Konstruktionsuppgift 2
G7006B
Sofi Isaksson
Lea-Friederike Koss
Henrik Silfvernagel
1
1. Assignment ......................................................................................................................... 3
2. Given conditions ................................................................................................................ 4
2.1 Task 1 ........................................................................................................................... 4
2.2 Task 2 ........................................................................................................................... 4
3. Assumptions ....................................................................................................................... 5
3.1 Task 1a ......................................................................................................................... 5
3.2 Task 1b ......................................................................................................................... 5
3.3 Task 2a ......................................................................................................................... 5
3.4 Task 2b ......................................................................................................................... 5
4. Calculations ........................................................................................................................ 6
4.1 Task 1a ......................................................................................................................... 6
4.2 Task 1b ......................................................................................................................... 7
4.3 Task 2a ......................................................................................................................... 8
4.4 Task 2b ......................................................................................................................... 9
5. Discussion ........................................................................................................................ 10
6. Appendix .......................................................................................................................... 11
6.1 Geotechnical data for task 1 ....................................................................................... 11
6.2 CPT for task 2 ............................................................................................................ 12
6.3 CPT for task 2 ............................................................................................................ 13
6.4 Data about HEB 400 for task 2 .................................................................................. 14
2
1. Assignment
The assignment is to calculate bearing capacity and settlements of piles. First task is to decide
the lenght of a shaft bearing concrete pile in clay. When the lenght is sorted out the
settlements are calculated for the given bearing capacity. The second task is to decide the
design bearing capacity for a shaft bearing steel pile in frictional and cohesive soil. In the
same task the design bearing capacity for a point bearing pile with a given lenght should be
calculated.
3
2. Given conditions
2.1 Task 1
o Security class 3 (SK3); Geotechnical class (GK3)
o Soil layering:
o Clay to a depth of 50 m.
o Ground water lever at 3 m below ground surface.
o Geotechnical data:
o Values for undrained shear strenght has been determined with CPT with
relatively good accuracy. See figure 1a. For depth deeper than 30 m the
undrained strenght can be assumed to be constant 60 kPa.
o In-situ vertical effective stress and preconsolidation pressure is given in figure
1b.
o At depth 20-30 m the compression modulus number m = 10 and for deeper
depth m = 15.
o For overconsolidated clay the compression modulus Mk = 6,5 MPa.
o Pile type: standardised concrete pile SP 3 (0,27 m * 0,27 m). Concrete K50
o Design pile load: 360 kN
2.2 Task 2
o Security class 3 (SK3); Geotechnical class (GK3)
o Soil layering: Determined with CPT down to 17,5 m depth.
o From 1 m (excavation bottom) down to 11,25 m mixed sand and silt layers
with some thin layers of clay (bearing capacity contribution assumed to be
zero)
o Between 11,25 m and 16 m depth there is clay.
o Below 16 m depth there are silt and sand.
o Geotechnical data:
o Tip resistance, friction angle (in sand and silt) and undrained shear strenght (in
clay) is found in given diagrams.
o Pile type: standardised steel profile HEB 400.
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3. Assumptions
3.1 Task 1a
 mm  1,3  1,6 .  mm is chosen to 1,4 because the undrained shear strenght was determined
with relatively good accuracy, se appendix 6.1.
 sm  1,6  2,0 .  sm is chosen to 1,7 because the undrained shear strenght was determined
with relatively good accuracy, se appendix 6.1.
3.2 Task 1b
When calculating Mmd,  m  1,3  1,6 .  m is chosen to 1,5 because the tests are relatively good.
Mk can be put to 6,5 MPa if the soil is overconsolidated. The figure 1b, see appendix 6.1,
shows that  0'   c' which gives that the soil is overconsolidated.
3.3 Task 2a
When looking at the CPT, appendix 6.2 it is assumed that the ground water table is at the
excavation bottom.
 mm  1,3  1,6 .  mm is chosen to 1,3 because the CPT was determined with good accuracy.
 sm  1,1  1,3 .  sm is chosen to 1,1 because the CPT was determined with good accuracy.
