- 1. TIN CY CA CC CNG TRNH TRN BIN TS. NGUYN VITm tt: Trong bi bo
trnh by s cn thit v phng php tnh ton cc cng trnh trn bin theo l
thuyt tin cy, nu ni dung v v d tnh ton vi kt qu c th xc nh tin cy
ca kt cu.I. T VN .Ngnh xy dng thu ni chung v xy dng cc cng trnh trn
bin ni ring c nhimv to ln l to c s h tng p ng nhu cu pht trin vn ra
bin ca nn kinh t qucdn. Hng lot cng trnh c quy m ln vi s vn u t hng
trm triu, thm ch hng tUSD v s c xy dng trn bin v ven bin. l cc cng
trnh thm d, khai thcdu kh, cc cng trnh cng ven b v xa b phc v vn
chuyn hng ho cho ni a v quct, cc o nhn to, cc cng trnh phc v quc
phng, du lch, Cc cng trnh trn bin lm vic trong iu kin khc nghit ca
mi trng xungquanh, chu tc ng ca sng, gi, dng chy, n mn, trong sut
thi hn phc v cachng, c th n 50-60 nm hoc lu hn. Chnh v vy, tin cy
ca chng l mi quan tmhng u khng ch ca cc nh chuyn mn, ca cc doanh
nghip, m ca c Nh nc vnhn dn c lin quan n s an ton ca cng trnh v hiu
qu u t. tin cy ca cngtrnh ph thuc vo cht lng cng tc kho st a cht,
thu hi vn, tnh ton - thit k, thicng v khai thc cng trnh, Trong cc
yu t nh hng n tin cy ca cng trnh thphng php tnh ton-thit k c vai tr
quan trng bc nht.Cc phng php tnh ton v thit k cc cng trnh xy dng
trong cc Tiu chun hinhnh c gi l phng php cc trng thi gii hn, c to
ra Lin X c v c sdng trong vng hn 50 nm nay. Cc phng php tng t cng c
s dng nhiu nckhc trn th gii di tn gi phng php bn xc sut, chng l c s
cho nhiu Tiuchun thit k ca Chu u v Tiu chun ISO [11]. Nhc im c bn
ca cc phng phptnh hin hnh l mu thun trong phng php lun, ngha l s
dng cc tham s tnh ton cbn cht ngu nhin trong thut ton vi cc quan h
hm s c tnh n tr v tin nh, cngnh khng xt n yu t thi gian. Nhiu kt qu
nghin cu c tin hnh trong 40-50 nmqua [1, 6, 8] khng nh: cc tham s
ca kt cu v ti trng c dng trong tnh ton cccng trnh khng phi l cc i
lng khng i m l cc i lng ngu nhin.V th, ngy nay trn th gii ngi ta s
dng tng i ph bin cc phng phpxc sut v tin cy trong tnh ton cc cng
trnh. y l h phng php tin tin tnhton cc kt cu xy dng, ang c p dng
nhiu nc pht trin trn th gii. cc ncnh Nga, M, Trung Quc, Nht Bn,...
u ban hnh cc Tiu chun theo hng ny [2,10, 12] dn thay th cc Tiu chun
c bin son theo cc phng php tin nh. ncta hin nay, vic nghin cu v p
dng h phng php tnh ton theo quan im tin cy thit k cc cng trnh,
trong c cc cng trnh trn bin, l ht sc cn thit v cp bch. II. TNH TON
CNG TRNH TRN BIN THEO L THUYT TIN CY.Khc vi cc Tiu chun hin hnh, cc
phng php thit k kt cu xy dng theo quanim xc sut ngh tiu ch mi v cht
lng l tin cy ca kt cu.Khi nim tin cy bao hm lng thng tin rt ln. tin
cy l tnh cht cacng trnh hay cc cu kin duy tr (theo thi gian trong
cc gii hn c thit lp) gi trca tt c cc tham s c trng cho kh nng hon
thnh chc nng yu cu trong ch khaithc c nh trc [2]. tin cy l tnh cht
phc tp v gm t hp cc tnh cht: tnh an ton (khng xy ras c), tnh lu di
(tui th hay thi hn phc v), tnh sa cha v tnh bo ton. Tuy nhin,ngi ta
coi c trng c bn ca tin cy ca cng trnh l xc sut lm vic an ton
(khng1
2. c s c, khng hng) ca n trong mt thi hn khai thc xc nh. S c (
ting Nga,failure ting Anh) l bin c ngu nhin ph hoi kh nng lm vic ca
cu kin hoc ca hthng. Khi nim s c rt gn vi khi nim trng thi gii hn
trong tnh ton tin nh.Nhng cng trnh khoa hc u tin v l thuyt tin cy
tnh ton kt cu l cccng trnh c cng b t nhng nm 1926-1929 ca M. Maier
v N. Ph. Khshianp [8],trong ph phn vic tnh kt cu theo phng php ng
sut cho php v nu ra tngtnh ton kt cu xy dng theo quan im xc sut.
