XAC D!NH DO LUN NEN CONG TRINH THEO THOfl GIAN CO x h o E N TINH BIEN THIEN CUA GRADIEN THUY LI/C BAN DAU
Nguyen Dufc Nguon
ABSTRACT The article presents method to solve consolidation problem of basement in one directly and theory on calculating settlement level of construction by the time with consideration the changes of seepage gradient. Huminity w/as calculated by the deepness of basement under construction
at the completedly consolidated point in the area of active zone.
TS Nguyen Dure Ngudn Trudng Dai hoc kien tnic Ha No!
Km 10 Nguyen Trai, Thanh Xuan, Ha Noi
I . O a t v a n d e
B i i t o i n ed ket ddi vdi nen di't set yeu no nude hien nay dugc g i i i tren CO sd ly thuyet ed ke't t h i m cua Terx- aghi. Ly thuye't n i y khdng xet de'n sp thay ddi g i i trj gradien t h i m ban d i u , do do nd chi phu hpp vdi nen di't yeu no nude khdng ed hoac ed g i i trj gradien t h i m ban d i u khdng d i n g ke. Trong khi d d gradien tham ban dau tdn tai trong hau het eie di't set no nude va ed gia trj c i n g Idn khi dp i'm d p c h i t eua di't c i n g cao, nd ed i n h hudng ra't Idn de'n thdi gian eung n h u g i i trj do ldn eda cdng trinh. Do d d viec nghien edu xem xet va giai b i i t o i n
ed ke't cd xet den sU thay ddi h,^, = x(H3Z eua gradien t h a m ban d i u ed
y nghia Idn trong edng t i e d p d o i n d p lun v i thdi gian ldn eua edng trinh x i y dpng tren nen d i t set no nude.
P-. = h,.«P. W (itjl^b (3)
PcK=Po-h,z,„-Pb (4) Pj^ - trpng lUpng rieng eua nUde;
hj^ji - i p lpc nude Id rdng d d p s i u z tai t h d i diem t dp lun t o i n phan eua Idp di't va nude sau t h d i gian t x i c djnh theo edng thdc sau:
J" 1 + 8,
-dz =
1 + £.
jPadz (5)
v i ke't hpp bieu thdc (4) vi{5) ta cd:
^[fP-d^-P^f^^.d^
(6)dang:
trong d d g i i trj h^^^^ theo [1] ed
^I
2. No! dung nghien cuTu 2.1. Xdc dinh do lun Theo p h u o n g trinh tinh toan bien dang thi dp lun eua Idp di't nguyen t d dz n i m tai chieu s i u z sau m d t k h o i n g thdi gian t ke t d khi c h i t t i i se cd dang:
4P 7 r p b ( 2 n - l )
n{2n-])[ 7i'(2n-l)
. (2n-1)7i xHa - ,sin-!^ ^ z + -^^-J-.e
H, (2n)7i
J _
' . 2n7i .sin z
H,
Thay gia tri h vao (6) g i i i bai
dS,„=-
a.P,t o i n n h i n dupe:
0,33a.PoH 0,24.a.Po.H .
Jr,i ~ e
-dz
1 + e., 1 + £„ (8)A 1 + ^c. , j ^ ) O day £^p- g i i tri he sd dp rdng t r u n g binh eho Idp di't;
a - he sd nen cua di't;
„ . . , , . 0,33a.Po.H ^ , , Bieu thi = \ ( 9 ) v i
1 + Scp
chia (8) eho S, ta n h i n duoc:
• i p lpc len hat di't.
d d i y : P^- ngoai t i i P^^ - ap lpc nude Id rdng.
(2) ( t )
^(k)
= 1-0,72.6""
d i t U j , p = 1 - 0 , 7 2 . e - " (10)
^ m i i t S l . 10.2009 69
c op -"k
^3c6S,„ = S,, d diy: S^ - dp U^p - Mdc dp cua Idp d i t .
