BTL Lý Thuyết Độ Tin Cậy CTB

Post on 16-Sep-2015

224 views 1 download

description

Bài tập lớn Lý thuyết độ tin cậyKhoa XD CTB- Dầu khí Đại học xây dựng

Transcript of BTL Lý Thuyết Độ Tin Cậy CTB

BTL L THUYT TIN CY KHOA XY DNG CNG TRNH BIN-DU KHMc Lc

A.S LIU BI.2S :3-H.21.S .22.S liu cng trnh.23.S liu ti trng,s liu sng.24.Thng s cng trnh.25.Yu cu.3B.PHNG PHP TNH.31.Chia ct lm 3 phn t.32.Xc nh ma trn M,K53.Xc nh tn s dao ng ring v cc dao ng.94.Xc nh cc ma trn cng K1,ma trn khi lng M1.105.Xc nh cc gi tr 1,2116.Trong khong v ,chia ra lm 20 khong u nhau,xc nh cc tn s ti cc im chia v cc ph sng tng ng.127.ng vi cc tn s sng ti cc im chia,xc nh lc ngang F ,F ,F138.Xc nh gi tr hm truyn H(i) ti cc im chia209.Xc nh phng sai chuyn v ti dng dao ng th I,II,III2110.Xc nh tin cy chuyn v nh ct23

A.S LIU BI. S :3-H.1.S . Cho ct thp ng t di bin chu tc dng ca ti trng sng nh hnh v sau:

2.S liu cng trnh.M sd(m)a(m)D(m)b(m)

314.531.150.2

3.S liu ti trng,s liu sng.M sHs(m)To(s)M(T)

H9.59.854

4.Thng s cng trnh.

Trng lng ring ca thp : = 78.5 (kN/m)

Trng lng ring ca nc bin : = 10.25 (kN/m)

Moduyn n hi ca thp : E = 2.1x10(kN/m)Notes:i. B qua khi lng nc km;ii. B qua cn nht;iii. Thuyt minh th hin trn giy A4.5.Yu cu. Xc nh tin cy v chuyn v ti nh ct khi bit chuyn v cho php:

[]= H=0.005H B.PHNG PHP TNH. Trnh t thc hin:1. Chia ct ra lm 3 phn t bng nhau;2. Xc nh ma trn M,K ca h;3. Xc nh tn s dao ng ring,cc dng dao ng ;4. Xc nh ma trn M,K;5.

Xc nh , ,cc tn s ng vi gi tr bng 0(zero)ca ph;6.

Trong khong ,,chia lm 20 khong u nhau,xc nh tn s ti cc im chia v cc ph sng tng ng;7.

ng vi cc tn s sng ti cc im chia,xc nh lc ngang F,F,F;8. Xc nh gi tr hm truyn H(i ) ti cc im chia;9. Xc inh phng sai chuyn v ti dao ng th I;10. Xc nh tin cy chuyn v nh ct; 1.Chia ct lm 3 phn t. S tnh:

Ta chia on tr ct chu ti sng thnh cc on nh bng nhau,trn mi on ny ta coi ti trng sng l phn b u.Ta xc nh ti trng phn b u theo cng thc Morison vi im cn xt l trung im ca cc on.i vi bi ton ny,ta chia tr ct lm 3 on bng nhau v bng H/3.

-Quy i khi lng,ti trng v cc nt: Vic quy i ti trng v khi lng v nt da trn phng php phn t hu hn.Coi vt th lin tc l mt tp hp nhiu phn t nh hn,c s lng kch thc hu hn.Gi thit rng cc phn t ch lin kt vi nhau cc nt.Khi cc ti trng(trong v ngoi)cng vi khi lng(bn thn ng,nc trong ng,nc km,...)s c quy i v cc nt ny. i vi mi phn t ring bit,ti trng nt c xc nh da trn s phn b ti trng theo chiu di phn t hoc cc on phn t .Tng t i vi khi lng cng theo chiu di phn t (khi lng bn thn) hoc theo cc on ca phn t (khi lng nc km,nc trong ng...) khi phn t na chm.Cui cng ti trng nt v khi lng nt c tnh nh tng cc lc nt v khi lng nt ca cc phn t quy t ti nt ang xt.*Quy i khi lng v nt. Xt h 3 bc t do,khi lng tp trung ti cc on (nh hnh v).B qua khi lng nc km, cn nht. Khi lng bn thn quy i ti cc on tng ng l q1,q2. Ta c:

q1= +

=>q1 = +

= 78.5**+=51.39(kN)

q2==*V

=

=78.5**= 46.86(kN)2.Xc nh ma trn M,K a)Xc nh ma trn khi lng M.

V biu momen xc nh phn lc V(m3) ,V (m2):

Cc php tnh thc hin:

Tnh V =?

: V*35/6-529.74*35/6-46.86*3*13/3-51.39*17/6*17/12=0

V=669.53(kN)

=(1)( 1)=1/EI(1/2*1*35/6)2/3*1+1/EI(1/2*1*35/6)2/3*1=35/9EI

==(1)(2)=1/EI(1/2*1*35/6)1/3*1=35/36EI

=(2)(2)=1/EI(1/2*1*35/6)2/3*1*2=35/9EI

=(1)(M)=1/EI(2/3*35/6*218.59)1/2*1*2=850.07/EI

= (2)(M) =1/EI(2/3*35/6*218.59*1/2*1)+1/EI(2/3*17/6*218.59*0.696)+1/EI(2/3*3*218.59*0.321 =833.06/EIPhng trnh chnh tc:

Thay cc gi tr vo biu thc trn ta c:

V biu Momen (M) t ta xc nh c cc phn lc ti gi A v B.Phn lc ta coi cng chnh l m2 v m3 cn tm. Suy ra ta c ma trn khi lng M:

M= (kg) b)Xc nh ma trn (K) cng ca kt cu. Gi s t mt lc P=1(kN) t ti cc nt ca ct (Hnh).Ct lm vic nh mt dm conxon ,s dng l thuyt kt cu ta tnh c chuyn v ti cc nt. Ta c:

==1/EI(1/2*17.5*17.5)*2/3*17.5

=

==

=1/EI(1/2*35/3*35/3)245/18 =

=( =1/EI(1/2*35/3*35/3)2/3*35/3

=

= = =1/EI(1/2*35/6*35/6)140/9

=

= = =1/EI(1/2*35/6*35/6)175/18

=

==1/EI(1/2*35/6*35/6)2/3*35/6

= Trong :

E =2.1x10 (kN/m)

I= = =0.087 (m)

Thay Ev I vo cc chuyn v ri vit di dng ma trn,ta c ma trn mm D:

D== (m) Ma trn K c xc nh bng biu thc:

K=D

(D l ma trn nghch o ca ma trn mm D) S dng phn mm Matlab ta tnh c ma trn nghch o D. Lnh Matlab: K=D^-1 Ta c ma trn K:

K= (1/m)3.Xc nh tn s dao ng ring v cc dao ng. Ta bit ma trn M v ma trn K,t d dng tm c tr ring(TR) v cc vc t ring (VR) ca K v M bng chng trnh Matlab,vi lnh: [phi,lamda]=eig(K,M) Ta c kt qu: Ma trn cc dao ng ring:

= Ma trn tr ring :

= Xc nh tn s dao ng ring:

= Lnh Matlab: Omega=sqrt(lamda) Ta c ma trn tn s ring ca kt cu:

= (rad/s) V cc dao ng ring: 4.Xc nh cc ma trn cng K1,ma trn khi lng M1. a)Tm ma trn cng K1 h ta suy rng Ma trn K1 c xc nh theo cng thc:

K1=*K* S dng Matlab tnh K1 ta c:

K1= (N/m) b)Tm ma trn khi lng M1 h ta suy rng:

M1= *M* S dng Matlab tnh M1 ta c:

M1= (kg) 5.Xc nh cc gi tr 1,2 Ta c ph song theo Pieron-Moskowiz:

S= () Vi:

A=4 =4* =1.2135 ( )

B= ==0.0538 ( ) Vy ta c ph song c dng sau:

S = () V th ph sng Pierson-Moskowitz: Ta s dng Matlab v,lnh l: W=0.1:0.2:4; y=1.2135*exp(-0.0538./w.^4)./w.^5; plot(w,y) gird on th ph sng Pierson-Moskowitz:

6.Trong khong v ,chia ra lm 20 khong u nhau,xc nh cc tn s ti cc im chia v cc ph sng tng ng.