Tuy nhin, vo thi gian cc tngny khng c ng h.N. X. Strenletsky c coi
l ngi t nn mng cho vic tnh ton cng trnh theo quanim thng k. Cc cng
trnh ca ng tr thnh c s cho phng php tnh kt cu xydng theo trng thi
gii hn vi vic s dng cc phng php thng k. ng l ngi utin nghin cu ng
thi phn b ca ti trng S v bn hoc kh nng chu ti R cakt cu, nu khi nim
m bo khng ph hoi v cng thc tnh, l i lng n ginv trc quan, cho php nh
gi tin cy ca kt cu. Tuy nhin, xc sut lm vic an tontnh theo m bo
khng ph hoi li qu cao v khng xt c tt c cc t hp c th cca R v S [8].T
nm 1952 tin cy ca kt cu c A. R. Rgianitsn nh ngha cht ch hn khing a
vo khi nim hm khng ph hoi (hnh 1): = RS. (1)K vng ton v phng sai i
vi phn b c biu th qua cc c trng tngng ca phn b ti trng v bn = R S ;
= R +S , 2 22(2) y , R, S k vng ton ca cc phn b tng ng; , R , S l
phng sai ca cc 2 2 2phn b. A. R. Rgianitsn a vo trong tnh ton i lng
c gi l c trng an tonca kt cu R S = = (3)R +S22 S R = Vng an tonVng
s c S S,R0 R = R S Hnh 1. Dn xut c trng an ton ca A. R. Rgianitsn R
rng, c trng an ton chnh l tin cy ca kt cu dng khng tng minh,n c
ngha nh cng c dng cng thc ton hc xc nh s c xc sut ri ca ccgi tr vo
vng khng an ton (hnh 1). i vi phn b chun, xc sut ny c tnh theocng
thc: 1Q = p( )d = ( ), (4) 22 3. 12 y ( ) =2 0 exp(2 )d l hm
Laplax, cc gi tr ca n c lp thnh bng. So vi xc sut s c Q th c trng
an ton c u im hn v n biu th bi mts khng ln, thng l ln hn 1, trong
khi xc sut s c Q l mt s thp phn rt nh.V d, khi = 1,28 th Q = 0,1 v
khi = 5 th Q = 2,9.107. Cc gi tr > 5 c th coi l rtln v tng ng vi
nhng gi tr cc k nh ca xc sut s c Q, khi xc nh Q theocng thc (4) s
rt kh khn.Trong Tiu chun ca Trung Quc [10] v ca Nht Bn nm 2007
[12], ngi ta u gic trng an ton l ch s tin cy , tc lR S =. (5) R +S2
2 Theo Tiu chun Trung Quc [10], cng trnh c thit k theo tin cy ph
thucdng ph hoi v mc an ton ca cng trnh (bng 1). Bng 1. Phn cp mc an
ton ca cng trnh [10]. Cp an ton Cp ICp II Cp III c trng ph (Rt
nghim trng) ( Nghim trng)(Khng nghim trng) hoi PS PS PS(ch s (xc
sut an (ch s tin (xc sut an (ch s (xc sut an tin cy) ton)cy)
ton)tin cy)ton) Ph hoi bin 3,7 0,9998900 3,2 0,9993189 2,7
0,996533hnh t tPh hoi t ngt4,20,99998665 3,7 0,9998900 3,20,9993189
Bng 2. Cc mc thit k kt cu theo [12]Phng trnh kim tra an Mc Khi
nimGhi chton M phng theo 3 Pf PfaXc sut s cMonte Carlo 2 a Ch s tin
cy Tiu chun ny R Rk S S kCc h s an tonPhng php thit k theo 1 b phn
trng thi gii hnTheo Tiu chun Nht Bn [12], cng trnh c thit k theo
tin cy mc 2 (xembng 2): i vi cu tu v cc cng trnh cng, ch s tin cy
cho php a c ly bng2 4; i vi cc dng chn sng khc nhau, ch s a = 2,04
3,60; i vi cc cng trnhrt quan trng, nh l phn ng ht nhn, ch s a c ly
bng 5 6.T nm 1986 Lin X c vic tnh ton thit k cc cng trnh cng theo l
thuyt tin cy c thc hin theo mc 3 [2]. Ngi ta xc nh khng phi xc sut
s c m lxc sut ngc vi n v ngha, l xc sut lm vic khng xy ra s c (xc
sut lmvic an ton) ca cc cu kin chu ti v ca c cng trnh.3 4. Xc nh
tin cy ca cu kin chu ti. i vi cc cng trnh m mt phn bxc sut ca cc
tham s kt cu v ti trng c bn tun theo quy lut phn b chun hoc rtgn vi
phn b chun [1, 7], c th xc nh xc sut lm vic an ton ca cc cu kin
chuti theo phng php tuyn tnh ho. Khi , vo thi im t bt k, xc sut lm
vic khngxy ra s c ca cu kin c xc nh theo cng thc: S R P = 1 , (6)S
+R 22 y S , S , R , R tng ng l k vng ton v lch chun ca hm ti trng S
v hm bn hay kh nng chu ti R ca cu kin; hm phn b chun.Tuy nhin,
