8/2/2019 Down drag
1/4
Prediction the settlement of pile folows the equivalent
foundation method
Equation: In which Ei :Modulus
S1= Si = ( zi aver.hi ) / Ei (1) ziaver :Vertical pressurehi
:Thickness soil layer
Due to the input data of pile is unknown, we assumped that
settlement of pile is 0,0cm.
Calculation settlement of each cohesive soil layers using
following equation:
when In which : Ccr : Recompress index
Cc :Compresstion index
pc :Preconsolidation pressure
po :Effective overburden pressure
when :Imposed pressure increase
BH - no1
Thickness
Symbol Description From To h (m) (T/m3)
S (m) S1 (m) (m)
1 0.0 -1.0 1.0 1.77 0.029 0.010 0.000 0.010
2 -1.0 -2.0 1.0 1.77 0.029 0.010 0.000 0.010
3 -2.0 -3.0 1.0 1.77 0.032 0.011 0.000 0.011
4 -3.0 -4.0 1.0 1.77 0.035 0.012 0.000 0.012
5 -4.0 -5.0 1.0 1.77 0.039 0.014 0.000 0.014
6 -5.0 -6.0 1.0 1.77 0.042 0.015 0.000 0.015
7 -6.0 -7.0 1.0 1.77 0.045 0.016 0.000 0.016
8 -7.0 -8.0 1.0 1.77 0.045 0.016 0.000 0.0169 -8.0 -9.0 1.0 1.77
0.044 0.015 0.000 0.015
10 -9.0 -10.0 1.0 1.71 0.041 0.014 0.000 0.014
11 -10.0 -11.0 1.0 1.71 0.038 0.013 0.000 0.013
12 -11.0 -12.0 1.0 1.71 0.036 0.012 0.000 0.012
13 -12.0 -13.0 1.0 1.71 0.033 0.012 0.000 0.012
14 -13.0 -14.0 1.0 1.71 0.032 0.011 0.000 0.011
15 -14.0 -15.0 1.0 1.82 0.023 0.008 0.000 0.008
No.
Layer Depth
Layer
Settlement
DOWNDRAG FORCE CALCULATIONPROJECT : apple - HAI PHONG
35% Layer
Settlement
Settlement
of pile
2. determine settlement of each soil layers.
3. Determine settlement
1. determine settlement of pile.
Unequal
settlement
Clay verry soft2
3
Clay , sandy clay
soft
( )lg 21
cr o
o o
C p ps H
e p
+ =
+
o c p p p+
( )lg lg 31 1
cr c c o
o o o c
C p C p ps H
e p e p
+ = +
+ +
o c p p p+ > p
S
16 -15.0 -16.0 1.0 1.82 0.022 0.008 0.000 0.008
17 -16.0 -17.0 1.0 1.82 0.021 0.007 0.000 0.007
18 -17.0 -18.0 1.0 1.82 0.020 0.007 0.000 0.007
19 -18.0 -19.0 1.0 1.82 0.019 0.007 0.000 0.007
20 -19.0 -20.0 1.0 1.82 0.018 0.006 0.000 0.006
21 -20.0 -21.0 1.0 1.82 0.017 0.006 0.000 0.006
22 -21.0 -22.0 1.0 1.78 0.020 0.007 0.000 0.007
23 -22.0 -23.0 1.0 1.78 0.019 0.007 0.000 0.007
T otal o f c al cu la ting th ic kn ess H (m) 23.0 0.244
Negative shaft resistence occures due to the settlement between
soil and pile. The amount of settlement between soil and pile
necessary to
mobilize the nagative shaft resistance is about 10 mm.
Therefore, nagative shaft resistance will occur on the pile shaft
in each soil layer or
portion of a soil layer with a settlement greater than 10
mm.
Therefore the pile segment in layer 2,3 will be subjected to the
negative shaft resistances (downdrag).
Diameter (cm) Perimeter (m)
30 0.942
Soil layer Depth D/b cu Rs (-)
(m) (T/m2) (T)Layer 2 6 20.00 1.3 7.3
Layer 3 5 16.67 1.26 5.9
13.3
4.Dertermine the pile length that will experience negative shaft
resistance
5. Determine downdrag force
Magnitude of downdrag force Rs(-)
Pile's parameter
4
5 Clay to firm
Clay stiff
( )lg 21
cr o
o o
C p ps H
e p
+ =
+
o c p p p+
( )lg lg 31 1
cr c c o
o o o c
C p C p ps H
e p e p
+ = +
+ +
o c p p p+ > p
S
8/2/2019 Down drag
3/4
Prediction the settlement of pile folows the equivalent
foundation method
Equation: In which Ei :Modulus
S1= Si = ( zi aver.hi ) / Ei (1) ziaver :Vertical pressurehi
:Thickness soil layer
Due to the input data of pile is unknown, we assumped that
settlement of pile is 0,0cm.