N='^V
4H'aK
un eudi eung ed ket trung(11) binh
(12)
(ngiy dem) Oi't set s o i l P=0,15.10Pa
OUa g i i tri N vao bieu thdc (10) v i t l m T :
T = - -In 0,72 Tl'C, 1 - U , -H
(13) Cdng thdc (13) eho thi'y r i n g , vdi mdc do ed ke't nhi't dinh U tdc l i vdi he sd ed ke't cv nao do, quan he thdi gian cd ket T vdi c i n b i e 2 cua chieu d i y v u n g hoat d p n g H^ l i h i m tuyen t i n h .
4 , 0,72
G i i tri -^—In la h i m nao
• 7r'C„ 1-U™
do x i c d|nh q u i trinh nen eua d i t . So s i n h edng thdc (13) vdi edng thdc ma g i i o su Kazarndpxki .B.D n h i n dUpc [4]:
t = b,ph^ + a,, (a) Trong edng thdc do, n g o i i h i m
p h i n anh tinh nen trong q u i trinh ed ket t h a m cdn ed h i m p h i n i n h t i n h nhdt eua di't. Tuy nhien, n h u c i c ke't q u i t h i nghiem eho thi'y (xem hinh 1) ddi vdi di't yeu no nude, g i i tri do nhdt a^p ri't nhd so vdi t h i n h phan nen ed ke't b^ph-^, d i e biet khi chieu dai d u d n g t h i m h cd gia trj Idn vi v i y cd the coi g i i trj dp nhdt a^p = 0. Nhu vay ddi vdi b i i t o i n thpe te, q u i trinh ed ket t h i m cd the mieu t i theo edng thdc [4]:
T= b,pH^ (b) cdng thdc d d thpe te gidng edng
t h d c da nhan dupc (13) neu gia tri -In 0,72
duong vdi b^p.
Ta bieu thi 4
la m d t ham tUOng
In 0,72
1-U,„ (14)
Bieu thdc (13) cd dang cua cdng thdc (b):
T=b,pH^ (15) Cdng thdc d d la eo sd ly thuye't
de mieu t i q u i trinh ed ket cua di't set eho nen d u d n g . Neu ta bieu t h j :
100 h'(em') Hinh 1. Xae djnh cac thong so a^p va b^^^ bang do thj theo ket qua thi nghiem co ket mau dat chieu cao khacnhau
Bw/p 0,6 0,5 0,4 0,3 0,2 0,1
-1,5
1,0
0,5
•
Awp
\ \
\ \
a=f(w)\ . b=f(w
)
Oi't set P=2,5.1 fPa
42 44 46 48 50 52 54 56 w%
Hinh 2. Quan he cac thong so co ket a va b vdi dp am cua dat
Cv 0,6
0,5
0,41-
•<o
<(u. 0,3 X
0,2
0,1
Oi't set
I P = 0 , 2 9 0 ; W L = 6 1 , 8 2 %
42 44 46 48 50 52 54 56 58 w % Op am cua di't Hinh 3. Quan he he so co ket vdi dp am cua dat
70 ^ l ^ n i i l ^ l . 10.2009
w%
Hinh 4. Difdng cong nen ciia dat
K„ ^ I n 4 , 0,72
Tt' 1 - U ™ (16)
Va thdc (16) va (15) vao bieu thdc (14) tase nhin dupe:
T„ K, -H
(17) Gii trj K^ thay ddi phu thude v i o mdc dp ed ke't trung binh.
Theo bieu thdc (15) v i (17) ed the xic dinh he sd dd cd ket c
c ' = ^
b,p (18) Cae thi nghiem eho thi'y ring
trong q u i trinh nen, thdng sd b^p.
Thay ddi phu thupc v i o dp i'm, dp chit cua di't. Gii trj b^p v i C^ cin xic djnh bang thuc nghiem. Quan he C^^
= f(w) v i b^p = f(w) xem hinh 2 v i 3.