trong nhiu trng hp, do s lng mu th hn ch cng nh do cc cim cng ngh
hoc cc nguyn nhn khc m cc quy lut phn b ca cc tham s c thch gn vi
phn b chun, chng c lch v nhn nht nh. iu c th dn n sais ln khi xc nh
xc sut lm vic an ton ca cu kin theo phng php tuyn tnh ho. Vth, tnh
tin cy ca cc cu kin chu ti ca cng trnh, hp l hn c l s dngphng php
bn bt bin tng qut ca Iu. A. Pavlp. C th xem chi tit cng thc tnh
theophng php ny trong [1, 8].Mc ch tnh ton cc cng trnh v tin cy l,
vi xc sut cao, khng cho phpny sinh s c trong cng trnh, trong cc cu
kin v nn ca n vo thi k xy dng v khaithc. Nhng i vi mt cu kin c th
xy ra mt s s c. V th, vo thi im t bt k,iu kin lm vic an ton ca cu
kin j theo dng s c i phi c tun th vi xc sut Pi jkhng c thp hn xc
sut tiu chun Ptc [2]: Pi j = P(Yi j , t ) Ptc , (7)vi Yi j = Ri j S
i j > 0, (8) y: Yi j d tr ca tham s c kim tra ca trng thi ng sut
hoc bin dng ca cukin j, i lng ngu nhin; Ri j gi tr gii hn ca tham s
c kim tra ca trng thing sut hoc bin dng ca cu kin j, i lng ngu
nhin; S i j gi tr thc t (c lytheo kt qu tnh ton xc sut kt cu) ca
tham s c kim tra ca trng thi ng sut hocbin dng ca cu kin j, i lng
ngu nhin; Ptc xc sut lm vic an ton tiu chun theodng s c i ca cu kin
j, tc tin cy tiu chun.Khi , xc sut lm vic an ton ca cu kin Pj theo
tp hp cc s c c th xy ra vin, c xc nh theo cng thc:T Pj = 1 (1 Pi
=1i j ), (9) y T s cc s c c th xy ra vi cu kin j. Xc nh tin cy ca
cng trnh. tin cy ca c cng trnh ph thuc vo cchlin kt v tc dng tng h
gia cc cu kin chu ti ca n [4, 8]. Vi tnh cht nh nhhng, chng ta xc
nh tin cy ca c cng trnh theo ch dn ca 31.31.35 85 [2].Khi , xc sut
lm vic an ton ca cng trnh Pc c xc nh xut pht t quan nim vcc s c ca
cc cu kin c chia ra, nh cc bin c ngu nhin c lp, ph hp vinh l nhn xc
sut theo cng thc: K Pc = Pj Pctc ,(10) j =1 y K s cu kin ca cng
trnh; Pctc tin cy tiu chun ca cng trnh.4 5. Cn phi lu rng, d thit k
cc cng trnh theo xc sut lm vic an ton Pj , Pchay ch s tin cy th u
phi tin hnh tnh ton xc sut kt cu xc nh cc ctrng thng k ca bn hay kh
nng chu ti R , R , 2( R ) , 3( R ) , 4( R ) ,... v ca ni lc dongoi
ti hay cc tc ng khc gy ra trong cc cu kin S , S , 2( S ) , 3( S ) ,
4( S ) ,... i vicc kt cu n gin v phn b ca cc tham s kt cu v ti trng
gn vi phn b chun cth s dng phng php tuyn tnh ho. Cn i vi cc kt cu
phc tp, m kh nng chuti v ni lc trong cu kin ph thuc vo mt s lng ln
cc bin ngu nhin, th tt nhtnn s dng cc phng php s tnh xc sut kt cu,
v d, phng php Monte Carlo.Lin quan n iu , tc gi nu Phng php m hnh
ho thng k tng bc tnhton xc sut cc kt cu xy dng, t xc nh c tin cy ca
cc cu kin v ca ccng trnh ni chung. Phng php c cng b nc ngoi v c
trnh by chi tittrong [4, 9]. V tin cy tiu chun ca cu kin v ca cng
trnh. tin cy tiu chun cthit lp trn c s kinh nghim thit k cc cng
trnh vi vic s dng cc phng php cal thuyt tin cy v cc tiu ch kinh t.
Vi tnh cht nh hng, trong quy phm [2] chophp ly tr s tin cy tiu chun
i vi cc cu kin chu ti ca cc cng trnh bn cngbin tu thuc dng s c ca
cu kin do ph hoi vt liu hay tng tc vi nn t. i vi cc cng trnh, tu
thuc quy m v mc quan trng ca chng m tin cytiu chun c ly vi cc gi tr
tng ng. Gi tr tin cy ca cng trnh c th thamkho trong [2, 8, 10,12],
tuy nhin, theo ngh ca mt s nh khoa hc [1, 2, 7], gi tr nykhng nn ly
nh hn 0,95, ngha l Pctc 0,95 hay a 1,645. y l vn vn ang cnghin cu
tip. Vic xc nh tin cy ca cng trnh c xt n yu t thi gian, tc dng n mn
cami trng, c xem xt trong cc cng trnh khc, v d trong [5]. III. TNH
