Tee m2464 En1997 Εφαρμογεσ 1

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Τεχνικό Επιμελητήριο Ελλάδος (ΤΕΕ) Τεχνικό Επιμελητήριο Ελλάδος (ΤΕΕ) Εκπαιδευτικό υλικό για τα Σεμινάρια Εκπαιδευτικό υλικό για τα Σεμινάρια Επιμόρφωσης Επιμόρφωσης των Ελλήνων Μηχανικών στους Ευρωκώδικε ς των Ελλήνων Μηχανικών στους Ευρωκώδικε ς Νοέμβριος Νοέμβριος Δεκέμβριος 2009 Δεκέμβριος 2009 Ε έ Ε ώδ 7( Ε έ Ε ώδ 7(EN 1997) EN 1997) Εφαρμογές του Ευρωκώδικα 7 ( Εφαρμογές του Ευρωκώδικα 7 (EN 1997) EN 1997) σε θέματα σχεδιασμού Γεωτεχνικών Έργων σε θέματα σχεδιασμού Γεωτεχνικών Έργων Μ. Καββαδάς, Αναπλ. Καθηγητής ΕΜΠ

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Transcript of Tee m2464 En1997 Εφαρμογεσ 1

  • () ()

    2009 2009

    7 ( 7 (EN 1997)EN 1997) 7 ( 7 (EN 1997)EN 1997)

    . , .

  • 7.1 ( EN 1997-1 )

    1. 2 2. 3. 4 4. , 5. ,

    6. 7. 8. 9. 10. 11. 12.

  • 6 :

    :

    : ( ) ( )

  • 6 :

    1. : ( GEO)

    ( GEO)

    ( EQU)

    ( GEO) ( GEO)

  • 1.1

    V RVd RvdVd = .

    , , . .

    Rvd = ( ).

    Rvd = (1/R) R (Fk , Xk)

    R (Fk , Xk) . 1997-1.

  • 1.1 Vd Rvd

    1997-1, 2* (DA-2*), :E V (Fk , Xk) (1/R) R (Fk , Xk) V (F Fk , Xk) (1/R) R (Fk , Xk)E ( k , k) ( R) ( k , k) (F k , k) ( R) ( k , k)

    1 + R2 :1 : F = = 1.35 1.50 R2 : R = 1.40

    : FS = F R 1.40 x 1.40 = 1.96

  • 1.2

    : :

    Hd Rd + Rp;dH ( ) Hd = ( ) . .

    Rd = .

    Rp;d = , .. ( ( ).

    2* (DA-2*)

    : d = E H (Fk , Xk) d = H (F Fk , Xk)

    , 2 (DA-2 )

    Rp;d 1 ( .3) :F = = 1.35 1.50

  • 1.2 Hd Rd + Rp;d

    Rp;d Rp;d = Rp / R R =1.40 ( .5)

    Rd :

    : R = (V tan ) /

    R 1.40 ( .5)

    : Rd = (V k tan k) / R;h : Rd = (Ac cu;k) / R;h

    V k = k = .

    , ('cv;k), (2/3 'cv;k) .

    Ac = cu;k =

    ( 5 ) R;h = ( .5, ). DA-2* : R;h = 1.1

    : FS = F R;h 1.40 x 1.10 = 1.54

  • 1.3 ( EQU )

    E < E E E

    EdstEstb

    Edst,d < Estb,d dst Edst,k < stb Estb,kEdst,d =

    Estb,d =

    2* (DA-2*). .1 1997-1 :

    : FS = dst / stb = 1.10 / 0.9 = 1.22

  • 1.4

    : , , , , .

    , , 11 1997-1, .