Phin tich edng thdc (15) eho thi'y ring, quan he thdi gian ed ke't T vdi binh phuong chieu diy vung hoat dpng l i tuye'n tinh, gdc nghieng eua dudng thing vdi true hoinh se bang
K
Tuy nhien thd nghiem eie mau d trong phdng ed chieu eao nhd eho tha'y ring dUdng thing eua quan he T=f(H^^) khdng c i t qua gdc eda do thj, m i eat true tung mpt g i i trj nio dd gpi l i a^p.
Khi dp doin tdc dp nen eua di't yeu, sU khic nhau neu tren cd the i n h hudng den thdi gian lun khi chieu d i i dUdng tham nhd hon 50em [4]. giio sU B.O.Kazarndpxki
Tai nen P.l O'Pa
giiithich ring, Sp sai khic dd do i n h hudng eua hien tUpng nhdt giy nen trong qua trinh ed ket eua mau thd.
Trong trUdng hpp tham 2 chieu g i i tri chieu d i i dUdng t h i m cin lay bing.
Td dd ta thi'y ring, khi dp doin thdi gian lun eua Idp di't thpe theo ke't qua thd nghiem eua eie mau cd chieu cao nhd, can phai xet den i n h hudng cua yeu td nhdt den dp d i i lun eua mau trong phdng thi nghiem, nghia l i cin phii tim g i i tri Jj^jj^ ddng thdi, phin tich g i i tri thpe eua thong sd b^p v i a^p ddi vdi di't yeu khi tai trpng gan 0.5.1 O^na die trUng cho nen dUdng tren dam liy cho thi'y ring, khi chieu dai dUdng t h i m thpe te H >50em, g i i tn a^p ri't nhd so vdi g i i trj H^^
Nhu viy, g i i trj b^p dUpe xic djnh tren co sd thpe nghiem bing bieu thdc sau:
t>.p-3;.p
(19) b,.p
h-'p
OUa bieu thUe dd vao edng thdc (15) ta se nhin dupe bieu thdc de dp doin dien bie'n dp lun theo thdi gian cua Idp di't thpe te theo ke't q u i thi nghiem trong phdng:
-r , wHa,^
' ' - % • (20) Td bieu thdc (20) ta thi'y rang xet i n h hudng cda yeu td nhdt de'n dp ldn theo thdi gian eua Idp da't thpe
-4r
Rio'TraH
\ CTz
z
Ha
Z
H.5. Bieu do phan bo iJng suat theo chieu sau vao thdi diem ket thuc CO ket
Bieu do thay doi dp am theo chieu sau
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 9S 100?o W J
,
^ ^
5g g 10 12 ,3 14
Z(M)
I
Gh
(
1
ch N 1
•
1
X
g i i gU
1 1 I I
' •
1
\
. ]/
71
>
"i
/
1'
gifl
/ J
/ '
li
Hinh 6. bieu do phan bo dp am cua dat theo chieu sau tai thdi diem ket thiic co ket (xae djnh difdi mong be chda xang dau 12A va 12B- kho xang dau Nha Be)
te dupe tien hanh b i n g c i c h g i i m g i i trj a^p theo thdi gian dat dupe dp lun X eua m i u di't t r o n g p h d n g . N h u v i y xet yeu t d nhdt ve thpe c h i t l i c i n thiet de ehuyen t d dieu kien t h i nghiem t r o n g p h d n g vdi mau di't ed chieu cao n h d (chieu d i i d u d n g t h i m nhd) m i ddi vdi nd yeu t d nhcit cd y nghTa Idn, sang viee d p d o i n thdi gian lun cua Idp di't thpe te.