TON TIN CY CA CNG TRNH TRN BIN. Vi tnh cht nh v d, di y trnh by cch
tnh tin cy ca cng trnh bn dngb cc cao mm, l dng kt cu c s dng rng
ri trong cc cng trnh trn bin. tincy ca cc cng trnh trn bin khc cng
c xc nh tng t. 1. Trnh t tnh ton. i vi cng trnh bn b cc cao mm, cc
cu kin chu ti c bn l (xem hnh 2):1- cc cc; 2- b cc; 3- mi dc gm bn;
4- khi t tc dng tng h vi kt cu. S b, cc dng s c sau y c th xy ra i
vi cng trnh bn b cc cao. P31 P42Bng 3. Ma trn xc sut lm vic an
tonca cng trnh bn b cc cao mm Cu kinP21P53 j1 23 4 P64 S c j i1P11
2 P21P113P314 P425 P53 Hnh 2. Cc s c c th xy ra ca6 P64cng trnh bn
b cc cao mm5 6. i vi cu kin 4: 6- mt n nh chung ca khi t cng vi cng
trnh vi xc sutP64 , xc sut n nh P64 = 1 P64 . i vi cu kin 1: 1- mt
kh nng chu ti ca cc cc theo nn t do ti trng ngvi xc sut P11 , xc
sut an ton P11 = 1 P11 ; 2- mt kh nng chu ti ca cc cc theo nnt do
ti trng ngang vi xc sut P21 , xc sut an ton P21 = 1 P21 ; 3- mt kh
nng chu tido ph hoi vt liu cc vi xc sut P31 , xc sut khng ph hoi
P31 = 1 P31 ; i vi cu kin 2: 4- ph hoi b cc vi xc sut P42 , xc sut
an ton P42 = 1 P42 ; i vi cu kin 3: 5- mt n nh cc b mi dc gm bn vi
xc sut P53 , xc sut nnh P53 = 1 P53 ; Ma trn xc sut lm vic an ton i
vi cng trnh bn b cc cao mm c dn ratrong bng 3 v tin cy hay xc sut
lm vic an ton ca cc cu kin chu ti ca b cccao c xc nh theo cc cng
thc (9):P1 = 1 [(1 P11 ) + (1 P21 ) + (1 P31 )]= P11 + P21 + P31 2
;P2 = P42 ; (11)P3 = P53 ;P4 = P64 . Khi , tin cy ca cng trnh b cc
cao mm c xc nh theo cng thc: Pc = P1 P2 P3 P4 .(12)Ngy nay k thut
tnh ton cho php tnh tin nh cng trnh dng b cc cao theo biton khng
gian. Hu ht cc nc u dng cc chng trnh mu dng SAP-2000 ca Mhoc cc
chng trnh tng t. Tuy nhin, trong nhiu trng hp, vic tnh ton cng
trnhc a v tnh cc bi ton phng. tnh tin cy ca kt cu b cc cao mm, cho
nnay cha c phng php tnh xc sut c th ng dng vo thc t. Ni chnh xc hn,
v mtl thuyt c th tnh xc sut cc kt cu theo phng php im nng, phng php
thnghim thng k, phng php Monte Carlo, nhng vic hin thc chng rt kh
khn vvic p dng cc phng php k trn ch t c trong nhng trng hp c bit.
Linquan n iu , tc gi nu phng php tnh ton xc sut kt cu b cc cao mm v
c cng b ti Hi ngh khoa hc thc tin ton Lin bang Nga v cng sng v cng
bin[3, 9]. Phng php c th c p dng tnh ton xc sut kt cu cho c bi ton
phngcng nh bi ton khng gian. Do tnh cht phc tp v khi lng tnh ton
ln, y ch nunguyn tc v kt qu tnh xc sut theo phng php nu trn i vi bi
ton phng, cn ivi bi ton khng gian cng tin hnh tng t.a) Thut ton tin
nh.Gi thit rng, phng php tnh ton cng trnh b cc cao mm ca V. X.
Xcuratp vN. I. Shapshnikp tin cy v c chn vi tnh cht l thut ton tin
nh xc nhcc tham s thng k ca cc ni lc trong cc cu kin chu ti. Coi kt
cu gm cc cc vcc dm n hi nh mt khung nhiu nhp thng thng vi cc gi n
hi, trong mi ntc chuyn v ng v chuyn v xoay c lp vi cc nt khc v mt
chuyn v ngang chungi vi tt c cc nt. V th, theo phng php ny, nu kt
cu c n1 nt th s chuyn v n cn tm l: n = 2n1 + 1 .(13)V d, khung
ngang ca mt cng trnh b cc cao mm trn hnh 3a c s chuyn v cntm l n =
2.5 + 1 = 11 , v h c bn ca kt cu c th hin trn hnh 3b. 6 7. a) b)
Z7Z8 Z9 Z10 Z11 123 45Z1Z2 Z3 Z4Z5Z6 Hnh 3. Tnh ton cng trnh b cc
cao mm: a) s tnh;b) h c bn ca phng php Xcuratp V. X. H c bn nhn c t
h cho bng cch a vo kt cu n lin kt b sung ngn cnn chuyn v c th ca cc
nt s tng ng vi h thc nu ni lc trong cc lin kt gibng 0. T iu kin ny
c th vit h phng trnh chnh tc sau y:r11 Z1 + r12 Z 2 + r13 Z 3 + ...