    , 3 (DA-3) 11

    Rd > Ed

    R ( F X / ) ( F X / )R (F Fk , Xk / M) > (F Fk , Xk / M)

  • : ( 2*, DA-2*)

    : : = 3m , : L = 4m : D = 1m

    :E ( ): = 20 kN/m3 ): = 20 kN/m3

    : ' = 28

    : c' = 10 kPa

  • :

    ( ) :

    : Gk = 1800 kN : Qk = 350 kN

    : Gk = 360 kN : Qk = 70 kN : = 0 kN : LGk = 0 kN : LQk = 0 kN

    ( ) : MG k = 1080 kNm ( ) G,k ( L) : MLG,k = 0 kNm ( ) : MQ,k = 210 kNm ( L) : MLQ k = 0 kNm ( L) : MLQ,k 0 kNm

    ( ) : qk = 0

  • :

    1. :

    E E ( F X )Ed = E (F Fk , Xk)

    .3 .5 1997-1 ( 1 2*). ( )

    : d = Vd = G Gk + Q Qk = 1.35 x 1800 + 1.50 x 350 = 2955 kN

    ( ) : d = G Gk + Q Qk = 1.35 x 360 + 1.50 x 70 = 591 kN

    ( ) : ( ) : Ld = G LGk + Q LQk = 0 kN

    ( ) : d = G MG,k + Q MQ,k = 1.35 x 1080 + 1.50 x 210 = 1773 kN

    ( L) : = M + M = 0 kNLd = G MLG,k + Q MLQ,k = 0 kN

  • 2. ( ) :

    1997-1 ( DA-2*) :

    Rd = (1/R) R (Fk , Xk)

    ( )

    ( ) :eBk = MBk / Vk = (1080+210) / (1800+350) = 0.6m

    : eLk = MLk / Vk = 0

    ( B) : tan ( ) = / V = (360+70) / (1800 + 350) = 0 20 = 11 31otan (Bk) = B / V = (360+70) / (1800 + 350) = 0.20 Bk = 11.31o

    ( L) : tan (Lk) = Lk / Vk = 0 Lk = 0

    : = 2 eBk = 3 2 x 0.60 = 1.80mL = L 2 eLk = 4m = L = 7.2 m2

  • 2. ( ) :

    (pu) :

    Rd = (1/R) R (Fk , Xk)

    1

    R = pu B L

    ( )1 212u c c c q q qp c N s i q D N s i B N s i = + + + i (' = 28) :

    Nq = e tan' tan2 (45.+ '/2) = 14.720( 1) / ' 2 803Nc = (Nq - 1) / tan ' = 25.803

    N = 2 (Nq- 1) tan ' = 14.590 :

    sq = 1 + (B' / L' ) sin ' = 1.2113s = 1 0 3 (B'/L' ) = 0 8650s = 1 0.3 (B'/L' ) = 0.8650sc = (sq Nq -1) / (Nq - 1) = 1.2267

  • 2. ( ) : (i i i ) :

    i = [1 H / (V + A' c' cot ' ) ]m

    (iq , i , ic) :m = mB = [2 + (B '/ L' )]/[1 + (B' / L' )] = 1.6897

    iq [1 H / (V + A c cot ) ]i = [1 H / (V + A' c' cot ' ) ]m+1ic = iq - (1 iq ) / ( Nc tan ' )

    : V = pu B L : H = V tan()

    (i , i , i )

    ( )1 212u c c c q q qp c N s i q D N s i B N s i = + + + (iq , i , ic) , pu ( ). : iq = 0.6949 , i = 0.5602 , ic = 0.6727

    : pu,k = pu = 588 kPa : Rk = Vu,k = pu B L = 588 x 1.80 x 4 = 4233 kN

    DA-2* :Rd = Rk / R = Vu,k / R,v = 4233 / 1.40 = 3024 kN

    (R,v) .5 ( R2 DA-2*) 1997-1.