Ta luu y t h e m r i n g he sd d p cd ke't n h i n dUpe khi g i i i b i i t o i n 1 chieu xet de'n g i i t n gradient thay ddi ed g i i tri cao hcfn so vdi he sd d d khi n h i n dugc t d ly thuye't Terxaghi - Gerxevandp vdi g i i tr| N n h u nhau, nghia la:
K h i n = 1
S I S n i R S l . 10.2009 71
Cd xet de'n J • U, = l - 0 , 7 2 e - "
H bp
Theo Terxaghi: U^^^ = 1-0,822e'~' (21 V i eudi eung l i thdi gian ed ke't eua nen edng trinh thpe te ed the x i c djnh theo edng thdc (15). Trong dd, he sd bj^p v i vl vay he sd ed ke't x i c djnh true tiep bang ket qua t h i nghiem cd ke't. Bang c i c h d d ed the t r i n h dupe kho k h i n trong viee xae dinh he sd t h i m , dieu nay g i i m dupe phdc tap trong edng t i e t h i nghiem va tinh t o i n .
2.2. Xdc dinh gid tri gradient thuy lUc ban dau va he so do rong cuoi ciing cua ddt
Trong qua trinh nen, dUdi t i c dung cua t i i trpng, dp c h i t cua di't t i n g v i ddng thdi t i n g tinh nhcit eua nude lien ket. Oieu d d dan den viee tang g i i tri gradien thuy lpc ban d i u J^.
Td Idi g i i i b i i toan cd ket 1 chieu [1 ] cd xet de'n J^ = f(£), g i i tri gradien ban d i u tai thdi diem ket thue ed ke't t = t^j^ phu thudc v i o chieu s i u ed the x i c djnh nhU sau.
Khi t = oo, ta cd:
h(z,oo) = x H,.z— ,
[ 2 )
(22)
Tai thdi diem nay sp t h o i t nude t d Id rdng eua di't ddng lai, gradi- ent nen t h o i t nude trong t i t c i c diem se b i n g gradien thuy lpc "ban dau"tUOng dng vdi t i i trpng t i c dung.
Td cdng thdc (22) khi z ?!: 0:
iu ^J„„„ = — = ( H a - z )
dz Ha (23)
Hoic:
2J„
J H , = ^ ( H , - Z )
•"' H, ' ^ (24) G i i tri gradien ban dau, x i c dinh
theo edng thdc (24) la gia trj d i t i n g trong q u i trinh cd ket tren t d n g dp s i u Idp d i t z.
Khiz = 0; J„ = JHp
d d i y : J"p - gradient thuy lpc ban dau t u o n g dng vdi t i i trpng (hay dp c h i t £p)
Tren eo sd ket q u i t h i nghiem
V = b [ e * " - = J - 1 ] - H j ^ (25)
^^'' t = W £, = ^K' quan he (25) ed dang
j ; - = b [ e * ' - ^ . ' - 1 ] - H j ^ (25') Td day cd the x i c djnh £^" (z ^ 0)
b i n g c i c h c i n b i n g bieu thdc (24) ) v i ( 2 5 ' ) :
-In
b.H • ( H a - z ) + 1 (26) trong do: £^'- he sd d p rdng eudi eung (t = °o) tai dp s i u z; £^ - he sd dp rdng ban dau (t=0);
b, c- c i c he sd x i c djnh bang thpe nghiem, php thude vao t i i trpng t i c dpng.
Trong dieu kien di't no nUde (G=l) he sd dp rdng cd the x i c djnh qua dp am:
•W
= w,
Pb z Po_
Pb
(27)
TN
w=
(28) trong do: W,^^- dp am eudi cung tai chieu s i u z; W.^^^ - dp i'm t p nhien eua di't; p^,- trpng lUpng rieng eua hat di't; p^- trpng lUpng rieng eda nUdc=1t/m^
Nhu v i y , cdng thdc (26) eho phep x i c djnh he sd dp rdng eua d i t (tuong dng l i dp am) d b i t ky chieu s i u nao trong vung chu ddng tai thdi diem ke't thue cd ke't.