+ r1n Z n + r1 p = 0 ;r21 Z1 + r22 Z 2 + r23 Z 3 + ... + r2 n Z n +
r2 p = 0 ;r31 Z1 + r32 Z 2 + r33 Z 3 + ... + r3n Z n + r3 p = 0 ;
(14)rn1 Z1 + rn 2 Z 2 + rn3 Z 3 + ... + rnn Z n + rnp = 0 ,trong
rij ni lc pht sinh trong lin kt i do chuyn v n v Z j = 1 gy ra;rip
phn lc pht sinh trong lin kt i do ti trng ngoi gy ra;Z i cc chuyn v
cn tm ca cc nt (i=1 n). Cc h s rij v s hng t do rip ca h phng trnh
chnh tc (14) cha cc tham sxut pht ca kt cu v ti trng ngoi. S dng cc
bng trong C hc kt cu, trong dnra cc kt qu tnh ton cc thanh siu tnh
mt nhp, c th xy dng cc biu mmen M 1 , M 2 ,..., M n do cc chuyn v n
v Z i ca cc lin kt c a thm vo v biu mmen M p do tc dng ca ti trng
ngoi ln h c bn. Sau , trn c s xt cc iu kincn bng ca cc nt hoc ca mt
phn h xc nh c tt c cc h s rij v s hng t dorip ca h phng trnh chnh
tc. Sau khi gii h phng trnh (14) s nhn c cc chuyn v cn tm Z i (i=1
n), t xc nh c cc ni lc trong cc cu kin chu ti ca kt cu. C th gii h
phng trnh (14) bng cc phng php khc nhau, tuy nhin, vi mcch m hnh ho
thng k kt cu chng ta gii (14) theo phng php ca K. Gauss.Phng php
gii ca K. Gauss da trn vic loi tr dn cc bin bng cch a vo mt hthng
cc k hiu rt thun li lm cho cch gii n gin hn. Biu din bin th nht qua
cc bin cn li t phng trnh u tin ca (14) r rrr1 p Z1 = 12 Z 2 13 Z 3
... 1n Z n . (15) r11 r11r11r11 7 8. a gi tr ca Z1 t (15) vo cc
phng trnh cn li ca h (14), sau xt n tnhi xng ca cc h s, ta nhn cr12
r12 r12 r13 r22 r Z 2 + r23 r Z 3 + ... 11 11 r r r12 r1 p + r2n 12
1n Z n + r2 p = 0; r11 r11 r r r r r23 12 13 Z 2 + r33 13 13 Z 3 +
... r11 r11 (16) r r r13 r1 p + r3n 13 1n Z n + r3 p = 0; r11 r11 .
. . r r r r r2 n 12 1n Z 2 + r3n 13 1n Z 3 + ... r11 r11 r r r1n r1
p + rnn 1n 1n Z n + rnp = 0. r11 r11 Chng ta a vo cc k hiur r r22
(1) = r22 12 12 ;r11(1) r r r23 = r23 12 13 ,r11(1) r r r2 n = r2 n
12 1n ; r11 (1)r r r33 = r33 13 13 , r11(1) r r r3n = r3n 13 1n
;(17)r11r13 r1 pr3 p (1) = r3 p ,r11. . .(1)r rrnn = rnn 1n 1n ,
r11r1n r1 p rnp (1) = rnp . r11 K hiu (1) pha trn th hin rng, bin
th nht c loi tr. By gi h (14) c bin i, trong cn li (n 1) phng trnh,
c dng:r22 (1) Z 2 + r23 (1) Z 3 + ... + r2 n (1) Z n + r2 p (1) =
0; r23 (1) Z 2 + r33 (1) Z 3 + ... + r3n (1) Z n + r3 p (1) =
0;(18) . . . r2 n (1) Z 2 + r3n (1) Z 3 + ... + rnn (1) Z n + rnp
(1) = 0. 8 9. T h phng trnh ny, sau khi loi tr bin th hai, chng ta
nhn c phng trnh kh ln th hair (1)r (1)r (1) r2 p (1) Z 2 = 23 (1) Z
3 24 (1) Z 4 ... 2 n (1) Z n (1) , (19)r22r22r22 r22v h phng trnh c
bin i ln th hai, trong cn li (n 2) phng trnh: (1) r23 (1) r23 (1)
(1) (1) r33 Z 3 + r34 (1) r23 r24 Z 4 + ...r22 (1) r22 (1) (1) r23
(1) r2 n (1) (1) r23 (1) r2 p (1) + r3n Z n + r3 p = 0; r22(1)
r22(1) (1) r23 r24(1) (1) (1) (1) r34 Z 3 + r44 (1) r24 r24 Z 4 +
... r22(1) r22(1) (20) (1) r24 (1) r2 n (1) r (1) r (1) + r4 n Z n
+ r4 p (1) 24 2 p = 0; r22(1) r22(1) . . . (1) r23 (1) r2 n (1)
(1)(1) r3n Z 3 + r4 n (1) r24 r2 n Z 4 + ... r22(1) r22(1) (1) r2 n
(1) r2 n (1) r(1)r(1) + rnn Z n + rnp (1) 2 n 2 p = 0. r22(1)
r22(1) Chng ta li a vo cc k hiu ( 2) (1)r23 (1) r23 (1)r33 = r33
;r22 (1)r23 (1) r24 (1) r34 ( 2) = r34 (1) , r22 (1)r23 (1) r2 n
(1) r3n ( 2) = r3n (1) ;r22 (1)( 2)(1) r23 (1) r2 p (1) r3 p= r3 p
,(21) r22 (1)(1) (1) ( 2)(1)r24 r24 r44 = r44 (1);r22 r24 (1) r2 n
(1) r4 n ( 2) = r4 n (1) , r22 (1) . . .r2 n (1) r2 n (1) rnn ( 2)
= rnn (1) ;r22 (1)9 10. ( 2) (1) r2 n (1) r2 p (1)rnp= rnp , (21)
