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 PresentationTRANSCRIPT
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6 : 7 . -. - . 2006 - 0725.10.2007
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: = i + c + s i = c = s = () K - () : , :(1) , ( )(2) . ( ) ( ). - ()
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, ( ) ( ) - , ( ). , . , . , - (SPT, CPT, , ). : , () ( ) () - :
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1. (SPT) : -
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1. (SPT) : - = 30 cm ( 15cm) 75kg 3 x 15cm : n = 6,8,9N = 8+9 = 1715cm15cm15cm = 17
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1. (SPT) : - , .
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1. (SPT) : - () SPT 2 (= ). 2
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1. (SPT) : - SPT : 60 = CER Nm : Nw = Cw Nm (, ) : Nn = Cn Nm , , (Dr), () : = CER Cw Cn Nm, (Nm) :
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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
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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
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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
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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
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4. =m ( Peck & Bazaraa, 1967) : vo < 71.8 kPa : vo > 71.8 kPa :
Chart2
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N' / N (Peck & Bazarra)
(kPa)
Sheet1
04
53.3085194376
102.8208744711
202.1786492375
301.7746228926
401.497005988
501.2944983819
601.1402508552
701.018848701
71.80.9996900961
71.81.000820673
800.9799118079
900.955566173
1000.9324009324
1100.9103322713
1200.8892841263
1300.8691873099
1400.8499787505
1500.8316008316
1600.814000814
1700.7971303308
1800.7809449434
1900.7654037505
2000.7504690432
2100.7361059993
2200.7222824124
2300.7089684509
2400.6961364427
2500.6837606838
2600.6718172657
2700.6602839221
2800.6491398896
2900.6383657836
3000.6279434851
Sheet1
N' / N (Peck & Bazarra)
(kPa)
Sheet2
Sheet3
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1.1. Alpan :i = ( cm)B, L = (B L)q = ( kPa) = SPT, Terzaghi & Peck ( Dr=100%)1. (SPT) : -
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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
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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 : -
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1.2. Schultze & Sherif :i = ( cm)B, L = (B L)D = ( ) = min ( , 2B) q = ( kPa)N = SPT, Terzaghi & Peck ( Dr=100%)f = 1. (SPT) : -
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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
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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) :
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: Terzaghi & Peck , () , (b) b, o o (q) : : , b , b = 0.305m (1), B ( ) :1.3. Terzaghi & Peck (1967) () (D=0) : : / 1 = 4
Chart1
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1.7773891288
2.2495080623
2.55949622
2.777291855
2.9383164879
3.0620694755
3.160090657
3.2396221079
3.305430224
3.3607769963
3.4079686721
3.4486811942
3.4841613992
3.5153558958
3.5429961892
3.5676564313
3.5897935323
3.6097756088
3.6279025379
3.6444210592
3.659536038
3.6734189823
3.6862145632
(m)
B / 1
Sheet1
/ 1
(m)
0.3051
0.60981.7773891288
0.91462.2495080623
1.21942.55949622
1.52422.777291855
1.8292.9383164879
2.13383.0620694755
2.43863.160090657
2.74343.2396221079
3.04823.305430224
3.3533.3607769963
3.65783.4079686721
3.96263.4486811942
4.26743.4841613992
4.57223.5153558958
4.8773.5429961892
5.18183.5676564313
5.48663.5897935323
5.79143.6097756088
6.09623.6279025379
6.4013.6444210592
6.70583.659536038
7.01063.6734189823
7.31543.6862145632
Sheet1
(m)
B / 1
Sheet2
Sheet3
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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) -
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1.4. Peck, Hanson & Thornburn (1974) : - B = (B = L)D = Dw = q = ( kPa)N = SPT Peck, Hanson & Thornburn (1974). 2 i = ( cm)Cw = :
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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) :
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1.5. Meyerhof :qa = (kPa) ( m), 25mm. N=60 SPT. D , (i cm) : - > 1.2m : < 1.2m :
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1.5. Meyerhof : < 1.2 m (i cm) (D=0), : > 1.2 m ()q = kPaB = = SPT i = cm - : (D), : [1+D/(3B)]
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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
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= . 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
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- 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 (.. )
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- 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)
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- 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
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2. (CPT) : - () qc
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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) : -
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Schmertmann : (Izp) Iz :vI = (z) : z = B , z = B/2 , : z. Iz Izp L/B==/2
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Schmertmann : Iz :qc = L/B
L / BE / qc1102.53.5
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: =2.6m, L=23m D=2m ( ). q=178 kPa. . (qc) CPT . Schmertmann :
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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. :
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Schmertmann : () :
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qc / N Robertson : (qc) SPT ( )
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qc / N (qc MPa) Burland and Burbridge
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(Dr)