T th ta suy ra gi tr v ng vi gi tr S()=0 l:

=0.25

=3.5 Khong cch gia 2 im chia:

= =0.1625 (rad/s) Gi tr tn s ng vi 21 im chia :

=( 0.25 0.4125 0.575 0.7375 0.9 1.0625 1.225 1.3875 1.55 1.7125 1.875 2.0375 2.2 2.3625 2.525 2.6875 2.85 3.0125 3.175 3.3375 3.5 ) (rad/s)

Gi tr ph sng S tng ng vi : S=( 0.0013 15.846 11.8 4.637 1.893 0.859 0.43 0.233 0.134 0.082 0.052 0.034 0.023 0.016 0.012 0.0086 0.006 0.0049 0.0038 0.0029 0.0023 )

7.ng vi cc tn s sng ti cc im chia,xc nh lc ngang F ,F ,F S kt cu cng trnh: Xc nh ti trng sng theo Morison

F(t)= CV+a(kN) Trong :

C= 0.5CD

Vi = 1044.85 (kg/)

C= 0.65 1.05 (theo tiu chun API ) , y ta ly C= 1 Thay vo ta tnh c :

=CA

Vi A = =3.14=1.039(m)

C=2 (vi tit din trn) Thay vo ta c :

tnh ti trng sng ta i xc nh ln lt cc thng s sau: a/Xc nh s sng k Ta c :

=>

S dng phn mm Matlab gii phng trnh trn tm gi tr k ng vi 21 im tn s

Lnh Matlab tnh K vi =0.25: k=solve(0.25^2=9.81*k*tanh(k*14.5)) Tnh tng t cho 20 gi tr cn li ta thu c kt qu sau:

=( 0.25 0.4125 0.575 0.7375 0.9 1.0625 1.225 1.3875 1.55 1.7125 1.875 2.0375 2.2 2.3625 2.525 2.6875 2.85 3.0125 3.175 3.3375 3.5 ) k = 0.0213 0.0361 0.0525 0.0714 0.0941 0.1219 0.1563 0.1975 0.2453 0.299 0.3584 0.4232 0.4934 0.569 0.6499 0.7363 0.828 0.9251 1.0276 1.1355 1.2487 ) b/ Xc nh vn tc v gia tc sng ti cc nt Ta c:

V=

a =

S dng phn mm Matlab tnh Vv atrn,lnh Matlab: Vx=w.*cosh(k.*z)./sinh(k.*d) ax=i.*w.^2.*cosh(k.*z)./sinh(k.*d)

+) Ti z=35/6 (m) ta c vn tc v gia tc tng ng l: V1= (0.8029 0.7699 0.7195 0.6521 0.5681 0.4723 0.3711 0.2764 0.1957 0.1322 0.0852 0.0524 0.0307 0.0171 0.0090 0.0046 0.0022 0.0010 0.0004 0.0002 0.0004 ) m/s

a1=( 0 + 0.2007i 0 + 0.3176i 0 + 0.4137i 0 + 0.4810i 0 + 0.5113i 0 + 0.5018i 0 + 0.4546i 0 + 0.3836i 0 + 0.3033i 0 + 0.2265i 0 + 0.1598i 0 + 0.1068i 0 + 0.0675i 0 + 0.0403i 0 + 0.0228i 0 + 0.0122i 0 + 0.0062i 0 + 0.0030i 0 + 0.0014i 0 + 0.0006i