r22 (1)tip theo, h phng trnh bin i ln th hai c dng:r33 ( 2) Z 3 +
r34 ( 2) Z 4 + ... + r3n ( 2) Z n + r3 p ( 2) = 0;r34 ( 2) Z 3 +
r44 ( 2) Z 4 + ... + r4 n ( 2) Z n + r4 p ( 2) = 0;. . . (22) r3n (
2) Z 3 + r4 n ( 2) Z 4 + ... + rnn ( 2) Z n + rnp ( 2) = 0. Tng t,
sau (n1) ln bin i th trong h cc phng trnh c bin i ch cn limt phng
trnh rnn ( n 1) Z n + rnp ( n 1) = 0 , (23)do ( n 1)rnpZ n = ( n 1)
.(24)rnn Lu rng, khng kh khn c th vit cc phng trnh kh (19) dng tng
qut:ri (i +1) (i 1)ri (i + 2) (i 1)rin (i 1)rip (i 1)Zi = ( i 1) Z
i +1 ( i 1)Z i + 2 ... (i 1) Z n (i 1) .(25) rii rii riirii Sau khi
tnh Z n , ch n cc phng trnh kh (25), t dn dn nhn c Z n 1 , Z n 2 ,,
Z 3 , Z 2 , v cui cng bng phng trnh (15) tm c Z1 . Ni lc trong cc
cu kin cng trnh c xc nh nh l hm ca cc chuyn v v ccni lc trong chnh
cc cu kin y do cc chuyn v n v Z i v do ti trng ngoi gy ratrong h c
bn. V d, mmen un trn mt ct ngang k ca cu kin no c dng M k = M 1k Z1
+ M 2 k Z 2 + ... + M nk Z n + M p k .(26) y M 1k , M 2 k ,..., M
nk , M p k tng ng l cc gi tr mmen trn mt ct k do cc chuynv n v Z i
=1 v ti trng ngoi gy ra trong h c bn. Chng ta xc nh kh nng chu ti
ca cc cu bc = 0,4mkin, v d, tnh kh nng chu ti ca cc 2-b (hnh 4, 5)
theo iu kin bn ca vt liu cc, kh nng chu tica cc cu kin khc v bn cng
c xc nh tng t. Kh nng chu un ca cc c xc nh bimmen bn: M 2b = Ra Fa
h0 (1 0,5 ) ,kn (27) h0 = 0,35ma = 0,005mhc = 0,40m Fa .Ratrong :=
,(28)bc .h0 R Hnh 4. B tr ct thp cc y Fa , Ra tng ng l din tch mt
ct ngang v bn tnh ton ca ct thp chu ko;R bn tnh ton ca b tng v nn
khi un; bc , h0 cc kch thc mt ct ngang cacc (hnh 4). b) Qu trnh m
hnh ho thng k tng bc. Nguyn tc chung m hnh ho thng k hm ca cc i lng
ngu nhin khi bit ccc trng thng k ca chng, cng nh phng php m hnh ho
thng k tng bc ctrnh by chi tit trong [3, 9]. y chng ta ch trnh by c
tnh cht ng dng phng php 10 11. m hnh ho thng k tng bc xc nh cc c
trng thng k ca cc ni lc trong cccu kin ca kt cu b cc cao mm, cng nh
ca bn cc cu kin. Chng ta bt u t cc h s rij v s hng t do rip ca h
phng trnh chnh tc (14).Cc h s rij l hm ca cc kch thc hnh hc v bn vt
liu ca dm v cc cc, cncc s hng t do rip l hm ca cc tham s va nu v ca
ti trng ngoi. Nh bit, cctham s tnh ton ca kt cu v ti trng u l cc i
lng ngu nhin v phn b cachng ch yu theo quy lut phn b chun.V d, i vi
khung c trnh by trn hnh 5b, h s r11 c dng4 E p J p 4 Ec J cr11 =+ ,
(29) lp Lc1trong E p , J p , l p , E c , J c , Lc1 l cc i lng ngu
nhin vi cc k vng ton E p , J p ,l p , E c , J c , Lc1 v lch chun Ep
, Jp , lp , Ec , Jc , Lc1 bit. Chng ta xc nh k vng ton v lch chun
ca i lng r11 theo trnh t sau: 1) to s ngu nhin chun Ep ; 2) tnh gi
tr E p = E p + Ep . Ep ; 3) to s ngu nhin chun Jp ; 4) tnh gi tr J
p = J p + Jp . Jp ; 5) to s ngu nhin chun lp ; 6) tnh gi tr l p = l
p + lp . lp ; 7) to s ngu nhin chun Ec ; 8) tnh gi tr E c = Ec + Ec
. Ec ; 9) to s ngu nhin chun Jc ; 10) tnh gi tr J c = J c + Jc . Jc
; 11) to s ngu nhin chun Lc1 ; 12) tnh gi tr Lc1 = Lc1 + Lc1 . Lc1
; 13) tnh v ghi li mt gi tr r11i (i=1 N) theo cng thc (29); 14) cc
thao tc 1 13 c lp li N ln, nhn c N gi tr ca r11 ; 15) cui cng, tnh
k vng ton v lch chun ca i lng r11 theo cng thc: 1 N r11 = r11 ; N i
=1 i1 N 21 N 2 r11 = r11i ( r11i ) . (30) N 1 i =1 N i =1 Tt c cc h
s rij v s hng t do rip ca h (14) u c m hnh ho theo nguyntc trn v s
nhn c k vng ton v lch chun ca chng rij rij , rip , rip (i=j=1n).
