Limit Load Analysis of suction anchors for cohesive materials

By Dr Amir RahimThe CRISP Consortium Ltd/South Bank University

14th CRISP User Group MeetingFriday 21ST September 2001

UNIVERSITYCOLLEGE LONDONDEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING

In association with

Introduction

Suction anchors (or suction piles) are deep water anchors for floating structures or offshore oil installations. The pile consists of a hallow cylinder which has a top cap and a relatively thin wall.

D

L

a

Problem Specifications

• Pile diameter, D= 6m• Pile penetration, L=13.5m• Attachment point positions are:

a= 6.75 m (or 1/2 depth L)

a= 9 m (or 2/3 of depth L)

a= 7.785 m (just below mid point,

considered to be optimum position)• Angle of inclination is 0o (horizontal), 15o, 26o, 30o, 45o, 90o

(vertical)

Problem Specifications

• Load applied consists of 50 unequal load increments with maximum load=10000kN

• The undrained shear strength, Cu, at mud-line is 3Kpa and it is assumed to increase with depth by a rate of 1.85.

• Young's modulus at mud-line is assumed to be 300 KPa and it

is assumed to increase with depth by a rate of 185.

• The pile shell's thickness is 3cm. The pile is also assumed to be very stiff, with E=2.1E8

• A von Mises undrained material is used for the soil, with Poisson's ratio=0.49

Finite Element Mesh

. This was done using Femap Basic edition. Due to symmetry, only half the model was considered due

X-Y elevation showing full mesh

Finite Element Mesh

Finite Element Mesh

Finite Element Mesh

Finite Element Mesh

Finite Element Results

The following charts show FE results for cases with different load attachment point and angle. A von Mises material is used for all of these cases

Load applied at 0 degreesFE mesh for half model, hence load is half actual

0

2000

4000

6000

8000

10000

12000

0 1 2 3 4 5

Resultant displacemnet (m)

App

lied

Load

(kN

)

Ur@1/2

Ur@below -mid

Ur@2/3 (9m deep)

Finite Element Results

Load applied at 15 degreesFE mesh for half model, hence load is half actual

0

2000

4000

6000

8000

10000

12000

0 1 2 3 4 5

Resultant displacemnet (m)

App

lied

Load

(kN

)

Ur@1/2

Ur@below -mid

Ur@2/3 (9m deep)

Finite Element Results

Load applied at 26 degreesFE mesh for half model, hence load is half actual

0

2000

4000

6000

8000

10000

12000

0 1 2 3 4 5

Resultant displacemnet (m)

App

lied

Load

(kN

)

Ur@1/2

Ur@below -mid

Ur@2/3 (9m deep)

Finite Element Results

Load applied at 30 degreesFE mesh for half model, hence load is half actual

0

2000

4000

6000

8000

10000

12000

0 1 2 3 4 5

resultant displacement (m)

App

lied

Load

(kN

)

Ur@1/2

Ur@below -mid

Ur@2/3 (9m deep)

Finite Element Results

Load applied at 45 degreesFE mesh for half model, hence load is half actual

0

2000

4000

6000

8000

10000

12000

0 1 2 3 4 5

Resultant displacemnet (m)

App

lied

Load

(kN

)

Ur@1/2

Ur@below -mid

Ur@2/3 (9m deep)

Finite Element Results

Load applied at 90 degreesFE mesh for half model, hence load is half actual

0

2000

4000

6000

8000

10000

12000

0 1 2 3 4 5

Resultant displacemnet (m)

App

lied

Load

(kN

)

Ur@below -mid

Finite Element Results

Variation of ultimate capacity

6000

7000

8000

9000

10000

0 20 40 60 80 100

Load angle (degrees)

Ult

imat

e L

oad

(K

N)

mid-level (1/2)

below mid-level

lower (2/3)

Finite Element Results

Load attachment 2/3 of depth from mud-linefor von Mises and Mohr Coulomb

0

2000

4000

6000

8000

10000

12000

0.0 1.0 2.0 3.0 4.0 5.0

Resultant displacemnet (m)

Ap

plie

d L

oa

d(k

N)

Ur@2/3 VM

Ur@2/3 MC

Analytical solution using upper bound theory

rigid

Fan zone

Pile

The equation for the limiting load for the above mechanism is given as:

}]]4/)sin{(2}[4/)cos{(42[ dcP uu

where d is the pile diameter, and cu is the undrained shear strengthThe pile soil adhesion is defined by:

us cf /sin

For a pile depth of 13.5m and assuming full adhesion, ie sin =1, we getPult=16225 KNIf we include the base resistance which isWe get Pult =17017 KNThis agrees reasonably with the FE solution for the case of =0o at a point below mid-depth which is 16400KN

Analytical solution using upper bound theory

The depth of the pile is 13.5m, therefore the total force isPu(total)=16225 KN

u2 cπr

ConclusionsThe FE solution for a middle attachment point loaded horizontally can be compared with the upper bound solution of laterally loaded pile. For this analysis the FE solution of =0o at a point below mid-depth is about 4% lower than the upper bound solution.

A more complex upper bound solution is necessary for cases in which the load is applied at an angle, due to the rotation of the pile.

Pull out capacity tends to reduce when the angle of load attachment approaches a horizontal level. This indicates that the vertical resistance is the predominant component for such deep piles.All the analyses above were carried out using reduced integration (ie 2x2x2 Gauss points). Use of full integration produced a slightly higher collapse load

Conclusions•Use of Mohr Coulomb is preferred over use of von Mises material. The Mohr-Coulomb yield surface in the -plane slightly smaller giving a lower collapse load.

C

von Mises

Mohr-Coulomb

CTriaxial CompressionTTriaxial Extension

Description of von-mises and Mohr-Coloumbyield surfaces in the -plane

T

Hardware specs

This problem contains about 30,000 d.o.f (about 10,000 nodes)The PC used is AMD Athelon, 1GHZ, with 512 MB RamEach analysis took about 4 hours per increment

In conclusion, the use of a fast computer with large memory is highly recommended as it would make a huge difference in computation time