0 + 0.0088i ) m/

+)Ti z=35/3 (m) ta c vn tc v gia tc tng ng l: V2=( 0.8215 0.8210 0.8200 0.8197 0.8196 0.8199 0.8160 0.8034 0.7767 0.7348 0.6794 0.6143 0.5436 0.4712 0.4005 0.3337 0.2729 0.2191 0.1727 0.1337 0.6105 ) m/s

a2=( 0 + 0.2054i 0 + 0.3386i 0 + 0.4715i 0 + 0.6045i 0 + 0.7376i 0 + 0.8711i 0 + 0.9996i 0 + 1.1147i 0 + 1.2039i 0 + 1.2584i 0 + 1.2738i 0 + 1.2516i 0 + 1.1960i 0 + 1.1132i 0 + 1.0112i 0 + 0.8968i 0 + 0.7777i 0 + 0.6600i 0 + 0.5483i 0 + 0.4463i

0 +12.8207i ) m/

+)Ti z=29/2 (m) ta c vn tc v gia tc tng ng l: V3=( 0.8350 0.8587 0.8959 0.9504 1.0257 1.1263 1.2516 1.3966 1.5525 1.7131 1.8751 2.0375 2.2000 2.3625 2.5250 2.6875 2.8500 3.0125 3.1750 3.3375 21.0000 ) m/s a3=( 0 + 0.0021i 0 + 0.0035i 0 + 0.0052i 0 + 0.0070i 0 + 0.0092i 0 + 0.0120i 0 + 0.0153i 0 + 0.0194i 0 + 0.0241i 0 + 0.0293i 0 + 0.0352i 0 + 0.0415i 0 + 0.0484i 0 + 0.0558i 0 + 0.0638i 0 + 0.0722i 0 + 0.0812i 0 + 0.0908i 0 + 0.1008i 0 + 0.1114i

0 + 4.4100i ) * (1.0e+02) m/

c/Xc nh lch chun ph vn tc Ta c :

=

=

S =

Ta i tm hm mt ph vn tc S S dng phn mm Matlab ta xc nh c hm mt ph vn tc: Svx=Vx.^2.*S T1=Svx(1)/2 T2=Svx(21)/2 T=T1+T2 for i=2:20 T=T+Svx(i) end

T ta tnh c lch chun ca ph vn tc bng Matlab,lnh l: D1=0.1625*T xicmavx=sqrt(T)

(xicmavx l k hiu lch chun ca ph vn tc ) Ta c kt qu nh sau:

=(1.7273 1.9827 2.2607 )

d)Xc nh ti trng sng q: S dng phn mm Matlab tnh cc ti trng sng q,lnh l: q1=CD*sqrt(8/pi).*xicmav1.*V1+C1.*a1 q2=CD*sqrt(8/pi).*xicmav2.*V2+C1.*a2 q3=CD*sqrt(8/pi).*xicmav3.*V3+C1.*a3 Ta c cc gi tr q l:

q1=( 1.3296 + 0.4358i 1.2750 + 0.6896i 1.1915 + 0.8983i 1.0799 + 1.0442i 0.9407 + 1.1101i 0.7821 + 1.0895i 0.6146 + 0.9871i 0.4578 + 0.8328i 0.3240 + 0.6585i 0.2190 + 0.4917i 0.1412 + 0.3470i 0.0868 + 0.2318i 0.0508 + 0.1465i 0.0283 + 0.0876i 0.0150 + 0.0496i 0.0075 + 0.0266i 0.0036 + 0.0135i 0.0016 + 0.0065i 0.0007 + 0.0030i 0.0003 + 0.0013i 0.0007 + 0.0191i ) *(1.0e+03 ) (N/m)