  • 2. ( ) : :

    Vd Rd Vd = 2955 kN 3024 kN = Rd (FS) (FS) :

    FS = R / V = 4233 / (1800 + 350) = 1.97

    FS (1 35 1 50) 1 40 1 40 1 40 1 96 : FSmin = F R = (1.35 1.50) x 1.40 1.40 x 1.40 = 1.96

  • 3. : Hd Rd + Rp;d ( ) ( ) : Rp;d = 0. ( ) : = + = 1 35 x 360 + 1 50 x 70 = 591 kNd = G Gk + Q Qk = 1.35 x 360 + 1.50 x 70 = 591 kN

    , ( 2*) :

    Rd = (Vk tan k) / R;hVk = Gk + Qk = = k = R;h = .5.

    DA-2*, R;h = 1.1 ( R2).

  • 3. :

    R = (V tan ) / R = R / Hd Rd + Rp;d

    Rd = (Vk tan k) / R;hVk = Gk + Qk = 1800 + 350 = 2150 kN

    Rp;d = Rp / R;e

    , () ('cv;k). : = 28 o, :cv;k Rd = ( 2150 x tan(28) ) / 1.10 = 1039.2 kN Rp;d = 0 / 1.40 = 0 : Hd Rd + Rp;d Hd = 591 1039.2 + 0 (FS) : FS = Rk / Hk = ( 2150 x tan(28) ) / (360 + 70) = 2.66 : FSmin = F R = (1.35 1.50) x 1.10 1.40 x 1.10 = 1.54 : FSmin F R (1.35 1.50) x 1.10 1.40 x 1.10 1.54

  • 2. : ( (, ) .

    . ( ).

    (.. , , ) , ). , ( ).

  • 7 : : : :

    GEO : -

    GEO :

    STR : ,

    GEO : GEO : (.. )

    :V

    HV

  • :Vd Rd

    1997-1, 2* (DA-2*), : V (F Fk , Xk) (1/R) R (Fk , Xk) 1 + R2 :1 : F = = 1.35 1.50 , R2 : R = 1.10 R =

    R R RR = Rs + Rb

  • ( Rk ) :

    7 7 , () (n) (n) (. (n) (. ).

    Rk : k1. (n) (..

    )

    2. (n)

    3. (n)

    4. (fsu) (qbu)

    : (4) (). , (m) (4) (4) .

  • 1 : ( Ru,k )

    ( ) * ( ) (n) * (.. ). (.. cu qc ) (f ) (q ) (fsu) (qbu). Ru. (n) Ru .

    * =

    AR = zfDR susu

    i (i = 1,2, n) : (fsu , qbu ) bubbu qAR =bRRR += busuu RRR +

    (Ru,mean) (Ru,min) ( ) :

    = n RR 1 { }uu RR minmin ==i

    umeanu RnR

    1, { }uu min,

  • 1 : ( Ru,k )

    iRR

    =

    4

    min,

    3

    ,, ,min

    umeanuku

    RRR :

    , 3 4 n :

  • .

    2 : ( Ru,k )

    n Ru :

    (Ru,mean) (Ru,min) ( ) : ( ) :

    = n umeanu RR , 1 { }uu RR minmin, =

    = min,, ,min

    umeanuku

    RRR :

    =in 1

    , ,

    21,, ku

    , 1 2 n :n :

    , 10% ( 1)

  • R = Rs + Rb

    4 : ( Ru,k )

    R :

    (.. SPT, CPT, PMT) / (.. UU, CU, DS)

    R Rs Rb

    / (.. UU, CU, DS) (.. cu , qc , N) (fsu,k ) ( qbu,k ).

    : :

    { } { }, ,1 1su bu su k b bu kR R R DL f A q = + = +m m 1997-1 : m = 1.30

    m = 1.30 :FS = F * R * m 1.40 x 1.10 x 1.30 = 2.0

    : FS = (Rsu + Rbu) / V Vd = Rd F V = (1/R) [ (1/m) (Rsu + Rbu) ]

  • :

    :

    :

    . D=0.8m, L=15m.

    (Vk) . 80% (permanent) 20% 80% (permanent) 20% (transient).