It is assumed that Km is the same as the earth pressure coefficient at rest, K0.
From CPT, appendix 6.3 the friction angle for 1-6 m is 38o and for 6-10 m 35o.
The water saturated density for the sand and silt is assumed to be 2,0 t/m3.
3.4 Task 2b
 mm  1,3  1,6 .  mm is chosen to 1,3 because the CPT was determined with good accuracy.
 sm  1,6  2,0 .  sm is chosen to 1,6 because the CPT was determined with good accuracy.
qcs is evaluated to 13,5 MPa in the CPT, se appendix 6.2.
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4. Calculations
4.1 Task 1a
Rd 
1
 Rd
Rmd  Rsd 
(1)
where:
Rd is the design bearing capacity
Rmd is the shaft bearing capacity
Rsd is the point bearing capacity
 Rd  1,7 is a constant given in “Grundläggningsteknik”
Rmd    i Ami
Cuki
(2)
 n  mm
where:
 i  0 for 0  z  3,0m
 i  1,0 for z  3,0m (There is no bearing capacity the first 3 m in cohesion soil).
Ami is the shaft area of the pile.
Cuki is the undrained strenght. Given by CPT.
 n =1,3 comes from the safety class (SK3)
C
Rsd  9 As uk
 n  ms
As is the area of the point.
Cuk is the undrained strenght at the depth of the point.
(3)
The design pile load is given to 360 kN which is the same as the design bearing capacity Rd.
To get a first idea about how long the pile is an assupmtion is made that the pile is over 30 m
long.
At 30 m and deeper the undrained strenght is constantly 60 kPa.
Then Rsd can be calculated as:
60
Rsd  9  0,27 2
 19,3kN
1,2  1,7
Then formula 1 can be rewritten to get Rmd:
Rmd  Rd  Rd  Rsd  592,7kN
The formula 2 is rewritten to:
(4)
Rmd  mm n   i Ami Cuki
To calculate Rmd the soil is divided into 3 layers because of the different undrained strenghts.
By doing that it is possible to calculate a lenght for the pile that goes down below 30 m.
60  16 

592,7  1,2  1,4  1,06  4  0,27 16  1,021  4  0,27 
  1,0L  4  0,27 60
2 

30,216  64,8L  L  0,47  0,5m
Therefore the pole will be 30,5 m long.
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4.2 Task 1b
The design settlements is obtained from formula 5:
s d  s pd   Rd s md  s sd 
(5)
where
spd is the compression of the pole
smd is the settlements at the ate the pile point due to load transmitted along the pile shaft
ssd is the settlement at the pile point caused by the load Qsd transmitted at the pile point
 Rd is 1,0
Lp
s pd  Qsd   s Qmd 
(6)
Ed Ap
where
R
Qsd is sd that was given in task 1a
1,7
R
Qmd is md that was given in task 1a
1,7
 s = 0,7 is a constant given in ”Grundläggningsteknik”
Lp is the lenght of the pile that was calculated in task 1a
Ed is the design value for the concrete.
Ap is the area of the point
Ek for concrete K50 is 34GPa
Ek
Ed 
(7)
 n  m
where
 is put to 1,0 for concrete
 m is put to 1,2 for concrete.
 n is 1,2 due to safetyclass.
Ed  23,61GPa
592,7 
30,5
 19,3
s pd  
 0,7
 0,00453m

1,7  23,61  10 6  0,27 2
 1,7
s md  12  r0
m
(8)
M md
r0 is the radius of the pile.
 m is the average shear stress transmitted along the pile shaft and can be calculated with
formula 9.
Mmd is the design value of the compression modulus of the soil in the zone influencing the
bearing capacity of the pile. Mmd is calculated using formula 10.
Q
(9)
 m  md
Am
 m  11,74
7
M md 
M mk
(10)
 m  n
Mmk is given in the conditions to be 6500 kPa if th.
