ΔΙΑΛΕΞΗ 11...
of 23
date post
10-Jan-2016Category
Documents
view
51download
0
Embed Size (px)
description
ΔΙΑΛΕΞΕΙΣ ΤΟΥ ΜΑΘΗΜΑΤΟΣ « ΘΕΜΕΛΙΩΣΕΙΣ » 7ο Εξ. ΠΟΛ-ΜΗΧ. ΕΜΠ - Ακαδ. Ετος 2007 - 08. ΔΙΑΛΕΞΗ 11 Θεμελιώσεις με πασσάλους : Καθιζήσεις πασσάλων. 06 .0 2 .200 8. 1. Κατηγορίες πασσάλων 2. Αξονική φέρουσα ικανότητα μεμονωμένου πασσάλου 2.1 Εμπηγνυόμενοι πάσσαλοι (πάσσαλοι εκτοπίσεως) - PowerPoint PPT Presentation
Transcript of ΔΙΑΛΕΞΗ 11...
11 : 7 . -. - . 2007 - 0806.02.2008
1. 2. 2.1 ( )2.2 ()2.3 73. 4. 4.1 4.2 5.
(Q) - () Q () : fs qp Qs Qp , : Q = Qs + Qp1. (. = ) :qpfs
(fs) , () = (0.4% - 1.2%) D = 4 15 mm (qp) , () = (4% - 10%) D = 30 - 100 mm fsu qpu (Q) - ()
: ( fs ) ( qp ) () , (Q ) Q . ,
QpQp QsQ : fs . , fs -.
: L=15m, B = 0.45m Ap = 0.159 m2 ( ) : fsu = 150 kPao . . qpu = 4 Mpa fs qp , .
:Qsu = B L fsu = 3.14 x 0.45 x 15 x 0.150 = 3.18 MNQpu = Ap qpu = 0.159 x 4 = 0.64 MNQu = Qsu + Qpu = 3.18 + 0.64 = 3.82 MN
: FS=2 :Q = Qu / FS = 3.82 / 2 = 1.9 MN
: : 3 mm
: 3 mm. : qp = 0.93 MPa : Qp = 0.15 MN : = Qp / Ap = 0.15 / 0.159 = 0.93 MPa = / b = 0.93 / 30000 = 0.000031
:Qm = 0.5 x (1.9 + 0.15) = 1.025 MN :m = Qm / Ap = 1.025 / 0.159 = 6.45 MPa : = m / Eb = 6.45 / 30000 = 0.00021 : = L = 0.00021 x 1500 cm = 3.2 mm
: 3 + 3.2 = 6.2 mm
: = Q / Ap = 11.9 MPa = / b = 11.9 / 30000 = 0.0004 :Qs = 3.14 x 0.45 x 15 x 85 = 1.8 MN 1.75 ( )
: Q , DIN 40141.1 ( fs ) sufs = fsu > su :Qsu = su = cm : : =
Q DIN 40141.2 ( qp ) MPa1.2.1 - (D = ) : qp MPa0.02 D0.03 D0.10 D
/ D (qc) CPT MPa1015202500.020.030.10> 0.1000.70.92.02.001.051.353.03.001.41.83.53.501.752.254.04.0
Q DIN 40141.2 ( qp ) MPa1.2.2 (D = ) : qp MPa0.02 D0.03 D0.10 D
/ D cu (kPa)10020000.020.030.10> 0.1000.350.450.80 MPa0.80 = qpu00.91.11.5 MPa1.5 = qpu
: DIN 4014 : : fsu = 45 kPa : qc = 0.5 N = 0.5 x 45 = 22.5 MPa fsu = 120 kPa qpu = 3.75 MPa : :Qsu = 3.14 x 0.80 x (45 x 12 + 120 x 3) = 1356.5 + 904.3 = 2261 kNAp = 3.14 x 0.82 / 4 = 0.5024 m2Qpu = 0.5024 x 3750 = 1884 kN : Qu = Qsu + Qpu = 2261 + 1884 = 4145 kN : , = 18 kN/m3 : cu = 125 kPa : , = 20 kN/m3 SPT N = 45 Q DIN 40141. : L = 20m, D = 0.8m :
Q DIN 4014 :Qsu = 2.261 MN su = 1.63 cm : Qs = min {( / su ) Qsu , Qsu }2.1 :2.2 : : qc = 0.5 N = 0.5 x 45 = 22.5 MPaAp = 0.5024 m22. Q - :
/ D(cm)qp(MPa)Qp(kN)00.020.030.10> 0.1001.62.48> 801.582.0253.753.750794101718841884
Q DIN 4014 : - FS = 2 :Qmax = Qu / 2 = 4145 / 2 = 2072 kN ( ) : = 11mm : ( = )
/ D - (cm)qp - (MPa)Qs - (kN)Qp - (kN)Q - (kN)00.020.02040.030.10> 0.1001.6su =1.632.4pu = 8> 801.581.602.0253.753.750221922612261226122610794802101718841884030133063327841454145
Chart1
000
1.61.61.6
1.631.631.63
2.42.42.4
888
101010
Qs
Qp
Qs+Qp
(kN)
(cm) .
Sheet1
(cm)QsQpQ
0000
1.622197943013
1.6322618023063
2.4226110173278
8226118844145
10226118844145
Sheet1
000
000
000
000
000
000
Qs
Qp
Qs+Qp
(kN)
(cm)
Sheet2
Sheet3
= P = , L = = Ip = (h) , (d) Poisson () (Poulos & Davis, 1980)
( Poulos & Davis, 1980) () : () : = = = d = 1 = Ri = = p = p = ps =
( Poulos & Davis, 1980)
( Poulos & Davis, 1980)
( Poulos & Davis, 1980)
4. QQsQsnQp (.. - ), ( - ) ( ). : <
4. QQsQsnQp -
4.
4.
