -48.704
4
ETD = P.Cd.D.1i0; = 1X1.025X1X5.54 3.14859
1
x0.5268 =2.3697
2 2
Tir (*) ta tim ducc qA nhu sau:
qA
= 2.3697* 0.5454 +48.704*0.1 3631 = 1.2924 + 6.6401
Gia tri tai tong song tai diem A cho tat ca cac tan so (1 1 gia tri dugc tinh theo MATLAB
nhu sau:
C1=1.025*2*pr5.5^2/4
CD=0.5*1.025* 1 *5.5 *sqrt(8/pi).*XIMA
q = CD*v+Cl*v1
Ta duce ket qua cac gia tri qA nhu sau:
qA = 1.2924 + 6.64051 1.0676 +10.42231 0.7155 +10.29371 0.3407 + 6.47721 0.1090 + 2.57731
0.0244 + 0.6890i 0.0039 + 0.1287i 0.0005 + 0.0172i 0.0000 + 0.00161 0.0000 + 0.0001i
0.0000 + 0.00001
Tuang to nhu vay ta co the tinh dugc gia tri tai tong cho qB, qc, qo theo chtrang trinh
sau:
TN& het tong MATLAB dua gia tri z = (14; 28; 29 m)
sau do dua chuang trinh tinh vao:
2.5266 0.7911 0.2845 0.1193 0.0565 0.0294 0.0165
S=0.0000 4.8595
0.0098 0.00611;
w=[0.2500 0.4750 0.7000 0.9250 1.1500 1.3750 1.6000 1.8250 2.0500
2.2750 2.5000];
k=0.0153 0.0317 0.0544 0.0883 0.1349 0.192 7 0.2610 0.3395 0.4284
0.5276 0.6371];
d=29;

8. % Tinh bien 0 van toc va gia toc song
v=w.*cosh(k.*z)./sinh(k.*d);
v1=1.*w.^2.*cosh(k.*z)./sinh(k.*d);
% Tinh gia tri g(x)
g=v.^2.*S;
% Tinh XIMA(Vx)
X=g(1)/2;
for j=2:10
X=X+4);
end
X=(X+g(11)/2)*(w(11)-w(1))/10;
XIMA=sqr100;
% Tinh tai trong q
CI=1.025*2*pi*5.5 ^2/4;
CD=0.5*1.025*1*5.5*sqrt(8/pi)*XIMA;
q = CD*v+CI*v1;
Ta se c6 gia tri q tai cac diem
vi
zB = 14 m
cls
1.5291 + 6.7934i 1.3583 +11.4656i 1.0793 +13.42621 0.7355 +12.08961
0.4264 + 8.71311 0.2103 + 5.1383i 0.0875 + 2.48791 0.0307 + 0.99651
0.0091 + 0.3314i 0.0023 + 0.0922i 0.0005 + 0.02151
zc=28m
qC
2.709+7.262*i 2.906+14.809 3.294+24.73*i 3.897+38.65*i 4.568+56.339
5.150+75.949 5.598+96.04*i 5.902+115.59 6.066+133.4*i 6.096+148.7*i
6.004+161.0%
42- 29 m
qD =
2.928+7.307*1 3.194+15.15*i 3.719+25.99*1 4.567+42.17*i 5.615+64.46*1

9. 7.806+124.7*1 8.904+162.2s 1 10.00+204.7*1 1.100+252.1*1
6.709+92.08s1
12.20+304.4s i
Sau khi da cO luc tac dung q A, qB, qc va qp, vOri gia thiet hie thkg la hang s6 trong
suat down dang xet, cO the tun duoc lye tac dung tai mit (khi be qua m6 men u6n,
chi clui 9 den lye thkg tac dung tai mit)
Luc tac dung qui ye mit la Fl, F2, F3, duoc tinh nhu sau:
Fl = (qA. 1)(1-0.5)/14
F2 = (qB. 1)(14-(1-0.5))/14 + qB.14/2
F3 = qC.14/2
Tren MATLAB se dtroc tinh nhu sau:
Fl = (qA* 1)(1-0.5)/14;
F2 = (qB* 1)(14-(1-0.5))/14 + qB*14/2;
F3 = qC*14/2 ;
Luc tac dung dm song len cac nut dm cot theo clang ma trail se la:
Fl
Qs = 14(0 = F2 g(t)
F3
Luc sang theo clang ma an a he toa do suy Ong:
Q*•ii(t) = ITT . itri(t) =CDT. 77(0,
trong do:
Fl .
F2
F3
Cmg vei 11 gia tri tan so song Mr 0.25 den 2.5, to c6 the tinh Q'MATLAB nhu saw
% QQ bieu thi bien Q*
% VR la clang dao doling 0
F = [F1 F2 F3];
QQ=VR'*F;
Gia tri chuyen vi x k(t) dtmc firth nhu sau:

10. o
xk o>=2, k 12a*i Hi (co)f(t)
1=1 co, m
trong do, k la bac to do thd k caa k6t cau (6 day la di6m thd k caa kat cau
i la dang dao dOng thd i (ta co 3 dang)
Tit do, nau song la QTNN dung, c6 lcY vong town bang 0, thi do lech chuan cua chuyin vi
kat cau se dtrec tinh theo bi6u thirc:
2 =
2
?k it 1
2
1 12 do ( ) y (ce , AH, (0)
mi 2if
trong do:
Hi (co) – Ham truyen a dang thd i len k6t cau do tat trong sang tan sow tac dOng;
co; – Tin s6 dao cIOng rieng caa k6t du a clang thd i;
mi* - kh6i Wong nut a he toa dO suy rang, arcing ling v6i clang thd i
(co) - Ph6 sang ting v6i tan s6 co.
Chti 1 d6n phd 2 phia (-co +co) ta (1) viat lai:
2
3
cx, 2 412 1
4 .
— It13 y Bri a))111,(co)124w
J
1
,. ( (2)
i=1 co; ma 7( 0
(13 , * )2 –
Gia tri !Dien dO tai trong song nhan vei lien hop caa chinh Bien dO d6:
(e)2 =0i(0 ).Q.( -i0))
Tinh tich phan nhu sau:
352/
I= IV3 /
, * 1.S,, (co )111,(co)1 2 dco = . RP,* f .S7 (a) )IH; (w)Izdw=
0
1)2 I/1 ro rA,
(— + + + r„, , +
N 2 2
trong d6:
(co )IH,(w)12
r(co) = (/),* y
ro= r(wo); = r(ioi)
Vilt (1)," , S,i (a)),IHi (o))1 2 img vei cac gia tri w = 0.25 .... 2.5 nhu sau:
)2
Tinh cho clang thd 1 (i=1) cho MATLAB
% P2 la ky hitt' cho ( Pes
%tinh r
P2 =double(QQ.*conj(QQ));
H1 = 1./(1-(w/OME(1,1)). ^ 2)
H2= H1.* conj(H1)
%tinh r
r=P2(1,:).*S.*H2;

11. ***************
Chti5): thing renh "double(...)" di lam gon kit qua trong qua trinh tinh wan. Hdy so sanh
2 tenh:
P2 =QQ.*conj(QQ);
va
P2 =double(QQ,conj(QQ));
st slay hieu qud cua lenh "double(...)"
**mamma
(7)
= 0.225
—
%Tinhtcpa,kMe:2
I=r(1)/2;
for j=2:10
I=I+r(j);
end
1=(I+011)/2) 4'0.225;
I1=I;
Tinh axing to nhu vay cho i=2, i=3 (dang dao Ong)
Tinh dO tech chart:
2
2 0,a2 ( 1 ) 1,
Cxk
i= Mt LA "
ax 2 = --eik1
rs
2
1 ) 1 . 1 1+ _A2 2 _7 ._.12
0 1 1 1 _7
0k32 1
2
1 ,
C014 mi (1)2 m 2 n a)3 m3
D0 tech chudn cua chuydn vi nhu sau:
cx, =
Kp 'tong man cha chuyen
xk = 0
Vi6t tren MATLAB nha sau:
XIMAX2=VR(1,1)^2/0ME(1,1)^4/(1v11(1,1)^2*pi)*11+VR( 1 ,2)^2/0 ME(2,2)^4/(M 1(2,2)^2* pi)*I2
^2*pi)*I3 +VR(1,3)^2/0ME4(,3)
XIMAX = sqrt(XLMAX2)
Ta tun duqc ket qua:
XIMAX2 = 7.7499e-007
XIMAX = 8.8034e-004

12. trong d6: XIMAX21a phuong sai cila chuyein vj tai dinh cOt
XIMAX la do lech chugn cua chuye'n vi tai dinh cOt
Ky vong toan chuye'n vi cho phep:
H 14*3 n , Tri‘ tl
cnieu cao cOt)
MA = = - V.Z1 m
200 200
Phuong sai chuyen vi cho phep:
vo = 0 (vi chuydn vi cho phep duct gia thiet a tien dinh).
Khi biet ductc ky vong man va phuung sai cua chuydn vi va chuye'n vi cho phep, to c6 thd
tun duqc chi s6 tin cay /3 nhu sau:
— mx _ 0.21— 0 _ 0.21 = lam trail: (3 =238
= 238.5449
+ Cfx2 VO +7.7499e -007 8.8034e - 004
Tit chi s6 tin cay (3 co the tun (bloc dO tin cay cua k6t cgu theo didu kien chuye'n vi dinh
cOt:
P = 0.5 +4)((3)= 0.5 + 0(238) = 0.5 +0.5 = 1
Nhu vay dO tin coy chuydn vi dinh cOt P=1 > Po