    : , = 18 kN/m3 : cu = 125 kPa

    : , = 20 kN/m3 SPT N = 45

    7, f DIN 1054 2005 fsu qpu. DIN 1054 : 2005

  • :

    4 : m= 1.30

    DIN 1054:2005 :

    : , = 18 kN/m3 : c 125 kPa : cu = 125 kPa : fsu = 45 kPa

    : , = 20 kN/m3 SPT N = 45 , qc = 0.5 N = 0.5 x 45 = 22.5 MPa. : fsu = 120 kPa qbu = 3.75 MPa

  • 2* (DA-2*) :

    :

    ( )

    :

    ( ) ( ) VVVQPF 3812050180351 =+=+= ( ) ( ) kkkkQkGd VVVQPF 38.12.050.18.035.1 =+=+= :

    fDR 1 = zfDR ksum

    ksu ,, :Rsu,k = (1/1.3) x [3.14 x 0.80 x (45 x 12 + 120 x 3)] = 1043.5 + 695.6 = 1739 kN

    kbubm

    kbu qAR ,,1= :

    A 3 14 0 82 / 4 0 5024 2 R (1/1 3) 0 5024 3750 1449 kNAb = 3.14 x 0.82 / 4 = 0.5024 m2

    :

    Rbu,k = (1/1.3) x 0.5024 x 3750 = 1449 kN

    14491.1

    117391.1

    111,,, +=+= kbu

    bRksu

    sRdu RRR kNR du 2898, =

    : dud RF , 1.38 Vk 2898 Vk 2100 kN

  • (FS=2) :

    :

    = zfDQ :

    zfDQ susu Qsu = 3.14 x 0.80 x (45 x 12 + 120 x 3) = 1356.5 + 904.3 = 2261 kN

    bubbu qAQ = :2261 kN

    Ab = 3.14 x 0.82 / 4 = 0.5024 m2

    Qbu = 0.5024 x 3750 = 1884 kN

    : Qu = Qsu + Qbu = 2261 + 1884 = 4145 kN

    : Qall / FS = 4145 / 2 = 2072 kN

    : FS=2

  • 8 : :

    , , .

    :

    (investigation tests), /. , .

    , :1 (s itabilit tests)1. (suitability tests)2. (acceptance tests).

    : , .

    , . (proof load), Pp, EN 1537:1999 ( ) .

  • :

    :

    : . , . EN 1537:1999 ( ). P 1 25 E P = 1.25 k Ek (). , (lock-off load) Po = (80-90%) k. .

  • Pd Ra;d :

    d a;d

    Pd = () . (.. )

    R = (R ) Ra;d = (Ra ) Ra;d = min { Ra;k / a , Rt;k / m }

    : Ra;k = (

    ). /

    a = , 3.3.4(1) A.12. DA-2*, : R2, : a = 1.10

    Rt;k = As y k = Rt;k As y,k , . As , y,k

    m = 1.15 () : Rt;d = Rt;k / m =

    : FS = F a2 a 1.4 x 1.0 x 1.1 = 1.54

  • ( Ra;k ) Ra;k :

    . . ( ( ).

    (Ra;k) , (a) : (n) : (Ra,1), (Ra,2), (Ra,n) .

    (Ra;k) :( ) ( ) min,,min mameanma RRR = 2 min1; ,min aa meankaR ( ) = n iRR 1 ( ) [ ]RRRR min=( )

    =iiameanma

    Rn

    R1

    ,, ( ) [ ]naaama RRRR ,2,1,min, ,....,min= (n) : 1 2 3

    1 20 1 15 1 10a1 = 1.20 1.15 1.10a2 = 1.20 1.10 1.00

  • :

    :

    .

    ( ) :

    : PGk = 320 kN : PQk =70 kN P 320 70 390 kN : Pk = 320 + 70 = 390 kN

    : , ( ). , (Pk) , (.. ).