q2=( 0.1561 + 0.0446i 0.1561 + 0.0735i 0.1559 + 0.1024i 0.1558 + 0.1312i 0.1558 + 0.1602i 0.1558 + 0.1891i 0.1551 + 0.2170i 0.1527 + 0.2420i 0.1476 + 0.2614i 0.1397 + 0.2732i 0.1291 + 0.2766i 0.1168 + 0.2718i 0.1033 + 0.2597i 0.0896 + 0.2417i 0.0761 + 0.2195i 0.0634 + 0.1947i 0.0519 + 0.1689i 0.0416 + 0.1433i 0.0328 + 0.1190i 0.0254 + 0.0969i 0.1160 + 2.7836i )*( 1.0e+04) ( N/m)

q3=( 0.0181 + 0.0045i 0.0186 + 0.0077i 0.0194 + 0.0112i 0.0206 + 0.0152i 0.0222 + 0.0200i 0.0244 + 0.0260i 0.0271 + 0.0333i 0.0303 + 0.0421i 0.0336 + 0.0522i 0.0371 + 0.0637i 0.0406 + 0.0763i 0.0442 + 0.0901i 0.0477 + 0.1051i 0.0512 + 0.1212i 0.0547 + 0.1384i 0.0582 + 0.1568i 0.0618 + 0.1764i 0.0653 + 0.1970i 0.0688 + 0.2189i 0.0723 + 0.2418i 0.4551 + 9.5750i ) *(1.0e+05) (N/m)

e/Xc nh ti trng ngang F: S quy v lc ti tp trung:

Ta quy i cc lc phn b q v ti trng tp trung ti cc nt ta c cc gi tr F1,F2,F3 (nh hnh v): F1=1/2*H/3*(q1+q2)=35/12*(q1+q2) F2=901*q3*H/7350+1/2*H/3*q2 F3=289/7350*H*q3 S dng phn mm Matlab ta tnh c cc gi tr ti trng tp trung,lnh l: F1=35/12*(q1+q2) F2=901/1260*q3+35/12*q2 F3=289/1260*q3 Ta c kt qu: F1=( 0.8432 + 0.2572i 0.8270 + 0.4156i 0.8021 + 0.5606i 0.7694 + 0.6874i 0.7288 + 0.7909i 0.6827 + 0.8694i 0.6317 + 0.9209i 0.5789 + 0.9488i 0.5251 + 0.9545i 0.4713 + 0.9403i 0.4178 + 0.9079i 0.3659 + 0.8602i 0.3162 + 0.8001i 0.2695 + 0.7305i 0.2264 + 0.6548i 0.1872 + 0.5756i 0.1523 + 0.4965i 0.1219 + 0.4198i 0.0960 + 0.3481i 0.0742 + 0.2830i 0.3387 + 8.1245i)*( 1.0e+04) N F2=( 0.0585 + 0.0162i 0.0588 + 0.0269i 0.0593 + 0.0379i 0.0602 + 0.0492i 0.0613 + 0.0610i 0.0629 + 0.0737i 0.0646 + 0.0871i 0.0662 + 0.1007i 0.0671 + 0.1136i 0.0673 + 0.1252i 0.0667 + 0.1353i 0.0656 + 0.1437i 0.0642 + 0.1509i 0.0627 + 0.1572i 0.0613 + 0.1630i 0.0602 + 0.1689i 0.0593 + 0.1754i 0.0588 + 0.1827i 0.0588 + 0.1912i 0.0591 + 0.2012i 0.3593 + 7.6588i) *(1.0e+05) N

F3=( 0.0042 + 0.0010i 0.0043 + 0.0018i 0.0045 + 0.0026i 0.0047 + 0.0035i 0.0051 + 0.0046i 0.0056 + 0.0060i 0.0062 + 0.0076i 0.0069 + 0.0096i 0.0077 + 0.0120i 0.0085 + 0.0146i 0.0093 + 0.0175i 0.0101 + 0.0207i 0.0109 + 0.0241i 0.0117 + 0.0278i 0.0126 + 0.0318i 0.0134 + 0.0360i 0.0142 + 0.0404i 0.0150 + 0.0452i 0.0158 + 0.0502i 0.0166 + 0.0555i 0.1044 + 2.1962i) *( 1.0e+05) N

8.Xc nh gi tr hm truyn H(i) ti cc im chia

() ( )

