ΔΙΑΛΕΞΗ 6 Καθιζήσεις Επιφανειακών Θεμελιώσεων :...

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ΔΙΑΛΕΞΕΙΣ ΤΟΥ ΜΑΘΗΜΑΤΟΣ « ΘΕΜΕΛΙΩΣΕΙΣ » 7 ο Εξ. ΠΟΛ-ΜΗΧ. ΕΜΠ - Ακαδ. Ετος 200 6 - 0 7. ΔΙΑΛΕΞΗ 6 Καθιζήσεις Επιφανειακών Θεμελιώσεων : Υπολογισμός καθιζήσεων σε αμμώδη εδάφη. 25 .1 0 .200 7. Υπολογισμός καθιζήσεων σε αμμώδη εδάφη. Συνολική καθίζηση : ρ = ρ i + ρ c + ρ s - PowerPoint PPT Presentation

Transcript of ΔΙΑΛΕΞΗ 6 Καθιζήσεις Επιφανειακών Θεμελιώσεων :...

  • 6 : 7 . -. - . 2006 - 0725.10.2007

  • : = i + c + s i = c = s = () K - () : , :(1) , ( )(2) . ( ) ( ). - ()

  • , ( ) ( ) - , ( ). , . , . , - (SPT, CPT, , ). : , () ( ) () - :

  • 1. (SPT) : -

  • 1. (SPT) : - = 30 cm ( 15cm) 75kg 3 x 15cm : n = 6,8,9N = 8+9 = 1715cm15cm15cm = 17

  • 1. (SPT) : - , .

  • 1. (SPT) : - () SPT 2 (= ). 2

  • 1. (SPT) : - SPT : 60 = CER Nm : Nw = Cw Nm (, ) : Nn = Cn Nm , , (Dr), () : = CER Cw Cn Nm, (Nm) :

  • m SPT :1. m ( 60% ) :N60 = CER Nm, , ( CER = 1 N60 = Nm ) 60% .1. (SPT) : m : SPT Nm > 15, , ( Nm) Terzaghi :N = 15 + 0.5 ( Nm 15) m < 15 , . : = 1.25 m

  • m , Terzaghi BS 8002 : Cn = Nn () / Nm () = / : (B=L=1.5m) z=2B=3m. (z=B=1.5m), 1.5x20=30 kPa ( ). 0 2 = 3m N60= m= 10. : Cn = 2.40 n = Cn Nm = 2.4 x 10 = 24 ( ) Terzaghi BS 8002 SPT ( ), . . Terzaghi, : 135 kPa

  • 3. m : v=100 kPa Cn ( Peck, Hanson & Thornburn, 1974) : Nn = Cn Nm : (B=L=1.5m) z=2B=3m. (z=B=1.5m), 1.5x20=30 kPa ( ). 0 2 = 3m N60= m= 10. : Cn = 1.35 n = Cn Nm = 1.35 x 10 = 13.5 ( ) SPT ( ), . . Peck, Hanson & Thornburn (1974), : 100 kPa

  • 3. m Dr = 100 % ( Terzaghi & Peck) : : (B=L=1.5m) z=2B=3m. (z=B=1.5m), 1.5x20=30 kPa ( ). 0 2 = 3m N60= m= 10. Dr = 67 %. 100%, = 31 ( ).Dr = 67%60

  • 4. =m ( Peck & Bazaraa, 1967) : vo < 71.8 kPa : vo > 71.8 kPa :

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    53.3085194376

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    N' / N (Peck & Bazarra)

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  • 1.1. Alpan :i = ( cm)B, L = (B L)q = ( kPa) = SPT, Terzaghi & Peck ( Dr=100%)1. (SPT) : -

  • m Alpan : Dr = 100 % ( Terzaghi & Peck) : : (B=L=1.5m) z=2B=3m. (z=B=1.5m), 1.5x20=30 kPa ( ). 0 2 = 3m N60= m= 10. Dr = 67 %. 100%, = 31 ( ).Dr = 67%60

  • Alpan : (L=3m, =2m) D=1m. q=300 kPa. =20 kN/m3 SPT N60 = 10. Alpan : : z = 2B = 4m : v = (2+1) x 20 =60 kPa v = 60 kPa 60 = 10 Dr = 60 %. 100%, = 25 ( ). = 25 = 0.13

    i = 0.94 cm = 9.4 mm1.1. Alpan : -

  • 1.2. Schultze & Sherif :i = ( cm)B, L = (B L)D = ( ) = min ( , 2B) q = ( kPa)N = SPT, Terzaghi & Peck ( Dr=100%)f = 1. (SPT) : -

  • Schultze & Sherif : (L=3m, =2m) D=1m. q=300 kPa. =20 kN/m3 SPT N60 = 10. Schultze & Sherif : = 2 = 4m H/B = 2 . H/B=2 L/B=1.5 f = 0.074 Terzaghi & Peck ( Dr=100%) : : z = 2B = 4m . : v = (2+1) x 20 =60 kPa v = 60 kPa 60 = 10 Dr = 58 %. 100%, = 25 ( ). = 25 :i = 0.88 cm = 8.8 mm