Mmd = 3869kPa
0,27 10,58
s md  12 
 0,0049m
2 3869
Qsd
s sd  2  r0
(11)
M sd A p
where
Msd is the design value of the compression modulus of the soil in the zone influencing the
bearing capacity og teh pile. Since the soil is overconsolidated Msd = Mmd.
0,27 11,353
s sd  2 
 0,0109m
2 3869  0,27 2
The total settlements will be
s d  0,0203m  20,3mm
4.3 Task 2a
When calculating task 2a the soil is divided into 2 layers. First layer goes from excavation
bottom, 1m in the CPT diagram, down to 6 m depth and second layer goes from 6 m depth
down to 11 m depth. In these calculations excavation bottom is 0 m.
1
Rmd  Rsd 
Rd 
(12)
 Rd
'
Rmd   d  om
Am
(13)
where
'
 om
is the mean value of the effective vertical overburden pressure along the pile shaft.
 d is the design value of the shaft bearing factor
Am is the area of the shaft along the pile. See appendix 6.4.
 Rd is 1,7
d 
k
 n  mm
(14)
 k  K m tan 
K m  1  sin  '
(15)
  0,7
(17)
(16)
'
Layer 1,
K m  0,38
   k  0,19   d  0,122
  26,6 
Layer 2,
K m  0,43
   k  0,196   d  0,126
  24,5 
When calculating the effective vertical overburden pressure ”Grundläggningsteknik” says that
it can only be calculated down to 20*diameter of the pile. In this case the diameter of the pile
is 300 mm wich gives a maximum depth of 6 m.
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Table 1. Effective vertical overburden pressure
Depth (m)  o (kPa)
0
0
5
100
6
120
10
200
u (kPa)
0
50
60
100
'
 om
(kPa)
0
50
60
100
Rmd  0,122  25  9,75  0,126  60  9,75  102,4kN
Rsd  N qd os' As
(18)
where
As is the area of the point of the pile, se appendix 6.4.
 os' is the value of the effective vertical overburden pressure at the point of the pile.
Nqd is the design value factor of the point bearing capacity factor which is based on the design
value of the internal friction angle.
tan  k
(19)
tan  d 
 n  ms
tan d  0,45
Bearing capacity factor Nqd is evaluated from Fig 11.2b in “Grundläggningsteknik”.
Nqd will be 7.
 os'  200kPa
Rsd  7  200  0,01078  15,1kN
The total bearing capacity for the pile is
Rd=69kN
4.4 Task 2b
Rd 
1
Rd 
1
( Rmd  Rsd )
(20)
 Rd
Since the pile is a point bearing pile the bearing capacity along the pile shaft should not be
included. Therefore equation 20 can be rewrewritten:
 Rd
Rsd
(21)
 Rd is 1,6
Rsd   s As
qcs
(22)
 n  ms
where
 s is a point bearing capacity factor which is 0,5 for a steel pile.
qcs is the point resistance of the pressure cone at the pile point.
13,5  10 3
Rsd  0,5  0,01078
 37,9kN
1,2  1,6
The total bearing capacity for the pile will be:
Rd  23,7kN
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5. Discussion
For task 1 we thought the results that we obtained were good and accurate. Since the task is to
calculate the length of a pile in cohesion soil and the design bearing capacity was pretty high
the pile needs to be long.
At first the settlements of the pile seemed to be to low but when considering we had a safety
class 3 the result seemed reasonable.
The design bearing capacity in task 2 seems low but since the pile is made of steel which has
a low roughness and a small area at the point the result seems accurate. Maybe the material
that should be used as a shaft bearing pile should have more roughness, such as concrete.
For the last task we only considered the bearing capacity for the point but it might be suitable
to add the bearing capacity of the shaft. The soil is a friction soil and therefore the bearing
capacity due to the friction between the shaft and the soil should not be neglected.
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6. Appendix
6.1 Geotechnical data for task 1
11
6.2 CPT for task 2
12
6.3 CPT for task 2
13
6.4 Data about HEB 400 for task 2
14