Tip theo, mi i lng trong h thng cc k hiu (17) (chng c th c gi l cch
s trung gian) l hm ca cc h s rij hoc l hm ca cc rij v cc rip . Khi
, cc rij vrip c coi l cc i lng ngu nhin vi cc c trng thng k c xc nh
trn rij ,11 12. rij , rip , rip . Theo phng php trnh by, chng ta xc
nh c cc k vng ton v lch chun ca cc h s trung gian rij (1) , rij (1)
, rip (1) , rip (1) , (i=j=2 n).n lt mnh, cc h s trung gian rij (
2) , rip ( 2) (i=j=3n) loi tr n s th hai (21)c coi l hm ca cc i lng
ngu nhin rij (1) , rip (1) m cc c trng thng k ca chng c xc nh rij
(1) , rij (1) , rip (1) , rip (1) . Tng t, chng ta tnh c cc k vng
ton v lch chun ca cc h s trung gian mc th hai rij ( 2) , rij ( 2 )
, rip ( 2) , rip ( 2 ) v ca cc h strung gian cc mc tip theo, v cui
cng tnh c cc k vng ton v lch chun cacc h s v s hng t do rnn ( n 1)
, r ( n 1) , rnp ( n 1) , r ( n 1) .nn npKhi , chuyn v Z n , c tnh
theo cng thc tin nh (24), c coi l hm ca ccbin ngu nhin rnn ( n 1) ,
rnp ( n 1) , m cc k vng ton v lch chun ca chng rnn ( n 1) ,r( n 1)
, rnp ( n 1) , r( n 1) c xc nh. Tng t, ta tnh c cc c trng thng k ca
nn npchuyn v-n s u tin Z n , Zn .Tip theo, trn c s phng trnh kh
(25) s m hnh ho thng k tt c cc chuyn vcn li Z n 1 , Z n 2 , , Z 3 ,
Z 2 , v trn c s phng trnh (15) s m hnh ho thng kchuyn v Z1 . Nh vy,
s nhn c k vng ton v lch chun ca tt c cc chuyn v Z i , Zi (i=1n) ca
cc nt.Cc ni lc trong cc cu kin l hm ca cc i lng ngu nhin: cc chuyn
v nt vcc tham s kt cu v ti trng. V d, hm mmen un trong mt ct 1 ca
dm 1-2 trn hnh5d c dng:4E p I p 2E p I p 6E p I p M 1 2 =Z1 + Z2 2
Z4 +lp lp lp(31) 6E p I pv12 + 2Z 5 + u1 2 P.lp lpBng phng php trn,
chng ta s nhn c khng ch k vng ton v lch chunm c cc c trng thng k cn
thit khc nh cc mmen trung tm ca cc ni lc trongcc cu kin cng trnh.
xc nh cc c trng thng k ca mmen bn cc, cn xc nh k vng ton v lch chun
ca i lng (tnh theo cng thc tin nh (28)). i lng c coi lhm ca cc bin
ngu nhin Fa , Ra , bc , h0 v Ru , m cc k vng ton v lch chun cachng
c a vo Fa , Fa , Ra , Ra , bc , bc , h0 , ho , Ru , Ru . Theo phng
php trnh by, chng ta xc nh c k vng ton v lch chun ca i lng , .n lt
mnh, mmen bn ca cc (cng thc (27)) c coi l hm ca cc ilng ngu nhin Ra
, Fa , h0 v m cc k vng ton v lch chun ca chng bitRa , Ra , Fa , Fa
, h0 , ho , , . Tng t nh trn, chng ta xc nh c k vng ton, lch chun v
cc mmen trung tm cn thit ca mmen bn ca cc M 2knb , M kn ,2 b2(
Mkn), 3( M kn ) , 4( M kn ) ,. 2 b2 b2 b 12 13. Khi , xc sut lm vic
an ton hay tin cy ca cc theo s c ph hoi vt liu ccc th c xc nh bng
cng thc (6): M 2 b M 2kn b P31 = 1 . M 2b + M 2knb 22 tin cy v cc
dng s c khc ca cc v cc cu kin cn li ca cng trnh cngc xc nh tng t nh
trn, cui cng, tin cy ca c cng trnh b cc cao mm theo tin cy ca cc cu
kin chu ti c tnh theo cng thc (12).c) Cc kt qu tnh ton c th. xc nh
cc tham s phn b ca bn v ni lc trong cc cu kin cng trnhdng b cc cao
mm m s tnh c th hin trn hnh 5b, cc tham s thng k sau yca kt cu v ti
trng c a vo tnh ton.a) b) c)d)Y 2,5 2,5u1 v1M1-2M2-1 P =100 TP P P
Z4Z5 2M1-aM2-b 1 2 12 Ep, Fp, Jp, lp E(xn1, yn1) (xn2, yn2)1Z2
Z312c2 Z1 Ec1 Fc2 Fc1 Jc2 Jc1 Lc2 Lc1ba a (xc1, yc1)b X a b a Ma-1
b (xc2, yc2) Mb-2Hnh 5. V d tnh ton kt cu b cc cao mm K vng ton v
lch chun ca cc kch thc mt ct ngang cc cc: hc =bc =0,40m, hc = bc =