: l s hng hng th I,ct th j ca ma trn khi lng M

: l tn s ring ca kt cu

: l tn s ring ca ti trng S dng phn mm Matlab tnh cc gi tr hm truyn, lnh l: H1=1./(33728.848^2).*1./(1-w./omega(1,1)).^2 H2=1./(37747.197^2).*1./(1-w./omega(2,2)).^2 H3=1./(66293.578^2).*1./(1-w./omega(3.3)).^2 Ta c kt qu: H1=(0.0901 0.4860 0.0260 0.0083 0.0040 0.0024 0.0016 0.0011 0.0008 0.0006 0.0005 0.0004 0.0003 0.0003 0.0002 0.0002 0.0002 0.0002 0.0001 0.0001 0.0000)* (1.0e-07) H2=(0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 0.0003 0.0003 0.0004 0.0006 0.0008 0.0012 0.0020 0.0042 0.0130 0.2280 0.0474 0.0079 0.0031 0.0000)*( 1.0e-05) H3=( 0.2422 0.2525 0.2634 0.2751 0.2876 0.3009 0.3152 0.3305 0.3470 0.3647 0.3839 0.4046 0.4270 0.4513 0.4777 0.5066 0.5381 0.5727 0.6108 0.6527 0.0915)*( 1.0e-09)

9.Xc nh phng sai chuyn v ti dng dao ng th I,II,III

Lc sng theo h ta suy rng l:

Chuyn v vi k l bc t do th k ca cng trnh bng cng thc sau:

i l dng dao ng th i ( y ta c dng)Trong nu sng l QTNN dng , c k vng ton bng 0;th lch chun ca chuyn v kt cu s c tnh theo cng thc:

H(iw) l hm truyn dng dao ng th I do ti trng sng tn s omega tc dng V l ph 2 pha nn ta c

l gi tr ti trng sng nhn lin hp chnh bin

Ta c :

Ta t P2=(Pi*)^2=Q*(iw).Q*(-iw) ,dung phn mm Matlab ta tnh c P2,lnh l: F=[F1;F2;F3] % Q*=QQ QQ=phit*F P2=double(QQ.*conj(QQ)) Ti dng dao ng ring th nht,ta i tnh tch phn I,s dng phn mm Matlab tnh,lnh l: H12=H11.*conj(H11) r1=P2(1,:).*S.*H12 I1=r1(1)/2 for j=2:20 I1=I1+r1(j) end I1=(I1+r1(21)/2)*(3.5-0.25)/21 Tng t vi 2 dao ng ring cn li ta tnh c 2 gi tr I tng ng I1=5.6633E-7 I2=1.0708E-7 I3=4.2961E-11 Vy ta xc nh c phng sai l:

S dng phn mm Matlab tnh,lnh l: Xicma2xk=(phi(1,1))^2*I1/(omega(1,1)^4*pi)+(phi(1,2))^2*I2/(omega(2,2)^4*pi)+(phi(1,3))^2*I3/(omega(3,3)^4*pi) Xicma2xk=1.0314E-5 Xicmaxk=sqrt(xicma2xk) Xicmaxk= 0.0032 10/Xc nh tin cy chuyn v nh ct

Ch s tin cy :

Trong

: l gi tr trung bnh ca chuyn v cho php ti nh ct

: l gi tr trung bnh chuyn v ti nh ct

: l phng sai ca chuyn v ti nh ct

: l phng sai ca chuyn v cho php

Sng l qu trnh ngu nhin dng,chun,trung bnh 0 v c tnh cht egodic v h kt cu ang xt l h tuyn tnh nn ta c:

Do chuyn v cho php tin nh nn

Thay s ta tnh c ch s tin cy Vy tin cy ca cng trnh:

BIN HC MNH MNG LY CHUYN CN LM BN MY XANH KHNG LI LY CH C DNG LN------------------------------------------------------------------------------------------------------------------------------------------SV.NGUYN VN LUN_MSSV:1356.56 Page 24 of 24