  • () (Pd) : , . , . .3 1997-1):

    Pd = G PGk + Q PQk = 1 35 x 320 + 1 50 x 70 = 537 kNPd G PGk + Q PQk 1.35 x 320 + 1.50 x 70 537 kN

    (Ra;k): : , :

    Ra;k = d L fsu,k = 3.14 x 0.12 x 8 x 200 = 603.2 kN , :

    Ra,1 = 640 kN, Ra,2 = 662 kN , Ra,3 = 700 kN. : (Ra,m)mean = 667.3 kN : (Ra,m)min = 640 kN. , : a1 = 1.10 a2 = 1.00. : ( ) ( )

    =

    2

    min,

    1

    ,; ,min

    mameanmaka

    RRR = 606.7 kN 21 aa

    : Ra;k = 603.2 kN

  • (Ra;d) :Ra;d = Ra;k / a = 603.2 / 1.1 = 548.4 kN

    (a) A.12. DA-2*, R2, : a = 1.10. (Ra d) (Ra;d) (Rt;d) : Rt;d = Rt;k / m = ( As y,k ) / mAs , y k s , y,k m = 1.15 (). , , 140 mm2 y,k = 1600 MPa. : Rt;k = As y,k = 3 x (140 x 10-6) x 1600 = 0.672 MN = 672 kN, :

    R = R / = 672 / 1 15 = 584 3 kNRt;d = Rt;k / m = 672 / 1.15 = 584.3 kN : Ra;d = 548.4 584.3 = Rt;d : Ra;d = 548.4 kN. : Pd Ra;d d a;d : Pd = 537 548.4 = Ra;d (FS) : FS = Rk / Pk = ( d L fsu,k) / (PG+PQ) =

    = (3.14 x 0.12 x 8 x 200) / (320 + 70) = 603.2 / 390 = 1.55

  • 9 : ( , , ) , :

    (gravity walls) (gravity walls) (embedded walls) ( )

    :

    :

    : ()

  • :

    : . (GEO)

    (wale)

    (GEO) ( ), , .

    , , (wale) (strut) . (STR) . .

    ( GEO). 11 1997-1.

    (hydraulic heave) (hydraulic heave) (piping). 10 1997-1.

    , .

    - . - .

  • :

  • :

  • :

  • :Rd > Ed ( > ) E

    E

    :

    d d ( )

    R

    .

    . , ,

    ( ( , ).

    ( DA-2* ).

    , , ( ). , .

  • DA-2* :

    :

    ( )

    EE

    Rd > Ed ( > ) R

    ( ) / R > F ( ) F ( ) 1.40 1.35 1.50 1.00

  • 1 : B0 q = 20 KPa

    : = 24KN/m3

    P

    = 20 KN/m3c = 0

    6

    m

    (, )

    x0

    W

    Pav = 37.5

    H

    =

    6

    = 2/3

    Pah

    a

    M

    T

    h

    p

    Pph 0: = 20 KN/m3

    h

    a

    B/2

    N: = 20 KN/m

    c = 10 KPa = = 30B/2

    1. ( ).

    2. (GEO) : Ed < Rd ( ) d dEd = 1.35 + 1.50 = 1.35 x 75.60 + 1.50 x 25.2 = 139.86 kN/m,ah gP ,ah qP

    ( )1 tanR N P = + R = 1.10 1.40 ( )tand phR

    R N P = + R

    : Ed < Rd 139.86 < 175.93 .

  • 3. (EQU) : Edst,d < Estb,d

    E =

    H

    =

    6

    m

    Edst = . 1.10 () 1.50 ()

    E tb = .

    h

    p

    h

    a

    Estb . 0.9 ( )

    Edst,d < Estb,d 109.08 < 496.89

  • 4. (GEO) : Ed < Rd ( ) 1N i D N i B N i

    6

    m

    ( )1 22u c c c q q qp c N s i q D N s i B N s i = + + +

    Vd = 454.29 1050.7 = Rd

    H

    =

    6

    h

    a

    .

    h

    p

  • 2 :

    . (D) .