  • 1.3. Terzaghi & Peck (1967) () (D=0) :q = ( kPa)N = SPT Terzaghi BS 8002i = ( cm) - : = 2.5m = 8.2 , q = 300 kPa , N () = 24 : q1 = 250 kPa : i = 2.5 x (300 / 250) = 3 cm1. q1 ( kPa), : = ( ) :2. ( cm) :

  • : Terzaghi & Peck , () , (b) b, o o (q) : : , b , b = 0.305m (1), B ( ) :1.3. Terzaghi & Peck (1967) () (D=0) : : / 1 = 4

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  • 1.4. Peck, Hanson & Thornburn (1974) :B = (B = L)D = Dw = q = ( kPa)N = SPT Peck, Hanson & Thornburn (1974)i = ( cm) -

  • 1.4. Peck, Hanson & Thornburn (1974) : - B = (B = L)D = Dw = q = ( kPa)N = SPT Peck, Hanson & Thornburn (1974). 2 i = ( cm)Cw = :

  • Peck, Hanson & Thornburn (1974) : (=L=2m) D=1m. q=300 kPa. =20 kN/m3 SPT N60 = 10. 1m ( : Dw = 2m) : : : z = 2B = 4m (2m ) :v = 2 x 20 + 1 x 10 = 50 kPa. Peck, Hanson & Thornburn : v = 50 kPa Cn = 1.15 Nn = Cn N60 = 1.234 x 10 =12.3D/ = 1/2 = 0.5 , = 2 / 0.305 = 6.55 , = 12.3 ( ) q1 = 130 kPaCw = 0.5 + 0.5 x 2 / (1 + 2) =0.83 : 1.4. Peck, Hanson & Thornburn (1974) :

  • 1.5. Meyerhof :qa = (kPa) ( m), 25mm. N=60 SPT. D , (i cm) : - > 1.2m : < 1.2m :

  • 1.5. Meyerhof : < 1.2 m (i cm) (D=0), : > 1.2 m ()q = kPaB = = SPT i = cm - : (D), : [1+D/(3B)]

  • i = ( cm) FS 3. FS < 3, - . FS = 2, FS=3.

    q = ( ) ( kPa)

    B = ( m), L B

    = SPT Z . m . - 1.6. Burland & Burbridge (1985) :Dw

  • = . SPT N , : = 2. , : < 4m : / = 0.80 = 0.80 > 25m : Z / B = 0.40 = 0.40 :

    fI = () . > : fI = 1. < : fI = (H/Z) [ 2 (H/Z)] - 1.6. Burland & Burbridge (1985) :Dw

  • - 1.6. Burland & Burbridge (1985) :fS = . L > B :fo = . : fo = 1 : fo = 1/3ft = . 3 ( t ) : : R3 = 0.3 , Rt = 0.2 : R3 = 0.7 , Rt = 0.8 : Burland & Burbridge (.. )

  • - 1.6. Burland & Burbridge (1985) : : = 20m , L = 30m , D = 3m, qo = 200 kPa. SPT : d = D = 3m Nm = 15, m (m = 2 ). . : (4m < B < 25m) : = 10m : d = 3m : Nm = 15 N = 15 d = Z = 13m : Nm = 15 + 2 x 10 = 35 = 25= 0.5 x (15 + 25) = 20 :N = 15 + 0.5 ( Nm 15)

  • - 1.6. Burland & Burbridge (1985) : () :fo = 1 , f = 1 , fS = 1.15 : : t = 30 : R3 = 0.5 , Rt = 0.5 : ft = 1+0.5 + 0.5 log (30/3) = 2

    : i = 2 x 3.4 = 6.8 cm :i = 0.171 x 1 x 1.15 x 1 x [ 200.7 / 201.4 ] 140 = 3.4 cm (D=3m) : q = qo D = 200 3 x 20 = 140 kPa

  • 2. (CPT) : - () qc

  • Schmertmann :i = ( zj)C1 = D ( = )vD = (D) :q = zj = (j). : zj = 0.1 0.2 Ej = (j)Ct = ( t ) (j) , zj :Izj = (j) (t=0.1 ) : Ct = 12. (CPT) : -

  • Schmertmann : (Izp) Iz :vI = (z) : z = B , z = B/2 , : z. Iz Izp L/B==/2

  • Schmertmann : Iz :qc = L/B

    L / BE / qc1102.53.5

  • : =2.6m, L=23m D=2m ( ). q=178 kPa. . (qc) CPT . Schmertmann :

  • 1.L / B = 23 / 2.6 = 8.85. (Iz) .2. Iz : Schmertmann :q vD = 178 15.7 x 2 = 147 kPavI = 15.7 x 2 + (15.7 10) x 2.6 = 47.6 kPa : Izp = 0.683. z ( 4 = 4 x 2.6 = 10.4m). 11 . (z) 2 . Iz ( ) 4 qc ( 5) = 3.5 qc C1 :C1 = 1 0.5 x (15.7 x 2) / 147 = 0.898. : Ct = 1.9. :

  • Schmertmann : () :

  • qc / N Robertson : (qc) SPT ( )

  • qc / N (qc MPa) Burland and Burbridge

  • (Dr)