Calculation settlement of each cohesive soil layers using
following equation:
when In which : Ccr : Recompress index
Cc :Compresstion index
pc :Preconsolidation pressure
po :Effective overburden pressure
when :Imposed pressure increase
BH -no13
Thickness
Symbol Description From To h (m) (T/m3)
S (m) S1 (m) (m)
0 1 Fillsand 4.20 - - - -
1 0.0 -1.0 1.0 1.77 0.029 0.010 0.000 0.010
2 -1.0 -2.0 1.0 1.77 0.029 0.010 0.000 0.010
3 -2.0 -3.0 1.0 1.77 0.032 0.011 0.000 0.011
4 -3.0 -4.0 1.0 1.77 0.035 0.012 0.000 0.012
5 -4.0 -5.0 1.0 1.77 0.039 0.014 0.000 0.014
6 -5.0 -6.0 1.0 1.77 0.042 0.015 0.000 0.015
7 -6.0 -7.0 1.0 1.77 0.045 0.016 0.000 0.016
8 -7.0 -8.0 1.0 1.77 0.045 0.016 0.000 0.016
9 -8.0 -9.0 1.0 1.77 0.044 0.015 0.000 0.015
10 -9.0 -10.0 1.0 1.71 0.041 0.014 0.000 0.014
11 -10.0 -11.0 1.0 1.71 0.038 0.013 0.000 0.013
12 -11.0 -12.0 1.0 1.71 0.036 0.012 0.000 0.012
13 -12.0 -13.0 1.0 1.71 0.033 0.012 0.000 0.012
14 -13.0 -14.0 1.0 1.71 0.032 0.011 0.000 0.011
15 -14.0 -15.0 1.0 1.82 0.030 0.010 0.000 0.010
Layer
Settlement
Settlement
of pile
Unequal
settlement
1. determine settlement of pile.
3. Determine settlement
2. determine settlement of each soil layers.
Clay , sandy clay
3
Clay firm to soft
DOWNDRAG FORCE CALCULATIONPROJECT : apple - HAI PHONG
No.
Layer Depth
35% Layer
Settlement
2 Clay verry soft
soft
4a
( )lg 21
cr o
o o
C p ps H
e p
+ =
+
o c p p p+
( )lg lg 31 1
cr c c o
o o o c
C p C p ps H
e p e p
+ = +
+ +
o c p p p+ > p
S
16 -15.0 -16.0 1.0 1.82 0.021 0.007 0.000 0.007
17 -16.0 -17.0 1.0 1.82 0.020 0.007 0.000 0.007
18 -17.0 -18.0 1.0 1.82 0.020 0.007 0.000 0.007
19 -18.0 -19.0 1.0 1.82 0.019 0.007 0.000 0.007
20 -19.0 -20.0 1.0 1.82 0.018 0.006 0.000 0.006
21 -20.0 -21.0 1.0 1.82 0.017 0.006 0.000 0.006
22 -21.0 -22.0 1.0 1.78 0.016 0.006 0.000 0.006
23 -22.0 -23.0 1.0 1.78 0.016 0.006 0.000 0.006
To ta l o f ca lc ulat in g thi ck nes s H (m) 23.0 0.244
Negative shaft resistence occures due to the settlement between
soil and pile. The amount of settlement between soil and pile
necessary to
mobilize the nagative shaft resistance is about 10 mm.
Therefore, nagative shaft resistance will occur on the pile shaft
in each soil layer or
portion of a soil layer with a settlement greater than 10
mm.
Therefore the pile segment in layer 2,3,4a will be subjected to
the negative shaft resistances (downdrag).
Diameter (cm) Perimeter (m)
30 0.942
Soil layer Depth D/b cu Rs (-)
(m) (T/m2) (T)Layer 2 5 16.67 1.31 6.2
Layer 3 5 16.67 1.26 5.9
Layer 4a 1 3.33 1.85 1.7
13.8
4.Dertermine the pile length that will experience negative shaft
resistance
Pile's parameter
Magnitude of downdrag force Rs(-)
5. Determine downdrag force
4 Clay stiff
( )lg 21
cr o
o o
C p ps H
e p
+ =
+
o c p p p+
( )lg lg 31 1
cr c c o
o o o c
C p C p ps H
e p e p
+ = +
+ +
o c p p p+ > p
S