  • fillfilla,h,qfillPa,h,q

    fillPa h ga,h,gfilla,h,g

    soila,h,q soila h gsoilPsoil p,h,g

    a,h,gPa,h,q

    soilPa,h,g

    soilPp,h,g

    1. GEO :

    . R . , () 20% .

  • fill ( filla,h,qfillPa,h,q

    fillPa h g

    ( q) .

    a,h,gfilla,h,g

    soila,h,q soilsoilsoil p h g soila,h,gsoilPa,h,q

    soilPa,h,g

    soilPp,h,g

    p,h,g

    D C :1 35 ( ) 1 50 ( ) ( ) / 1 401.35 x ( . ) + 1.50 ( . q) = ( . ) / 1.40

    D=6m (

    20%

    ( , , R) Rk = 1696 kN/m. 20%, , : D = 1.20 * 6.0 = 7.20m

  • 2. STR :

    () . ( ( ) () , :

    1. : (g k) ( g,k)

    2. ( + ) : (g+q,k)

    : :q,k = g+q,k - g,k

    : d = 1.35 *g,k +1.50* q,k

    (.. )

  • 10 :

    UPL ( lift) UPL : (uplift) . ( ( ).

    HYD : . . , , (boiling).

    UPL HYD

  • 1. UPL ( )

    : V G + R : Vdst,d Gstb;d + Rd G,dst Gdst + Q,dst Qdst G;stb (Gstb + R)

    Vdst,d = ,

    Gstb;d = Rd = ( Rd (.

    ) - ()

    : FS = 1 / = 1 / 0 9 = 1 10 : FS = 1 / G;stb = 1 / 0.9 = 1.10

    RRd

    Vdst,d

  • : = 200 kPa ( ) =150 kPa ( ) G =150 kPa Q =50 kPa ( ). =12m () () .stb;d = G;stb stb;k = G;stb G = 0.9 x 150 = 135 kPaudst;d = G;dst udst;k = G;dst (W H ) = 1.0 x 10 x 12 = 120 kPa : udst;d stb;d 120 < 135 .

  • 1. HYD ( ) : S G

    HYD : Sdst;d G stb;dSdst;d = Gstb;d =

    Sdst,d Gstb;d

    : FS = G;dst G;stb = 1.35 / 0.90 = 1.50

  • 11 : : , ,

    , (cut or buried structures), .

    , 3 (DA-3)

  • 3 (DA-3) :

    :(1) ( ) (1) ( ), (2) ( ),

    ( ( , , ) .

  • ,

    3 (DA-3) :

    ( : cd , d : cu;d) (m), :

    kc tan1tan kuc ; 1 25 1 40mM

    kdc = kmMd

    = tantanmM

    kuduc

    ;; = = 1.25 1.40

    :(1) : m = 1.1.

    (FS) : (FS) :FS = m = 1.25 x 1.1 = 1.38 FS = m = 1.40 x 1.1 = 1.54

    (2) 1(2) : m = 1. (FS) :FS = m = 1.25 x 1 = 1.25 FS = = 1 40 x 1 = 1 40 FS = m = 1.40 x 1 = 1.40

  • : 15 2:3 (:) ( ) : ( ) : : = 20 kN/m3 , () : c = 15 kPa () : = 25 , : q = 20 kPa

    qd = F q = 1.30 x 20 = 26 kN/m2

    :c = c / 1 25 = 15 / 1 25 = 12 kPacd = c / 1.25 = 15 / 1.25 = 12 kPad = arctan (tan / 1.25) = arctan (tan 25 / 1.25 ) = 20.46o

  • ( 3)

    = 20 kN/m3

    cd = 12 kPad = 20 46od 20.46

    > 1.0, .

  • : 1998-5 :(1)

    (F = = 1).

    (2) ( )(2) () (R) ( = R = 1).

    (3) (m) ( ) (m) . .

    (4) 2 DA-2* (4) 2 DA-2 , 3 (DA-3) 3 (DA 3).

    , : 1. : FS = 12. : FS = 1.303. : FS = 14. : FS = 15. : FS = 1.1 ( )