Oe x i c djnh d p am cua di't ed the s d d u n g edng thdc sau:
Qw
Qek (29) trong dd: Qw - trpng lUpng nude
chda trong Id rdng eua di't
Qek - trpng lupng hat di't (pha r i n ) trong m i u di't dd.
X i c dinh sp thay ddi d p i'm eua di't theo thdi gian trong t h i nghiem ri't de dang v i chi c i n sd dung nhdng thie't bj don g i i n .
Ke't q u i nghien edu, quan t r i e ve dp am eua di't set theo chieu s i u neu trong eie edng trinh cua N.N.
Maxidp [5], Skempton A.W., HenKel D.S. [3], E.M. Dobrdp [6]... eho thi'y r i n g , dp am eua d i t trong trang t h i i t p nhien theo chieu s i u ed g i i trj khdng ddi mac du c i n g t i n g chieu s i u i p Ipc t p nhien P^^ = Y H c i n g Idn.
Tuy nhien khi c h i t t h e m t i i trpng, sp c i n b i n g dp am do bi p h i vd. Op i'm thay ddi lien tpc trong qua trinh cd ke't, theo g i i trj d d ed the suy xet dugc nhieu quy l u i t k h i e . Khi cd ket, trong vung chu ddng, dp am tai mdi diem v i mdi thdi diem cd m d t g i i trj
nhi't djnh v i thay ddi lien tue theo sp thay ddi dp c h i t eua da't. Trong d d tai thdi diem ke't thue ed ke't d mdi dp s i u deu ed m p t g i i trj nhi't djnh t u o n g dng vdi dng s u i t t r o n g di't tai diem dd. Tren eo sd quy luat thay ddi d p am - d p c h i t theo chieu s i u trong dieu kien t p nhien v i t d ket l u i n cua Idi g i i i b i i t o i n d i trinh b i y cd the x i c dinh quy l u i t bien ddi dp am (dp chat) theo chieu s i u tai thdi diem eudi cung, nghia l i thdi diem ke't thue q u i trinh cd ke't ( x e m hinh 4, 5 , 6 ) .
III. KET LUAN:
1. Chieu d i y vung chiu nen phu thude v i o g i i tri t i i trpng, g i i trj gradien t h i m ban dau t i n g dan trong q u i trinh ed ke't phu thude vao d p am - d p c h i t cda di't.
2. Oe g i i i b i i t o i n cd ket nen edng trinh trong dieu kien b i i t o i n 1 chieu ed xet de'n sU bie'n ddi eua gradien thuy lpc ban d i u ed the thpe hien theo 2 phuong p h i p :
a/Tren eo sd Idi g i i i ly thuye't sd d u n g so do Terxaghi-Gerxevandp.
b/ Tren eO sd t h d nghiem trpc tiep vdi viee x i c djnh eie t h d n g sd cdketb^pvaa^p.
3. Cd the x i c djnh dp am cua di't theo chieu s i u nen cdng trinh tai t h d i diem ed ke't h o i n t o i n trong gidi han vung chju nen:
a/ Theo g i i tri dng s u i t trong khdi di't theo quan he W=f(P)
b/Theo edng thdc (26).
TAI LIEU THAM KHAO
1. Nguyen Ddc Nguon - Giai bai toan co ket mpt chieu xet den si/bien thien cua gradien tham ban dau, tap chi dia ky thuat sd...
2. Nguyen Dufc Ngudn- Nghien cufu thdc nghiem tai hien trudng sir tang dp chat- dp am cua nen dat set yeu no nddc theo chieu sau dddi mdng mem dang trdn, tap chi xay di/ng so...
3. Skempton.A.W. Henkel.D.S. Test on London clay from deep borings at Padington Victoria and the sout Bank. Proc. nO 4inter Conf. On soil mech.
and found eng. VI,1957, London.
72 a i g n H I B i l . 10.2009