0,005 m, v ca cc kch thc mt ct ngang dm: h p = 0,25 m, hp =0,003m;
b p = 3,0 m, bp = 0,03m.B tng dm ngang v cc cc mc 300 c mun n hi Ep
= Ec = 3,15.106 T/m2.K vng ton v lch chun ca cng nn ca b tng khi
un: R =160/2, Ru = 32/2. Ct thp ch ca cc gm 6 thanh loi A-III ng
knh 25 mm (hnh 4), ck vng ton v lch chun ca din tch mt ct ngang: Fa
= 29,45 cm2, Fa = 4,42 cm2, vca gii hn chy: Ra = 3400 /2; Ra = 272
/2. K vng ton v lch chun ca cc ta u di v u trn ca cc cc (m): xc1 =
0; xc1 =0; y c1 = 0; yc1 = 0.x n1 = 0; xn1 = 0; y n1 = 10; yn1 =
0,10.xc 2 = 5,0; xc 2 = 0,05; y c 2 = 0; yc 2 = 0.x n 2 = 5,0; xn 2
= 0,05; y n 2 = 10; yn 2 = 0,10.K vng ton v lch chun ca ti trng tp
trung v cc kch thc: P = 100T; P = 5T; u1 = 2,5 m; u1 = 0,03 m; v1 =
2,5 m; v1 = 0,03 m.S ln th nghim: N = 10 000.13 14. Cc kt qu tnh
ton kt cu theo xc sut v tin nh c dn ra trong bng 4. Nh cc v d minh
ha, trn hnh 6 dn ra cc biu thc nghim v cc c trngthng k phn b ca
mmen M 1 2 ca u dm 1-2 v mmen M 1 a ca u cc 1-a vi sth nghim N = 10
000 ln, cn trn hnh 7 cc biu thc nghim v cc c trng thngk phn b ca lc
dc N1 trong cc 1-a v N 2 trong cc 2-b. Bng 4. Kt qu tnh ton kt cu b
cc cao mm Theo phng php m hnh ho thng kTheo phng php tin nh tng bc:
N =10 000 K vng ton lch chun-Cc chuyn v:Z 1 = 0,008213Z 1 =
0,008226 Z 1 = 0,000474Z 2 = 0,008213Z 2 = 0,008221 Z 2 =
0,000527Z3 = 0Z3 = 0 Z3 = 0Z 4 = 0,001389 (m)Z 4 = 0,001393 Z 4 =
0,000105Z 5 = 0,001389 (m)Z 5 = 0,001386 Z 5 = 0,000108-Cc mmen un
(Tm):M 1 2 = 22,0766 TmM 12 = 21,9932 M 12 = 6,6770M 21 = 22,0766
Tm. M 21 = 6,9967M 21 = 21,9934M 1 a = 22,0766 Tm M 1a = 1,6961M 1a
= 22,1497M a 1 = 11,0383 Tm Ma 1 = 0,8503M a 1 = 11,0708M 2b =
22,0766 Tm M 2b = 1,8238M 2b = 22,1050M b 2 = 11,0383 Tm Mb2 =
0,9119M b 2 = 11,0525-Lc dc trong cc cc (T): N 1 = 50,00 N1 =
3,9770N1 = 50,1719N 2 = 50,00 N 2 = 4,1090-Mmen kh nng ca N 2 =
49,9395 cc 2-b (Tm):knM 2b = 27,2128 M 2knb = 26,7443 M kn =
5,35362 b Hnh 6. Biu thc nghim v cc c trng thng k phn b ca mmen dm
M 1 2 v mmen u cc M 1 a14 15. Hnh 7. Biu thc nghim v cc c trng thng
k phn b ca lc dc N1 trong cc 1-a v N 2 trong cc 2-bKhi , t kt qu
tnh ton xc sut trong bng 4 c th xc nh c tin cy ca cctheo s c ph hoi
vt liu cc: 22,1050 26,7443 M 2 b M 2 b kn P31 = 1 = 1 = 0,7939. M 2
b + M kn 22 (1,8238) + (5,3536) 222 b Nh vy, tin cy ca cc qu thp,
cn phi iu chnh cc c trng tnh ton ca cc nng cao tin cy ca cc, m bo
iu kin (7).TI LIU THAM KHO[1]. . . . .: , 1987. 223 .[2].
31-31-35-85. . .:/ , 1986.[3]. Nguyn Vi. . .: . - .. 23-24 2002, c.
116 129.[4]. Nguyn Vi. . : , 4, 2003.[5]. Nguyn Vi. . : , 4,
2004.[6]. Nguyn Vi. Tnh ton cc cng trnh bn cng theo l thuyt tin cy.
Tp ch Giao thng vnti, s 9-1996, H Ni.[7]. Nguyn Vi. nh hng s dng
quy phm v khi tho quy phm mi thit k cc cng trnhcng v ng thu. Tp ch
Giao thng vn ti, s 4-2008, H Ni.[8]. Nguyn Vi. tin cy ca cc cng
trnh bn cng. NXB Giao thng vn ti, H Ni, 2009. 184trang.[9]. Nguyn
Vi. Phng php m hnh ha thng k tng bc trong tnh ton tin cy ca cc
cngtrnh cng. NXB Giao thng vn ti, H Ni, 2009. 228 trang.[10]. JB
5015392, Beijing, China.[11]. International Standards Organization
(ISO). General Principles for the Verification of theSafety of
Structures, ISO-2394. 1973.[12]. New Standards for Port and Habour
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